ACI- SP-17M(09) ACI REINFORCED CONCRETE DESIGN HANDBOOK A Companion to ACI 318-05 SI system

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4

ACI DESIG N HANDBOOK-S P-17M(09)

Appendix C $=0.57+67e,

0.003

H

--r·-;

:

"'41,....

j ,,,./'

Reinforcement closest to the tension face

.,· Ld ~

E1 = 0.005

~~ ,, :

\._)

~ i j·

/,.

,, .: i

;

Chapter 9 cj>=0.48+83E, J.....................L............L......... E, = 0.002 E1 = 0.005

- - i'

f-.........

!

,

t

inimum permitted for beams E1 = 0.004 min. strain permitted for pure flexure

Fig. 1.2-Strength reduction ()factors for Grade 420 reinforcement.

Aids Flexure 1 through Flexure 4, included at the end of the chapter, were developed using this condition. Accordingly, T=C

=

Asfl' 0.85fd b

(1 -8)

( J-1)

Asfy = 0.85/~ P1cb

p

Pie =

o.85/c' P1c df)'

M 11 -- bd2[1

(1 -2)

(1-3)

Pfv ] pf

-~·~

uJ:

)'

\.J (1 -9)

M11 = bd2KII

(1 -10)

pf,. ] .f' Kn = [ 1 - l.7fd P1 y

(1 -11)

where where

As

(1 -4)

p = bd

The c/d ratio in Eq. (1 -3) can be written in terms of the steel strain Es illustrated in Fig. 1. 1. For sections with single layer tension reinforcemen t, d = d 1 and Es = E1. The c/d ratio for this case becomes

Flexure 1 through 4 contain ~K11 values computed by Eq. (1-11 ), where the ~-factor is obtained from Fig. 1.2 for selected values of i::1 listed in the design aids. Flexure Examples 1 through 4 illustrate the application of Flexure 1 through 4.

1.2.2 Rectangular sections with compression reinforce0.003 - = - = -d 0.003 + e, d, C

p =

C

o.85fc' p1 0.003 0.003 + E 1 fv

(1 -5)

(1-6)

Equation (1-6) was used to generate the values for reinforcement ratio p (%) in Flexure 1 through 4 for sections with single layer tension reinforcemen t. For other sections, where the centroid of tension reinforcemen t does not coincide with the centroid of extreme tension layer, multiply the p values given in Flexure 1 through 4 by d/d. Compute the nominal moment strength from the internal force couple as shown as follows

M ,, = A s~,(d From Eq. (1 -2),

\.J

Pt)

(1-7)

\._,I

ment-Gener ally, flexural members are designed for only tension reinforcement . Any additional moment strength required in the section is usually provided by increasing the section s ize or the amount of tens ion reinforcemen t. However, the cross-sectional dimensions of some applications can be limi ted by architectural or functional requirements, and the extra moment strength may have to be provided by additional tension and compression reinforcement . The extra steel generates an internal force couple, adding to the sectional moment strength without changing the section's ductility. In such cases, the total moment s trength consists of two components: i) moment due to the tension reinforcemen t that balances the compression concrete, and ii) moment generated by the internal steel force couple consisting of compression reinforcement and an equal amount of additional tension reinforcemen t, as illustrated in Fig. 1.3.

M11=M1 +M2

(1 -12)

M 1 =K11bd2

(1-13)

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ACI DESIGN HANDBOOK-SP-17M(09)

6

b

- - ___ ___, ---· +-- ·- -

d

-

= A .: I

~,._,-.---,-.--=

Cf

71

T=!~-r;= :-~

--~ -

l J~?~--L ~~-=- -~-~~-~ l" ~ ~ b

hr

l_

-

- -

:

·-

iI

: : A :L_·__sf1:

M

(

+

nr

bw

r-

----~ ---- -, , ~

n.a.

1

I)

.:

"_,.~

Cw

-+-

\,_/

~-~-~-~(- ' '_:_~

M nw

A

I

SW

___.. ,._ Tr

••

- - Tw

Fig. 1.4- T-section behavior. Asw

P..-

(1-23)

bwd

Moment components M,if and Mmv can be obtained from Flexure 1 through 4 when the tables are entered with Pf and Pw values. For design, p1 needs to be found first and this can be done from the equilibrium of internal forces for the portion of total tension steel balancing the overhang concrete. This is illustrated as follows. T1 =Cr

(1-24)

A,J/2, = 0.85f: hj(b - bw)

(1-25)

0.85fc'

PJ

= -fy

'::J d

(1-26)

Equation (1-26) was used to generate Flexure 7 and 8. Flexure Examples 6, 7, and 8 illustrate the use of Flexure 7 and 8. When T-section flanges are in tension, part of the flexural tension reinforcement is required to be distributed over an effective area, as illustrated in Flexure 6, or a width equal to 1/10 the span, whichever is smaller (Section 10.6.6). This requirement is intended to control cracking that can result from widely spaced reinforcement. When 1/10 of the span is smaller than the effective width, additional reinforcement should be provided in the outer portions of the flange to minimize wide cracks in these regions.

1.3-Minimum flexural reinforcement Reinforced concrete sections that are larger than required for strength, for architectural and other functional reasons, may need to be protected against a brittle failure immediately after cracking by a minimum amount of tension reinforcement. Reinforcement in a section is effecti ve only after the cracking of concrete. When the reinforcement area is too small to generate a sectional strength that is less than the cracking moment, the section cannot sustain its strength upon cracking. To safeguard against such brittle failures, ACl 318M requires a minimum area of tension reinforcement in positive and negative moment regions (Section 10.5.1).

As, min

0.25 Jf: bwd 4f,.

but not less than l.4bwdffy

(1 -27)

The aforementioned requirement is indicated in Flexure 1 through 4 by a horizontal line above which the reinforcement ratio p is less than that for minimum reinforcement. For statically determinate members, when the T-section flange is in tension, the minimum reinforcement required to have a sectional strength above the cracking moment is approximately twice that required for rectangular sections. Therefore , Eq. (1-27) is used with bw replaced by 2bw or the flange width, whichever is smaller (refer to Section 10.5.2). When the steel area provided in every section of a member is high enough to provide at least 1/3 greater flexural strength than required by analysis, then the minimum steel require ment need not apply (refer to Section 10.5.3). This exception prevents the use of excessive reinforcement in very large members that have sufficient reinforcement. For structural slabs and footings, minimum reinforcement in the direction of the span is the same as that used for shrinkage and temperature control (refer to Section 10.5.4). The minimum area of such reinforcement is 0.0018 times the gross area of concrete for Grade 420 deformed bars (refer to Section 7 .12.2.1). Where higher grade reinforcement is used, with yield stress measured at 0.35 % strain, the minimum reinforcement ratio is proportionately adjusted as (0.0018 x 420)//2,· The maximum spacing of this reinforcement is limited to three times the slab or footing thickness, or 450 mm, whichever is smaller (refer to Section 10.5.4).

'._/

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1.4-Placement of reinforcement in sections Flexural reinforcement is placed in a section with due considerations given to reinforcement spacing, crack control, and concrete cover. lt is usually preferable to use a sufficient number of small bars, as opposed to fewer large bars, while respecting spacing requirements. 1.4.1 Minimum spacing of longitudinal reinforcementLongitudinal reinforcement should be placed with sufficient spacing to allow proper placement of concrete. Flexure 9 shows the minimum spacing requirement for beam reinforcement. 1.4.2 Concrete protection for reinforcement-Flexural reinforcement should be placed to maximize the lever arm between internal fo rces for increased moment strength. This implies that the main longitudinal reinforcement should be placed as close to the concrete surface as possible. The reinforcement should be protected against corrosion and aggressive environments by a sufficiently thick concrete cover (refer to Section 7.7), as indicated in Flexure 9. The concrete cover should satisfy the requirements for fire protection (referto Section 7.7 .7) .

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8

ACI DESIGN HANDBOOK-SP-17M(09}

1.5-Flexure examples Flexure Example 1: Calculation o,f tension reinforcement area for a rectangular tension-controlled cross section

\,_/'

For a rectangular section subjected to a factored bending moment M u, detennine the required tension reinforcement area for the dimensions given. Assume interior construction not exposed to weather. .GiY.m;

b h

fJ fy Mu

= = = = =

250mm 510mm 28 MPa 420MPa 122 kN·m

ACI318M-05 section Procedure 7.7.1 Estimate d by allowing for clear cover, the radius of longitudinal reinforcement, and stirrup diameter.

