Reinforced Concrete Design: A Practical Approach, Instructor’s Solution Manual [1 ed.] 0132001888

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INSTRUCTOR'S SOLUTIONS MANUAL

Reinforced Concrete Design: A Practical Approach Svetlana Brzev British Columbia Institute of Technology

John Pao Bogdonov Pao Associates Ltd.

Toronto 0-13-200188-8

Copyright © 2006 Pearson Education Canada Inc., Toronto, Ontario. Pearson Prentice Hall. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Reinforced Concrete Design: A Practical Approach, by Svetlana Brzev and John Pao, to post this material online only if the use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and provided the reproduced material bears this copyright notice.

Chapter 1 - Solutions __________________________________________________________________________ 1.1. a) 2.4 kPa (NBC 2005 Table 4.1.5.3) b) 2.4 kPa (NBC 2005 Table 4.1.5.3) c) 1.9 kPa (NBC 2005 Table 4.1.5.3) d) 7.2 kPa (stack rooms, NBC 2005 Table 4.1.5.3) ___________________________________________________________________________ 1.2. NBC 2005 Table 4.1.5.3 prescribes occupancy live load of 2.4 kPa for classrooms with or without fixed seats. To determine the actual occupancy live load, add together the weight for all students in the classroom and divide by the classroom plan area. ___________________________________________________________________________ 1.3. a) Beam properties: width b = 350mm depth h = 700mm unit weight γ w = 24kN ⁄ m

3

Load analysis: - self-weight

kN kN b × h × γ w = 0.35m × 0.7m × 24 ------3- = 5.9 ------m m ____________________________

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kN DL ≅ 6.0 ------m LL = 15kN ⁄ m

- Live load

kN Total load w = DL + LL = 6.0 + 15.0 = 21.0 ------m

2

w×l M s = -------------- (simply supported beam, see Table A.16) 8 kN 2 21.0 ------- ( 8.0 m ) m = --------------------------------------------- = 168kNm 8 M s = 168kNm b) w f = 1.25DL + 1.5LL (NBC 2005 Table 4.1.3.2)

kN = 1.25 × 6.0 + 1.5 × 15.0 = 30.0 ------m 2 kN 30 ------- ( 8.0m ) 2 wf × l m M f = ---------------- = --------------------------------------- = 240kNm 8 8

M f = 240kNm ___________________________________________________________________________ 1.4. a) DL = 10.0kPa

LL = 5.0kPa Load on beam B1: Tributary width s = 3.0m

w f = ( 1.25DL + 1.5LL ) × s

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= ( 1.25 × 10.0 + 1.5 × 5.0 ) × 3.0 = 60kN ⁄ m

Load on girder G1: Tributary width s = 8m

w f = ( 1.25DL + 1.5LL ) × s kN = ( 1.25 × 10.0 + 1.5 × 5.0 ) × 8.0 = 160 ------m

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b) Tributary area for the column C1

A = 8m × 9m = 72m

2

Factored axial load

P f = ( 1.25DL + 1.5LL )A = ( 1.25 × 10.0 + 1.5 × 5.0 ) ( 72.0 ) = 1, 440kN ___________________________________________________________________________ 1.5.

Cross sectional area 2

A = 1100 × 100 + 300 × 600 = 290, 000mm = 0.29m

2

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Unit weight γ w = 24.0kN ⁄ m

3

Beam self-weight:

kN kN kN 2 w = A × γ w = ( 0.29m ) 24.0 ------3- = 6.96 ------- ≅ 7.0 ------m m m Hence, kN w = 7.0 ------m ___________________________________________________________________________ 1.6. a) LL = 2.4kPa (Garage for passenger cars, NBC 2005 Table 4.1.5.3) b) A typical T-beam - Tributary width s = 4500mm = 4.5m - Cross sectional area 6

2

A = 4500 × 200 + 700 × 450 = 1.215 × 10 mm = 1.2m

2

kN - Unit weight γ w = 24 ------3m

- Beam self-weight

kN kN kN 2 w = A × γ w = ( 1.2m ) 24.0 ------3- = 28.8 ------- ≅ 29.0 ------m m m c) Load analysis for a typical T-beam

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- Self weight - Superimposed dead load

kN = 29.0 ------m kN 0.3kPa × s = 0.3kPa × 4.5m = 1.35 ------m _________________________________

kN DL ≅ 30.4 ------m - Live load

kN kN LL = 2.4kPa × s = 2.4kPa × 4.5m = 10.8 ------- ≅ 11 ------m m Total factored load (NBC 2005 Table 4.1.3.2): Case 1:

kN w f = 1.4DL = 1.4 × 30.4 = 42.6 ------m

Case 2:

w f = 1.25DL + 1.5LL = 1.25 × 30.4 + 1.5 × 11.0 = 54.5kN ⁄ m

It can be concluded that Case 2 load combination gives larger value and it governs, that is,

kN w f = 54.5 ------- . m ___________________________________________________________________________ 1.7.

P D = 1000kN P L = 800kN P S = 500kN Consider all possible load combinations from NBC 2005 Table 4.1.3.2. Case 1: P f = 1.4P D = 1.4 × 1000 = 1400kN Case 2: P f = 1.25P D + 1.5P L + 0.5P S

= 1.25 × 1000 + 1.5 × 800 + 0.5 × 500 = 2700kN Case 3: P f = 1.25P D + 1.5P S + 0.5P L

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= 1.25 × 1000 + 1.5 × 500 + 0.5 × 800 = 2400kN It can be concluded that Case 2 governs, that is,

P f = 2700kN .

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Chapter 2 - Solutions __________________________________________________________________________ 2.1. a) Concrete mix proportion is usually expressed as the ratio, by volume (or more accurately by mass), of cement to fine aggregate to coarse aggregate, in that order. An example of concrete mix proportion is 1:2:4 (cement: fine aggregate: coarse aggregate). In addition to the above, a w/c ratio also needs to be specified to define the concrete mix. b) The four main requirements which the concrete mix proportion needs to meet are: 1) compressive strength (based on w/c ratio), 2) durability (including the requirements for air entrainment related to a particular exposure class, compressive strength, and cement type), 3) slump (based on minimum workability requirements for method of placement), and 4) maximum aggregate size (limited by section dimensions and reinforcement spacing). _________________________________________________________________________ 2.2. The two most important mechanical properties of hardened concrete are: - compressive strength and - tensile strength The compressive strength generally depends on the type of concrete mix, the properties of aggregate, the time and quality of curing, etc. The tensile strength depends on the compressive strength, the type of test used to determine the tensile strength, etc. __________________________________________________________________________ 2.3. a) It may be expected that the beam deflections after 10 years are on the order of 2.5 to 3 times the initial deflections.

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b) Deflections in reinforced concrete structures increase over time as a result of creep. Creep is inelastic time-dependent deformation of concrete under the sustained stress. c) If the load is completely removed, the deflections will decrease up to a certain extent, however some residual deflections will still exist. This is illustrated in Figure 2.7, which shows the variation of creep strain over time. __________________________________________________________________________ 2.4.

2

2

πd π ( 100mm ) 2 A = --------- = ------------------------------ = 7854mm 4 4 3

P 300 × 10 Nf c ′ = ---- = --------------------------= 38.2MPa 2 A 7854mm Hence, the specified compressive strength ( f c ′ ) is 38MPa . Next, find the modulus of elasticity ( E c ), as follows

E c = 4500 f c ′ (CSA A23.3 Eq.8.2) = 4500 38MPa = 27740MPa ≅ 28000MPa Modulus of elasticity ( E c ) is equal to 28000 MPa.

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__________________________________________________________________________ 2.5. Use the equation

E c = 4500 f c ′ (CSA A23.3 Eq.8.2) f c ′ ( MPa ) E c ( MPa ) 20 25 30 35 40

20125 22500 24648 26622 28460

__________________________________________________________________________ 2.6. a)

The modulus of rupture ( f r ) can be obtained by setting

ft = fr where

M × yt f t = --------------Ig h 150 y t = --- = --------- = 75mm 2 2

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3

3

bh 6 4 ( 150mm ) ( 150mm ) I g = --------- = -------------------------------------------------- = 42.2 × 10 mm 12 12 Set M = 1.70kNm (maximum value) 6 My t ( 1.7 × 10 Nmm ) × 75mm - = 3.0MPa f r = --------- = --------------------------------------------------------------6 4 Ig 42.2 × 10 mm

b) Let us compare the f r value obtained in part a) with the CSA A23.3 prescribed value. Use f c ′ = 25MPa

λ = 1.0 (normal-density concrete) f r = 0.6λ f c ′ (CSA A23.3 Eq.8.3) = 0.6 × 1.0 × 25MPa = 3.0MPa The f r value obtained experimentally is the same as the value obtained using the CSA A23.3 equation. _____________________________________________________________________________ 2.7. 2

a) 25M bar area = 500mm (Table A.1) 3 - 25M : A = 3 × 500 = 1500mm

2

The bars are subjected to the tensile force T as shown below.

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T = 300kN Normal stress in the bars is equal to 3

T 300 × 10 N= 200MPa f = ---- = --------------------------2 A 1500mm The strain can be obtained from the Hooke’s Law as f 200MPa ε = ---- = -------------------------------- = 0.001 E 200 ,000M Pa The above calculation assumes that the steel is elastic, that is, yielding has not taken place. This assumption needs to be verified. Stress-strain diagram for steel is shown below.

fy 400MPa ε y = ----- = -------------------------------- = 0.002 200 ,000M Pa Es Since ε = 0.001 < ε y = 0.002 it follows that the bars have not yielded. b) The maximum force which the bars can resist corresponds to the yield stress f y , that is,

T = f×A where

f = f y = 400MPa so

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2

3

T y = f y × A = ( 400MPa ) ( 1500mm ) = 600 × 10 N = 600kN The maximum force in the bars is equal to 600 kN. __________________________________________________________________________ 2.8. 20M bar: area = 300mm

2

25M bar: area = 500mm

2

30M bar: area = 700mm

2

a) A req = 1 ,650m m

(Table A.1)

2

Use 6 - 20M bars : A = 6 × 300 = 1800mm or 4 - 25M bars : A = 4 × 500 = 2 ,000m m b) A req = 2320mm

2

2

2

Use 8 - 20M bars : A = 8 × 300 = 2 ,400m m or 4 - 30M bars : A = 4 × 700 = 2 ,800m m c) A req = 3300mm

2

2

A = 8 × 500 = 4000mm

Use 8 - 25M bars: d) A req = 4 ,700m m

e) A req = 5 ,450m m

2

2

A = 10 × 500 = 5000mm

Use 10 - 25M bars:

Use 8 - 30M bars:

2

2

2

A = 8 × 700 = 5600mm

2

__________________________________________________________________________ 2.9. a) Shrinkage strain is approximately equal to 0.0002 – 0.0003mm ⁄ mm , hence the total shrinkage deformation for a 70 m long floor is on the order of 14 mm to 21 mm.

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b) There should be an ample amount of reinforcement perpendicular to the expected crack direction in order to ensure that the cracks are uniformly distributed in the form of small cracks (microcracks). c) If the total shrinkage deformation is roughly 20 mm, then the number of cracks can be estimated as

20mm -------------------- = 60cracks 0.33mm It can be concluded that, on average, there is one crack per each 1 to 1.2 m of the slab length.

d) Based on part c), there are 60 cracks in total. Consequently, 5% of the total number of cracks corresponds to 3 cracks with an approximate width of 1.6 mm. These cracks are expected to be visible to a naked eye. These cracks may be acceptable in case the concrete slab is not exposed and the structural capacity is not affected by the cracking; however, provisions for crack repair should be included in the maintenance schedule. If these cracks are indeed visible to a naked eye, the occupants might have an uncomfortable perception that the structure is unsafe. e) Any cracks exposed in a parking structure must be repaired as soon as possible in order to preserve the durability of this structure. Otherwise, if water enters into cracks and is allowed to freeze during the winter, the structure may deteriorate due to rusting of rebars and spalling of concrete from freeze-thaw cycles. This is not a good design, and special provisions should be included in the construction procedure to minimize the chances for cracking. f) Shrinkage cracking will not take place when unrestrained shrinkage in the slab is permitted. Given this principle, the cracking can be significantly minimized if unrestrained shrinkage can be permitted over a period of time. i) Introduce a pour strip somewhere in the middle of the slab.

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This allows the two halves of the slab to shrink towards the ends freely. If the construction schedule permits, the pour of the infill strip should be delayed by at least 1 to 2 months. ii) Split the slab into 3 segments. Pour segment (1) first, and delay the pouring of segment (2) by as long as possible.

iii) First, pour the walls in the N-S direction. Subsequently, pour the entire slab. Finally, after the shrinking of the slab has taken place, pour the walls in the E-W direction. Note that a temporary slab support may be required until the walls in the E-W direction have been poured; this allows a reasonably unrestrained slab shrinkage in both directions.

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Chapter 3 - Solutions __________________________________________________________________________ 3.1. The five basic assumptions for flexural design of reinforced concrete members according to the CSA A23.3 code are: 1) Plane sections remain plane: sections perpendicular to the axis of bending that are plane before bending remain plane after bending (Cl.10.1.2). 2) Strain compatibility: the strain in reinforcement is equal to the strain in concrete at the same location. 3) Stress-strain relationship: the stresses in the concrete and the reinforcement can be computed from the stress-strain diagrams for concrete and steel, however Cl.10.1.7 permits the use of equivalent rectangular stress block instead of the actual stress distribution for concrete. 4) Concrete tensile strength has been neglected in flexural strength calculations (Cl.10.1.5) 5) The maximum compressive strain in concrete is equal to 0.0035 (Cl.10.1.3). _________________________________________________________________________ 3.2. a) The four stages of flexural behaviour are: 1) elastic uncracked, 2) elastic cracked, 3) yielding, and 4) failure. b) Elastic cracked stage. c) Failure.

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_________________________________________________________________________ 3.3. a) The two basic failure modes characteristic for reinforced concrete flexural members are: •

steel-controlled mode

•

concrete-controlled mode

b) Steel-controlled mode is preferred in the design of reinforced concrete flexural members. It is considered a good practice to design reinforced concrete flexural members to fail in the steel-controlled mode that is initiated by yielding of steel reinforcement and characterized by ductile behaviour. __________________________________________________________________________ 3.4. a) Find the depth of the compression zone ( a ).

β 1 = 0.9

α 1 = 0.8

ε y = 0.002 0.0035 --c- = -------------------------- = 0.64 d 0.0035 + ε y c = 0.64d a = β 1 c = 0.9 × 0.64d = 0.58d α 1 φ c f c ′ab A sb = ------------------------φs fy

[3.21]

Find the balanced reinforcement ratio ρ b .

A sb α1 φc fc ′ a ρ b = -------- = ------------------ --φs fy d bd

[3.22]

0.8 × 0.65 × 30MPa 0.58d = ------------------------------------------------- × ------------0.85 × 400MPa d = 0.027 = 2.7 %

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b) The balanced condition is characterized by the simultaneous crushing of concrete and yielding of the tension steel. Therefore, the strain in concrete ( ε cmax ) is 0.0035 while the strain in steel is equal to ε y . c) Balanced reinforcement ratio ( ρ b ) is used to predict potential failure mode for a reinforced concrete flexural member with the reinforcement ratio ( ρ ) , as follows •

If ρ > ρ b

concrete-controlled failure mode

•

If ρ < ρ b

steel-controlled failure mode

•

If ρ = ρ b

balanced failure mode

_________________________________________________________________________ 3.5. a) α 1 = 0.8

β 1 = 0.9

Find the depth of the compression zone ( a ). 3-25M bars: A s = 3 × 500 = 1500mm

2

(Table A.1)

2 φs fy As 0.85 × 400MPa × 1500mm a = --------------------- = --------------------------------------------------------------------------- = 82mm 0.8 × 0.65 × 30MPa × 400mm α 1 φ c f c ′b

82 a c = ----- = ------- ≅ 91mm 0.9 β1 c 0.0035 --- = --------------------------d 0.0035 + ε s

[3.20]

or

91mm 0.0035 ------------------- = --------------------------650mm 0.0035 + ε s Hence,

ε s = 0.0215

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b) Check whether ε s > ε y

fy 400MPa ε y = ----- = -------------------------------- = 0.002 200 ,000M Pa Es ε s = 0.0215 > ε y = 0.002

properly reinforced beam

A properly reinforced beam is expected to fail in the steel-controlled failure mode. (Note that the beam is expected to fail in the concrete-controlled mode when ε s < ε y .) c) 2

T r = φ s f y A s = 0.85 × 400MPa × 1500mm = 510kN 3 a 82 M r = T r d – --- = ( 510 × 10 ) 650 – -----2 2

[3.9] [3.13]

= 310.6kNm Hence,

M r ≅ 310kNm __________________________________________________________________________ 3.6. a) Check whether the section is properly reinforced. i) Find the balanced reinforcement area ( A sb ) . Note: α 1 = 0.8

β 1 = 0.9

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Use the following proportion based on Eqn (3.20)

cb 0.0035 ----- = -------------------------d 0.0035 + ε y where

fy 400MPa ε y = ----- = -------------------------------- = 0.002 200 ,000M Pa Es Hence,

cb 0.0035 ----- = ------------------------------------ = 0.636 d 0.0035 + 0.002 Since d = 650mm (given), it follows that

c b = 0.636d = 0.636 × 650 = 413.6mm a = β 1 c = 0.9 × 413.6 = 372mm Use the equation of equilibrium

Tr = Cr

[3.10]

or

φ s A sb f y = α 1 φ c f c ′ab Hence,

α 1 φ c f c ′ab 0.8 ( 0.65 ) ( 20MPa ) ( 372mm ) ( 500mm ) A sb = ------------------------- = -----------------------------------------------------------------------------------------------φs fy 0.85 × 400MPa = 5692mm

[3.21]

2

2

Actual reinforcement: 4 - 20M bars A s = 4 × 300 = 1200mm (Table A.1) Since 2

A s = 1200mm < A sb = 5692mm

2

it follows that the section is properly reinforced. The same statement is valid for parts ii), iii), and iv) of this problem.

f c ′ = 20MPa

φs As fy 0.85 ( 1200 ) ( 400 ) a 20 = --------------------- = ----------------------------------------------------------------------- = 78mm α 1 φ c f c ′b 0.8 ( 0.65 ) ( 20MPa ) ( 500mm )

[3.12]

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a 20 M r = φ s A s f y d – ------2

[3.14]

78 = 0.85 × 1200 × 400 650 – ------ = 249kNm 2 ii) f c ′ = 25MPa

20MPa a 25 = a 20 ------------------- = 63mm 25MPa a 25 M r = φ s A s f y d – ------2

[3.14]

63 = 0.85 × 1200 × 400 650 – ------ = 252kNm 2 iii) f c ′ = 30MPa

20MPa a 30 = a 20 ------------------- = 52mm 30MPa a 30 M r = φ s A s f y d – ------2

[3.14]

52 = 0.85 × 1200 × 400 650 – ------ = 255kNm 2 iv) f c ′ = 35MPa

20MPa a 35 = a 20 ------------------- = 45mm 35MPa a 35 M r = φ s A s f y d – ------2

[3.14]

45 = 0.85 × 1200 × 400 650 – ------ = 256kNm 2

b) A variation in the concrete compressive strength has a very small effect on the M r value. When concrete strength increases from 20 to 35 MPa (by approximately 75%), the corresponding M r value increases by only 3%, that is,

256 – 249----------------------≅ 3% 249 This increase is not significant from design perspective.

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__________________________________________________________________________ 3.7. a) f c ′ = 35MPa

a = 45.0mm (see Problem 3.6.d)

a M r = φ s × A s × f y d – --2

45.0 2 = 0.85 × 1200mm × 400MPa 590 – ---------2

= 231.5kNm b) From Problem 3.6.d): M r = 256kNm for d = 650mm From Problem 3.7.a): M r = 231.5kNm for d = 590mm It can be concluded that the decrease in M r value is equal to

256 – 231.5 ---------------------------- = 10.5% 231.5 Hence, a 10% decrease in d value (from 650 mm to 590 mm) results in 10% decrease in the M r value (from 256 kNm to 231.5 kNm). __________________________________________________________________________ 3.8. a) f c ′ = 31.5MPa d = 590mm

a = 45mm (Problem 3.7.a) 35MPa a = 45 × ----------------------- = 50mm 31.5MPa

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a 50 M r = φ s A s f y d – --- = 0.85 × 1200 × 400 590 – -----2 2

[3.14]

= 230.5kNm b) A 10% decrease in the f c ′ value (from 35 MPa to 31.5 MPa) has a very small effect on M r value, that is,

231.5 – 230.5 --------------------------------- = 0.4% 230.5 However, a 10% error in the placement of reinforcement (from 650 mm to 590 mm) has a significant effect on M r value (on the order of 10%). __________________________________________________________________________ 3.9. a) Find the depth of concrete compression zone a

d = 600 + 100 – 110 = 590mm 8 - 30M bars: A s = 8 × 700 = 5600mm

2

(Table A.1)

Note: α 1 = 0.8 Use the equation of equilibrium

Tr = Cr

[3.10]

where

Tr = φ s As fy

[3.9]

C r can be determined similar to equation [3.8] as C r = α 1 φ c f c ′A c A c denotes the area of the concrete compression zone (shown hatched on the sketch), and it can be determined from the equation of equilibrium [3.10] as φ s A s f y = α 1 φ c f c ′A c

[3.10]

or 2 φs As fy 0.85 × 5600mm × 400MPa 2 A c = ------------------ = --------------------------------------------------------------------- = 122051mm 0.8 × 0.65 × 30MPa α1 φc fc ′

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A c = 600 × a – 2 × 100 × 100 = 600a – 20 ,000 122 ,051 + 20 ,000 a = ------------------------------------------- = 237mm 600 b) Find the centroid of the compression zone ( a ).

100 137 600 × 137 × 100 + --------- + 400 × 100 × --------2 2 a = ---------------------------------------------------------------------------------------------------------------- = 130mm 600 × 137 + 400 × 100 c) Find M r

M r = φ s A s f y ( d – a ) = 0.85 × 5600 × 400MPa ( 590 – 130 ) = 876kNm Hence,

M r = 876kNm

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__________________________________________________________________________ 3.10. a) 8-35M bars:

A s = 8 × 1000 = 8000mm

Note: α 1 = 0.8

2

(Table A.1)

β 1 = 0.9

In order to find M r , the same calculations need to be repeated as in Problem 3.9., hence 2 φs As fy 0.85 × 8000mm × 400MPa 2 A c = ------------------ = --------------------------------------------------------------------- = 174 ,359m m 0.8 × 0.65 × 30MPa α1 φc fc ′

or

A c = 600a – 20 ,000 It follows that

a = 324 mm

Find the centroid of the compression zone ( a ).

100 224 600 × 224 × 100 + --------- + 400 × 100 × --------2 2 a = ---------------------------------------------------------------------------------------------------------- = 175mm 174 ,359 Find M r as

M r = φ s A s f y ( d – a ) = 0.85 × 8000 × 400 ( 590 – 175 ) = 1128.8kNm ≅ 1129kNm Hence,

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M r = 1129kNm b) a = 324mm

a 324 c = ----- = --------- = 360mm β1 0.9 Use the proportion:

0.0035 --c- = -------------------------d 0.0035 + ε s

[3.20]

Since

d = 590mm it follows that

ε s = 0.0024 Since

fy 400MPa ε y = ----- = -------------------------------- = 0.002 200 ,000M Pa Es and

ε s = 0.0024 > ε y = 0.002 it follows that the beam is expected to fail in the steel-controlled mode. __________________________________________________________________________ 3.11. a) 8 - 30M bars: A s = 5600mm Note: α 1 = 0.8

2

(same as Problem 3.9.)

β 1 = 0.9

d = 440mm Repeat the same calculations as performed in Problem 3.9., hence 2 φs As fy × 5600mm × 400MPa 0.85 -------------------------------------------------------------------- = 122 ,051m m 2 ----------------Ac = = 0.8 × 0.65 × 30MPa α1 φc fc ′

Since

A c = 600a – 20000

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it follows that

a = 237mm and

a = 130mm (Same as Problem 3.9.) Find M r as 2

M r = φ s A s f y ( d – a ) = 0.85 × 5600mm × 400MPa × ( 440 – 130 ) = 590kNm b) a = 237mm

237 a c = ----- = --------- = 263mm 0.9 β1 Find ε s from the following proportion

0.0035 --c- = -------------------------d 0.0035 + ε s

[3.20]

where

d = 440mm It follows that

ε s = 0.0024 Since

ε y = 0.002 and

ε s = 0.0024 > ε y = 0.002 It follows that the beam is expected to fail in the steel-controlled mode. c) A 43% increase in the amount of reinforcement from 8-30M to 8-35M (Problems 3.9. and 3.10.) has resulted in 30% increase in the M r value. On the other hand, 25% decrease in the effective depth ( d ) value (see Problems 3.9. and 3.11.) results in approximately 30% decrease in the M r value.

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__________________________________________________________________________ 3.12. a) In order to find the area of reinforcement, it is required to find the neutral axis depth ( c ). Use the following proportion based on Eqn (3.20)

0.0035 --c- = -------------------------d 0.0035 + ε y Since

fy 400MPa ε y = ----- = -------------------------------- = 0.002 200 ,000M Pa Es it follows that

0.0035 --c- = ------------------------------------ = 0.636 0.0035 + 0.002 d Since

d = 600mm (given) c = 0.636 × d = 382mm a = β 1 c = 0.9 × 382 = 344mm where

β 1 ≅ 0.9

Next, determine the width of the compression zone ( y ). Use the following proportion

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200mm x ------------------- = ----------------700mm 700 – a Hence,

200 x = ( 700 – 344 ) --------- = 101.7mm ≅ 102mm 700 and

y = 600 – x = 498mm ≅ 500mm Next, find the area of the concrete compression zone ( A c ) as

400 + y 400 + 500 2 A c = ------------------ × a = ------------------------ × 344 = 154 ,800m m 2 2 Next, find the area of reinforcement corresponding to the balanced condition ( A sb ). Use the equation of equilibrium

Tr = Cr

[3.10]

or

φ s A sb f y = α 1 φ c f c ′A c Hence,

α 1 φ c f c ′A c 0.8 × 0.65 × 25MPa × 154800 2 A sb = ------------------------ = -------------------------------------------------------------------------- = 5920mm 0.85 × 400MPa φs fy

[3.21]

Note: α 1 = 0.8 Use 6-35M bars: A s = 6 × 1000 = 6000mm

2

(Table A.1)

b) Find the centroid of the compression zone ( a ).

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344 344 × 100 2 400 × 344 × --------- + ------------------------ × --- × 344 2 2 3 a = ----------------------------------------------------------------------------------------------- = 178mm 154 ,800 Find M r as 2

M r = φ s A s f y ( d – a ) = 0.85 × 6000mm × 400MPa ( 600 – 178 ) = 860.9kNm ≅ 861kNm Hence,

M r = 861kNm _________________________________________________________________________ 3.13. a) d = 170mm (given) Reinforcement: 15M@250 2

For 15M bars, A b = 200mm (Table A.1)

s = 250mm Hence, 2

1000 1000 mm A s = A b × ------------ = 200 × ------------ = 800 -----------s 250 m

[3.28]

Use the equation of equilibrium [3.10]

Tr = Cr where 2

mm T r = φ s f y A s = 0.85 × 400MPa × 800 ------------ = 272kN ⁄ m m C r = α 1 φ c f c ′ab

[3.9] [3.8]

α 1 ≅ 0.8 For a unit slab strip: b = 1000mm It follows from Eqn [3.10] that 3 Tr 272 × 10 N ⁄ m ----------------------------------------------------------------------------- = 20.9mm ≅ 21mm -------------------= a = 0.8 × 0.65 × 25MPa × 1000mm α 1 φ c f c ′b

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3-15

Find M r as

21 kNm kNm a 3N M r = T r d – --- = 272 × 10 ---- 170 – ------ = 43.4 ------------ ≅ 43 -----------m 2 m m 2

[3.13]

Hence,

kNm M r = 43 -----------m b)

__________________________________________________________________________ 3.14. a) d = 120mm (given) Reinforcement: 15M@300 2

For 15M bars, A b = 200mm (Table A.1)

s = 300mm Hence, 2

1000 1000 mm A s = A b × ------------ = 200 × ------------ = 667 -----------s 300 m Find T r as

[3.28]

2

mm kN T r = φ s f y A s = 0.85 × 400MPa × 667 ------------ ≅ 227 ------m m For a unit slab strip: b = 1000mm

α 1 ≅ 0.8 Find a from Eqn (3.10) as 3 Tr 227 × 10 N ⁄ m a = --------------------- = ------------------------------------------------------------------------------ = 15mm 0.8 × 0.65 × 30MPa × 1000mm α 1 φ c f c ′b

Copyright © 2006 Pearson Education Canada Inc.

3-16

Find M r as

a 15 3N M r = T r d – --- = 227 × 10 ---- 120 – ------ = 25.6kNm ⁄ m ≅ 26kNm ⁄ m 2 m 2

[3.13]

Hence,

M r = 26kNm ⁄ m b) l = 5m (given) For a simple slab: 2

wf × l M f = ---------------- (Table A.16) 8 Set

Mf = M r it follows that

8 × Mr w f = --------------2 l 8 × 25.6kNm ⁄ m kN - = 8.2 ------= ---------------------------------------2 2 ( 5.0m ) m Consider the following load combination from NBC 2005 Table 4.1.3.2: w f = 1.25DL + 1.5LL Find self-weight DL as

kN kN DL = h × γ w = 0.15m × 24.0 ------3- = 3.6 ------2m m where h = 150mm = 0.15m slab thickness Then LL can be determined as

w f – 1.25DL LL = -----------------------------1.5 8.2 – 1.25 × 3.6 kN = ------------------------------------- ≅ 2.5 ------2- = 2.5kPa 1.5 m Therefore, the slab can carry specified live load of 2.5kPa in addition to its self-weight.

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3-17

__________________________________________________________________________ 3.15. Reinforcement: 2-30 M

2

A s = 2 × 700 = 1400mm (Table A.1)

d = 530mm (given) Note: α 1 = 0.8 Find the depth of the equivalent rectangular stress block ( a ) and confirm that the neutral axis is in the flange.

b = b f = 1500mm 2

T r = φ s f y A s = 0.85 × 400MPa × 1400mm = 476kN

[3.9]

3 Tr 476 × 10 N a = --------------------- = ------------------------------------------------------------------------------ = 20mm 0.8 × 0.65 × 30MPa × 1500mm α 1 φ c f c ′b

Since

a = 20mm < h f = 100mm it follows that the neutral axis is in the flange. Find M r as 3 a 20 M r = T r d – --- = 476 × 10 N 530 – ------ = 247.5kNm ≅ 247kNm 2 2

[3.13]

Hence,

M r = 247kNm __________________________________________________________________________ 3.16.

d = 590mm (given) Note: α 1 = 0.8 Reinforcement: 7-35M bars

2

A s = 7 × 1000 = 7000mm (Table A.1)

Copyright © 2006 Pearson Education Canada Inc.

3-18

First, determine the depth of the compression stress block ( a ). 2

T r = φ s f y A s = 0.85 × 400MPa × 7000mm = 2380kN

[3.9]

Set

b = b f = 1000 mm (note that b denotes the width of the compression zone) Hence, can be determined from the equation of equilibrium [3.10] as 3 Tr 2380 × 10 N a = --------------------- = ------------------------------------------------------------------------------ = 152.6mm ≅ 153mm α 1 φ c f c ′b 0.8 × 0.65 × 30MPa × 1000mm

Since

a = 153mm > h f = 100mm It follows that the neutral axis is within the web. Next, determine the area of the concrete compression zone ( A c ) as 2 φs fy As × 400MPa × 7000mm 0.85 2 -----------------------------------------------------------------------------------Ac = = = 152564mm 0.8 × 0.65 × 30MPa α1 φc fc ′

Depth of the compression block ( a ) can be determined as:

A f = h f × b f = 100mm × 1000mm = 100000mm

2

Ac – Af 152 ,564 – 100 ,000 a = h f + ----------------- = 100mm + --------------------------------------------- = 217mm bw 450 Web area under compression:

A w = b w ( a – h f ) = 450 ( 217 – 100 ) = 52650mm

2

Location of the centroid of the concrete compression zone ( a′ ) :

a – hf hf A f ----- + A w h f + -------------2 2 a′ = -------------------------------------------------------Ac 100 217 – 100 100 ,000 × --------- + 52650 100 + -----------------------2 2 = ------------------------------------------------------------------------------------------------------ = 87mm 152 ,564

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3-19

Finally, factored moment resistance ( M r ) can be determined as

M r = T r ( d – a′ ) 3

= 2380 × 10 N × ( 590 – 87 ) = 1196kNm Hence,

M r = 1196kNm __________________________________________________________________________ 3.17.

30 a) d = 600 – 50 – ------ = 535mm 2 8- 30M bars: A s = 8 × 700 = 5600mm Note: α 1 = 0.8

2

(Table A.1)

β 1 = 0.9

First, let us find a . Note that b = 400mm (width of the compression zone). 2

T r = φ s f y A s = 0.85 × 400MPa × 5600mm = 1904kN

[3.9]

Find a from Eqn (3.10) as 3 Tr 1904 × 10 N a = --------------------- = --------------------------------------------------------------------------- = 366mm 0.8 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

Find c as

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3-20

366 a c = ----- = --------- = 407mm 0.9 β1 Use the following proportion

c 0.0035 --- = --------------------------d 0.0035 + ε s

[3.20]

Hence,

d 535 ε s = 0.0035 --- – 1 = 0.0035 --------- – 1 = 0.0011 c 407 fy 400MPa ε y = ----- = ------------------------------- = 0.002 200000MPa Es Since

ε s = 0.0011 < ε y = 0.002 It follows that the beam is expected to fail in the concrete-controlled mode.

b)

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3-21

Find d′ as

25 d′ = 40 + ------ = 52mm 2 The steel-controlled failure occurs if the following condition has been met:

0.0035 0.0035 --c- ≤ -------------------------- = ----------------------------------= 0.636 d 0.0035 + ε y 0.0035 + 0.002 Hence,

c ≤ 0.636d = 0.636 × 535 = 340 mm and

a = β 1 c = 0.9 × 340 = 306mm Use the following equation of equilibrium

Cr + Cr ′ = Tr

[3.37]

where

C r = α 1 φ c f c ′ab = 0.8 × 0.65 × 25MPa × 306mm × 400mm = 1591.2kN T r = 1904kN

[3.8]

(same as part a)

C r ′ = φ s f s ′A s ′ Assume that the compression steel has yielded, hence f s ′ = f y . Then,

Cr + φs fy As ′ = Tr

[3.37]

and 3 3 Tr – Cr 1904 × 10 N – 1591 × 10 N 2 A s ′ = ----------------- = -------------------------------------------------------------------- = 920mm φs fy 0.85 × 400

Use 2-25M bars for compression steel: A s ′ = 2 × 500 = 1000mm

2

(Table A.1)

Therefore, 2

C r ′ = φ s f y A s ′ = 0.85 × 400MPa × 1000mm = 340kN Find a from Eqn (3.37) as 3 3 Tr – Cr ′ × 10 – 340 × 10 1904 ------------------------------------------------------- = 301mm -------------------= = a α 1 φ c f c ′b 0.8 × 0.65 × 25 × 400

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3-22

and

a 301 c = ----- = --------- = 334mm β1 0.9 Check whether the compression steel has yielded. Find ε s ′ as

d′ 52 ε s ′ = 0.0035 1 – ---- = 0.0035 1 – --------- = 0.003 c 334 Since

ε s = 0.003 > ε y = 0.002 it follows that the compression steel has yielded. Check whether the tension steel has yielded. Find ε s as

d 535 ε s = 0.0035 --- – 1 = 0.0035 --------- – 1 = 0.0021 c 334 Since

ε s = 0.0021 > ε y = 0.002 it follows that the tension steel has yielded. Hence, it is required to provide 2-25M bars in the compression zone to ensure the steel-controlled failure. c)

Find M r as

a M r = C r ′ ( d – d′ ) + C r d – --2

[3.39]

3 3 301 = 340 × 10 N ( 535 – 52 ) + ( 1591 × 10 ) 535 – --------2

= 776kNm

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3-23

Hence,

M r = 776kNm __________________________________________________________________________ 3.18. 2

a) Tension steel: 6- 25M

A s = 6 × 500 = 3000mm

Compression steel: 2- 20M

A s ′ = 2 × 300 = 600mm

2

(Table A.1) (Table A.1)

d = 510mm (given) Note: α 1 = 0.8

β 1 = 0.9

Tension steel reinforcement ratio: 2 As 3000mm ρ = ------ = -------------------------------------------- = 0.015 = 1.5% 400mm × 510mm bd

[3.1]

Compression steel reinforcement ratio: 2 As ′ 600mm ρ′ = ------- = -------------------------------------------- = 0.003 = 0.3% bd 400mm × 510mm

[3.35]

b) Find the neutral axis location.

Use the equation of equilibrium [3.37] (assume that both tension and compression steel have yielded):

Cr + Cr ′ = Tr

[3.37]

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3-24

where

C r ′ = φ s f s ′A s ′ = 0.85 × 400 × 600 = 204kN T r = φ s f y A s = 0.85 × 400 × 3000 = 1020kN

[3.9]

Find C r from Eqn (3.37) as

C r = T r – C r ′ = 1020 – 204 = 816kN

[3.37]

Since

C r = α 1 φ c f c ′ab

[3.8]

it follows that 3 Cr 816 × 10 N a = --------------------- = --------------------------------------------------------------------------- = 131mm 0.8 × 0.65 × 30MPa × 400mm α 1 φ c f c ′b

and

131 a c = ----- = --------- = 145mm 0.9 β1 Check whether the compression steel has yielded. Find ε s ′ as

d′ 50 ε s ′ = 0.0035 1 – ---- = 0.0035 1 – --------- = 0.0023 c 145 Since

ε s ′ = 0.0023 > ε y = 0.002 it follows that the compression steel has yielded. Check whether the tension steel has yielded. Find ε s as

d 510 ε s = 0.0035 --- – 1 = 0.0035 --------- – 1 = 0.0088 c 145 Since

ε s = 0.0088 > ε y = 0.002 it follows that the tension steel has yielded. Strain distribution in the section is shown on the sketch below.

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3-25

c) Since the tension steel has yielded ( ε s = 0.0088 > ε y ) , it follows that the beam is expected to fail in the steel-controlled mode. d) Find M r as

a M r = C r ′ ( d – d′ ) + C r d – --2

[3.39]

3 3 131 = 204 × 10 N × ( 510 – 50 ) + 816 × 10 N × 510 – --------2

= 456.5kNm ≅ 456kNm Hence,

M r = 456kNm

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3-26

Chapter 4 - Solutions __________________________________________________________________________ 4.1. 3

3

bh ( 400mm ) ( 500mm ) 9 4 a) I g = --------- = -------------------------------------------------- = 4.17 × 10 mm 12 12 b)

- Find f r

λ = 1.0 (normal-density concrete) f r = 0.6λ f c ′ (A23.3 Eq. 8.3)

[2.1]

= 0.6 × 1.0 30MPa = 3.3MPa - Find y t

h 500 y t = --- = --------- = 250mm 2 2 - Find M cr 9 4 fr × Ig ( 3.3MPa ) ( 4.17 × 10 mm ) M cr = --------------- = -------------------------------------------------------------------- = 55kNm yt 250mm

[4.1]

c) Effective depth

d = 430mm (given) - Find the modular ratio ( n )

E c = 4500 f c ′ (A23.3 Eq.8.2) [2.2]

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4-1

= 4500 30MPa = 24648MPa E s = 200000MPa Es 200000MPa n = ------ = ------------------------------- = 8.1 ≅ 8.0 24648MPa Ec Find the transformed section area ( A tr ) 6-25M bars: for a 25M bar, Area = 500mm

A s = 6 × 500 = 3000mm

2

(Table A.1)

2

nA s = 8 × 3000 = 24000mm

2

A tr = b × h + nA s = 400 × 500 + 24000 = 224000mm

2

- Find the location of the centroid of the transformed section ( y )

h ( bh ) --- + ( nA s ) × d 2 y = --------------------------------------------A tr 500 ( 400 × 500 ) --------- + ( 24000 ) × 430 2 = ------------------------------------------------------------------------------------- = 269mm 224000 Find the moment of inertia ( I gtr ) for the transformed section. 3

bh 2 2 I gtr = --------- + bh ( y – y t ) + nAs ( d – y ) 12 3 400 × ( 500 ) 2 2 = ------------------------------- + 400 × 500 ( 269 – 250 ) + 24000 ( 430 – 269 ) 12 9

= 4.9 × 10 mm

4

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4-2

9

I gtr ≅ 5.0 × 10 mm

4

d) Find the cracked moment of inertia ( I cr ) 3

by 2 I cr = -------- + nA s ( d – y ) 3

[4.10]

3

400 × ( 269 ) 2 9 4 = ------------------------------- + ( 24000 ) ( 430 – 269 ) = 3.22 × 10 mm 3 9

I cr = 3.2 × 10 mm

4

_______________________________________________________________________ 4.2. a) Width of a unit strip: b = 1000mm

3

3

bh ( 1000mm ) ( 200mm ) I g = --------- = ----------------------------------------------------12 12 8

= 6.7 × 10 mm

4

b) Find f r

λ = 1.0 (normal-density concrete) f r = 0.6 × 1.0 25MPa = 3.0MPa (A23.3 Eq.8.3)

[2.1]

- Find y t

200 y t = --------- = 100mm 2 - Find M cr

M cr

8 4 fr Ig ( 3.0MPa ) ( 6.7 × 10 mm ) = --------- = ----------------------------------------------------------------- = 20.1kNm ⁄ m 100mm yt

[4.1]

M cr ≅ 20kNm ⁄ m c) Effective depth

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4-3

d = 170mm (given) - Find the modular ratio ( n )

E c = 4500 f c ′ (A23.3 Eq.8.2)

[2.2]

= 4500 25MPa = 22500MPa E s = 200000MPa Es 200000MPa n = ------ = ------------------------------- = 8.9 ≅ 9.0 22500MPa Ec - Find the transformed section area ( A tr ) Slab reinforcement: 15M@300 15M bars: Area = 200mm

2

(Table A.1)

Spacing s = 300mm

1000 1000 2 A s = A b × ------------ = 200 × ------------ = 667mm ⁄ m s 300

[3.28]

2

nA s = 9 × 667 ≅ 6000mm ⁄ m

A tr = b × h + nA s = 1000 × 200 + 6000 = 206000mm

2

- Find the location of the centroid of the transformed section ( y )

h ( bh ) --- + ( nA s )d 2 y = --------------------------------------A tr 200 ( 1000 × 200 ) × --------- + ( 6000 ) ( 170 ) 2 = ------------------------------------------------------------------------------------- = 102mm 206000 - Find ( I gtr )

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4-4

I gtr

3

b×h 2 2 = --------------- + b × h × ( y – y t ) + ( nA s ( d – y ) ) 12 3

1000 × ( 200 ) 2 2 = ---------------------------------- + ( 1000 × 200 ) ( 102 – 100 ) + 6000 ( 170 – 102 ) 12 8

= 696 × 10 mm

4

8

I gtr ≅ 7 × 10 mm

4

d) Find the cracked moment of inertia ( I cr ) 3

b×y 2 I cr = -------------- + nA s ( d – y ) [4.10] 3 3 1000mm × ( 102mm ) 2 2 8 4 = ------------------------------------------------------ + ( 6000mm ⁄ m ) ( 170 – 102 ) = 3.81 × 10 mm 3 8

I cr = 3.8 × 10 mm

4

________________________________________________________________________ 4.3.

a) - Find the cross sectional area

A = A 1 + A 2 = 600 × 200 + 500 × 300 = 270000mm

2

- Find the centroid ( y )

A1 y1 + A2 y2 y = -----------------------------A

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4-5

600 × 200 × 100 + 500 × 300 × 450 = ------------------------------------------------------------------------------------- = 294mm 270000 3

3

b1 × h1 b 2 × h2 2 2 I g = ----------------- + ----------------- + b 1 × h 1 ( y – y 1 ) + b 2 × h 2 ( y – y 2 ) 12 12 3

3

600 × ( 200 ) 300 × ( 500 ) 2 2 = ------------------------------- + ------------------------------- + ( 600 × 200 ) ( 294 – 100 ) + ( 300 × 500 ) ( 450 – 294 ) 12 12 8

4

8

4

= 116.9 × 10 mm I g ≅ 117 × 10 mm b) Effective depth

d = 600mm (given) Find the modular ratio ( n )

E c = 4500 f c ′ = 4500 25MPa = 22500MPa

(A23.3 Eq.8.2) [2.1]

E s = 200000MPa Es 200000MPa n = ------ = ------------------------------- = 8.9 ≅ 9.0 22500MPa Ec - Find the transformed section area ( A tr ) 6-25M bars: For a 25M bar, Area = 500mm

A s = 6 × 500 = 3000mm

2

(Table A.1)

2

nA s = 9 × 3000 = 27000mm

2

A tr = A + nA s = 270000 + 27000 = 297000mm

2

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4-6

- Find the centroid of the transformed section ( y )

A × y + ( nAs ) × d y = ------------------------------------------A tr 270000 × 294 + 27000 × 600 = ---------------------------------------------------------------------- = 322mm 297000 - Find I gtr 2

I gtr = I g + A ( y – y ) + nA s ( d – y )

2

2

8

2

10

= 117 × 10 + 270000 ( 294 – 322 ) + 27000 ( 600 – 322 ) = 1.4 × 10 mm 10

I gtr = 1.4 × 10 mm 8

4

4 4

10

4

c) I g = 117 × 10 mm = 1.2 × 10 mm (see part a) 10

I gtr = 1.4 × 10 mm

4

(see part b)

The difference is on the order of 17%, that is, 10 10 I gtr – I g × 10 – 1.2 × 10 ------------------- = 1.4 ------------------------------------------------------- = 0.167 ≅ 17 % 10 Ig 1.2 × 10

d) A 17% difference is significant enough from design perspective. If this were an actual design project it would be preferred to use I gtr value than I g value. __________________________________________________________________________ 4.4. a) - Find f r

λ = 1.0 (normal-density concrete) f r = 0.6 × λ f c ′ = 0.6 × 1.0 × 35MPa = 3.55MPa (A23.3 Eq.8.3) [2.1] - Find y t

y = 294mm (see Problem 4.3.) h = 700mm

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4-7

y t = h – y = 700 – 294 = 406mm

- Find M cr 8 4 fr × Ig 3.55MPa × ( 117 × 10 mm ) M cr = --------------- = ----------------------------------------------------------------------- = 102.3kNm [4.1] 406mm yt

y = 322mm (see Problem 4.3.) 3

300 ( 322 – 200 ) 2 2 I cr = ---------------------------------------- + ( 600 × 200 ) ( 322 – 100 ) + ( 27000 ) ( 600 – 322 ) 3 9

= 8.2 × 10 mm

4

Note: I cr is the moment of inertia around the axis x-x; only a cracked portion of the section above the axis x-x and the tension reinforcement are considered.

M cr c) I e = I cr + ( I g – I cr ) --------Ma

3

(A23.3 Eq. 9.1)

[4.11]

where 9

4

10

4

I cr = 8.2 × 10 mm

I g = 1.2 × 10 mm

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4-8

M cr = 102.3kNm The given M a values and the corresponding I e values calculated from Eqn (4.11) are summarized in the table below. Note that, according to CSA A23.3 Cl.9.8.2.3, I e ≤ I g ; hence, the I e values to be used in the further calculations based on this condition are framed in the table.

M a ( kNm )

4

4

I e ( mm )

I g ( mm ) 10

50

4.1 × 10

100

1.22 × 10

200 300

1.2 × 10

10

10

1.2 × 10

10

8.7 × 10

9

1.2 × 10

10

8.4 × 10

9

1.2 × 10

10

d)

It can be observed from the above diagram that the I e value decreases with an increase in bending moment ( M a ) value. Also, note the following: •

When M a ≤ M cr , then I e = I g

• When M a > M cr , then I e < I g _________________________________________________________________________ 4.5.

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4-9

Given:

M cr = 50kNm 10

I g = 10 × 10 mm

4

10

4

10

4

+

= 2 × 10 mm

–

= 4 × 10 mm

I cr I cr

Span 1

i) The effective moment of inertia at point 1 (positive moment region):

M a = 50kNm

M cr I em = I cr + ( I g – I cr ) --------Ma = 2 × 10

10

+ ( 10 × 10 10

= 10 × 10 mm

10

3

(A23.3 Eq. 9.1)

10 50kNm – 2 × 10 ) ------------------50kNm

[4.11]

3

4

Hence, 10

I em = I g = 10 × 10 mm

4

ii) The effective moment of inertia at point 2 (negative moment region):

M a = – 50kNm

M cr I ec = I cr + ( I g – I cr ) --------Ma

3

(A23.3 Eq. 9.1)

[4.11]

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4-10

= 4 × 10

10

+ ( 10 × 10

10

10 50kNm – 4 × 10 ) ------------------50kNm

3

10

= 10 × 10 mm

4

Hence, 10

I ec = I g = 10 × 10 mm

4

iii) Average moment of inertia for the span with one continuous end (CSA A23.3 Cl.9.8.2.4):

I eavg = 0.85I em + 0.15I ec (A23.3 Eq. 9.4)

[4.19]

= 0.85I g + 0.15I g = I g Note that, in reality, effective moment of inertia values for the positive and negative moment regions are less than the gross moment of inertia; however, the maximum positive and maximum negative bending moment values are not given. Therefore, it can be concluded that the I e value can be used for the design, that is, 10

I e = I g = 10 × 10 mm

4

Span 2

The effective moment of inertia values for points 1 and 2 are the same as for Span 1, that is, 10

4

I em = 10 × 10 mm = I g (point 1) 10

4

I e1 = I e2 = 10 × 10 mm = I g (point 2) The average moment of inertia value for Span 2 (both ends continuous) is

I eavg = 0.7I em + 0.15 ( I e1 + I e2 ) (A23.3 Eq. 9.3)

[4.18]

= 0.7I g + 0.15 ( I g + I g ) = I g Therefore, the average effective moment of inertia value 10

I e = I g = 10 × 10 mm

4

can be used for the Span 2 design. In conclusion, gross moment of inertia ( I g ) can be used for deflection calculations along the beam length (note that Span 3 has the same properties as Span 1).

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4-11

__________________________________________________________________________ 4.6. a)

Span l = 9m - Find bending moment at the midspan - Due to dead load

M DL

kN 2 7.5 ------- × ( 9m ) DL × l m = ------------------ = ------------------------------------- = 76kNm 8 8 2

- Due to live load

M LL

kN 2 7.5 ------- × ( 9m ) LL × l m = ------------------ = ------------------------------------- = 76kNm 8 8 2

- Total bending moment

M t = M DL + M LL = 152 kNm - Find f r

λ = 1.0 (normal-density concrete) f r = 0.6λ f c ′ (A23.3 Eq. 8.3)

[2.1]

= 0.6 × 1.0 30MPa = 3.3MPa - Find M cr

3

9 4 b×h 300 × ( 600 ) 3 I g = --------------- = ----------------------------- = 5.4 × 10 mm 12 12 h 600mm y t = --- = ------------------- = 300mm 2 2

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4-12

fr × Ig M cr = --------------yt

[4.1]

9

4

3.3MPa ( 5.4 × 10 mm ) = ------------------------------------------------------------ = 60kNm 300mm Since

M DL = 76kNm > 60kNm it follows that the cracking has taken place due to the dead load only.

