Solution Manual for Mechanics of Aircraft Structures 2'nd edition C.T. Sun [Second ed.]

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Name: Mohamed Naleer Abdul Gaffor Email: [email protected] IP: 184.162.144.24

Solution Manual for

Mechanics of Aircraft Structures 2nd edition, 3rd printing, 2007

C.T. Sun School of Aeronautics and Astronautics Purdue University W. Lafayette, Indiana U.S.A.

Prepared with the assistance of Mr. Hsin-Haou Huang, graduate student in School of Aeronautics and Astronautics, Purdue University

June, 2007

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

Name: Mohamed Naleer Abdul Gaffor Email: [email protected] IP: 184.162.144.24

Mechanics of Aircraft structures C.T. Sun

1.1

The beam of a rectangular thin-walled section (i.e., t is very small) is designed to carry both bending moment M and torque T. If the total wall contour length L = 2( a + b) (see Fig. 1.16) is fixed, find the optimum b/a ratio to achieve the most efficient section if M = T and σ allowable = 2τ allowable . Note that for closed thin-walled sections such as the one in Fig.1.16, the shear stress due to torsion is

τ=

Figure 1.16

T 2abt

Closed thin-walled section

Solution: (1) The bending stress of beams is σ =

My , where y is the distance from the neutral I

axis. The moment of inertia I of the cross-section can be calculated by considering the four segments of thin walls and using the formula for a rectangular section with height h and width w. I = ∑ (

1 wh 3 + Ad 2 ) in which A is the 12

cross-sectional area of the segment and d is the distance of the centroid of the segment to the neutral axis. Note that the Parallel Axis Theorem is applied. The result is I = 2 ⋅

1 3 1 b tb 2 tb + 2 ⋅ [ ⋅ at 3 + (at ) ⋅ ( ) 2 ] ≈ (3a + b) , assuming that t is 12 12 2 6

very small. (2) The shear stress due to torsion for a closed thin-walled section shown above is

τ=

T . 2abt

1.1.1

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Mechanics of Aircraft structures C.T. Sun

(3) Two approaches are employed to find the solution. (i) Assume that the bending stress reaches the allowable σ allowable first and find the corresponding bending maximum bending moment. Then apply the stated loading condition of T = M to check whether the corresponding τ max has exceeded the allowable shear stress τ allowable . If this condition is violated, then the optimized b/a ratio is not valid.

b M⋅ My 3M 2 = = 2 (a) σ | b = y= I tb(3a + b) tb 2 (3a + b) 6 When given L = 2( a + b) as a constant, a can be expressed in terms of b and L as a = S=

L − b . Then we can minimize 2

tb(3a + b) tb(3L − 4b) in order to maximize σ , i.e., = 3 6

∂S t 3L L L , so a = − b = = 0 ⇒ (3L − 8b) = 0 ⇒ b = ∂b 6 8 2 8

where the optimum ratio is

b =3 a

3M 3M 32M = = tb(3a + b) t ⋅ (3L / 8) ⋅ (3 ⋅ L / 8 + 3L / 8) 3tL2 (b) Check τ max with T = M and b/a = 3 and check whether τ max is within the allowable shear stress τ allowable . Thus, σ max =

τ max =

T M 32M = = = σ max = σ allowable 2abt 2 ⋅ ( L / 8) ⋅ (3L / 8) ⋅ t 3tL2

> τ allowable =

σ allowable 2

The result above means that under this assumption, shear stress τ would reach the allowable stress τ allowable before σ reaches σ allowable . Consequently, the optimal ratio obtained is not valid and different assumption needs to be made. (ii) Assume now that failure is controlled by shear stress. We assume that τ max = τ allowable is reached first and then find the corresponding bending stress according to the loading condition M = T . T 2abt Again we minimize S = 2abt = ( L − 2b)bt in order to maximize τ , i.e.,

(a) τ =

1.1.2

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Mechanics of Aircraft structures C.T. Sun

∂S L L L = 0 ⇒ ( L − 4b) = 0 ⇒ b = , so a = − b = ∂b 4 2 4

and the optimum ratio is

b =1 a

T T 8T = = 2 2abt 2 ⋅ ( L / 4) ⋅ ( L / 4) ⋅ t tL (b) Then corresponding σ max under the optimum condition stated above can and τ max =

be obtained using M = T . We have 3M 3T 12T 3 3 = = 2 = τ max = τ allowable tb(3a + b) t ⋅ ( L / 4) ⋅ (3 ⋅ L / 4 + L / 4) tL 2 2 < σ allowable = 2τ allowable

σ max =

This means that when the structure fails in shear, the bending stress is still within the allowable stress level. Thus the optimum ratio

b = 1 is a

valid.

(4) In conclusion,

b = 1 achieves the most efficient section for the stated conditions. a

--- ANS

1.1.3

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Mechanics of Aircraft structures C.T. Sun

1.2

Do problem 1.1 with M = αT where α = 0 to ∞ .

Figure 1.16

Closed thin-walled section

Solution: My , where y is the distance from the neutral I axis. The moment of inertia I of the cross-section can be calculated by considering the four segments of thin walls and using the formula for a rectangular section 1 with height h and width w. I = ∑ ( wh 3 + Ad 2 ) in which A is the 12 cross-sectional area of the segment and d is the distance of the centroid of the segment to the neutral axis. Note that the Parallel Axis Theorem is applied. The 1 1 b tb 2 result is I = 2 ⋅ tb 3 + 2 ⋅ [ ⋅ at 3 + (at ) ⋅ ( ) 2 ] ≈ (3a + b) , assuming that t is 12 12 2 6 very small.

(1) The bending stress of beams is σ =

(2) The shear stress due to torsion for a closed thin-walled section shown above is T . τ= 2abt (3) Two approaches are employed to find the solution. (i) Assume that the bending stress reaches the allowable σ allowable first and find the corresponding bending maximum bending moment. Then apply the stated loading condition of M = αT to check whether the corresponding τ max has exceeded the allowable shear stress τ allowable . If this condition is violated, then the optimized b/a ratio is not valid. b M⋅ My 3M 2 = 2 = (a) σ | b = y= I tb(3a + b) tb 2 (3a + b) 6 When given L = 2( a + b) as a constant, a can be expressed in terms of b 1.2.1

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Mechanics of Aircraft structures C.T. Sun

and L as a = S=

L − b . Then we can minimize 2

tb(3a + b) tb(3L − 4b) in order to maximize σ , i.e., = 3 6

∂S t 3L L L , so a = − b = = 0 ⇒ (3L − 8b) = 0 ⇒ b = ∂b 6 8 2 8

where the optimum ratio is

b =3 a

3M 3M 32M = = tb(3a + b) t ⋅ (3L / 8) ⋅ (3 ⋅ L / 8 + 3L / 8) 3tL2 (b) Check τ max with M = αT and b/a = 3 and check whether τ max is within the allowable shear stress τ allowable . 32M 1 T M /α = = = σ max τ max = 2 α 2abt 2 ⋅ ( L / 8) ⋅ (3L / 8) ⋅ t 3αtL 1 2 = σ allowable = τ allowable Thus, σ max =

α

α

We have τ max ≤ τ allowable ⇒

2

τ allowable ≤ τ allowable α Î α ≥ 2 (since τ allowable > 0 is always satisfied) (ii) Assume now that failure is controlled by shear stress. We assume that τ max = τ allowable is reached first and then find the corresponding bending stress according to the loading condition M = αT . T 2abt Again we minimize S = 2abt = ( L − 2b)bt in order to maximize τ , i.e.,

(a) τ max =

∂S L L L = 0 ⇒ ( L − 4b) = 0 ⇒ b = , so a = − b = ∂b 4 2 4 b =1 a T T 8T and τ max = = = 2 2abt 2 ⋅ ( L / 4) ⋅ ( L / 4) ⋅ t tL (b) Then corresponding σ max under the optimum condition stated above can be obtained using M = αT . We have 3M 3αT 12αT 3 σ max = = = = ατ max tb(3a + b) t ⋅ ( L / 4) ⋅ (3 ⋅ L / 4 + L / 4) 2 tL2

and the optimum ratio is

σ 3 3 3 = ατ allowable = α ⋅ ( allowable ) = ασ allowable 2 2 2 4 3 Since σ max ≤ σ allowable ⇒ ασ allowable ≤ σ allowable 4 4 Î α≤ (since σ allowable > 0 is always satisfied) 3 1.2.2

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Mechanics of Aircraft structures C.T. Sun

(4) From the above two approaches, we have the conclusions. 4 (i) For 0 < α ≤ , the failure is controlled by shear and the optimum ratio 3 b of = 1 achieves the most efficient section.. a (ii) For α ≥ 2 , the failure is controlled by bending and the optimum ratio of b = 3 achieves the most efficient section. a 4 (iii) For < α < 2 , the optimal ratio lies between 1 and 3. The most 3 straightforward way in finding the best ratio for a given α in this range is to calculate the maximum bending moments and torques for different values of b/a ratios between 1 and 3 and pick the ratio that produces the greatest minimum failure load, either T or M. --- ANS

1.2.3

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Mechanics of Aircraft structures C.T. Sun

1.3

The dimensions of a steel (300M) I-beam are b = 50 mm, t = 5 mm, and h = 200 mm (Fig. 1.17). Assume that t and h are to be fixed for an aluminum(7075-T6) I-beam. Find the width b for the aluminum beam so that its bending stiffness EI is equal to that of the steel beam. Compare the weights-per-unit length of these two beams. Which is more efficient weightwise?