I•

b

•I

hT[lf Calculation Considering a minimum clear cover of 38 mm for interior exposure, allow 65 mm to the cenu-oid of main reinforcement.

Compute q>Kn = M J(bd2 ) . Select p from Flexure 1.

d = 510 - 65 = 445 mm q>K11 = 122 x 10°/(250 x (445)2] = 2.46 MPa For q>Kn = 2.46 MPa, p = 0.70%

Compute required steel area: As= pbd.

As= pbd = 0.0070 x 250 x 445 = 778 rnm 2

Design aid Flexure 9 \._,I

Flexure 1

'.._,/

Use three No. 19; (A 5 )prov = (3)(284) = 7.6.1 3.3.2

Detennine the provided steel area.

10.3.4 9.3.2

For reinforcement placement, refer to Flexure Example 9.

852 mm 2

Flexure 9

Note: Three No. 19 can be placed within a 250 mm width. (p)prov= (852)/((250)(445)] = 0.75% Note: for (p)prov = 0. 75% &1 ::: 0.0163 c; = 0.0163 > 0.005 (tension-controlled section) and ¢ = 0.9

Flexure 1

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10

ACI DESIGN HANDBOOK-SP-17M(09)

Flexure Example 3: Calculation of tension reinforcement area for a rectangular cross section in the transition zane For a rectangular section subjected to a factored bending moment Mu , determine the required area of tension reinforcement for the dimensions given. Assume interior construction not exposed to weather. .Gi.Y.en;_

h

.r;

As =

ACI318M-05 section Procedure 7.7.1 Estimated by allowing for clear cover, the radius of longitudinal reinforcement, and stirrup diameter.

•I

I'

Calculation Design aid Considering a minimum clear cover of 40 mm for interior Flexure 9 exposure, allow 65 mm to the centroid of main reinforcement

Compute ~K11 =MJ(bd ) . Select p from Flexure 1.

d = 660 - 65 = 595 mm ~Kn= 660 x 106 /[360 X (595 )2] = 5.18 MPa For ~K11 = 5.18 MPa, p = 1.59%

Compute As = pbd.

A s= pbd = 0.0 159 x 360 x 595 = 3406 mm2

Determine the steel area provided.

Try No. 25 bars; 3406/510 = 6.7.

2

7.6.1 3.3.2

b

- I•

25 mm maximum aggregate size b = 360mm h = 660mm = 28MPa = 420MPa .fy Mu = 660k.N·m

'\_,,

Need seven No. 25 bars in a single layer, but seven No. 25 bars cannot be placed in a single layer within a 360 mm width without violating spacing limits. Try placing in two layers.

\J Flexure 1

'-..,..I

Flexure 9

Allow 90 mm from the extreme tension fiber to the centroid of two layers of reinforcement. Revise d = 660 - 90 = 570 mm ~K11

= 660 x 106/[360 x (570)2] = 5.64 MPa

\._,I

Flexure 1

For ~K,, = 5.64 MPa, p = 1.77% As= pbd = 0.01 77 x 360 x 570 = 3632 mm 2

Try No. 25 bars; 3632/510 =7 .1 7 .6.1 3 .3.2

Select eight No. 25 bars in two layers (four No. 25 bars in each layer). Note that four No. 25 bars can be placed within a 360 mm width.

Flexure 9

(AJpro v = (8)(510) = 4080 mm2

(p)prov = 4080/((360)(570)]

10.3.4 9.3 .2

= 0.020

Note: For (PJprov = 0.020, ¢Kn = 5.76 MPa ci = 0.0042 st= 0.0042 < 0.005 (transition zane) ¢ = 0.83 and ¢Mn > Mu

Flexure 1

Flexure 1

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ACI DESIGN HANDBOOK- SP-17M(09)

12

Flexure Example 5: Calculation of tension and compression reinforcement area for a rectangular beam section subjected to positive bending

~

For a rectangular section subjected to a factored positive moment Mu, determine the required tension and compression reinforcement area for the dimensions given as follows. Given: b = h = d' =

1; Jy Mu

= =

=

360mm 620mm 65mm 28 MPa 420MPa 786kN·m

ACI318M-05 Procedure section Estimate d by allowing for clear 7.7.1 cover, the radius of longitudinal reinforcement, and stirrup diameter. Compute q>K11 = Mul(bd2 ). Select p from Flexure l.

I•

ctr:

b

'" I

i

::11ct·

Design aid Calculation Flexure 9 Considering a minimum clear cover of 40 mm for interior exposure, allow 65 mm to the centroid of main reinforcement. °--._I

d = 620 - 65 = 555 mm q>K11 = 786 x 106/(360 x (555)1) = 7.09 MPa q>K11 = 7.09 MPa is outside the range of Flexure 1. This

Flexure 1

indicates that the amount of steel needed exceeds the maximum allowed when only tension steel is provided. Therefore, compression steel is needed.

7.6. J 3.3.2

Compute (As -A;). Select a reinforcement ratio close to the maximum allowed to reach the full strength of compression concrete. Select p = 1.8% (slightly below Pmax= 2.06% so that when the bars are placed, Pmax is not exceeded).

Select p = 0.018 (f:1 =0.005) 2 As -A.: = pbd = 0.ol8 x 360 x 555 = 3596 mm Try No. 25 bars; 3596/510 =7 .05. Select eight No. 25 bars for (As - A;). However, eight No. 25 bars cannot be placed in a single layer. Try two layers.

Flexure 1

\._I Flexure 9

Allow 90 mm from the extreme tension fiber to the centroid of two layers of No. 25 bars.

10.3.4 9.3.2

Revised = 620 - 90 = 530 mm. As - A; = pbd = 0.018 X 360 x 530 = 3434 mm 2 Try No. 25 bars; 3434/510 = 6.73 Select seven No. 25 bars for (A 5 -A;) to be placed in two layers. (As - A;) = (7)(510) = 3570 mm 2 Corresponding p = 3570/((360)(530)] OK = 0 .0187 < Pmax = 0.0206 For p = 0.0187, K11 = 5.75 MPa, f: 1 =0.0047, and qi= 0.87 Compute moment to be resisted q>M11 = q>K11 bd2 by compression concrete and corresponding tension steel (A, -A; ). q>M11 = 5.75 x 360(530)2/10 6 = 58 1 kN·m Compute moment to be resisted by the steel couple (with an equal tension and compression steel area of A; ).

\.._I

Flexure 1

q>M,; = Mu - q>M11 q>M,; = 786 - 581 =205 kN-m

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ACI DESIGN HANDBOOK- SP-17M(09)

14

a Flexure Example 6: Calculation of tension reinforcement area for a T-secrion subjected to positive bending, behaving as rectangular section For a T-section subjected to a factored bending moment Mu, determine the required tension reinforcement area for the dimensions given. Given: b = bw = d = = ht = Iv = Mu =

t:

760mm 360 mm 480 mm 65mm 28MPa 420MPa 312 k.N·m

ACI318M-0S section

d =480 mm

As

••

f------+j

b.... = 360 mm

Procedure Assume tension-controlled section (= 0.9). Determine if the section behaves as a T- or rectangular section. When M,, > [0.85/d bhfd - hJ'2)]

L-section, otherwise rectangular section behavior. Compute the amount of steel that balances compression concrete in the flange o verhang from Flexure 7 . Find the moment amo unt resisted by P.r from Flexure 1.

7.6. J 3.3.2

'-"

dlht = 835/75 = 1 1. 13

Flexure 7

Pt = 0.51 % For Pf = 0.51 % Kn = 1. 83 MPa, and q> = 0 .90

Mi = q,K,/b - b....,)d2

Flexure 1

= 1.83(900 - 550)(835)2 = 447 kN-m q>Mw = M 11 - M1= 2400 - 447 = 1953 kN-m Determine the amount of steel q,K,, = q>M,,J[(bw)(d)2] req uired to resist the remaining 6 moment. This additional moment is q,K,, = 1953 x 10 /((550)(835 )2] = 5.09 MPa Flexure 1 For Kn= 5.09 MPa, Pw = 1.56% to be resisted by the web, Pw· rolled). Note: ¢ = 0.90 (tension-cont Ai= pfb - bw)d = 0.005 1(900 - 550)(835) = 1490 mmz Compute the total area of tension 2 Aw= Pnhwd = 0 .0156(550)(83 5) = 7 164 mm reinforcemen t. 2 As= At+ Aw = 1490 + 7 164 = 8654 mm Flexure 9 needed. are bars 29 Select No. 29 bars; fo urteen No. Fourteen No. 29 bars cannot be placed in a single layer. Therefore , use two layers of reinforcement and revise the design. d = 900 - 90 = 8 10 mm Recalculate the effective depth d and revise design. Asswne cover of 90 mm Note: Reduced d will result in increased area of steel and the beam will continue behaving as a T-section (no to the centroid of two layers of need to check again). reinforcement.