- Find the modular ratio ( n )

E c = 4500 f c ′ (A23.3 Eq.8.2) [2.2] = 4500 30MPa = 24650MPa E s = 200000MPa Es 200000MPa n = ------ = ------------------------------- = 8.1 ≅ 8.0 24650MPa Ec - Find the location of the centroid of the cracked section ( y ). 4-25M bars: for a 25M bar, Area = 500mm

A s = 4 × 500 = 2000mm

2

(Table A.1)

2

2

nA s = 8 ( 2000mm ) = 16000mm

2

2 As 2000mm ρ = ------------ = -------------------------------------------- = 0.0126 b×d 300mm × 530mm

nρ = 8 × 0.0126 = 0.1 2

y = d ( nρ + 2nρ – nρ )

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4-13

2

= 530mm ( ( 0.1 ) + 2 ( 0.1 ) – 0.1 ) = 190mm - Find I cr

3

b×y 2 I cr = -------------- + nA s ( d – y ) [4.10] 3 300mm × ( 190mm ) 3 3 2 = ------------------------------------------------- + 16000mm ( 530mm – 190mm ) 3 9

I cr = 2.5 × 10 mm

4

Find the I cr ⁄ I g ratio as

9 4 I cr × 10 mm ------ = 2.5 ---------------------------------- = 0.46 ≅ 50% 9 4 Ig 5.4 × 10 mm - Find the effective moment of inertia ( I e ).

M a = 152kNm M cr = 60kNm

M cr I e = I cr + ( I g – I cr ) --------Ma

3

(A23.3 Eq. 9.1)

9 4 9 4 60kN ⋅ m = 2.5 × 10 mm + ( 5.4 – 2.5 ) × 10 mm ----------------------152kNm

[4.11] 3

9

= 2.7 × 10 mm

4

- Find the deflection ( ∆ t ) due to the total load. 2

5 Ma l ∆ t = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 152 × 10 Nmm ) ( 9000mm ) - = 19.3mm ≅ 20mm = 1.0 ------ -----------------------------------------------------------------------48 ( 24650MPa ) ( 2.7 × 10 9 mm 4 ) ∆ t = 20mm Note that k = 1.0 (simply supported beam, see Figure 4.10) b) - Find the effective moment of inertia ( I e ).

M a = 76kN ⋅ m M cr = 60kN ⋅ m M cr I e = I cr + ( I g – I cr ) --------Ma

3

(A23.3 Eq. 9.1)

[4.11]

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4-14

60kNm 9 4 9 4 = 2.5 × 10 mm + ( 5.4 – 2.5 ) × 10 mm × ------------------76kNm 9

= 3.9 × 10 mm

3

4

- Find the deflection ( ∆ DL ) due to the dead load only. 2

∆ DL

5 Ma l = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 76 × 10 Nmm ) × ( 9000mm ) = ( 1.0 ) ------ ---------------------------------------------------------------------------48 24650MPa ( 3.9 × 10 9 mm 4 ) = 7mm

Note that k = 1.0 (simply supported beam, see Figure 4.10) c) Find the deflection ( ∆ LL ) due to the live load only.

∆ LL = ∆ t – ∆ DL = 20mm – 7mm = 13mm d) Find the total long-term deflection ( ∆ t ) based on the following equation

∆t = ζs ∆i where

s ζ s = 1 + -------------------1 + 50ρ′

(A23.3 Eq.9.5)

[4.15]

= 1+s = 1+2 = 3 Note:

ρ′ = 0 (no compression reinforcement s = 2 (5 years or more, see Table 4.1) The sustained portion of the load is due to dead load. Based on parts a), b) and c), it follows that

∆ i = ∆ DL = 7mm (immediate deflection due to dead load only) ∆ tDL = ζ s ∆ DL = 3 ( 7mm ) = 21mm (long-term deflection due to dead load only) ∆ LL = 13mm (immediate live load deflection)

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4-15

Hence, the total long-term deflection is equal to

∆ t = ∆ tDL + ∆ LL = 21mm + 13mm = 34mm __________________________________________________________________________ 4.7. a) Suppose that the architectural elements are installed immediately after the formwork has been stripped off. Therefore, the initial deflection due to dead load will have occurred prior to this installation. Note that the following deflections were determined in Problem 4.6. as:

∆ i = ∆ DL = 7mm (immediate deflection due to dead load only) ∆ tDL = ζ s ∆ DL = 3 ( 7mm ) = 21mm (long-term deflection due to the sustained dead load only) ∆ LL = 13mm (immediate live load deflection) The total incremental deflection ( ∆ t ) after stripping of the formwork can be determined as

∆ t = ∆ tDL – ∆ DL + ∆ LL = 21mm – 7mm + 13mm = 27mm Next, find the allowable deflections per CSA A23.3-04. For the part of total deflection occurring after attachment of nonstructural elements, Table 9.3 of CSA A23.3 (Cl.9.8.5.3) prescribes the following deflection limit:

ln ∆ ≤ --------480 where

l n ≅ l = 9000mm (see Problem 4.6.) and

ln 9000mm -------- = --------------------- = 19mm 480 480 Since

∆ t = 27mm > 19mm It follows that the beam deflections exceed the CSA A23.3-prescribed limits. b) Find the total incremental deflection one year after the construction has been completed.

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The incremental deflection can be determined from the following equation

∆ t = ∆ tDL – ∆ DL1 + ∆ LL where

∆ LL = 13mm (immediate live load deflection, see part a) ∆ tDL = 21mm (long-term deflection due to sustained dead load over a period of 5 years or more, see part a) s ζ s1 = 1 + -------------------1 + 50ρ′

(A23.3 Eq.9.5)

[4.15]

= 1 + s = 1 + 1.4 = 2.4 ρ′ = 0 (no compression reinforcement s = 1.4 (load sustained for 1 year, see Table 4.1) Hence, the long-term deflection when the dead load has been sustained for 1 year can be determined as

∆ DL1 = ζ s1 ∆ DL = 2.4 × 7mm = 17mm Finally, the total deflection can be determined as

∆ t = ∆ tDL – ∆ DL1 + ∆ LL = 21mm – 17mm + 13mm = 17mm Since

∆ t = 17mm < 19mm It can be concluded that the deflection is within the CSA A23.3 prescribed limits discussed in part a) of this problem. c) The incremental deflection affecting the architectural elements can be decreased by almost 40% by delaying the installation of architectural elements and allowing a large portion of the creep deflection to occur prior to the installation. In general, live load is considered a transient load, hence the focus should be on trying to reduce the incremental creep deflection due to sustained dead load. The decrease in creep deflection can be significant when the installation of architectural elements is delayed by a year, as discussed below.

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The total deflection after a 1-year period when the architectural elements are installed immediately (no delay) is equal to:

∆ t = ζ s ∆ DL – ∆ DL = 3 ( 7mm ) – 7mm = 14mm The total deflection after a 1-year period when the installation of architectural elements has been delayed is equal to:

∆ t = ζ s ∆ DL – ∆ DL1 = 3 ( 7mm ) – 17mm = 4mm The difference is equal to

14mm – 4mm ----------------------------------- = 0.71 ≅ 70 % 14mm It can be concluded that the decrease in sustained deflections can be on the order of 70% when the installation of architectural elements has been delayed. __________________________________________________________________________ 4.8. According to CSA A23.3 Cl.9.8.2.1, detailed deflection calculations are not required provided that flexural members have minimum thickness prescribed by Table 9.2 of CSA A23.3. For simply supported beams, Table 9.2 prescribes that

ln ---- ≤ 16 h where

l n = l = 9000 mm (a conservative assumption, since the exact clear span length is not given) Hence,

ln 9000mm h min = ------ = ---------------------- = 562mm 16 16 Since

h = 600mm (given) and

600mm > 562mm it can be concluded that detailed deflection calculations according to CSA A23.3 are not required.

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_______________________________________________________________________ 4.9. a) The following properties were used in developing the beam model:

f c ′ = 30MPa 9

4

9

4

I g = 5.4 × 10 mm I cr = 2.5 × 10 mm M cr = 60kNm

1) First, the beam was modeled using gross section properties throughout the span ( I g ). Bending moment diagram for the total service load was developed and all areas with M > M cr were identified. The beam was analyzed using a computer structural analysis software. 2) Next, a new model was created using gross section properties ( I g ) in the region where M ≤ M cr , and using I cr where M > M cr . The corresponding bending moment diagram was developed to check whether additional beam segments cracked, however no further changes were noticed. Hence, further iterations were not required. Bending moment diagram for the second model is shown below.

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Finally, the deflection diagram was plotted for the second beam model. The diagram is shown below. Maximum deflection of 20.7 mm occurs at the midspan.

b) Beam deflections for the model analyzed using iterative procedure discussed in part a) were on the order of 20.7 mm. This value compares very well with the value of 20 mm obtained using the CSA A23.3 prescribed simplified deflection procedure followed in Problem 4.6.a. Hence, for prismatic simply supported beams, the CSA A23.3 prescribed approximate procedure is acceptable, and there is no need to carry out the computer aided iterative procedure. ________________________________________________________________________ 4.10. a) - Total service load

w = DL + LL = 7.5 + 7.5 = 15kN ⁄ m Given beam properties (see Problem 4.6.):

M cr = 60kNm 9

I g = 5.4 × 10 mm

4

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9

I cr = 2.5 × 10 mm

4

n = 8 E c = 24650MPa The beam model was analyzed using a structural analysis software package. Gross sectional properties were used ( I g ) and the beam was subjected to the total service load w = 15kN ⁄ m . Bending moment diagram for the beam is shown below (note that the values have been reduced at the face of the support).

Span 1 - properties

- Effective moment of inertia at the midspan:

M cr 3 I em = I cr + ( I g – I cr ) --------- (A23.3 Eq. 9.1) [4.11] Ma 60 3 9 4 9 4 = 2.5 + ( 5.4 – 2.5 ) --------× 10 mm = 3.1 × 10 mm 100 - Effective moment of inertia at the interior support:

M cr 3 I ec = I cr + ( I g – I cr ) --------- (A23.3 Eq. 9.1) [4.11] Ma 60 3 9 4 9 4 = 2.5 + ( 5.4 – 2.5 ) --------× 10 mm = 2.6 × 10 mm 175 - Average moment of inertia for Span 1 (one continuous end):

I eavg = 0.85I em + 0.15I ec (A23.3 Eq. 9.4) 9

4

[4.19] 9

= [ ( 0.85 ) ( 3.1 ) + 0.15 ( 2.6 ) ] × 10 mm = 3.0 × 10 mm

4

Span 2 - properties

- Effective moment of inertia at the midspan:

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M cr I em = I cr + ( I g – I cr ) --------Ma 60 = 2.5 + ( 5.4 – 2.5 ) -----90

3 3

(A23.3 Eq. 9.1) 9

4

[4.11] 9

× 10 mm = 3.4 × 10 mm

4

- Effective moment of inertia at the right support:

M cr 3 I ec = I cr + ( I g – I cr ) --------- (A23.3 Eq. 9.1) [4.11] Ma 60 3 9 4 9 4 = 2.5 + ( 5.4 – 2.5 ) --------× 10 mm = 2.6 × 10 mm 170 - Average moment of inertia for Span 2 (two continuous ends):

I eavg = 0.7I em + 0.15 ( I e1 + I e2 ) (A23.3 Eq. 9.3) 9

4

[4.18] 9

= [ ( 0.7 ) ( 3.4 ) + 0.15 ( 2.6 + 2.6 ) ] × 10 mm = 3.2 × 10 mm

4

Span 1 - deflections due to total service load

l n = l – 0.2m = 10m – 0.2m = 9.8m clear span Mo 88 k = 1.2 – 0.2 --------- = 1.2 – 0.2 --------- = 0.83 100 Mm kN 2 15 ------- × ( 10m ) w×l m M o = -------------- = -------------------------------------- = 188kNm 8 8 2

M a = M m = 100kNm 2

∆ 1t

5 Ma l = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 100 × 10 Nmm ) × ( 9800mm ) = 0.83 ------ ------------------------------------------------------------------------------= 11mm 48 24650MPa × 3.1 × 10 9 mm 4 Span 2 - deflections due to total service load l n = l – 0.2m = 12m – 0.2m = 11.8m clear span Mo 270 k = 1.2 – 0.2 --------- = 1.2 – 0.2 --------- = 0.6 90 Mm kN 2 15 ------- × ( 12m ) w×l m M o = -------------- = -------------------------------------- = 270kNm 8 8 2

M a = M m = 90kNm

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2

∆ 2t

5 Ma l = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 90 × 10 Nmm ) × ( 11800mm ) = 0.6 ------ ------------------------------------------------------------------------------= 10mm 48 24650MPa × 3.2 × 10 9 mm 4 Note that Spans 3 and 4 are mirror images for Spans 1 and 2. b) The bending moment diagram due to dead load only obtained by means of elastic analysis is shown below.

Span 1 - properties

- Effective moment of inertia at the midspan: 9

I em = I g = 5.4 × 10 mm

4

- Effective moment of inertia at the interior support:

M cr I ec = I cr + ( I g – I cr ) --------Ma 60 = 2.5 + ( 5.4 – 2.5 ) -----88

3 3

(A23.3 Eq. 9.1) 9

[4.11]

4

9

× 10 mm = 3.4 × 10 mm

4

- Average moment of inertia for Span 1 (one continuous end):

I eavg = 0.85I em + 0.15I ec (A23.3 Eq. 9.4) 9

4

[4.19] 9

= [ ( 0.85 ) ( 5.4 ) + 0.15 ( 3.4 ) ] × 10 mm = 5.1 × 10 mm

4

Span 2 - properties

- Effective moment of inertia at the midspan: 9

I em = I g = 5.4 × 10 mm

4

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- Effective moment of inertia at the support: 9

I ec ≅ 3.4 × 10 mm

4

- Average moment of inertia for Span 2 (two continuous ends):

I eavg = 0.7I em + 0.15 ( I e1 + I e2 ) (A23.3 Eq. 9.3) 9

[4.18]

4

9

= [ ( 0.7 ) ( 5.4 ) + 0.15 ( 3.4 + 3.4 ) ] × 10 mm = 4.8 × 10 mm

4

Span 1 - deflections due to dead load

k = 0.83 (same as part a) M a = 50kNm ∆ 1DL

2

5 Ma l = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 50 × 10 Nmm ) × ( 9800mm ) = 0.83 ------ ---------------------------------------------------------------------------= 4mm 48 24650MPa × 5.1 × 10 9 mm 4 Span 2 - deflections due to dead load k = 0.6 (same as part a) M a = 45kNm ∆ 2DL

2

5 Ma l = k ------ -----------48 E c I e

[4.12]

6

2

5 ( 45 × 10 Nmm ) × ( 11800mm ) = 0.6 ------ ------------------------------------------------------------------------------= 3.3mm 48 24650MPa × 4.8 × 10 9 mm 4 c) Live load deflection Span 1

∆ 1LL = ∆ 1t – ∆ 1DL = 11 mm – 4mm = 7mm Span 2

∆ 2LL = ∆ 2t – ∆ 2DL = 10 mm – 3.3mm ≅ 7mm d) Find the total long-term deflection ( ∆ t ) based on the following equation

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∆t = ζs ∆i where

s ζ s = 1 + -------------------1 + 50ρ′

(A23.3 Eq.9.5)

[4.15]

= 1+s = 1+2 = 3 Note:

ρ′ = 0 (no compression reinforcement s = 2 (5 years or more, see Table 4.1) Span 1

The sustained portion of the load is due to dead load:

∆ i = ∆ DL = 4mm (immediate deflection due to dead load only) ∆ tDL = ζ s ∆ DL = 3 ( 4mm ) = 12mm (long-term deflection due to dead load only) ∆ LL = 7mm (immediate live load deflection) Hence, the total long-term deflection is equal to

∆ t = ∆ tDL + ∆ LL = 12mm + 7mm = 19mm Span 2

The sustained portion of the load is due to dead load:

∆ i = ∆ DL = 3.3mm (immediate deflection due to dead load only) ∆ tDL = ζ s ∆ DL = 3 ( 3.3mm ) = 10mm (long-term deflection due to dead load only) ∆ LL = 7mm (immediate live load deflection) Hence, the total long-term deflection is equal to

∆ t = ∆ tDL + ∆ LL = 10mm + 7mm = 17mm

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_____________________________________________________________________ 4.11. a) The following properties were used in developing the beam model:

f c ′ = 30MPa 9

4

9

4

I g = 5.4 × 10 mm I cr = 2.5 × 10 mm M cr = 60kNm

1) First, the beam was modeled using gross section properties throughout the span ( I g ). Bending moment diagram for the total service load was developed and all areas with M > M cr were identified. The beam was analyzed using a computer structural analysis software. 2) Next, a new model was created using gross section properties ( I g ) in the region where M ≤ M cr , and using I cr where M > M cr . The corresponding bending moment diagram was developed to check whether additional beam segments cracked, however no further changes were noticed. Hence, further iterations were not necessary.

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4-26

Bending moment diagram for the second model is shown below.

Finally, the deflection diagram was plotted for the second beam model, and the diagram is shown below.

b) A comparison of deflection calculations obtained by the CSA A23.3 approximate procedure (see Problem 4.10.) and the computer-aided iterative procedure are summarized in the table below. Maximum deflections (mm) Span

A23.3 approximate procedure

Computer-aided iter- Difference ative procedure

(Problem 4.10.) 1

11

13.2

20 %

2

10

12.6

26 %

It can be noticed from the table that the difference in deflection values obtained using these two procedures is on the order of 20%, however the results still compare reasonably well. Hence, for

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continuous beams with prismatic sections where adjacent span lengths are not significantly different, the CSA A23.3 approximate procedure is acceptable and there is no need to use the computer-aided iterative procedure for deflection calculations. __________________________________________________________________________ 4.12. a) Find the long-term deflection due to total dead and live load.

∆ D = 10mm (immediate deflection due to dead load only) ∆ tD = ζ s ∆ D = 3 ( 10mm ) = 30mm (long-term deflection due to the sustained dead load only) where

ζ s = 3 (assume deflection sustained for 5 years or more, see Table 4.1) ∆ L = 3mm (immediate live load deflection) The total long-term deflection ( ∆t ) can be determined as

∆ t = ∆ tD + ∆ L = 30mm + 3mm = 33mm Next, find the allowable deflections per CSA A23.3-04. Table 9.3 of CSA A23.3 (Cl.9.8.5.3) prescribes the following deflection limit for the case of “roof construction supporting or attached to non-structural elements likely to be damaged by large deflections”:

ln ∆ ≤ --------480 where

l n = 10000mm (given) and

ln 10000mm -------- = ------------------------ = 21mm 480 480 Since

∆ t = 33mm > 21mm it follows that long-term deflections exceed the CSA A23.3-prescribed limits. Hence, it can be concluded that the damage to architectural elements is possible if these elements are installed prior to stripping off the formwork.

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b) In reality, architectural elements cannot be installed until at least the formwork has been stripped off. First, suppose that the formwork has been stripped off and the architectural elements have been installed immediately thereafter. At that stage, the floor slab supports the dead load, which causes the immediate dead load deflection. The incremental deflection after the installation of architectural elements can be determined as the total deflection (considering that the dead load has been sustained for 5 years or more) less the immediate deflection due to dead load, that is,

∆ inc = ∆ t – ∆ D = ∆ tD + ∆ L – ∆ D = 30mm + 3mm – 10mm = 23mm Note that ∆ t and ∆ D were determined in part a). Since

∆ inc = 23mm > 21mm it follows that the incremental deflection exceeds the CSA A23.3-prescribed limit. Next, suppose that the installation of the architectural elements was performed 3 months after the formwork has been stripped off. Long-term deflection for the load sustained over 3 month period can be determined as follows:

s ζ s1 = 1 + -------------------1 + 50ρ′

(A23.3 Eq.9.5)

[4.15]

= 1+s = 1+1 = 2 ρ′ = 0 (no compression reinforcement) s = 1 (load sustained for 3 months, see Table 4.1) Hence, the long-term deflection when the dead load has been sustained for 3 months can be determined as

∆ D3months = ζ s ∆ D = 2 × 10mm = 20mm The incremental deflection for the architectural elements installed 3 months after the formwork has been stripped off can be determined as

∆ inc = ∆ tD + ∆ L – ∆ D3months = 30mm + 3mm – 20mm = 13mm Since

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∆ inc = 13mm < 21mm it follows that the incremental deflection is within the CSA A23.3-prescribed limit. It should be noted that, when architectural elements were installed immediately after stripping off the formwork, the incremental deflection was equal to 23 mm; this is very close to the allowable limit of 21 mm. Therefore, had the installation of architectural elements been performed just 2 weeks after stripping off the formwork, the deflection would be within the permissible limits - this is more suitable from the aspect of construction schedule as well. c) If the beam was constructed with an initial camber of 20 mm, then the total long-term dead load deflection after a period of 5 years or more can be determined as

∆ = ∆ tD – camber = 30mm – 20mm = 10mm where ∆ tD was determined in part a). In this case, long-term deflection due to dead load would be on the order of 10 mm. When this deflection is expressed as a fraction of the beam span, that is,

ln 10000mm 10mm = ------------ = ------------------------1000 1000 This deflection is well below the CSA A23.3 prescribed limit of 21 mm for a beam with 10 m span; also, small deflections are very favourable for their appearance. d) When both the dead and live load are sustained, the long-term deflection can be determined as follows:

∆ = ζ s ( ∆ D + ∆ L ) = 3 ( 10mm + 3mm ) = 39mm ≅ 40mm Determine the magnitude of 2% positive slope at the midspan:

l 10000mm slope = 0.02 × --- = 0.02 × ------------------------- = 100mm 2 2 In order to ensure that a 2% slope at the midspan is maintained in the long-term, the beam would need to be constructed with a positive camber of

camber = slope + ∆ = 100 mm + 40mm = 140mm

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____________________________________________________________________________ 4.13.

- Compute the effective tension area per bar ( A ).

h = 700mm (given) d = 630mm (given) d c = h – d = 700 – 630 = 70mm d s = d c = 70mm A e = b ( 2d s ) = 300mm ( 2 × 70mm ) = 42000mm

2

There are three bars, hence 2 Ae 42000mm 2 A = ------ = --------------------------- = 14000mm 3 3

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Determine the stress in reinforcement ( f s )

f s = 0.6 × f y = 0.6 × 400MPa = 240MPa (this is based on CSA A23.3 Cl.10.6.1 in lieu of detailed calculations) Determine the value of z factor.

z = f s 3 d c A (A23.3 Eq.10.6)

[4.20] 2

= ( 240MPa ) 3 ( 70mm ) ( 14000mm ) = 23839N ⁄ mm ≅ 24000N ⁄ mm The beam is of exterior exposure, therefore

max z = 25000N ⁄ mm Since

z = 24000N ⁄ mm < max z = 25000N ⁄ mm it follows that the CSA A23.3 cracking control requirements for this beam are satisfied, considering the exterior exposure. _________________________________________________________________________ 4.14.

- Compute the effective tension area per bar ( A ).

h = 300mm (given) d = 250mm (given) d c = h – d = 300 – 250 = 50mm d s = d c = 50mm Copyright © 2006 Pearson Education Canada Inc.

4-32

The effective tension area per bar ( A ) will be computed as follows

A = s × ( 2d s ) = 250mm × ( 2 × 50mm ) = 25000mm

2

Determine the stress in reinforcement ( f s )

f s = 0.6f y = 0.6 × 400MPa = 240MPa (this is based on CSA A23.3 Cl.10.6.1 in lieu of detailed calculations) Determine the value of z factor

z = fs 3 dc A

(A23.3 Eq.10.6)

[4.20]

= ( 240MPa ) 3 ( 50 ) ( 25000 ) = 25853N ⁄ mm ≅ 26000N ⁄ mm The slab is of exterior exposure, therefore

max z = 30000N ⁄ mm Since

z = 26000N ⁄ mm < max z = 30000N ⁄ mm it follows that the CSA A23.3 cracking control requirement for this slab have been satisfied.

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Chapter 5 - Solutions __________________________________________________________________________ 5.1. a) Calculate the factored moment resistance.

α 1 = 0.8

β 1 = 0.9 .

5-30M bars: A s = 5 × 700 = 3500mm

2

(Table A.1)

2 φs fy As 0.85 × 400MPa × 3500mm --------------------------------------------------------------------------------------------- = 191mm a = = 0.8 × 0.65 × 30MPa × 400mm α 1 φ c f c ′b

a M r = φ s A s f y d – --2

[3.12]

[3.14]

191 2 = 0.85 × 3500mm × 400MPa 830 – --------- = 874kNm 2 b) Check the whether the amount of tension reinforcement meets the minimum CSA A23.3 requirements. i) Minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2)

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 30MPa 2 = ------------------------------- × 400mm × 900mm = 986mm 400MPa (Note: b t = b rectangular sections) Since

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5-1

2

A s = 3500mm > A smin = 986mm

2

okay

ii) Maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2) 2 As 3500mm ρ = ------ = -------------------------------------------- = 0.01 = 1% 400mm × 830mm bd

[3.1]

Balanced reinforcement ratio:

ρ b = 0.027 ( f c ′ = 30MPa , see Table A.4) Since

ρ = 0.001 < ρ b = 0.027 okay minimum and maximum CSA A23.3 reinforcement requirements are satisfied. ________________________________________________________________________ 5.2. a) Determine the required clear cover. The beam is of interior exposure, so cover = 30mm (Table A.2, Class N) b) Determine the clear spacing between the bars. CSA A23.1 requirement (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 30mm = 42mm (30M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 42mm governs (largest value). Find the actual bar spacing.

300 – 2 × 30 – 2 × 10 – 3 × 30 s = ---------------------------------------------------------------------- = 65mm 2 Since

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5-2

s = 65mm > 42mm okay

c) Determine the effective depth.

30 d 1 = 500 – 30 – 10 – ------ = 445mm 2 30 d 2 = d 1 – 42 – ------ = 388mm 2 3 × d 1 + 2 × d 2 3 × 445 + 2 × 388 d = ----------------------------------- = ------------------------------------------- = 422mm 5 5

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5-3

d) Determine the effective depth for the revised design.

d 1 = 445mm (same as part c) d 2 = 388mm 3 × d1 + 3 × d2 d = ----------------------------------6 3 × 445 + 3 × 388 = ------------------------------------------- = 416mm 6 e) Calculate the difference in the effective depth determined in parts c) and d). The difference in d value for design in part c) (5-30M bars) and design in part d) (6-30M bars) is 6 mm, or

422 – 416 ------------------------ = 1.4% 416 This difference is not significant from design perspective.

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_________________________________________________________________________ 5.3. a) Calculate the factored moment resistance for the beam section.

2

4-20M bars: (20M bar area = 300mm , see Table A.1)

A s = 4 × 300 = 1200mm φs fy As a = --------------------α 1 φ c f c ′b

2

[3.12]

2

0.85 × 400MPa × 1200mm = --------------------------------------------------------------------------- ≅ 87mm 0.8 × 0.65 × 30MPa × 300mm a 87 2 M r = φ s A s f y d – --- = 0.85 × 1200mm × 400MPa 300 – ------ = 116.9kNm [3.14] 2 2 M r ≅ 117kNm b) Find the factored moment resistance if the effective depth is reduced to 300 mm.

d = 300mm a = 87mm (same as part a) 2 a 87 M r = φ s A s f y d – --- = 0.85 × 1200mm × 400MPa 300 – ------ = 104.7kNm [3.14] 2 2

M r ≅ 105kNm c) Find the factored moment resistance if the 28-day compressive strength is 27 MPa.

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f c ′ = 27MPa d = 330mm 30MPa a = 87mm × ------------------- = 96.7mm ≅ 97mm 27MPa 2 a 97 M r = φ s A s f y d – --- = 0.85 × 1200mm × 400MPa × 330 – ------ = 114.9kNm [3.14] 2 2 M r ≅ 115kNm d) Find the factored moment resistance if the effective depth is reduced to 300 mm and the 28-day compressive strength. Since d = 300mm and f c ′ = 27MPa it follows that a = 97mm (same as part c).

a 97 2 M r = φ s A s f y d – --- = 0.85 × 1200mm × 400MPa × 300 – ------ = 102.6kNm [3.14] 2 2 M r ≅ 103kNm e) A comparison Design

d ( mm )

f c ′ (MPa)

M r (kNm)

5.3.a)

330

30

117

5.3.b)

300

30

105

5.3c)

330

27

115

5.3.d)

300

27

103

i) The effect of variation in the f c ′ value is not significant. A decrease in the f c ′ value of 10%, that is, from 30 MPa (part a) to 27 MPa (part c), results in an decrease in the M r from 117 kNm (part a) to 115 kNm (part c), that is, by approximately 2%; this is not a significant decrease from design perspective. ii) The effect of variation in d value is more significant. A 10 % decrease in the d value (from 330 mm in part a) to 300 mm in part c) results in a 10% decrease in the M r value (from 117 kNm to 105 kNm); this is a significant decrease. iii) When both the d and f c ′ values decreased by 10% between the designs in parts a) and d), the corresponding M r value decreased from 117 kNm to 103 kNm, that is, by approximately 12%.

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5-6

In conclusion, the effect of bar placement on moment resistance is much more significant then the effect of concrete strength. f) Discuss the effect of various construction errors. i) A lower f c ′ value than specified by the design can be tolerated when a 5% reduction in the M r value is considered acceptable. ii) A misplacement of rebars resulting in 10% drop in the d value can be tolerated when a 10% reduction in the M r value is considered acceptable. g) It is recommended to have a reserve in M r value on the order of 10% while performing design of reinforced concrete flexural members. Since it is hard to predict possible construction errors while designing a structure, it is recommended to perform design in a conservative manner in order to account for possible construction errors. __________________________________________________________________________ 5.4. a) Find the minimum required number of 25M rebars for the beam section required to support the given load.

Perform the load analysis. Dead load:

kN - self-weight = b × h × γ w = 0.4m × 0.6m × 24.0 ------3- = 5.8kN ⁄ m m

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= 5kN ⁄ m

- superimposed dead load

_______________________

DL = 10.8kN ⁄ m Live load:

LL = 35kN ⁄ m

Factored load (NBC 2005 Table 4.1.3.2)

kN w f = 1.25DL + 1.5LL = 1.25 × 10.8 + 1.5 × 35.0 ≅ 66.0 ------m Span l = 6m

kN 2 66 ------- ( 6.0m ) 2 wf × l m M f = ---------------- = --------------------------------------- = 297.0kNm ≅ 300kNm 8 8 - Estimate d value, assuming one layer of reinforcement. - interior exposure: cover = 30mm (Table A.2) - stirrup diameter: d s = 10mm (10M bar) - bar diameter: d b = 25mm (25M bar)

db 25 d = h – cover – d s – ----- = 600 – 30 – 10 – ------ = 547mm 2 2 Estimate a value:

a = 0.3d = 0.3 × 547 = 164mm 6 Mf 300 × 10 Nmm 2 2 --------------------------------------------------------------------------------------------------As = = = 1897mm ≅ 2000mm [5.5] 164 a 547 – --------- × 0.85 × 400MPa d – --- φ s f y 2 2 2 φs fy As 0.85 × 400MPa × 2000mm --------------------------------------------------------------------------------------------= = 109mm [3.12] a = 0.8 × 0.65 × 30MPa × 400mm α 1 φ c f c ′b

The difference between the initial estimate ( a = 164mm ) and the new one ( a = 109mm ) is significant. Therefore, estimate A s value using a = 109mm in the equation [5.5]: 6 Mf 300 × 10 2 A s = --------------------------- = ------------------------------------------------------------- = 1817mm [5.5] 109 a 540 – --------- × 0.85 × 400 d – --- φ s f y 2 2

φs fy As 0.85 × 400 × 1817 a = --------------------- = --------------------------------------------------- = 99mm α 1 φ c f c ′b 0.8 × 0.65 × 30 × 400

[3.12]

Copyright © 2006 Pearson Education Canada Inc.

5-8

The difference between the A s values from the last two iterations is

2000 – 1817----------------------------= 0.10 = 10% 1817 Let us perform one more iteration using a = 99mm in equation [5.5]: 6 Mf 2 300 × 10 ----------------------------------------------------------------------------------As = = 1800mm [5.5] = 99 a 540 – ------ × 0.85 × 400 d – --- φ s f y 2 2 2

The A s value obtained in the last iteration ( 1800mm ) is very similar to the previous value 2

( 1817mm ) , hence there is no need to perform further iterations. Use 4-25M bars, hence A s = 4 × 500 = 2000mm

φs fy As 0.85 × 400 × 2000 a = --------------------- = ------------------------------------------------ = 109mm α 1 φ c f c ′b 0.8 ( 0.65 ) ( 30 ) ( 400 )

2

[3.12]

a 109 M r = φ s A s f y d – --- = 0.85 × 2000 × 400 540 – --------- = 330.0kNm [3.14] 2 2 Since

M r = 330kNm > M f = 300kNm okay Check whether the beam is properly reinforced. Find c .

109 a c = ----- = --------- = 121mm 0.9 β1 where

β 1 = 0.9 Use the following proportion:

c 0.0035 --- = -------------------------d 0.0035 + ε y and also

d 530 ε s = 0.0035 --- – 1 = 0.0035 --------- – 1 = 0.012 c 121 Since

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5-9

fy 400MPa ε y = ----- = ------------------------------- = 0.002 200000MPa Es and

ε s = 0.012 > ε y = 0.002 It can be concluded that the beam is properly reinforced. The assumption that f s = f y used in M r calculation is correct; this also indicates that the amount of reinforcement is less than the maximum value. - Check CSA A23.3 minimum tension reinforcement requirement (Cl.10.5.1.2)

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 30MPa 2 = ------------------------------- × 400mm × 600mm = 657mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 2000mm > 657mm okay - Check whether 4-25M bars can fit in one layer. CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 25mm = 35mm (25M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 35mm governs (largest value). - Find actual bar spacing.

400 – 2 × 30 – 2 × 10 – 4 × 25 s = ------------------------------------------------------------------------ = 73mm 3 Since

s = 73mm > 35mm okay it follows that 4 bars can fit in 1 layer.

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5-10

b) Provide a design summary.

_________________________________________________________________________ 5.5. a) Estimate beam depth based on CSA A23.3 Cl.9.8.2.1(see Table A.3 in Appendix A). Clear span: l n = 5m For a cantilever beam:

l n 5000mm h ≥ ---- = ---------------------- = 625mm 8 8 Round to h = 700mm . Note that b = 400mm . b) Design the required tension reinforcement for the beam such that it can carry design loads. Perform load analysis. - Self-weight - Dead load

kN kN = b × h × γ w = 0.4 m × 0.7m × 24.0 ------3- = 6.7 ------m m = 15.0kN ⁄ m _______________________

DL = 21.7kN ⁄ m - Live load:

LL = 25kN ⁄ m

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5-11

- Factored load (NBC 2005 Table 4.1.3.2)

kN kN w f = 1.25DL + 1.5LL = 64.6 ------- ≅ 65 ------m m - Find factored bending moment for a cantilever beam (see Table A.16 in Appendix A) 2

2 wf × l (----------------------------65.0 ) ( 5.0 ) --------------Mf = = = 812.5kNm 2 2

M f ≅ 810kNm - Estimate d value.

d = h – 70mm = 700 – 70 = 630mm

- Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 810 × 10 2 2 = 0.0015 ( 30MPa ) ( 400mm ) 630mm – ( 630mm ) – ------------------------------------------------ = 4678mm ( 30MPa ) ( 400mm ) Use 10-25M bars. Since 2

A s = 10 × 500 = 5000mm > 4678mm

2

okay

- Check whether 10 - 25M bars can fit in 2 layers. cover = 30 mm (Table A.2, beam not exposed) CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 25mm = 35mm (25M bar)

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5-12

1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 35mm governs (largest value). - Check whether 5 bars can fit in one layer.

b min = 5d b + 4s min + 2 × cover + 2 × d s = 5 × 25 + 4 × 35 + 2 × 30 + 2 × 10 = 345mm Since

b min = 345mm < b = 400mm it follows that 5 bars can fit in 1 layer.

- Calculate the actual effective depth ( d ).

s min 35 d = h – cover – d s – d b – --------- = 700 – 30 – 10 – 25 – -----2 2 = 617mm ≅ 620mm - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 30MPa 2 = ------------------------------- × 400mm × 700mm = 767mm 400MPa (Note: b t = b rectangular sections)

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5-13

Since 2

2

A s = 5000mm > 657mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 5000mm ------------------------------------------- = 0.021 = 2.1% [3.1] ----ρ = = 400mm × 620mm bd

Balanced reinforcement ratio:

ρ b = 0.027 ( f c ′ = 30MPa , see Table A.4) Since

ρ = 0.021 < ρ b = 0.027 okay - Check whether the strength requirement has been satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 5000mm --------------------------------------------------------------------------------------------= = 272mm [3.12] a = 0.8 × 0.65 × 30MPa × 400mm α 1 φ c f c ′b

a 272 2 M r = φ s A s f y d – --- = 0.85 ( 5000mm ) ( 400MPa ) 620 – --------- = 823kNm [3.14] 2 2 Since

M r = 823kNm > M f = 810kNm okay c) Provide a design summary.

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5-14

_________________________________________________________________________ 5.6. a) Estimate beam depth based on CSA A23.3 Cl.9.8.2.1(see Table A.3 in Appendix A). Find clear span ( l n ). Span l = 8m and support width is 400 mm, hence

l n = 8000 – 400 = 7600mm For a simply supported beam:

l n 7600mm h ≥ ------ = ---------------------- = 475mm 16 16 In theory, a depth of 500 mm could be used in this case. However, considering that the width b = 400mm and h ⁄ b aspect ratio

h 600--- = -------= 1.5 b 400 is within the recommended range, use h = 600mm Since

h = 600mm > 475mm okay In general, deeper beams require less tension reinforcement. b) Design the required tension reinforcement for the beam such that it can carry the design loads. Perform load analysis. - Self-weight - Dead load

kN kN = b × h × γ w = 0.4m × 0.6m × 24 ------3- = 5.8 ------m m = 20.0kN ⁄ m _______________________

DL = 25.8kN ⁄ m - Factored dead load (NBC 2005 Table 4.1.3.2)

kN w DLf = 1.25DL = 1.25 × 25.8 = 32.3 ------m

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5-15

- Factored live load (NBC 2005 Table 4.1.3.2)

P LLf = 1.5 × P L = 1.5 × 50kN = 75kN

Find the maximum bending moment (use the principle of superposition): First, find the factored bending moment M 1f due to point loads (see Table A.16).

8m l M 1f = P LLf × --- = ( 75kN ) × -------- = 200.3kNm 3 3

Then, find the factored bending moment M 2f due to uniform load (see Table A.16).

M 2f

kN 2 2 32.3 ------- × ( 8.0m ) w DLf × l m = ---------------------- = -------------------------------------------- = 258.4kNm 8 8

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5-16

Finally, find the total factored bending moment at the midspan using the principle of superposition:

M f = M 1f + M 2f = 200.3 + 258.4 = 458.7kNm ≅ 460kNm - Find the required area of tension reinforcement A s . Since

M r ≥ M f = 460kNm Set M r = 460kNm Note that b = 400mm (given). Estimate effective depth d (assume 1 layer of reinforcement) as follows:

d = h – 70 = 600 – 70 = 530mm Find the reinforcement area A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2 3.85 × 460 × 10 2 = 0.0015 ( 25MPa ) ( 400mm ) 530mm – ( 530 ) – ---------------------------------------- = 3117mm 25 × 400 2

Use 7 - 25M bars (25M bar area = 500mm ):

A s = 7 × 500 = 3500mm

2

Since 2

2

A s = 3500mm > 3117mm okay - Check bar spacing. cover = 30 mm (Table A.2, beam not exposed) CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 25mm = 35mm (25M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size)

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5-17

30mm It follows that s min = 35mm governs (largest value). - Check whether 10 - 25M bars can fit in 2 layers (5 bars per layer):

b min = 2 × 30 + 2 × 10 + 5 × 25 + 4 × 35 = 345mm Since

b min = 345mm < 400mm it follows that 5 bars can fit in 1 layer.

- Find actual d value. Since

25 d 1 = 600 – 30 – 10 – ------ = 547mm 2 25 25 d 2 = d 1 – ------ – 35 – ------ = 487mm 2 2 it follows that

5 × d1 + 2 × d2 d = ----------------------------------- = 530mm 7

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5-18

- Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h [5.7] fy 0.2 25MPa 2 = ------------------------------- × 400mm × 600mm = 600mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 3500mm > 657mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 3500mm ρ = ------ = -------------------------------------------- = 0.017 [3.1] bd 400mm × 530mm

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , Table A.4) Since

ρ = 0.017 < ρ b = 0.022 okay - Check whether the strength requirement has been satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 3500mm a = --------------------- = --------------------------------------------------------------------------- = 203mm [3.12] 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and

a 203 2 M r = φ s A s f y d – --- = 0.85 ( 3500mm ) ( 400MPa ) 530 – --------- ≅ 510kNm [3.14] 2 2 Since

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5-19

M r = 510kNm > M f = 460kNm okay c) Provide a design summary.

_________________________________________________________________________ 5.7. a) Find the minimum required beam depth and the amount of tension reinforcement. Estimate beam dimensions. - Width: b = 400mm (given) - Estimate beam depth ( h ) based on CSA A23.3 Cl.9.8.2.1(see Table A.3 in Appendix A). Find clear span ( l n ). Span l = 8m and support width is 400 mm, hence

l n = 8000 – 400 = 7600mm For a simply supported beam:

l n 7600mm h ≥ ------ = ---------------------- = 475mm 16 16 Estimate h from the aspect ratio:

h --- = 2 b or

h = 2 × b = 800mm

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5-20

Use h = 800mm > 475mm

h 600--- = -------= 1.5 b 400 - Estimate d (assume 1 layer of reinforcement):

d = h – 70mm = 800 – 70 = 730mm Perform load analysis. Dead load:

kN DL = 40 ------- (including self-weight) m Live load:

kN LL = 40 ------m - Factored load (NBC 2005 Table 4.1.3.2):

kN w f = 1.25DL + 1.5LL = 1.25 × 40 + 1.5 × 40 = 110 ------m - Factored bending moment (simply supported beam, see Table A.16)

kN 2 2 110 ------- × ( 8.0m ) wf × l m M f = ---------------- = ------------------------------------------- = 880kNm 8 8 Set M r = M f - Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × ( 880 × 10 ) 2 = 0.0015 ( 25MPa ) ( 400mm ) 730 – ( 730 ) – --------------------------------------------- = 4341mm 25 × 400 2

- Check whether ρ < 0.75ρ b (recommended upper limit). Try 10 - 25M bars A s = 10 × 500 = 5000mm

2

As 5000 ρ = ------ = ------------------------ = 0.017 [3.1] bd 400 × 730 Since f c ′ = 25MPa

ρ b = 0.022 (Table A.4)

Copyright © 2006 Pearson Education Canada Inc.

5-21

and

0.75ρ b = 0.17 In this case ρ ≅ 0.75ρ b . It is recommended to increase overall depth h in order to reduce the ρ value, so try

h = 900mm . Estimate d (assume 2 reinforcement layers):

d = h – 110 = 790mm - Find A s using the direct procedure.

2 3.85 × M r A s = 0.0015f c ′b d – d – -----------------------f c ′b

[5.4]

6

2 3.85 × 880 × 10 2 = 0.0015 × 25MPa × 400mm 790 – ( 790 ) – ---------------------------------------- = 3838mm 25MPa × 400

- Use 8 - 25M bars in 2 layers.

A s = 8 × 500 = 4000mm

2

2 As 4000mm ρ = ------------ = -------------------------------------------- = 0.013 [3.1] b×d 400mm × 790mm

Since

ρ = 0.013 < 0.75ρ b = 0.017 okay - Check bar spacing. cover = 30 mm (Table A.2, beam not exposed) CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 25mm = 35mm (25M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 35mm governs (largest value).

Copyright © 2006 Pearson Education Canada Inc.

5-22

Check whether 4 bars can fit in 1 layer:

b min = 2 × 30 + 2 × 10 + 4 × 25 + 3 × 35 = 285mm Since

b min = 285mm < 400mm it follows that 4 bars can fit in 1 layer. Find actual d value:

35 d = 900 – 30 – 10 – 25 – ------ = 817mm ≅ 820mm 2 - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ - bth A smin = ----------------[5.7] fy 0.2 25MPa 2 = ------------------------------- × 400mm × 900mm = 900mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 4000mm > 900mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 4000mm ρ = ------ = -------------------------------------------- = 0.012 bd 400mm × 820mm

[3.1]

Balanced reinforcement ratio:

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5-23

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.012 < ρ b = 0.022 okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 4000mm --------------------------------------------------------------------------------------------= = 232mm [3.12] a = 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and 2 a 232 M r = φ s A s f y d – --- = 0.85 ( 4000mm ) ( 400MPa ) 820 – --------- ≅ 957kNm [3.14] 2 2

Since

M r = 957kNm > M f = 880kNm okay The design is satisfactory, since ρ < 0.75ρ b and the beam is able to resist design loads. b) Provide a design summary.

Copyright © 2006 Pearson Education Canada Inc.

5-24

________________________________________________________________________ 5.8. a) Find the amount of tension steel required at the section x = 2.67m

kN w f = 110 ------- (see Problem 5.7.) m l - Find M x = --- . 3 Find support reactions:

wf × l R A = R B = ------------2 Develop the bending moment equation:

wf × l x M ( x ) = ------------- x – ( wf × x ) --2 2 Since x = 2.67m it follows that 2 kN kN 110 ------- ( 8.0m ) ( 2.67m ) 110 ------- ( 2.67m ) m m M ( x ) = ------------------------------------------------------------- – --------------------------------------------- = 783kNm 2 2

- Set M r = M f = 783kNm - Find the tension reinforcement area ( A s ) - use the direct procedure.

b = 400mm h = 900mm (see Problem 5.7.) d = 820mm (estimate)

Copyright © 2006 Pearson Education Canada Inc.

5-25

2 3.85 × M r A s = 0.0015f c ′b d – d – -----------------------f c ′b

[5.4]

6

2 3.85 × ( 783 × 10 ) 2 = 0.0015 ( 25MPa ) ( 400mm ) 820 – ( 820 ) – --------------------------------------------- = 3165mm 25MPa × 400mm

Use 7-25M bars: A s = 7 × 500 = 3500mm

2

(see Table A.1)

Find actual d . Since

25 d 1 = 900 – 30 – 10 – ------ = 847mm 2 25 25 d 2 = d 1 – ------ – 35 – ------ = 787mm 2 2 it follows that

5 × d1 + 2 × d2 d = ----------------------------------- ≅ 830mm 7 (Note: cover and clear bar spacing are the same as Problem 5.7)

- Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2). 2

A smin = 900mm (same as Problem 5.7.) Since 2

2

A s = 3500mm > 900mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 3500mm ρ = ------ = -------------------------------------------- = 0.011 [3.1] bd 400mm × 830mm

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5-26

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.011 < ρ b = 0.022 okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 3500mm a = --------------------- = --------------------------------------------------------------------------- = 203mm [3.12] 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and

a 203 2 M r = φ s A s f y d – --- = 0.85 ( 3500mm ) ( 400MPa ) 830 – --------- ≅ 867kNm [3.14] 2 2 Since

M r = 867kNm > M f = 783kNm okay b) Find the amount of tension steel required at the section x = 2m

kN 110 ------- ( 2m ) wf × x m M ( x ) = --------------- ( l – x ) = ----------------------------------- ( 8m – 2m ) = 600kNm 2 2 Set M r = M ( x ) = 660 kNm . Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 660 × 10 2 2 = 0.0015 × 25MPa × 400mm 820 – ( 820 ) – -------------------------------------------- = 2598mm ≅ 2600mm 25MPa × 400mm 2

Use 6-25M bars: A s = 6 × 500 = 3000mm

2

Note: the original beam from Example 5.7. has two reinforcement layers of 4 bars, that is, 8 bars in total. It is common to cutoff rebars in the upper layer; this should be considered when detailing of this beam is being carried out. - Find the actual d value.

Copyright © 2006 Pearson Education Canada Inc.

5-27

25 d = 900 – 30 – 10 – ------ = 847mm ≅ 850mm 2

There is no need to check whether 5-25M bars can fit in one layer; this was already checked in part a) of this problem. - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2). 2

A smin = 900mm (same as Problem 5.7.) Since 2

2

A s = 2500mm > 900mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 2500mm ρ = ------ = -------------------------------------------- = 0.007 bd 400mm × 850mm

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.007 < ρ b = 0.022 okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 2500mm a = --------------------- = --------------------------------------------------------------------------- = 145mm [3.12] 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and

a 145 2 M r = φ s A s f y d – --- = 0.85 ( 2500mm ) ( 400MPa ) 850 – --------- ≅ 661kNm [3.14] 2 2 Copyright © 2006 Pearson Education Canada Inc.

5-28

Since

M r = 661kNm > M f = 660kNm okay c) Find the amount of tension steel required at the section x = 1m

kN 110 ------- ( 1m ) wf × x m M ( x ) = --------------- ( l – x ) = ----------------------------------- ( 8m – 1m ) = 385kNm 2 2 Set M r = M ( x ) = 385 kNm (assume d = 850mm ). - Find A s using the direct procedure 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × ( 385 × 10 ) 2 = 0.0015 × 25MPa × 400mm 850 – ( 850 ) – --------------------------------------------- = 1383mm 25MPa × 400mm 2

A s = 3 × 500 = 1500mm

Use 3-25M bars:

2

(Table A.1)

Note: d = 850mm (same as part b) - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2). 2

A smin = 900mm (same as Problem 5.7.) Since 2

2

A s = 1500mm > 900mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 1500mm ρ = ------ = -------------------------------------------- = 0.004 bd 400mm × 850mm

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.004 < ρ b = 0.022 okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3).

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5-29

2 φs fy As 0.85 × 400MPa × 1500mm --------------------------------------------------------------------------------------------= = 87mm [3.12] a = 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and

a 87 2 M r = φ s A s f y d – --- = 0.85 ( 1500mm ) ( 400MPa ) 850 – ------ ≅ 411kNm [3.14] 2 2 Since

M r = 411kNm > M f = 385kNm okay - Provide a design summary.

d) Find the amount of tension steel required at the section x = 0.2m

kN 110 ------- × 0.2m wf × x m M ( x ) = --------------- ( l – x ) = ------------------------------------ ( 8.0m – 0.2m ) = 86kNm 2 2 Set M r = M ( x ) = 86kNm Assume d = 850mm (1 layer of reinforcement). Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 86 × 10 2 2 = 0.0015 × 25MPa × 400mm 850mm – ( 850mm ) – -------------------------------------------- = 296mm 25MPa × 400mm - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2). 2

A smin = 900mm (same as Problem 5.7.) Since Copyright © 2006 Pearson Education Canada Inc.

5-30

2

A smin = 900mm > 296mm

2

Set

A s = A smin = 900mm

2

Use 2-25M bars: A s = 2 × 500 = 1000mm

2

(Table A.1)

- Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 1000mm ρ = ------ = -------------------------------------------- = 0.003 400mm × 850mm bd

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.003 < ρ b = 0.022 okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 1000mm a = --------------------- = --------------------------------------------------------------------------- = 58mm [3.12] 0.9 × 0.65 × 25MPa × 400mm α 1 φ c f c ′b

and

a 58 2 M r = φ s A s f y d – --- = 0.85 ( 1000mm ) ( 400MPa ) 850 – ------ ≅ 279kNm [3.14] 2 2 Since

M r = 279kNm > M f = 86kNm okay Provide a design summary.

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5-31

________________________________________________________________________ 5.9. a) Design a T-beam section in terms of the overall depth and reinforcement. Find the factored bending moment ( M f ).

M f = 880kNm (same as Problem 5.7.)

- Estimate beam depth ( h ) based on CSA A23.3 Cl.9.8.2.1(see Table A.3 in Appendix A). Clear span: l n = 7600mm For a simply supported beam:

l n 7600mm h ≥ ------ = ---------------------- = 475mm (same as Problem 5.7.) 16 16 Estimate h from the aspect ratio:

h --- = 1.5 b or

h = 1.5 × b = 600mm where

b = 400mm Use h = 600mm > 475mm - Estimate d (assume 2 layers of reinforcement):

d = h – 110mm = 490mm - Estimate effective width b f for T - beam. For a simply supported beam (Table 5.4): Copyright © 2006 Pearson Education Canada Inc.