Figure 1.17

Dimensions of the cross-section of an I-beam

Solution: (1) The expression of area moment of inertia I for an I-beam is: I=

t b h (h − t ) 3 + [ t 3 + (bt )( ) 2 ] × 2 , 12 12 2

by applying Parallel Axis Theorem.

(2) First obtaining the area moment of inertia of the steel (300M) I-beam with given b, t, and h. 5 50 200 2 (200 − 5) 3 + [ ⋅ 5 3 + (50 ⋅ 5)( ) ] × 2 = 8090573mm 4 12 12 2 (3) For the given condition ( EI ) Alu min um = ( EI ) Steel I Steel =

E St 200 I St = × 8090573 = 22790000mm 4 E Al 71

we have I Al =

which allows to calculate the width b for the aluminum beam with the following result: 5 b 200 2 ( 200 − 5 )3 + [ ⋅ 5 3 + ( b ⋅ 5 )( ) ]×2 12 12 2 = 3089531.3 + 100020.8b = 22790000

I Al =

and b = 197 mm ---- ANS (4) Then we compare the weights-per-unit length of these two beams. 1.3.1

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The weights-per-unit length is defined as w = ρ ⋅ A , where ρ = density , and A = cross-sectional area (i) For the Steel beam

ρ St = 7.8( g / cm 3 ) = 7.8 × 10 −3 ( g / mm 3 ) ASt = (200 − 5) × 5 + 2 × 50 × 5 = 1475( mm 2 )

wSt = ρ St ⋅ ASt = 7.8 × 10 −3 × 1475 = 11.5( g / mm ) (ii) For the Aluminum beam

ρ Al = 2.78( g / cm 3 ) = 2.78 × 10 −3 ( g / mm 3 ) AAl = ( 200 − 5 ) × 5 + 2 × 196.97 × 5 = 2945( mm 2 ) w Al = ρ Al ⋅ AAl = 2.78 × 10 −3 × 2944.7 = 8.2( g / mm ) For a unit length of both materials, the aluminum beam is much lighter than the steel beam. It means that the ALUMINUM BEAM IS MORE EFFICIENT! --- ANS

1.3.2

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Mechanics of Aircraft structures C.T. Sun

1.4

Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in Problem 1.3. Compare its weight with that of the aluminum beam.

Figure 1.17

Dimensions of the cross-section of an I-beam

Solution: Proceed in the same manner as that of problem 1.3. (1) The expression of area moment of inertia I for a I-beam is: I=

t b h (h − t ) 3 + [ t 3 + (bt )( ) 2 ] × 2 12 12 2

(2) First, obtain the area moment of inertia of the steel (300M) I-beam with given b, t, and h. We have I Steel =

5 50 200 2 (200 − 5) 3 + [ ⋅ 5 3 + (50 ⋅ 5)( ) ] × 2 = 8090573mm 4 12 12 2

(3) For the condition ( EI ) Composite = ( EI ) Steel we have I Com =

E St 200 I St = × 8090573 = 11558000mm 4 ECom 140

The moment of inertia of the composite beam is given by 5 b 200 2 ( 200 − 5 )3 + [ ⋅ 5 3 + ( b ⋅ 5 )( ) ]×2 12 12 2 = 3089531.3 + 100020.8b = 11558000

I Com =

Thus the width of the cross-section is obtained as b = 84.7 mm ---- ANS (4) Then, we compare the weights-per-unit length of these two beams. The weights-per-unit length is defined as where ρ = density , and A = cross-sectional area w=ρ⋅A ,

1.4.1

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Mechanics of Aircraft structures C.T. Sun

(i) For the composite beam

ρ Com = 1.55( g / cm 3 ) = 1.55 × 10 −3 ( g / mm 3 ) ACom = ( 200 − 5 ) × 5 + 2 × 84.67 × 5 = 1822( mm 2 ) wCom = ρCom ⋅ ACom = 1.55 × 10 −3 × 1821.7 = 2.8( g / mm ) (ii) Compare the weights per unit length with that of the aluminum beam wCom = 2.8( g / mm ) < wAl = 8.2( g / mm ) This indicates that the AS4/3501-6 CARBON/EPOXY COMPOSITE BEAM IS MORE EFFICIENT than the aluminum beam! --- ANS

1.4.2

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Mechanics of Aircraft structures C.T. Sun

1.5

Derive the relations given by (1.4) and (1.5). (1.4) : V x = τ ⋅ t ⋅ a Remark: (1.5) : V y = τ ⋅ t ⋅ b

Solution: (1) Consider a very small section within the curved panel with thickness t and length δL . τ is the constant shear stress, so we have the shear force ΔV = τ ⋅ ( δL ⋅ t ) acting on the cross section. (2) It is possible to take apart the shear force into x and y direction shown in the figure, where ΔV x = ΔV ⋅ cos θ = τ ⋅ δL ⋅ t ⋅ cos θ = τ ⋅ t ⋅ (δL ⋅ cos θ ) = τ ⋅ t ⋅ Δx similarly, ΔV y = τ ⋅ t ⋅ δy (3) Now consider the length to be extremely small, therefore ΔV x → dV x as well as ΔV y → dV y . The horizontal component and the vertical component of the shear

force V x , V y can be verified as following: a

V x = ∫ dVx = ∫ τ ⋅ t ⋅ dx = τ ⋅ t ⋅ a 0

b

V y = ∫ dV y = ∫ τ ⋅ t ⋅ dy = τ ⋅ t ⋅ b 0

1.5.1

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Mechanics of Aircraft structures C.T. Sun

1.6

The sign convention (positive direction of resultants) used in the beam theory depends on the coordinate system chosen. Consider the moment-curvature relation d 2w M = − EI 2 dx in reference to the coordinate system shown in Fig. 1.18. If w is regarded as a positive displacement (or deflection) in the positive y-direction, find the positive direction of the bending moment. State the reason.

Figure 1.18

Coordinate system for a beam

Solution: d 2w d 2w gives that is always dx 2 dx 2 opposite in sign to M. (It is quite obvious that both E and I are always positive.). (2) We can assume a moment M applying to the beam as shown below, which makes (1) The moment-curvature relation M = − EI

the beam concave upwards. It is not difficult to observe that the slope increases with increasing x and thus a positive

dw dx

d 2w . dx 2

(3) By applying the statement (1), it is concluded that the deformation described in (2) is produced by a negative moment while a positive moment makes the beam concave downward as shown below..

1.6.1

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Mechanics of Aircraft structures C.T. Sun

1.7

Compare the load-carrying capabilities of two beams having the respective cross-sections shown in Fig. 1.19. Use bending rigidity as the criterion for comparison. It is given that a = 4 cm, t = 0.2 cm, and the two cross-sections have the same area.