\....-,I

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ACI DESIGN HANDBOOK-S P-17M(09)

18

Flexure Example 9: Placement of reinforcemen t in the recrangular beam section designed in Flexure Example 1 cover Select and place flexural beam reinforcemen t in the section provided below, with due considerations given to spacing and weather. to exposed not requirements. Assume interior construction Given: 20 mm maximum aggregate size No. 10 stirrups 2 = 787 mm As b = 250mm h = 500mm Jy = 420MPa

ACI318M-05 section

7.7.1

7.6

b

I•

•I

{ r As

=

Calculation Procedure Determine bar size and number of bars. Select No. 19 bars; No. of bars = 787 /284 = 2.8. Use three No. 19 bars. Considering minimum clear cover of 40 mm on each Determine bar spacing. side for interior exposure and allowing two srirrup bar diameters, s = [250 - 2(40) - 2(10)- 3(20))/2 = 45 mm Check against minimum spacing.

\.._.,I

Design aid

Flexure 9 \._,I

Flexure 9 (s),nin = {db;( 1 ~) amax; 25 mm} (s)min = {20 mm; ( 1~) (20 mm); 25 mm}= 20 mm

OK

s = 50 mm > 20 mm 10.6.4

Check against maximum spacing as governed by crack control.

(S)111ax = 380(280ifs) - 2.5cc ~ 300(280ifs) fs = 2/3fv = 2/3(420 MPa) = 280 MPa Cc= (40·+ 10) = 50 mm (s)max = 380(1) - 2.5(50) = 255 mm s = 50 mm < 300 mm OK

Eq.(1-28)

Final bar placement.

Provide three No. 19 as indicated below.

Flexure 9

40 mm

_,

\.._,I

\._,I

1 0mm

,. 20mm

40mmT

I

I

' ' 50mm

250mm

.1

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ACI DESIGN HANDBOOK-SP-17M(09)

20

1.6-Flexure design aids

~ r

Flexure 1: Flexural coefficients for rectangular beams with tension reinforcement; fy = 420 MPa 2 P =A/bd

Nonualweight concrete b d

Vu= 258 kN

$Ve= 111.7 kN

-

d = 500 m

Face of support

ACI318M-05 section Procedure Step 1-Detennine, at d from face of 11.1.3.1 support, the value of maximum Vu = Vu-wud/12. 11.3.1.1 Step 2-Detennine the value Ve= 0. l 7( Ji: )bd or use design aid for b = 330 mm and 9.3 .2.3 h = 572 mm; Kvc = 146 kN Step 3- Detennine required Vu each side of Pui· Left of Pu1, vu= Vu- WuX1 Right of Pui, change in Vu= Pu1 11. J.l Step 4-Determine spacing s 1 11.5.7.2 required for No. 10 U-stirrups at face of support. A v= 2(7 1) = 142 mm2 s 1 = Avfyt d/(maximum Vu - q> Ve)

11.5.5.1 11.5.7.2 11.1.1

X

Step 5- Since maximum spacing S 111ax = d/2 with d = 250 mm, determine value of Vu = Ve+ Avfy1dls.

X1=1 .4m

tv1

Calculation Maximum Vu = 258 kN - 67 kN/m(500 mm) x 10- 3 = 224.5 kN Vc = 0.17(0.75)( J 28 MPa)330 mm(500 mm) = 111.3 kN Ve = 0.75KfeKvc = 0.75(1.00) 149 = 111.7 kN

Design aid \....._,/

Shear 2 Table 2(c)

Left Pui, Vu= 258 kN - 67 kN/m(l.4 m) = 164.2 kN Right Pu 1, Vu= 164.2 - 67 = 97.2 kN

_ (0.75)142 mm\ 420 MPa)(500 mm) x 10- 3 (224.5 - 111.7) = 198 mm Maximum spacing s111ax = 500 mm/2 = 250 mm

\,_I

SI -

Shear 4.2

\....._,/

0

V "

= lll. 7 kN + 0.75(142 mm- )(420 MPa)(500 mm) 250 mm

= 201 kN

11.5.6.1

or use design aid for Vs when s = 250mm Vu = Vc + Vs

Vs= 119 kN for No. 10 stirrups at 250 mm spacing

Conclude: uses = 175 mm until Vu < 201 kN and uses= 250 mm until Vu < 0 .5Vc

From face of support, use 75 mm space then fi ve spaces at 175 mm (890 mm) and fi ve spaces at 250 mm (1 250 mm); 2160 mm > 2032 mm

Vu= 11 1.7kN + 0.75(1 19 kN) = 201 kN Step 6- Detennine distance x from face of support to point at which Vu = 201 kN x = (change in shear)/wu x = (258 kN - 201 kN)/67 kN/m = 0.85 m = 850 mm Step 7-Determine distance 1\,i, distance beyond x I at which no stirrups are required. t;,1 = (97.2 kN - 111.7 kN/2)/67 kN/m =0.61 7 m = 617 mm Find l;,1 = (Vu - Vcf2)lwu X] + f;v ] = 1400 + 617 = 2017 mm Compute x 1 + evl

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ACI DESIGN HANDBOOK-SP-17M(09)

40

Shear Example 6: Detennination of thickness required for perimeter shear strength ofa flat slab at an interior rectangular column

-"-../

Given: = 400mm = 600 mm J; = 35 MPa f yt = 420MPa Vu = 792 kN Normalweight concrete Refer to Shear Example 5 for diagram of shear perimeter and Code clauses be he

ACI318M-05 section 11.12.2.1 9.3.2.3 11.12.1.2 9.1.1

Procedure Step 1-Set up expression for ~Ve· ~Ve= 0.33~( Jf: )b0 d = 0.33q>( )2(he + d + b C + d)d Step 2-Equate Vu to qiVe and solve ford.

JI:

Calculation Vc= 0.33(0.75)(,/35 MPa)2(600 mm+ d + 400 mm + d)dmm 792(1 O-' N) = 1.46 N/mnl" (1 OOOd + 2d°1.) mm1 271,233 mm 2 = (1000d + 2d2 ) mm2 135,616 + 62,500 = 62,500 + 250d + d 2 d = (J198,116 ) - 250 = 195 mm h 5 = 195 mm + 20 mm + 16 mm= 231 mm, say, h 5 = 230mm

Step 3- Allow for 20 mm clear cover of tension bars to make h5 = d + 20 + bar diameter (estimated) ALTERNATE METHOD using Design aid Shear 5.1 9.3.2.3 Step I-Compute minimum Vn = V j v,, = 792 kN/0.75 = 1056 kN (he + be)= (600 mm+ 400 mm)= 1000 mm and compute (he + be) 11.12.1.2 1056 kN is between: Step 2- Withfj = 35 MP a and V11 = 1056 kN Vn = 887 kN and V,, = 1 I 83 kN Use (he + be)= 1000 mm, interpolate K1K2 = 150 + 50(10 56 - 887 ) = 178.5 MPa K1K2 (1183-887) 7 .7.1

11.12.2.l

7.7.1

Step 3- Compute J3e = he/be

Design aid

J3e = 600 mm/400 mm = 1.5 < 2, so K2 = 8.2 and Kl = KIK2 MPa/K2 = 21.77 MPa

\J

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Table 5. l (c)

Table 5.l(b)

Step 4- Table 5.1 (a) with (he+ be) = Table 5.l (a) 1000 and KI = 21.77 MPa Interpolate for d d= 150mm+ 16mm = 166mm Step 5- Allow for 20 mm clear cover h 5 = 189.88 mm + 20 mm + 16 mm= 225.88 mm slab, of tension bars to make h 5 = d + 20 + say, h 5 = 230 mm bar diameter (estimated)