5-32

l n 7600mm 1) b T ≤ ---- = ---------------------- = 1520mm ≅ 1500mm 5 5 2) b T ≤ 12h f = 12 × 200mm = 2400mm

lw 6000 – 400 3) b T ≤ ---- = --------------------------- = 2800mm 2 2 Use b T = 1500mm (smallest value governs).

b f = b w + 2b T = 400 + 2 × 1500 = 3400mm - Find the required area of tension reinforcement ( A s ) . Assume that the neutral axis is in the flange, that is, a < h f . Set

M f = M r = 880kNm and

b = b f = 3400mm Use the direct procedure to determine A s . 2 3.85 × M r A s = 0.0015f c ′b d – d – -----------------------f c ′b

[5.4]

6

3.85 × 880 × 10 2 = 0.0015 × 25MPa × 3400mm 490 – ( 490 ) – ---------------------------------------- = 5421mm 25 × 3400 2

Try 8 - 30M bars (in 2 layers): A s = 8 × 700 = 5600mm

2

Check whether ρ < 0.75ρ b

Copyright © 2006 Pearson Education Canada Inc.

5-33

As 5600 ρ = ------ = --------------------------- = 0.0034 bd 3400 × 490 ρ b = 0.022

( f c ′ = 25MPa , Table A.4)

Since

0.75ρ b = 0.0165 > ρ = 0.0034 okay - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h [5.7] fy 0.2 25MPa 2 = ------------------------------- × 400mm × 600mm = 600mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 5600mm > 600mm okay - Check bar spacing. cover = 30 mm (Table A.2, beam not exposed) CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 30mm = 42mm (30M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 42mm governs (largest value). - Can 4 bars fit in 1 layer? Since

b min = 2 × 30 + 2 × 10 + 4 × 30 + 3 × 42 = 326mm < 400mm and

b min = 326mm < 400mm

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it follows that 4 bars can fit in 1 layer.

- Find the actual d value.

42 d = 600 – 30 – 10 – 30 – ------ = 509mm ≅ 510mm 2 - Confirm that the strength requirement is satisfied (A23.3 Cl.8.1.3). 2 φs fy As 0.85 × 400MPa × 5600mm a = --------------------- = ------------------------------------------------------------------------------ = 38mm 0.9 × 0.65 × 25MPa × 3400mm α 1 φ c f c ′b

[3.12]

Since a = 38mm < b f = 200mm , the assumption that the neutral axis is in the flange is correct.

a 38 2 M r = φ s A s f y d – --- = 0.85 × ( 5600mm ) ( 400MPa ) 510 – ------ ≅ 935kNm [3.14] 2 2 Since

M r = 935kNm > M f = 880kNm okay - Provide a design summary.

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b) Compare the beam designs from Problems 5.7. and 5.9. in terms of the amount of reinforcement and cross sectional dimensions. The beams in Problems 5.7. and 5.9. have the same span and are subjected to the same load. •

The beam in Problem 5.7. is of rectangular section, with the width of 400 mm and the

•

overall depth of 900 mm. The reinforcement consists of 8-25M bars (area 4000mm ) and the reinforcement ratio is 1.2%. The beam in Problem 5.9. is of T-section, with the width of 400 mm and the overall depth

2

2

of 600 mm. The reinforcement consists of 8-30M bars (area 5600mm ) and the reinforcement ratio is 0.34%. It can be concluded that the beam of rectangular section (Problem 5.7.) is by 50% deeper than the beam with T-section (Problem 5.9.). The reinforcement ratio corresponding to the rectangular beam (1.2%) is 4 times larger than the reinforcement ratio corresponding to the T-beam (0.3%). Finally, it can be concluded that a beam with T-section makes more efficient use of material as compared to the rectangular beam section. ________________________________________________________________________ 5.10. a) Design the beam section with tension and compression reinforcement. Find the beam depth h . From Problem 5.7., it follows that the overall depth h = 670mm . However, it is required to restrict the depth to 0.75h of the value obtained in Problem 5.7., that is,

h = 0.75 × 900 = 675mm

h = 670mm

- Find the factored bending moment M f : M f = 880kNm (same as Problem 5.7.) - Estimate the effective depth ( d ) (assume 1 layer of reinforcement).

d = h – 70 = 600mm Note: b = 400mm (same as Problem 5.7.) - Find the required area of tension reinforcement ( A s ) 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

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6

3.85 × 880 × 10 2 = 0.0015 × 25MPa × 400mm 600 – ( 600 ) – ---------------------------------------- = 6816mm 25 × 400 2

Try 10-30M bars: A s = 10 × 700 = 7000mm

2

(Table A.1)

- Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 7000mm ------------------------------------------- = 0.029 ----ρ = = 400mm × 600mm bd

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

ρ = 0.029 > ρ b = 0.022 it follows that the beam is overreinforced. It is required to add the compression steel in order to prevent the occurrence of concrete-controlled failure. - Estimate the amount of compression steel. It is a good practice when the compression steel area is less than 50% of the tension steel area, that is,

A s ′ < 0.5A s = 3500mm

2

Try 4-30M bars: A s ′ = 4 × 700 = 2800mm

2

Calculate the reinforcement ratio for the compression steel: 2 As ′ 2800mm ρ′ = ------------ = -------------------------------------------- = 0.012 b×d 400mm × 600mm

Check the beam failure mode (steel-controlled or concrete-controlled), as follows:

ρ – ρ′ = 0.029 – 0.012 = 0.017 Since

ρ b = 0.022

f c ′ = 25MPa (see Table A.4)

and

ρ = 0.017 < 0.022 it follows that the beam is expected to fail in the steel-controlled mode. - Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

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0.2 f c ′ A smin = ------------------ b t h [5.7] fy 0.2 25MPa 2 = ------------------------------- × 400mm × 670mm = 670mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 7000mm > 670mm okay - Check bar spacing. cover = 30 mm (Table A.2, beam not exposed) CSA A23.1 (Cl.6.6.5.2) specifies that the clear spacing between the bars ( s min ) should be at least equal to

1.4 × d b = 1.4 × 30mm = 42mm (30M bar) 1.4 × a max = 1.4 × 20mm = 28mm (20 mm maximum aggregate size) 30mm It follows that s min = 42mm governs (largest value). - Can 5-30M bars fit in one layer?

b min = 2 × 30 + 2 × 10 + 5 × 30 + 4 × 42 = 398mm Since

b min = 398mm < 400mm it follows that 5-30M bars can fit in one layer.

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- Find the actual d value.

42 d = 670 – 30 – 10 – 35 – ------ = 574mm ≅ 570mm 2 Calculate M r . Assume that f s = f y and f s ′ = f y , and set d′ = 70mm . - Find the depth of compression stress block ( a ). The force in tension steel ( T r ) : 2

T r = φ s f y A s = 0.85 × 400MPa × 7000mm = 2380kN The resultant force in the compression steel ( C r ′ ) : 2

C r ′ = φ s f y A s ′ = 0.85 × 400MPa × 2800mm = 952kN Calculate a :

Tr – Cr ′ a = --------------------α 1 φ c f c ′b

[3.38]

3

3

2380 × 10 N – 952 × 10 N = --------------------------------------------------------------------------- = 274.6mm ≅ 275mm 0.8 × 0.65 × 25MPa × 400mm Check whether the compression steel has yielded.

a 275 c = ----- = --------- = 305mm β1 0.9 -Find the strain in compression steel ( ε s ′ )

d′ 70mm ε s ′ = 0.0035 1 – ---- = 0.0035 1 – ------------------- = 0.0027 c 305mm - Calculate the yield strain

fy 400MPa ε y = ----- = ------------------------------- = 0.002 200000MPa Es Since

ε s ′ = 0.0027 > ε y = 0.002

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it follows that the compression steel has yielded.

- Calculate the factored moment resistance M r .

C r = α 1 φ c f c ′ab = 0.8 × 0.65 × 25MPa × 275mm × 400mm = 1430kN C r ′ = 952kN a M r = C r ′ ( d – d′ ) + C r d – --2

[3.39]

275 3 3 = ( 1430 × 10 N ) ( 570 – 70 ) + ( 952 × 10 N ) 570 – --------- = 1127kNm 2

Confirm that the strength requirement is satisfied (A23.3 Cl.8.1.3).

M f = 880kNm Since

M r = 1127kNm > M f = 880kNm it follows that the strength requirement is satisfied. Note that the amount of reinforcement could be further reduced, since there is a 28% reserve in moment resistance.

Copyright © 2006 Pearson Education Canada Inc.

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- Provide a design summary.

b) Compare the beams from Problems 5.7. and 5.10. The beams designed in Problems 5.7. and 5.10. have the same span and are subjected to the same load. •

The beam in Problem 5.7. is of rectangular section, with the width of 400 mm and the overall depth of 900 mm. The tension reinforcement consists of 8-25M bars (area 2

•

4000mm ) and the reinforcement ratio is 1.2%. The beam in Problem 5.10. is of a rectangular section, with the width of 400 mm and the overall depth of 670 mm. The tension reinforcement consists of 10-30M bars (area 2

2

7000mm ), plus 4-30M compression rebars (area 2800mm ). The reinforcement ratio 2

( ρ – ρ′ ) is 1.7%. In total, 14-30 M bars are required (area 9800mm ). It can be concluded that the beam reinforced with tension steel only (Problem 5.7.) is by 35% deeper than doubly reinforced beam of Problem 5.10. The reinforcement ratio corresponding to singly reinforced beam (1.2 %) is less than the reinforcement ratio (1.7 %) corresponding to dou2 bly reinforced beam. The total amount of steel in doubly reinforced beam (area 9800mm ) is 2 over twice the amount of steel in singly reinforced beam (area 4000mm ). _______________________________________________________________________ 5.11. a) Find the minimum required slab thickness. Estimate beam depth ( h ) based on CSA A23.3 Cl.9.8.2.1(see Table A.3 in Appendix A). Clear span ( l n ): l n ≅ l = 6000 mm (width of support beams at A and B not given) For a simply supported one-way slab:

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l n 6000mm h ≥ ------ = ---------------------- = 300mm 20 20 Note that the slab thickness has been determined based on simply supported span AB. Use h = 300mm b) Draw the bending moment diagram. Perform load analysis. - Dead load: DL = 6.0kPa - Live load: LL = 5.0kPa - Factored load (NBC 2005 Table 4.1.3.2)

w f = 1.25DL + 1.5LL = 15.0kPa However, this is one-way slab and the design will be based on 1 m wide strip, hence

Shear force and bending moment diagrams are shown below.

M min = – 30kNm ⁄ m M max ≅

+ 53 kNm ⁄ m

c) Design the required tension reinforcement for the slab at the location of the maximum negative bending moment. Copyright © 2006 Pearson Education Canada Inc.

5-42

M min = – 30kNm ⁄ m Set M r = M min = – 30 kNm ⁄ m

M min = 30kNm ⁄ m Note : b = 1000mm ! - Estimate effective depth d . - cover: = 20mm (not exposed slab, Table A.2) - Use 15M bars, hence

15 d = 300 – 20 – ------ = 272mm ≅ 270mm 2 - Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

2 3.85 × 30 × 10 mm = 0.0015 ( 25MPa ) ( 1000mm ) 270 – ( 270 ) – ------------------------------------- = 326 -----------25 × 1000 m

- Find the required bar spacing. For 15M bars: A b = 200mm

2

(Table A.1)

1000 1000 s ≤ A b × ------------ = 200 ------------ = 613mm [3.28] As 326 - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max =

3h = 3 × 300 = 900mm 500mm

Hence

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s max = 500mm (smaller value governs) Use 15 M@500. - Determine the reinforcement area A s .

2

1000 1000 mm A s = A b × ------------ = 200 × ------------ = 400 -----------s 500 m - Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1). 5

A g = b × h = 1000 × 300 = 3 × 10 mm

2

Since 2

2

mm mm A smin = 0.002A g = 600 ------------ > 333 -----------m m

[5.16]

it follows that the minimum reinforcement area governs, hence

A s = A smin

2

mm = 600 -----------m

- Find the required bar spacing.

1000 1000 s = A b × ------------ = 200 × ------------ = 333mm [3.28] As 600 Use 15M@300. 2

1000 1000 mm A s = A b ------------ = 200 ------------ = 667 -----------300 300 m

- Check whether the CSA A23.3 maximum reinforcement requirement is satisfied.

As 667 ρ = ------ = ----------------------------------------------- = 0.0024 bd 1000mm × 275mm and

ρ b = 0.022

( f c ′ = 25MPa , Table A.4)

Since ρ = 0.0024 < ρ b okay - Check whether the strength requirement is satisfied (A23.3 Cl.8.1.3). Find a . 2 φs fy As 0.85 × 400MPa × 667mm -----------------------------------------------------------------------------------------------= = 17mm [3.12] a = 0.8 × 0.65 × 25MPa × 1000mm α 1 φ c f c ′b

Copyright © 2006 Pearson Education Canada Inc.

5-44

a M r = φ s f y A s d – --2

[3.14]

17 2 = 0.85 × 400MPa × 667mm 270 – ------ = 59kNm ⁄ m 2 Since

M r = 59kNm > M f = 30kNm okay d) Design the required tension reinforcement for the slab at the location of the maximum positive bending moment. Set M f = M max = 53kNm ⁄ m Use d = 270mm . -Find A s (use the direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

2 3.85 × 53 × 10 mm = 0.0015 ( 25MPa ) ( 1000mm ) 270 – ( 270 ) – ------------------------------------- = 587 -----------25 × 1000 m

Note that 2

mm A smin = 600 ------------ (see part c) m Since

A smin > A s it follows that A smin governs, hence 2

mm A s = A smin = 600 -----------m Use the same reinforcement as in part c), that is, 15M@300. There is no need to check the minimum and maximum reinforcement requirements (this was done in part c). Since

M r = 59kNm ⁄ m (see part c) and

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M f = 53kNm < M r = 59kNm ⁄ m okay e) Design the temperature and shrinkage reinforcement. - Find the required area of shrinkage reinforcement.

A smin

2

mm = 600 ------------ (part c) m

- Determine the spacing of shrinkage reinforcement:

s max =

5h = 5 × 300 = 1500mm 500mm

Hence

s max = 500mm (smaller value governs) Use 15M@300. Since 2

2

mm mm A s = 667 ------------ > 600 ------------ okay m m f) Sketch a vertical section through the slab and the corresponding reinforcement layout.

_____________________________________________________________________ 5.12. a) Design the reinforcement for the section with M f = 10kNm ⁄ m and h = 200mm . - Estimate d . - cover = 20mm (not exposed slab, Table A.2) - 15M bars

Copyright © 2006 Pearson Education Canada Inc.

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15 d = 200 – 20 – ------ = 172mm ≅ 170mm 2 Set b = 1000mm (width of a unit strip). - Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

mm 2 3.85 × 10 × 10 = 0.0015 ( 25MPa ) ( 1000mm ) 170 – ( 170 ) – ------------------------------------- = 172 -----------( 25 ) ( 1000 ) m Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1).

A g = b × h = 1000 × 200 = 200000mm 2

mm A smin = 0.002A g = 400 -----------m

2

[5.16]

Since

A smin

2

2

mm mm = 400 ------------ > A s = 172 -----------m m

it follows that A smin governs, hence

A s = A smin

2

mm = 400 -----------m

Find the required bar spacing. 2

For 15M bars: A b = 200mm (Table A.1)

1000 1000 s ≤ A b ------------ = 200 ------------ = 500mm As 400

[3.28]

Use 15M@500. - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max =

3h = 3 × 200 = 600mm 500mm

Hence

s max = 500mm (smaller value governs) Use 15M@500.

kNm - Design the reinforcement for the section with M f = 20 ------------ and h = 400mm . m

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- Estimate effective depth d :

15 d = 400 – 20 – ------ = 372mm ≅ 370mm 2 Set b = 1000mm (width of a unit slab strip). - Find A s using the direct procedure 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

3.85 × 20 × 10 mm = 0.0015 ( 25MPa ) ( 1000mm ) 370 – ( 370 ) – --------------------------------------------------------- = 157 -----------( 25MPa ) × ( 1000mm ) m 2

- Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1).

A g = b × h = 1000 × 400 = 400000mm A smin

2 2

mm = 0.002A g = 0.002 ( 400000mm ) = 800 ------------ [5.16] m 2

Since 2

2

mm mm A smin = 800 ------------ > 157 -----------m m

it follows that the minimum reinforcement area governs, hence 2

mm A s = A smin = 800 -----------m - Find the required bar spacing (assume 15M bars).

1000 1000 s ≤ A b ------------ = 200 ------------ = 250mm As 800

[3.28]

Use 15M@250. - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max =

3h = 3 × 400 = 1200mm 500mm

Hence

s max = 500mm (smaller value governs) Use 15M@250. b) Comment on the solution from part a).

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For a section with M f = 10kNm ⁄ m use 15M@500. For a section with M f = 20kNm ⁄ m use 15M@250. If this was a problem from design practice, it would be reasonable to prescribe larger amount of reinforcement (15M@250) throughout the negative moment region (above the support). It would not be considered practical to increase bar spacing over a very limited length. Note that the negative moment reinforcement may be discontinued once the bending moment switches into the positive range.

Copyright © 2006 Pearson Education Canada Inc.

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Chapter 6 - Solutions __________________________________________________________________________ 6.1. a) Failure modes characteristic for reinforced concrete beams subjected to flexure and shear • • •

Flexural failure Diagonal tension failure Shear compression failure

b) Identify the failure mode which is most desirable. Flexural failure is most desirable, because it is associated with a ductile behaviour, which is characterized by significant structural deformations before failure takes place. Note that the above description refers to steel-controlled flexural failure. _____________________________________________________________________________ 6.2. Shear failure in reinforced concrete beams can be avoided by designing a beam in such a way that flexural failure will occur first. To achieve that, the load that causes flexural failure should be less than the load that causes a shear failure. Once the flexural failure has been initiated, further load increase is not possible due to yielding of steel reinforcement. ___________________________________________________________________________ 6.3. a) Sketch stress trajectories for uniformly distributed load

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6-1

b) Sketch stress trajectories for the point load.

____________________________________________________________________________ 6.4. Find the concrete shear resistance V c . - Find d and d v . 10M stirrups: d s = 10mm 25M rebars: d b = 25mm cover = 40mm

db 25 d = h – cover – d s – ----- = 800 – 40 – 10 – ------ = 737mm 2 2 0.9d = 0.9 × 737 = 660mm 0.72h = 0.72 × 800 = 576mm d v ≅ 660mm (larger value governs) dv =

- Determine β (assume that the shear reinforcement exceeds the minimum prescribed by CSA A23.3).

β = 0.18 (Table 6.1) - Find V c .

b w = b = 600mm V c = φ c λβ f c ′ b w d v

(A23.3 Eq.11.6) [6.12] Copyright © 2006 Pearson Education Canada Inc.

6-2

= 0.65 × 1.0 × 0.18 25MPa × 600mm × 660mm = 231.7kN V c ≅ 232kN -Find V s . 2

4-legged 10M stirrups ( area = 100mm ):

A v = 4 × 100 = 400mm

2

Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq. 11.7) [6.9] s 0.85 × 400mm × 400MPa × 660mm × 1.43 = ---------------------------------------------------------------------------------------------------------- = 320.9kN 400mm V s ≅ 321kN Find V r

V r = V c + V s = 232 + 321 = 553kN ____________________________________________________________________________ 6.5. Find concrete shear resistance V c . - Find d and d v . 10M stirrups: d s = 10mm 25M rebars: d b = 25mm cover = 30mm

db 25 d = h – cover – d s – ----- = 600 – 30 – 10 – ------ = 547mm 2 2 0.9d = 0.9 × 547 = 492mm 0.72h = 0.72 × 600 = 432mm d v ≅ 490mm (larger value governs) dv =

Copyright © 2006 Pearson Education Canada Inc.

6-3

- Determine β (assume that the shear reinforcement exceeds the minimum value prescribed by CSA A23.3).

β = 0.18 (Table 6.1) -Find V c .

b w = b = 300mm V c = φ c λβ f c ′ b w d v (A23.3 Eq. 11.6) [6.12] = 0.65 × 1.0 × 0.18 25MPa × 300mm × 490mm = 86kN Since

V c = 86.0kN < V f = 300kN it follows that the shear reinforcement is required. - Find the required V s . Since

V f = 300kN set

V r = V f = 300kN V s ≥ V r – V c = 300 – 86 = 214kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq. 11.7) [6.9] s - Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 2

- Assume 2 - legged 10M stirrups ( area = 100mm )

A v = 2 × 100 = 200mm φ s A v f y d v cot θ s = --------------------------------Vs

2

[6.16]

2

0.85 × 200mm × 400MPa × 490mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 223mm 214000N Copyright © 2006 Pearson Education Canada Inc.

6-4

Round to s = 200mm . Use 10M@200. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 490 = 343mm

s max = 343mm (smaller value governs) Since

s = 200mm < 343mm okay - Check the CSA A23.3 minimum stirrup requirements (Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------(A23.3 Eq.11.1) [6.18] fy 2 300mm × 200mm = 0.06 25MPa -------------------------------------------- = 45mm 400MPa Since 2

2

A vmin = 45mm < A v = 200mm okay Check maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 300mm × 490mm = 597kN - Find actual V r .

V c = 86kN φ s A v f y d v cot θ V s = --------------------------------- [6.9] s 2 0.85 × 200mm × 400MPa × 490mm × 1.43 -----------------------------------------------------------------------------------------------------------= ≅ 238kN 200mm V r = V c + V s = 86 + 238 = 324kN Since

V r = 324kN < maxV r = 597kN okay

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6-5

Use 10M@200. _____________________________________________________________________________ 6.6. a) Design the required spacing of 10M stirrups for this beam according to CSA A23.3 requirements.

- Find the effective depth d . 10M stirrups: d s = 10mm 30M stirrups: d b = 30mm cover = 40mm

db 30 d = h – cover – d s – ----- = 800 – 40 – 10 – ------ = 735mm 2 2 - Find effective shear depth d v .

0.9d = 0.9 × 735 = 661mm 0.72h = 0.72 × 800 = 576mm d v ≅ 660mm (larger value governs) dv =

- Find clear span l n .

400 400 l n = 6000 – --------- – --------- = 5600mm 2 2 - Find the factored load w f (NBC 2005 Table 4.1.3.2).

kN DL = 120 ------- (given) m LL = 60kN ⁄ m (given) Copyright © 2006 Pearson Education Canada Inc.

6-6

kN w f = 1.25DL + 1.5LL = 1.25 × 120 + 1.5 × 60 = 240 ------m - Find the maximum V f at the support.

kN 240 ------- ( 5.6m ) wf × ln m V f = ---------------- = ---------------------------------------- = 672kN 2 2

V fmidspan

ln kN 5.6m 1.5 × 60 ------- -----------1.5LL ---2 m 2 = -------------------------- = ---------------------------------------------------- = 63kN 4 4

-Find the design shear force at distance d v from the support ( V fd v ) .

V fdv

ln 5.6---- – d v -----– 0.66 2 2 = V fmidspan + --------------- ( V f – V fmidspan ) = 63.0 + ------------------------ ( 672 – 63.0 ) = 528.5kN ln 5.6--------2 2

V fdv ≅ 530kN - Sketch the shear envelope diagram.

- Find concrete shear resistance ( V c ) . Determine β (assume that the shear reinforcement is less than the minimum amount prescribed by CSA A23.3).

230 230 β = ----------------------- = --------------------------- = 0.14 (Table 6.1) 1000 + d v 1000 + 660

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6-7

-Find V c .

b w = b = 600mm V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.14 25MPa × 600mm × 660mm = 180.2kN ≅ 180kN - Is shear reinforcement required? Since

V fdv = 530kN > V c = 180kN it follows that the shear reinforcement is required wherever V f > V c (A23.3 Cl.11.2.8.1). - Find the required V s .

V s ≥ V f – V c = 530 – 180 = 350kN - Find the required stirrup spacing. 2

2 - legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35° (CSA A23.3 Simplified Method)

cot ( θ ) = 1.43

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

2

0.85 × 200mm × 400MPa × 660mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 183mm ≅ 180mm 350000N Use 10M@180. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 660 = 462mm

Copyright © 2006 Pearson Education Canada Inc.

6-8

s max = 462mm (smaller value governs) Since

s = 180mm < 462mm okay - Check minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bws A vmin = 0.06 f c ′ -------fy

(A23.3 Eq.11.1) [6.18]

2 600mm × 180mm = 0.06 25MPa -------------------------------------------- = 81.0mm 400MPa

Since 2

A vmin = 81.0mm < A v = 200mm

2

okay

- Check the maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 600mm × 660mm = 1609kN - Find actual V r value.

V c = 180kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 2 0.85 × 200mm × 400MPa × 660mm × 1.43 -----------------------------------------------------------------------------------------------------------= = 356kN 180mm V r = V c + V s = 180 + 356 = 536kN Since

maxV r = 1609kN > V r = 536kN okay - Determine the region where shear reinforcement is not required.

V f x = V c = 180kN

V fx

ln ---- – x 2 ------------ ( V f – V fmidspan ) = V fmidspan + ln ---2 Copyright © 2006 Pearson Education Canada Inc.

6-9

5.6 ------- – x 2 180kN = 63kN + ---------------- ( 672 – 63 ) 5.6 ------2 x = 2.26m

Sketch a beam diagram showing the stirrup arrangement.

b) Develop an alternative solution using two different stirrup spacings. Try 2 different stirrup spacings (see Table 6.2) Region 0: V f ≤ V c stirrups not required Region 1: V c < V f < V rmin Region 2: V rmin < V f < V fmax - Determine V rmin . Maximum permitted stirrup spacing (A23.3 Cl.11.3.8.1):

s max = 462mm ≅ 460mm (see part a) Steel shear resistance V s :

Copyright © 2006 Pearson Education Canada Inc.

6-10

φ s A v f y d v cot θ 0.85 × 200 × 400 × 660 × 1.43 V s = --------------------------------- = ------------------------------------------------------------------------- ≅ 140kN s 460

[6.9]

Concrete shear resistance V c :

V c = 180kN V rmin = V c + V s = 180 + 140 = 320kN - Define Region 1 (find x 1 ).

V c = 180kN < V f < V rmin = 320kN

V f x1

ln ---- – x 1 2 = V fmidspan + ---------------- ( V f – V fmidspan ) ln ---2

V fx 1 = V rmin = 320kN 5.6m ------------ – x 1 2 320kN = 63kN + ------------------------ ( 672kN – 63kN ) 5.6m -----------2 x 1 ≅ 1.62m

Copyright © 2006 Pearson Education Canada Inc.

6-11

- Sketch the stirrup arrangement.

c) Which solution would you use if faced with a similar problem in a design office? The first stirrup arrangement prescribes the same stirrup spacing ( s = 210mm ) along the beam length (except in the region where stirrups are not required). The second stirrup arrangement prescribes two stirrup spacings: • ( s = 180mm ) in the support region ( 0 < x < 1620mm ), and • s = 460mm in the region ( 1620mm < x < 2260mm ) . As a result, only 2 sets of stirrups with spacing s = 460mm are required on each side of the beam. If this was a problem from a design practice, it would not be considered practical to change stirrup spacing. Note that the first stirrup arrangement requires 26 stirrup sets in total, while the second arrangement requires 22 stirrup sets in total. In conclusion, the first stirrup arrangement with uniform stirrup spacing would be considered more practical than the second one.

Copyright © 2006 Pearson Education Canada Inc.

6-12

_____________________________________________________________________________

6.7.

- Find d and d v . Section A-A:

h = 1200mm 10M stirrups: d s = 10mm 25M rebars: d b = 25mm cover= 40mm

db 25 d = h – cover – d s – ----- = 1200 – 40 – 10 – ------ = 1137mm 2 2 0.9d = 0.9 × 1137 = 1023mm 0.72h = 0.72 × 1200 = 864mm d v ≅ 1000mm (larger value governs) dv =

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6-13

Section B-B:

h = 600mm db 25 d = h – cover – d s – ----- = 600 – 40 – 10 – ------ = 537mm 2 2 0.9 × d = 0.9 × 537 = 483mm 0.72h = 0.72 × 600 = 432mm d v ≅ 480mm (larger value governs) dv =

Find clear span l n .

400 400 l n = 4000 – --------- – --------- = 3600mm 2 2 - Find the factored load ( w f ) (NBC 2005 Table 4.1.3.2).

DL = 150kN ⁄ m LL = 60kN ⁄ m w f = 1.25DL + 1.5LL = 1.25 × 150 + 1.5 × 60 = 278kN ⁄ m - Design shear force V f at the support:

wf × ln 278 × 3.6m V f = ---------------- = --------------------------- = 500.4kN ≅ 500kN 2 2

- Design shear force V fmidspan at the midspan:

V fmidspan

ln kN 3.6m 1.5 60 ------- -----------1.5LL ---2 m 2 = -------------------------- = ---------------------------------------------- = 40kN 4 4

- Design shear force at distance d v from the support ( V fdv ) :

ln 3.6---- – d v -----– 0.48 2 2 V fdv = V fmidspan + --------------- ( V f – V fmidspan ) = 40kN + ------------------------ ( 500 – 40 ) = 377kN ln 3.6--------2 2 Note that d v = 480mm corresponds to Section B-B. This is a conservative design - in reality, d v value varies in the range from d v = 480mm (Section B-B) to d v = 1000mm (Section A-A).

Copyright © 2006 Pearson Education Canada Inc.

6-14

- Sketch the shear envelope diagram.

- Find concrete shear resistance V c . Determine β (assume that shear reinforcement is below the minimum value prescribed by CSA A23.3):

230 β = ----------------------- (Table 6.1) 1000 + d v Section A-A:

d v = 1000mm 230 β = ------------------------------ = 0.115 1000 + 1000 Section B-B:

d v = 480mm 230 β = --------------------------- = 0.155 1000 + 480 Use the average β value:

0.115 + 0.155 β = --------------------------------- = 0.135 ≅ 0.13 2 Use β = 0.13 .

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6-15

Note that β = 0.18 could be used if shear reinforcement exceeds the minimum amount prescribed by CSA A23.3 Cl.11.2.8.2. - Find V c for Section A-A ( d v = 1000mm ) .

b w = b = 600mm V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6) [6.12] = 0.65 × 1.0 × 0.13 25MPa × 600mm × 1000mm = 253.5kN - Find V c for Section B-B ( d v = 480mm ) .

V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6) [6.12] = 0.65 × 1.0 × 0.13 25MPa × 600mm × 480mm = 121.7kN ≅ 122kN - Is shear reinforcement required? Section B-B:

V fdv = 377kN > V c = 122kN Section A-A:

V fdv = 377kN > V c = 253kN Shear reinforcement is required wherever V f > V c (A23.3 Cl.11.2.8.1). - Find the required V s . Section A-A:

V s ≥ V f – V c = 500 – 253 = 247kN Section B-B:

V s ≥ 500 – 122 = 378kN ← governs Design shear reinforcement based on Section B-B parameters:

d v = 480mm V s = 378kN - Find the required stirrup spacing.

Copyright © 2006 Pearson Education Canada Inc.

6-16

2

2-legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

2

0.85 × 200mm × 400MPa × 480mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 123.5mm ≅ 120mm 378000 Use 10M@120. - Check the maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 480 = 336mm

s max = 336mm (smaller value governs) Since

s = 120mm < 336mm okay - Check CSA A23.3 minimum stirrup requirements (Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

2 600mm × 120mm = 0.06 25MPa -------------------------------------------- = 54mm 400MPa

Since 2

A vmin = 54mm < A v = 200mm

2

okay

- Check the maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v = 0.25 × 0.65 × 25MPa × 600mm × 480mm = 1170kN

Copyright © 2006 Pearson Education Canada Inc.

[6.17]

6-17

- Find actual V r .

V c = 122kN

(Section B-B)

2 φ s A v f y d v cot θ 0.85 × 200mm × 400MPa × 480mm × 1.43 V s = --------------------------------- = ------------------------------------------------------------------------------------------------------------ = 389kN [6.9] s 120mm

V r = V c + V s = 122 + 389 = 511kN Since

maxV r = 1170kN > 511kN okay - Determine the region where shear reinforcement is not required.

V fx = V c = 122kN

V fx

ln ---- – x 2 ------------ ( V f – V fmidspan ) = V fmidspan + ln ---2

3.6 ------- – x 2 122kN = 40kN + ---------------- ( 500 – 40 ) 3.6 ------2 x = 1.48m ≅ 1.5m

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6-18

- Sketch a beam diagram showing the stirrup arrangement.

Notes: 1. In this case, it would be considered more practical to use the same stirrup spacing ( s = 120mm ) throughout the beam length. There is a very small region of 600 mm length where stirrups are not required. 2. As a point of interest, it should be noted that the required stirrup spacing at the left support (Section A-A) is equal to 180 mm. Therefore, a different stirrup spacing might be used in the tapered portion of the beam ( s = 180mm ) as compared to the right-hand portion with the constant depth ( s = 120mm ) . This would have resulted in certain material savings, but would complicate the field specifications. ____________________________________________________________________________ 6.8.

- Find d and d v . cover = 30mm 10M stirrups: d s = 10mm

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6-19

25M rebars: d b = 25mm

db 25 d = h – cover – d s – ----- = 800 – 30 – 10 – ------ = 747mm 2 2 0.9d = 0.9 × 747 = 672mm 0.72h = 0.72 × 800 = 576mm d v ≅ 670mm (larger value governs) dv =

- Find clear span l n . Set 400 400 l n = 7500 – --------- – --------- = 7100mm 2 2 - Find factored shear force at the support.

wf × ln ( 75kN ⁄ m ) ( 7.1m ) V f = ---------------- = --------------------------------------------- = 266kN 2 2 V fmidspan = 0 (given) - Find V fdv .

V fdv

ln ---- – d v 2 = --------------- V f = ln ---2

7.1-----– 0.67 2 ------------------------ ( 266kN ) ≅ 216kN 7.1-----2

- Sketch the shear envelope diagram.

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6-20

- Find V c . Find β (assume less than minimum stirrups prescribed by A23.3 cl.11.3.6.3).

230 230 β = ----------------------- = --------------------------- = 0.14 (Table 6.1) 1000 + d v 1000 + 670 V c = φ c λβ f c ′b w d v

(A23.3 Eq.11-6) [6.12]

= 0.65 × 1.0 × 0.14 25MPa × 600mm × 670mm = 182.9kN ≅ 183kN Since

V fdv = 216kN > V c = 183kN it follows that the stirrups are required, however it is expected that the maximum permitted stirrup spacing per CSA A23.3 Code will be satisfactory for this design. 2

Assume 2 - legged 10M stirrups ( area = 100mm ):

A v = 2 × 100 = 200mm

2

- Find the maximum stirrup spacing according to A23.3 Cl.11.3.8.1.

s max =

600mm 0.7d v = 0.7 × 670 = 469mm

s max = 469mm (smaller value governs) - Find the maximum stirrup spacing corresponding to A vmin (A23.3 Cl.11.2.8.2).

A v = A vmin = 200 mm

2

and 2 Av fy 200mm × 400MPa s max = --------------------------- = ------------------------------------------------- = 444mm 0.06 25 × 600 0.06 f c ′b w Therefore, the maximum stirrup spacing is the smallest of:

s max =

600mm 469mm 444mm

Hence,

s max = 444mm ≅ 440mm Copyright © 2006 Pearson Education Canada Inc.

6-21

- Find the corresponding V s value.

θ = 35° (CSA A23.3 Simplified Method) cot ( θ ) = 1.43 φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 0.85 × 200 × 400 × 670 × 1.43 = ------------------------------------------------------------------------- ≅ 148kN 440 V r = V c + V s = 183 + 148 = 331kN Since

V r = 331kN > V fdv = 216kN okay - Determine the region where shear reinforcement is not required.

V fx = V c = 183kN

V fx

ln ---- – x 2 ------------ Vf = ln ---2

7.1 ------- – x 2 183kN = ---------------- ( 266kN ) 7.1 ------2 x = 1.10m

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6-22

- Sketch the stirrup arrangement.

Note that, in this case, only 3 sets of stirrups are required on each side of the beam. _____________________________________________________________________________ 6.9. - Perform load analysis.

DL = 100kN ⁄ m LL = 40kN ⁄ m w f = 1.25DL + 1.5LL = 185kN ⁄ m (NBC 2005 Table 4.1.3.2) P DL = 500kN P LL = 100kN P f = 1.25P DL + 1.5P LL = 1.25 × 500 + 1.5 × 100 = 775kN Use the principle of superposition.

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6-23

V fmax = V f1 + V f2 = 659 + 518 = 1177kN Find d and d v .

cover = 40mm 15M stirrups: d s = 15mm 30M rebars: d b = 30mm

db 30 d = h – cover – d s – ----- = 800 – 40 – 15 – ------ = 730mm 2 2 0.9d = 0.9 × 730 = 657mm 0.72h = 0.72 × 800 = 576mm d v = 660mm (larger value governs) dv =

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6-24

- Sketch the shear envelope diagram.

It follows from the shear envelope diagram that the maximum shear force occurs in the region AC, that is,

V f = 1177kN Maximum shear force in the region BD also needs to be considered in the design, that is,

V f = 634kN Find V fd v . Region AC:

V fdv = 1047kN Region BD:

2800 – 660 V fdv = 116 + --------------------------- ( 634 – 116 ) = 512kN 2800 - Find V c . Determine β assume that the shear reinforcement is less than the minimum required according to CSA A23.3.

230 230 β = ----------------------- = --------------------------- = 0.14 (Table 6.1) 1000 + d v 1000 + 660 Use b w = b = 600mm

V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6) [6.12] = 0.65 × 1.0 × 0.14 25MPa × 600mm × 660mm = 180.2kN ≅ 180kN - Is shear reinforcement required?

Copyright © 2006 Pearson Education Canada Inc.

6-25

Region AC:

V fdv = 1047kN > V c = 180kN Region BD:

V fdv = 512kN > V c = 180kN It follows that shear reinforcement is required (A23.3 Cl.11.2.8.1). Region AC:

- Find the required V s value.

V s ≥ V f – V c = 1047 – 180 = 867kN Find the required stirrup spacing. 2

For 2 - legged 15M stirrups ( area = 200mm ) :

A v = 2 × 200 = 400mm

2

- Calculate the stirrup spacing ( s ).

θ = 35° (CSA A23.3 Simplified Method) cot ( θ ) = 1.43 φ s A v f y d v cot θ s = --------------------------------[6.16] Vs 0.85 × 400 × 400 × 660 × 1.43 = ------------------------------------------------------------------------- = 148mm 867000 Use 15M@140. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 660 = 462mm

s max = 462mm (smaller value governs) Since

s = 140mm < 462mm okay - Check the minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

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6-26

600mm × 140mm 2 = 0.06 25MPa -------------------------------------------- = 63mm 400MPa Since 2

A vmin = 63mm < A v = 400mm

2

okay

- Check the maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 600mm × 660mm = 1609kN Since

maxV r = 1609kN > V f = 1047kN Beam dimensions are okay. - Find the actual V s value.

θ = 35° (CSA A23.3 Simplified Method) cot ( θ ) = 1.43 φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 0.85 × 400 × 400 × 660 × 1.43 = ------------------------------------------------------------------------- ≅ 916kN 140 Find actual V r value.

V r = V c + V s = 180 + 916 = 1096kN Since

V r = 1096kN > V fdv = 1047kN okay Region BD

-Find the required V s value.

V s ≥ V f – V c = 512 – 180 = 332kN Find the required stirrup spacing

φ s A v f y d v cot ( θ ) s = -------------------------------------Vs

[6.16]

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6-27

0.85 × 400 × 400 × 660 × 1.43 = ------------------------------------------------------------------------- = 387mm ≅ 380mm 332000 Use 15M@380. It is not required to perform the Code checks related to shear reinforcement requirements, since these checks have already been performed for the region AC. - Determine the region where shear reinforcement is not required. Shear envelope for the left half of the beam:

2800 – x 180kN = 116kN + --------------------- ( 272 – 116 ) 2100 x = 1938mm ≅ 2000mm Shear envelope for the right half of the beam:

2800 – x 180kN = 116kN + --------------------- ( 634 – 116 ) 2800 x = 2454mm

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6-28

It should be noted that the required stirrup spacing in the region CD should be based on shear force V f = 272kN . Hence,

V s ≥ V f – V c = 272 – 180 = 92kN Since

φ s A v f y d v cot ( θ ) s = -------------------------------------- [6.16] Vs 0.85 × 400 × 400 × 660 × 1.43 = ------------------------------------------------------------------------- = 1395mm 92000 In this case, maximum stirrup spacing per CSA A23.3 code governs, that is,

s max = 462mm however, it would not be practical to use 3 different stirrup spacings for this beam. Therefore, use 15M @ 380 throughout the region CD. Also, the region where the shear reinforcement is not required is rather small and can be neglected. - Sketch the stirrup arrangement.

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6-29

_____________________________________________________________________________ 6.10. Perform load analysis.

P DL = 1000kN P LL = 400kN P f = 1.25P DL + 1.5P LL = 1.25 × 1000 + 1.5 × 400 = 1850kN (NBC 2005 Table 4.1.3.2) - Find the beam reactions.

1 tan α = --2 R A = 617kN

α = 26.57° R c = 1233kN

R cv = R c cos α = 1103kN ≅ 1100kN - Sketch the shear force diagram.

In this design, regions AB and BC are characterized by different shear forces and different depths, therefore a separate shear design needs to be performed for each region.

Copyright © 2006 Pearson Education Canada Inc.

6-30

Region AB:

h = 800mm

- Find d and d v . cover = 40mm 15M stirrups: d s = 15mm 30M rebars: d b = 30mm

db 30 d = h – cover – d s – ----- = 800 – 40 – 15 – ------ = 730mm 2 2 0.9d = 0.9 × 730 = 657mm 0.72h = 0.72 × 800 = 576mm d v ≅ 660mm (larger value governs) dv =

-The shear force is constant throughout the region AB.

V f = 617kN - Find V c . In this case, assume that the minimum shear reinforcement has been provided throughout the region AB, and so

β = 0.18 (Table 6.1) V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6) [6.12] = 0.65 × 1.0 × 0.18 25MPa × 500mm × 660mm = 193kN Is shear reinforcement required? Since

V f = 617kN > V c = 193kN it follows that shear reinforcement is required where V f > V c (in this case, throughout the region AB) (A23.3 Cl.11.2.8.1). - Find the required V s .

V s ≥ V f – V c = 617 – 193 = 424kN

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6-31

-Find the required stirrup spacing ( s ). 2

2-legged 15M stirrups ( area = 200mm ) :

A v = 2 × 200 = 400mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

0.85 × 400 × 400 × 660 × 1.43 = ------------------------------------------------------------------------- ≅ 300mm 424000 Use 15M@300. -Check the maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 660 = 462mm

s max = 462mm (smaller value governs) Since

s = 300mm < 462mm okay - Check minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

500mm × 300mm 2 = 0.06 25MPa -------------------------------------------- = 112mm 400MPa Since 2

A vmin = 112mm < A v = 400mm

2

okay

- Check maximum permitted shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 500mm × 660mm = 1340kN

Copyright © 2006 Pearson Education Canada Inc.

6-32

Since

maxV r = 1340kN > 617kN Beam dimensions are okay. - Find actual V r value.

V c = 193kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 2

0.85 × 400mm × 400MPa × 660mm × 1.43 = ------------------------------------------------------------------------------------------------------------ ≅ 428kN 300mm V r = V c + V s = 193 + 428 = 621kN Since

V r = 621kN > V f = 617kN okay In this case, shear reinforcement will be required throughout the region AB (since V f > V c ), according to A23.3 Cl.11.2.8.1. Region BC:

h = 800 cos α = 716mm - Find d and d v .

db 30 d = h – cover – d s – ----- = 716 – 40 – 15 – ------ = 646mm 2 2

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6-33

0.9d = 0.9 × 646 = 581mm 0.72h = 0.72 × 716 = 515mm d v ≅ 580mm (larger value governs) dv =

- The shear force is constant throughout the region BC, that is,

V f = 1100kN - Find V c . In this case, assume that the minimum shear reinforcement has been provided throughout the region BC, and so β = 0.18 (Table 6.1).

V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6) [6.12] = 0.65 × 1.0 × 0.18 25MPa × 500mm × 580mm ≅ 170kN Is shear reinforcement required? Since

V f = 1100kN > V c = 170kN Shear reinforcement is required where V f > V c - in this case, throughout the region BC (A23.3 Cl.11.2.8.1). - Find the required V s .

V s ≥ V f – V c = 1100 – 170 = 930kN - Find the required stirrup spacing ( s ). 2

For 2 - legged 15M stirrups ( area = 200mm ) :

A v = 400mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

0.85 × 400 × 400 × 580 × 1.43 = ------------------------------------------------------------------------- = 121mm 930000

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6-34

Use 15M@120. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 580 = 406mm

s max = 406mm (smaller value governs) Since

s = 120mm < 406mm okay - Check minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

2 500mm × 120mm = 0.06 25MPa -------------------------------------------- = 45mm 400MPa

Since 2

A vmin = 45mm < A v = 400mm

2

okay

- Check the maximum permitted shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 500mm × 580mm = 1178kN Find the actual V r value.

V c = 170kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 0.85 × 400 × 400 × 580 × 1.43 = ------------------------------------------------------------------------- = 940kN 120 V r = V c + V s = 170 + 940 = 1110kN Since

maxV r = 1178kN > V r = 1110kN okay Note that the maxV r and V r values are rather similar.

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6-35

- Sketch the stirrup arrangement. Note that the shear reinforcement is required throughout the region BC since V f > V c ..

___________________________________________________________________________ 6.11. Perform load analysis.

DL = 100kN ⁄ m LL = 40kN ⁄ m w f = 1.25DL + 1.5LL = 185kN ⁄ m (NBC 2005 Table 4.1.3.2) P LL = 500kN P f = 1.5P LL = 1.5 × 500kN = 750kN (NBC 2005 Table 4.1.3.2)

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6-36

The shear force diagram for this beam is presented below. Note that, in this design, shear force diagram has been developed considering a full factored load over the beam length; alternatively, live load might be applied over half-span length in order to obtain the shear envelope diagram.

Design shear force at point C:

V fmax = 473kN - Find d and d v .

cover = 40mm 10M stirrups: d s = 10mm 30M rebars: d b = 30mm

db 30 d = h – cover – d s – ----- = 700 – 40 – 10 – ------ = 635mm 2 2

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6-37

0.9d = 0.9 × 635 = 571mm 0.72h = 0.72 × 700 = 504mm d v ≅ 570mm (larger value governs) dv =

- Find concrete shear resistance ( V c ) . Determine β - assume that the shear reinforcement is less than the minimum value.

230 230 β = ----------------------- = --------------------------- = 0.15 1000 + d v 1000 + 570

(Table 6.1)

- Find V c .

b w = b = 500mm V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.15 25MPa × 500mm × 570mm = 139kN - Is shear reinforcement required? Since

V fmax = 473kN > 139kN Shear reinforcement is required where V f > V c (A23.3 Cl.11.2.8.1). Support B

- Find the required V s value.

V s ≥ V f – V c = 473 – 139 = 334kN -Find the required stirrup spacing. 2

For 2 - legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ s = --------------------------------- [6.16] Vs 0.85 × 200 × 400 × 570 × 1.43 = ------------------------------------------------------------------------- = 166mm ≅ 160mm 334000

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6-38

Use 10M@160. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 570 = 399mm

s max = 399mm (smaller value governs) Since

s = 160mm < 399mm okay - Check minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

2 500mm × 160mm = 0.06 25MPa -------------------------------------------- = 60mm 400MPa

Since 2

A vmin = 60mm < A v = 200mm

2

okay

- Check maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 500mm × 570mm = 1158kN Since

maxV r = 1158kN > V f = 473kN okay Support A

V f = 372kN Find V fd v .

1500 – 570 V fdv = 94 + --------------------------- ( 372 – 94 ) = 266kN 1500 - Find the required V s value.

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6-39

V s ≥ V fd v – V c = 266 – 139 = 127kN

Find the required stirrup spacing ( s ).

φ s A v f y d v cot θ s = --------------------------------- [6.16] Vs 0.85 × 200 × 400 × 570 × 1.35 = ------------------------------------------------------------------------- = 412mm ≅ 400mm 127000 Note that s max = 400mm (see the stirrup design at support B). - Determine region in which the spacing s = 400mm is adequate. - Find the corresponding V r value.

V c = 139kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 0.85 × 200 × 400 × 570 × 1.43 = ------------------------------------------------------------------------- ≅ 139kN 400 V r = V c + V s = 139 + 139 = 278kN Find the distance ( x ) from point B where V fx = 278kN .

V fx = 278kN Copyright © 2006 Pearson Education Canada Inc.

6-40

2783.0m – ( x + 0.5m ) -------= ---------------------------------------------473 3.0m x ≅ 0.8m Therefore, the two different stirrup spacings will be used in this design: 1) s 1 = 400mm (from the support A to the section located at the distance x ≅ 0.8m from point B) 2) s 2 = 160mm (for the remaining portion of the beam) - Sketch the stirrup arrangement.

___________________________________________________________________________ 6.12. - Find the clear span ( l n ) :

l n = 4800mm

- Calculate the factored load ( w f ) .

= 6.0kPa

dead load:

(given)

kN = h × γ w = 0.2 × 24.0 ------3- = 4.8kPa m ________________________________________ self-weight:

DL = 10.8kPa w f ′ = 1.25DL + 1.5LL = 1.25 × 10.8 + 1.5 × 2.0 = 16.5kPa (NBC 2005 Table 4.1.3.2) Consider a unit strip of the slab:

b w = 1.0m

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w f = w f ′ × b = 16.5 kPa × 1.0m = 16.5kN ⁄ m - Find the maximum shear force at the face of the support.

wf × ln 16.5kN ⁄ m × 4.8m V f = ---------------- = ---------------------------------------------- = 39.6kN ⁄ m 2 2 - Find the effective depth ( d ). 15M flexural rebars: d b = 15mm cover = 20mm

db 15 d = h – cover – ----- = 200 – 20 – ------ = 172mm 2 2 - Find the effective shear depth ( d v ) .

0.9d = 0.9 × 172 = 155mm 0.72h = 0.72 × 200 = 144mm d v = 155mm (larger value governs) dv =

- Determine the β value. Since h = 200mm < 350mm , it follows that

β = 0.21

(A23.3 Cl.11.3.6.2)

V c = φ c λβ f c ′b w d v (A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.21 25MPa × 1000mm × 155mm = 105.8kN ⁄ m Since

V f = 39.6kN ⁄ m < V c = 105.8kN ⁄ m It follows that the shear resistance is adequate (A23.3 Cl.11.2.8.1); therefore, it is not required to provide shear reinforcement in this slab.

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_____________________________________________________________________________ 6.13.

- Find effective depth d . 10M stirrups: d s = 10mm 25M rebars: d b = 25mm cover = 40mm

d 25 d = h – cover – d s – ----b- = 600 – 40 – 10 – ------ = 537mm 2 2 Find the effective shear depth d v .

0.9d = 0.9 × 537 = 483mm 0.72h = 0.72 × 600 = 432mm d v = 480mm (larger value governs) dv =

- Find clear span l n .

400 400 l n = 3000 – --------- – --------- = 2600mm 2 2 - Find the factored load w f .

= 100kN ⁄ m (given) self-weight b × h × γ w = 0.4m × 0.6m × 24.0kN ⁄ m = 5.8kN ⁄ m __________________________________________________

DL ≅ 106kN ⁄ m

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LL = 40kN ⁄ m

Live load:

- Factored load ( w f ).

(NBC 2005 Table 4.1.3.2)

w f = 1.25DL + 1.5LL = 1.25 × 106 + 1.5 × 40 ≅ 192kN ⁄ m - Find shear force at the support.

Vf = RA = RB

ln kN w f × ---192 ------- × 2.6m 2 m = ---------------------- = ------------------------------------ = 249.6kN ≅ 250kN 2 2

- Find shear force at the midspan.

V fmidspan

ln kN 2.6m 1.5 40 ------- -----------( 1.5LL ) ---2 m 2 = ------------------------------ = ---------------------------------------------- = 19.5kN ≅ 20kN 4 4

- Find shear force at distance d v from the support ( V fdv ) .

V fdv

ln 2.6 ------- – 0.48 ---- – d v 2 2 = V fmidspan + --------------- ( V f – V fmidspan ) = 20.0 + ------------------------ ( 250 – 20 ) = 165kN ln 2.6 ---------2 2

- Sketch the shear envelope diagram.