Figure 1.19

Cross-sections of two beams

Solution: When using the bending rigidity ( EI ) as a criterion for comparison, Young’s modulus E and the area moment of inertia I should be estimated. (1) Young’s modulus E : Assume the Young’s modulus of the beam having the left-hand-side cross-section and the right-hand-side cross-section are

El

and

Er

,

respectively. (2) Moment of inertia I : (i) Left cross-section: Il =

1 4 1 a = × 4 4 = 21.33cm 4 12 12

(ii) Right cross-section: Ir =

b b−t 3 ( a + 2b) 3 − a 12 12

or { I r =

--- (a)

t 3 1 a b a + [ b4 + b2 ⋅ ( + )2 ]× 2 } 12 12 2 2

where b remains unknown. There is another condition, two cross-section have the same area, which will help to solve b. Al = a 2 = 4 2 = 16cm 2 , Ar = 2 ⋅ b 2 + a ⋅ t = 2 ⋅ b 2 + 4 ⋅ 0.2

1.7.1

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let Al=Ar => b = 2.7568cm , then we have Ir =

2.7568 (4 + 2 × 2.7568 )3 − (2.7568 − 0.2 ) ⋅ 4 3 = 184 cm 4 12 12

or { I r =

0 .2 3 1 4 2.7568 2 4 + [ 2.7568 4 + 2.7568 2 ⋅ ( + ) ] × 2 = 184 cm 4 } 12 12 2 2

(3) Performance: The ratio of the moments of inertia of the two cross-sections can be expressed as ( EI )l EI 21.33 El El E = l l = = = 0.12 l ( EI )r Er I r 184.18 Er 8.635 Er Er The cross-section to the right is much better if the same material is used for both beams. (i) If Er < 0.12 El The left cross-section outperforms the right one. (ii) If Er = 0.12 El They are equivalent. (iii) If Er > 0.12 El The right cross-section outperforms the left one. --- ANS

1.7.2

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Mechanics of Aircraft structures C.T. Sun

2.1 Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loads are applied, the displacements are given by u = αx v = βy

w=0 (a) Sketch the deformed shape for α = 0.03 , β = −0.01 . (b) Calculate the six strain components. (c) Find the volume change ΔV [ ΔV = V (the volume after deformation) – V0 (the original volume) ] for this unit cube. Show that ε xx + ε yy + ε zz ≈ ΔV .

Solution: (a) Since w = 0 , there is no deformation in the z-direction and the deformation can be represented in the x-y plane. The new position of point B after deformation is given by x ' = 1 + u | x =1, y = 0 = 1 + 0.03 ⋅ 1 = 1.03 y ' = 0 + v | x =1, y =0 = 0 + ( −0.01) ⋅ 0 = 0

New coordinates of B’ = ( 1.03 , 0 ) Similarly, new positions of A, C, D can be obtained as follows: A’ = ( 0 , 0 ) C’ = ( 1.03, 0.99 ) D’ = ( 0 , 0.99 ) y

D, 1

C

D’ 0.99

A=(0,0) A’=(0,0) B=(1,0) B’=(1.03,0) C=(1,1) C’=(1.03,0.99) D=(0,1) D’=(0,0.99)

C’

x A, A’

B 1

B’ 1.03 --- ANS 2.1.1

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Mechanics of Aircraft structures C.T. Sun

(b) Strain components are: Normal strain components,

ε xx =

∂u = α = 0.03 ∂x

ε yy =

∂v = β = −0.01 ∂y

ε zz =

∂w =0 ∂z

Shear strain components, ∂u ∂v γ xy = γ yx = + =0 ∂y ∂x

γ xz = γ zx =

∂u ∂w + =0 ∂z ∂x

γ yz = γ zy =

∂v ∂w + =0 ∂z ∂y --- ANS

(c) The volume change is defined by ΔV = V − V0 = x' ⋅ y' ⋅z' −1 = 1.03 × 0.99 × 1 − 1 × 1 × 1 = 0.0197 Also, ε xx + ε yy + ε zz = 0.03 + ( −0.01) + 0 = 0.02 ≈ ΔV = 0.0197 --- ANS It can also be verified by:

ΔV = V − V0 = ( 1 + ε xx ) ⋅ 1 ⋅ ( 1 + ε yy ) ⋅ 1 ⋅ ( 1 + ε zz ) ⋅ 1 − 1 ⋅ 1 ⋅ 1 = V0 ( 1 + ε xx + ε yy + ε zz + ε xxε yy + ε xxε zz + ε yyε zz + ε xxε yyε zz − 1 ) = V0 ( ε xx + ε yy + ε zz + ε xxε yy + ε xxε zz + ε yyε zz + ε xxε yyε zz ) Since the deformation is very small, we have ε higher order terms can be dropped. Therefore we have ΔV ≈ V0 (ε xx + ε yy + ε zz ) = ε xx + ε yy + ε zz ,

since V0 = 1

2.1.2

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Mechanics of Aircraft structures C.T. Sun

2.2

Consider the following displacement field: u = αy

v = −αx w=0 Sketch the displaced configuration of a unit cube with the faces originally perpendicular to the axes, respectively. This displacement field does not yield any strains; it only produces a rigid body rotation. Show that the angle of rotation is 1 ⎛ ∂v ∂u ⎞ ⎜ − ⎟ = −α 2 ⎜⎝ ∂x ∂y ⎟⎠ Solution: (a) Consider a unit cube, the coordinates before deformation corresponding to each corner are: A(0,0,0), E(0,0,1),

B(1,0,0), F(1,0,1),

C(1,1,0), G(1,1,1),

D(0,1,0), H(0,1,1)

y

D

C

H

G A

B

E

x

F

z After deformation, we have the coordinates for each point as follows: For point A’ (‘ denotes the point after deformation) : x' = x + u | x =0, y =0, z =0 = 0 + α ⋅ 0 = 0 y ' = y + v | x =0, y = 0, z = 0 = 0 + (−α ⋅ 0) = 0 z ' = z + w | x =0 , y =0 , z =0 = 0 + 0 = 0 ,

Thus, we have A’(0,0,0)

Another example for G’, we have 2.2.1

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x' = x + u | x =1, y =1, z =1 = 1 + α ⋅ 1 = 1 + α y ' = y + v | x =1, y =1, z =1 = 1 + ( −α ⋅ 1) = 1 − α z ' = z + w | x =1, y =1, z =1 = 1 + 0 = 1 ,

G’(1+ α ,1- α ,1)

Thus,

Similarly, we can verify other points B’(1,- α ,0), C’(1+ α ,1- α ,0), E’(0,0,1), F’(1,- α ,1), (1) In x-y plane A(0,0,0), A’(0,0,0),

B(1,0,0), B’(1,- α ,0),

D’( α ,1,0), H’( α ,1,1)

C(1,1,0), C’(1+ α ,1- α ,0),

D(0,1,0) D’( α ,1,0)

y

D

D’

C C’

α A,A’

B

α

x B’

(2) In y-z plane A(0,0,0), A’(0,0,0),

D(0,1,0), D’( α ,1,0),

H(0,1,1), H’( α ,1,1),

E(0,0,1) E’(0,0,1)

z

E E’

H

A,A’

H’

D D’

y

2.2.2

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Mechanics of Aircraft structures C.T. Sun

(3) In z-x plane A(0,0,0), A’(0,0,0),

E(0,0,1), E’(0,0,1,)

B(1,0,0), B’(1,- α ,0),

F(1,0,1) F’(1,- α ,1)

x

B B’

F

A,A’

F’

E E’

z

--- ANS (b) Then we verify the strains are zero under this circumstance: ∂v ∂u ∂w ε yy = =0, ε xx = = 0, ε zz = =0 ∂x ∂y ∂z

γ xy =

∂u ∂v + = α −α = 0 , ∂y ∂x

γ yz =

∂v ∂w + = 0, ∂z ∂y

γ xz =

∂u ∂w + =0 ∂z ∂x

--- ANS (c) If we denote the counterclockwise rotation to be positive, we have the angle of rotation equals to β avg =

y

1 (β 1 − β 2 ) . 2

If the deformation is small enough, we ∂u ∂v have β 1 = =α = −α , and β 2 = ∂x ∂y

β2

Therefore the angle of rotation is

β avg = =

β1

1 (β1 − β 2 ) = 1 ⎛⎜⎜ ∂v − ∂u ⎞⎟⎟ 2 2 ⎝ ∂x ∂y ⎠

x

1 (− α − α ) = −α 2

--- ANS

2.2.3

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2.3

Consider the displacement field in a body: u = 0.02x + 0.02y – 0.01z cm v= + 0.01y – 0.02z cm w = -0.01x + 0.01z cm Find the locations of the two points (0,0,0) and (5,0,0) after deformation. What is the change of distance between these two points after deformation? Calculate the strain components corresponding to the given displacement field. Use the definition of ε xx to estimate the change of distance between the two points. Compare the two results.