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_;

FOREWORD

This edition of the AC! Design Handbook was prepared in a new format based on the feedback received from handbook users over the years. The advances in programmable and nonprogrammable electronic calculators and personal computers have enabled the elimination of numerous design aids that were intended to carry out relatively simple design calculations. Instead, explanatory materials have been added to each chapter, while maintaining essential design aids and illustrative examples. This is the first edition of the handbook after the introduction of explicit strain limits for tension- and compression-controlled sections in flexure, with variable strength reduction factors q> within the transition zone. This necessitated the development of a new set of design aids for members subjected to flexure. The increased use of higher strength concretes and higher grade reinforcement has been recognized and the design aids were developed for concrete strengths of up to 70 MPa for beams and 85 MPa for columns; and steel Grades 420 and 520. The column interaction diagrams of Chapter 3 were developed for nominal quantities and in a non-dimensional form to facilitate the use of interaction diagrams with any system of units and any strength reduction factor. The use of nominal quantities also facilitates column investigations. The chapters of the handbook were developed by individual authors, as indicated on the first page of each chapter. The authors were long-standi ng members of the fonner ACT Committee 340, Design Aids for Building Codes, which had the mandate to develop the previous editions of the handbook in accordance with the ACI 318 Building Code. Committee 340 was first established in 1958 and was discharged in 2003. The first handbook was developed on the basis of the 1963 edition of the ACI Building Code. The committee subsequently engaged in revising the handbook in the years to come, and developed updated editions of the handbook for the 1971, 1977, 1983 and 1995 ACI Codes. The current edition is based on ACI 318M-05. Although the handbook chapters were developed by individual authors, the first draft of the document was reviewed by the former members of ACI Committee 340. Their input is gratefully acknowledged. Furthermore, many individuals and former members of ACI Committee 340 contributed to the earlier editions of the handbook which formed the basis for the current edition. Their contributions, as well as the administrative and technical assistance received from ACI Staff, are gratefully acknowledg ed.

Murat Saatcioglu Editor

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42

ACI DESIGN HANDBOOK-SP-17M(09)

Shear Example 8: Determination of required thickness of a footing to satisfy perimeter shear strength at a rectangular column be = 400 mm h, = 400 mm 21MPa normalweight concrete = fd �= 420MPa Pu = 1165 kN Interior column, a.s = 40 Normalweight concrete

ACl318M-05 section

1 1.12.2.1 9. 3.2. 3

Procedure Step I-Determine net bearing pressure under factored load P11 f br = Pj(footing area) Step 2-Express V11 = fb,.(footing area- prism area) = fbr[2100 X 2100 - (400 + d)] Step 3-Express 10.5 kN·m, ties Threshold torsion=0.25(42.2 kN-m) =10.5 kN·m for torsion are required. Step 3-ls section large enough? fv =271 kNI(400 mm x 546 mm)= 1.24 Nlmm2 Compute!,, = Vj(b11.d) f,,1=72 kN-m(l640 mm)l[l.7(158,000 mm2)2 ] Computefvt =TuP,/(l.7A 0/) =2.78 Nlmm2 Compute limit Limit= 0.75(0.17 + 0.66](J3s MPa) =3.68 Nlmm2 =(j>[(0.17Jj; +U66Jj;)l1000 =

9.3.2.3

Calculation

ls +fv�] < limit? Therefore, section is large enough. Step 4-Compute (b wd)]l(J;.d ) A v is= [Vu - 0.17

J[f.;

JI:

Compute A,fs = T/[2A 0J;,cot0] Compute (A)s + 2A 1 ls)

J[I.24 + 2.78 ) =3.04 Nlm.m 2 < 3.68 Nlmm 2 2

2

3 - 0.17(0.75)(J35 MPa)(400)(510) A 1• Is=271 ( 10 ) kN(0.75)(4 20 MPa)(510)

=0.73 mm 2lmm

72( 10 ) N-mm =0.85 mm2 lmm A 1Is= ( 2 )(0.75)(134,385)(4 20)cot45 AJs + 2A,fs=0.73+ 2(0.85) =2.43 mm 6

Use No. 13 ties for which (A v + 2A 1 )1 s =258 mm, and compute s= s =2.5812.43=106.2 mm 285 mm! s + 2A,fs). Use 100 mm ls 0.062( Jt 'Jc' )b,)fy, < (A 11 ls + 2A,ls)? 0.062( )4001420 =0.349 mm YES

J3s

Design aid

48

ACI DESIGN HANDBOOK-SP-17M(09)

Shear Example 13: Determination of closed ties required for the beam of Example 12 to resistflexural shear and indetemunaie torque � Use the same data as that for Shear Example 12, except that the required torsion estimate of 72 kN-m is based on an indeterminate analysis, not an equilibrium requirement. = 35 MPa normalweight concrete f/ bw = 400 mm vu = 271 kN = !yr =420MPa = 600 mm h Tu = 72 kN-m (based on indeterminate analysis)

.t;,

ACl318M-05 section Procedure 11.3.1.l Step 1-Look up parameters for 11.5.7.2 Jc' 35 MPa, Grade 420 reinforcement, bw=400 mm, h = 600 mm l l.6.2.2a 11.6.3.1 11.6.3.6

Calculation = (1.118)193 kN = 215.8 kN KfcK vc

=

l l.6.2.2a

Kvs=224.7kN/m K1�1=(1.118)(113.7kN-m)=127.1 kN-m K_r�icr= (1.118)(50.3 kN·m) = 56.2 kN·m K1s 56.44 k:N-m/mm =

Step 2-If indeterminate T11 > q>KfcK rc, (0.75)56.2 k:N·m=42.2 kN·m, q>Kf�tcr , q>KfcKrcr can be used as Tu. Tu 72 k:N·m > 42.2 k:N-m Use T11 =42.2 kN·m 0.25 , 42.2 k:N·m > 0.25(56.2 kN·m)=14.1 kN·m. = 42.2 > KfcKtcr Step 3-If T,, ties are required. Therefore, ties are required. Step 4-Section is large enough if 2 71 kN 4 2 . 2 kN·m 2 2 = J£(V)(5�K1,K .,)] + [T.l(�K 1,K,)] < I =

=

11.6. la

,

{ 5(0.75)(215.8)} + { 0.75 x 1 27.I kN-m

2 J{ 0.335} 2 + { 0.443}

11.6.3.7 11.6.5.3

Step 5-Compute A)s = (V11 /q> - Kf�vc)/ Kvs Compute 2A,Js= T,,f(q>K,s) Compute (A vis+ 2A,Js) No. 13 ties provide 258 mm2 /mm Compute s< 258/(A/s+2A/s) Step 6---Compute Ph 2(b + h - 7). From Step 4, 2A,ls=1 .0 rnm 2 Compute At = (A,Js)(phcot 2 45) Is At> At .min 0.42(Jj: )Ac/fy -p0 1 ls? =

=

r

= = 0.555 < 1 Therefore, section is large enough. A,Js = (271 kN/0.75 -215.8)/224.7 kN/m = 0.6S nun:;/mm 2A,Js=42.2 k:N·m/[0.75(56.44)] =1.0 nun2/mm (A 11 /s + 2A,Js) = 1.65 mm2/mm s< 258/1.65= 156.4 mm. Use 150 mm spacing. Ph= 2(400 mm+ 600 mm -190 mm)= 1620 mm Thus A,ls =(1/2)(1 .0)= 0.5 mm A e = (0.5 mm)(1620 mm)(1.0) = 810 mm 2 A e .min = 0.42( ,.J35 MPa)240,000 mm 2 /420 MPa 2 2 - 1620(0.5) mm = 609.8 mm YES

11.6.6.1

Ae ,min =810 mm 2 governs. (No. 10 bars provide sufficient area, but select No. 13 as a minimum. Maximum spacing of longitudinal bars=pJ8 or 300 mm

1 l .6.6.2

For 10 positions, use No. 13 bars. Place (See Shear Example 12) three No. 13 in bottom and two No. 13 in each side face. Excess flexural capacity in top at d from support can serve in place of three No. 13 in top.

pJ8 = 1620/8 = 203 mm

Design aid Shear 2 Table 2(a) & 2(c) Shear 2(a) & 2(b) Shear 6.l(a) Shear 6.1(b) Shear 6.2(b)

50

ACI DESIGN HANDBOOK-SP-17M(09)

Shear 2: Shear strength coefficients Kfc, Kvc• and Kvs Reference: Sections 11.3. l. l and 11.5.7.2 Ve = 0.17(jJ:)bwd=0.9KrA,.d Vs Vn

=

=

Kfc = Kvs = K11, =

A.jy 150 mm Table 2(b) Values K.,, kN/m

Beamh,mm 250 300 350 400 450 500 550 600 650 700 750

800 850 900 950 1000 1050 1100 1150 1200

f.v orJy,, MPa

280 51.8 65.8 79.8 93.8 107.8 121.8 135.8 149.8 163.8 177.8 189.0

203.0 217.0 231.0 245.0 259.0 273.0 287.0 301.0 315.0

420 77.7 98.7 I 19.7 140.7 161.7 182.7 203.7 224.7 245.7 266.7 283.5 340.5 325.5 346.5 367.5 388.5 409.5 430.5 451.5 472.5

70 1.581

Ji: , but keep fer 10.56 s Ji: .