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In this case, design shear force cannot be reduced at the support A. In case of a hanging support that introduces tension, Cl.11.3.2 of CSA A23.3 prescribes that the critical section needs to be taken at the face of the support. Therefore, the shear design will be performed by considering the design shear force

V f = 250kN - Find concrete shear resistance ( V c ) . Determine β (assume that the shear reinforcement is less than the minimum amount prescribed by CSA A23.3).

230 230 β = ----------------------- = --------------------------- = 0.155 ≅ 0.15 (Table 6.1) 1000 + d v 1000 + 480 - Find V c .

b w = b = 400mm V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.15 25MPa × 400mm × 480mm = 93.6kN ≅ 94kN - Is shear reinforcement required? Since

V f = 250kN > V c = 94kN Shear reinforcement is required wherever V f > V c (A23.3 Cl.11.2.8.1). - Find the required V s .

V s ≥ V f – V c = 250 – 94 = 156kN - Find the required stirrup spacing. 2

For 2-legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

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2

0.85 × 200mm × 400MPa × 480mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 299mm ≅ 300mm 156000N Use 10M@300. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 480 = 336mm

s max = 336mm (smaller value governs) Since

s = 300mm < 336mm okay - Check CSA A23.3 minimum stirrup requirement (Cl.11.2.8.2).

bw s A vmin = 0.06 f c′ -------fy

[6.18]

400mm × 300mm 2 = 0.06 25MPa -------------------------------------------- = 90mm 400MPa Since 2

A vmin = 90mm < A v = 200mm

2

okay

- Check the maximum permitted shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 25MPa × 400mm × 480mm = 780kN -Find actual V r value.

V c = 94kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 2 0.85 × 200mm × 400MPa × 480mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 155.6kN 300mm V r = V c + V s = 94 + 155.6 ≅ 250kN Since

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maxV r = 780kN > V r = 250kN beam dimensions are adequate. Determine the region where shear reinforcement is not required.

V fx = V c = 94kN

V fx

ln ---- – x 2 = V fmidspan + ------------- ( V f – V fmidspan ) ln ---2

1.3 – x 94 = 20 + ---------------- ( 250 – 20 ) 1.3 x ≅ 0.9m It can be concluded that shear reinforcement is required over the major portion of the beam length, except for the 800 mm long region around the midspan. Since the beam span length is rather limited ( l n = 2600mm ), it is not considered practical to discontinue the stirrups; therefore, stirrups are provided throughout the beam length. - Sketch the stirrup arrangement.

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___________________________________________________________________________

6.14. -Find the shear force distribution.

In this design, regions AB and BC are characterized by different depths and need to be designed separately. Region AB

h = 1000mm -Find d and d v . cover = 40mm 10M stirrups: d s = 10mm 30M rebars: d b = 30mm

db 30 d = h – cover – d s – ----- = 1000 – 40 – 10 – ------ = 935mm 2 2 0.9d = 0.9 × 935 = 841mm 0.72h = 0.72 × 1000 = 720mm d v ≅ 840mm (larger value governs) dv =

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- Find shear force at a distance d v from the face of the support.

- Distance from the support centreline:

400mm = 840mm + ------------------- = 1040mm 2 3.0 – 1.04 V fdv = 500kN + ------------------------ ( 575 – 500 ) = 549kN 3.0 V fdv ≅ 550kN Note: in this case, the difference in shear force values at points A and B is insignificant and so the V fdv calculation could have been avoided. - Find V c . In this case, we will assume that the minimum shear reinforcement prescribed by CSA A23.3 will be provided throughout the region AB, and so

β = 0.18 (Table 6.1) V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.18 30MPa × 500mm × 840mm = 269kN Is shear reinforcement required? Since

V f = V fd = 550kN > V c = 269kN v

according to A23.3 Cl.11.2.8.1, shear reinforcement is required where V f > V c (in this case, throughout the region AB). - Find the required V s .

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V s ≥ V f – V c = 550 – 269 = 281kN - Find the required stirrup spacing. 2

For 2 - legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43

φ s A v f y d v cot θ s = --------------------------------Vs

(CSA A23.3 Simplified Method)

[6.16]

2

0.85 × 200mm × 400MPa × 840mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 290mm 281000N Use 10M@290. - Check maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 840 = 588mm

s max = 588mm (smaller value governs) Since

s = 290mm < 588mm okay - Check minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

500mm × 290mm 2 = 0.06 30MPa -------------------------------------------- = 119mm 400MPa Since 2

A vmin = 119mm < A v = 200mm

2

- Check maximum shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

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maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 30MPa × 500mm × 840mm = 2047kN - Find actual V r value

V c = 269kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 2

0.85 × 200mm × 400MPa × 840mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 282kN 290 V r = V c + V s = 269 + 282 = 551kN Since

maxV r = 2047kN > V r = 551kN okay Since V f > V c , shear reinforcement is required throughout the region AB (A23.3 Cl.11.2.8.1). Region BC

h = 500mm

Find d and d v .

db 30 d = h – cover – d s – ----- = 500 – 40 – 10 – ------ = 435mm 2 2 0.9d = 0.9 × 435 = 391mm 0.72h = 0.72 × 500 = 360mm d v ≅ 390mm (larger value governs) dv =

- Find shear force at a distance d v from the face of the support B.

Distance from the support centreline:

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400mm = 390mm + ------------------- = 590mm 2 4.5 – 0.59 V fdv = 302kN + ------------------------ ( ( 415 – 302 ) ≅ 400kN ) 4.5 - Find V c .

β = 0.18 (same as region AB) V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.18 30MPa × 500mm × 390mm = 125kN - Is shear reinforcement required? Since

V f = V fdv = 400kN > V c = 125kN Cl.11.2.8.1 of CSA A23.3 prescribes that shear reinforcement is required throughout the region BC (see the shear force diagram). - Find the required V s value.

V s ≥ V f – V c = 400 – 125 = 275kN - Find the required stirrup spacing. 2

For 2 - legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s ).

θ = 35°

cot ( θ ) = 1.43 (CSA A23.3 Simplified Method)

φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

0.85 × 200 × 400 × 390 × 1.43 = ------------------------------------------------------------------------- = 138mm ≅ 140mm 275000 Use 10M@140. - Check the maximum stirrup spacing permitted by CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 390 = 273mm Copyright © 2006 Pearson Education Canada Inc.

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s max = 273mm (smaller value governs) Since

s = 140mm < 273mm okay - Check CSA A23.3 minimum stirrup requirements (Cl.11.2.8.2).

bw s A vmin = 0.06 f c ′ -------fy

[6.18]

500mm × 140mm 2 = 0.06 30MPa -------------------------------------------- = 57mm 400MPa Since 2

A vmin = 57mm < A v = 200mm

2

okay

- Check maximum permitted shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v

[6.17]

= 0.25 × 0.65 × 30MPa × 500mm × 390mm = 951kN - Find actual V r value.

V c = 125kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 0.85 × 200 × 400 × 390 × 1.43 = ------------------------------------------------------------------------- = 271kN 140 V r = V c + V s = 125 + 271 = 396kN Since

maxV r = 951kN > V r = 396kN beam dimensions are adequate Since V f > V c , shear reinforcement is required throughout the region BC.

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- Sketch the stirrup arrangement.

_____________________________________________________________________________ 6.15. a) Determine the area of shear friction reinforcement required at the beam supports. Factored shear force required to be transferred at each support:

w f × l 100kN ⁄ m × 6m V f = ------------- = ---------------------------------------- = 300kN 2 2 i) Left support (wall A) roughened to at least 5 mm amplitude The factored shear resistance across an interface:

v r = λφ c ( c + µσ ) + φ s ρ v f y cos α f

(A23.3 Eq.11-24) [6.20]

where

α f = 0°

cos α f = 1

c = 0.5

µ = 1.0

(case b), Table 6.3)

λ = 1.0 (normal-density concrete) σ = 0 effective normal stress Calculate the area of the concrete section resisting shear transfer:

h = 600mm b = 300mm

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A cv = b × h = 300 × 600 = 180000mm

2

Vf 300000N v r = -------- = -----------------------------2- = 1.667 A cv 180000mm When the above values are substituted in equation [6.20], it follows that 1.667 = 0.65 × 0.5 + 0.85ρ v ( 400 )

[6.20]

so

ρ v = 0.0039 where

A vf ρ v = -----------b×h and

A vf = ρ v × b × h = 0.0039 × 300 × 600 = 710mm Use 4-15M bars:

2

A vf = 4 × 200mm = 800mm

2

2

ii) Right support (wall B) - hardened but not roughened surface

c = 0.25

(case a), Table 6.3)

v r = λφ c ( c + µσ ) + φ s ρ v f y cos α f

(A23.3 Eq.11-24) [6.20]

1.667 = 0.65 × 0.25 + 0.85ρ v ( 400 ) ρ v = 0.0044 A vf = ρ v × b × h = 0.0044 × 300 × 600 = 796mm

2

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Hence, use 4-15M bars (same solution as for the left support):

2

A vf = 4 × 200mm = 800mm

2

b) Do you believe that this support is good for the right support? Since the beam is not subjected to axial load, it is expected that a shrinkage deformation might take place. It is probable that the bond at the right support is weaker than that at the left support. Due to the shrinkage effects, a gap might develop at the right support, thereby weakening the beam shear capacity. It is believed that this solution is not good for the right support. c) Which measures would you take to ensure that the aggregate interlock mechanism develops at the left support? The following measures can be taken during the site inspection: 1. The inspector can take a straight edge and measure the amplitude of roughness by placing the straight edge randomly over the roughened surface. 2. Also, the inspector must ensure that the holes drilled for the 4-15M dowels are deep enough to develop the full strength of the 15M bars. 3. Finally, the inspector must ensure that 15M bars are sufficiently long to be completely inserted into the hole and extended into the new beam by a length at least equal to their development length. ______________________________________________________________________________ 6.16. This design is not acceptable. If epoxy adhesive requires 250 mm length in order for 15M bars to develop their full strength, then the provision of longer than required development length, that is, 350 mm, is not going to result in an increased capacity for 15M bars. On the other hand, if 15M bars are inserted into 150 mm deep holes, then epoxy adhesive is not sufficient to fully develop the bond between the rebars and the existing concrete; hence, these bars will not develop their full strength. As a result, one-half of the bars with 150 mm deep holes can only develop partial strength, approximately equal to 60% of full strength (since 150/250 = 0.6), while the remaining bars with 350 mm deep holes cannot develop their full strength. If the design requires that each 15M bar must develop its full strength in order to be effective, then the total capacity of this design is below the required capacity. If one-half of the bars are able to develop 60% of the required strength while the remaining bars are able to develop their full strength, this design is going to result in 80% of the required strength (based on 0.5 × 0.6 + 0.5 × 1.0 = 0.8 ).

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Chapter 7 - Solutions _________________________________________________________________________ 7.1. 1. Find the distribution of factored shear forces and torsional moment. - Find the factored load ( w f ) (NBC 2005 Table 4.1.3.2)

DL = 20kN ⁄ m LL = 50kN ⁄ m w f = 1.25DL + 1.5LL = 1.25 × 20 + 1.5 × 50 = 100kN ⁄ m - Determine the shear force distribution.

kN R A = w f × l = 100 ------- × 3.0m = 300kN m - Sketch the shear force diagram

- Determine the torsional moment distribution. Eccentricity ( e ):

e = 100mm = 0.1m

Distributed moment ( m t ) :

kN kNm m t = w f × e = 100 ------- × 0.1m = 10 -----------m m

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Sketch the torsional moment distribution along the beam length and the corresponding moment diagram.

2. Find the design torsional moment ( T f@d ) at a distance d v from the support. Find the effective depth ( d ) 10M stirrups: d s = 10mm 25M flexural rebars: d b = 25mm Cover to the stirrups: cover = 30mm

db 25 d = h – cover – d s – ----- = 700 – 30 – 10 – ------ = 647mm 2 2 Find the effective shear depth ( d v )

dv =

0.9d = 0.9 × 647 = 582mm 0.72h = 0.72 × 700 = 504mm

so

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d v ≅ 580mm (larger value governs) Determine torsional moment ( T f@dv )

l – dv 3.0 – 0.58 T f@dv = T f ------------ = 300 ------------------------ = 24.2kNm ≅ 24kNm l 3.0 Torsional moment envelope:

3. Find the design shear force at a distance d v from the support ( V f@d ) .

l – dv 3.0 – 0.58 V f@dv = V f ------------ = 300 ------------------------ = 242kN ≅ 240kN 3.0 l - Sketch the shear force envelope.

4. Check whether shear reinforcement is required (A23.3 Cl.11.2.8.1). Find the V c value. Assume that the shear reinforcement will be less than the minimum CSA A23.3 requirement, so

230 230 β = ----------------------- = --------------------------- = 0.15 (see Table 6.1) 1000 + d v 1000 + 580 λ = 1 (normal-density concrete) V c = φ c λβ f c ′b w d v (A23.3 Eq. 11.6)

[6.12]

= 0.65 × 1.0 × 0.15 30MPa × 400 × 580 = 123.9kN ≅ 124kN

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7-3

Note that b w = b = 400mm Since

V f@dv = 240kN > V c = 124kN it follows that the shear reinforcement is required. - Find the required V s value.

V s ≥ V f – V c = 240 – 124 = 116kN 2

- Find the required stirrup spacing for 2-legged 10M stirrups ( area = 100mm ) :

A v = 2 × 100 = 200mm

2

- Calculate the stirrup spacing ( s )

θ = 35°

cot ( θ ) = 1.43

φ s A v f y d v cot θ s = --------------------------------Vs

[6.16]

0.85 × 200 × 400 × 580 × 1.43 = ------------------------------------------------------------------------- = 486 ≅ 480mm 116000 Use 15M@480. - Check CSA A23.3 maximum permitted stirrup spacing (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 580 = 406mm

s max = 406mm (smaller value governs) Since

s = 480mm > s max = 406mm it follows that the stirrup spacing needs to be reduced to

s = 400mm Use 10M@400. - Check CSA A23.3 minimum stirrup requirements (Cl.11.2.8.2)

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7-4

bw s A vmin = 0.06 f c ′ -------[6.18] fy 400mm × 400mm 2 = 0.06 30MPa -------------------------------------------- = 131mm 400MPa Since 2

2

A v = 200mm > A vmin = 131mm okay - Check the CSA A23.3 maximum permitted shear resistance ( maxV r ) (Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v = 0.25 × 0.65 × 30MPa × 400mm × 580mm = 1131kN -Find actual V r .

V c = 124kN

φ s A v f y d v cot θ V s = --------------------------------- (A23.3 Eq.11.7) [6.9] s 2 0.85 × 200mm × 400MPa × 580mm × 1.43 = ------------------------------------------------------------------------------------------------------------ = 141kN 400mm V r = V c + V s = 124 + 141 = 265kN Since

V r = 265kN < maxV r = 1131kN Beam dimensions are adequate for shear. 5. Check whether torsional effects need to be considered in the design (A23.3 Cl.11.2.9.1). - Calculate the torsional resistance of concrete T cr .

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A c = b × h = ( 400mm ) × ( 700mm ) = 280000mm

2

p c = 2 ( b + h ) = 2 ( 400mm + 700mm ) = 2200mm 2

T cr

Ac = ------ 0.38λφ c f c ′ (A23.3 Eq.11.2) pc

[7.6]

2 2

( 280000mm ) = ------------------------------------- 0.38 × 1.0 × 0.65 30MPa = 48.2kNm 2200mm - Check whether torsional effects need to be considered in the design. According to A23.3 Cl.11.2.9.1, torsional effects need to be considered in the design when

T f > 0.25T cr Since

T f = T f@d = 24kNm and

0.25T cr = 0.25 × 48.2 = 12kNm T f = 24kNm > 0.25T cr = 12kNm Therefore, torsional effects need to be considered in the design. 6. Check whether the beam dimensions are okay for torsional effects (A23.3 Cl.11.3.10.4).

- Find A oh .

ds 10 b h = b – 2 cover + ---- = 400mm – 2 30 + ------ = 330mm 2 2

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7-6

ds 10 h h = h – 2 cover + ---- = 700 – 2 30 + ------ = 630 2 2 p h = 2 ( b h + h h ) = 2 ( 330 + 630 ) = 1920mm A oh = b h × h h = ( 330mm ) ( 630mm ) = 207900mm

2

- Check the CSA A23.3 requirement.

Tfph + ---------------2 1.7A oh

2

240000N -------------------------------------------400mm × 580mm

2

Vf ----------bw d v

2

and

≤ 0.25φ c f c ′ (A23.3 Eq. 11.19) [7.13] 6

( 24 × 10 Nmm ) × 1920mm + ------------------------------------------------------------------2 2 1.7 × ( 207900mm )

2

= 1.21MPa

0.25φ c f c ′ = 0.25 × 0.65 × 30MPa = 4.9MPa Since

1.21MPa < 4.9MPa it follows that the beam cross sectional dimensions are okay for torsion. 7. Determine the required torsional resistance ( T r ) . Set T r = T f@d = 24kNm 8. Find the required stirrup spacing based on the torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Since

At φs fy T r = 2A o --------------- cot θ (A23.3 Eq. 11.17) [7.10] s it follows that

At φs fy s = 2A o --------------- cot θ Tr Find A o (the area calculated by the centreline of the tube walls) 2

A o = 0.85A oh = 0.85 × 207900mm = 176715mm

2

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7-7

- Find A t (area of one leg of closed stirrup reinforcement) 2

10M stirrups ( area = 100mm ): A t = 100mm

2

- Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 - Find spacing ( s ). 2 At φs fy 2 ( 100mm ) ( 0.85 ) ( 400MPa ) - × 1.43 = 716mm s = 2A o --------------- cot θ = 2 × 176715mm -------------------------------------------------------------------6 Tr 24 × 10 Nmm

9. Check the maximum stirrup spacing requirement (A23.3 Cl.11.2.8.2) - Consider the combined shear and torsional reinforcement.

At Av ′ b ----- + -------- ≥ 0.03 f c ′ -----ws fy s 200 2 A v ′ = --------- = 100mm area of shear reinforcement per leg 2 Try spacing s = 400mm based on shear requirements: 2

A t = 100mm , so

A t A v ′ 100mm 2 100mm 2 ----- + -------- = --------------------- + --------------------- = 0.50 s s 400mm 400mm

bw 400mm 0.03 f c ′ ------ = 0.03 30MPa ---------------------- = 0.16 400MPa fy Since

0.50 > 0.16 okay

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7-8

Use 10M@400. 10. Calculate the area of longitudinal torsional reinforcement. Use the following equation

ph At A l = --------------------2s ( tan θ )

[7.11]

2

( 1920mm ) × ( 100mm ) - = 979mm 2 = ---------------------------------------------------------2 ( 400mm ) ( tan 35 ) - Check the bar diameter (A23.3 Cl.11.2.7). s The code requires that the nominal bar diameter should not be less than ------ . Since 16 s 400 ------ = -------- = 25mm 16 16 one 25M bar needs to be placed in each corner. 2

2

A l = 4 × 500 = 2000mm > 979mm o.k. Use 4-25M bars (one in each corner). It is recommended that the longitudinal bar spacing should not exceed 300 mm. Therefore, provide nominal reinforcement (10M bars) in the vertical direction. 11. Provide a design summary. The summary shows the beam cross section and the arrangement of transverse reinforcement.

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_______________________________________________________________ 7.2. 1. Find the factored shear and torsional effects.

-Find the factored beam load ( w f ) according to NBC 2005 Table 4.1.3.2.

DL = 15kN ⁄ m kN kN w f = 1.25 × DL = 1.25 × 15.0 = 18.7 ------- ≅ 19 ------m m - Determine the shear and torsional effects transferred to the column.

a = 600mm – 250mm = 350mm (see the sketch above) kN T fc = V fj × a + M fj = 60 ------- × ( 0.35m ) + 40.0kNm ⁄ m = 61kNm ⁄ m m V fc = w f + V fj = 19.0kN ⁄ m + 60kN ⁄ m = 79kN ⁄ m

- Eccentricity (column-to-beam centreline): e = 50mm Copyright © 2006 Pearson Education Canada Inc.

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Next, find the shear & torsion effects at the centreline of the beam (point C). Torsion:

m tf = T fc – V fc × e = 61.0kNm ⁄ m – ( 79kN ⁄ m ) ( 0.05m ) = 57.0kNm ⁄ m Shear:

w fB = V fc = 79kN ⁄ m

- Find the factored torsional moment in the beam.

500 500 Clear span: l n = 8000mm – --------- – --------- = 7500mm 2 2 m tf × l n ( 57.0kNm ⁄ m ) ( 7.5m ) T f = ------------------ = ------------------------------------------------------ = 214kNm 2 2

- Find the factored shear force in the beam.

w fB × l n ( 79kN ⁄ m ) ( 7.5m ) V f = ------------------- = --------------------------------------------- = 296kN 2 2

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7-11

2. Find the design torsional moment ( T f@d ) at a distance d v from the support. - Effective depth ( d ). 15M stirrups: d s = 15mm 25M flexural rebars: d b = 25mm cover to the stirrups: cover = 30 mm

db 25 d = h – cover – d s – ----- = 500 – 30 – 15 – ------ = 442mm 2 5 - Calculate the effective shear depth ( d v ) .

0.9d = 0.9 × 442 = 398mm 0.72h = 0.72 × 500 = 360mm d v ≅ 400mm (larger value governs) dv =

- Find the design torsional moment ( T f@d ) :

T f@d = T f

ln 7.5---- – d v -----– 0.4 2 2 --------------- = 214kNm × -------------------- = 191.2kNm ≅ 191 kNm ln 7.5 ---------2 2

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- Sketch the torsional moment envelope.

3. Find the design shear force at a distance d v from the support ( V f@d ) .

V f@d = V f

ln ---- – d v 2 --------------- = 296kN ln ---2

7.5-----– 0.4 2 --------------------- = 264kN 7.5 ------2

- Sketch the shear force envelope.

4. Check whether shear reinforcement is required (A23.3 Cl.11.2.8.1). - Find the V c value. Assume that the shear reinforcement will be less than the minimum CSA A23.3 requirement

230 230 β = ----------------------- = --------------------------- = 0.16 (see Table 6.1) 1000 + d v 1000 + 400 λ = 1 (normal-density concrete) V c = φ c λβ f c ′b w d v

(A23.3 Eq. 11.6) [6.12]

= 0.65 × 1.0 × 0.16 30MPa × 600mm × 400mm = 136.7kN ≅ 137kN Note that b w = b = 600mm Since

V f@dv = 264kN > V c = 137kN Copyright © 2006 Pearson Education Canada Inc.

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it follows that the shear reinforcement is required. - Find the required V s value.

V s ≥ V f – V c = 264 – 137 = 127kN - Find the required stirrup spacing. 2

For 2-legged 15M stirrups ( area = 200mm ) :

A v = 2 × 200 = 400mm

2

- Calculate the stirrup spacing ( s )

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 φ s A v f y d v cot θ 0.85 × 400 × 400 × 400 × 1.43 s = --------------------------------- = ------------------------------------------------------------------------- = 612mm Vs 127000 - Check maximum permitted stirrup spacing per CSA A23.3 (Cl.11.3.8.1).

s max =

600mm 0.7d v = 0.7 × 400 = 280mm

Since

s max = 280mm (smaller value governs) it follows that the stirrup spacing needs to be reduced to

s = 280mm Use 15M@280. - Check CSA A23.3 minimum stirrup requirements (A23.3 Cl.11.2.8.2).

bw s -------[6.18] fy 2 600mm × 280mm = 0.06 30MPa -------------------------------------------- = 138mm 400MPa

A smin = 0.06 f c ′

Since 2

2

A smin = 138mm < A v = 200mm okay - Check the maximum permitted shear resistance ( maxV r ) (A23.3 Cl.11.3.3).

maxV r = 0.25φ c f c ′b w d v = 0.25 × 0.65 × 30MPa × 600mm × 400mm = 1170kN

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- Find actual V r .

V c = 137kN

φ s A v f y d v cot θ 0.85 × 400 × 400 × 1.43 V s = --------------------------------- = --------------------------------------------------------- = 278kN (A23.3 Eq.11.7) [6.9] s 280 V r = V c + V s = 137 + 278 = 415kN Since

V r = 415kN < maxV r = 1170kN Beam dimensions are adequate for shear. 5. Check whether the torsional effects need to be considered in the design (A23.3 Cl.11.2.9.1). Calculate the torsional resistance of concrete T cr .

A c = b × h = ( 600mm ) × ( 500mm ) = 300000mm

2

p c = 2 ( b + h ) = 2 ( 600 + 500 ) = 2200mm 2

T cr

Ac = ------ 0.38λφ c f c ′ (A23.3 Eq.11.2) pc

[7.6]

2 2

( 300000mm ) = ----------------------------------- 0.38 × 1.0 × 0.65 30MPa = 55.3kNm 2200mm - Check whether torsional effects need to be considered in the design. According to A23.3 cl.11.2.9.1, torsional effects need to be considered in the design if

T f > 0.25T cr T f = T f@d = 191kNm 0.25T cr = 0.25 × 55.3 = 13.8 ≅ 14kNm Since

T f = 191kNm > 0.25T cr = 14kNm torsional effects need to be considered in the design.

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6. Check whether the beam dimensions are okay for torsional effects (A23.3 Cl.11.3.10.4).

Find A oh .

ds 10 b h = b – 2 cover + ---- = 600mm – 2 30 + ------ = 530mm 2 2 ds 10 b h = h – 2 cover + ---- = 500 – 2 30 + ------ = 430mm 2 2 p h = 2 ( b h + h h ) = 2 ( 530 + 430 ) = 1920mm A oh = b h × h h = ( 530mm ) ( 430mm ) = 227900mm

2

- Check the CSA A23.3 requirement:

Tfph + ---------------2 1.7A oh

2

264000N -------------------------------------------600mm × 400mm

2

Vf ----------bw d v

2

and

≤ 0.25φ c f c ′ (A23.3 Eq. 11.19) [7.13] 6

( 191 × 10 Nmm ) ( 1920mm ) + --------------------------------------------------------------------2 2 1.7 × ( 227900mm )

2

= 4.3MPa

0.25φ c f c ′ = 0.25 × 0.65 × 30MPa = 4.9MPa Since

4.3MPa < 4.9MPa it follows that the beam cross sectional dimensions are okay for torsion. 7. Determine the required torsional resistance ( T r ) . Set T r = T f@dv = 191kNm 8. Find the required stirrup spacing based on the torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Copyright © 2006 Pearson Education Canada Inc.

7-16

Since

At φs fy T r = 2A o --------------- cot θ (A23.3 Eq. 11.17) s

[7.10]

it follows that

At φs fy s = 2A o --------------- cot θ Tr Find A o .

A o = 0.85A oh = 0.85 × 227900 = 193715mm

2

Find A t (area of one leg of closed stirrup reinforcement). 2

15M stirrups ( area = 200mm ): A t = 200mm

2

- Determine θ .

θ = 35° (CSA A23.3 Simplified method) cot ( θ ) = 1.43 - Find spacing ( s ). 2 At φs fy ( 200mm ) ( 0.85 ) ( 400MPa ) - × 1.43 ≅ 190mm s = 2A o --------------- cot θ = 2 × 193715 -------------------------------------------------------------------6 Tr 191 × 10 Nmm

9. Check the CSA A23.3 maximum stirrup spacing requirement (Cl.11.2.8.2). - Consider the combined effect of shear and torsional reinforcement

At Av ′ b ----- + -------- ≥ 0.03 f c ′ -----w[7.20] s fy s 2 400 A v ′ = --------- = 200mm (area of shear reinforcement per leg) 2 Try spacing s = 190mm based on torsional requirements, so 2 2 At Av ′ 200mm ----- + -------- = --------------------- + 200mm --------------------- = 2.1 s s 190mm 190mm

bw 600mm 0.03 f c ′ ------ = 0.03 30MPa ---------------------- = 0.25 400MPa fy Since

2.1 > 0.25 okay 10. Calculate the area of longitudinal torsional reinforcement.

Copyright © 2006 Pearson Education Canada Inc.

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ph At A l = --------------------2s ( tan θ )

[7.11]

( 1920 ) ( 200 ) 2 = -----------------------------------2- = 4122mm ( 190 ) ( tan 35 ) Use 9 - 25M bars. Since 2

2

A l = 9 × 500 = 4500mm > 4122mm okay The reinforcement should be distributed uniformly around the beam section. - Check the bar diameter (A23.3 Cl.11.2.7).

s The Code required that the nominal bar diameter should not be less than ------ , that is, 16 s190 ----= --------- = 11.9 ≅ 12mm 16 16 25M bars here have been used, hence this requirement is satisfied. 11. Provide a design summary. The summary shows the beam cross section and the arrangement of transverse reinforcement.

Copyright © 2006 Pearson Education Canada Inc.

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__________________________________________________________________ 7.3. 1. Find the distribution of torsional and shear loads.

- Factored wind load (NBC 2005 Table 4.1.3.2):

w f = 1.5WL = 1.5 × 3.0 = 4.5kPa - Uniform load along the depth of the sign board:

w f ′ = w f × 0.9m = 4.5kPa × 0.9m = 4.05kN ⁄ m - Resultant forces at the face of the post:

kN kN R B = w f ′ × 2.7m = 4.05 ------- × 2.7m = 10.9 ------m m 2.7m kN 2.7m kNm M B = w f ′ × ------------ = 4.05 ------- × ------------ = 5.5 -----------2 m 2 m

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- Forces at the centroid of the post:

400mm Eccentricity e = ------------------- = 200mm 2 m t = R B × e + M B = ( 10.9kN ⁄ m ) ( 0.2m ) + ( 5.5kNm ⁄ m ) = 7.7kNm ⁄ m V f ′ = R B = 10.9kN ⁄ m - Find the torsional moment distribution.

T f = ( 7.7kNm ⁄ m ) × ( 0.9m ) = = 6.93kNm ≅ 7.0kNm

- Find the shear force distribution over the post height.

V f = 10.9kN ⁄ m × 0.9m = 9.81kN ≅ 10kN

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2. Check whether shear reinforcement is required (A23.3 Cl.11.2.8.1). - Find effective depth d and d v . 10M stirrups: d s = 10mm 25M flexural rebars: d b = 25mm cover to the stirrups: cover = 40mm

db 25 d = h – cover – d s – ----- = 400 – 40 – 10 – ------ = 337mm 2 2 0.9d = 0.9 × 337 = 303mm 0.72h = 0.72 × 400 = 288mm d v ≅ 300mm (larger value governs) dv =

Assume that the shear reinforcement is less than the minimum per CSA A23.3 Code.

230 230 β = ----------------------- = --------------------------- = 0.18 (See Table 6.1) 1000 + d v 1000 + 300 λ = 1 (normal-density concrete) V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.18 30MPa × 400mm × 300mm ≅ 77kN Since

V f = 10kN < V c = 77kN Copyright © 2006 Pearson Education Canada Inc.

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it follows that the shear reinforcement is not required. 3. Check whether torsional effects need to be considered in the design (A23.3 Cl.11.2.9.1). Calculate the torsional resistance of concrete T cr .

A c = b × h = 400mm × 400mm = 160000mm

2

p c = 2 ( b + h ) = 2 ( 400 + 400 ) = 1600mm 2

T cr

Ac = ------ 0.38λφ c f c ′ (A23.3 Eq.11.2) pc

[7.6]

2 2

( 160000mm ) = ------------------------------------- × 0.38 × 1.0 × 0.65 30MPa = 21.6kNm 1600mm

- Check whether torsional effects need to be considered in the design. According to A23.3 Cl.11.2.9.1, torsional effects need to be considered in the design when

T f > 0.25T cr T f = 7.0kNm 0.25T cr = 0.25 × 21.6 = 5.4kNm Since

T f = 7.0kNm > 0.25T cr = 5.4kNm it follows that torsional effects need to be considered in the design.

Copyright © 2006 Pearson Education Canada Inc.

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4. Check whether the post dimensions are okay (A23.3 Cl.11.3.10.4).

ds 10 b h = h h = h – 2 cover + ---- = 400mm – 2 40 + ------ = 310mm 2 2 p h = 2 ( b h + h h ) = 2 ( 310 + 310 ) = 1240mm A oh = b h × h h = 310mm × 310mm = 96100mm

2

- Check the CSA A23.3 requirement.

Tf ph ---------------- ≤ 0.25φ c f c ′ (A23.3 Eq.11.19) [7.13] 2 1.7A oh 6

( 7 × 10 Nmm ) ( 1240mm ) ---------------------------------------------------------------- = 0.55MPa 2 1.7 ( 96100 ) 0.25φ c f c ′ = 0.25 × 0.65 × 30MPa = 4.9MPa Since

0.55MPa < 4.9MPa it follows that the beam cross sectional dimensions are okay. 5. Determine the required torsional resistance ( T r ) . Set T f = T r = 7.0kNm . 6. Find the required stirrup spacing based on the torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Since

At φs fy T r = 2A o --------------- cot θ s

(A23.3 Eq.11.17) [7.10]

it follows that

At φs fy s = 2A o --------------- cot θ Tr Copyright © 2006 Pearson Education Canada Inc.

7-23

-Find A o .

A o = 0.85A oh = 0.85 × 96100 = 81685mm

2

- Find A t . 2

10M stirrups ( area = 100mm ): 2

A t = 100mm (area of 1 stirrup leg) - Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 -Find spacing ( s ). 2

2 100mm × 0.85 × 400MPa - × 1.43 = 1135mm s = 2 × 81685mm × ----------------------------------------------------------------6 7 × 10 7. Check the maximum stirrup spacing requirement (A23.3 Cl.11.2.8.2).

At b ----- ≥ 0.03 f c ′ -----ws fy

[7.20]

2 At fy 100mm 400MPa-------------------------------------------------------- --------------------s≤ = = 608mm b 0.03 f c ′ w 0.03 30MPa 400mm Therefore, stirrup spacing should be reduced to s = 600mm .

Use 10M@600. 8. Calculate the area of longitudinal torsional reinforcement.

ph At A l = --------------------2s ( tan θ )

[7.11]

2

( 1240mm ) ( 100mm ) 2 - = 421mm = ---------------------------------------------------2 ( 600mm ) ( tan 35 ) - Use 4 - 25M bars (one at each corner). 2

2

A = 4 × 500 = 2000mm > 421mm okay - Check whether the bar diameter is okay (A23.3 Cl.11.2.7) CSA A23.3 requires that the nominal bar diameter should not be less than

Copyright © 2006 Pearson Education Canada Inc.

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s600 ----= --------- ≅ 37mm 16 16 hence, bar size needs to be increased to 35M. Use 4-35M bars, or keep 25M bars and reduce stirrup spacing to 400 mm, that is,

s400 ----= --------- = 25mm 16 16 It is recommended that the longitudinal bar spacing should not exceed 300 mm. This requirement is satisfied. 9. Sketch a typical cross section showing the reinforcement.

_______________________________________________________________________ 7.4. 1. Find the factored torsional moment T f .

- Find the torsional moment around the centroid of the beam. Eccentricity:

Copyright © 2006 Pearson Education Canada Inc.

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0.6 e = 1.0 + ------- = 1.3m 2 Torsional moment:

T f = w f × e = ( 15kN ⁄ m ) ( 1.3m ) = 19.5kNm T f ≅ 20kNm

2. Find the effective depths d and d v . cover = 40mm 10M stirrups: d s = 10mm 25M flexural rebars: d b = 25mm

db 25 d = h – cover – d s – ----- = 600 – 40 – 10 – ------ = 537mm 2 2 0.9d = 0.9 × 537 = 483mm 0.72h = 0.72 × 600 = 432mm d v ≅ 480mm (larger value governs) dv =

3. Check whether torsional effects need to be considered in the design (A23.3 Cl.11.2.9.1). Calculate the torsional resistance of concrete T cr .

A c = b × h = 600mm × 600mm = 360000mm

2

Copyright © 2006 Pearson Education Canada Inc.

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p c = 2 ( b + h ) = 2 ( 600 + 600 ) = 2400mm 2

T cr

Ac = ------ 0.38λφ c f c ′ (A23.3 Eq.11.2) pc

[7.6]

2 2

( 360000mm ) = ----------------------------------- × 0.38 × 1.0 × 0.65 30MPa = 73.1kNm 2400mm Check whether torsional effects need to be considered in the design. According to A23.3 Cl.11.2.9.1, torsional effects need to be considered in the design when

T f > 0.25T cr T f = 20kNm 0.25T cr = 0.25 × 73.1 = 18.3kNm Since

T f = 20kNm > 18.3kNm it follows that torsional effects need to be considered in the design. 4. Check whether the beam dimensions are okay (A23.3 Cl.11.3.10.4).

Find A oh .

ds 10 h h = b h = b – 2 cover + ---- = 600 – 2 40 + ------ = 510mm 2 2 p h = 2 ( b h + h h ) = 2 ( 510 + 510 ) = 2040mm A oh = b h × h h = 510mm × 510mm = 260100mm

2

- Check the CSA A23.3 requirement.

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Tf ph ---------------- ≤ 0.25φ c f c ′ (A23.3 Eq.11.19) [7.13] 2 1.7A oh 6

( 20 × 10 Nmm ) ( 2040mm ) ------------------------------------------------------------------- = 0.35 2 2 1.7 × ( 260100mm ) 0.25φ c f c ′ = 0.25 × 0.65 × 30MPa = 4.9 Since

0.35 < 4.9 It follows that the beam cross sectional dimensions are okay (note that shear has been ignored). 5. Determine the required torsional resistance T r . Set T r = T f = 20kNm 6. Find the required stirrup spacing based on the torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Since

At φs fy T r = 2A o --------------- cot θ (A23.3 Eq.11.17) [7.10] s it follows that

At φs fy s = 2A o --------------- cot θ Tr - Find A o .

A o = 0.85A oh = 0.85 × 260100 = 221085mm

2

- Find A t . 2

10M stirrups (area = 100mm ):

A t = 100mm

2

- Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 - Find spacing ( s ).

At φs fy s = 2A o --------------- cot θ Tr

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2

2 ( 100mm ) ( 0.85 ) ( 400MPa ) - × 1.43 = 1075mm = 2 × ( 221085mm ) -------------------------------------------------------------------6 ( 20 × 10 Nmm ) 7. Check the maximum stirrup spacing requirement (A23.3 Cl.11.2.8.2).

At b ----- ≥ 0.03 f c ′ -----wfy s At s ≤ --------------------0.03 f c ′

[7.20]

2 f 100mm 400MPa -----y- = ---------------------------------- --------------------- = 406mm bw 0.03 30MPa 600mm

Therefore, stirrup spacing should be reduced to s = 400mm . Use 10M@400. 8. Calculate the area of longitudinal torsional reinforcement.

ph At A l = --------------------2s ( tan θ )

[7.11]

2

( 2040mm ) ( 100mm ) 2 - = 1040mm = ---------------------------------------------------2 ( 400mm ) ( tan 35 ) - Use 4 - 25M bars (one at each corner). 2

2

A = 4 × 500 = 2000mm > 1040mm okay - Check whether the bar diameter is okay (A23.3 Cl.11.2.7) CSA A23.3 requires that the nominal bar diameter should not be less than

s 400 ------ = -------- = 25mm 16 16 hence, it is okay to use 25M bars. Use 4-25M bars. It is recommended that the longitudinal bar spacing should not exceed 300 mm. Therefore, provide additional 10M bars as required.

Copyright © 2006 Pearson Education Canada Inc.

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9. Sketch a typical beam cross section.

___________________________________________________________________________ 7.5. 1. Determine the torsional moment distribution.

- Find the distributed torsional moment ( m t ) around the girder centroid C. Eccentricity:

900mm e = 500mm + ------------------- = 950mm 2 Distributed moment ( m t ) :

m t = w f × e = ( 30kN ⁄ m ) ( 0.95m ) = 28.5kNm ⁄ m Find the factored torsional moment.

Copyright © 2006 Pearson Education Canada Inc.

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m×l ( 28.5kNm ⁄ m ) ( 20m ) T f = ------------ = ----------------------------------------------------- = 285kNm 2 2 - Sketch the torsional moment distribution along the girder..

2. Find the design torsional moment T f@d at a distance d v from the support. Find the effective depth ( d ). 15M stirrups: d s = 15mm 35M flexural rebars: d b = 35mm Cover to the stirrups:

cover = 40mm

db 35 d = h – cover – d s – ----- = 1500 – 40 – 15 – ------ = 1427mm 2 2 Find d v .

0.9d = 0.9 × 1427 = 1284mm 0.72h = 0.72 × 1500 = 1080mm d v ≅ 1300mm (larger value governs). dv =

- Find the design torsional moment ( T f@dv ) .

T f@dv

20m--l- – d v ---------– 1.3m 2 2 = T f -------------- = ( 285kNm ) ----------------------------- = 248kNm ≅ 250kNm 20--l----2 2

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- Sketch the torsional moment envelope.

3. Check whether torsional effects need to be considered in the design (A23.3 Cl.11.2.9.1).

Calculate the torsional resistance of concrete T cr . 6

A c = b × h = ( 900mm ) ( 1500mm ) = 1.35 × 10 mm

2

p c = 2 ( b + h ) = 2 ( 900 + 1500 ) = 4800mm 2

T cr

Ac = ------ 0.38λφ c f c ′ (A23.3 Eq.11.2) pc 6

[7.6]

2 2

( 1.35 × 10 mm ) = --------------------------------------------- × 0.38 × 1.0 × 0.65 30MPa = 513.7kNm ≅ 514kNm 4800mm - Check whether torsional effects need to be considered in the design. According to A23.3 cl.11.2.9.1, torsional effects need to be considered in the design if

T f > 0.25T cr T f = T f@d = 250kNm 0.25T cr = 0.25 × 514 = 128.5kNm Since

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T f = 250kNm > 128.5kNm it follows that torsional effects need to be considered in the design. 4. Check whether the girder cross sectional dimensions are okay (A23.3 Cl.11.3.10.4).

- Find A oh .

ds 15 b h = b – 2 cover + ---- = 900 – 2 40 + ------ = 805mm 2 2 ds 15 h h = h – 2 cover + ---- = 1500 – 2 40 + ------ = 1405mm 2 2 p h = 2 ( b h + h h ) = 2 ( 805 + 1405 ) = 4420mm 6

A oh = b h × h h = 805mm × 1405mm = 1.13 × 10 mm

2

- Check the CSA A23.3 requirement

Tf ph ---------------- ≤ 0.25φ c f c ′ (A23.3 Eq.11.19) 2 1.7A oh

[7.13]

6

( 250 × 10 Nmm ) ( 4420mm ) ---------------------------------------------------------------------- = 0.51MPa 6 2 2 1.7 ( 1.13 × 10 mm ) and 0.25φ c f c ′ = 0.25 × 0.65 × 30MPa = 4.9MPa Since

0.51MPa < 4.9MPa

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it follows that the girder cross sectional dimensions are okay. 5. Find the required torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Set T r = T f@d = 250kNm . 6. Find the required stirrup spacing based on the torsional resistance ( T r ) (A23.3 Cl.11.3.10.3). Since

At φs fy T r = 2A o --------------- cot θ s

(A23.3 Eq. 11.17) [7.10]

it follows that

At φs fy s = 2A o --------------- cot θ Tr Find A o . 6

5

A o = 0.85A oh = 0.85 × 1.13 × 10 = 9.6 × 10 mm

2

Find A t . 2

15M stirrups (area = 200mm ):

A t = 200mm

2

- Determine θ .

θ = 35° (CSA A23.3 Simplified Method) cot θ = 1.43 - Find spacing ( s ).

At φs fy s = 2A o --------------- cot θ Tr

2

5 2 ( 200mm )0.85 × 400MPa - × 1.43 = 747mm = 2 × ( 9.6 × 10 mm ) ---------------------------------------------------------------6 250 × 10 Nmm

7. Check the maximum stirrup spacing requirement (A23.3 Cl.11.2.8.2). Since

At b ----- ≥ 0.03 f c ′ -----wfy s

[7.20]

it follows that

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2 fy At 200mm 400MPa-------------------------------------------------------- --------------------s≤ == = 541mm b 0.03 f c ′ 0.03 30MPa 900mm w Therefore, stirrup spacing should be reduced to s = 500mm .

Use 15M@500. 8. Identify the region in the girder where the torsional reinforcement is required. The torsional reinforcement is not required where

T f ≤ 0.25T cr 0.25T cr = 128.5kNm ≅ 130kNm Perform a linear interpolation to locate the distance from the left support ( x ) where

T f@x = 0.25T cr = 130kNm l --- – x 2 T f@x = T f -----------l --10 – x 130kNm = (2285kNm ) --------------10 x = 5.44m ≅ 5.5m

9. Calculate the area of longitudinal torsional reinforcement.

ph At A l = --------------------2s ( tan θ )

[7.11]

2

( 4420mm ) ( 200mm ) - = 3606mm 2 = ---------------------------------------------------2 ( 500mm ) ( tan 35 ) Copyright © 2006 Pearson Education Canada Inc.

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Use 4 - 35M bars (one at each corner of the girder cross section). 2

A = 4 × 1000 = 4000mm > 3606mm

2

- Check whether the bar diameter is okay (A23.3 Cl.11.2.7) The code requires the nominal bar diameter should not be less than

s500 ----= --------- = 31mm 16 16 it follows that 35M bars are okay. It is recommended that the longitudinal bar spacing should not exceed 300 mm, therefore provide additional 15M bars as required. 10. Sketch the girder and the reinforcement arrangement.

It should be noted that the requirement related to transverse reinforcement might change when shear effects are taken into account.

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Chapter 8 - Solutions _________________________________________________________________________ 8.1. a) Concentrically loaded column (zero eccentricity) The maximum axial load capacity of the column is reached when the compressive strain in concrete ( ε c ) reaches the maximum value ( ε cmax ) of 0.0035 ( ε c = ε cmax ) . The maximum axial load capacity of a concentrically loaded column ( P ro ) is the sum of the concrete contribution and the steel contribution. The failure of a concentrically loaded column occurs after yielding of the longitudinal reinforcement has taken place. The behaviour characteristic for concentrically loaded columns is “concrete-controlled”. b) Eccentrically loaded column - small eccentricity The axial load resistance in eccentrically loaded columns with small eccentricity is less than the corresponding load resistance of concentrically loaded columns. The decrease in the load capacity depends on the magnitude of eccentricity: the larger the eccentricity, the smaller is the load capacity. Eccentrically loaded columns with small eccentricity behave similarly to concentrically loaded columns - they are characterized by “concrete-controlled” behaviour. c) Eccentrically loaded column - infinitely large eccentricity At a large load eccentricity ( e ), the effect of bending moment ( M = P e ) becomes significant, and the column behaves like a beam subjected to flexure. When the eccentricity is indefinitely large ( e = ∞ ) , the axial load becomes insignificant and can be taken as zero, that is, ( P ≅ 0 ) ; this is denoted as a pure bending condition. Behaviour of eccentrically loaded columns under large eccentricity is “steel-controlled”. In an eccentrically loaded column with large eccentricity, an increase in the axial load results in an increased flexural resistance. ___________________________________________________________________________ 8.2. a) The balanced condition occurs when the strain in steel at the tension face reaches the yield strain ( ε s = ε y ) simultaneously as the strain at the concrete compression face reaches the maximum value ( ε c = ε cmax ) . This condition represents the threshold between the small and large eccentricity load conditions. The load eccentricity corresponding to the balanced condition is called the balanced eccentricity ( e b ) . b) When a load eccentricity is larger than the balanced eccentricity, the column is considered to be subjected to large eccentricity and is expected to demonstrate concrete-controlled failure. On the contrary, when a load eccentricity is less than the balanced eccentricity, the column is con-

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8-1

sidered to be subjected to small eccentricity and is expected to demonstrate steel-controlled failure. __________________________________________________________________________ 8.3. a) The two main failure modes characteristic of eccentrically loaded reinforced concrete columns are: 1) Steel-controlled failure 2) Concrete-controlled failure. b) In eccentrically loaded reinforced concrete columns, the mode of failure depends on the load eccentricity, which is equal to the ratio of the factored bending moment and axial load. If the eccentricity exceeds the value corresponding to the balanced eccentricity, steel-controlled failure mode is expected to take place, otherwise the concrete-controlled mode of failure will take place. __________________________________________________________________________ 8.4. In case of reinforced concrete beams, the mode of failure depends on the reinforcement ratio relative to the balanced reinforcement ratio. For the beam reinforced with a moderate amount of reinforcement corresponding to a reinforcement ratio that is less than the balanced ratio, steelcontrolled mode of failure can be expected (otherwise the concrete-controlled mode of failure is expected to take place). In case of reinforced concrete columns, the mode of failure (steel-controlled versus concrete-controlled) depends on the load eccentricity, as discussed in Problem 8.3. __________________________________________________________________________ 8.5. a) Determine the factored moment resistance and the axial load resistance at the balanced condition.

α 1 ≅ 0.8

β 1 ≅ 0.9

f c ′ = 30MPa f y = 400MPa Column cross sectional dimensions:

b = 350mm

h = 350mm

Copyright © 2006 Pearson Education Canada Inc.

8-2

A g = b × h = 350 × 350 = 122500mm A st = 4 × 300mm = 1200mm

2

2

P rco = ( α 1 φ c f c ′ ) ( A g – A st )

[8.3]

2

(4-20M bars)

= 0.8 ( 0.65 ) ( 30 ) ( 122500 – 1200 ) = 1892kN P rso = φ s f y A st

[8.4] 2

= 0.85 ( 400MPa ) ( 1200mm ) = 408kN P ro = P rco + P rso

[8.2]

= 1892kN + 408kN = 2300kN b) Derive the column interaction diagram. Find the distance from the compression fibre to reinforcement layers 1 and 2:

20mm d 1 = 40mm + 10mm + ---------------- = 60mm 2 20mm d 2 = 350mm – 40mm – 10mm – ---------------- = 290mm 2 The reinforcement information for this column (4-20M bars in total) is summarized below. Reinforcement

A si

di mm

Layer #

Rebars

mm

1

2-20M

600

60

2

2-20M

600

290

2

Neutral axis depth for balanced condition ( c b ):

ε cmax c b = ------------------------ d 2 ε y + ε cmax

[8.18]

0.0035 = ------------------------------------ ( 290 ) = 185mm 0.002 + 0.0035

Copyright © 2006 Pearson Education Canada Inc.

8-3

Let us develop an interaction diagram defined by 6 points: point 1 corresponds to the concentrically loaded column, whereas points 2 to 6 correspond to varying degrees of eccentricity. In order to determine these points, c values need to be estimated (similar to Example 8.3 in the text). Point

c (mm)

1

∞

2

275

3

225

4

185

5

100

6

50

Strain distributions corresponding to points 2 to 6 are summarized below.

The following equations are used to develop this interaction diagram:

C r = α 1 β 1 φ c f c ′cb

[8.22]

= 0.8 ( 0.90 ) ( 0.65 ) ( 30 ) × c × 0.35 = 4.86c F rs1 = φ s f s1 A s1 = ( 0.85 )f s1 ( 600 )10

–3

= 0.51f s1

F rs2 = φ s f s2 A s2 = ( 0.85 )f s2 ( 600 )10

–3

= 0.51f s2

n

Pr = Cr +

F rsi

[8.23]

[8.24]

1

Copyright © 2006 Pearson Education Canada Inc.

8-4

n

M r = M rc +

M rsi

[8.25]

1 –3 h β1 c 350 × 10 0.9c M rc = C r --- – -------- = C r -------------------------- – ---------- = C r ( 0.175 – 0.45c ) 2 2 2 2 h 350 –3 M rs1 = F rs1 --- – d 1 = F rs1 --------- – 60 10 = ( 0.115 )F rs1 2 2

h 350 –3 M rs2 = F rs2 --- – d 2 = F rs2 --------- – 290 10 = – ( 0.115 ) F rs2 2 2 Calculations related to the development of the interaction diagram are presented in the table below. Point

1

2

3

4

5

6

c ( mm )

275

225

185

100

50

ε s1

> 0.002 (Y)

> 0.002 (Y)

> 0.002 (Y)

0.0014

-0.0007

ε s2

-0.002 (Y)

-0.0010

-0.002

> – 0.002 > – 0.002 (Y) (Y)

f s1 (MPa)

400

400

400

280

-140

f s2 (MPa)

-40

-200

-400

-400

-400

Cr

1336

1093

900

486

243

F rs1

204

204

204

143

-71

F rs2

-20

-102

-204

-204

-204

1520

1195

900

425

-32

M rc

69

81

83

63

37

M rs1

23

23

23

16

-8

M rs2

2

12

23

23

23

Pr

2300

P rmax

1840

Copyright © 2006 Pearson Education Canada Inc.