Solution: (a) Consider the point (0,0,0), after deformation : x' = x + u | x =0, y =0, z =0 = 0 y ' = y + v | x =0, y =0, z =0 = 0 z ' = z + w | x =0, y =0, z =0 = 0

So the corresponding location after deformation is (0,0,0) (b) Consider the point (5,0,0), after deformation : x' = x + u | x =5, y = 0, z = 0 = 5 + 0.02 × 5 = 5.1 y ' = y + v | x = 5, y = 0 , z = 0 = 0 z ' = z + w | x =5, y = 0, z = 0 = 0 − 0.01 × 5 = −0.05

(c) The change of distance between these two points after deformation. (1) before deformation: distance between (0,0,0) and (5,0,0), Dbefore = 5 (2) after deformation: distance between (0,0,0) and ( 5.1 , 0 , -0.05 ), Dafter =

(5.1 − 0)2 + (0 − 0)2 + (− 0.05 − 0)2

= 5.100245

(3) change of distance δD :

δD = Dafter − Dbefore = 5.100245 − 5 = 0.100245 (d) Calculate the strain components 2.3.1

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ε xx =

∂u = 0.02 , ∂x

γ xy =

∂u ∂v + = 0.02 + 0 = 0.02 , ∂y ∂x

ε yy =

∂v = 0.01 , ∂y

γ yz =

∂v ∂w + = −0.02 + 0 = −0.02 , ∂z ∂y

ε zz =

∂w = 0.01 ∂z

γ xz =

∂u ∂w + = −0.01 − 0.01 = −0.02 ∂z ∂x

(e) The normal strain in the x-direction is ε xx =

∂u = 0 ,02 . The change of distance ∂x

between the two points can be estimated by => ΔD x = ε xx ⋅ ( x 2 − x1 ) = 0.02 ⋅ (5 − 0) = 0.1 (f) Compare the two results From the displacement field calculation, we have δD = 0.100245 , And directly from the strain calculation, we have ΔD = 0.1 They are basically the same when the strain components are small.

2.3.2

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2.4

Consider the problem of simple shear in Example 2.1 and Fig. 2.5. From the deformed shape, find the normal strain for material along the line CB by comparing the deformed length C ' B' and undeformed length CB . Set up new coordinates (x’,y’) so that the x’-axis coincides with CB , and y’ is perpendicular to the x’-axis. The relation between (x,y) and (x’,y’) is given by x' = x cos θ + y sin θ y ' = − x sin θ + y cos θ where θ = 45 o is the angle between x’ and the x-axis. Write the displacements u’ and v’ in the x’ and y’ directions, respectively, in terms of the new coordinates x’ and y’. The relation between (u’,v’) and (u,v) is the same as between (x’,y’) and (x,y). Then calculate the strains using u’ and v’, i.e.,

ε ' xx =

∂u ' ∂x'

ε ' yy =

∂v' ∂y '

γ ' xy =

∂u ' ∂v' + ∂y ' ∂x'

Compare ε ' xx with the normal strain (along CB ) obtained earlier.

Solution: (a) The result of example 2.1 gives the new positions of A, B, C, D, which are A’(0.01,0), B’(1.01,1.015), C’(0,0), D’(1,0.015). From which we obtain

CB = C ' B' =

(1 − 0)2 + (1 − 0)2

= 2 = 1.414214

(1.01 − 0)2 + (1.015 − 0)2

= 1.431896

and the normal strain along the line CB is C ' B ' − CB 1.431896 − 1.414214 = = 0.0125 ε= 1.414214 CB --- ANS (b) The relation between (x,y) and (x’,y’) is given by x' = x cos θ + y sin θ y ' = − x sin θ + y cos θ which can be written in matrix form as

2.4.1

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⎧ x '⎫ ⎧x⎫ ⎨ ⎬ = [β ]⎨ ⎬ ⎩ y '⎭ ⎩ y⎭

--- (1)

⎡ cos θ sin θ ⎤ where the transformation matrix is [β ] = ⎢ , and θ = 45 o ⎥ ⎣− sin θ cos θ ⎦ From this equation we can get the equivalent form: ⎧x⎫ −1 ⎧ x ' ⎫ T ⎧ x'⎫ --- (2) ⎨ ⎬ = [β ] ⎨ ⎬ = [β ] ⎨ ⎬ ⎩ y⎭ ⎩ y '⎭ ⎩ y '⎭ ⎡cos θ − sin θ ⎤ −1 T Here it is easy to prove that [β ] = [β ] = ⎢ ⎥ ⎣ sin θ cos θ ⎦ Since displacements transform like coordinates, we can write ⎧u '⎫ ⎧u ⎫ ⎨ ⎬ = [β ]⎨ ⎬ ⎩v'⎭ ⎩v ⎭ From example 2.1 we have the displacement field : u = 0.01 y , where we can also write in matrix form, 0.01⎤ ⎧ x ⎫ ⎧u ⎫ ⎡ 0 ⎨ ⎬=⎢ ⎥⎨ ⎬ ⎩v ⎭ ⎣0.015 0 ⎦ ⎩ y ⎭

--- (3)

v = 0.015x ,

--- (4)

So the displacements u’ and v’ can be derived, from equation (3) and (4), and by applying (2), set θ = 45 o , as follows: 0.01⎤ ⎧ x ⎫ 0.01⎤ T ⎧ x'⎫ ⎧u '⎫ ⎧u ⎫ ⎡ 0 ⎡ 0 ⎨ ⎬ = [β ]⎨ ⎬ = [β ]⎢ ⎨ ⎬ = [β ]⎢ ⎥ ⎥[β ] ⎨ y '⎬ ⎩v'⎭ ⎩v ⎭ ⎣0.015 0 ⎦ ⎩ y ⎭ ⎣0.015 0 ⎦ ⎩ ⎭ ⎡ 0.025 sin θ cos θ =⎢ 2 2 ⎣0.015 cos θ − 0.01sin θ ⎡0.0125 − 0.0025⎤ ⎧ x'⎫ =⎢ ⎥⎨ ⎬ ⎣0.0025 − 0.0125⎦ ⎩ y '⎭

0.01 cos 2 θ − 0.015 sin 2 θ ⎤ ⎧ x'⎫ ⎥⎨ ⎬ − 0.025 sin θ cos θ ⎦ ⎩ y '⎭

--- ANS (c) Strains in the new transformed coordinate and displacements are:

ε ' xx =

∂u ' = 0.0125 ∂x '

ε ' yy =

∂v' = −0.0125 ∂y '

γ ' xy =

∂u ' ∂v' + = −0.0025 + 0.0025 = 0 ∂y ' ∂x' --- ANS

Note that ε ' xx is the same as the strain measured along CB .

2.4.2

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2.5 A cantilever beam of a rectangular cross-section is subjected to a shear force V as shown in Fig. 2.17. The bending stress is given by Mz I where M = −V ( L − x ) . Assume a state of plane stress parallel to the x-z plane,

σ xx =

i.e., σ yy = τ xy = τ yz = 0 . Find the transverse shear stress τ xz (= τ zx ) by integrating the equilibrium equations over the beam thickness and applying the h . 2

boundary conditions τ xz = 0 at z = ± Hint: From the equilibrium equation

∂σ xx ∂τ xz + =0 ∂x ∂z we have

∂τ xz ∂σ z ∂M = − xx = − ∂x ∂z I ∂x

Figure 2.17

Cantilever beam subjected to a shear force

Solution: (a) Bending moment is M = −V ( L − x ) ,

so

∂M =V ∂x

(b) From the equilibrium equation for a state of plane stress parallel to the x-z plane, we have

∂τ xz ∂σ z ∂M V = − xx = − =− z ∂z ∂x I ∂x I V ⎛ V ⎞ Therefore, τ xz = ∫ ⎜ − z ⎟dz = − z 2 + C 2I ⎝ I ⎠