ACI DESIGN HANDBOOK-SP-17M(09)

52

Shear 3: Minimum beam height to provide development length required for No. 19, No. 22, and No. 25 Grade 420 stirrups Reference: Sections 12.13.2.1 and 12.13.2.2. Section 11.5.2 states "Design yield strength of shear reinforcement (bars) shall not exceed 420 MPa, ...."

0.17 do(420)

--Ir; h

Minimum beam height h = 2[0.17 db (420)/ Ji[+ 40] in millimeters.

Concrete J; , MPa

21 28 35 40 55 70

Minimum beam height h, mm Stirrup size No.19 No.22

660 582 528 490 434 396

757 666 605 559 493 450

No. 25

856 752 681 627 554 503

Nore: Values shown in the rable are for 40 mm clear cover over srirmps. For cover greater than 40 mm, add 2/cover-40 mm) to tabulated values.

Example: Determine whether a beam 600 mm high (h = 600 mm) with 35 MPa concrete will provide sufficient development length for No. 19 Grade 420 vertical stirrups. Solution: With/; = 35 MPa, for No. 19 stirrups, minimum beam height reads 528 mm for beams with 40 mm clear cover over stirrups. Because h = 600 mm, the beam is deep enough.

54

ACI DESIGN HANDBOOK-SP-17M(09)

Shear 4.2: Shear strength Vs with Grade 420 U-stirrups Reference: Sections 11.5.5.3 and 11.5.6.3

V5 = V11

-

Ve = A vfw(dls)

Ma.ximumbw = AJy,l(0.35s) whenfj :$ 31 MPa Ma.ximumhw = Avfy/(0.062s Ji[) whenfj 2 31 MPa

Table 4.2(a) Stirrup Beam d. size mm 200 250 300 350 400 450 500 550 No.10 U600 stirrups 650 700 750 800 850 900 950 1000 Maximumbw (mm) for f/ d/2 is not allowed

-

-119 128 136 145 153 162 170

500

,_

___

-

I I _] I I I

=--cc....

--

-�--.�. -1 -·-

-- --

-

-----·-

I

_I

--

-II

119 127 134 142 149

119 126 133

I 19

487

426

379

341

350

400

450

500

--

Table 4.2(b) Valu�s uf V,, kN

Sti_rrup size

No. 13 U-

stirrups

Beam d,

mm 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 JOO

Maxi1num b u .

(mm) for f/ d/2 is not allowed

I I

-· �---

--

- -

- ----- �

I I I

���1 I I I _I - I

I ·- I 217 619

56

ACI DESIGN HANDBOOK-SP-17M(09)

Shear 5.2: Shear strength of slabs based on perimeter shear at interior round columns when no shear reinforcement is used Reference: Sections 11.12.1.2 and 11.12.2.1 V11 =Ve= (K3) Jf: (kN) he = column diameter (mm) d = slab depth (mm) K3 = 0.33nd(d + he) when he< 5.37d K3 = 0.17nd(hc + 7.37d) when he> 5.37d for which Table 5.2(a) values are in bold type

Table 5.2(a) Column h,mm 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 l000

Values K3 (x 10 3 ). mm2 75 21 25 29 33 37 40 42

44

46 48 50 52 54 56 58 60 62

100 31 36 41 47 52 57 62

125 42 49 55 62 68 75 81 87 94 100

69 71 74 77 79 82 85 87 90 93

108 112 115 118 122 125 128

150 54 62 70 78 86 93 10) 109 117 124 )32 140 148

175 68 77 86 95 104 113 122 132 141 150 159 168 177 186 195

157 161 165 169

209 214

d,mm

200 83 93 104 114 124 135 145 156 166 176 187 197 207 218 228 238 249

230 l03 114 126 138 150 162 174 186 198 210 222 234 246 258 269 281 293

250 117 130 143 156 168 181 194 207 220 233 246 259 272 285 298 311 324

300 156 171 187 202 218 233 249 264 280 295 311 327 342 358 373 389 404

350 200 218 236 254 272 290 308 327 345 363 38] 399 417 435 454 472 490

400 249 270 290 311 332 352 373 394 415 435 456 477 498 518 539 560 581

450 303 327 350 373 397 420 443 467 490 5)3 537 560 583 606 630 653 676

Table 5.2(b) Values Vn � V,, 1$, kN

Jc', MPa

K3 (x 10 3 ), rnm 2

21

40 80 120 160 200 250 300 400 500 600 800 900 1000

28

183 367 550 733 917 1146 1375 1833 2291 2750 3666 4124 4583 5041 5499 5957 6416 6874 7332 8249

35

212 423 635 847 1058 1323 1587 2ll7 2646 3175 4233 4762 5292 5821 6350 6879 7408 7937 8466 9525

237 473 710 947 1183 1479 1775 2366 2958 3550 4733 5324 5916 6508 7099 769) 8283 8874 9466 10,649

I JOO

1200 1300 1400 1500 1600 1800

40 253 506 759 1012 1265 1581 1897 2530 3162 3795 5060 5692 6325 6957 7589 8222 8854 9487 10,119 11,384

55

297 593 890 l )87 1483 1854 2225 2966 3708 4450 5933 6675 7416 8158 8899 9641 10,383 11,124 11,866 13,349

70 335 669 1004 1339 1673 2092 2510 3347 4183 5020 6693 7530 8367 9203 10,040 10,877 11.713 12,550 13,387 15,060

500 363 389 415 441 467 492 518 544 570 596 622 648 674 700 726 752 778

58

ACI DESIGN HANDBOOK-SP-17M(09)

Shear 6.2: Shear and torsion coefficients K1 s Reference: Section 11.6.3.6 T,, = (2A 0 Arfyfs)cot0 = 2K1s(A/s) kN-m A 0 = O.85(h - 3.5)(b - 3.5) 0 = 45 degrees Table 6.2(a) Values K1s (kN-mlmm) with Grade 280 ties Beam/z, mm 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000

250 6.09 8.00 9.90 11.80 13.71 15.61 17.52 19.42 21.32 23.23 25.13 27.04 28.94 30.84 32.75 34.65

300 8.00 10.50 12.99 15.49 17.99 20.49 22.99 25.49 27.99 30.49 32.99 35.49 37.98 40.48 42.98 45.48

350 9.90 12.99 16.09 19.18 22.28 25.37 28.46 31.56 34.65 37.75 40.84 43.93 47.03 50.12 53.22 56.31

400 11.80 15.49 19.18 22.87 26.56 30.25 33.94 37.63 41.32 45.01 48.69 52.38 56.07 59.76 63.45 67.14

450 13.71 17.99 22.28 26.56 30.84 35.13 39.41 43.70 47.98 52.26 56.55 60.83 65.12 69.40 73.68 77.97

Beam b, mm 500 15.61 20.49 25.37 30.25 35.13 40.01 44.89 49.77 54.64 59.52 64.40 69.28 74.16 79.04 83.92 88.80

550 17.52 22.99 28.46 33.94 39.41 44.89 50.36 55.83 61.31 66.78 72.26 77.73 83.20 88.68 94.15 99.63

600 19.42 25.49 31.56 37.63 43.70 49.77 55.83 61.90 67.97 74.04 80.11 86.18 92.25 98.32 104.39 110.46