8-5

Mr

0

92

112

125

99

51

The interaction diagram for 350 mm by 350 mm column with 4-20M vertical rebars is shown below. Note that the points on the diagram are defined by P r and M r values highlighted in the above table.

_________________________________________________________________________

8.6. a) Determine the factored axial load resistance.

P ro = 2300kN (from Problem 8.5.). b) If M f = 40kNm , then the corresponding P r ≅ 2100kN (estimated from the column interaction diagram). c) If M f gradually increases to 80kNm , then P r ≅ 1750kN (from the column interaction diagram). d) When the column failure is initiated by concrete crushing (concrete-controlled failure), an increase in applied bending moment results in a decrease in the axial load capacity.

Copyright © 2006 Pearson Education Canada Inc.

8-6

__________________________________________________________________________ 8.7. a) Pure bending condition. In a reinforced concrete flexural member where the compression and tension steel areas are the same, and the reinforcement is symmetrical with regards to the column horizontal axis, the moment resistance can be estimated as 2

M r = T × z = ( 0.85 × f y × A s ) × z = ( 0.85 × 400 × 600mm ) × 230mm = 47kNm

b) Determine the factored moment resistance corresponding to P f = 200kN . Refer to the column interaction diagram in Problem 8.5:

P f = 200kN and the corresponding M r ≅ 80kNm c) Determine the factored moment resistance corresponding to P f = 400kN .

P f = 400kN and the corresponding M r ≅ 95kNm d) When the column failure is initiated by steel yielding (steel-controlled failure), an increase in the axial load magnitude results in an increase in the moment resistance. __________________________________________________________________________ 8.8. Design parameters:

α 1 ≅ 0.8

β 1 ≅ 0.9

Copyright © 2006 Pearson Education Canada Inc.

8-7

f c ′ = 30MPa f y = 400MPa b = 350mm

h = 350mm

Find the distance from the compression fibre to the reinforcement layers 1, 2, and 3:

25mm d 1 = 40mm + 10mm + ---------------- ≅ 60mm 2 350 d 2 = --------- = 175mm 2 25mm d 3 = 350mm – 40mm – 10mm – ---------------- ≅ 290mm 2 Neutral axis depth for balanced condition:

c b = 185mm (same as Problem 8.5.) The reinforcement information for this column (8-25M bars in total) is summarized below. Reinforcement

A si

di mm

Layer #

Rebars

mm

1

3-25M

1500

60

2

2-25M

1000

175

3

3-25M

1500

290

2

The following equations are used to develop this interaction diagram:

C r = α 1 β 1 φ c f c ′cb

[8.22]

= 0.8 ( 0.9 ) ( 0.65 ) ( 30 ) × c × 0.35 = 4.86c

Copyright © 2006 Pearson Education Canada Inc.

8-8

F rs1 = φ s f s1 A s1 = ( 0.85 )f s1 ( 1500 )10

–3

= 1.275f s1

F rs2 = φ s f s2 A s2 = ( 0.85 )f s2 ( 1000 )10

–3

= 0.85f s2

F rs3 = φ s f s3 A s3 = ( 0.85 )f s3 ( 1500 )10

–3

= 1.275f s3

n

Pr = Cr +

F rsi

[8.23]

[8.24]

1 n

M r = M rc +

M rsi

[8.25]

1

h β1 c M rc = C r --- – -------2 2 h M rs1 = F rs1 --- – d 1 2 h M rs2 = F rs2 --- – d 2 2 h M rs3 = F rs3 --- – d 3 2

–3

350 × 10 0.9c = C r -------------------------- – ---------- = C r ( 0.175 – 0.45c ) 2 2 350 –3 = F rs1 --------- – 60 10 = ( 0.115 )F rs1 2 350 –3 = F rs2 --------- – 175 10 = 0 2 350 –3 = F rs3 --------- – 290 10 = – ( 0.115 ) F rs3 2

Let us develop an interaction diagram defined by 6 points: point 1 corresponds to the concentrically loaded column, whereas points 2 to 6 correspond to varying degrees of eccentricity. In order to determine these points, c values need to be estimated (same values as in Problem 8.5 will be used herein). Calculations related to the development of the interaction diagram are presented in the table below. Point

1

2

3

4

5

6

c ( mm )

275

225

185

150

100

ε s1

0.0020 (Y)

0.0020 (Y)

0.0020 (Y)

0.0020 (Y)

0.0014

ε s2

0.0013

0.0008

0.0002

-0.0006

-0.0020 (Y)

ε s3

-0.0002

-0.0010

-0.0020 (Y)

-0.0020 (Y)

-0.0020 (Y)

f s1

400

400

400

400

280

f s2

255

156

38

-117

-400

f s3

-38

-202

-400

-400

-400

Copyright © 2006 Pearson Education Canada Inc.

8-9

Cr

1336

1093

899

729

486

F rs1

510

510

510

510

357

F rs2

216

132

32

-99

-340

F rs3

-49

-258

-507

-510

-510

2014

1478

935

630

-7

M rc

68

81

82

78

63

M rs1

59

59

59

59

41

M rs2

0

0

0

0

0

M rs3

6

30

58

59

59

133

169

199

196

163

P

r

P rmax

Mr

3209

2567

0

The interaction diagram for 350 mm by 350 mm column with 8-25M vertical rebars is shown below. Note that the points on the diagram are defined by P r and M r values highlighted in the above table.

Copyright © 2006 Pearson Education Canada Inc.

8-10

_________________________________________________________________________

8.9. This problem can be solved either by using the computer spreadsheet provided in the companion CD or own spreadsheet. Both methods will be illustrated below. Development of own spreadsheet/table to obtain the interaction diagram

Design parameters:

α 1 ≅ 0.8

β 1 ≅ 0.9

f c ′ = 30MPa f y = 400MPa b = 350mm

h = 350mm

Find the distance from the compression fibre to the reinforcement layers 1, 2, and 3:

20mm d 1 = 40mm + 10mm + ---------------- = 60mm 2 350 d 2 = --------- = 175mm 2 20mm d 3 = 350mm – 40mm – 10mm – ---------------- = 290mm 2 Neutral axis depth for balanced condition:

c b = 185mm (same as Problem 8.5.) The reinforcement information for this column (8-20M bars in total) is summarized below. Reinforcement

A si

di mm

Layer #

Rebars

mm

1

3-20M

900

60

2

2-20M

600

175

3

3-20M

900

290

2

Copyright © 2006 Pearson Education Canada Inc.

8-11

The following equations are used to develop this interaction diagram:

C r = α 1 β 1 φ c f c ′cb

[8.22]

= 0.8 ( 0.9 ) ( 0.65 ) ( 30 ) × c × 0.35 = 4.86c F rs1 = φ s f s1 A s1 = ( 0.85 )f s1 ( 900 )10

–3

= 0.765f s1

F rs2 = φ s f s2 A s2 = ( 0.85 )f s2 ( 600 )10

–3

= 0.51f s2

F rs3 = φ s f s3 A s3 = ( 0.85 )f s3 ( 900 )10

–3

= 0.765f s3

n

Pr = Cr +

F rsi

[8.23]

[8.24]

1 n

M r = M rc +

M rsi

[8.25]

1

h β1 c M rc = C r --- – -------2 2 h M rs1 = F rs1 --- – d 1 2 h M rs2 = F rs2 --- – d 2 2 h M rs3 = F rs3 --- – d 3 2

–3

350 × 10 0.9c = C r -------------------------- – ---------- = C r ( 0.175 – 0.45c ) 2 2 350 –3 = F rs1 --------- – 60 10 = ( 0.115 )F rs1 2 350 –3 = F rs2 --------- – 175 10 = 0 2 350 –3 = F rs3 --------- – 290 10 = – ( 0.115 ) F rs3 2

Copyright © 2006 Pearson Education Canada Inc.

8-12

Let us develop the interaction diagram defined by 6 points: point 1 corresponds to the concentrically loaded column, whereas points 2 to 6 correspond to varying degrees of eccentricity. In order to determine these points, c values need to be estimated (same values as in Problem 8.5 will be used herein). Calculations related to the development of the interaction diagram are presented in the table below. Point

1

2

3

4

5

6

c ( mm )

275

225

185

125

80

ε s1

0.0020 (Y)

0.0020 (Y)

0.0020 (Y)

0.0018

0.0009

ε s2

0.0013

0.0008

0.0002

-0.0014

-0.0020 (Y)

ε s3

-0.0002

-0.0010

-0.0020 (Y)

-0.0020 (Y)

-0.0020 (Y)

f s1

400

400

400

364

175

f s2

255

156

38

-280

-400

f s3

-38

-202

-400

-400

-400

Cr

1336

1093

899

607

389

F rs1

306

306

306

278

134

F rs2

130

79

19

-143

-204

F rs3

-29

-155

-304

-306

-306

1743

1324

920

437

13

M rc

68

81

82

72

54

M rs1

35

35

35

32

15

M rs2

0

0

0

0

0

M rs3

3

18

35

35

35

107

134

153

139

105

Pr

2690

P rmax

2152

Mr

0

Copyright © 2006 Pearson Education Canada Inc.

8-13

Computer-generated interaction diagram

Copyright © 2006 Pearson Education Canada Inc.

8-14

__________________________________________________________________________ 8.10. Design parameters:

α 1 ≅ 0.8

β 1 ≅ 0.9

f c ′ = 30MPa f y = 400MPa b = 350mm

h = 350mm

Find the distance from the compression fibre to the reinforcement layers 1, 2, 3, and 4:

20mm d 1 = 40mm + 10mm + ---------------- = 60mm 2 350 d 2 = --------- = 175mm 2 20mm d 3 = 350mm – 40mm – 10mm – ---------------- = 290mm 2 20mm d 4 = 350mm – 40mm – 10mm – ---------------- = 290mm 2 Neutral axis depth for balanced condition:

c b = 185mm (same as Problem 8.5.) The reinforcement information for this column (12-20M bars in total) is summarized below. Reinforcement

A si

di mm

Layer #

Rebars

mm

1

4-20M

1200

60

2

2-20M

600

135

3

2-20M

600

215

4

4-20M

1200

290

2

Copyright © 2006 Pearson Education Canada Inc.

8-15

The following equations are used to develop this interaction diagram:

C r = α 1 β 1 φ c f c ′cb

[8.22]

= 0.8 ( 0.9 ) ( 0.65 ) ( 30 ) × c × 0.35 = 4.86c F rs1 = φ s f s1 A s1 = ( 0.85 )f s1 ( 1200 )10

–3

= 1.02f s1

F rs2 = φ s f s2 A s2 = ( 0.85 )f s2 ( 600 )10

–3

= 0.51f s2

F rs3 = φ s f s3 A s3 = ( 0.85 )f s3 ( 600 )10

–3

= 0.51f s3

F rs4 = φ s f s4 A s4 = ( 0.85 )f s4 ( 1200 )10 n

Pr = Cr +

F rsi

–3

[8.23]

= 1.02f s4

[8.24]

1 n

M r = M rc +

M rsi

[8.25]

1

h β1 c M rc = C r --- – -------2 2 h M rs1 = F rs1 --- – d 1 2 h M rs2 = F rs2 --- – d 2 2 h M rs3 = F rs3 --- – d 3 2 h M rs4 = F rs4 --- – d 4 2

–3

350 × 10 0.9c = C r -------------------------- – ---------- = C r ( 0.175 – 0.45c ) 2 2 350 –3 = F rs1 --------- – 60 10 = 0.115F rs1 2 350 –3 = F rs2 --------- – 135 10 = 0.04F rs2 2 350 –3 = F rs3 --------- – 215 10 = – ( 0.04 ) F rs3 2 350 –3 = F rs4 --------- – 290 10 = – ( 0.115 ) F rs4 2

Let us develop the interaction diagram defined by 6 points: point 1 corresponds to the concentrically loaded column, whereas points 2 to 6 correspond to varying degrees of eccentricity. In order to determine these points, c values need to be estimated (same values as in Problem 8.5 will be

Copyright © 2006 Pearson Education Canada Inc.

8-16

used herein). Calculations related to the development of the interaction diagram are presented in the table below. Point

1

2

3

4

5

6

c ( mm )

275

225

185

125

100

ε s1

0.0020 (Y)

0.0020 (Y)

0.0020 (Y)

0.0018 (Y)

0.0014 (Y)

ε s2

0.0018

0.0014

0.0009

-0.0003

-0.0013

ε s3

0.0008

0.0002

-0.0005

-0.0020 (Y)

-0.0020 (Y)

ε s4

-0.0002

-0.0010

-0.0020 (Y)

-0.0020 (Y)

-0.0020 (Y)

f s1

400

400

400

364

280

f s2

351

274

182

-67

-259

f s3

158

37

-106

-400

-400

f s4

-38

-202

-400

-400

-400

Cr

1336

1093

899

607

486

F rs1

408

408

408

371

286

F rs2

179

140

93

-34

-132

F rs3

80

19

-54

-204

-204

F rs4

-39

-206

-405

-408

-408

1965

1454

940

332

27

M cr

68

81

82

72

63

M rs1

47

47

47

43

33

M rs2

7

5

4

-1

-5

M rs3

-3

-1

2

8

8

Pr

3079

P rmax

2463

Copyright © 2006 Pearson Education Canada Inc.

8-17

M rs4 Mr

0

4

24

47

47

47

124

156

182

168

146

The interaction diagram for 350 mm by 350 mm column with 12-20M vertical rebars is shown below. Note that the points on the diagram are defined by P r and M r values highlighted in the above table.

_________________________________________________________________________

8.11. Interaction diagrams for columns discussed in Problems 8.5 to 8.10 are shown below.

Copyright © 2006 Pearson Education Canada Inc.

8-18

a) 4- 20M b) 4-20M c) 4-20M d) 12-20M e) 12-20M f) 4-20M g) 8-25M It is recommended that columns be designed with some reserve in loadbearing capacity wherever practical, hence designs under c) and f) may wish to consider the use of 8-20M vertical rebars.

Copyright © 2006 Pearson Education Canada Inc.

8-19

Copyright © 2006 Pearson Education Canada Inc.

8-20

Chapter 9 - Solutions __________________________________________________________________________ 9.1. a) 15M top bars, regular uncoated reinforcement The development length will be determined according to A23.3 Cl.12.2.3. As the beam is reinforced with the minimum shear reinforcement, it can be classified as Case 1.

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc′ where (see Table 9.1)

[9.2]

k 1 = 1.3 top bar k 2 = 1.0 regular uncoated bar k 3 = 1.0 normal-density concrete k 4 = 0.8 15M bars (less than 25M) d b = 16mm 15M bar (Table A.1) So,

400MPa l d = 0.45 × 1.3 × 1.0 × 1.0 × 0.8 ----------------------- 16mm = 547mm 30MPa l d = 547mm∴ Table A.5: l d = 550mm ( f c ′ = 30MPa , top bar) b) 20M top bars, epoxy-coated reinforcement

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b [9.2] fc ′ where (see Table 9.1) k 1 = 1.3 top bar k 2 = 1.2 (cover not specified) k 3 = 1.0 normal-density concrete

Copyright © 2006 Pearson Education Canada Inc.

9-1

k 4 = 0.8 20M bar (less than 25M) d b = 19.5 ≅ 20mm 20M bars (Table A.1) 400MPa l d = 0.45 × 1.3 × 1.2 × 1.0 × 0.8 ----------------------- 20mm = 820mm 30MPa l d = 820mm∴ Table A.5: l d = 675 × 1.2 = 810mm ( f c ′ = 30MPa , top bar) Note that the l d value from Table A.5 needs to be multiplied by 1.2 to account for epoxy-coated bar ( k 2 = 1.2 ) . c) 30M bottom bars, regular uncoated reinforcement

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b [9.2] fc ′ where (see Table 9.1) k 1 = 1.0 bottom bar k 2 = 1.0 regular uncoated reinforcement k 3 = 1.0 normal-density concrete k 4 = 1.0 30M bar d b = 29.9mm ≅ 30mm (see Table A.1) So,

400MPa l d = 0.45 × 1 × 1 × 1 × 1 ----------------------- 30mm = 986mm 30MPa l d = 986mm ≅ 1000mm∴ The same result can be obtained from Table A.5, that is, l d = 1000mm ( f c ′ = 30MPa , bottom bar).

Copyright © 2006 Pearson Education Canada Inc.

9-2

__________________________________________________________________________ 9.2. a) Determine the point of maximum stress in the reinforcement.

M max

2

wl = -------- (at point A) 2

Hence, the maximum stress develops at point A. b) Determine the minimum required development length for straight bars per CSA A23.3. According to CSA A23.3 Cl.12.2.3, this is a Case 1 application, since clear spacing between bars being developed is not less than 2 d b ; in this case, d b = 15mm (15M bars), and the clear spacing between the bars is

db 300 – 2 × ----- = 285mm > 2d b = 2 × 15 = 30mm 2 Therefore, development length can be determined from Eqn (9.2) as

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.2]

k 1 = 1.3 top bar k 2 = 1.0 regular uncoated bars k 3 = 1.0 normal-density concrete k 4 = 0.8 15M bars Copyright © 2006 Pearson Education Canada Inc.

9-3

d b = 15mm 15M bars So,

400MPa l d = 0.45 × 1.3 × 1.0 × 1.0 × 0.8 ----------------------- 15mm = 562mm 25MPa Therefore, the required development length for straight 15M rebars is l d = 562mm ≅ 560mm

c) Hooked anchorage design according to A23.3 Cl.12.5.1 and 12.5.2:

MF = 1.0 (see Table 9.2) db l dh = 100 ---------- × MF fc′ 15mm = 100 ----------------------- × 1.0 = 300mm 25MPa

Copyright © 2006 Pearson Education Canada Inc.

9-4

__________________________________________________________________________ 9.3. a) Determine the largest beam rebar size that can be used without a hook. In this case, the available development length is equal to 560 mm. Since the minimum stirrups are provided in the anchorage zone, this can be considered a Case 1 application according to A23.3 Cl.12.2.3. Development length can be determined from the following equation

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.2]

k 1 = 1.3 top bar k 2 = 1.0 regular uncoated bars k 3 = 1.0 normal-density concrete k 4 = 0.8 assume bar size less than 25M 400MPa l d = 0.45 × 1.3 × 1.0 × 1.0 × 0.8 × ----------------------- × d b = 31.6d b 35MPa

However,

l d ≤ 560mm consequently,

Copyright © 2006 Pearson Education Canada Inc.

9-5

31.6d b ≤ 560mm and

d b ≤ 17.6mm In conclusion, the maximum bar size that could be used in this design (provided that only straight bars are to be used) is 15M ( d b ≅ 16mm ). b) Is it feasible to use straight bar anchorage in this design? In general, 15M bars are considered to be too small for use as flexural reinforcement in beams and so this solution would not be considered feasible; larger bar sizes (at least 20M) would need to be used, and so hooked anchorage would be required. __________________________________________________________________________ 9.4.

The available bar length (AB) is equal to

1200 – 200 – 2 × 75 a = ------------------------------------------------ = 425mm 2 Determine the full development length for 20M bars according to CSA A23.3 Cl.12.2.3. This is a Case 1 application since the reinforcement spacing is greater than 2 d b , where d b = 20mm . Clear bar spacing is equal to

db 400 – 2 × ----- = 380mm > 2d b = 40mm 2 Therefore, development length can be determined from Eqn (9.2) as

Copyright © 2006 Pearson Education Canada Inc.

9-6

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.2]

k 1 = 1.0 bottom bar k 2 = 1.0 regular uncoated bar k 3 = 1.0 normal-density concrete k 4 = 0.8 20M bars d b = 20mm 20M bar (see Table A.1) 400MPa l d = 0.45 × 1.0 × 1.0 × 1.0 × 0.8 ----------------------- 20mm = 576mm 25MPa The required development length is l d = 576mm however, the available length is

a = 425mm < 576mm This indicates that the rebars will not be able to develop their full strength. As a result, the stress developed in the bars can be calculated as

a f sactual = f y × ---ld

425mm = 400MPa × ------------------- = 295MPa 576mm

where f y = 400MPa Therefore, the bars are able to develop around 75% of their full strength, since

f sactual 295MPa ----------------- = --------------------- = 0.74 = 74% 400MPa fy

Copyright © 2006 Pearson Education Canada Inc.

9-7

_________________________________________________________________________ 9.5. a) Develop the factored shear force and bending moment diagrams for the beam. Based on the bending moment diagram, it follows that

M max = 1600kNm (span AB) M min = – 900kNm (support B)

b) Design the flexural reinforcement. i) Positive bending (span AB)

M f = M max = 1600kNm Estimate d (assume 2 layers of reinforcement):

d = h – 110 = 600 – 110 = 490mm (see Section 5.4.1)

Copyright © 2006 Pearson Education Canada Inc.

9-8

Set M r = M f = 1600kNm .

Use direct procedure to find A s . 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × ( 1600 × 10 Nmm ) 2 = 0.0015 ( 30MPa ) ( 1200mm ) 490 – ( 490 ) – --------------------------------------------------------------- = 12277mm ( 30MPa ) ( 1200mm ) 2

Use 20-30M bars in 2 layers. Since 2

A s = 20 × 700 = 14000mm > 12277mm

2

okay

Determine the actual effective depth ( d ).

cover = 30mm (not exposed)

d b = 30mm (30M bars) d s = 10mm (10M stirrups)

s min ≥

1.4d b = 1.4 × 30 = 42mm 1.4a max = 1.4 × 20 = 28mm 30mm

Copyright © 2006 Pearson Education Canada Inc.

9-9

s min = 42mm (governs) So,

s min 42 d = h – cover – d s – d b – --------- = 600 – 30 – 10 – 30 – ------ = 509mm ≅ 510mm 2 2 Check the minimum reinforcement requirement (A23.3 Cl.10.5.1.2).

0.2 f c ′ 0.2 30MPa 2 A smin = ------------------ b t h = ------------------------------- × 1200mm × 600mm = 1972mm fy 400MPa

[5.7]

Note that b t = b = 1200mm Since 2

2

A smin = 1972mm < A s = 14000mm okay Check the maximum reinforcement requirement (A23.3 Cl.10.5.2). Find the actual reinforcement ratio. 2 As 14000mm ------------------------------ = 0.023 [3.1] ρ = = bd 1200 × 510

Find the balanced reinforcement ratio.

f c ′ = 30MPa

ρ b = 0.027 (Table A.4)

Since

ρ = 0.023 < ρ b = 0.027 okay Find the actual M r value.

2 φs fy As 0.85 × 400MPa × 14000mm -----------------------------------------------------------------------------------------------= = = 254mm a 0.8 × 0.65 × 30MPa × 1200mm α 1 φ c f c ′b

[3.12]

2 a 254 M r = φ s A s f y d – --- = 0.85 × ( 14000mm ) ( 400MPa ) 510 – --------- = 1823kNm [3.14] 2 2

Since

M r = 1823kNm > M f = 1600kNm okay ii) Negative bending: support B Set M r = M f = M min = – 900kNm

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9-10

Estimate effective depth d :

d = h – 70 = 600 – 70 = 530mm (assume 1 layer of reinforcement, see Section 5.4.1)

Use direct procedure to find A s . 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2 3.85 × ( 900 × 10 Nmm ) 2 = 0.0015 ( 30MPa ) ( 1200mm ) 530 – ( 530 ) – ------------------------------------------------------------ = 5416mm 30MPa × ( 1200mm )

Use 8-30M bars in 1 layer. Since 2

A s = 8 × 700 = 5600mm > 5416mm

2

okay

Find the actual effective depth ( d ).

db 30 d = h – cover – d s – ----- = 600 – 30 – 10 – ------ = 545mm 2 2

Check the CSA A23.3 minimum reinforcement requirement (Cl.10.5.1.2).

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9-11

Since 2

2

A s = 5600mm > A smin = 1972mm okay Check the CSA A23.3 maximum reinforcement requirement (Cl.10.5.2).

As 5600 ρ = ------ = --------------------------- = 0.009 bd 1200 × 545 Since

ρ = 0.009 < ρ b = 0.027 okay - Find the actual M r value.

φs fy As 0.85 × 400 × 5600 a = --------------------- = -------------------------------------------------------------------- = 102mm [3.12] α 1 φ c f c ′b 0.8 × 0.65 × 30MPa × 1200 2 a 102 M r = φ s A s f y d – --- = 0.85 ( 5600mm ) ( 400MPa ) 545 – --------- = 941kNm 2 2

[3.14]

Since

M r = 941kNm > M f = 900kNm okay c) Design 50% bar cutoffs for the top reinforcement. 50% bar cutoff means that 4 (out of 8) bars will be discontinued where no longer required. Set M fmax = M r8 = 900kNm So

4 M r4 = --- × 900 = 450kNm 8 Determine the locations along the negative moment region where

M f = M r4 = 450kNm Find x 1 . Since

2

( 3 – x1 ) M x = w f --------------------- = 450kNm 2 x 1 can be obtained as

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9-12

x 1 = 0.88m ≅ 0.9m Find x 2 .

2

( 3 + x2 ) M x = w f ---------------------- – 1600x 2 = 450kNm 2 200 --------- ( 3 + x 2 ) 2 – 1600x 2 – 450 = 0 2

or 2

100x 2 – 1000x 2 + 450 = 0 x 2 can be obtained as x 2 = 0.47m

Determine the bar development length (A23.3 Cl.12.2.3). The beam is reinforced with the minimum shear reinforcement, so it can be considered as a Case 1 application. Development length can be obtained from the following equation

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9-13

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.2]

k 1 = 1.3 top bars k 2 = 1.0 regular uncoated bars k 3 = 1.0 normal-density concrete k 4 = 1.0 30M bars d b = 30mm 30M bars 400MPa l d = 0.45 × 1.3 × 1.0 × 1.0 × 1.0 × ----------------------- × 30mm = 1282mm 30MPa l d ≅ 1300mm Determine the bar cutoffs according to CSA A23.3 requirements. This design problem will be solved by following the rules summarized in Checklist 9.1. First, consider the reinforcement on the right-hand side of support B. Terminating bars A (4-30M)

i) Rule 1: actual cutoff point (Cl.11.3.9.1) The bars need to be extended beyond point 1 (theoretical cutoff point) by • 1.3d = 1.3 × 545 = 708mm ← governs or • h = 600mm Let us denote point 4 as the actual cutoff point for bars A.

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9-14

ii) Rule 2: general anchorage requirement (A23.3 Cl.12.1.1) The total distance from the point of maximum stress (point 3 on the sketch) to the actual cutoff point for bars A is

l 3 – 4 = l 3 – 1 + l 1 – 4 = 900 + 710 = 1610mm On the other hand, the development length ( l d ) for these bars is equal to

l d = 1300mm < 1610mm okay Continuing bars B (4-30M)

i) Rule 3: Continuing reinforcement (A23.3 Cl.12.10.4) The continuing reinforcement needs to be extended beyond point 1 (theoretical cutoff point) by the greater of • l d + d = 1300 + 545 = 1845mm ← governs or • l d + 12d b = 1300 + 12 × 30 = 1660mm However, the available bar length from point 1 to the end of the beam (considering a 40 mm side cover) is

l 3 – 5 = 3000mm – 900mm – 40mm = 2060mm Since

l d + d = 1845mm < 2060mm it means that the continuing bars can have straight anchorage. Next, let us consider top reinforcement on the left-hand side of support B. Terminating bars A (4-30M)

i) Rule 1: actual bar cutoff point (A23.3 Cl.11.3.9.1). The bars need to be extended beyond point 2 (theoretical cutoff point) by

1.3d = 708mm

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9-15

Let us denote point 6 as actual cutoff point on the sketch below. .

ii) Rule 2: general anchorage requirement (A23.3 Cl.12.1.1). The total distance from the point of maximum stress (point 3 on the sketch) to the actual cutoff point is

l 3 – 6 = l 3 – 2 + l 2 – 6 = 470 + 710 = 1180mm On the other hand, the development length ( l d ) for these bars is equal to

l d = 1300mm > 1180mm To ensure the full development of continuing bars, point 6 needs to be moved to a distance of 1300 mm from point 3. Continuing bars B (4-30M)

i) Rule 3: Continuing reinforcement (A23.3 Cl.12.10.4). The continuing reinforcement needs to be extended beyond point 2 (theoretical cutoff point) by the distance of

l d + d = 1845mm In this case, the inflection point (IP) (point of zero moment) is located at a distance of 1 m from the support B. Another rule needs to be checked related to bar extension beyond the IP (see Section 9.7.7). ii) Bar extension beyond the IP (A23.3 Cl.12.12.2)

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9-16

The bar embedment length beyond IP should be larger than • •

d = 545mm ← governs 12d b = 12 × 30 = 360mm

ln 9000 – 400 ----- = --------------------------- = 537mm 16 16 Therefore, the continuing bars need to be extended beyond IP by the length of 545 mm. •

The length of continuing bars B to the left-hand side of point 3 is equal to the larger of • 470 + 1845 = 2315mm or • 1000 + 545 = 1545mm Therefore, the length is equal to 2315 mm.

The sketch showing bar arrangement and lengths for top reinforcement is presented below.

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9-17

d) Confirm whether it is necessary to provide a hooked anchorage in this case. In this case, it is not required to provide hooked anchorage for the top reinforcement continuing up to point C. Based on the solution presented in part c), straight bar anchorage is adequate (Rule 3: continuing reinforcement, A23.3 Cl.12.10.4). e) Design 50% bar cutoffs for the bottom reinforcement When 50% bar cutoff is assumed, it means that 10 (out of 20) bars will be discontinued where no longer required. Set M fmax = M r20 = 1600kNm so

10 M r10 = ------ × 1600 = 800kNm 20 Determine the location along the positive moment region where

M f = M r10 = 800kNm

Find distances x 1 and x 2 . 2

x M x = R A × x – w f × ----- = 800kNm 2 2 x 800x – 200 ----- – 800 = 0 2 2

x – 8x + 8 = 0 The solutions of this quadratic equation are

x 1 = 1.17m ≅ 1120mm

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9-18

x 2 = 6.83m ≅ 6800mm The objective of this design is to design bar cutoffs at point 2.

Determine the bar development length (A23.3 Cl.12.2.3) - Case 1 application

fy l d = 0.45 × k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.2]

k 1 = 1.0 bottom bars k 2 = 1.0 regular uncoated bars k 3 = 1.0 normal-density concrete k 4 = 1.0 30M bars d b = 30mm 30M bars So,

400MPa l d = 0.45 × 1 × 1 × 1 × 1 ----------------------- 30mm = 986mm 30MPa l d ≅ 1000mm Determine bar cutoffs according to CSA A23.3 requirements. This problem will be solved by following the rules summarized in Checklist 9.1. Terminating bars A (10-30M)

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9-19

i) Rule 1: actual cutoff point (A23.3 Cl.11.3.9.1) The bars need to be extended beyond point 2 (theoretical cutoff point) by • 1.3d = 1.3 × 510 = 663mm ← governs or • h = 600mm Let us denote point 4 as actual cutoff point for bars A.

ii) Rule 2: general anchorage requirement (A23.3 Cl.12.1.1). The total distance from the point of maximum stress (point 3 on the sketch) to the actual cutoff point for bars A is

l 3 – 4 = l 3 – 2 + l 2 – 4 = 2800 + 660 = 3460mm On the other hand, the development length ( l d ) for these bars is equal to

l d = 1000mm < 3460mm okay Continuing bars B (10-30M)

i) Rule 3: continuing reinforcement (A23.3 Cl.12.10.4) The continuing reinforcement needs to be extended beyond point 2 (theoretical cutoff point) by the greater of • l d + d = 1000 + 510 = 1510mm ← governs or •

l d + 12d b = 1000 + 12 × 30 = 1360mm

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9-20

ii) Rule 4: supports (A23.3 Cl.12.11.1) - support B At least one-third of the total tension reinforcement should extend into support by at least 150 mm. In this case, 10 (out of 20) bars are extended into the support. The available bar length, from point 2 up to point 5 (extending into support by 150 mm) is equal to

400 l 2 – 5 = 2200 – --------- + 150 = 2150mm 2 On the other hand, the required bar length beyond point 2 is 1510 mm. Since 2150mm > 1510mm , it follows that 10-30M bars can be extended into support by 150 mm. iii) Rule 5: zero moment locations (A23.3 Cl.12.11.3) The factored moment resistance based on the bars continuing into the support B is

M f = M r10 = 800kNm The factored shear force at the IP (see the shear force diagram) is

V f = 800kN In this case, the provision related to IPs of continuous beams will be used, that is (see Figure 9.18),

Mr l d ≤ ------- + l a (A23.3. Eq.12.6) Vf

[9.8]

where

la =

d = 510mm ← governs 12d b = 12 × 30 = 360mm

Mr ------- + l a = 800kNm ---------------------- + 0.51m = 1.51m = 1510mm 800kN Vf Since

l d = 1000mm < 1510mm okay

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9-21

The sketch showing bar arrangement and lengths - bottom reinforcement is shown below.

f) Check whether bottom reinforcement is required for span BC according to the CSA A23.3 flexural requirements. According to the flexural design requirements, the bottom reinforcement is not required in the span BC of the beam. Bending moment is negative along the span BC and hence only the top reinforcement is required. CSA A23.3 does not formally require reinforcement in this region, hence the bottom reinforcement need not be extended beyond the support B.

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9-22

_________________________________________________________________________ 9.6. a) Develop the factored bending moment diagram.

b) Design the bar cutoffs. Set M r6 = M fmax = 800kNm 50 % bar cutoff means that 3 (out of 6) bars will be discontinued where no longer required, hence

3 M r3 = --- × 800 = 400kNm 6 - Determine the locations of theoretical bar cutoff points. Find the location along the beam length where the factored bending moment is equal to

M f = M r3 = 400kNm The following 2 cases need to be considered in this design: • Case 1: 0 < x < 2m • Case 2: 2m < x < 10m Case 1:

M x = R A × x 1 = ( 400kN )x 1 where

R A = 400kN When

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9-23

M x = 400kNm then

x 1 = 1m

Case 2:

M x = R A × x 2 – ( 500kN ) ( x 2 – 2 ) M x = 400x 2 – 500x 2 + 1000 M x = – 100x 2 + 1000 When

M x = 400kNm then

x 2 = 6m

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9-24

Determine the bar development length (A23.3 Cl.12.2.3). The beam is reinforced with the minimum shear reinforcement (Case 1 application).

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b [9.2] fc ′ where (see Table 9.1) k 1 = 1.0 bottom bars k 2 = 1.0 regular uncoated bars k 3 = 1.0 normal-density concrete k 4 = 1.0 25M bars d b = 25mm 25M bar 400MPa l d = 0.45 × 1.0 × 1.0 × 1.0 × 1.0 ----------------------- ( 25mm ) = 900mm 25MPa (Note that l d = 925mm from Table A.5) Determine the bar cutoffs according to CSA A23.3 requirements. This design problem will be solved by following the rules summarized in Checklist 9.1. First, consider the reinforcement on the right-hand side of point load. Terminating bars A (3-25M)

i) Rule 1: actual cutoff point (A23.3 Cl.11.3.9.1) The bars need to be extended beyond point 2 (theoretical cutoff point) by the length greater of

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9-25

• 1.3d = 1.3 × 530 = 689mm ≅ 700mm ← governs or • h = 600mm Let us denote point 3 as actual cutoff point for bars A.

ii) Rule 2: general anchorage requirement (A23.3 Cl.12.1.1) The total distance from the point of maximum stress (point 4 on the sketch) to the actual cutoff point (3) for bars A is

l 4 – 3 = l 4 – 2 + l 2 – 3 = 4000mm + 700mm = 4700mm On the other hand, development length ( l d ) for these bars is equal to

l d = 900mm Since

l d = 900mm < 4700mm , it follows that the Rule 2 has been satisfied. Continuing bars B (3-25M)

i) Rule 3: continuing reinforcement (A23.3 Cl.12.10.4) The total available length for the continuing bars B, from point 2 to the exterior beam face (at support B), is equal to

400 4000mm + --------- = 4200mm (where 400 mm is the support width) 2 When a 40 mm side cover is taken into account, then the available length is

4200 – 40 = 4160mm

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9-26

According to A23.3 Cl.12.10.4, the continuing bars need to be extended beyond point 2 (theoretical cutoff point) by the length greater of • l d + d = 900 + 530 = 1430mm ← governs or • l d + 12d b = 900 + 12 × 25 = 1200mm Since

1430mm < 4160mm okay The application of Rule 3 is illustrated below.

ii) Rule 4: supports (A23.3 Cl.12.11.1) - support B At least one-third of the total tension reinforcement should extend into support by at least 150 mm. In this case, 3 (out of 6) bars are extended into the support. The support width = 400mm side cover = 40mm Therefore, the total extension length is equal to

400mm – 40mm = 360mm > 150mm okay iii) Rule 5: zero moment locations (A23.3 Cl.12.11.3) The factored moment resistance based on the bars continuing into the support B is

M f = M r3 = 400kNm The factored shear force at the support B is

V f = R B = 100kN Embedment length beyond the support centreline is equal to

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9-27

400 l a = --------- – 40 = 160mm = 0.16m 2 check whether

Mr l d ≤ 1.3 ------- + l a [9.7] Vf 400kNm l d ≤ 1.3 ---------------------- + 0.16m = 5.36m = 5360mm 100kN Since

l d = 900mm < 5360mm okay Now, let us consider the reinforcement at the left hand side of the point load. Terminating bars A (3-25M)

i) Rule 1: actual cutoff point (A23.3 Cl.11.3.9.1) The bars need to be extended beyond the point 1 (theoretical cutoff point) by the distance greater of • 1.3d = 700mm ← governs or • h = 600mm Let us denote point 5 as actual cutoff point for bars A. It is obvious from the sketch that the actual cutoff point for bars A is only 300 mm away from the centreline of support A. Therefore, it is not practical to terminate bars A before the support.

If all 6 bars are extended into support A, the available length from the point load to the bar end is equal to

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9-28

400mm l 4 – 6 = 2000mm + ------------------- – 40mm = 2160mm 2 where a 40 mm side cover is taken into account. ii) Check general anchorage requirement (A23.3 Cl.12.1.1) Since

l d = 900mm < l 4 – 6 = 2160mm okay iii) Rule 4: supports (A23.3 Cl.12.11.1) - support A This rule had already been checked with regard to support B. In this case, all 6 bars are continuing into supports. iv) Rule 5: zero moment locations (A23.3 Cl.12.11.3)

M f = M r6 = 800kNm (all 6 bars are extended into the support) V f = R A = 400kN (factored shear force at the support) l a = 160mm = 0.16m Mr l d ≤ 1.3 ------- + l a Vf

[9.7]

800kNm l d ≤ 1.3 ---------------------- + 0.16m = 2.76m = 2760mm 400kN Since

l d = 900mm < 2760mm okay c) A sketch showing the beam elevation and the bar arrangement and lengths is presented below,

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9-29

_________________________________________________________________________ 9.7. a) Determine the cutoff locations for 50 % of the bottom reinforcement. In this design, beam depth along the span is variable. As a result, moment resistance ( M r ) decreases with a decrease in d value. Section 1-1: d 1 – 1 ≅ 800mm

1 Section 2-2: d 2 – 2 ≅ 400mm = --- d 1 – 1 2

Note that

a M r = T r d – --2 When T r is constant (same reinforcement area throughout the span), it follows that the M r value is proportional to d value. Consequently, 2–2

Mr

1 1–1 ≅ --- M r 2 1–1

where M r tively.

2–2

and M r

are factored moment resistance values at sections 1-1 and 2-2 respec-

If 10-20M bars are required at the midspan (Section 1-1), then 5-20M bars (50% reinforcement) is required at Section 2-2. The location ( x ) of Section 2-2 can be determined from the parabolic function 2

f ( x ) = ax + bx + c The following constraints apply:

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9-30

f( x) = 0

x = 0

f ( x ) = 0.8m f( x) = 0

c = 0

(1)

x = 5m

(2)

x = 10m

(3)

Equations (2) and (3) can be reformulated as 2

a ( 5 ) + b ( 5 ) = 0.8

b + 5a = 0.16

(2)

and 2

a ( 10 ) + b ( 10 ) = 0

b = – 10a

(3)

When b value is substituted from equation (3) into equation (2), coefficient a can be determined as

– 10a + 5a = 0.16 or

a = – 0.032 and

b = – 10a = 0.32 Therefore, the function can be expressed as 2

f ( x ) = – 0.032x + 0.32x When f ( x ) = 0.4 , x value can be determined as the solution of the quadratic equation 2

– 0.032x + 0.32x = 0.4 or

x = 1.46m ≅ 1.5m Therefore, theoretical bar cutoff point for 50% bars is at the distance of 1.5 m from the support centreline. Note that CSA A23.3 requirements summarized in Checklist 9.1 would need to be checked at the detailed design stage. b) Check whether a hooked anchorage is required.

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9-31

According to the CSA A23.3 requirement related to the reinforcement extended into the supports (Cl.12.11.1) - Rule 4, Checklist 9.1. In this case, 100% reinforcement (10-20M bars) extends into the support; this is more than onethird of the total reinforcement, which is okay. The bars should be extended into supports by the length of at least 150 mm. The support width is equal to 600 mm. The total extended length is close to 600 mm (side cover should be taken into account), which is more than 150 mm - this is okay. Therefore, it is not required to provide a hooked anchorage for the bars extending into the supports. __________________________________________________________________________ 9.8. a) 20M bars, tension lap splice - 50% reinforcement to be spliced

A srequired --------------------- = 0.6 A sprovided According to A23.3 Cl.12.15.2, Class B splice should be used (see Figure 9.24). Class B splice: l p = 1.3l d Find development length ( l d ) for 20M bars. According to A23.3 Cl.12.2.3 (assume Case 2 application):

fy l d = 0.6k 1 k 2 k 3 k 4 ---------- d b fc ′ where (see Table 9.1)

[9.3]

k 1 = 1.3 top bar k 2 = 1 regular uncoated k 3 = 1 normal-density concrete k 4 = 0.8 20M bars d b = 20mm 20M bars So,

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9-32

400MPa l d = 0.6 × 1.3 × 1.0 × 1.0 × 0.8 ----------------------- 20mm = 998mm 25MPa l p = 1.3l d = 1.3 × 998 = 1297mm ≅ 1300mm Therefore, lap length is equal to 1300 mm. b) Compression lap splice, 20M bars - 100 % reinforcement to be spliced According to CSA A23.3 Cl.12.16.1, the minimum length of compression lap splice should not be less than • 0.073f y d b • 300 mm where

d b = 20mm

( f y ≤ 400MPa )

(20M bar)

f y = 400MPa lp ≥

0.073f y d b = 0.073 × 400MPa × 20mm = 584mm 300mm

In this case, l p = 584mm governs (larger value), so

l p ≅ 600mm∴ __________________________________________________________________________ 9.9. a) Find the lap length ( l p ) for 4-30M dowels according to CSA A23.3 requirements. The column is under compression. According to CSA A23.3 Cl.12.16.1, the minimum length of compression lap splice should not be less than • 0.073f y d b • 300 mm where

d b = 30mm

( f y ≤ 400MPa )

(30M dowels)

f y = 400MPa lp ≥

0.073f y d b = 0.073 × 400MPa × 30mm = 876mm 300mm Copyright © 2006 Pearson Education Canada Inc.

9-33

In this case, l p = 876mm governs (larger value), so

l p ≅ 900mm∴ b) Consider hooked dowels. The available bar length is equal to

a = 650 – 75 = 575mm Find the development length for hooked bars in compression ( l db ) according to A23.3 Cl.12.3.2, as follows:

fy l db = 0.24 ---------- d b ≤ 0.044f y d b [9.4] fc ′ Since 400MPa l db = 0.24 ----------------------- × 30mm = 576mm 25MPa and 0.044 × f y d b = 0.044 × 400MPa × 30mm = 528mm so

l db = 576mm > 528mm not okay However, in this case the provided reinforcement is in excess of the required amount, since

A srequired = 2500mm

2

A sprovided = 4 × 700 = 2800mm

2

(4-30M bars, see Table A.1)

2 A srequired 2500mm l dbrequired = l db ------------------------ = 576 mm × -----------------------2- = 515mm A sprovided 2800mm

Since

l dbrequired = 515mm < 528mm okay Therefore, use

l db = l dbrequired = 515 mm Since

l db = 515mm < a = 575mm Copyright © 2006 Pearson Education Canada Inc.

9-34

It means that the available hook length is sufficient.

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9-35

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9-36

Chapter 10 - Solutions ___________________________________________________________________________ 10.1. a) Maximum bending moment for a beam subjected to a triangular-shaped distributed load: 2

2

M max = 0.0642w × l = 0.0642 ( 20kN ⁄ m ) ( 10m ) = 128kNm The distance defining the location of the maximum bending moment relative to the left support is

x = 0.5774l = 0.5774 × 10m = 5.8 m

b) Maximum bending moment for a beam subjected to a uniformly distributed load of magnitude

20 w = ------ = 10kN ⁄ m 2 2 2 w×l ( 10kN ⁄ m ) × ( 10m ) M max = -------------- = ---------------------------------------------------- = 125kNm 8 8 The distance defining the location of the maximum bending moment relative to the left support is

x = 5m

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10-1

c) Maximum bending moment values are almost identical; the difference is on the order of 2% (128 kNm versus 125 kNm). The location of the maximum bending moment is also very similar in both cases - the difference is by 0.8 m or approximately 8% of the span length. Therefore, the approach taken in part b) gives a reasonable approximation of the maximum bending moment determined in part a). ___________________________________________________________________________

10.2. a) Shear force and bending moment diagrams corresponding to load configurations (1) to (3) are presented below. Load configuration (1)

Load configuration (2)

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10-2

Load configuration (3)

b) Shear force and bending moment diagrams

c) Maximum bending moment and shear force values are very similar for all 3 configurations. d) A comparison Load configuration (1):

•

Maximum bending moment values are very similar (maximum difference on the order of 4%). • Maximum shear forces are the same. Load configuration (2):

• •

Maximum bending moments are identical. However, the maximum moment due to point load extends over the middle half of the span. This approximation is acceptable provided that the flexural steel is not cut off prematurely. Maximum shear force values are the same.

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10-3

Load configuration (3):

• •

Differences in maximum bending moment values are very small. In this case, it would be conservative to use maximum shear force due to uniformly distributed load. In any case, a designer should be cautious when designing the middle portion of the span for shear. e) A comparison of shear force diagrams Load configuration (1)

There is no underdesign in bending moment values. There are 6 localized regions with underdesign in shear force values (as shown with * in the diagram below).

Load configuration (2)

There are two localized regions with underdesign in bending moment and shear force values (as shown with * in the diagrams below).

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Load Configuration (3)

There is no underdesign in bending moment values, however middle portion of the beam is underdesigned for shear (as shown with * in the diagram below).

_________________________________________________________________________ 10.3.

w f × l ( 15kN ⁄ m 2 ) × 6 m V CL ≅ ------------- = -------------------------------------------- = 45kN ⁄ m 2 2 a = 400mm (beam width) a M 1 = M CL – V CL × --2

0.4m = – 55kNm ⁄ m – 45kN ⁄ m × ------------ = – 46kNm ⁄ m 2

_________________________________________________________________________ 10.4. a) Perform load analysis.

w DL, f = 15kN ⁄ m × 1.25 = 18.8kN ⁄ m w LL, f = 20kN ⁄ m × 1.5 = 30kN ⁄ m w f = w DL, f + w LL, f = 48.8 kN ⁄ m ≅ 50kN ⁄ m

Copyright © 2006 Pearson Education Canada Inc.

10-5

Clear span: l n = 7.2m – 0.4m = 6.8m Factored bending moments and shear forces obtained using the CSA A23.3 approximate method for frame analysis (Cl.9.3.3) are summarized in the table below. Span

1

2

ln( m )

6.8

6.8

Location

Top

Bottom

Top

Top

Bottom

Top

M f (see Figure 10.14b)

0

w f l n2 ---------11

w f l n2 – ---------10

w f l n2 – ---------11

w f l n2 ---------16

w f l n2 – ---------11

0

210

-231

-210

145

-210

-

1.15w f l n --------------------2

wf ln ---------2

-

(see Figure 10.14c)

w f ln ---------2

w f ln ---------2

V f ( kN )

170

-

195

170

-

170

Mf

( kNm )

Vf

b) Factored bending moments and shear forces obtained using the elastic analysis (beam diagrams in Table A.17 of Appendix A) are summarized in the table below.

Span

1

2

l( m)

7.2

7.2

Location

Top

Bottom

Mf

0

0.077w f l 2 – 0.107w f l 2

– 0.107w f l 2 0.036w f l 2 – 0.071w f l 2

0

200

-277

-277

Mf

( kNm )

Top

Top

Bottom

93

Top

-184

M 1 ( kNm ) (at girder face)

0

-233

-238

-151

Vf

0.393w f l -

0.607w f l

0.536w f l -

0.464w f l

V f ( kN )

141

218

193

167

V 1 ( kN ) (at girder face)

131

208

183

157

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10-6

Note that M 1 and V 1 denote the reduced bending moment and shear force values at the face of the girder. Bending moment at the face of the support ( M 1 ) can be obtained as follows:

a M 1 = M CL – V CL × --2 where M CL denotes the bending moment at the girder centreline, V CL denotes shear force at the girder centreline, and a denotes the girder width. For example, bending moment at the face of right support for span 1 has been calculated as

0.4 M 1 = – 277 – 218 ------2

= – 233 kNm

Shear force at the face of the support ( V 1 ) can be determined as

a V 1 = V CL – w f --2 For example, shear force at the face of right support for span 1 has been calculated as

0.4 V 1 = 218 – 50 ------- = 208kN 2 c) The CSA A23.3 approximate frame analysis procedure (Cl.9.3.3) incorporates pattern loading for live load and is more realistic as compared to elastic analysis that uses 100% dead and live load only. Elastic analysis based on full factored load either underestimates or overestimates bending moments and shear forces in reinforced concrete continuous members. In conclusion, the CSA A23.3 approximate frame analysis procedure should be used in this design. __________________________________________________________________________ 10.5. a) Perform the load analysis. 2

w DL, f = 6kN ⁄ m × 1.25 = 7.5kN ⁄ m 2

w LL, f = 5kN ⁄ m × 1.5 = 7.5kN ⁄ m w f = w DL, f + w LL, f = 15 kN ⁄ m

2

2

2

This is an one-way slab and the design is based on a unit strip (1 m wide), so 2

w f ′ = 15kN ⁄ m × 1m = 15kN ⁄ m

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10-7

Clear span: l n = 6.0m – 0.3m = 5.7m Factored bending moments and shear forces obtained using the CSA A23.3 approximate method for frame analysis (Cl.9.3.3) are summarized in the table below. Span

1

2

ln( m )

5.7

5.7

Location

Top

Bottom

Top

Top

Bottom

Top

Mf

0

w f l n2 ---------11

w f l n2 – ---------10

w f l n2 – ---------11

w f l n2 ---------16

w f l n2 – ---------11

M f ( kNm ⁄ m )

0

44

-49

-44

30

-44

Vf (see Figure 10.14c)

w f ln ---------2

1.15w f l n --------------------2

wf ln ---------2

w f ln ---------2

V f ( kN ⁄ m )

43

49

43

43

(see Figure 10.14b)

b) Factored bending moments and shear forces obtained using the elastic analysis (beam diagrams in Table A.17 of Appendix A) are summarized in the table below.