2.5.1

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(c) Applying the B.C.: τ xz = 0 at z = ±

h 2

we have V − 2I

2

⎛h⎞ ⎜ ⎟ +C = 0 ⎝2⎠

So τ xz = − => τ xz =

=>

(

V 2 V 2 z +C = h − 4z 2 2I 8I

C=

Vh 2 8I

)

also I =

bh 3 12

V 2 3V ( (h 2 − 4 z 2 ) h − 4z 2 ) = 8I 2bh 3

--- ANS

2.5.2

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2.6

The state of stress in a body is uniform and is given by

σ xx = 4MPa ,

τ xy = 2MPa ,

τ xz = 0

σ yy = 3MPa ,

τ yz = 0 ,

σ zz = 0

Find the three components of the stress vector t on the surface ABCD as shown in Fig. 2.18. Find the normal component σ n of the stress vector. Hint: From the equilibrium equation

Figure 2.18

Shape of a wedge

Solution:

{ }

(a) The stress vector t can be expressed as {t } = [σ ] n , in which σ ji = σ ij

⎧t x ⎫ {t i } = ⎪⎨t y ⎪⎬ is the stress vector on surface ABCD, ⎪t ⎪ ⎩ z⎭

[σ ] ij

⎡σ xx ⎢ = ⎢τ yx ⎢ τ zx ⎣

τ xy τ xz ⎤ ⎥ σ yy τ yz ⎥ are the stress components associated with the coordinate τ zy σ zz ⎥⎦

x-y-z,

⎧n x ⎫ ⎪ ⎪ and {n j } = ⎨n y ⎬ is the normal vector to the surface ABCD, ⎪n ⎪ ⎩ z⎭ 2.6.1

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(b) Calculate the normal vector to the surface ABCD, {n j } Assume that the positions of point B, C, D are B(1,0,1), C(1,0,0), D(0,1,0) We have BC = (0 ,0 ,−1) and CD = (− 1,1,0 ) thus {n} = T

BC × CD

⎛ 1 1 ⎞ =⎜ , ,0 ⎟ BC × CD ⎝ 2 2 ⎠

[ ]{n }, we have

(c) Now applying {t } = σ

⎧t x ⎫ ⎡4 2 0 ⎤ ⎧1 / 2 ⎫ ⎧6 / 2 ⎫ ⎧4.24 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎥⎪ ⎨t y ⎬ = ⎢2 3 0 ⎥ ⎨1 / 2 ⎬ = ⎨5 / 2 ⎬ = ⎨3.54 ⎬MPa ⎪t ⎪ ⎢0 0 0 ⎥ ⎪ 0 ⎪ ⎪ 0 ⎪ ⎪ 0 ⎪ ⎩ z⎭ ⎣ ⎦⎩ ⎭ ⎭ ⎩ ⎭ ⎩ --- ANS (d) The normal component σ n = {t} ⋅ {n}, T

⎧ 6 σn = ⎨ ⎩ 2

5 2

⎧1 / 2 ⎫ ⎪ 11 ⎫⎪ 0⎬⎨1 / 2 ⎬ = = 5.5MPa ⎭⎪ ⎪ 2 ⎩ 0 ⎭ --- ANS

2.6.2

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2.7

Find the principal stresses and corresponding principal directions for the stresses given in Problem 2.6. Check the result with other methods such as Mohr’s circle.

Solution: (a) The stress given in problem 2.6 is

⎡ 4 2 0⎤ [σ ij ] = ⎢⎢2 3 0⎥⎥ , ⎢⎣0 0 0⎥⎦

[ ]

To find the principal stresses, we require that σ ij − σ [I ] = 0 , or

4 −σ 2

2 3−σ

0

0

0 0 =0 −σ

(

)

Expanding the determinant yields − σ σ 2 − 7σ + 8 = 0 , the solutions of σ are

σ = 0 , or σ =

7 ± 17 , (which are 1.43845 and 5.56155) 2

--- ANS (i)

When σ 1 = 0 We have the equations

⎧4 n x + 2 n y = 0 ⎪ 2 2 2 ⎨ 2n x + 3n y = 0 , and also we have (n x ) + (n y ) + (n z ) = 1 ⎪ 0⋅n = 0 z ⎩ So the solutions can be obtained uniquely as

⎧n x = 0 ⎪ ⎨n y = 0 , and ⎪n = 1 ⎩ z

⎧n x ⎫ ⎪ ⎪ ⎨n y ⎬ ⎪n ⎪ ⎩ z⎭

(1)

⎧0⎫ ⎪ ⎪ = ⎨0⎬ is the corresponding principal direction ⎪1⎪ ⎩ ⎭

--- ANS (ii)

When σ 2 = 1.43845 We have the equations

⎧2.56155n x + 2n y = 0 ⎪ 2 2 2 ⎨ 2n x + 1.56155n y = 0 , and also we have (n x ) + (n y ) + (n z ) = 1 ⎪ − 1.43845n = 0 z ⎩

2.7.1

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Therefore we have the corresponding principal direction ⎧n x ⎫ ⎪ ⎪ ⎨n y ⎬ ⎪n ⎪ ⎩ z⎭

( 2)

⎧ 0.61541 ⎫ ⎪ ⎪ = ⎨− 0.78821⎬ ⎪ ⎪ 0 ⎩ ⎭

--- ANS (iii) When σ 3 = 5.56155 We have the equations

⎧− 1.56155n x + 2n y = 0 ⎪ 2 2 2 ⎨ 2n x − 2.56155n y = 0 , and also we have (n x ) + (n y ) + (n z ) = 1 ⎪ − 5.56155n = 0 z ⎩ Therefore we have the corresponding principal direction ⎧n x ⎫ ⎪ ⎪ ⎨n y ⎬ ⎪n ⎪ ⎩ z⎭

( 3)

⎧0.78821⎫ ⎪ ⎪ = ⎨0.61541⎬ ⎪ 0 ⎪ ⎩ ⎭

--- ANS (b) Comparing with Mohr’s circle Since the stresses associated with z are all zero, we know one principal stress is 0, ⎧n x ⎫ ⎪ ⎪ and its corresponding principal direction is ⎨n y ⎬ ⎪n ⎪ ⎩ z⎭

(1)

⎧0⎫ ⎪ ⎪ = ⎨0⎬ . So here we can use the ⎪1⎪ ⎩ ⎭

2D analysis on the x-y plane just for other principal values.

τ (4,2)

σ

σ max

σ min

(3,-2) According to the Mohr’s circle, we have the radius of the circle 2.7.2

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r=

1 2

((4 − 3)

2

+ (2 + 2 )

2

) = 2.06155 ,

⎛4+3 2−2⎞ The central coordinate of the circle is ( xc , y c ) = ⎜ , ⎟ = (3.5,0 ) 2 ⎠ ⎝ 2 Therefore we have the maximum and minimum stresses, respectively, σ max = xc + r = 3.5 + 2.06155 = 5.56155 σ min = xc − r = 3.5 − 2.06155 = 1.43845

These are the same as we obtained above. --- ANS

2.7.3

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2.8

A state of hydrostatic stress is given by

⎡σ 0 [σ ij ] = ⎢⎢ 0 ⎢⎣ 0

0

σ0 0

0⎤ 0 ⎥⎥ σ 0 ⎥⎦

Show that on any surface the force (or stress vector) is always perpendicular to the surface and that the magnitude of the stress vector is equal to σ 0 . Solution: (a) Assume any arbitrary plane surface with its normal unit vector {n} = {n x , n y , n z }. T

The stress vector acting on this surface, from equation (2.29) in the textbook,

⎧t x ⎫ {t} = ⎪⎨t y ⎪⎬ = σ ⎪t ⎪ ⎩ z⎭

⎧ nx ⎫ {n} = σ 0 ⎪⎨n y ⎪⎬ . ⎪n ⎪ ⎩ z⎭

[ ]

Since σ 0 is a scalar, the stress vector on this arbitrary surface is always parallel to the normal vector of this surface. This leads to the conclusion that the stress vector is always perpendicular to the surface. --- ANS (b) The magnitude of this stress vector t is

{t} = (t x )2 + (t y )2 + (t z )2 = σ 0 {n} = σ 0 . --- ANS

2.8.1

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2.9 An isotropic solid with Young’s modulus E and Poisson’s ratio υ is under a state of hydrostatic stress as given in Problem 2.8. Find the corresponding strain components.