650 21.32 27.99 34.65 41.32 47.98 54.64 61.31 67.97 74.64 81.30 87.96 94.63 101.29 107.96 114.62 121.28

700 23.23 30.49 37.75 45.01 52.26 59.52 66.78 74.04 81.30 88.56 95.82 103.08 110.34 117.60 124.85 132.11

750 25.13 32.99 40.84 48.69 56.55 64.40 72.26 80.11 87.96 95.82 103.67 111.53 119.38 127.23 135.09 142.94

600 29.13 38.23 47.34 56.44 65.55 74.65 83.75 92.86 101.96 111.06 120.17 129.27 138.37 147.48 156.58 165.68

650 31.99 41.98 51.98 61.98 71.97 81.97 91.96 101.96 111.96 121.95 l 31.95 141.94 151.94 161.94 171.93 181.93

700 34.84 45.73 56.62 67.51 78.40 89.29 100.17 111.06 121.95 132.84 143.73 154.62 165.51 176.39 187.28 198.17

750 37.70

Table 6.2(b) Values K1s (kN-m/mm) with Grade 420 ties Beam h. mm 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000

250 9.14 12.00 14.85 17.71 20.56 23.42 26.28 29.13 31.99 34.84 37.70 40.56 43.41 46.27 49.12 51.98

300 12.00 15.74 19.49 23.24 26.99 30.74 34.49 38.23 41.98 45.73 49.48 53.23 56.98 60.73 64.47 68.22

350 14.85 19.49 24.13 28.77 33.42 38.06 42.70 47.34 51.98 56.62 61.26 65.90 70.54 75.18 79.83 84.47

400 17.71 23.24 28.77 34.31 39.84 45.37 50.91 56.44 61.98 67.51 73.04 78.58 84.11 89.64 95.18 100.71

450 20.56 26.99 33.42 39.84 46.27 52.69 59.12 65.55 71.97 78.40 84.82 91.25 97.68 104.10 110.53 116.95

Beam b, mm 500 23.42 30.74 38.06 45.37 52.69 60.01 67.33 74.65 81.97 89.29 96.60 103.92 111.24 118.56 125.88 133.20

550 26.28 34.49 42.70 50.91 59.12 67.33 75.54 83.75 91.96 100.17 108.39 116.60 124.81 133.02 141.23 149.44

49.48

61.26 73.04 84.82 96.60 108.39 120.17 l 31.95 143.73 155.51 167.29 179.07 190.85 202.63 214.41

60

ACI DESIGN HANDBOOK-SP-17M(09)

-� h

i::=i; f

Cross-Section

t .,y

i:,=0.005 Strain Distribution

Stress Distribution

Fig. 3.2-Column section analysis.

arranged in a circle have been published in The AC! Struc­ tural Journal (Everard 1997). The interaction diagrams contained in SP-7 were subsequently published in SP-17A (ACI Committee 340 1970). The related equations were derived considering the following: (a) For rectangular and square columns having steel bars placed on the end faces only, the reinforcement was assumed to consist of two equal thin strips parallel to the compression face of the section; (b) For rectangular and square columns having steel bars equally distributed along all four section faces, the reinforcement was considered to consist of a thin rectangular or square tube; and (c) For square and circular sections having steel bars arranged in a circle, the reinforcement was considered to consist of a thin circular tube. The interaction diagrams were developed using the rectan­ gular stress block, provided in Section 10.2.7. In all cases, for reinforcement within the compressed portion of the depth perpendicular to the compression face of the concrete (a = fk), the compression stress was reduced by 0.85/; to account for the concrete area displaced by the reinforcement bars within the compression stress block. The interaction diagrams were plotted in nondimensional form. The vertical coordinate [Kil = P11 /(f;Ag)] represents the nondimensional form of the nominal axial load strength of the section. The horizontal coordinate [R11 = M,,l(fd Ai)J represents the nondimensional nominal bending moment strength of the section. The nondirnensional forms were used so that the interaction diagrams could be used with any system of units (SI or in.-lb units). Because ACI 3 l 8M-05 contains different -factors in Chapter 9, Chapter 20, and Appendix C, the strength reduction factor was considered as 1.0 so the nominal values in the interaction diagrams could be used with any set of -factors. It is important to note that the -factors provided in Chapter 9 are based on the strain values in the tension reinforce­ ment farthest from the compression face of a member, or at the centroid of the tension reinforcement. Code Section 9.3.2 references Sections 10.3.3 and 10.3.4 where the strain values for tension conrrol and compression conrrol are defined. It should also be noted that the eccentricity ratios (elh = MIP), sometimes included as diagonal lines, are not included in the interaction diagrams. Using the eccentricity ratio as a

coordinate with Kil or R,, can lead to inaccuracies because the e/h lines converge rapidly at the lower ends of the diagrams. Straight lines for tension steel stress ratios !,.If;, have been plotted to assist in designing splices for the reinforcement. Further, the ratio fsf/2, = 1.0 represents steel strain E:y = f;JE, which is the boundary point for the compression control ­ factor, and the beginning of the transition zone for linear increase of the -factor to that for tension control. To provide interpolation for the -factor, other strain lines were plotted. The strain line for f;1 = 0.005, the beginning of the tension control zone, has been plotted on all diagrams. The intermediate strain line for E:, = 0.035 has been plotted for steel yield strength 420 MPa. The intermediate strain line for E:1 = 0.038 has been plotted for steel yield strength 520 MPa. Note that all strains refer to the reinforcement bar or bars farthest from the compression face of the section. Discussions and tables related to the strength reduction factors a.re contained in two articles in Concrete International (Everard 2002a,b). To illustrate designs prohjbited by Section 10.3.5, strain lines for f;1 = 0.004 have also been plotted. Designs between the lines for i::, = 0.004 and K11 < 0.10 are not permitted by ACI 318M-05. This includes tension axial loads, with K,, negative. Tension axial loads are not included in the interaction diagrams. However, the interaction diagram lines for tension axial loads are nearly linear from Kil = 0.0 to R,, = 0.0 with [Kil= A51f..lifd Ag)]. This is discussed in the next section. Straight lines for Kmax are also provided on each interac­ tion diagram. Here, Kmax refers to the maximum permissible nominal axial load on a column laterally reinforced with ties conforming to Section 7.10.5. Defining Ko as the theoretical axial compression strength of a member with R11 = 0.0, K111ax = 0.80K0 or considering Eq. (10-2) without the -factor, Pn,max = 0.8[0.85/d (Ag -As1 ) + f,.A51]

(3-5)

Then, (3-6) For columns with spirals conforming to Section 7.10.4, multiply K1110x values from the interaction diagrams by 0.85/ 0.80 ratio. The number of longitudinal reinforcement bars that can be contained is not limited to the number shown on the interaction

62

ACI DESIGN HANDBOOK-SP-17M(09)

pll •

;:-�ilure �rfac:e

Fig. 3.7-Faillfre surface S3, which is reciprocal ofsurface S 1. Fig. 3.5-Failure surface S 1.

Fig. 3.8-Graphical representation of reciprocal load method. Fig. 3.6-Failure surface S 2 .

pni

1 P,u

1

1

pn_v

Po

= -+---

(3-10)

where

approximation of nominal axial load strength at eccentricities ex and ey nominal axial load strength for eccentricity ey , P v: along the y-axis only (x-axis is axis of bending) P11Y = nominal axial load strength for eccentricity ex along the x-axis only (y-axis is axis of bending) Po = nominal axial load strength for zero eccentricity For design purposes, when

--

_ '

0

1 .2

•••

• I I

1.8

= 0.70 Nominal axial load Pn= 2936/0.70 =4194kN Nominal moment Mn= 298/0.70 =426 kN·m

E E

Materials Compressive strength of concrete Jc'= 28 MPa Yield strength of reinforcement.t;, = 420 MPa Normalized maximum size of aggregate is 25 mm

..c:

Design conditions Column section size h = b =450 mm Slenderness effects may be neglected because kljh isknown to be below critical value ACl318M-05 section

Procedure Determine reinforcement ration p8 usingknown values of variables on appropriate interaction diagrams and compute required cross section area Ast of longitudinal reinforcement.

Kn =

Mn B) Compute R n =-Jc' A 8 h

Rn =

E) Read Pg for Kn and R,, values.