Span

1

2

l( m)

6.0

6.0

Location

Top

Bottom

Top

Top

Mf

0

0.077w f l 2 – 0.107w f l 2

– 0.107w f l 2 0.036w f l 2 – 0.071w f l 2

M f ( kNm ⁄ m )

0

42

-58

-58

M 1 ( kNm ⁄ m ) (at support face)

0

-50

-51

-32

Vf

0.393w f l -

0.607w f l

0.536w f l -

0.464w f l

V f ( kN ⁄ m )

35

55

48

42

V 1 ( kN ⁄ m ) (at support face)

33

53

46

40

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Bottom

19

Top

-38

10-8

Note that M 1 and V 1 denote the reduced bending moment and shear force values at the face of the support. Bending moment at the face of the support ( M 1 ) can be obtained as follows:

a M 1 = M CL – V CL × --2 where M CL denotes the bending moment at the support centreline, V CL denotes shear force at the support centreline, and a denotes the support width. For example, bending moment at the face of right support for span 1 has been calculated as

0.3 M 1 = – 58 – 55 ------2

= – 50 kNm ⁄ m

Shear force at the face of the support ( V 1 ) can be determined as

a V 1 = V CL – w f --2 For example, shear force at the face of right support for span 1 has been calculated as

0.3 V 1 = 55 – 15 ------- = 53kN ⁄ m 2 c) The CSA A23.3 approximate frame analysis procedure (Cl.9.3.3) incorporates pattern loading for live load and is more realistic as compared to elastic analysis that uses 100% dead and live load only. Elastic analysis based on full factored load either underestimates or overestimates bending moments and shear forces in reinforced concrete continuous members. In conclusion, the CSA A23.3 approximate frame analysis procedure should be used in this design.

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10-9

__________________________________________________________________________ 10.6. a) The following factored loads are considered in this design:

DL f = 50kN ⁄ m LL f = 50kN ⁄ m In total, five different load patterns (LP1 to LP5) need to considered in this design, as illustrated in the sketch that follows. There are a few more load patterns, however they represent mirror images of patterns LP1 to LP5 and need not be considered.

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10-10

When the analysis has been performed for each load pattern, moment envelope diagram can be developed by overlaying bending moment diagrams corresponding to various load patterns. Moment envelope diagram is presented below.

b) Moment envelope diagram gives maximum bending moment values for all load patterns, whereas load pattern 1 (LP1) corresponds to the full factored load on all spans (including the dead and live load). The following conclusions can be drawn by comparing these two diagrams: •

Positive bending moment values corresponding to LP1 are significantly lower than corresponding the envelope values, particularly so in spans 2 and 3 (the difference is on the order of 30 to 40%). • Negative bending moment values corresponding to LP1 are lower than the envelope values, however the difference is not very significant (on the order of 5 to 15%). c) It is recommended to obtain bending moment values from the moment envelope diagram.

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10-11

__________________________________________________________________________ 10.7. a) The following factored loads are considered in this design:

DL f = 80kN ⁄ m LL f = 20kN ⁄ m In total, five different load patterns (LP1 to LP5) need to considered in this design, as illustrated in the sketch that follows. There are a few more load patterns, however they represent mirror images of patterns LP1 to LP5 and need not be considered.

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10-12

When the analysis has been performed for each load pattern, moment envelope diagram can be developed by overlaying bending moment diagrams corresponding to various load patterns. Moment envelope diagram is presented below.

b) Moment envelope diagram gives larger bending moment values than load pattern 1 (LP1) corresponding to the full factored load on all spans (including the dead and live load). This difference is on the order of 10% (with the exception of positive bending moments in interior spans). Obviously, moment envelope values will always be greater then the corresponding bending moments derived based on full factored loads; however, a 10% difference in bending moment values can be considered acceptable, considering inelastic behaviour of reinforced concrete flexural members (discussed in Section 10.7). c) When live load is small in comparison to dead load, it may not be necessary to determine moment envelopes if the designer makes a provision for the reserve in flexural resistance.

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10-13

__________________________________________________________________________

10.8. a) The factored bending moment diagram is presented below.

b) Determine the required reinforcement. Find the effective depth.

h = 900mm so

d = h – 70 = 900 – 70mm = 830mm Determine the required amount of tension reinforcement. Positive reinforcement

Set

M r = M f = 530kNm (maximum negative bending moment) Find the reinforcement area using the direct method. 2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

3.85 ( 530 ×10 ) 2 = 0.0015 ( 30 ) ( 600 ) 830 – 830 – ------------------------------------ = 1926mm 30 ( 600 ) 2

Use 4-25M bars: A s = 2000mm

2

Negative reinforcement

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10-14

Set

M r = M f ≅ 1000kNm (maximum positive bending moment) Find the reinforcement area using the direct method. 2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

2 3.85 ( 1000 ×10 ) 2 = 0.0015 ( 30 ) ( 600 ) 830 – 830 – --------------------------------------- = 3801mm 30 ( 600 )

Use 8-25M bars: A s = 4000mm

2

c) Alternative (2) is the most suitable based on elastic analysis (8-25M bars at the top and 4-25M bars at the bottom). d) Find the moment gradient for spans 1 and 2. Span 1:

M g1

Span 2:

M g2

2

2 wf × l × ( 8m ) ( 120kN ⁄ m ) = ---------------- = ---------------------------------------------------- = 960kNm 8 8 2

2 wf × l ( 120kN ⁄ m ) ( 10m ) × = ---------------- = ------------------------------------------------------- = 1500kNm 8 8

Check whether Alternatives (1) and (3) are safe, considering the amount of reinforcement obtained in part b). Alternative (1)

Find M r for the section with 6-25M bars (top/bottom). 2

2

A s = 6 × 500mm = 3000mm (6-25M, see Table A.1)

2 φs fy As 0.85 × 400MPa × 3000mm --------------------------------------------------------------------------------------------= = 109mm a = 0.8 × 0.65 × 30MPa × 600mm α 1 φ c f c ′b

[3.12]

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10-15

a M r = φ s A s f y d – --2

[3.14]

109 2 = 0.85 × 3000mm × 400MPa 830 – --------- = 791kNm 2

Find the combined moment resistance for positive and negative bending moments. Span 1:

0 + 791 M rcomb = 791 + ------------------ = 1187kNm 2 Since

M g1 = 960kNm and

M rcomb = 1187kNm > 960kNm this reinforcing arrangement is safe. Span 2:

791 + 791 M rcomb = 791 + ------------------------ = 1582kNm 2 Since

M g2 = 1500kNm and

M rcomb = 1582kNm > 1500kNm this reinforcing arrangement is safe. Alternative (3)

Find M r for the section with 4-25M top rebars (negative bending moment). 2

2

A s = 4 × 500mm = 2000mm (4-25M, see Table A.1)

2 φs fy As 0.85 × 400MPa × 2000mm a = --------------------- = --------------------------------------------------------------------------- = 73mm 0.8 × 0.65 × 30MPa × 600mm α 1 φ c f c ′b a M r = φ s A s f y d – --[3.14] 2 73 2 = 0.85 × 2000mm × 400MPa 830 – ------ = 540kNm 2

[3.12]

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10-16

Find M r for the section with 8-25M bottom rebars (positive bending moment). 2

2

A s = 8 × 500mm = 4000mm (8-25M, see Table A.1)

2 φs fy As 0.85 × 400MPa × 4000mm a = --------------------- = --------------------------------------------------------------------------- = 145mm 0.8 × 0.65 × 30MPa × 600mm α 1 φ c f c ′b a M r = φ s A s f y d – --[3.14] 2 145 2 = 0.85 × 4000mm × 400MPa 830 – --------- = 1030kNm 2

[3.12]

Find the combined moment resistance for positive and negative bending moments: Span 1:

Since

M g1 = 960kNm and

M r = 1030kNm > 960kNm this arrangement is safe. Span 2:

540 + 540 M rcomb = 1030 + ------------------------ = 1570kNm 2 Since

M g2 = 1500kNm and

M rcomb = 1570kNm > 1500kNm this arrangement is okay from a safety perspective. e) Let us redistribute bending moments in each span as permitted by CSA A23.3 Cl.9.2.4.

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10-17

First, decrease bending moments at supports by 20% (in absolute values), and increase bending moments at midspan regions by 20%. Bending moment diagram after redistribution is shown below. The resulting moment distribution would be okay for Alternative (1).

Next, redistribute bending moments such that the bending moments at supports are increased by 20% (in absolute values), while bending moments at midspans are decreased by 20%. The resulting moment distribution would be okay for Alternative (3).

Alternative (3) is a probable solution from the strength perspective, however, it is likely that the support regions would experience excessive cracking. A detailed analysis would be required to estimate crack widths at the supports.

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10-18

Chapter 11 - Solutions ________________________________________________________________________ 11.1. a) Develop the factored moment envelope diagram.

1. Check whether the criteria for the CSA A23.3 approximate frame analysis are satisfied. The criteria for the approximate frame analysis according to CSA A23.3 Cl.9.3.3 are as follows: • There are more than two continuous spans; • The length of each span is within 20% of the adjacent spans; The ratio between span lengths for spans 2 and 1 is equal to

l2 7m ---- = ------------ = 1.27 5.5m l1 This is indicates that the length of span 2 exceeds the length of span 1 by 27%, hence the approximate frame analysis according to A23.3 Cl.9.3.3 cannot be used in this design. Therefore, a linear elastic analysis will be performed by using a computer analysis program. 2. Determine the factored loads. - Determine the beam self-weight. Only a portion of the beam below the slab will be considered for the self-weight calculation (the slab portion has been already taken into account), hence 3

DL w = b × h web × γ w = 0.3m × 0.48m × 24kN ⁄ m = 3.5kN ⁄ m where 3

γ w = 24kN ⁄ m unit weight of concrete

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11-1

(note that the slab is 220 mm thick)

- Determine the tributary width ( a ). The tributary width for the beam is equal to (see the sketch below)

5.5m a = ------------ + 0.15m = 2.9m 2

- Find the factored dead load for the beam.

g DLf = w DL, f × a + 1.25 × DL w = 7.85kPa × 2.9m + 1.25 × 3.5kN ⁄ m ≅ 28kN ⁄ m - Find the factored live load for the beam.

g LLf = w LL, f × a = 7.2kPa × 2.9m ≅ 21kN ⁄ m 3. Identify all possible load patterns. According to CSA A23.3 Cl.9.2.3.1, the following load patterns should be considered in the design of the four-span beam under consideration:

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11-2

• LP1: full dead load and live load on all spans • LP2: full dead load on all spans, plus live load on spans 1 and 3 • LP3: full dead load on all spans, plus live load on spans 1 and 2 • LP4: full dead load on all spans, plus live load on spans 2 and 3 Note that this beam is symmetrical with regard to the vertical axis. Span 1 is a mirror image of span 4, while span 2 is a mirror image of span 3. Therefore, there are a few other possible load patterns, but due to symmetry they would give the same results as patterns LP1 to LP4. 4. Develop the moment envelope diagram and determine the design bending moments. First, moment diagrams need to be developed for each load pattern. Subsequently, bending moment diagrams need to be superimposed on the same chart, as shown on the diagram below. The moment envelope diagram is constructed from portions of bending moment diagrams corresponding to different load patterns.

b) Design the beam for flexure. - Determine the effective depth ( d ). It is expected that the reinforcement will be placed in one layer, hence the d value can be estimated as (see Section 5.4.1)

d = h – 70mm = 700mm – 70mm = 630 mm The same effective depth ( d ) will be used for the positive and negative moment areas. - Determine the section width. Although this beam behaves as a T-beam, the slab contribution will be neglected for the sake of simplicity. Hence, the beam will be considered as a rectangular beam with the width

b = 300mm

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11-3

both for negative and positive moment areas. Positive bending

Set

M r = M f = 135kNm (largest positive bending moment) - Determine the required area of tension reinforcement. Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 135 × 10 2 2 = 0.0015 ( 30MPa ) ( 300mm ) 630mm – ( 630mm ) – ------------------------------------------------ = 643mm ( 30MPa ) ( 300mm ) Use 4-15M bars:

A s = 4 × 200 = 800mm

2

- Find the minimum tension reinforcement area per CSA A23.3 (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 30MPa 2 = ------------------------------- × 300mm × 700mm = 575mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 800mm > 575mm okay - Find the maximum amount of tension reinforcement per CSA A23.3 (Cl.10.5.2). 2 As 800mm ρ = ------ = -------------------------------------------- = 0.0042 [3.1] bd 300mm × 630mm

Balanced reinforcement ratio:

ρ b = 0.027 ( f c ′ = 30MPa , see Table A.4) Since

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11-4

ρ = 0.0042 < ρ b = 0.027 okay Negative bending

Set

M r = M f = – 190k Nm (largest negative bending moment) - Determine the required area of tension reinforcement. Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 190 × 10 2 2 = 0.0015 ( 30MPa ) ( 300mm ) 630mm – ( 630mm ) – ------------------------------------------------ = 920mm ( 30MPa ) ( 300mm ) Use 4-20M bars:

A s = 4 × 300 = 1200mm

2

- Find the minimum tension reinforcement area per CSA A23.3 (Cl.10.5.1.2). 2

A smin = 575mm (same as above) Since 2

2

A s = 1200mm > 575mm okay - Find the maximum amount of tension reinforcement per CSA A23.3 (Cl.10.5.2). 2 As 1200mm ρ = ------ = -------------------------------------------- = 0.0063 [3.1] bd 300mm × 630mm

Balanced reinforcement ratio:

ρ b = 0.027 ( f c ′ = 30MPa , see Table A.4) Since

ρ = 0.0063 < ρ b = 0.027 okay

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11-5

Sketch a design summary for the flexural reinforcement.

c) Design the beam for shear. First, develop the shear envelope diagram. The diagram is constructed in a similar manner to moment envelope diagram. The same load patterns have been used and the resulting diagram is shown below.

Span 1 - shear design

The following steps need to be followed to design shear reinforcement for span 1: 1. Calculate the design shear force ( V f @dv ) at a distance d v from the face of the support (A23.3 Cl.11.3.2) In order to determine V f @dv , the following will have to be calculated: i) Effective shear depth ( d v )

d v = effective shear depth taken as the greater of dv =

0.9d = 0.9 × 630mm = 567mm 0.72h 0.72 × 700mm 504mm

Hence, d v ≅ 570mm (larger value governs) ii) Centre-to-centre span:

l = 5.5m Copyright © 2006 Pearson Education Canada Inc.

11-6

iii) Maximum shear force at the midspan ( V fmidspan ):

l 5.5m LL f --( 21kN/m ) × -----------2 2 V fmidspan = ------------------ = ------------------------------------------------- = 14.4kN ≅ 15kN 4 4 iv) The factored shear force ( V f @dv ) at a distance ( d v ) from the support A can be determined as V f @dv

b l --- – d v + ----s 2 2 = V fmidspan + ------------------------------- ( V f – V fmidspan ) l --2

5.5m0.4m ----------– 0.57m + -----------2 2 = 15kN + ---------------------------------------------------------- ( 110kN – 15kN ) ≅ 83kN 5.5m -----------2 The maximum shear force envelope shows shear forces at the centreline of the supports. In order bs to determine the V f @dv , we need to find the shear force at a distance d v + ---- from the support 2 centreline, where b s is the width of the supporting column equal to 400 mm. v) The factored shear force ( V f @dv ) at a distance ( d v ) from the support B can be determined as

5.5m0.4m ----------– 0.57m + -----------2 2 V f @dv = 15kN + ---------------------------------------------------------- ( 172kN – 15kN ) ≅ 128kN 5.5m -----------2 vi) Sketch the shear envelope diagram for span 1.

2. Determine the concrete shear resistance ( V c ) (A23.3 Cl.11.3.4).

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11-7

Consider that a larger than the minimum shear reinforcement will be provided (as per A23.3 Cl.11.2.8.2), hence β = 0.18 (A23.3 Cl.11.3.6.3) and

V c = φ c λβ f c ′b w d v = 0.65 ( 1 ) ( 0.18 ) 30MPa ( 300mm ) ( 570mm ) = 109kN [6.12] 3. Determine whether shear reinforcement is required and identify the region(s) in which the reinforcement is required (A23.3 Cl.11.2.8.1). If V f < V c , then the shear reinforcement is not required. i) At the support A: V f = 83kN < 109kN ii) At the support B: V f = 128kN > 109kN Therefore, it can be concluded that shear reinforcement is required at support B. Considering that shear reinforcement is not required at support A, check the concrete shear resistance at support A using β values for members without shear reinforcement:

230 230 β = ----------------------- = --------------------------- = 0.15 (A23.3 Eq. 11-9) [6.13] 1000 + d v 1000 + 570 So,

V c = φ c λβ f c ′b w d v = 0.65 ( 1 ) ( 0.15 ) 30MPa ( 300mm ) ( 570mm ) = 91kN Since V f = 83kN < 91kN shear reinforcement at support A is not required. 4. Calculate the stirrup spacing according to the CSA A23.3 requirements (A23.3 Cl.11.2.8.2 and Cl.11.3.8.1). i) Let us use 2-legged 10M stirrups, therefore A v = (#of legs)(Area of each leg) 2

2

= 2 × 100mm = 200mm ii) Maximum spacing of shear reinforcement can be determined based on the following equation (A23.3 Cl.11.2.8.2) Av fy s = --------------------------[6.19] 0.06 f c ′b w 2

200mm × 400MPa s = ---------------------------------------------------------- = 811mm 0.06 30MPa ( 300mm ) iii) Check whether the spacing is within the maximum limits prescribed by A23.3 Cl.11.3.8.1, as follows: 600mm 600mm = = 600mm s max ≤ 0.7d v 0.7 × 570mm 399mm Copyright © 2006 Pearson Education Canada Inc.

11-8

Hence, the maximum permitted spacing s max is the smallest of

811mm 600mm 399mm

Hence, s ≅ 400mm 5. Determine the region where stirrups are required (support B only).

V c = 109kN V f = 172kN (at column B centreline) Use the proportion from the shear envelope diagram for span 1 (see the sketch below).

2.75 – x 109kN = 15 + ( 172 – 15 ) ------------------2.75 so, x value can be determined as

x = 1.1m from the column centreline

6. Check whether the strength requirement is satisfied (A23.3 Cl.11.3.1). i) Find the shear capacity ( V s ) of shear reinforcement (A23.3 Cl.11.3.5.1). As this design has been performed according to the CSA A23.3 Simplified Method, Cl.11.3.6.3 prescribes that θ = 35° and cot θ = 1.43 φ s A v f y d v cot θ V s = --------------------------------(A23.3 Eq.11-7) [6.9] s 2

0.85 × 200mm × 400MPa × 570mm × cot ( 35 ) = --------------------------------------------------------------------------------------------------------------------- = 138kN 400mm ii) Find the factored shear resistance ( V r ) (A23.3 Cl.11.3.3). Vr = Vc + Vs + Vp (A23.3 Eq.11-4)

[6.11]

= 109kN + 138kN + 0 = 247kN iii) Check whether the CSA A23.3 strength requirement is satisfied, that is,

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11-9

Vr ≥ Vf

(A23.3 Eq.11-3)

[11.1]

At the support B:

V r = 247kN > V f @dv = 128kN It can be concluded that the strength requirement is satisfied and the shear reinforcement is adequate. 7. Check the maximum factored shear resistance ( V r ) (A23.3 Cl.11.3.3). The factored shear resistance for the beam section is equal to

V r = 247kN According to A23.3 Cl.11.3.3, the maximum factored shear resistance ( maxV r ) is limited to

maxV r = 0.25φ c f c ′b w d v

(A23.3 Eq.11-5)

[6.17]

= 0.25 × 0.65 × 30MPa × 300mm × 570mm = 834kN Since

V r = 247kN < 834kN section dimensions are adequate. h) Sketch a beam elevation showing the distribution of shear reinforcement for Span 1.

Span 2 - shear design

The following steps need to be followed to design shear reinforcement for span 2: 1. Calculate the design shear force ( V f @dv ) at a distance d v from the face of the support (A23.3 Cl.11.3.2) In order to determine V f @dv , the following will have to be calculated: Copyright © 2006 Pearson Education Canada Inc.

11-10

i) Effective shear depth ( d v )

d v ≅ 570mm (same as Span 1) ii) Centre-to-centre span:

l = 7.0m iii) Maximum shear force at the midspan ( V fmidspan ):

l 7.0m LL f --( 21kN/m ) × -----------2 2 V fmidspan = ------------------ = ------------------------------------------------- ≅ 18kN 4 4 iv) The factored shear force ( V f @dv ) at a distance ( d v ) from the support B can be determined as V f @dv

b l --- – d v + ----s 2 2 = V fmidspan + ------------------------------- ( V f – V fmidspan ) --l2

7.0m 0.4m ------------ – 0.57m + ----------2 2 = 18kN + ---------------------------------------------------------- ( 182kN – 18kN ) ≅ 146kN 7.0m----------2 The maximum shear force envelope shows shear forces at the centreline of the supports. In order bs to determine the V f @dv , we need to find the shear force at a distance d v + ---- from the support 2 centreline, where b s is the width of the supporting column equal to 400 mm. Since the maximum shear force values at supports B and C are the same, V f @dv values are the same as well. v) Sketch the shear envelope diagram for span 2.

2. Determine the concrete shear resistance ( V c ) (A23.3 Cl.11.3.4). Consider that a larger than the minimum shear reinforcement will be provided (as per A23.3 Cl.11.2.8.2), hence β = 0.18 (A23.3 Cl.11.3.6.3) and

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11-11

V c = φ c λβ f c ′b w d v = 0.65 ( 1 ) ( 0.18 ) 30MPa ( 300mm ) ( 570mm ) = 109kN [6.12] 3. Determine whether shear reinforcement is required and identify the region(s) in which the reinforcement is required (A23.3 Cl.11.2.8.1). If V f < V c , then the shear reinforcement is not required. Shear force at the supports B and C: V f = 146kN > 109kN Therefore, it can be concluded that shear reinforcement is required at both supports. 4. Calculate the stirrup spacing according to the CSA A23.3 requirements (A23.3 Cl.11.2.8.2 and Cl.11.3.8.1). i) Let us use 2-legged 10M stirrups, therefore A v = (#of legs)(Area of each leg) 2

2

= 2 × 100mm = 200mm ii) Maximum spacing of shear reinforcement (A23.3 Cl.11.2.8.2) s = 811mm (same as Span 1) iii) Check whether the spacing is within the maximum limits prescribed by A23.3 Cl.11.3.8.1, as follows: The maximum permitted spacing s max is the smallest of

811mm 600mm 399mm

(same as Span 1)

So, s ≅ 400mm 5. Determine the region where stirrups are required. V c = 109kN

V f = 182kN (at column centreline) Use the proportion from the shear envelope diagram for span 2 (see the sketch below).

3.5 – x 109kN = 18 + ( 182 – 18 ) ---------------3.5 so, x value can be determined as

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11-12

x = 1.56m from the column centreline

6. Check whether the strength requirement is satisfied (A23.3 Cl.11.3.1). i) Find the shear capacity ( V s ) of shear reinforcement (A23.3 Cl.11.3.5.1).

V s = 138kN (same as Span 1) ii) Find the factored shear resistance ( V r ) (A23.3 Cl.11.3.3).

V r = 247kN (same as Span 1) iii) Check whether the CSA A23.3 strength requirement is satisfied, that is, Vr ≥ Vf (A23.3 Eq.11-3)

[11.1]

Since

V r = 247kN > V f @dv = 146kN It can be concluded that the strength requirement is satisfied and the shear reinforcement is adequate. 7. Check the maximum factored shear resistance ( V r ) (A23.3 Cl.11.3.3). The factored shear resistance for the beam section is equal to V r = 247kN According to A23.3 Cl.11.3.3, the maximum factored shear resistance ( maxV r ) is limited to

maxV r = 834kN (same as Span 1) Since

V r = 247kN < 834kN section dimensions are adequate.

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11-13

8. Sketch a beam elevation showing the distribution of shear reinforcement for Span 2.

d) Check the crack control parameter ( z ) (A23.3 Cl.10.6.1). Find the actual effective depth as 15 d = 700 – 40 – 10 – ------ ≅ 640mm 2 (based on 40 mm clear cover, 10M stirrups, and 15M bars) 1. Compute the effective tension area per bar ( A ). In order to determine A , the following will have to be calculated: i) Distance between the centroid of the tensile reinforcement and tensile face of the concrete section ( d s ) (see the sketch below):

d s = h – d = 700 – 640 = 60mm ii) The total effective tension area for all bars ( A e ):

A e = b ( 2d s ) = ( 300mm ) ( 2 ⋅ 60mm ) = 36000mm 2 iii) The effective tension area per bar ( A ) for the bottom reinforcement (4-15M): For 4 bars in total: N = 4 Hence,

Ae 36000mm 2 A = ------ = --------------------------- = 9000mm 2 N 4

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11-14

2. Determine the stress in steel reinforcement ( f s ) under the service load level. The stress in steel reinforcement can be computed based on an approximate estimate permitted by the CSA A23.3 Cl.10.6.1 in lieu of the detailed calculations, as follows

f s = 0.6f y = 0.6 ⋅ 400MPa = 240MPa

[4.21]

3. Determine the value of z factor.

d c = d s = 60mm (one reinforcement layer) z = fs 3 dc A

(A23.3 Eq.10-6)

[4.20]

2

= 240MPa 3 ( 60mm ) ( 9000mm ) = 19540N ⁄ mm Since

z = 19540N ⁄ mm < 30000N ⁄ mm okay It follows that the CSA A23.3 cracking control requirements for the interior exposure are satisfied.

Copyright © 2006 Pearson Education Canada Inc.

11-15

________________________________________________________________________ 11.2.

Slab thickness: h = 125mm a) The criteria for the approximate frame analysis according to CSA A23.3 Cl.9.3.3 are as follows: • There are more than two continuous spans; • The length of each span is within 20% of the adjacent spans; 2.5m Since ------------ = 1.19 okay 2.1m • The loads are uniformly distributed; • The factored live load does not exceed twice the factored dead load (including the slab selfweight) - this will be checked later; • The slab is prismatic (all sections are of uniform thickness). Hence, the approximate frame analysis according to CSA A23.3 Cl.9.3.3 can be used in this design. b) Determine the factored bending moment and shear force distribution for this slab using the CSA A23.3 approximate frame analysis method. 1. Determine the factored loads. Dead load: First, let us calculate the slab self-weight, as follows 3

DL w = h × γ w = 0.125 m × 24kN ⁄ m = 3.0kPa where 3

γ w = 24kN ⁄ m unit weight of concrete The superimposed dead load is given as

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11-16

DL s = 1.0kPa Finally, the total factored dead load is equal to

w DL, f = 1.25 ( DL w + DL s ) = 1.25 ( 3.0kPa + 1kPa ) = 5.0kPa Calculate the factored live load

w LL, f = 1.5 × LL = 1.5 × 5.0kPa = 7.5kPa The total factored load is equal to

w f = w DL, f + w LL, f = 5.0kPa + 7.5kPa ≅ 12.5kPa This is an one-way slab and the design is based on the unit strip (1 m wide), therefore 2

w f = 12.5kN ⁄ m × 1m = 12.5kN ⁄ m per metre Note that the factored live load w LL, f = 7.5kPa is less than the factored dead load, that is, w DL, f = 5.0kPa . Apparently, the factored live load does not exceed twice the factored dead load, hence the condition for the application of the CSA A23.3 coefficient method has been met. 2. Determine the factored bending moments at critical sections. Use the moment coefficients prescribed by A23.3 Cl.9.3.3 (see Figure 10.14). Span

1

2

3

4

l(m )

2.1

2.1

2.1

2.5

ln( m )

1.7

1.7

1.7

2.1

Location

Left

Right

– M f (negative bending)

0

w f l n2 ---------10

– M f ( kNm ⁄ m ) +M f (positive bending) +M f ( kNm ⁄ m )

-3.6

Left

w f l n2 ---------11 -3.3

Right

Left

Right

Left

Right

w f l n2 ---------11

w f l n2 ---------11

w f l n2 ---------11

w f l n2 ---------11

w f l n2 ---------11

-3.3

-3.3

-5

-5

-3.3

w f l n2 ---------11

w f l n2 ---------16

w f l n2 ---------16

w f l n2 ---------16

3.3

2.3

2.3

3.4

c) Determine the flexural reinforcement.

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11-17

1. Determine the effective depth ( d ). As the slab is of interior exposure, the cover can be determined from Table A.2 as cover = 20mm As 10M bars are to be used in this design, it follows that d b = 10mm (see Table A.1) Therefore, the effective depth can be determined as

db 10mm d = h – cover – ----- = 125mm – 20mm – ---------------- = 100mm 2 2 The same effective depth ( d ) will be used for the positive and negative moment areas. 2. Determine the required area of tension reinforcement. Since the maximum positive and negative bending moment values are similar, design the reinforcement for the largest positive/negative moment (compare absolute values). Consequently, set

M r = M f = – 5k Nm ⁄ m The required area of tension reinforcement can be determined using the direct procedure according to the following equation 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 5 × 10 2 2 = 0.0015 ( 25 ) ( 1000 ) 100mm – ( 100mm ) – --------------------------------------------------- = 147mm ⁄ m ( 25MPa ) ( 1000mm ) Confirm that the minimum reinforcement requirement for slab is satisfied. 3

A g = h × 1000mm = 125mm × 1000mm = 125 × 10 mm 3

2

2

A smin = 0.002A g = 0.002 ( 125 × 10 mm ) = 250mm 2 ⁄ m Determine the required bar spacing ( s ). The spacing is determined assuming 10M rebars, with the bar area of

A b = 100mm

2

(see Table A.1)

Therefore,

1000 1000mm 2 - = 400mm s ≤ A b ------------ = ( 100mm ) × ----------------------------2 As 250mm ⁄ m

[3.29]

Check the CSA A23.3 maximum reinforcement requirements (Cl.10.5.2). Since the minimum reinforcement requirements govern, there is no need to perform this check. Copyright © 2006 Pearson Education Canada Inc.

11-18

Confirm that the maximum bar spacing requirement is satisfied (CSA A23.3 Cl.7.4.1.2). First, calculate the maximum bar spacing ( s max ) as the lesser of: • 3 × h = 3 × 125 mm = 375mm • 500mm The smaller value governs, therefore s max = 375mm Since s = 400mm > 375mm Use s = 375mm Top and bottom reinforcement: 10M@375 3. Design shrinkage and temperature reinforcement (A23.3 Cl.7.8.1 and 7.8.3). Determine the minimum area of shrinkage and temperature reinforcement: A smin = 250mm 2 ⁄ m Determine the maximum bar spacing ( s max ).

s max is the lesser of: • 5 × h = 5 × 125 mm = 625mm • 500mm The smaller value governs, therefore s max = 500mm Determine the required bar spacing. Based on the minimum area requirement using 10M bars, the required bar spacing is s = 400mm Temperature and shrinkage reinforcement: 10M@400 4. Sketch the reinforcement arrangement for the slab.

d) Check the shear resistance of the slab according to the CSA A23.3 Simplified Method for shear design. The following steps need to be followed in order to determine the shear resistance for this slab: 1. Determine the factored shear forces.

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11-19

The shear forces are determined according to the CSA A23.3 Simplified Frame Method (Cl.9.3.3), see Figure 10.14. The results are summarized in table below (note that w f = 12.5kN ⁄ m ). Span

1

2

3

4

l(m )

2.1

2.1

2.1

2.5

ln ( m )

1.7

1.7

1.7

2.1

Location

Left

Right

Left

Right

Left

Right

Left

Right

Vf

wf ln ---------2

1.15w f l n --------------------2

wf ln ---------2

w f ln ---------2

wf ln ---------2

w f ln ---------2

wf ln ---------2

w f ln ---------2

V f ( kN ⁄ m )

10.6

12.2

10.6

10.6

10.6

10.6

13.1

13.1

2. Determine the concrete shear resistance ( V c ) (A23.3 Cl.11.3.4). i) Find the effective shear depth ( d v ).

d v = effective shear depth taken as the greater of d v = 0.9d = 0.9 × 100mm = 90mm 0.72h 0.72 × 125mm 90mm Hence, d v = 90mm . ii) Determine the β value. Since

h = 125mm < 350mm According to A23.3 Cl.11.3.6.2, β = 0.21 can be used for slabs with an overall thickness not greater than 350 mm. iii) Finally, the V c value can be determined as

V c = φ c λβ f c ′b w d v

(A23.3 Eq.11-6)

[6.12]

= 0.65 × 1.0 × 0.21 × 25MPa × 1000mm × 90mm = 61kN ⁄ m 3. Determine whether shear reinforcement is required (A23.3 Cl.11.2.8.1) If V f < V c , then shear reinforcement is not required. In this example, the concrete shear resistance greatly exceeds the factored shear force in each span presented in the above table, that is,

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11-20

V f < V c = 61kN ⁄ m hence it can be concluded that the shear reinforcement does not need to be provided in the slab. e) Find the crack control parameter ( z ) (A23.3 Cl.10.6.1). 1. Compute the effective tension area per bar ( A ). In order to determine A , the following will have to be calculated: i) Calculate the distance ( d s ) from the centroid of the tensile reinforcement to the tensile face of the concrete section:

d s = h – d = 125mm – 100mm = 25mm ii) Determine the effective tension area per bar ( A ). In this case, the A value can be determined directly for one bar with a spacing ( s ), as follows (see the sketch below)

A = s ( 2d s ) = ( 375mm ) ( 2 × 25mm ) = 18750mm

2

2. Determine the stress in steel reinforcement ( f s ) under the service load level. An approximate estimate for f s permitted by the A23.3 Cl.10.6.1 will be used in lieu of the detailed calculations, as follows

f s = 0.6f y = 0.6 ⋅ 400MPa = 240MPa

[4.21]

3. Determine the value of z factor.

d c = d s = 25mm z = fs 3 dc A

(A23.3 Eq.10-6)

[4.20]

2

= 240MPa 3 ( 25mm ) ( 18750mm ) = 18640N ⁄ mm Since

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11-21

z = 18640N ⁄ mm < 30000N ⁄ mm It follows that the CSA A23.3 cracking control requirements for the interior exposure are satisfied. ________________________________________________________________________ 11.3. a) Find the factored load (NBC 2005 Table 4.1.3.2) as

w f = 1.25DL + 1.5LL = 1.25 × 15 + 1.5 × 30 ≅ 65kN ⁄ m Next, determine the factored bending moments at critical sections. Use the moment coefficients prescribed by A23.3 Cl.9.3.3 (see Figure 10.14). Span

1

2

l(m )

6.0

7.2

ln( m )

5.6

6.8

Location

Left

Right

–Mf (negative bending)

0

w f l n2 ---------10

0

-204

– M f ( kNm )

Left

Right

w f l n2 ---------11 -273

w f l n2 ---------11 -273

+M f (positive bending)

w f l n2 ---------11

w f l n2 ---------16

+M f ( kNm )

185

188

b) Design the beam for flexure. - Determine the effective depth ( d ). It is expected that the reinforcement will be placed in one layer, hence the d value can be estimated as (see Section 5.4.1)

d = h – 70mm = 600mm – 70mm = 530 mm The same effective depth ( d ) will be used for the positive and negative moment areas.

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11-22

The beam has a rectangular section with the width

b = 500mm (given) - Determine the required area of tension reinforcement. Design the reinforcement for the largest positive/negative moment (compare absolute values). Consequently, set

M r = M f = 273kNm (largest positive bending moment) - Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 273 × 10 2 2 = 0.0015 ( 25MPa ) ( 500mm ) 530mm – ( 530mm ) – ------------------------------------------------ = 1619mm ( 25MPa ) ( 500mm ) Use 6-20M bars:

A s = 6 × 300 = 1800mm

2

- Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 25 2 = ------------------ ( 500 ) ( 600 ) = 750mm 400 (Note: b t = b rectangular sections) Since 2

2

A s = 1800mm > 750mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 1800mm ρ = ------ = -------------------------------------------- = 0.0068 [3.1] bd 500mm × 530mm

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

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11-23

ρ = 0.0068 < ρ b = 0.022 okay Use 6-20M bars (top and bottom). c) Determine the theoretical cut off points - use Figure 10.15:

Span

1

2

l(m )

6.0

7.2

ln( m )

5.6

6.8

Location

Left

Right

Left

0.28 l n

0.24 l n

1.6m

1.6m

1.6m

0.11 l n

0.1 l n

0.1 l n

0.6m

0.7m

0.7m

0.07 l n

0.07 l n

0.15 l n

0.15 l n

0.4m

0.4m

1.0m

1.0m

0.2 l n

0.2 l n

0.25 l n

0.25 l n

1.1m

1.1m

1.7m

1.7m

(-) inflection point

(-) 50% cut-off

(+) inflection point

(+) 50% cut-off

Right

- Development length ( l d ) (see Table A.5): 20M bars, f c ′ = 25MPa Bottom reinforcement: l d = 575mm Top reinforcement: l d = 750mm Follow the CSA A23.3 bar cutoff requirements per Checklists 11.1 and 11.2. Rules P-1 and N-1 - actual point of cutoff (Cl.12.10.3): Bar length extensions (beyond the theoretical cut off point) are • 1.3d = 1.3 × 530 = 690mm (governs) • h = 600mm Rules P-3 and N-3 - continuing reinforcement (Cl.12.10.4): • l d + d = 750 + 530 = 1280mm (governs)

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11-24

• l d + 12d b = 750 + 12 ( 20 ) = 990mm Rules P-5 and N-5 - inflection points (Cl.12.11.3) Span 1: • d = 530mm (governs) • 12d b = 12 ( 20mm ) = 240mm

ln ----- = 350mm 16 Span 2: • d = 530mm (governs) • 12d b = 12 ( 20mm ) = 240mm •

ln ----- = 485mm 16 Rebar lengths are summarized in the table below. •

Span

1

2

l(m )

6.0m

7.2m

Long (-)

1.6+.53=2.2m

1.6+.53=2.2m

=2.2m

Short (-)

.6+.69=1.4m

.7+.69=1.4m

=1.4m

3.4+.15=3.6m

=3.6m

(3.4-1.7) +.69=2.4m

=2.4m

Long (+)

2.8+.15=3.0m

Short (+)

(2.8-1.1) +.69=2.4m

Top steel Bottom steel

3.0m =2.4m

6-20Mx3600 STAG 800 2-20M x 5500 2-20M x 6000 ALT

6-20Mx3600 STAG 800 4-20M x 6000 STAG

A sketch showing the reinforcement arrangement for this beam is presented below.

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11-25

________________________________________________________________________ 11.4. a) The following factored loads are considered in this design:

DL f = 1.25 × 15 kN ⁄ m ≅ 19kN ⁄ m LL f = 1.5 × 30 kN ⁄ m = 45kN ⁄ m In total, four different load patterns (LP1 to LP4) need to considered in this design, as illustrated in the sketch that follows. There are a few more load patterns, however they represent mirror images of patterns LP1 to LP4 and need not be considered.

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11-26

When the analysis has been performed for each load pattern, moment envelope diagram can be developed by overlaying bending moment diagrams corresponding to various load patterns. Moment envelope diagram is presented below.

b) Bending moment values obtained using the CSA A23.3 approximate frame analysis (Problem 11.3.) and the elastic analysis (moment envelope) performed in this problem are summarized below.. Span

1

2

Location

Left

Right

Left

Right

– M f ( kNm ) (approximate frame analysis)

0

-204

-273

-273

– M f ( kNm ) (moment envelope)

0

-257 (reduced)

-257 (reduced)

-276

+M f ( kNm ) (approximate frame analysis)

185

188

+M f ( kNm ) (moment envelope)

216

217 (Span 3)

The following conclusions can be drawn after comparing the bending moment values: 1) Negative bending moments after reduction at the face of support are almost identical at the support C and are within 6% at the support B (273 kNm versus 257 kNm).

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11-27

2) Positive bending moments obtained from moment envelope are higher by about 15% (216 kNm versus 185 kNm). 3) The minimum positive bending moment in the spans 2 and 4 are in the negative range for load pattern 2 (LP2); this is not accounted for in the CSA A23.3 approximate frame method. 4) The approximate frame analysis does not recognize item 3) above. 5) Ratio between the factored live load and dead load LL f ⁄ DL f = 2.4 exceeds permissible limit of 2 specified by A23.3 Cl.9.3.3. Therefore, this beam span cannot be designed using the CSA A23.3 approximate frame analysis method. c) Let us perform moment redistribution per CSA A23.3 Cl.9.2.4 for bending moment diagram corresponding to load pattern 2 (LP2). After redistribution, maximum positive bending moments may be reduced to a similar level like the moments obtained by CSA A23.3 approximate frame analysis. However, negative bending moments at midspans (Spans 2 and 4) cannot be eliminated by this method (see the diagrams below).

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11-28

________________________________________________________________________ 11.5. a) Find the factored bending moment and shear force distribution. Factored uniform load:

w f = 200kN ⁄ m

Determine the factored bending moments at critical sections. Use the moment coefficients prescribed by A23.3 Cl.9.3.3 (see Figure 10.14). Span

1

2

l(m )

10

10

ln( m )

9.6

9.6

Location

Left

Right

Left

Right

– M f (negative bending)

0

w f l n2 – ---------9

w f l n2 – ---------9

0

-2048

-2048

– M f ( kNm ) +M f (positive bending)

w f l n2 ---------11

w f l n2 ---------11

+M f ( kNm )

1675

1675

b) Design the top and bottom reinforcement. Estimate the effective depth as

d = 830mm

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11-29

and

b = 600mm (given) Positive bending

Set

M r = M f = 1675kNm - Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 1675 × 10 2 2 = 0.0015 ( 25MPa ) ( 600mm ) 830mm – ( 830mm ) – ------------------------------------------------ = 7225mm ( 25MPa ) ( 600mm ) Use 16-25M bars:

A s = 16 × 500 = 8000mm

2

- Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2).

0.2 f c ′ A smin = ------------------ b t h fy

[5.7]

0.2 25MPa 2 = ------------------------------- × 600mm × 900mm = 1350mm 400MPa (Note: b t = b rectangular sections) Since 2

2

A s = 8000mm > 1350mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 8000mm ρ = ------ = -------------------------------------------- = 0.0161 [3.1] bd 600mm × 830mm

Balanced reinforcement ratio:

ρ b = 0.022 ( f c ′ = 25MPa , see Table A.4) Since

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11-30

ρ = 0.0161 < ρ b = 0.022 okay Negative bending

Set

M r = M f = – 2048k Nm (largest negative bending moment) - Find A s (use direct procedure). 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

3.85 × 2048 × 10 2 2 = 0.0015 ( 25MPa ) ( 600mm ) 830mm – ( 830mm ) – ------------------------------------------------ = 9600mm ( 25MPa ) ( 600mm ) Use 14-30M bars:

A s = 14 × 700 = 9800mm

2

- Check the minimum CSA A23.3 tension reinforcement requirement (Cl.10.5.1.2). 2

A smin = 1350mm (same as above) Since 2

2

A s = 9800mm > 1350mm okay - Check the maximum CSA A23.3 tension reinforcement requirement (Cl.10.5.2). 2 As 9800mm ----------------------------------------------- = 0.0197 [3.1] ρ = = bd 600mm × 830mm

Since

ρ = 0.0197 < ρ b = 0.022 okay However, ρ > 0.75ρ b , which is rather large. The designer should consider using a wider beam section or increasing concrete strength (for example, f c ′ = 30MPa . Moment envelope for a typical beam span is shown below.

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11-31

The locations of inflection points can be determined from Figure 10.15.

- Development length ( l d ) (see Table A.5), f c ′ = 25MPa : •

30M top steel: l d = 1400mm

•

25M bottom steel: l d = 925mm

Follow the CSA A23.3 bar cutoff requirements per Checklists 11.1 and 11.2. Rules P-1 and N-1 - actual point of cutoff (Cl.12.10.3): Bar length extensions (beyond the theoretical cut off point) are • 1.3d = 1.3 × 830 = 1080mm (governs) • h = 900mm Rules P-3 and N-3 - continuing reinforcement (Cl.12.10.4): • l d + d = 1400 + 830 = 2230mm (governs)

• l d + 12d b = 1400 + 12 ( 30 ) = 1760mm Rules P-5 and N-5 - inflection points (Cl.12.11.3) • d = 830mm (governs) • 12d b = 12 ( 30mm ) = 360mm

ln 9600 ----- = ----------- = 600mm 16 16 Rebar lengths are summarized in the table below. •

Span Location (-) inflection point

1 Left

2 Right

Left

0.33 l n

0.33 l n

3.2+0.83 =4.03m*

Right

=4.03m

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11-32

Span

1

(-) 50% cut-off

(+) inflection point

(+) 50% cut-off

Top steel Bottom steel

2 0.13 l n

0.13 l n

1.25+1.08 =2.33m

=2.33m

0.07 l n

0.07 l n

0.7m

0.7m

0.2 l n

0.2 l n

1.9m

1.9m

14-30Mx6400 STAG 1700 8-25Mx8700 8-25Mx9900 ALT

8-25Mx8700 8-25Mx9900 ALT

*Check continuing bar: 1.25 + 2.23 < 4.03 3.48 < 4.03 okay A sketch showing the reinforcement arrangement for this beam is presented below.

c) Reinforcement arrangement developed using the stock length approach is summarized below. Span Location Top steel Bottom steel

1 Left

2 Right

Left

Right

14-30Mx7200 STAG 1200 8-25Mx9000 8-25Mx10000 ALT

8-25Mx9000 8-25Mx10000 ALT

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11-33

A sketch showing the reinforcement arrangement for this beam is presented below.

_________________________________________________________________________ 11.6. a) Design the top and bottom reinforcement. Slab thickness:

h = 200mm Effective depth:

15 d = 200 – 20 – ------ ≅ 170mm 2 (assume 20 mm clear cover and 15M bars)

1. Design the top reinforcement (negative bending). Set –

Mr = Mf

= – 48 kNm ⁄ m (bending moment at the face of the support)

Find A s using the direct procedure.

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11-34

2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

mm 2 3.85 × 48 × 10 = 0.0015 ( 25MPa ) ( 1000mm ) 170 – ( 170 ) – ------------------------------------- = 875 -----------25 × 1000 m - Find the required bar spacing. For 15M bars: A b = 200mm

2

(Table A.1)

1000 1000 s ≤ A b × ------------ = 200 ------------ = 228mm [3.28] As 875 - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max =

3h = 3 × 200 = 600mm 500mm

Hence

s max = 500mm (smaller value governs) Since s = 228mm < 500mm set s = 225mm Use 15M@225. - Determine the reinforcement area A s .

2

1000 1000 mm A s = A b × ------------ = 200 × ------------ = 890 -----------s 225 m - Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1). 5

A g = b × h = 1000 × 200 = 2 × 10 mm

2

Since 2

2

mm mm A smin = 0.002A g = 400 ------------ < 890 -----------m m

okay

- Check whether the CSA A23.3 maximum reinforcement requirement is satisfied.

As 890 ρ = ------ = ----------------------------------------------- = 0.0052 bd 1000mm × 170mm and

ρ b = 0.022

( f c ′ = 25MPa , Table A.4)

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11-35

Since ρ = 0.0052 < ρ b okay 2. Design the bottom reinforcement (positive bending). Maximum positive bending moment at midspan:

Mf

+

2

2 wf × l 15 ( 6 ) ----------------------------= – 55kNm ⁄ m = – 55 ≅ 13kNm ⁄ m 8 8

Set

M r = M f = 13kNm ⁄ m (bending moment at the face of the support) Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

3.85 × 13 × 10 mm = 0.0015 ( 25MPa ) ( 1000mm ) 170 – ( 170 ) – ------------------------------------- = 225 -----------25 × 1000 m 2

- Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1). Since

A smin

2

2

mm mm = 400 ------------ > 225 -----------m m

Use 2

mm A s = 400 -----------m

- Find the required bar spacing. For 15M bars: A b = 200mm

2

(Table A.1)

1000 1000 s ≤ A b × ------------ = 200 ------------ = 500mm [3.28] As 400 - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max = 500mm (same as above) Hence, use s = s max = 500 mm Use 15M@500. - Check whether the CSA A23.3 maximum reinforcement requirement is satisfied.

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11-36

Since the minimum reinforcement requirement governs, there is no need to check this requirement. Note that, in this case, alternative CSA A23.3 minimum reinforcement requirement (Cl.10.5.1.1) could be used (see Section 5.6.1 for more details). In order to satisfy this criterion, the designer should ensure that

M r ≥ 1.2M cr where fr Ig M cr = --------yt

(A23.3 Eq.10-3)

[5.6] [4.1]

f r = 0.6λ f c ′ (MPa)

(A23.3 Eq.8-3)

[2.1]

λ = 1.0 (normal-density concrete) 200 h y t = --- = --------- = 100mm 2 2 3

3

b×h 1000 × 200 8 4 I g = --------------- = ----------------------------- = 6.7 ×10 mm 12 12 Therefore, the cracking moment can be calculated from equation (4.3) as 8

4

( 0.6 × 1.0 × 25MPa ) × ( 6.7 × 10 mm ) M cr = ----------------------------------------------------------------------------------------------------- = 20kNm 100mm Check the minimum reinforcement requirement: 1.2M cr = 1.2 × 20kNm = 24kNm Therefore, the minimum moment resistance this section should be designed for is M r = 1.2M cr = 24kNm When this M r value is substituted into equation (5.4), the corresponding reinforcement area is 2

mm A s = 421 -----------m This reinforcement area is slightly larger than A smin value obtained using A23.3 Cl.7.8.1, however the latter value governs. b) Detail the flexural reinforcement by applying 50% bar cutoff where permitted. Bottom reinforcement: Since the area of reinforcement is at the minimum, it is not possible to cut off reinforcement. Top reinforcement: 50% bars to be cut off Find the reinforcement area corresponding to 50% cut offs: Copyright © 2006 Pearson Education Canada Inc.

11-37

2

2

A s = ( 890mm ⁄ m ) × 0.5 = 445mm ⁄ m Find the corresponding moment resistance M r50 :

2 φs fy As 0.85 × 400MPa × 445mm a = --------------------- = ------------------------------------------------------------------------------ = 12mm [3.12] 0.8 × 0.65 × 25MPa × 1000mm α 1 φ c f c ′b

a M r50 = φ s f y A s d – --2

[3.14]

12 2 = 0.85 × 400MPa × 445mm 170 – ------ ≅ 25kNm ⁄ m 2 Note that

M f = 25kNm ⁄ m > 1.2M cr = 24kNm ⁄ m however the two values are rather close, that is, 50% bar cut off corresponds to the minimum reinforcement required for this slab.

Shear force at the left support:

wf × l ( 15kN ⁄ m ) × 6m V f = ------------- = ----------------------------------------- = 45kN ⁄ m 2 2 Find the distance x from the left support where

M x = M r50 = – 25 kNm (negative bending) 2

(x) M x = – 55 + 45 ( x ) – 15 ---------- = – 25 2 or 2

7.5x – 45x + 30 = 0 Copyright © 2006 Pearson Education Canada Inc.

11-38

2

45 ± 45 – 4 ( 25 ) ( 30 ) x = -------------------------------------------------------2 ( 25 ) The solution to this quadratic equation is

x = 0.76m ≅ 0.8m (use smaller value) - Development length ( l d ) (see Table A.5), f c ′ = 25MPa : 15M top steel: l d = 600mm Follow the CSA A23.3 bar cutoff requirements per Checklist 11.2. Rule N-1 - actual point of cutoff (Cl.12.10.3): Bar length extensions (beyond the theoretical cut off point) are • 1.3d = 1.3 × 170 = 221mm (governs) • h = 200mm

Rule N-3 - continuing reinforcement (Cl.12.10.4): • l d + d = 600 + 170 = 770mm

•

l d + 12d b = 600 + 12 ( 15 ) = 780mm (governs)

Rule N-5 - inflection points (Cl.12.11.3) • d = 170mm • 12d b = 12 ( 15mm ) = 180mm

•

l n 5.7m ----- = ------------ = 356mm ≅ 360mm (governs) 16 16

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11-39

Rebar arrangement is summarized on the sketch below.

Note that the top reinforcement may be changed to 15M@225x3600 STAG 900 in order to use stock lengths. ________________________________________________________________________ 11.7. a) Redesign the positive and negative reinforcement using 50% bar cutoffs.