⎡σ 0 0 Recall: [σ ij ] = ⎢⎢ 0 σ 0 ⎢⎣ 0 0

0⎤ 0 ⎥⎥ σ 0 ⎥⎦

Solution: (a) Three dimensional stress-strain relations can be expressed as:

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

(b) When the material is isotropic, a ij can be obtained individually as: a11 = a 22 = a33 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

a 44 = a 55 = a 66 =

1 , G

and others are zero.

υ E

,

(c) For a state of hydrostatic stress, we can obtain strain components with matrix multiplication: ⎡1 ⎢E ⎢ ⎧ε xx ⎫ ⎢ ⎪ε ⎪ ⎢ ⎪ yy ⎪ ⎢ ⎪⎪ε zz ⎪⎪ ⎢ ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ ⎪γ xz ⎪ ⎢ ⎪ ⎪ ⎢ ⎩⎪γ xy ⎭⎪ ⎢ ⎢ ⎢ ⎣

−

υ

E 1 E

symm

− −

υ E

υ

E 1 E

0

0

0

0

0

0

1 G

0 1 G

⎤ 0⎥ ⎧σ 0 ⎥ ⎪ 0 ⎥ ⎧σ 0 ⎫ ⎪ E ⎥ ⎪σ ⎪ ⎪σ 0 ⎪ 0⎪ 0 ⎥ ⎪σ ⎪ ⎪ E ⎥ ⎪ 0 ⎪ ⎪σ ⎥⎨ 0 ⎬ = ⎨ 0 0 ⎥⎪ ⎪ ⎪ E ⎥⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎪⎩ 0 ⎪⎭ ⎪ ⎥ ⎪⎩ 1⎥ ⎥ G⎦

⎫ (1 − 2υ )⎪ ⎪ (1 − 2υ )⎪ ⎪ ⎪ (1 − 2υ )⎬ ⎪ 0 ⎪ ⎪ 0 ⎪ ⎪⎭ 0 --- ANS

2.9.1

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2.10

For small strains, the volume change ΔV

V

is identified to be equal to

ε xx + ε YY + ε zz . The bulk modulus K of an isotropic solid is defined as the ratio of the average stress and the volume change, i.e., 1 (σ xx + σ yy + σ zz ) = K ΔV 3 V

Derive K in terms of E and υ . Solution: (a) Three dimensional stress-strain relations can be expressed as:

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

(b) If the material is isotropic, a ij are given as a11 = a 22 = a33 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

a 44 = a 55 = a 66 =

1 , G

and the rest are zero.

υ E

,

(c) For arbitrary stresses, we can obtain strain components with matrix multiplication: ⎫ ⎧ 1 ⎤ 0⎥ ⎪ E (σ xx − υσ yy − υσ zz ) ⎪ E E ⎪ ⎪1 ⎥ 1 υ 0 0 0 ⎥ ⎧σ xx ⎫ ⎪ (−υσ xx + σ yy − υσ zz )⎪ − ⎪ E E ⎥ ⎪σ yy ⎪ ⎪ E ⎪ ⎪ ⎪1 1 ⎪ ⎥ 0 0 0 ⎪σ ⎪ ⎪ (−υσ xx − υσ yy + σ zz )⎪ ⎥ ⎪ ⎪ zz E E ⎬ ⎥ ⎨τ ⎬ = ⎨ τ yz 1 ⎪ 0 0 ⎥ ⎪ yz ⎪ ⎪ G G ⎪ ⎪ ⎪ ⎪ τ ⎥ xz τ xz 1 ⎪ ⎪ ⎪ ⎪ ⎥ 0 ⎪τ xy ⎪ ⎪ symm ⎪ ⎭ ⎩ ⎥ G G ⎪ ⎪ ⎥ 1 τ xy ⎪ ⎪ ⎥ G⎦ G ⎭ ⎩ (d) With the definition of bulk modulus K, we have ⎡1 ⎢E ⎢ ⎧ε xx ⎫ ⎢ ⎪ε ⎪ ⎢ ⎪ yy ⎪ ⎢ ⎪⎪ε zz ⎪⎪ ⎢ ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ ⎪γ xz ⎪ ⎢ ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢ ⎢ ⎢ ⎣

−

υ

−

υ

0

0

1 (σ xx + σ yy + σ zz ) = K ΔV = K (ε xx + ε YY + ε zz ) 3 V

Substituting the strains in terms of stresses in the equation above, we obtain 2.10.1

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[

]

1 (σ xx + σ yy + σ zz ) = K σ xx (1 − 2υ ) + σ yy (1 − 2υ ) + σ zz (1 − 2υ ) 3 E K (1 − 2υ ) (σ xx + σ yy + σ zz ) = E

Thus, K =

E 3(1 − 2υ ) --- ANS

2.10.2

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2.11 A block of elastic solid is compressed by normal stress σ xx as shown in Fig. 2.19. The containing walls are rigid and smooth (frictionless). Find the values of k for plane strain and plane stress conditions, respectively, in the stress-strain relation obtained from the compression test above. σ xx = kε xx Assume that E = 70GPa and υ = 0.3 .

σ xx = −σ 0

y

σ xx = −σ 0 x

Figure 2.19

Solid between two smooth rigid walls

Solution: Recall: Three dimensional stress-strain relations can be expressed as:

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎬ , where c ij are elastic constants. 6×6 ⎨γ ⎪ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎨ ⎬ = cij ⎪τ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

or

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

If the material is isotropic, c ij and a ij are given as

c11 = c 22 = c33 = λ + 2G , c12 = c13 = c 23 = c 21 = c 31 = c32 = λ , c 44 = c55 = c66 = G , and the rest are zero. where λ =

υE E and G = (1 + υ )(1 − 2υ ) 2(1 + υ ) 2.11.1

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or, a11 = a 22 = a33 = a 44 = a 55 = a 66 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

υ E

,

1 , and the rest are zero. G

(a) Plane strain problem: In plane strain problems, we have

ε zz = γ yz = γ xz = 0 In this problem, we also have the following constraint condition

ε yy = 0

[ ] {ε }

Therefore, the relation {σ } = cij

6×6

can be used to obtain

⎧σ xx ⎫ ⎡λ + 2G 0 0 0 ⎤ ⎧ ε xx ⎫ λ λ ⎪σ ⎪ ⎢ ⎪ ⎪ 0 0 0 ⎥⎥ ⎪ε yy = 0⎪ λ + 2G λ ⎪ yy ⎪ ⎢ ⎪⎪σ zz ⎪⎪ ⎢ λ + 2G 0 0 0 ⎥ ⎪⎪ε zz = 0 ⎪⎪ ⎨ ⎬=⎢ ⎥⎨ ⎬ G 0 0 ⎥ ⎪γ yz = 0⎪ ⎪τ yz ⎪ ⎢ ⎪τ xz ⎪ ⎢ symm G 0 ⎥ ⎪γ xz = 0 ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ G ⎥⎦ ⎪⎩ γ xy ⎪⎭ ⎪⎩τ xy ⎪⎭ ⎢⎣ Expanding the matrix multiplication, we have ⎧σ xx = (λ + 2G )ε xx ⎪ σ = λε yy xx ⎪ ⎪⎪ σ zz = λε xx ⎨ τ yz = 0 ⎪ ⎪ τ xz = 0 ⎪ ⎪⎩ τ xy = Gγ xy Comparing to the problem statement, we have (1 − υ ) E (1 − 0.3) ⋅ 70 = 94.23 GPa k = λ + 2G = = (1 + υ )(1 − 2υ ) (1 + 0.3)(1 − 2 ⋅ 0.3) --- ANS (b) Plane stress problem: In plane stress problems, we have

σ zz = τ yz = τ xz = 0 In this problem, we also have

ε yy = 0 2.11.2

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[ ] {σ }

Therefore, the relation {ε } = aij ⎡1 ⎢E ⎢ ⎧ ε xx ⎫ ⎢ ⎪ε = 0⎪ ⎢ ⎪ yy ⎪ ⎢ ⎪⎪ ε zz ⎪⎪ ⎢ ⎨ ⎬=⎢ ⎪ γ yz ⎪ ⎢ ⎪ γ xz ⎪ ⎢ ⎪ ⎪ ⎢ ⎪⎩ γ xy ⎪⎭ ⎢ ⎢ ⎢ ⎣

−

υ

E 1 E

− −

υ E

υ

E 1 E

6×6

0

0

0

0

0

0

1 G

0

symm

can be used to obtain:

1 G

⎤ 0⎥ ⎥ 0 ⎥ ⎧ σ xx ⎫ ⎥ ⎪ σ yy ⎪ ⎪ ⎪ 0 ⎥ ⎪σ = 0⎪ ⎥ ⎪ zz ⎪ ⎥ ⎨τ = 0 ⎬ ⎪ 0 ⎥ ⎪ yz ⎪ ⎪ = 0 τ ⎥ xz ⎪ ⎪ ⎥ 0 ⎪ τ xy ⎪ ⎭ ⎥⎩ ⎥ 1 ⎥ G⎦

Expanding the matrix multiplication, we have 1 ⎧ ⎪ ε xx = E (σ xx − υσ yy ) ⎪ 1 ⎪ε yy = (−υσ xx + σ yy ) = 0 E ⎪ ⎪ ε = − υ (σ + σ ) zz xx yy ⎨ E ⎪ γ yz = 0 ⎪ ⎪ γ xz = 0 ⎪ 1 γ xy = τ xy ⎪ G ⎩ Solving the first two equations leads to,

ε xx

1 1−υ 2 = (σ xx − υσ yy ) = σ xx E E

Thus, k =

E 70 = = 76.92GPa under plane stress condition. 2 1−υ 1 − 0 .3 2

--- ANS

2.11.3

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2.12

An aluminum 2024 T3 bar of unit cross-sectional area is subjected to a tensile force in the longitudinal direction. If the lateral surface of the bar is confined and not allowed to contract during loading, find the force that is needed to produce a 1 percent longitudinal strain. Compare this with the corresponding load for the bar under simple tension.

Solution: Recall: Three dimensional stress-strain relations can be expressed as:

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎨ ⎬ = cij ⎪τ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎬ , where c ij are elastic constants. 6×6 ⎨γ ⎪ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

[ ]

If the material is isotropic, c ij and a ij are given by c11 = c 22 = c33 = λ + 2G , c12 = c13 = c 23 = c 21 = c 31 = c 32 = λ , c 44 = c55 = c66 = G , and the rest are zero. where λ =

υE E and G = (1 + υ )(1 − 2υ ) 2(1 + υ )

or, a11 = a 22 = a33 = a 44 = a 55 = a 66 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

υ E

,

1 , and the rest are zero. G

(a) Aluminum alloys are usually considered isotropic, so the above three-dimensional stress-strain relations can be utilized. Also, we have the mechanical properties for aluminum 2024 T3: E = 72GPa , υ = 0.33 . (b) If the lateral surface of the bar is not allowed to contract during loading, we have the conditions: ε zz = ε yy = 0 . Also, we need to produce a 1 percent longitudinal

2.12.1

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strain, which means ε xx = 0.01 and all shear strains vanish.

[ ] {ε }

So we can use the equation {σ } = cij

6×6

to obtain stresses.

⎧σ xx ⎫ ⎡λ + 2G λ λ 0 0 0 ⎤ ⎧ε xx = 0.01⎫ ⎪σ ⎪ ⎢ ⎪ ⎪ λ + 2G λ 0 0 0 ⎥⎥ ⎪ ε yy = 0 ⎪ ⎪ yy ⎪ ⎢ ⎪⎪σ zz ⎪⎪ ⎢ λ + 2G 0 0 0 ⎥ ⎪⎪ ε zz = 0 ⎪⎪ ⎨ ⎬=⎢ ⎥⎨ ⎬ G 0 0 ⎥ ⎪ γ yz = 0 ⎪ ⎪τ yz ⎪ ⎢ ⎪τ xz ⎪ ⎢ symm G 0 ⎥ ⎪ γ xz = 0 ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ G ⎥⎦ ⎪⎩ γ xy = 0 ⎪⎭ ⎪⎩τ xy ⎪⎭ ⎢⎣

Calculate λ and G for aluminum 2024-T3, we have: 0.33 ⋅ 72 υE λ= = = 52.5 GPa (1 + υ )(1 − 2υ ) (1 + 0.33)(1 − 2 ⋅ 0.33) G=

72 E = = 27.1 GPa 2(1 + υ ) 2(1 + 0.33)

we have

⎧σ xx = (λ + 2G )ε xx = (52.5431 + 2 × 27.0677) ⋅ 0.01 = 1.07 GPa ⎪ σ yy = λε xx = 52.5431× 0.01 = 0.525 GPa ⎨ ⎪ σ zz = λε xx = 52.5431 × 0.01 = 0.525 GPa ⎩ With a unit cross-sectional area, we have the axial force P = σ xx × 1 = 1070 × 10 6 N to produce the required strain.

--- ANS (c) Compare with the corresponding load for the bar under simple tension. Under simple tension, we have σ zz = σ yy = 0 and the axial stress is

σ xx = Eε xx = 72 × 0.01 = 0.72 GPa . The required axial force is P = σ xx A = 0.72GPa × 1 = 720 × 106 N . --- ANS Note that for the material (Al 2024-T3) used in this illustration, the stresses produced for a 1% strain for both cases have exceeded the yield stress of the material. However, the results still reveal the fact that the longitudinal stiffness of a bar increases if its lateral displacements are suppressed.

2.12.2

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2.13

Compare the axial stiffnesses of aluminum alloy 2024-T3 under plane strain and plane stress conditions, respectively.

Solution: Recall: Three dimensional stress-strain relations can be expressed as:

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎨ ⎬ = cij ⎪τ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎬ , where c ij are elastic constants. 6×6 ⎨γ ⎪ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

[ ]

If the material is isotropic, c ij and a ij are given by c11 = c 22 = c33 = λ + 2G , c12 = c13 = c 23 = c 21 = c 31 = c 32 = λ , c 44 = c55 = c66 = G , and the rest are zero. where λ =

υE E and G = (1 + υ )(1 − 2υ ) 2(1 + υ )

a11 = a 22 = a33 = a 44 = a 55 = a 66 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

υ E

,

1 , and the rest are zero. G

(a) For aluminum 2024 T3, E = 72GPa , υ = 0.33 υE E => λ = = 52 GPa , G= = 27 GPa (1 + υ )(1 − 2υ ) 2(1 + υ ) (b) Plane strain condition In plane strain problem, we have

ε zz = γ yz = γ xz = 0 For axial loading alone, we have

γ xy = 0 and σ yy = 0 2.13.1

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Expanding ⎧σ xx ⎫ ⎡λ + 2G λ λ 0 0 0 ⎤ ⎧ ε xx ⎫ ⎪σ ⎪ ⎢ ⎪ ⎪ λ + 2G λ 0 0 0 ⎥⎥ ⎪ ε yy ⎪ ⎪ yy ⎪ ⎢ ⎪⎪σ zz ⎪⎪ ⎢ λ + 2G 0 0 0 ⎥ ⎪⎪ε zz = 0 ⎪⎪ ⎨ ⎬=⎢ ⎥⎨ ⎬ G 0 0 ⎥ ⎪γ yz = 0⎪ ⎪τ yz ⎪ ⎢ ⎪τ xz ⎪ ⎢ symm G 0 ⎥ ⎪γ xz = 0 ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ G ⎥⎦ ⎪⎩γ xy = 0⎪⎭ ⎪⎩τ xy ⎪⎭ ⎢⎣

we have ⎧σ xx = (λ + 2G )ε xx + λε yy ⎪σ = λε + (λ + 2G )ε xx yy ⎪ yy ⎪⎪ σ zz = λ (ε xx + ε yy ) ⎨ τ yz = 0 ⎪ ⎪ τ xz = 0 ⎪ τ xy = 0 ⎪⎩

Applying σ yy = 0 , we have

− λε xx 4G (λ + G ) = ε xx (λ + 2G ) (λ + 2G ) 4(27.0677)(52.5431 + 27.0677) = ε xx = 80.8ε xx GPa (52.5431 + 2 ⋅ 27.0677)

σ xx = (λ + 2G )ε xx + λ ⋅

(c) Plane stress condition In plane stress problem, we have

σ zz = τ yz = τ xz = 0 For this problem we also have

γ xy = 0 and σ yy = 0

[ ] {σ }

Therefore, the {ε } = aij

6×6

can be used to obtain:

2.13.2

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⎡1 ⎢E ⎢ ⎧ε xx ⎫ ⎢ ⎪ε ⎪ ⎢ ⎪ yy ⎪ ⎢ ⎪⎪ε zz ⎪⎪ ⎢ ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ ⎪γ xz ⎪ ⎢ ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢ ⎢ ⎢⎣