Design aid

Mn =426kN-m h=450 mm b=450 mm A�= bx h= 450 x 450 = 2.03 x 105 mm2

A) Compute Kn = pn Jc' A s

. - 125 C) Estimate y "" h - -h D) Determine the appropriate interaction diagrams. 10.2 10.3

Calculation P11 = 4194kN

4194 X 10 =0.74 5 X 10 ) 28)(2.03 ( 3

6

426

X

]0

(28)(2.03

X

J0 )(450)

5

=0.17

450- 125 =0.72 450 For a square spiral column,/) =28 MPa,JY = 420 MPa, and an estimated y of 0.72, use interaction diagram S28-420.7 and S28-420.8. For K11 = 0.73 and Rn= 0.16, and for Pg= 0.035 y = 0.70: Pg = 0.031 y =0.80: Pg = 0.034 y =0.72: 5 As1 = 0.034 x 2.03 x 10 mm2 = 6902 mm2 Y ""

Columns 3.20.2 (S28-420.7) and 3.20.3 (S28-420.8)

ACI DESIGN HANDBOOK-SP-17M(09)

64

3.4-Columns examples Columns Example 1: Determination of required steel area for a rectangular tied column with bars on four faces with slender­ ness ratio below critical value

For a rectangular tied column with bars equally distributed along four faces, find steel area. Given: Loading Pu = 2491 kN, and Mu = 443 kN·m Assume =0.70 Nominal axial load P11 =249110.70 =3559 kN Nominal moment M11 =443/0.70 =633 kN-m Materials Compressive strength of concrete Jc' =28 MPa Yield strength of reinforcementfy =420 MPa Normalized maximum size of aggregate is 25 mm Design conditions Short column braced against sidesway ACI318M-05 section

p + f A

A) Compute K,, =

c

8

M B) Compute R,, = --"-

f,.' A 8 h

. h-5 C) Estnnate y :::: --

K,, = R,, =

3559 (28)(2

X X

10

3

10

5

)

633 X 10 (28)(2

X

5

=0.64

6

J0 )(500)

=0.23

500- 125 = 0.75 500 D) Determine the appropriate inter- For a rectangular tied column with bars along four faces, action diagrams. fd =28 MPa,fv =420 MPa, and an estimated y of 0.75, enter diagram R28-420.7 and R28-420.8 with K11 =0.64 and R11 = 0.23, respectively. E) Read Pg for K11 and R,, values from Read Pg = 0.041 for y =0.7 and Pg= 0.039 for y =0.8. Columns 3.2.2 Interpolating pR =0.040 for y = 0.75 appropriate interaction diagrams. (R28-420.7) F) Compute required A51 from A 51 = Required A 51 = 0.040 x 2 x 10::, mm2 = 8000 mm and 3.2.3 PgAg · (R28-420.8) h

10.2 10.3

Design aid

Procedure Calculation Determine column section size. Given: h = 500 mm and b =400 mm Determine reinforcement ratio Pg P,, = 3559 kN using known values of variables on M,, =633 kN·m appropriate interaction diagrams and h= 500 mm compute required cross section area b = 400 mm A 51 of longitudinal reinforcement. A £ =b x h = 500 x 400 =2 x 1o 5 mm2

y::::

L

62

ACI DESIGN HANDBOOK-SP-17M(09)

pll •

;:-�ilure �rfac:e

Fig. 3.7-Faillfre surface S3, which is reciprocal ofsurface S 1. Fig. 3.5-Failure surface S 1.

Fig. 3.8-Graphical representation of reciprocal load method. Fig. 3.6-Failure surface S 2 .

pni

1 P,u

1

1

pn_v

Po

= -+---

(3-10)

where

approximation of nominal axial load strength at eccentricities ex and ey nominal axial load strength for eccentricity ey , P v: along the y-axis only (x-axis is axis of bending) P11Y = nominal axial load strength for eccentricity ex along the x-axis only (y-axis is axis of bending) Po = nominal axial load strength for zero eccentricity For design purposes, when d/2 is not allowed

-

-119 128 136 145 153 162 170

500

,_

___

-

I I _] I I I

=--cc....

--

-�--.�. -1 -·-

-- --

-

-----·-

I

_I

--

-II

119 127 134 142 149

119 126 133

I 19

487

426

379

341

350

400

450

500

--

Table 4.2(b) Valu�s uf V,, kN

Sti_rrup size

No. 13 U-

stirrups

Beam d,

mm 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 JOO

Maxi1num b u .

(mm) for f/ d/2 is not allowed

I I

-· �---

--

- -

- ----- �

I I I

���1 I I I _I - I

I ·- I 217 619

ACI DESIGN HANDBOOK-SP-17M(09)

52

Shear 3: Minimum beam height to provide development length required for No. 19, No. 22, and No. 25 Grade 420 stirrups Reference: Sections 12.13.2.1 and 12.13.2.2. Section 11.5.2 states "Design yield strength of shear reinforcement (bars) shall not exceed 420 MPa, ...."

0.17 do(420)

--Ir; h

Minimum beam height h = 2[0.17 db (420)/ Ji[+ 40] in millimeters.

Concrete J; , MPa

21 28 35 40 55 70

Minimum beam height h, mm Stirrup size No.19 No.22

660 582 528 490 434 396

757 666 605 559 493 450

No. 25

856 752 681 627 554 503

Nore: Values shown in the rable are for 40 mm clear cover over srirmps. For cover greater than 40 mm, add 2/cover-40 mm) to tabulated values.

Example: Determine whether a beam 600 mm high (h = 600 mm) with 35 MPa concrete will provide sufficient development length for No. 19 Grade 420 vertical stirrups. Solution: With/; = 35 MPa, for No. 19 stirrups, minimum beam height reads 528 mm for beams with 40 mm clear cover over stirrups. Because h = 600 mm, the beam is deep enough.

50

ACI DESIGN HANDBOOK-SP-17M(09)

Shear 2: Shear strength coefficients Kfc, Kvc• and Kvs Reference: Sections 11.3. l. l and 11.5.7.2 Ve = 0.17(jJ:)bwd=0.9KrA,.d Vs Vn

=

=

Kfc = Kvs = K11, =

A.jy 150 mm Table 2(b) Values K.,, kN/m

Beamh,mm 250 300 350 400 450 500 550 600 650 700 750

800 850 900 950 1000 1050 1100 1150 1200

f.v orJy,, MPa

280 51.8 65.8 79.8 93.8 107.8 121.8 135.8 149.8 163.8 177.8 189.0

203.0 217.0 231.0 245.0 259.0 273.0 287.0 301.0 315.0

420 77.7 98.7 I 19.7 140.7 161.7 182.7 203.7 224.7 245.7 266.7 283.5 340.5 325.5 346.5 367.5 388.5 409.5 430.5 451.5 472.5

70 1.581

Ji: , but keep fer 10.56 s Ji: .

48

ACI DESIGN HANDBOOK-SP-17M(09)

Shear Example 13: Determination of closed ties required for the beam of Example 12 to resistflexural shear and indetemunaie torque � Use the same data as that for Shear Example 12, except that the required torsion estimate of 72 kN-m is based on an indeterminate analysis, not an equilibrium requirement. = 35 MPa normalweight concrete f/ bw = 400 mm vu = 271 kN = !yr =420MPa = 600 mm h Tu = 72 kN-m (based on indeterminate analysis)

.t;,

ACl318M-05 section Procedure 11.3.1.l Step 1-Look up parameters for 11.5.7.2 Jc' 35 MPa, Grade 420 reinforcement, bw=400 mm, h = 600 mm l l.6.2.2a 11.6.3.1 11.6.3.6

Calculation = (1.118)193 kN = 215.8 kN KfcK vc

=

l l.6.2.2a

Kvs=224.7kN/m K1�1=(1.118)(113.7kN-m)=127.1 kN-m K_r�icr= (1.118)(50.3 kN·m) = 56.2 kN·m K1s 56.44 k:N-m/mm =

Step 2-If indeterminate T11 > q>KfcK rc, (0.75)56.2 k:N·m=42.2 kN·m, q>Kf�tcr , q>KfcKrcr can be used as Tu. Tu 72 k:N·m > 42.2 k:N-m Use T11 =42.2 kN·m 0.25 , 42.2 k:N·m > 0.25(56.2 kN·m)=14.1 kN·m. = 42.2 > KfcKtcr Step 3-If T,, ties are required. Therefore, ties are required. Step 4-Section is large enough if 2 71 kN 4 2 . 2 kN·m 2 2 = J£(V)(5�K1,K .,)] + [T.l(�K 1,K,)] < I =