1. Design the top reinforcement (negative bending). Set –

Mr = Mf

= – 39k Nm ⁄ m (bending moment at the face of the support)

Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

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11-40

6

2

mm 2 3.85 × 39 × 10 = 0.0015 ( 25MPa ) ( 1000mm ) 170 – ( 170 ) – ------------------------------------- = 700 -----------25 × 1000 m - Find the required bar spacing. For 15M bars: A b = 200mm

2

(Table A.1)

1000 1000 s ≤ A b × ------------ = 200 ------------ = 285mm [3.28] As 700 - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max =

3h = 3 × 200 = 600mm 500mm

Hence

s max = 500mm (smaller value governs) Since s = 285mm < 500mm set s = 275mm Use 15M@275. - Determine the reinforcement area A s .

2

1000 1000 mm A s = A b × ------------ = 200 × ------------ = 730 -----------s 275 m - Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1). Since 2

2

mm mm A smin = 400 ------------ < 730 -----------m m

okay

- Check whether the CSA A23.3 maximum reinforcement requirement is satisfied.

As 730 ρ = ------ = ----------------------------------------------- = 0.0043 bd 1000mm × 170mm and

ρ b = 0.022

( f c ′ = 25MPa , Table A.4)

Since ρ = 0.0043 < ρ b okay 2. Design the bottom reinforcement (positive bending).

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11-41

Maximum positive bending moment at midspan:

Mf

+

2

2 wf × l 15 ( 6 ) = ---------------- – 46kNm ⁄ m = = ---------------- – 46 = 22kNm ⁄ m 8 8

Set

M r = M f = 22kNm ⁄ m (bending moment at the face of the support) Find A s using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

6

2

mm 2 3.85 × 22 × 10 = 0.0015 ( 25MPa ) ( 1000mm ) 170 – ( 170 ) – ------------------------------------- = 385 -----------25 × 1000 m - Check the CSA A23.3 minimum reinforcement requirement (Cl.7.8.1). Since

A smin

2

2

mm mm = 400 ------------ > 385 -----------m m

Use 2

mm A s = 400 -----------m

- Find the required bar spacing. For 15M bars: A b = 200mm

2

(Table A.1)

1000 1000 s ≤ A b × ------------ = 200 ------------ = 500mm [3.28] As 400 - Check the CSA A23.3 maximum bar spacing requirement (Cl.7.4.1.2).

s max = 500mm (same as above) Hence, use s = s max = 500 mm Use 15M@500. - Check whether the CSA A23.3 maximum reinforcement requirement is satisfied. Since the minimum reinforcement requirement governs, there is no need to check this requirement.

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11-42

Bottom reinforcement: since the area of reinforcement is at the minimum, it is not possible to cut off reinforcement. Top reinforcement: 50% bars to be cut off Find the reinforcement area corresponding to 50% cut offs: 2

2

A s = ( 730mm ⁄ m ) × 0.5 = 365mm ⁄ m Find the corresponding moment resistance M r50 :

2 φs fy As 0.85 × 400MPa × 365mm a = --------------------- = ------------------------------------------------------------------------------ = 10mm [3.12] 0.8 × 0.65 × 25MPa × 1000mm α 1 φ c f c ′b

a M r50 = φ s f y A s d – --2

[3.14]

10 2 = 0.85 × 400MPa × 365mm 170 – ------ ≅ 20kNm ⁄ m 2 Since

M f = 20kNm ⁄ m < 1.2M cr = 24kNm ⁄ m 50% bar cut off would result in the reinforcement area below the CSA A23.3 minimum reinforcement required for this slab, hence there will be no bar cut offs in this case. Rebar arrangement is summarized on the sketch below.

b) Difference between the amount of positive moment reinforcement determined in Problems 11.6 and 11.7. There is an approximately 20% decrease in the amount of top reinforcement when the redistributed bending moment values are used in the design instead of the bending moments determined from linear elastic analysis in Problem 11.6; the corresponding reinforcement areas are 730 and 890 mm2/m.

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11-43

The amount of bottom reinforcement is same in both cases. This is due to the fact that positive moment obtained from the linear elastic analysis is rather small and so the bottom reinforcing is governed by CSA A23.3 minimum reinforcement requirement. c) Bending moment diagram to be used when the design objective is to use the least amount of reinforcement: Based on the discussion from part b), redistributed bending moments should be used when the design requires the use of least amount of reinforcement.

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11-44

Chapter 12 - Solutions _________________________________________________________________________ 12.1. a) Footing thickness is governed by the shear design criteria. b) The amount of flexural reinforcement is governed by the flexural requirements. c) The allowable soil pressure governs the footing plan dimensions (width and length). _________________________________________________________________________ 12.2. a) Determine the footing plan dimensions and the soil pressure.

P DL = 500kN P LL = 500kN q all = 200kPa Service Load ( P s )

P s = P DL + P LL = 500 + 500 = 1000kN Footing plan area ( A ):

Ps 1000kN 2 A ≥ -------- = -------------------- = 5.0m q all 200kPa

[12.1a]

A square footing plan has been specified, hence

b≥ A =

2

5.0m = 2.24m ≅ 2.3m

Since 2

2

2

A = b = ( 2.3m ) = 5.29m > 5.0m

2

okay

The factored soil pressure:

P 1.25P DL + 1.5P LL q f = -----f = -------------------------------------------A A

[12.2]

1.25 ( 500kN ) + 1.5 ( 500kN ) = -------------------------------------------------------------------≅ 260kPa 2 5.29m b) Consider the effect of surcharge. Net allowable soil pressure:

q allnet = q all – ( q s + γw h s ) [12.5]

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12-1

where

q s = 4.8kPa

specified service load

h s = 300mm + 250mm = 550mm = 0.55m kN γ w = 24 ------3- (concrete weight density) m kN q allnet = 200kPa – 4.8kPa + 24.0 ------3- × 0.55m = 182kPa m Ps 2 1000kN A ≥ -------- = -------------------- = 5.5m q all 182kPa For a square footing:

b≥ A =

2

5.5m = 2.35m ≅ 2.4m

Since 2

2

2

A = b = ( 2.4 ) = 5.76m > 5.5m

2

okay

The factored soil pressure:

Pf 1.25P DL + 1.5P LL 1.25 ( 500kN ) + 1.5 ( 500kN ) q f = ----- = -------------------------------------------- = -------------------------------------------------------------------= 239kPa 2 A A 5.76m c) Comment on the effect of the surcharge. When the soil surcharge is considered, the design requires a 4% increase in plan dimensions (from 2.3 m to 2.4 m) and approximately 8% decrease in the factored soil pressure (from 260 kPa to 239 kPa). This difference is not significant from the design perspective. _________________________________________________________________________ 12.3. 1. Check whether the footing plan dimensions are okay. In this case, footing width is given as

l = 1.5m and the length is

b = 1.0m Footing plan area:

A = l × b = 1.5m × 1m = 1.5m

2

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12-2

Soil pressure under the service load is

Ps P DL + P LL 140kN + 150kN - = 193kPa q = ----- = -------------------------- = --------------------------------------2 A A 1.5m Since

q = 193kPa < q all = 200kPa okay 2. Find the factored soil pressure ( q f ) .

P f = 1.25P DL + 1.5P LL = 1.25 × ( 140kN ⁄ m ) + 1.5 ( 150kN ⁄ m ) = 400kN ⁄ m The factored soil pressure is equal to

Pf 400kN - = 267kPa ≅ 270kPa q f = ----- = ---------------2 A 1.5m This value will be used in the design. 3. Determine the required footing thickness based on the shear design requirements. cover = 75mm (Table A.2) Try the minimum thickness

h = 250mm d b = 20mm

(20M bars)

db 20 d = h – cover – ----- = 250mm – 75mm – ------ = 165mm 2 2

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12-3

The factored shear force at the critical section

l–t V f = q f --------- – d × b 2 1.5m – 0.3m V f = 270kPa ------------------------------- – 0.165m × 1m = 117.5kN ⁄ m 2 Find the concrete shear resistance ( V c ) .

dv =

0.9d = 0.9 × 165 = 149mm 0.72h = 0.72 × 250 = 180mm ← governs

d v = 180mm β = 0.21 for h < 350mm (CSA A23.3 Cl.11.3.6.2) V c = φ c λβ f c ′b w d v (A23.3 Eq. 11-6)

[6.12]

= 0.65 × 1 × 0.21 25MPa × 1000mm × 180mm = 122.8kN ⁄ m where

b w = b = 1000mm Since

V f = 117.5kN ⁄ m < V c = 122.8kN ⁄ m shear reinforcement is not required (A23.3 Cl.11.2.8.1). 4. Determine the required flexural reinforcement based on the flexural design requirements.

l–t M f = q f --------2

l–t --------- b 4

1.5 – 0.3 = ( 270kPa ) --------------------2

1.5 – 0.3 --------------------- ( 1.0m ) = 48.6kNm ⁄ m 4

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12-4

Find the required reinforcement area using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

2 3.85 ( 48.6 × 10 Nmm ) 2 = 0.0015 ( 25MPa ) ( 1000mm ) 165 – ( 165 ) – ------------------------------------------------------- = 919mm ⁄ m ( 25MPa ) ( 1000mm )

5. Confirm that the minimum reinforcement requirement is satisfied (CSA A23.3 Cl.7.8.1). 3

A g = h × 1000 = 250 × 1000 = 250 × 10 mm 3

2

2

2

A smin = 0.002A g = 0.002 × ( 250 × 10 mm ) = 500mm ⁄ m Since 2

2

A s = 919mm ⁄ m > 500mm ⁄ m okay 6. Confirm that the maximum reinforcement requirement is satisfied (CSA A23.3 Cl.10.5.2).

f c ′ = 25MPa ρ b = 0.022 (Table A.4) 2 As 919mm ρ = ------ = ----------------------------------------------- = 0.006 < 0.022 okay bd 1000mm × 165mm

7. Determine the required bar spacing for flexural reinforcement. 20M bars A b = 300mm

2

(Table A.1)

1000 1000mm 2 - = 326mm ≅ 300mm [3.29] s ≤ A b ------------ = ( 300mm ) ----------------------------2 As 919mm ⁄ m Check the maximum permitted bar spacing ( s max ) according to A23.3 Cl.7.4.1.2 and 13.10.4.

s max ≤

3h = 3 × 250 = 750mm 500mm ← governs

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12-5

Since

s = 300mm < 500mm okay Use 20M @ 300. 8. Design the minimum reinforcement in the longitudinal direction (A23.3 Cl.7.8.1 and 7.8.3).

A g = l × t = 1500mm × 250mm = 375000mm 2

2

A smin = 0.002A g = 0.002 × 375000mm = 750mm

2

2

Use 3-20M bars ( A b = 300mm ) , so

A s = 3 × 300 = 900mm

2

The maximum permitted bar spacing ( s max ) is equal to the lesser of

s max ≤

5h = 5 × 250 = 1250mm 500mm ← governs

hence

s max = 500mm 9. Provide a design summary.

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12-6

__________________________________________________________________________ 12.4. 1. Determine the footing width.

P DL = 100kN ⁄ m P LL = 100kN ⁄ m P s = P DL + P LL = 100 + 100 = 200kN ⁄ m b = 1.0m (unit strip) q all = 200kPa Ps 200kN ⁄ m l ≥ ----------------- = ------------------------------------------- = 1.0m b × q all 1.0m × ( 200kPa ) Use l = 1.0m

2. Determine the factored soil pressure ( q f ) .

A = l × b = 1.0m × 1.0m = 1m

2

Pf 1.25P DL + 1.5P LL 1.25 × 100 + 1.5 × 100 - = 275kPa ≅ 280kPa q f = ----- = -------------------------------------------- = -----------------------------------------------------2 A A 1.0m Use

q f = 280kPa 3. Determine the required footing thickness (h) based on the shear design requirements (A23.3 Cl.11.3.4). cover = 75mm (Table A.2) Try minimum footing thickness

h = 250mm d b = 15mm (15M bars)

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12-7

db 15 d = h – cover – ----- = 250 – 75 – ------ ≅ 165mm 2 2 The factored shear force at the critical section:

l–t V f = q f --------- – d × b 2 1.0 – 0.2 = 280kPa --------------------- – 0.165 × 1.0m ≅ 66.0kN ⁄ m 2

Find the concrete shear resistance ( V c ) . Find d v as

dv =

0.9d = 0.9 × 165mm = 149mm 0.72h 0.72 × 250mm 180mm

so,

d v = 180mm (larger value governs) β = 0.21 for h < 350mm (CSA A23.3 Cl.11.3.6.2) V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6)

[6.12]

= 0.65 × 1 × 0.21 25MPa × 1000mm × 180mm = 122.8kN ⁄ m

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12-8

Since

V f = 66kN ⁄ m < V c = 122.8kN ⁄ m Hence, shear reinforcement is not required (A23.3 Cl.11.2.8.1). 4. Determine the required flexural reinforcement based on the flexural design requirements.

l–t M f = q f --------2

l–t – 0.2 --------- b = ( 280kPa ) 1--------------4 2

1 – 0.2 ---------------- ( 1.0m ) = 22.4kNm ⁄ m 4

Find the required reinforcement area using the direct procedure. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

2 3.85 × ( 22.4 × 10 Nmm ) 2 = 0.0015 ( 25MPa ) ( 1000mm ) 165 – ( 165 ) – ------------------------------------------------------------- = 405mm ⁄ m 25MPa × 1000mm

5. Confirm that the minimum reinforcement requirement is satisfied (A23.3 Cl.7.8.1). 3

A g = h × 1000mm = 250mm × 1000mm = 250 × 10 mm

2

Then,

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12-9

3

2

2

A smin = 0.002A g = 0.002 × ( 250 × 10 mm ) = 500mm ⁄ m Since 2

2

A s = 405mm ⁄ m < 500mm ⁄ m use 2

A s = 500mm ⁄ m 6. Confirm that the maximum reinforcement requirement is satisfied (A23.3 Cl.10.5.2). In this case, the maximum reinforcement requirement is satisfied by default since the amount of reinforcement is governed by the minimum reinforcement requirement. 7. Determine the required bar spacing for flexural reinforcement. 15M bars: A b = 200mm

2

(Table A.1)

1000 1000mm 2 - = 400mm s ≤ A b ------------ = ( 200mm ) × ----------------------------2 As 500mm ⁄ m

[3.29]

Use

s = 400mm According to A23.3 Cl.7.4.1.2 and 13.10.4

s max ≤

3h = 3 × 250 = 750mm 500mm ← governs

Since

s = 400mm < 500mm okay Use 15M@400. 8. Determine the minimum reinforcement in the longitudinal direction (A23.3 Cl.7.8.1 and 7.8.3). 3

2

3

2

A g = h × l = 250 × 1000 = 250 × 10 mm

A smin = 0.002A g = 0.002 × ( 250 × 10 mm ) = 500mm

2

2

Use 3-15M bars ( A b = 200mm ):

A s = 3 × 200 = 600mm

2

The maximum bar spacing

s max ≤

5h = 5 × 250 = 1250mm 500mm ← governs

s max = 500mm

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12-10

9. Provide a design summary.

__________________________________________________________________________ 12.5. a) Check whether the footing designed in Problem 12.4 is able to resist the design loads, when the allowable soil pressure is 200 kPa.

Find the soil pressure distribution under the footing. Eccentricity e :

l 1000 e = --- – 200mm = ------------ – 200 = 300mm = 0.3m 2 2 Service load P s :

P s = P DL + P LL = 100kN ⁄ m + 100kN ⁄ m = 200kN ⁄ m Service moment M s :

M s = P s × e = ( 200kN ⁄ m ) × 0.3m = 60kNm ⁄ m Footing area A :

A = b × l = 1m × 1m = 1m

2

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12-11

Soil pressure due to axial load ( q p )

Ps 200kNq p = ----- = ---------------= 200kPa 2 A 1m Pressure due to bending moment ( q M )

M q M = ----S 2

2

b×l 1m × ( 1m ) 3 S = ------------- = ----------------------------- = 0.167m 6 6 60kNm ⁄ m - = 359kPa q M = -------------------------3 0.167m q max = q P + q M = 200 + 359 = 559kPa (compression) q min = q P – q M = 200 – 359 = – 159kPa (tension) Since

q max = 559kPa > q all = 200kPa it can be concluded that this footing cannot carry the design loads when subjected to eccentric loading.

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12-12

b) If the answer to the question posed in part a) is negative, find the maximum dead load and the maximum live load which this footing is able to resist. Consider a triangular soil pressure distribution - boundary case ( q min = 0 ).

2 × P smax q max = ---------------------b×l

[12.18]

Set

q max = q all = 200kPa , thus

q max × b × l 200kPa × 1m × 1m P smax = ---------------------------- = ------------------------------------------------ = 100kN ⁄ m 2 2 P s = P DL + P LL When

P DL = P LL (a design assumption) then

100kN ⁄ m P DLmax = P LLmax = ------------------------- = 50kN ⁄ m 2

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12-13

_________________________________________________________________________ 12.6. a) Determine the required footing dimensions. i) Consider the boundary case:

l e = --6

and q min = 0

l e = --- – 0.3m 3 1. Find the footing width ( l ). Service load:

P DL = 100kN ⁄ m P LL = 100kN ⁄ m P s = P DL + P LL = 100 + 100 = 200kN ⁄ m b = 1000mm = 1m (unit strip) Maximum soil pressure: Since

2 × Ps q max = -------------- ≤ q allow = 200kPa b×l

[12.18]

it follows that

2 × Ps 2 × 200kN ⁄ m l ≥ ----------------------- = -------------------------------------- = 2m b × q allow 1.0m × 200kPa Load eccentricity:

l 2m e = --- – 0.3m = -------- – 0.3m = 0.37 m 3 3

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12-14

2. Determine the factored soil pressure ( q f ) . Factored axial load:

P f = 1.25P DL + 1.5P LL = 1.25 × 100 + 1.5 × 100 = 275kN ⁄ m Factored maximum soil pressure:

2P f 2 × 275kN q fmax = ---------- = -------------------------- = 275kPa b×l 1m × 2m

[12.18]

3. Determine the required footing thickness based on the shear design requirements (A23.3 Cl.11.3.4). Critical section for one-way shear: Find a (see the sketch below).

a = 2000 – 200 – 200 = 1600mm = 1.6m Find the soil pressure at the critical section.

a 1.6m q f = q fmax --- = 275kPa ------------ = 220kPa l 2.0m Find the factored shear force.

qf × a 220kPa × 1.6m V f = -------------- × b = -------------------------------------- × 1.0m = 176kN ⁄ m 2 2

Find the concrete shear resistance ( V c ) .

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12-15

Assume footing thickness h = 350mm . Effective depth:

d ≅ h – 100 = 250mm dv =

0.9d = 0.9 × 250mm = 225mm 0.72h 0.72 × 350mm 252mm

so,

d v ≅ 250mm (larger value governs) β = 0.21 for h ≤ 350mm (A23.3 Cl.11.3.6.2) b w = b = 1000mm (unit strip) V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6)

[6.12]

= 0.65 × 1.0 × 0.21 25MPa × 1000mm × 250mm = 170.6kN ⁄ m Since

V c = 170.6kN ⁄ m < V f = 176kN ⁄ m either the shear reinforcement needs to be provided, or the footing thickness need to be increased. Let us increase the footing thickness to

h = 500mm so

d = h – 100 = 400mm dv =

0.9d = 0.9 × 400mm = 360mm 0.72h 0.72 × 500mm 360mm

so,

d v = 360mm 230 230 β = ----------------------- = --------------------------- = 0.17 1000 + d v 1000 + 360 V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6)

(A23.3 Eq.11-9) [6.13] [6.12]

= 0.65 × 1.0 × 0.17 25MPa × 1000mm × 360mm = 199kN ⁄ m Since

V c = 199kN ⁄ m > V f = 176kN ⁄ m shear reinforcement is not required. Use h = 500mm .

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12-16

4. Sketch the footing section and dimensions.

ii) Consider uniform soil pressure distribution.

Set

b = 1.0m (unit strip). 1. Find the footing width ( l ).

P s = P DL + P LL = 100 + 100 = 200kN ⁄ m Since

Ps Ps q max = ----- = ---------- [12.18] b×l A and

q max ≤ q all = 200kPa it follows that

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12-17

Ps ---------= q all b×l The footing width can be determined as

Ps 200kN l ≥ ----------------- = -------------------------------------- = 1.0m b × q all 1.0m × 200kPa therefore, use

l = 1m 2. Determine the factored soil pressure ( q f ) .

Pf 1.25 × P DL + 1.5P LL 1.25 × 100 + 1.5 × 100 q f = ----- = -------------------------------------------------- = ------------------------------------------------------- = 275kPa b×l 1.0m × 1.0m A 3. Find the required footing thickness based on the shear design requirements (A23.3 Cl.11.3.4). Critical section for one-way shear: Find a (see the sketch below).

a = 1000 – 200 – 200 = 600mm = 0.6m Find the factored shear force.

V f = ( q f × a ) × b = ( 275kPa × 0.6m ) × 1.0m = 165kN ⁄ m

Try footing thickness h = 500mm . Estimate d as Copyright © 2006 Pearson Education Canada Inc.

12-18

d ≅ h – 100 = 400mm dv =

0.9d = 0.9 × 400mm = 360mm 0.72h 0.72 × 500mm 360mm

so,

d v = 360mm 230 230 β = ----------------------- = --------------------------- = 0.17 (A23.3 Eq. 11-9) [6.13] 1000 + d v 1000 + 360 Find V c .

V c = φ c λβ f c ′b w d v (A23.3 Eq.11-6)

[6.12]

= 0.65 × 1.0 × 0.17 25MPa × 1000mm × 360mm = 199kN ⁄ m Since

V c = 199kN ⁄ m > V f = 165kN ⁄ m it follows that the shear reinforcement is not required. 4. Sketch the footing section and dimensions.

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b) Sketch the bending moment diagrams for the footing and the wall. Bending moment diagram - triangular soil pressure distribution

Design loads:

M max

2

2

q ( 1.7m ) 234 ( 1.7m ) = ----------------------- = ----------------------------- = 113 kNm ⁄ m 6 6

P f = 275kN ⁄ m e = 0.37m Pf × e ( 275kN ⁄ m ) ( 0.37m ) T = – C = -------------- = --------------------------------------------------- ≅ 34.0kN ⁄ m h ( 3.0m ) Bending moment diagram - uniform soil pressure distribution

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12-20

2

M max

2 q f ( 0.7m ) 275 ( 0.7m ) = ------------------------ = ----------------------------- = 67 kNm ⁄ m 2 2

P f = 275kN ⁄ m l e = --- – 0.3m = 0.5 – 0.3 = 0.2m 2 P f × e ( 275kN ⁄ m ) ( 0.2m ) T = – C = -------------- = ------------------------------------------------ = 18.3kN ⁄ m h 3.0m _________________________________________________________________________ 12.7. Based on the design performed in Problem 12.6, a triangular soil pressure distribution results in larger footing plan dimensions (width of 2 m) as compared to the uniform soil pressure distribution (width of 1 m). Also, bending moment diagrams presented in Problem 12.6c) show that the triangular soil pressure distribution results in larger maximum bending moment value ( 113kN ⁄ m ) as compared to the uniform soil pressure distribution ( 67kN ⁄ m ) . As a result, the forces developed in the slabs in case of triangular stress distribution ( 34kN ⁄ m ) are larger than the corresponding forces developed when the uniform stress distribution is assumed ( 18kN ⁄ m ) . __________________________________________________________________________ 12.8. 1. Determine the footing plan dimensions.

q all = 400kPa P DL = 1000kN P LL = 500kN P s = P DL + P LL = 1000 + 500 = 1500kN P s 1500kN 2 A ≥ ------- = -------------------- = 3.75m q all 400kPa b≥ A =

2

3.75m = 1.94m ≅ 2m

2. Determine the factored soil pressure ( q f ) . 2

2

A = b = ( 2m ) = 4m

2

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12-21

Pf 1.25P DL + 1.5P LL 1.25 × 1000kN + 1.5 × 500kN - = 500kPa q f = ----- = -------------------------------------------- = ------------------------------------------------------------------------2 A A 4m The factored soil pressure ( q f ) will be used for the flexural and shear design.

3. Determine the required footing thickness ( h ) based on the shear design requirements. •

Check two-way shear requirements (A23.3 Cl.13.3.3 and 13.3.4).

Find the factored shear force.

P f = 1.25P DL + 1.5P LL = 1.25 × 1000 + 1.5 × 500 = 2000kN Set

V f = P f = 2000kN •

Determine the concrete shear resistance for two-way shear ( V c ) .

v c = 0.38λφ c f c ′ = 0.38 ( 1 ) ( 0.65 ) 30MPa = 1.35MPa (A23.3 Eq.13- 7) [12.10] Use

ν c = 1.35MPa for further calculations. Determine the d value based on the two-way shear requirements.

Vc = νc × bo × d

[12.7]

Set

V c = V f = 2000kN Vc 2 2000kN b o d = ------ = ---------------------- = 1.48m 1350kPa νc Shear perimeter ( b o ) :

bo = 4 ( t + d ) where

t = 0.4m (column width)

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12-22

b o d = 4 ( 0.4m + d )d = 1.48m

2

or 2

4d + 1.6d – 1.48 = 0 This quadratic equation can be solved for d as follows 2

– 1.6 ± ( 1.6 ) + 4 × 4 × 1.48 d = ------------------------------------------------------------------------ = 0.44m 2×4 Note that the negative solution has been discarded. The shear perimeter ( b o ) is

b o = 4 × ( t + d ) = 4 ( 0.4m + 0.44m ) = 3.36m

Check the other two ν c criteria.

2 ν c = 1 + ----- 0.19λφ c f c ′ βc

(A23.3 Eq.13-5)

[12.8]

2 = 1 + --- × 0.19 ( 1.0 ) ( 0.65 ) 30MPa = 2.0MPa 1 where

t β c = - = 1.0 t

α s = 4 (interior column)

and

αs d ν c = --------- + 0.19 λφ c f c ′ bo

(A23.3 Eq.13-6)

[12.9]

4 × 0.44m = ------------------------ + 0.19 × 1.0 × 0.65 30MPa = 2.54MPa 3.36m

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12-23

In this case, ν c = 1.35MPa is the smallest value and it governs in the design. Consequently,

d = 0.44m = 440mm Determine the overall footing depth ( h ).

d b = 20mm

20M bars

cover = 75mm (Table A.2)

db 20mm h = d + ----- + cover = 440mm + ---------------- + 75mm = 525mm ≅ 550mm 2 2 Use

h = 550mm then

db 20 d = h – cover – ----- = 550 – 75 – ------ = 465 ≅ 460mm 2 2 •

Check the one-way shear requirements (A23.3 Cl.11.3.4).

Find the factored shear force ( V f ) at the critical section at a distance ( d ) away from the face of the column.

b–t 2 – 0.4 V f = q f b ---------- – d = ( 500kPa ) ( 2.0m ) ---------------- – 0.46 = 340kN 2 2

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Find the concrete shear resistance ( V c ) at the critical section. Find d v as

dv =

0.9d = 0.9 × 460mm = 414mm 0.72h 0.72 × 550mm 396mm

so

d v ≅ 410mm (larger value governs) 230 β = ----------------------1000 + d v

(A23.3 Eq. 11-9)

[6.13]

230 = --------------------------- = 0.163 ≅ 0.16 1000 + 410 V c = φ c λβ f c ′b w d v

(A23.3 Eq. 11-6) [6.12]

= 0.65 × 1.0 × 0.16 30MPa × ( 2000mm ) ( 410mm ) = 467kN Since

V c = 467kN > V f = 340kN it follows that the shear reinforcement is not required. 4. Determine the required flexural reinforcement based on the flexural design requirements. Find the factored bending moment at the critical section (face of the column), as follows:

b–t M f = q f ---------2

b–t – 0.4 ---------- b = ( 500kPa ) 2.0 --------------------4 2

2.0 – 0.4 --------------------- ( 2.0m ) = 320kNm 4

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Find the required area of flexural reinforcement using the direct procedure as 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 × ( 320 × 10 ) 2 2 = 0.0015 ( 30MPa ) ( 2000mm ) 460mm – ( 460 ) – ----------------------------------------------- = 2060mm 30MPa × 2000mm 5. Confirm that the minimum reinforcement requirement is satisfied (A23.3 Cl.7.8.1). 6

A g = b × h = 2000mm × 550mm = 1.1 × 10 mm 6

2

2

A smin = 0.002A g = 0.002 × ( 1.1 × 10 mm ) = 2200mm

2

Since 2

A s = 2060mm < A smin = 2200mm

2

Use 2

A s = 2200mm (minimum reinforcement requirement governs) Use 8-20M bars:

A s = 8 × 300 = 2400mm

2

6. Confirm that the CSA A23.3 bar spacing requirements are satisfied. Consider 100 mm end spacing

2000 – 2 × 100 s = ------------------------------------ = 257mm 7 Maximum bar spacing according to A23.3 Cl.7.4.1.2 and 13.10.4:

s max =

3h = 3 × 550 = 1650mm 500mm

Since

s max = 500mm > 257mm okay 7. Confirm that the maximum reinforcement requirement is satisfied (A23.3 Cl.10.5.2). This requirement is satisfied by default - minimum reinforcement requirement governs. 8. Check the development length for the flexural reinforcement (A23.3 Cl.12.2.3 and 12.2.4).

fy l d = 0.45k 1 k 2 k 3 k 4 ---------- d b fc ′

[9.2]

400MPa = 0.45 × 1 × 1 × 1 × 0.8 ----------------------- × 20mm = 526mm 30MPa

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where (see Table 9.1)

k 1 = k 2 = k 3 = 1 (bottom bar, regular uncoated reinforcement) k 4 = 0.8

(20M bar)

Consider a 100 mm cover at the edge of the footing. The available length of the reinforcing bars from the face of the footing to the face of the column is equal to

b–t 2000 – 400 l = ---------- – 100m = --------------------------- – 100 = 700mm 2 2 Since

l = 700mm > 526mm okay it follows that the bars can develop their full strength.

9. Provide a design summary.

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__________________________________________________________________________ 12.9. a) Determine the footing capacity P r for the scenarios i), ii) and iii). Scenario i)

Stress distribution in an eccentrically loaded footing:

Mr Pr q max = ------- + ----- ≤ q f A S where

Mr = Pr × e e = 0.1m 2

3

( 2m ) b(b ) 3 S = ------------- = --------------- = 1.33m 6 6 A = b × b = 2 × 2 = 4m

2

Find load factor α corresponding to P DL = 1000kN and P LL = 500kN :

1.25P DL + 1.5P LL 1.25 × 1000 + 1.5 × 500 α = -------------------------------------------- = ---------------------------------------------------------- = 1.33 P DL + P LL 1000 + 500 Since

q all = 400kPa it follows that

q f = αq all = 1.33 × 400 = 532kPa So,

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Pr × e Pr 0.1m e 1 1 q max = -------------- + ----- = P r --- + ---- = P r -----------------3- + ---------2- = 0.325P r S S A A 1.33m 4m and

q max ≤ q f = 532kPa it follows that

532 P r ≤ ------------- = 1637kN 0.325 So, the footing capacity for scenario i) is

P r = 1637kN Scenario ii) The overall footing depth ( h ) is reduced by 75 mm, that is,

h = 550 – 75 = 475mm and

d = h – 100 = 375mm

Footing depth is governed by the shear design requirements. Therefore, the shear design performed in Problem 12.8 needs to be revisited. Check two-way shear requirements (A23.3 Cl.13.3.3 and 13.3.4).

v c = 0.38λφ c f c ′ (A23.3 Eq.13- 7)

[12.10]

= 0.38 × 1.0 × 0.65 30MPa = 1.35 MPa = 1350kPa Shear perimeter:

b o = 4 ( t + d ) = 4 ( 0.4m + 0.375m ) = 3.1m Concrete shear resistance for two-way shear:

Vc = νc × bo × d

[12.7]

= 1350kPa × 3.1m × 0.375m = 1569kN Copyright © 2006 Pearson Education Canada Inc.

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Since

V r = V c = 1569kN (no shear reinforcement) The footing capacity corresponding to the two-way shear is

P r = V r ≅ 1570kN Check the one-way shear requirements (A23.3 Cl.11.3.4).

q f = 532kPa Find V f .

b–t 2 – 0.4 V f = q f b ---------- – d = ( 532kPa ) ( 2.0m ) ---------------- – 0.375 = 452kN 2 2 Find V c . First, find d v as

dv =

0.9d = 0.9 × 375mm = 337mm 0.72h 0.72 × 475mm 342mm

so,

d v ≅ 340mm (larger value governs) 230 230 β = ----------------------- = --------------------------- = 0.17 1000 + d v 1000 + 340 V c = φ c λβ f c ′b w d v

[6.13]

(A23.3 Eq. 11-6) [6.12]

= 0.65 × 1.0 × 0.17 30MPa × ( 2000mm ) ( 340mm ) = 412kN Since

V f = 452kN > V c = 412kN Set

V f = V c = 412kN Find the factored soil pressure as

Vf 412kN q f = ---------------------------- = ------------------------------------------------------ = 485kPa 2 – 0.4 b–t ( 2.0 ) ---------------- – 0.375 b ---------- – d 2 2 Since

Pf q f = ----A so

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2

P f = q f × A = ( 485kPa ) ( 4m ) = 1940kN Therefore, the footing capacity corresponding to the one-way shear is

P r = P f = 1940kN Since the footing capacity corresponding to the two-way shear (1570 kN) is less than the capacity corresponding to the one-way shear (1940 kN), the former value governs, that is,

P r = 1570kN Scenario iii)

h = 550mm (Problem 12.8) d = 450mm (Problem 12.8) d = 450 – 50 = 400mm Check shear design requirements. Check the two-way shear requirements (A23.3 Cl.13.3.3 and 13.3.4)

v c = 1350kPa Shear perimeter:

b o = 4 ( t + d ) = 4 ( 0.4m + 0.4m ) = 3.2m Vc = νc × bo × d

[12.7]

= 1350kPa × 3.2m × 0.4m = 1728kN So,

V r = V c = 1728kN (no shear reinforcement) Therefore, the footing capacity corresponding to scenario iii) is

P r = 1730kN Check the one-way shear requirements (A23.3 Cl.11.3.4).

q f = 532kPa Find V f .

b–t 2 – 0.4 V f = q f × b × ---------- – d = ( 532kPa ) ( 2.0m ) ---------------- – 0.4 = 426kN 2 2 Find V c . First, find d v as

dv =

0.9d = 0.9 × 400mm = 360mm 0.72h 0.72 × 550mm 396mm

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so

d v = 396mm 230 230 β = ----------------------- = --------------------------- = 0.16 [6.13] 1000 + d v 1000 + 396 V c = φ c λβ f c ′b w d v

(A23.3 Eq. 11-6) [6.12]

= 0.65 × 1.0 × 0.16 30MPa ( 2000mm ) ( 396mm ) = 451kN Since

V r = V c = 451kN (no shear reinforcement) and

V f = 426kN < V r = 451kN Use

V f = 451kN (maximum factored shear force for this footing) Find the factored soil pressure as

Vf 451kN q f = ---------------------------- = ----------------------------------------------------- = 564kPa 2 – 0.4 b–t ( 2.0m ) ---------------- – 0.4 b ---------- – d 2 2 Since

Pf q f = ----A it follows that 2

P f = q f × A = ( 564kPa ) ( 4m ) = 2256kN Since the footing capacity corresponding to the two-way shear (1730 kN) is less than the capacity corresponding to the one-way shear (2256 kN), the former value governs, that is,

P r = 1730kN b) Check whether the footing capacity for the construction scenarios in part a) is adequate to withstand the design loads from Problem 12.8. Factored load considered in Problem 12.8:

P f = 1.25P DL + 1.5P LL = 1.25 × 1000 + 1.5 × 500 = 2000kN Scenario i):

P r = 1637kN < P f = 2000kN (deficient by 18%)

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The capacity is not adequate. Scenario ii):

P r = 1570kN < P f = 2000kN (deficient by 22%) The capacity is not adequate. Scenario iii):

P r = 1730kN < P f = 2000kN (deficient by 14%) The capacity is not adequate. __________________________________________________________________________ 12.10. Considering that a designer generally does not have a control over the possible construction inaccuracies, it is recommended to design footings with a reserve capacity. Based on the possible undesirable construction scenarios discussed in Problem 12.9, the designer should consider a 20% reserve in the factored capacity ( P f ) . ____________________________________________________________________________ 12.11.

Find the gravity load resultant. Footing weight: 3

W f = ( 2m × 2m × 0.6m ) ( 24.0kN ⁄ m ) = 57.6kN Soil weight (surcharge):

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3

W s = 2m × 2m × ( t – 0.6 ) × ( 17.0kN ⁄ m ) = 68 ( t – 0.6 ) W = W f + W s = 57.6kN + 68 ( t – 0.6 ) = 16.8 + 68t Find the wind load resultant:

H = 2kPa × 1m × 1m = 2kN Bending moment at the base of the footing due to wind load:

M = H ( 6 + t ) = 2kN ( 6 + t ) Soil pressure at the base of the footing:

M P q max = ----- + ---S A where 2

3

b(l) ( 2m ) 3 S = ------------ = --------------- = 1.33m 6 6 2

A = b × b = ( 2m ) = 4m

2

M = 2(6 + t) P = W = 16.8 + 68t So,

M P 2 ( 6 + t ) 16.8 + 68t q max = ----- + ---- = ------------------- + -----------------------S A 1.33 4 Since

q max ≤ q all = 150kPa it follows that

1.5 ( 6 + t ) + 0.25 ( 16.8 + 68t ) = 150 Finally, t value can be obtained as

t = 7.4m ≅ 7.5m Check overturning resistance

M or F.S. = --------- ≥ 1.5 Mo l b 2.0 M r = W × --- = ( 16.8 + 68t ) --- = ( 16.8 + 68 × 7.5 ) ------- = 527.0kNm 2 2 2 M o = H × ( 6 + t ) = ( 2kN ) ( 6 + 7.5 ) = 27kNm M r 527 F.S. = ------- = --------- = 19.5 » 1.5 okay 27 Mo

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In conclusion, the required foundation depth is

t = 7.5m (note that the depth is governed by the bearing pressure criterion.) __________________________________________________________________________ 12.12.

Footing plan area (based on 2 m square dimension): 2

A = ( 2m ) = 4m

2

1. Find the maximum axial load based on the flexural resistance. Effective depth (assume 75 mm cover and 25M bar size):

d = 400mm – 75 – 25 = 300mm Unit width:

b = 1000mm Reinforcement: 25M@200 Reinforcement area per metre width can be determined as

1000 2 1000mm 2 A s = A b × ------------ = 500 mm × ---------------------- = 2500mm ⁄ m s 200mm Reinforcement ratio:

As 2500 ρ = ------------ = --------------------------- = 0.0083 b×d 1000 × 300 Find the moment resistance. Find a . 2 φs fy As 0.85 ( 400MPa ) ( 2500mm ) a = --------------------- = -------------------------------------------------------------------------- = 65mm [3.12] 0.8 ( 0.65 ) ( 25MPa ) ( 1000mm ) α 1 φ c f c ′b

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2 a 65 M r = φ s A s f y d – --- = 0.85 ( 2500mm ) ( 400MPa ) 300 – ------ = 227kNm ⁄ m [3.14] 2 2

Set

M f = M r = 227kNm ⁄ m and 2

x M f = q f × ----2 where

x = 0.85m so

q f = 628kPa Finally, the corresponding maximum P f value can be determined as 2

P f = q f × A = ( 628 kPa ) × ( 4m ) = 2512kN

2. Find the maximum axial load based on the one-way shear resistance.

dv =

0.9d = 0.9 × 300mm = 270mm 0.72h 0.72 × 400mm 288mm

so,

d v ≅ 290mm (larger value governs) 230 230 β = ----------------------- = --------------------------- = 0.18 1000 + d v 1000 + 290 V c = φ c λβ f c ′b w d v

[6.13]

(A23.3 Eq. 11-6) [6.12]

= 0.65 ( 1 ) ( 0.18 ) 25 ( 2000mm ) ( 290mm ) = 339kN where

b w = 2000mm footing plan dimension

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Set

V f = V c = 339kN and

V f = q f × y × bw where

y = 0.55m so

q f = 308kPa Finally, the corresponding maximum P f value can be determined as 2

P f = q f × A = 308 kPa × 4m = 1232kN

3. Find the maximum axial load based on two-way shear resistance. Find concrete shear resistance based on two-way shear criteria.

v c = 0.38λφ c f c ′ = 0.38 ( 1 ) ( 0.65 ) 25MPa = 1.24MPa (A23.3 Eq.13- 7) [12.10] V c = v o × b o × d = 1.24MPa × 2400mm × 300mm = 892kN where

b o = 4 × ( t + d ) = 4 × ( 0.3m + 0.3m ) = 2.4m Finally, the corresponding maximum P f value can be determined as

P f = V c = 892 kN 4. Find the maximum axial load resistance based on the bearing resistance. Bearing at the column base: 4

2

B rc = 0.85φ c f cc ′A 1 = 0.85 ( 0.65 ) ( 25MPa ) ( 9 × 10 mm ) = 1243kN [12.28] where

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4

2

A 1 = 300mm × 300mm = 9 ×10 mm column cross-sectional area 6

2

A 2 = 2000mm × 2000mm = 4 × 10 mm footing plan area Since

A2 ------ = A1

6

2

× 10 mm 4----------------------------- = 6.7 > 2 4 2 9 ×10 mm

The footing bearing resistance is equal to 4

B rfmax = 1.7 φ c f cf ′A 1 = 1.7 × 0.65 × 25 × ( 9 × 10 ) = 2486kN The smaller of the column and footing resistances governs, that is,

P f = B rc = 1243kN 5. Find the governing axial load value. The following values should be considered:

P f = 2512kN (based on the flexural resistance) P f = 1232kN (based on one-way shear resistance) P f = 892kN (based on two-way shear resistance) P f = 1243kN (based on the bearing resistance) The maximum factored axial load is governed by two-way shear (smallest value) and so P f = 892kN ≅ 900kN _______________________________________________________________________ 12.13. 1. Perform the load analysis. Factored axial load (NBC 2005 Table 4.1.3.2):

P f = 700 × 1.25 + 500 × 1.5 = 1625kN Service load:

P s = 700 + 500 = 1200kN 2. Find the footing dimensions. Width: b = 4m Find the length l . Set the equation of equilibrium of vertical forces at the service load level:

Ps = q × A where

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A = ( b – 1m ) ( l ) (footing plan area in contact with the soil) P s = 1200kN so

l = 4m Hence, use 4 m by 4 m square footing, where

A = ( 4 – 1 ) ( 4 ) = 12m

2

3. Check the one-way shear resistance. Find the effective depth. Since

d = 600 – 100 = 500mm and

dv =

0.9d = 0.9 × 500mm = 450mm 0.72h 0.72 × 600mm 430mm

so,

d v = 450mm (larger value governs) 230 230 β = ----------------------- = --------------------------- = 0.16 1000 + d v 1000 + 450 V c = φ c λβ f c ′b w d v

[6.13]

(A23.3 Eq. 11-6) [6.12]

V c = 0.65 ( 1 ) ( 0.16 ) 25 ( 4000mm ) ( 450mm ) = 936kN Pf 1625kN V f = ----- = -------------------- = 812kN 2 2 Since

V f = 812kN < 936kN okay

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4. Check the two-way shear resistance.

v c = 0.38λφ c f c ′ = 0.38 ( 1 ) ( 0.65 ) 25MPa = 1.24MPa (A23.3 Eq.13- 7) [12.10] V c = v c × b o × d = 1.24MPa × 3600mm × 500mm = 2230kN where

b o = 4 × ( 400 + 500 ) = 3600mm Since

V c = 2230kN > 1625kN okay

5. Design for flexure in the longitudinal direction. Find the factored soil pressure.

Pf 1625kN= 135kPa q f = ----- = ------------------2 A 12m Find the factored bending moment at the critical section (face of the column).

1.5 M f = 135kPa × 4m × 1.5m × 1.8 – ------- = 850kNm (see the sketch above) 2 Find the area of reinforcement using the direct method. 2 3.85M f A s = 0.0015f c ′b d – d – ----------------f c ′b

[5.4] 6

2 3.85 ( 850 ×10 ) 2 = 0.0015 ( 25 ) ( 4000 ) 500 – ( 500 ) – ------------------------------------ = 5080mm 25 ( 4000 )

Use 11-25M bars: A s = 11 × 500 = 5500mm

2

Since 2

A s = 5500mm > 5080mm

2

okay

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Check the minimum reinforcement requirement (A23.3 Cl.7.8.1). 3

A g = h × l = 600 × 4000 = 2400 × 10 mm 3

2 2

A smin = 0.002A g = 0.002 × ( 2400 × 10 mm ) = 4800mm

2

Since 2

A smin = 4800 < 5500mm okay Check the maximum reinforcement requirement (A23.3 Cl.10.5.2).

f c ′ = 25MPa ρ b = 0.022 (Table A.4) 2 As 5500mm ρ = ------ = ----------------------------------------------- = 0.003 < 0.022 okay 4000mm × 500mm bd

6. Design for flexure in the transverse direction.

Find the factored bending moment at the critical section (face of the column). 2

( 1.8m ) M f = 135kPa × 3m × ------------------- = 656kNm 2 Find the area of reinforcement using the direct method. 6

3.85 ( 656 ×10 ) 2 A s = 0.0015 ( 25 ) ( 4000 ) 500 – 500 – ------------------------------------ = 3890mm 25 ( 4000 ) 2

2

Use 8-25M bars: A s = 8 × 500mm = 4000mm

2

Since 2

A s = 4000mm > 3890mm

2

okay

Check the minimum reinforcement requirement (A23.3 Cl.7.8.1).

A smin = 4800mm

2

(same as longitudinal direction)

Since

A smin = 4800 > 3890mm

2

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Set

A s = 4800mm

2

Use 10-25M bars: A s = 5000mm

2

Since the minimum reinforcement requirement governs, there is no need to check the maximum reinforcement requirement. In this case, there are two practical reinforcement arrangements: 10-25M each way or 11-25M each way.

_________________________________________________________________________ 12.14.

1. Find the footing length. Footing width: b = 4500mm Total service load

P s = 2500 + 3500 = 6000kN Allowable soil pressure

q = 200kPa Footing length can be determined as Copyright © 2006 Pearson Education Canada Inc.

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Ps 6000kN l = ------------ = -------------------------------------- = 6.6m q × b 200kPa × 4.5m Find the location of the centroid of applied service load resultant (distance from the left end of the footing).

2500 × 1m + 3500 × 5m x = ----------------------------------------------------------- = 3.3m 2500 + 3500 Note that the centroid of the footing is in line with the centroid of applied load resultant. 2. Develop the shear force and bending moment diagram for the factored load. Load factor:

α = 1.4 (given) Find the factored load for column 1:

P f 1 = α × P s1 = 1.4 × 2500 = 3500kN Find the factored load for column 2:

P f 2 = α × P s2 = 1.4 × 3500 = 4900kN Total factored load:

P f = P f 1 + P f 2 = 3500 + 4900 = 8400kN Find the factored soil pressure:

Pf 8400kN q f = ----- = ------------------------------- = 280kPa A 6.6m × 4.5m Factored soil pressure per metre:

q f ′ = q f × b = 280kPa × 4.5m = 1260kN ⁄ m Shear force and bending moment diagrams are presented below.

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3. Determine the footing depth based on two-way shear requirements. Set

V f = 4900kN (larger column reaction) Find concrete shear resistance.

v c = 0.38λφ c f c ′ = 0.38 ( 1 ) ( 0.65 ) 25MPa = 1.24MPa (A23.3 Eq.13- 7) [12.10] Estimate effective depth as

d = 800mm so

V c = vc bo d = vc × 4 ( t + d ) × d

[12.7]

= 1.24 × 4 ( 500 + 800 ) × 800 = 5158kN Since

V c = 5158kN > 4900kN okay Use footing thickness h = 900mm , so

d = 900 – 75 – 25 = 800mm Note that the clear cover for the bottom steel is 75 mm (see Table A.2). Assume 25M rebars. 4. Determine the footing depth based on one-way shear requirements. Maximum shear at the distance d from the face of column 2 is at point A (see the shear force diagram). Distance x from the column centreline can be determined as

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a 500 x = --- + d = --------- + 800 = 1050mm = 1.05m 2 2 where

a = 500mm is the column width V fA = 2884kN – ( 1260kN ⁄ m ) ( 1.05m ) = 1561kN dv =

0.9d = 0.9 × 800mm = 720mm 0.72h 0.72 × 900mm 650mm

d v = 720mm (larger value governs) 230 230 β = ----------------------- = --------------------------- = 0.13 1000 + d v 1000 + 720 V c = φ c λβ f c ′b w d v

[6.13]

(A23.3 Eq. 11-6) [6.12]

= 0.65 ( 1 ) ( 0.13 ) ( 25 ) ( 4500mm ) ( 720mm ) = 1370kN Since

V c = 1370kN < 1561kN concrete shear resistance is not sufficient, so shear reinforcement needs to be provided. Alternatively, footing depth can be increased. Try

d = 900mm and

h = 1000mm dv =

0.9d = 0.9 × 900mm = 810mm 0.72h 0.72 × 1000mm 720mm

so,

d v = 810mm (larger value governs) 230 230 β = ----------------------- = --------------------------- ≅ 0.13 1000 + d v 1000 + 810 V c = φ c λβ f c ′b w d v

[6.13]

(A23.3 Eq. 11-6) [6.12]

= 0.65 ( 1 ) ( 0.13 ) ( 25 ) ( 4500mm ) ( 810mm ) = 1540kN Since

V c = 1540kN ≅ 1561kN

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concrete shear resistance is deficient, but the difference is not significant. Keep the selected footing dimensions (alternatively, increase concrete strength). 5. Design the footing for flexure in the longitudinal direction. Top steel (midspan region):

d = 1000 – 50 – 25 = 925mm Set

M r = M f ≅ 1360kNm (see the bending moment diagram) 2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

3.85 ( 1360 ×10 ) 2 = 0.0015 ( 25 ) ( 4500 ) 925 – ( 925 ) – --------------------------------------- = 4304mm 25 ( 4500 ) 2

Check the minimum reinforcement requirement (A23.3 Cl.7.8.1). 3

2

3

2

A g = b × h = 4500 × 1000 = 4500 × 10 mm

A smin = 0.002A g = 0.002 × ( 4500 × 10 mm ) = 9000mm

2

Since 2

A smin = 9000mm > 4304mm

2

Set

A s = A smin = 9000mm

2 2

Use 18-25M bars: A s = 18 × 500mm = 9000mm

2

Column 2 - bottom reinforcement

d = 1000 – 75 – 25 = 900mm a --- = 250mm 2 Bending moment at the column face (distance a ⁄ 2 from the column centreline): 2

( 0.25m ) M f = 1612kNm – ( 1260kNm ) ---------------------- = 1572kNm ≅ 1570kNm 2 Find the reinforcement area using the direct method. Set

M r = M f = 1570kNm

Copyright © 2006 Pearson Education Canada Inc.

12-46

2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

3.85 ( 1570 ×10 ) 2 = 0.0015 ( 25 ) ( 4500 ) 900 – ( 900 ) – --------------------------------------- = 5123mm 25 ( 4500 ) 2

Since 2

A s = 5123mm < A smin Set

A s = A smin = 9000mm

2 2

Use 18-25M bars: A s = 18 × 500mm = 9000mm

2

Column 1 - bottom reinforcement

Bending moment at the column face (distance a ⁄ 2 from the column centreline):

( 0.25 ) 2 M f = 630 – ( 1260 ) --------------- = 590kNm 2 Find the reinforcement area using the direct method. Set

M r = M f = 590kNm 2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

2 3.85 ( 590 ×10 ) 2 = 0.0015 ( 25 ) ( 4500 ) 900 – 900 – ------------------------------------ = 1904mm 25 ( 4500 )

Since 2

A s = 1904mm < A smin Set

A s = A smin = 9000mm

2

Since the CSA A23.3 minimum reinforcement requirement governs, there is no need to check the CSA A23.3 maximum reinforcement requirement (Cl.10.5.2). 2

Use 18-25M bars: A s = 18 × 500mm = 9000mm

2

Determine the bar spacing. Take 100 mm end spacing, hence

Copyright © 2006 Pearson Education Canada Inc.

12-47

4500mm – ( 2 × 100mm ) s = ------------------------------------------------------------- ≅ 250mm 17 According to A23.3 Cl.7.4.1.2 and 13.10.4, the maximum permitted bar spacing ( s max ) is equal to the lesser of

3h 500mm

Since 3h = 3 ( 1000mm ) = 3000mm It follows that s max = 500mm Since s = 250mm < s max = 500mm okay Check development length for 25M bars to the left hand side of the column 1.

For 25M bars: l d = 925mm > 700 mm The ratio between the required development length and the available length is

925-------≅ 1.3 700 however the actual provided bottom reinforcement area at column 1 (9000 mm2) is significantly larger than the required area (approximately 1900 mm2), that is, there is an overstrength associated with the amount of reinforcement, with the ratio of

9000 ------------ = 4.7 1900 Considering a significantly larger overstrength ratio as compared to the deficiency in the development length for straight bars, it can be concluded that the available length of 700 mm is sufficient to develop straight rebars in tension. 6. Design the footing for flexure in the transverse direction. Consider a unit strip, that is,

b = 1000mm Find the bending moment at the face of the column as (see the next sketch) Copyright © 2006 Pearson Education Canada Inc.

12-48

4.5m – 0.5m 2 280kPa × ------------------------------2 M f = ----------------------------------------------------------------- = 560kNm ⁄ m 2

Find the reinforcement area using the direct method. 2 3.85M r A s = 0.0015 ( f c ′ ) ( b ) d – d – -----------------f c ′b

[5.4] 6

3.85 ( 560 ×10 ) 2 = 0.0015 ( 25 ) ( 1000 ) 900 – ( 900 ) – ------------------------------------ = 1850mm ⁄ m 25 ( 1000 ) 2

Check the minimum reinforcement requirement (A23.3 Cl.7.8.1). 3

2

3

2

A g = b × h = 1000 × 1000 = 1000 × 10 mm

2

A smin = 0.002A g = 0.002 × ( 1000 × 10 mm ) = 2000mm ⁄ m Since 2

A smin = 2000mm > 1850mm

2

use 2

A s = A smin = 2000mm ⁄ m Use 25M bars: A b = 500mm

2

(Table A.1)

1000 1000mm 2 - = 250mm s ≤ A b ------------ = ( 500mm ) × -------------------------------2 As 2000mm ⁄ m

[3.29]

Use

s = 250mm

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12-49

According to A23.3 Cl.7.4.1.2 and 13.10.4, the maximum permitted bar spacing ( s max ) is equal to the lesser of

3h 500mm

In this case, s max = 500mm governs. Since s = 250mm < s max = 500mm okay Use 25M@250 or 26-25M transverse bars. 7. Provide a design summary.

__________________________________________________________________________ 12.15. Compressive strengths: footing:

f cf ′ = 25MPa

column:

f cc ′ = 30MPa

1. Determine the footing bearing strength.

Column:

A 1 = 0.4m × 0.4m = 0.16m

2

Footing:

A 2 = 4m × 4m = 16m

2

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Since

A2 ------ = A1

2

16m - = 10 > 2 ----------------(A23.3 Cl.10.8.1) 2 0.16m

then 6

2

B rf = 1.7φ c f cf ′A 1 = 1.7 × 0.65 × 25MPa × ( 0.16 × 10 mm ) = 4420kN [12.30] Since

B rf = 4420kN > P f = 3500kN it follows that the footing bearing strength is satisfactory. 2. Determine the column bearing strength.

B rc = 0.85φ c f cc ′A 1

[12.28] 6

2

= 0.85 × 0.65 × 30MPa × ( 0.16 × 10 mm ) = 2652kN Since

B rc = 2652kN < P f = 3500kN it follows that the column bearing resistance is not satisfactory, and so dowel reinforcement is required. 3. Design the dowel reinforcement. Since

B rd ≥ P f – B rc = 3500kN – 2652kN = 848kN and

B rd = A d φ s f y

[12.32]

the dowel area can be obtained as 3 B rd 848 × 10 N 2 A d = --------- = -------------------------------------- = 2494mm 0.85 × 400MPa φs fy

According to A23.3 Cl.15.9.2.1, 6

2

A d ≥ 0.005A 1 = 0.005 ( 0.16 × 10 mm ) = 800mm

2

Use 4-30M bars (to match column reinforcement). Since 2

2

A s = 4 × 700 = 2800mm > 2494mm okay The column bearing resistance including dowels:

B rc = 0.85φ c f c ′A 1 + A d φ s f y 3

[12.34]

2

= 2562 × 10 N + ( 2800mm × 0.85 × 400MPa ) = 3514kN

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Since

B rc = 3514kN > P f = 3500kN okay Finally, let us determine the dowel development length (straight bars in compression) (A23.3 Cl.12.3).

d b = 30mm

(30M bars)

Footing:

fy 400MPa l dbf = 0.24 ------------ d b = 0.24 ----------------------- × 30mm = 576mm 25MPa f cf ′

[9.4]

Column:

fy 400MPa l dbc = 0.24 ------------ d b = 0.24 ----------------------- × 30mm = 526mm 30MPa f cc ′

[9.4]

The ratio of required and provided areas of reinforcement is

A drequired --------------------- = 2494 ------------ = 0.89 2800 A dprovided Therefore, the required development length can be reduced as follows. Footing: l dbf = 576 × 0.89 = 513mm ≅ 520mm Column: l dbc = 526 × 0.89 = 468mm ≅ 470mm The available straight bar development length into footing is equal to

l = h – cover = 600 – 75 = 525mm cover = 75mm (Table A.2) Since

l dbf = 520mm < 525mm okay 4. Sketch the reinforcement arrangement.

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_________________________________________________________________________ 12.16.

Copyright © 2006 Pearson Education Canada Inc.

12-53

Copyright © 2006 Pearson Education Canada Inc.

12-54

Chapter 13 - Solutions __________________________________________________________________________ 13.1.

First, check whether the wall meets criteria related to the Empirical Method for bearing wall design (A23.3 Cl.14.2), as follows a) Solid rectangular cross section constant along the wall height. b), c) Wall is loaded concentrically. d) The wall is supported against lateral displacement at the top and bottom. In conclusion, the wall meets the above criteria and so the Empirical method can be used in the design. 1. Determine the wall thickness.

h u = 4500mm Wall thickness is restricted based on A23.3 Cl.14.1.7.1, as follows:

h u 4500 t ≥ ------ = ------------ = 180mm 25 25 t ≥ 150mm Therefore, the 180 mm value governs, however use

t = 200mm 2. Check the bearing resistance.

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13-1

The contact area ( A 1 ) is

A 1 = 200mm × 200mm = 40000mm

2

Bearing resistance (A23.3 Cl.10.8.1) can be determined as

B r = 0.85φ c f c ′A 1

[12.28] 2

= 0.85 × 0.65 × 25MPa × 40000mm = 552kN Since

B r = 552kN > P f = 300kN okay the wall bearing resistance is adequate. 3. Determine the factored axial load resistance. •

Determine the gross area of the wall section ( A g ) .

Determine l b according to A23.3 Cl.14.1.3.1. i) a = 200mm bearing width

l b = a + 2 × 9t = 200 + 2 × 9 × 200 = 3800mm ii) l b will be determined by drawing the lines sloping downward from each side of the bearing area - the slope is 2:1. Note that the l b value is limited by the intersection with the lines corresponding to the adjacent point loads.

l b = 2500mm (based on the intersecting points) iii) l b ≤ s = 2500mm The smallest value governs, hence

l b = 2500mm

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13-2

Let us determine the gross area

A g = l b × t = 2500mm × 200mm = 500000mm •

2

Determine the factored axial load resistance.

k = 1.0 (pin supported wall) Use α 1 ≅ 0.8 [3.7]

kh u 2 P r = --- α 1 φ c f c ′A g 1 – --------3 32t

2

(A23.3 Eq.14.1) [13.13]

2 1 × 4500mm 3 2 = --- × 0.8 × 0.65 × ( 25 MPa ) × ( 500 × 10 mm ) 1 – ------------------------------3 32 × 200

2

= 2191kN

Since

P r = 2191kN > P f = 300kN okay the wall is adequate for the given loads. 4. Determine the distributed and concentrated wall reinforcement. • Distributed horizontal reinforcement i) Find the area of reinforcement. 3

A g = 1000mm × t = 1000mm × 200mm = 200 × 10 mm

2

Minimum required area of horizontal reinforcement (A23.3 Cl.14.1.8.6) can be determined as 3

2

2

A hmin = 0.002A g = 0.002 ( 200 × 10 mm ) = 400mm ⁄ m ii) Determine the required bar spacing. 2

A b = 200mm (15M bars, Table A.1) 1000 2 1000 - = 500mm s ≤ A b ------------ = 200mm × ----------------------------2 As 400mm ⁄ m

[3.29]

s = 500mm Find the maximum permitted bar spacing according to A23.3 Cl.14.1.8.4.

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

In this case,

s = s max = 500mm Horizontal reinforcement: 15M@500. iv) Check whether one layer of reinforcement is adequate. One layer is adequate when t < 210mm (A23.3 Cl.14.1.8.3). Since Copyright © 2006 Pearson Education Canada Inc.

13-3

t = 200mm < 210mm okay • Distributed vertical reinforcement i) Determine the area of vertical reinforcement. 3

A g = 1000mm × t = 1000mm × 200mm = 200 × 10 mm

2

Minimum area of vertical reinforcement (A23.3 Cl.14.1.8.5): 3

2

2

A vmin = 0.0015A g = 0.0015 ( 200 × 10 mm ) = 300mm ⁄ m [13.1] ii) Required bar spacing 2

A b = 200mm (15M bars, Table A.1) 1000 2 1000mm - = 666mm s ≤ A b ------------ = 200mm ----------------------------2 As 300mm ⁄ m

[3.29]

iii) Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

Since

s = 666mm > s max = 500mm use s = 500mm Vertical reinforcement: 15M@500 iv) Concentrated reinforcement (A23.3 Cl.14.1.8.8): 2-15M bars at each end 5. Sketch a design summary.

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_________________________________________________________________________ 13.2.

First, check whether the wall meets criteria related to the Empirical Method for bearing wall design (A23.3 Cl.14.2), as follows a) Solid rectangular cross section constant along the wall height b), c) The wall is concentrically loaded d) The wall is supported against lateral displacement at the top and bottom In conclusion, the wall meets the above criteria and the Empirical Method can be used in this design. 1. Determine the wall thickness.

h u = 4000mm h u 4000mm t ≥ ------ = ---------------------- = 160mm (A23.3 Cl.14.1.7.1) 25 25 Since

t ≥ 150mm use t = 160mm . 2. Determine the factored axial load.

DL = 80kN ⁄ m LL = 40kN ⁄ m w f = 1.25DL + 1.5LL

(NBC 2005 Table 4.1.3.2)

kN = 1.25 × 80kN ⁄ m + 1.5 × 40 ------- = 160kN ⁄ m m

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13-5

3. Determine the factored axial load resistance. Find the gross area of the wall section ( A g ) (consider a strip of unit length l b = 1000mm ). 3

A g = l b × t = 1000mm × 160mm = 160 × 10 mm

2

k = 1.0 (pin supported wall) α 1 ≅ 0.8 [3.7] kh u 2 P r = --- α 1 φ c f c ′A g 1 – --------3 32t

2

(A23.3 Eq.14.1)

[13.13]

2 1 × 4000mm 3 2 = --- × 0.8 × 0.65 × 25MPa × ( 160 × 10 mm ) 1 – ------------------------------3 32 × 160mm

2

= 540 kN ⁄ m

Since

P r = 540.3kN ⁄ m > w f = 160kN ⁄ m okay The wall axial resistance is adequate for this design. 4. Determine the distributed and concentrated wall reinforcement. • Distributed horizontal reinforcement. i) Find the minimum horizontal reinforcement area (A23.3 Cl.14.1.8.6). 3

A g = 1000 × t = 1000 × 160 = 160 × 10 mm 3

2

2 2

A hmin = 0.002A g = 0.002 ( 160 × 10 mm ) = 320mm ⁄ m ii) Find the required bar spacing.

A b = 200mm

2

(15M bars, Table A.1)

1000 2 1000mm - = 625mm s ≤ A b ------------ = 200mm ----------------------------2 As 320mm ⁄ m

[3.29]

iii) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)

s max ≤

3t = 3 × 160 = 480mm 500mm

← governs

Use s = 480mm Horizontal reinforcement: 15M@480 iv) Check whether one layer of reinforcement is adequate. One layer is adequate when t < 210mm , so

Copyright © 2006 Pearson Education Canada Inc.

13-6

t = 160mm < 210mm okay • Distributed vertical reinforcement i) Find the minimum required area (A23.3 Cl.14.1.8.5). 3

A g = 160 × 10 mm

2

(same as Step 4) 3

2

2

A vmin = 0.0015A g = 0.0015 ( 160 × 10 mm ) = 240mm ⁄ m ii) Required bar spacing

A b = 200mm

2

(15M bars, Table A.1)

1000 1000mm 2 - = 833mm s ≤ A b × ------------ = 200mm × ----------------------------2 As 240mm ⁄ m

[3.29]

iii) Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 160 = 480mm 500mm

← governs

Since

s = 833mm > s max = 480mm Use s = 480mm Vertical reinforcement: 15M@480 iv) Concentrated reinforcement at wall ends (A23.3 Cl.14.1.8.8): 2-15M bars at each end 5. Sketch a design summary.

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__________________________________________________________________________ 13.3. 1. Find the shear force and bending moment distribution in the wall.

Load analysis (NBC 2005 Table 4.1.3.2):

p 1f = 1.5p 1 = 1.5 × 20 = 30kPa p 2f = 1.5p 2 = 1.5 × 60 = 90kPa Find the average pressure:

p 1f + p 2f 30 + 90 p of = -------------------- = ------------------ = 60kPa 2 2 Use the equivalent wall model loaded by a uniform pressure of 60 kPa (see the sketch below).

2

2 p of ( l ) ( 60kPa ) ( 4m ) M f = ---------------- = ------------------------------------- = 120kNm ⁄ m 8 8

p of × l 60kPa × 4m V f = --------------- = ------------------------------- = 120kN ⁄ m 2 2 Note that very similar results would be obtained by using the “actual” pressure distribution.

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2. Design the wall for flexure. i) Find the effective depth d . Use 20M bars ( d b = 20mm ) cover = 20mm (interior exposure, Table A.2)

t = 300mm (wall thickness) db 20 d = t – cover – ----- = 300 – 20 – ------ = 270mm 2 2 ii) Find the required moment resistance. Set M r = M f = 240kNm ⁄ m iii) Find the required area of vertical tension reinforcement. Design the wall for flexure based on a strip of unit width b = 1000mm . Find the reinforcement area - use the direct method. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 × ( 120 × 10 ) 2 2 = 0.0015 × 25MPa × 1000mm 270 – ( 270 ) – --------------------------------------------------- = 1377mm ⁄ m ( 25MPa ) ( 1000mm ) iv) Select the amount of reinforcement in terms of size and spacing. Use 20M bars

2

( A b = 300mm ) see Table A.1

1000 2 1000mm s ≤ A b × ------------ = ( 300mm ) -----------------------2- = 218mm As 1377mm

[3.29]

v) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)

s max ≤

3t = 3 × 300 = 900mm 500mm ← governs

Since

s = 218mm < s max = 500mm Use s = 200mm . Vertical reinforcement: 20M@200 vi) Confirm that the CSA A23.3 maximum tension reinforcement (Cl.10.5.2) requirement has been satisfied. Actual reinforcement area:

Copyright © 2006 Pearson Education Canada Inc.

13-9

1000 2 1000mm 2 A s = A b ------------ = ( 300mm ) ---------------------- = 1500mm ⁄ m s 200mm Reinforcement ratio: 2 As 1500mm ρ = ------ = ----------------------------------------------- = 0.0055 1000mm × 270mm bd

[3.1]

Balanced reinforcement ratio:

ρ b = 0.022

( f c ′ = 25MPa, Table A.4 )

Since

ρ = 0.0055 < ρ b = 0.022

okay

vii) Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 300mm = 300 × 10 mm A vmin = 0.0015A g

2

[13.1] 3

2

2

= 0.0015 ( 300 × 10 mm ) = 450mm ⁄ m Since 2

2

A s = 1500mm ⁄ m > 450mm ⁄ m okay The vertical reinforcement is adequate. 3. Design the wall for shear. Design shear force is

V f = 120kN ⁄ m a) Concrete shear resistance ( V c ) . Find the effective shear depth ( d v ) :

dv =

0.9d = 0.9 × 270mm = 243mm 0.72t 0.72 × 300mm 216mm

So,

d v = 243mm ≅ 240mm (larger value governs) Set b w = 1000mm (unit strip) Find β (A23.3 Cl.11.3.6.3b).

230 β = ----------------------- (A23.3 Eq.11.9) [6.13] 1000 + d v

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13-10

230 = --------------------------- = 0.185 ≅ 0.18 1000 + 240 Find V c .

V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6) [6.12]

= 0.65 ( 1.0 ) ( 0.18 ) 25MPa ( 1000mm ) ( 240mm ) = 140kN ⁄ m b) Is shear reinforcement required? According to A23.3 Cl.11.2.8.1, shear reinforcement is not required when

Vf ≤ Vc Since

V f = 120kN ⁄ m < V c = 140kN ⁄ m it follows that the shear reinforcement is not required, however CSA A23.3 Code requires minimum provision of horizontal reinforcement in walls. c) Determine the minimum distributed horizontal reinforcement. i) Reinforcement area (A23.3 Cl.14.1.8.6) 3

A g = 1000mm × t = 1000mm × 300mm = 300 × 10 mm

2

A hmin = 0.002A g 3

2

2

= 0.002 ( 300 × 10 mm ) = 600mm ⁄ m ii) Determine the required bar spacing. 2

20M bars: ( A b = 300mm , Table A.1 )

1000 2 1000mm s ≤ A b ------------ = 300mm × ---------------------2- = 500mm [3.29] As 600mm s = 500mm iii) Check whether the spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 300 = 900mm 500mm ← governs

Since

s = s max = 500mm okay Horizontal reinforcement: 15M@500 iv) Check whether one layer of reinforcement is adequate. According to A23.3 Cl.14.1.8.3, one layer of reinforcement is adequate when the wall thickness is less than 210 mm, that is, t < 210mm . In this case, t = 300mm . Therefore, two layers of rein-

Copyright © 2006 Pearson Education Canada Inc.

13-11

forcement are required. Instead of using 20M bars for horizontal reinforcement, it would be better to use 15M bars. For 2-15M bars at 500 mm spacing, the area is

A b = 2 × 200 = 400mm

2

1000 1000mm 2 2 A s = A b ------------ = ( 400mm ) × ---------------------- = 800mm ⁄ m s 500mm Since 2

2

A s = 800mm ⁄ m > 600mm ⁄ m Use 2 layers of horizontal reinforcement, that is, 15M@500. 4. Sketch a design summary.

Note that one layer of vertical reinforcement has been specified at the exterior wall face (15M@500) in order to satisfy the requirement for 2 layers of reinforcement. The main vertical tension reinforcement for this wall is located at the interior face.

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_________________________________________________________________________ 13.4. a) Design the wall for the given reinforcement.

1. Determine the design bending moments and shear forces in the wall. Load analysis (NBC 2005 Table 4.1.3.2):

kN kN p f = 1.5 × 30 ------2- = 45 ------2m m Find the sum of moments about the support B as

ΣM B

kN 45 ------2- × 4m m = ---------------------------2

4m -------- – A ( 3m ) = 0 3

So, the reaction at point A is equal to

A = 40kN The resultant of lateral load can be calculated as (see the sketch above)

kN 45 ------2- × 4m m R = ---------------------------------- = 90kN 2 Find the reaction at point B from the equilibrium of lateral forces as

ΣH = A + B – R = 0 or

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13-13

B = R – A = 90 kN – 40kN = 50kN Sketch the shear force diagram.

The point of zero shear (located at a distance x from the top of the wall) can be determined from the equilibrium of lateral forces on a free-body diagram of the top portion of the wall (see the sketch above):

pf p x = ---- × x = 11.25x 4 Equation of equilibrium of horizontal forces for the free-body diagram: 2

px × x ΣH = ----------------- – A = 0 2 or 2

11.25x ------------------- = 40kN 2 The distance x can be found from the above equation as

x = 2.667m Develop the bending moment diagram. The following critical moment values need to be determined: • Bending moment at support A (negative):

kN 2 ( 11.25 ------2- × ( 1m ) m M = – ------------------------------------------------- = – 1.87kNm ≅ – 2kNm 6 •

Maximum bending moment at the distance x = 2.67m from the top of the wall:

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13-14

2 kN 11.25 ------2- × ( 2.67m ) m M = – -------------------------------------------------- + ( 40kN ) ( 1.67m ) = 53kNm 6

2. Design the wall for flexure. Find the effective depth d . Estimate d = 100mm (reinforcement located in the middle of the wall - 1 layer). Section at the midspan (maximum positive bending moment) Find the required moment resistance.

Set M r = M f = 53kNm ⁄ m Find the required area of vertical tension reinforcement. Design the wall for flexure based on a strip of unit width b = 1000mm . Find the reinforcement area - use the direct method. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 ( 53 × 10 ) = 0.0015 ( 25MPa ) ( 1000mm ) 100mm – 100mm – ---------------------------------------------25MPa ( 1000mm ) 2

2

= 2142mm ⁄ m Select the amount of reinforcement in terms of size and spacing. Use 20M bars

2

A b = 300mm see Table A.1

1000 2 1000mm s ≤ A b × ------------ = ( 300mm ) -----------------------2- = 140mm As 2142mm

[3.29]

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iii) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

Since

s = 140mm < s max = 500mm Use s = 140mm . Vertical reinforcement: 20M@140 Confirm that the CSA A23.3 maximum tension reinforcement requirement (Cl.10.5.2) has been satisfied. Reinforcement ratio: 2 As 2142mm ρ = ------ = ----------------------------------------------- = 0.0214 bd 1000mm × 100mm

[3.1]

Balanced reinforcement ratio:

ρ b = 0.022

( f c ′ = 25MPa, Table A.4 )

Since

ρ = 0.0214 < ρ b = 0.022 This amount of reinforcement is acceptable per CSA A23.3, however it is considered to be large as it is very close to the balanced reinforcement. Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 200mm = 200 × 10 mm A vmin = 0.0015A g

2

[13.1] 3

2

2

= 0.0015 ( 200 × 10 mm ) = 300mm ⁄ m Since 2

2

A s = 2143mm ⁄ m > 300mm ⁄ m okay Section at the support A (negative bending moment) Find the required moment resistance.

Set M r = M f = 2kNm ⁄ m Find the reinforcement area - use the direct method. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4]

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6

2 2 3.85 ( 2 × 10 ) A s = 0.0015 ( 25 ) ( 1000 ) 100 – 100 – --------------------------------- = 58mm ⁄ m 25 ( 1000 )

Select the amount of reinforcement in terms of size and spacing. Use 15M bars

2

A b = 200mm see Table A.1

1000 2 1000mm - = 3448mm s ≤ A b × ------------ = ( 200mm ) --------------------2 As 58mm

[3.29]

s max = 500mm (A23.3 Cl.14.1.8.4, same as for other section) Since

s = 3448mm > s max = 500mm Use s = 500mm . Find the actual reinforcement area.

1000 1000mm 2 2 A s = A b ------------ = ( 200mm ) × ---------------------- = 400mm ⁄ m s 500mm Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 200mm = 200 × 10 mm A vmin = 0.0015A g

2

[13.1] 3

2

2

= 0.0015 ( 200 × 10 mm ) = 300mm ⁄ m Since 2

2

A s = 400mm ⁄ m > 300mm ⁄ m okay Vertical reinforcement: 15M@500. 3. Design the wall for shear. Design shear force:

V f = 50kN ⁄ m •

Concrete shear resistance ( V c ) .

Find the effective shear depth ( d v ) :

dv =

0.9d = 0.9 × 100mm = 90mm 0.72t 0.72 × 200mm 144mm

So,

d v ≅ 140mm (larger value governs) Copyright © 2006 Pearson Education Canada Inc.

13-17

Set b w = 1000mm (unit strip) Find β (A23.3 Cl.11.3.6.3b).

230 β = ----------------------- (A23.3 Eq.11.9) [6.13] 1000 + d v 230 = --------------------------- = 0.2 1000 + 140 Find V c .

V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6) [6.12]

= 0.65 ( 1.0 ) ( 0.2 ) 25MPa ( 1000mm ) ( 140mm ) = 91kN ⁄ m • Is shear reinforcement required? According to A23.3 Cl.11.2.8.1, shear reinforcement is not required when

Vf ≤ Vc Since

V f = 50kN ⁄ m < V c = 91kN ⁄ m it follows that the shear reinforcement is not required, however CSA A23.3 Code requires minimum provision of horizontal reinforcement in walls. • Determine the minimum distributed horizontal reinforcement. i) Reinforcement area (A23.3 Cl.14.1.8.6) 3

A g = 1000mm × t = 1000mm × 200mm = 200 × 10 mm

2

A hmin = 0.002A g 3

2

2

= 0.002 ( 200 × 10 mm ) = 400mm ⁄ m ii) Determine the required bar spacing. 2

15M bars ( A b = 200mm , Table A.1 )

1000 2 1000mm s ≤ A b ------------ = 200mm × ---------------------2- = 500mm [3.29] As 400mm s = 500mm iii) Check whether the spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

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13-18

s = s max = 500mm okay Horizontal reinforcement: 15M@500 4. Sketch a design summary.

b) Comment on the effectiveness of placing the reinforcement in the middle of the wall. It is convenient to place vertical reinforcement in the middle of the wall; in this way, the wall is able to resist both positive and negative bending moments without adjusting the location of reinforcing steel to suit the moment diagram. This is a good approach to save labour cost on site, as well as to reduce complexities of reinforcement installation and minimize chances for field errors. c) Does this reinforcement strategy represent the most cost-effective solution? The required amount of vertical reinforcement placed in the middle of the wall is very large (very close to the balanced reinforcement). In order to reduce the amount of reinforcement, an alternative solution can be considered wherein the reinforcement is placed close to the inside wall face. This will result in an increased effective depth which will in turn result in a reduced amount of vertical reinforcement. When the reinforcement is moved to the inside face of the wall, the effective depth can be estimated as

d ≅ 200mm – 35mm = 165mm Consider the section with the maximum bending moment, where

M r = M f = 53kNm ⁄ m Find the required area of vertical tension reinforcement- use the direct method.

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2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 ( 53 × 10 ) = 0.0015 ( 25MPa ) ( 1000mm ) 165 – 165mm – ---------------------------------------------25MPa ( 1000mm ) 2

2

= 1009mm ⁄ m Select the amount of reinforcement in terms of size and spacing. Use 15M bars

2

( A b = 200mm , see Table A.1)

1000 2 1000mm s ≤ A b × ------------ = ( 200mm ) -----------------------2- = 198mm As 1009mm

[3.29]

Use 15M@200. This alternative solution results in significantly smaller amount of vertical reinforcement - there is a 50% reduction in the total reinforcement area (1009 mm2 versus 2142 mm2). Vertical wall elevation and the reinforcement arrangement is presented below.

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13-20

_________________________________________________________________________

13.5. a) Draw the bending moment diagram for the wall and the footing. Gravity load effects Factored axial load (NBC 2005 Table 4.1.3.2)

P f = 1.25P = 1.25 × 150 = 188kN ⁄ m Eccentricity

1m 0.25m e = -------- – --------------- = 0.2m 3 2 Factored bending moment:

kNm kN M f = P f × e = 188 ------- × 0.2m = 38 -----------m m Axial forces in the slabs (see the sketch above):

Mf 38kNm ⁄ m T = C = ------- = --------------------------- = 12.5kN ⁄ m h 3m Bending moment diagram is shown on the sketch below.

Lateral load effects Factored soil pressure (NBC 2005 Table 4.1.3.2)

p f = 1.5 × 20kPa = 30kPa

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13-21

2

2 pf × h ( 30kPa ) × ( 3m ) M f = ---------------- = ------------------------------------------- = 33.8kNm ⁄ m 8 8

Bending moment diagram is shown on the sketch below.

Combined gravity and lateral loads Shear force diagram is obtained using the principle of superposition, by combining the corresponding shear force values for gravity and lateral loads (see below).

Maximum bending moment develops at the point of zero shear (distance 1.08 m from the top of the wall). The corresponding value can be obtained as follows 2

( 1.08m ) M f = ( 32.5kN ⁄ m ) × ( 1.08m ) – ( 30kN ⁄ m ) × ---------------------- = 17.6kNm ⁄ m 2

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13-22

Bending moment diagram due to combined gravity and lateral loads is shown below.

Find the inflection point located at a distance x from the base of the wall (see the sketch above). Bending moment M x at a distance x from the base of the wall is equal to zero, that is, 2

(x) M x = – 38 + 57.5 ( x ) – 30 ---------- = 0 2 or 2

15x – 57.5x + 38 = 0 The solution of this quadratic equation is

x = 0.85m b) Design the flexural reinforcement for the footing and the exterior wall face. Find the effective depth d . Use 20M bars ( d b = 20mm ) Footing: cover = 75mm (Table A.2)

t = 300mm (footing thickness) db 20 d = t – cover – ----- = 300 – 75 – ------ = 215mm 2 2 Wall: cover = 50mm (exterior exposure, Table A.2)

t = 250mm (wall thickness)

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db 20 d = t – cover – ----- = 250 – 50 – ------ = 190mm 2 2 Use

d = 190mm for the design (smaller value). Find the required moment resistance. Set M r = M f = 38kNm ⁄ m Find the required area of vertical tension reinforcement. Design the wall for flexure based on a strip of unit width b = 1000mm . Find the reinforcement area - use the direct method. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 × ( 38 × 10 ) 2 = 0.0015 × 25MPa × 1000mm 190 – ( 190 ) – --------------------------------------------------- = 603mm ⁄ m ( 25MPa ) ( 1000mm ) 2

Select the amount of reinforcement in terms of size and spacing. Use 20M bars

2

( A b = 300mm ) see Table A.1

1000 2 1000mm s ≤ A b × ------------ = ( 300mm ) ---------------------2- = 497mm ≅ 450mm As 603mm

[3.29]

Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)

s max ≤

3t = 3 × 250 = 750mm 500mm ← governs

Since

s = 450mm < s max = 500mm Use s = 450mm . Vertical reinforcement: 20M@450 Confirm that the CSA A23.3 maximum tension reinforcement requirement (Cl.10.5.2) has been satisfied. Actual reinforcement area:

1000 2 1000mm 2 A s = A b ------------ = ( 300mm ) ---------------------- = 667mm ⁄ m s 450mm Reinforcement ratio:

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2 As 667mm -------------------------------------------------- = 0.0035 ρ = = bd 1000mm × 190mm

[3.1]

Balanced reinforcement ratio:

ρ b = 0.022

( f c ′ = 25MPa, Table A.4 )

Since

ρ = 0.0035 < ρ b = 0.022

okay

Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 250mm = 250 × 10 mm A vmin = 0.0015A g

2

[13.1] 3

2

2

= 0.0015 ( 250 × 10 mm ) = 375mm ⁄ m Since 2

2

A s = 667mm ⁄ m > 375mm ⁄ m okay The vertical reinforcement is adequate. Determine the reinforcement length for the wall (above the footing-wall interface). Development length for 20M bars in this example (bottom bars, regular uncoated bars, normaldensity concrete) can be determined from Table A.5 as

l d = 575mm A23.3 Cl.12.11.3 (as applied to continuous beams) states that (see Section 9.7.8)

Mr l d ≤ ------- + l a Vf

(A23.3 Eq.12-6)

[9.8]

where

d 190mm = 12d b 12 × 20mm = 240mm l a = 240mm (greater value governs) la =

So,

38kNm l d ≤ ------------------- + 0.24m = 0.66m + 0.24m = 0.9 m = 900mm 57.5kN Since

l d = 575mm < 900mm okay Therefore, the total wall reinforcement length above the wall-footing interface can be obtained as

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l = 0.85m + 0.575m = 1.425m ≅ 1.5m c) Design the vertical and horizontal reinforcement at the interior wall face. 1. Design the wall for flexure. Find the effective depth d . cover = 20mm (interior exposure, Table A.2)

t = 250mm (wall thickness) db 20 d = t – cover – ----- = 250 – 20 – ------ = 220mm 2 2 Find the required moment resistance. Set M r = M f ≅ 18kNm ⁄ m Find the required area of vertical tension reinforcement. Design the wall for flexure based on a strip of unit width b = 1000mm . Find the reinforcement area - use the direct method. 2 3.85M r A s = 0.0015f c ′b d – d – -----------------f c ′b

[5.4] 6

3.85 × ( 18 × 10 ) 2 2 = 0.0015 × 25MPa × 1000mm 220 – ( 220 ) – --------------------------------------------------- = 240mm ⁄ m ( 25MPa ) ( 1000mm ) Select the amount of reinforcement in terms of size and spacing. Use 15M bars

2

( A b = 200mm , see Table A.1)

1000 2 1000mm s ≤ A b × ------------ = ( 200mm ) ---------------------2- = 833mm As 240mm

[3.29]

Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)

s max ≤

3t = 3 × 250 = 750mm 500mm ← governs

Since

s = 833mm > s max = 500mm Use s = 500mm . Vertical reinforcement: 15M@500

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Confirm that the CSA A23.3 maximum tension reinforcement requirement (Cl.10.5.2) has been satisfied. Actual reinforcement area:

1000 2 1000mm 2 A s = A b ------------ = ( 200mm ) ---------------------- = 400mm ⁄ m s 500mm Reinforcement ratio: 2 As 400mm -------------------------------------------------- = 0.0018 ρ = = bd 1000mm × 220mm

[3.1]

Balanced reinforcement ratio:

ρ b = 0.022

( f c ′ = 25MPa, Table A.4 )

Since

ρ = 0.0018 < ρ b = 0.022

okay

Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 250mm = 250 × 10 mm A vmin = 0.0015A g

2

[13.1] 3

2

2

= 0.0015 ( 250 × 10 mm ) = 375mm ⁄ m Since 2

2

A s = 400mm ⁄ m > 375mm ⁄ m okay The vertical reinforcement is adequate. 2. Design the wall for shear. Design shear force:

V f = 57.5kN ⁄ m •

Concrete shear resistance ( V c ) .

Find the effective shear depth ( d v ) :

dv =

0.9d = 0.9 × 220mm = 198mm 0.72t 0.72 × 250mm 180mm

So,

d v ≅ 200mm (larger value governs) Set b w = 1000mm (unit strip) Find β (A23.3 Cl.11.3.6.3b).

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230 β = ----------------------- (A23.3 Eq.11.9) [6.13] 1000 + d v 230 = --------------------------- = 0.19 1000 + 200 Find V c .

V c = φ c λβ f c ′b w d v

(A23.3 Eq.11.6) [6.12]

= 0.65 ( 1.0 ) ( 0.19 ) 25MPa ( 1000mm ) ( 200mm ) = 123.5kN ⁄ m • Is shear reinforcement required? According to A23.3 Cl.11.2.8.1, shear reinforcement is not required when

Vf ≤ Vc Since

V f = 57.5kN ⁄ m < V c = 123.5kN ⁄ m it follows that the shear reinforcement is not required, however CSA A23.3 Code requires minimum provision of horizontal reinforcement in walls. • Determine the minimum distributed horizontal reinforcement. i) Reinforcement area (A23.3 Cl.14.1.8.6) 3

A g = 1000mm × t = 1000mm × 250mm = 250 × 10 mm

2

A hmin = 0.002A g 3

2

2

= 0.002 ( 250 × 10 mm ) = 500mm ⁄ m ii) Determine the required bar spacing. 2

15M bars ( A b = 200mm , Table A.1 )

1000 2 1000mm s ≤ A b ------------ = 200mm × ---------------------2- = 400mm [3.29] As 500mm s = 400mm iii) Check whether the spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 250 = 750mm 500mm ← governs

Since

s = 400mm < s max = 500mm Use s = 400mm . Horizontal reinforcement: 15M@400 Copyright © 2006 Pearson Education Canada Inc.

13-28

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______________________________________________________________________ 13.6.

1. Determine the factored axial load resistance. •

Determine the gross area of the wall section ( A g ) .

Determine l b according to A23.3 Cl.14.1.3.1. i) a = 300mm bearing width

l b = a + 2 × 9t = 300 + 2 × 9 × 300 = 5700mm ii) l b will be determined by drawing the lines sloping downward from each side of the bearing area - the slope is 2:1. Note that the l b value is limited by the intersection with the lines corresponding to the adjacent point loads.

l b = 1200mm (based on the intersecting points, see the sketch above) iii) l b ≤ s = 1200mm The smallest value governs, hence

l b = 1200mm Let us determine the gross area

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A g = l b × t = 1200mm × 300mm = 360000mm •

2

Determine the factored axial load resistance

k = 1.0 (pin supported wall) Use α 1 ≅ 0.8 [3.7]

kh u 2 P r = --- α 1 φ c f c ′A g 1 – --------3 32t

2

(A23.3 Eq.14.1) [13.13]

2 1 × 4000mm 3 2 = --- × 0.8 × 0.65 × ( 25 MPa ) × ( 360 × 10 mm ) 1 – ------------------------------3 32 × 300

2

= 2578kN

Since

P r = 2578kN > P f = 500kN okay the wall is adequate for the given loads. 2. Determine the distributed and concentrated wall reinforcement. • Check whether one layer of reinforcement is adequate. One layer is adequate when t < 210mm (A23.3 Cl.14.1.8.3). Since

t = 300mm > 210mm use 2 layers of reinforcement • Distributed horizontal reinforcement i) Find the area of reinforcement. 3

A g = 1000mm × t = 1000mm × 300mm = 300 × 10 mm

2

Minimum required area of horizontal reinforcement (A23.3 Cl.14.1.8.6) can be determined as 3

2

2

A hmin = 0.002A g = 0.002 ( 300 × 10 mm ) = 600mm ⁄ m ii) Determine the required bar spacing. 2

A b = 200mm (15M bars, Table A.1) Use 2 layers 2 1000 1000 - = 666mm s ≤ A b ------------ = ( 2 × 200 mm ) × ----------------------------2 As 600mm ⁄ m

[3.29]

Find the maximum permitted bar spacing according to A23.3 Cl.14.1.8.4.

s max ≤

3t = 3 × 300 = 900mm 500mm ← governs

Since

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s = 666mm > s max = 500mm Use

s = s max = 500mm Horizontal reinforcement: 15M@500 (2 layers). • Distributed vertical reinforcement i) Determine the area of vertical reinforcement. 3

A g = 1000mm × t = 1000mm × 300mm = 300 × 10 mm

2

Minimum area of vertical reinforcement (A23.3 Cl.14.1.8.5): 3

2

2

A vmin = 0.0015A g = 0.0015 ( 300 × 10 mm ) = 450mm ⁄ m [13.1] ii) Required bar spacing 2

A b = 100mm (10M bars, Table A.1) Use 2 layers 2 1000mm 1000 - = 444mm s ≤ A b ------------ = ( 2 × 100 mm ) ----------------------------2 As 450mm ⁄ m

[3.29]

iii) Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 300 = 900mm 500mm ← governs

Since

s = 444mm < s max = 500mm use s = 400mm Vertical reinforcement: 10M@400 (2 layers) • Concentrated reinforcement (A23.3 Cl.14.1.8.8): 2-15M bars at each end

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3. Sketch a design summary.

_______________________________________________________________________ 13.7. 1. Determine the design axial loads, bending moments, and shear forces. Wind load: w = 150kN ⁄ m Factored wind load (NBC 2005 Table 4.1.3.2):

w f = 1.4 × w = 1.4 × 150 = 210kN ⁄ m Find the factored bending moment. 2

2 wf × l ( 210kN ⁄ m ) ( 7.0m ) M f = ---------------- = -------------------------------------------------- = 5145kNm 2 2

Find the factored shear force.

kN V f = w f × l = 210 ------- ( 7.0m ) = 1470kN m Find the factored axial load at the base of the wall. P f = 1.25P 2 + 1.25P 1 = 1.25 × 700 + 1.25 × 500 = 1500kN

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Develop the bending moment, shear force, and axial force diagrams.

The critical section is at the base of the wall, where

M f = 5145kNm ≅ 5100kNm V f = 1470kN ≅ 1500kN P f = 1500kN 2. Design the wall for shear. Design shear force:

V f = 1500kN Wall length: l w = 5000mm Wall thickness: t = 200mm The effective shear depth ( d v ) is

d v = 0.8 × l w = 0.8 × ( 5000mm ) = 4000mm Find β (A23.3 Cl.11.3.6.3), assuming a minimum shear reinforcement:

β = 0.18 Find V c

V c = φ c λβ f c ′td v (A23.3 Eq. 11.6)

[6.12]

= 0.65 × 1.0 × 0.18 30MPa × 200mm × 4000mm = 512.7kN ≅ 510kN Since

V f = 1500kN > V c = 510kN it follows that the shear reinforcement is required. Find the required spacing of shear reinforcement

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V s ≥ V f – V c = 1500kN – 510kN = 990kN where

φ s A h f y d v cot θ V s = ---------------------------------s

[6.9]

Use 15M bars, hence A h = 200mm

2

Find θ (A23.3 cl.11.3.6.3) as

θ = 35° cot θ = 1.43 The spacing of horizontal reinforcement can be determined from equation (6.9) as

φ s A h f y d v cot θ s = ---------------------------------Vs 2

0.85 × ( 200mm ) ( 400MPa ) ( 4000mm )1.43 - = 393mm = ----------------------------------------------------------------------------------------------------------3 990 × 10 N So,

s = 393mm ≅ 350 mm Check whether the spacing is within the CSA A23.3 prescribed limits (Cl.14.1.8.4).

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

Since

s = 350mm < 500mm okay Horizontal reinforcement: 15M@350 Check whether one layer of reinforcement is adequate (A23.3 Cl.14.1.8.3). Since t = 200mm < 210mm one layer of reinforcement is adequate. 3. Design the wall for flexure and axial load. Design axial and flexural load effects:

P f = 1500kN M f = 5100kNm •

Determine the minimum vertical reinforcement (A23.3 Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 200 = 200 × 10 mm

2

A vmin = 0.0015A g

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3

2

2

= 0.0015 × ( 200 × 10 mm ) = 300mm ⁄ m Set 2

A s = A vmin = 300mm ⁄ m Determine the required bar spacing. 2

15M bars ( A b = 200mm , Table A.1 )

1000 1000 s ≤ A b ------------ = 200 ------------ = 666mm As 300

[3.29]

Check whether the spacing is within the limits prescribed by the CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 200 = 600mm 500mm ← governs

Since

s = 666mm > s max = 500mm Hence, use

s = 500mm Vertical reinforcement: 15M@500 Actual area:

Ab 200 2 A v = ------ × 1000 = --------- × 1000 = 400mm ⁄ m s 500 •

Determine the moment resistance ( M r ) .

i) Determine the total vertical reinforcement area ( A vt ) along the wall length as 2

A vt = A s × l w = ( 400mm ⁄ m ) ( 5m ) = 2000mm

2

c ii) Calculate the parameters ω , α , and ---- . lw α 1 ≅ 0.8 , β 1 ≅ 0.9

[3.7]

φ s f y A vt ω = ------------------φ c f c ′l w t

[13.9] 2

0.85 × 400MPa × 2000mm = ----------------------------------------------------------------------------------------- = 0.035 0.65 × 30MPa × 5000mm × 200mm 3 Pf × N 1500 10 α = ------------------- = ----------------------------------------------------------------------------------------- = 0.077 [13.10] 0.65 × 30MPa × 5000mm × 200mm φ c f c ′l w t

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c ω+α 0.035 + 0.077 ---- = -------------------------------- = ------------------------------------------------- = 0.142 [13.8] lw 2ω + α 1 × β 1 2 × 0.035 + 0.8 × 0.9 iii) Find the M r value.

Pf c M r = 0.5φ s f y A vt l w 1 + ----------------- 1 – ---lw φ s f y A vt

[13.7] 3

1500 × 10 N 2 = 0.5 × 0.85 × 400MPa × ( 2000mm ) ( 5000mm ) 1 + --------------------------------------------------------------------2- × 0.85 × 400MPa × 2000mm × ( 1 – 0.142 ) = 4676kNm Since

M r = 4676kNm < M f = 5100kNm it follows that the moment resistance is not adequate. However, let us take into account the effect of concentrated reinforcement as well. Assume the minimum concentrated reinforcement according to A23.3 Cl.14.1.8.8, that is, 2-15M bars at each end of the wall section.

2-15M bars: A s = 2 × 200 = 400mm

x = 4850mm

2

(see the sketch above)

Moment resistance provided by the concentrated reinforcement can be determined as

∆M r = T × x = ( φ s f y A s )x 2

= ( 0.85 × 400MPa × 400mm ) ( 4850mm ) = 659.6kNm ≅ 660kNm The total wall moment resistance is total

Mr

= M r + ∆M r = 4676 + 660 = 5336kNm

Since total

Mr

= 5336kNm > M f = 5100kNm

it can be concluded that the total moment resistance provided by the combined distributed and concentrated wall reinforcement is adequate.

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4. Sketch a design summary.

__________________________________________________________________________ 13.8. 1. Determine the design axial loads, bending moments, and shear forces. Factored wind loads (NBC 2005 Table 4.1.3.2):

H 1f = 1.4 × H 1 = 1.4 × 400kN = 560kN H 2f = H 1f = 560kN H 3f = 1.4H 3 = 1.4 × 200kN = 280kN Factored gravity loads (NBC 2005 Table 4.1.3.2):

P 1f = 1.25P 1 = 1.25 × 500 = 625kN P 2f = P 1f = 625kN P 3f = 1.25 × 600 = 750kN Develop the bending moment, shear force, and axial force diagrams.

The critical wall section is at the base of the wall, where

M f = 8820kNm ≅ 8800kNm V f = 1400kN Copyright © 2006 Pearson Education Canada Inc.

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P f = 2000kN 2. Design the wall for shear. Design shear force:

V f = 1400kN Wall length: l w = 6000mm Wall thickness: t = 250mm The effective shear depth ( d v ) is

d v = 0.8l w = 0.8 × 6000mm = 4800mm Find β (A23.3 Cl.11.3.6.3), assuming a minimum shear reinforcement

β = 0.18 Find V c .

V c = φ c λβ f c ′td v (A23.3 Eq.11.6)

[6.12]

= 0.65 × 1.0 × 0.18 30MPa × 250mm × 4800mm = 769kN Since

V f = 1400kN > V c = 769kN it follows that the shear reinforcement is required in this case. Find the required spacing of shear reinforcement

V s ≥ V f – V c = 1400 – 769 = 631kN where

φ s A h f y d v cot θ V s = ---------------------------------s

[6.9]

Use 10M bars in 2 layers (because the wall thickness is larger than 210 mm according to A23.3 Cl.14.1.8.3), hence

A h = 2 × 100 = 200mm

2

Find θ (A23.3 cl.11.3.6.3) as

θ = 35° cot θ = 1.43 The spacing of horizontal reinforcement can be determined from equation (6.9) as

φ s A h f y d v cot θ s = ---------------------------------s

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2

0.85 × ( 200mm ) ( 400MPa ) ( 4800mm )1.43 - = 740mm = ----------------------------------------------------------------------------------------------------------3 631 × 10 N Check whether the spacing is within the CSA A23.3 prescribed limits (Cl.14.1.8.4).

s max ≤

3t = 3 × 250 = 750mm 500mm ← governs

Since

s = 740mm > s max = 500mm Use s = 500mm Horizontal reinforcement: 10M@500 (2 layers) In this design, two layers of reinforcement are required according to A23.3 Cl.14.1.8.3 (wall thickness larger than 210 mm). 3. Design the wall for flexure and axial load. Design axial and flexural load effects:

P f = 2000kN M f = 8800kNm •

Determine the minimum vertical reinforcement (A23.3 Cl.14.1.8.5). 3

A g = 1000mm × t = 1000mm × 250mm = 250 × 10 mm

2

A vmin = 0.0015A g 3

2

2

= 0.0015 × ( 250 × 10 mm ) = 375mm ⁄ m Assume 2

A s = A vmin = 375mm ⁄ m Determine the required bar spacing. 2

10M bars ( Area = 100mm , Table A.1 ) Assume 2 layers of reinforcement, so A b = 2 × 100 = 200mm

1000 1000 s ≤ A b ------------ = 200 × ------------ = 533mm As 375

2

[3.29]

Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).

s max ≤

3t = 3 × 250 = 750mm 500mm ← governs

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Since

s = 533mm > 500mm use

s = 500mm Vertical reinforcement: 10M@500 (2 layers) •

Determine the moment resistance ( M r ) .

i) Determine the total vertical reinforcement area ( A vt ) along the wall length as 2 Ab 200mm -------------------- × 1000 = 400mm 2 ⁄ m ----As = × 1000 = 500mm s 2

A vt = A s × l w = ( 400mm ⁄ m ) ( 6m ) = 2400mm

2

c ii) Calculate the parameters ω , α , and ---- . lw α 1 ≅ 0.8

β 1 ≅ 0.9

φ s f y A vt ω = ------------------φ c f c ′l w t

[3.7] [13.9] 2

0.85 × 400MPa × 2400mm = ----------------------------------------------------------------------------------------- = 0.028 0.65 × 30MPa × 6000mm × 250mm 3 Pf 2000 × 10 N ---------------------------------------------------------------------------------------- = 0.068 -----------------= α = 0.65 × 30MPa × 6000mm × 250mm φ c f c ′l w t

c ω+α 0.028 + 0.068 ---- = ------------------------------- = ------------------------------------------------- = 0.12 lw 2ω + α 1 × β 1 2 × 0.028 + 0.8 × 0.9

[13.10]

[13.8]

iii) Find the M r value.

Pf c M r = 0.5φ s f y A vt l w 1 + ----------------- 1 – ---lw φ s f y A vt

[13.7] 3

2000 × 10 N 2 = 0.5 × 0.85 × 400MPa × ( 2400mm ) ( 6000mm ) 1 + --------------------------------------------------------------------20.85 × 400MPa × 2400mm × ( 1 – 0.12 ) = 7434kNm Since

M r = 7434kNm < M f = 8800kNm it follows that the moment resistance is not adequate (deficiency on the order of 15%).

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Let us increase the amount of vertical reinforcement by decreasing the bar spacing to s = 300mm . Use 2 layers of 10M@300 vertical rebars:

1000 2 1000 2 A v = A b ------------ = 200mm ------------ = 666mm ⁄ m s 300 2

2

A vt = A s × l w = ( 666mm ⁄ m ) × ( 6m ) = 3996mm ≅ 4000mm

2

c Recalculate ω , α , and ---- as follows lw 4000 ω = ------------ × 0.028 = 0.047 2400 α = 0.068 (same as before) c ω+α 0.047 + 0.068 ---- = ------------------------------- = ------------------------------------------------= 0.141 lw 2ω + α 1 × β 1 2 × 0.047 + 0.8 × 0.9

[13.8] 3

2000 × 10 N 2 M r = 0.5 × 0.85 × 400MPa × ( 4000mm ) ( 6000mm ) 1 + --------------------------------------------------------------------20.85 × 400MPa × 4000mm × ( 1 – 0.14 ) = 8669kNm Since

M r = 8669kNm < M f = 8800kNm the moment resistance is still not adequate. However, let us take into the account the effect of concentrated wall reinforcement as well. Assume the minimum concentrated reinforcement according to A23.3 Cl.14.1.8.8, that is, 2-15M bars at each end of the wall section.

2-15M bars: A s = 2 × 200 = 400mm

2

x = 5850mm Moment resistance provided by the concentrated reinforcement can be determined as 2

∆M r = T × x = ( φ s f y A s )x = ( 0.85 × 400MPa × 400mm ) ( 5850mm ) = 796kNm ≅ 800kNm

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The total wall moment resistance is total

Mr

= M r + ∆M r = 8669 + 800 = 9469kNm

Since total

Mr

= 9469kNm > M f = 8800kNm

it can be concluded that the total moment resistance provided by the distributed and concentrated wall reinforcement is adequate. 4. Sketch a design summary.

Copyright © 2006 Pearson Education Canada Inc.

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