=

11.6. la

,

{ 5(0.75)(215.8)} + { 0.75 x 1 27.I kN-m

2 J{ 0.335} 2 + { 0.443}

11.6.3.7 11.6.5.3

Step 5-Compute A)s = (V11 /q> - Kf�vc)/ Kvs Compute 2A,Js= T,,f(q>K,s) Compute (A vis+ 2A,Js) No. 13 ties provide 258 mm2 /mm Compute s< 258/(A/s+2A/s) Step 6---Compute Ph 2(b + h - 7). From Step 4, 2A,ls=1 .0 rnm 2 Compute At = (A,Js)(phcot 2 45) Is At> At .min 0.42(Jj: )Ac/fy -p0 1 ls? =

=

r

= = 0.555 < 1 Therefore, section is large enough. A,Js = (271 kN/0.75 -215.8)/224.7 kN/m = 0.6S nun:;/mm 2A,Js=42.2 k:N·m/[0.75(56.44)] =1.0 nun2/mm (A 11 /s + 2A,Js) = 1.65 mm2/mm s< 258/1.65= 156.4 mm. Use 150 mm spacing. Ph= 2(400 mm+ 600 mm -190 mm)= 1620 mm Thus A,ls =(1/2)(1 .0)= 0.5 mm A e = (0.5 mm)(1620 mm)(1.0) = 810 mm 2 A e .min = 0.42( ,.J35 MPa)240,000 mm 2 /420 MPa 2 2 - 1620(0.5) mm = 609.8 mm YES

11.6.6.1

Ae ,min =810 mm 2 governs. (No. 10 bars provide sufficient area, but select No. 13 as a minimum. Maximum spacing of longitudinal bars=pJ8 or 300 mm

1 l .6.6.2

For 10 positions, use No. 13 bars. Place (See Shear Example 12) three No. 13 in bottom and two No. 13 in each side face. Excess flexural capacity in top at d from support can serve in place of three No. 13 in top.

pJ8 = 1620/8 = 203 mm

Design aid Shear 2 Table 2(a) & 2(c) Shear 2(a) & 2(b) Shear 6.l(a) Shear 6.1(b) Shear 6.2(b)

46

ACI DESIGN HANDBOOK-SP-17M(09)

Shear Example 12: Determination of closed ties required for the beam shown to resist flexural shear and determinate torque

Given: J; = 35 MPa norrnalweight concrete Grade 420 reinforcement 1111 = 271 kN T11 = 72 kN·m determinate Wu = 69 kN/m = 420MPa

• • • d =546m

600mm

(v

• • I

40 mm clear cover 1-- (typical)

i-•--4-'-00 _m_m_-·�1

ACI318M-05 section Procedure 11.0 Step I-Determine section properties for torsion, allowing 6 mm as radius of ties. A cp=hwh A oh = (bw - 90)(h - 90)

11.6.1

11.6.3.1

11.5.6.2 11.6.3.6

11.6.5.2

A cp =400 mm(600 mm) = 240,000 mm2 A 0h =(400 mm - 90 mm)(600 mm - 90 mm) =158,100 mm2 A0 =0.85(158,100 mm 2)= 134,385 mm2 Pep = 2(400 mm+ 600 mm) = 2000 mm Ph=2(400 mm- 90 mm+600 mm - 90 mm)= 1640 mm

A 0= 0.85A 0h Pep 2(bw+ h) Ph =2(bw - 90+ h - 90) Step 2--Compute cracking torsion Tcr Tc,.=0.33(0.75)(,/35 MPa)(240,000 mm 2)2/2000 mm Tc r=0.33(JJ: )A c/lPcp =42,164, 817 N·mm Compute threshold torsion=0.25Tcr · Tc ,-= 42,169,817 N-mm/106 =42.2 kN-m Since T11=72 kN·m > 10.5 kN·m, ties Threshold torsion=0.25(42.2 kN-m) =10.5 kN·m for torsion are required. Step 3-ls section large enough? fv =271 kNI(400 mm x 546 mm)= 1.24 Nlmm2 Compute!,, = Vj(b11.d) f,,1=72 kN-m(l640 mm)l[l.7(158,000 mm2)2 ] Computefvt =TuP,/(l.7A 0/) =2.78 Nlmm2 Compute limit Limit= 0.75(0.17 + 0.66](J3s MPa) =3.68 Nlmm2 =(j>[(0.17Jj; +U66Jj;)l1000 =

9.3.2.3

Calculation

ls +fv�] < limit? Therefore, section is large enough. Step 4-Compute (b wd)]l(J;.d ) A v is= [Vu - 0.17

J[f.;

JI:

Compute A,fs = T/[2A 0J;,cot0] Compute (A)s + 2A 1 ls)

J[I.24 + 2.78 ) =3.04 Nlm.m 2 < 3.68 Nlmm 2 2

2

3 - 0.17(0.75)(J35 MPa)(400)(510) A 1• Is=271 ( 10 ) kN(0.75)(4 20 MPa)(510)

=0.73 mm 2lmm

72( 10 ) N-mm =0.85 mm2 lmm A 1Is= ( 2 )(0.75)(134,385)(4 20)cot45 AJs + 2A,fs=0.73+ 2(0.85) =2.43 mm 6

Use No. 13 ties for which (A v + 2A 1 )1 s =258 mm, and compute s= s =2.5812.43=106.2 mm 285 mm! s + 2A,fs). Use 100 mm ls 0.062( Jt 'Jc' )b,)fy, < (A 11 ls + 2A,ls)? 0.062( )4001420 =0.349 mm YES

J3s

Design aid

44

ACI DESIGN HANDBOOK-SP-17M(09)

Shear Example 10: Determination of thickness required for a flat slab based on required perimeter shear strength at an inte­ rior round column Determine the thickness required for a two-way slab to resist an ultimate shear force of Vu= 676 kN, based on perimeter shear strength at an interior circular column of 500 mm diameter when fc' = 28MPa for the normalweight slab concrete.

/0\ \,,! ,//

Gjven:

vn

O.s

ho

vu

he f:

= =

0.083(a.sdlb0 + 2)(§: )brfl �4(§: )b0d 40 for interior column, 30 for edge column, and 20 for corner column = perimeter of shear prism= n(hc+ d) = 676 kN = 500 mm diameter = 28MPa

/

a, = 40 for interior column, 30 for edge column, 20 for comer column

.,,,.----

..... '

'

I

Mechanism of shear prism representing perimeter shear strength surface V, =0.083(a,d/b0 + 2)(,ff; )bod S 0.33(�)b0d

/

I ·.

b0 = perimeter of shear prism =x(hc + d) Slab h•

I

____

d/2

Diameter he

.,,,,..

d/2

Plan at column ·•

J .· I

··.\

I ·._ .

r-.

· ___.___,, 1

1 I.·

Averaged .·1 .-· I ( I

.

·

•

Section at column

ACl318M-05 section Procedure Calculation Design aid 11.12.1.2 Step 1-Set up equation V11 = 0.33(0.75)( Ji8 MPa)n(500 mm+ d mm)d 11.12.2.l q> V11 = 0.33( J1:) rt(hc + a) X d. = 4.109 MPa(500d + d 2) mm2 9.3.2.3 11.12.2.1 Step 2-Equate Vu to V11 and 676,000 N = 4.109 MPa(500d + d 1') mm' solve for d. 1.0d2 + (500)d (mm)+ [(0.5)(500)) 2 = 164,517+ [(0.5)(500))2 (d + 250) 2= 227,017 d= ( J227,0l 7 mm2) - 250 mm= 476.5 mm - 250 mm= 226.5 mm 7.7.lc Step 3-Make hs deep enough for hs = 226.5 mm+ 20 mm+ 22 mm= 268.5 mm 20 mm concrete cover plus diameter Use hs = 270 mm or top bars. Estimate No. 22 bars. ALTERNATEMETHOD with Design aid Shear 5.2 Table 5.2(b) 9.3.2.3 v11= 676//(0.75) = 901 kN Step I-Compute V11 = Vu /. 11.12.1.2 Withf; =28MPa and V11=901 kN, K3 = 160 x 10 3 + ( 200 x l0 3 - 160 x ]0 3 )