−

υ

E 1 E

− −

υ E

υ

E 1 E

0

0

0

0

0

0

1 G

0

symm

1 G

⎤ 0⎥ ⎥ 0 ⎥ ⎧ σ xx ⎫ ⎥ ⎪σ yy = 0⎪ ⎪ ⎪ 0 ⎥ ⎪σ = 0 ⎪ ⎪ ⎥ ⎪ zz ⎥ ⎨τ = 0 ⎬ ⎪ 0 ⎥ ⎪ yz ⎪ ⎪ τ = 0 ⎥ xz ⎪ 0 ⎥ ⎪τ xy = 0 ⎪⎪ ⎭ ⎥⎩ 1⎥ G ⎥⎦

Expanding the matrix multiplication, we have 1 ⎧ ⎪ ε xx = E σ xx ⎪ υ ⎪ε yy = − σ xx E ⎪ ⎪ υ ⎨ε zz = − σ xx E ⎪ ⎪ γ yz = 0 ⎪ γ =0 xz ⎪ ⎪⎩ γ xy = 0 Thus, σ xx = Eε xx = 72ε xx GPa

(d) The above analysis indicates that the axial stiffness of an aluminum bar under the plane strain condition is (80.8-72)/72 = 12% greater than that under the plane stress condition. --- ANS

2.13.3

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2.14

Show that the state of stress of a solid body of any shape placed in a pressured chamber is a state of hydrostatic stress. Neglect the effect of the gravitational force.

Solution: Assume a solid body (without internal voids) with an arbitrary shape is placed in a pressured chamber with a pressure σ 0 as shown the figure below.

σ0

We will show that the hydrostatic stress listed below is the solution.

⎡− σ 0 [σ ij ] = ⎢⎢ 0 ⎢⎣ 0

0 −σ0 0

0 ⎤ 0 ⎥⎥ − σ 0 ⎥⎦

--- (1)

First the constant stress field given by (1) satisfies the 3D equilibrium equations (2.21 – 2.23). Second, this stress field satisfies the boundary condition, i.e., the traction at any point on the surface of the body is given by

⎡− σ 0 {t} = [σ ij ]{n} = ⎢⎢ 0 ⎢⎣ 0

0 −σ0 0

0 ⎤ ⎧n x ⎫ ⎪ ⎪ 0 ⎥⎥ ⎨n y ⎬ = −σ 0 {n} − σ 0 ⎥⎦ ⎪⎩n z ⎪⎭

---(2)

⎧n x ⎫ ⎪ ⎪ where {n} = ⎨n y ⎬ is the unit normal vector to the surface at the point of interest. ⎪n ⎪ ⎩ z⎭ Last, the constant stress field given by (1) implies that the corresponding strain field is also constant and the compatibility equations are also satisfied. Thus, the hydrostatic stress field given by (1) is the solution. 2.14.1

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2.15

Write the strain energy density expression in terms of stress components by using (2.95) for isotropic solids and show that the Poisson’s ratio is bounded by -1 and 0.5.

Solution: From equation (2.95), we have the strain energy density: 1 1 W = {ε }T [c ]{ε } = {σ }T [ a ]{σ } 2 2

(2.15.1)

For isotropic material, the stress-strain relationship can be expressed in terms of elastic constants or elastic compliances:

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎬ , where c ij are elastic constants. 6×6 ⎨γ ⎪ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎨ ⎬ = cij ⎪τ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

c11 = c 22 = c33 = λ + 2G , c12 = c13 = c 23 = c 21 = c 31 = c 32 = λ , c 44 = c55 = c66 = G , and the rest are zero. where λ =

υE E and G = (1 + υ )(1 − 2υ ) 2(1 + υ )

⎧ε xx ⎫ ⎪ε ⎪ ⎪ yy ⎪ ⎪⎪ε zz ⎪⎪ ⎨ ⎬ = aij ⎪γ yz ⎪ ⎪γ xz ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ ⎬ , where aij are elastic compliances. 6×6 ⎨τ ⎪ yz ⎪ ⎪τ xz ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭

[ ]

a11 = a 22 = a33 =

1 , E

a12 = a13 = a 23 = a 21 = a 31 = a32 = −

a 44 = a 55 = a 66 =

1 , G

and the rest are zero.

υ E

,

The strain energy density in terms of stress components can be derived from equation (2.15.1) as

2.15.1

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W =

=

1 {σ xx 2

σ yy σ zz τ yz τ xz

⎧σ xx ⎫ ⎪σ ⎪ ⎪ yy ⎪ ⎪⎪σ zz ⎪⎪ τ xy } aij 6×6 ⎨ ⎬ ⎪τ yz ⎪ ⎪ τ xz ⎪ ⎪ ⎪ ⎩⎪τ xy ⎭⎪

[ ]

1 2 2 2 2 2 2 {σ xx + σ yy + σ zz − 2υ (σ xxσ yy + σ xxσ zz + σ yyσ zz ) + 2(1 + υ )(τ xy + τ xz + τ yz )} 2E

If we choose to use the principal directions as the coordianate system, then

σ xx = σ 1, σ yy = σ 2 , σ zz = σ 3 , and τ xy = τ xz = τ yz = 0 , where σ 1, σ 2 ,σ 3 are principal stresses. The strain energy density can be expressed in the form W =

1 2 2 2 {σ 1 + σ 2 + σ 3 − 2υ (σ 1σ 2 + σ 1σ 3 + σ 2σ 3 )} 2E

(2.15.2)

Note that the strain energy of an isotropic material compounds of two parts, dilatational and distortional effect, i.e., W = Wv + W d (2.15.3) where Wv is the strain energy density associated with the volume dilatation and Wd is the strain energy density associated with the shape distortion. We know that the dilatation is given by σ 0 = Kε 0 1 ΔV where σ 0 = (σ 1 + σ 2 + σ 3 ) is the average stress, ε 0 = ε 1 + ε 2 + ε 3 = is 3 V

unit volume change, K =

E is the bulk modulus. 3(1 − 2υ )

1 1 Hence, Wv = σ 0 ε 0 = σ 02 . 2 2K

The expression of Wd is obtained by subtracting Wd from the total strain energy W given by equation (2.15.2). We obtain Wd =

1 J2 2G

where 1 J 2 = [(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ] 6

2.15.2

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Thus, W =

1 1 σ 02 + J2 2K 2G

(2.15.4)

It is evident that W is always positive and vanishes only when all stresses components vanish. To ensure that W be always positive, it is clear that K and G must be positive. Noting the relations E E K= > 0 and G = >0 3( 1 − 2υ ) 2( 1 + υ ) We conclude that − 1 < υ < 0.5 --- ANS

2.15.3

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2.16

Derive the compatibility equation for plane elasticity problems in terms of stresses, i.e.,

∇ 2 (σ xx + σ yy ) = 0

Solution: (a) For simplicity, we will use the following notations for differentiations with respect ∂ 2τ xy ∂ 2σ xx ∂ 2σ xx = , σ xx , yy = , and τ xy , xy = and so on. The ∂x∂y ∂x 2 ∂y 2

to x and y: σ xx , xx

same differentiation notations are applied to the strain components (b) For 2D problems, the compatibility equation is given by

ε xx , yy + ε yy , xx = γ xy , xy (c) The stress-strain relations for isotropic solids are υ ⎡ 1 ⎤ − 0⎥ ⎢ E E ⎧ε xx ⎫ ⎧σ ⎫ ⎥ ⎪ xx ⎪ 1 ⎪ ⎪ ⎢ υ 0 ⎥ ⎨σ yy ⎬ ⎨ε yy ⎬ = ⎢− E ⎥⎪ ⎪ ⎪γ ⎪ ⎢ E 1 ⎥ ⎩σ xy ⎭ ⎩ xy ⎭ ⎢ 0 0 ⎢⎣ G ⎥⎦ Substituting the strain components into the compatibility equation, we have 1 1 1 [ (σ xx − υσ yy )], yy + [ ( −υσ xx + σ yy )], xx = ( τ xy ), xy E E G

=>

1 1 1 2(1 + υ ) τ xy , xy (σ xx , yy − υσ yy , yy ) + (−υσ xx , xx + σ yy , xx ) = τ xy , xy = E E G E

--- (1)

(d) Now we consider the equilibrium equations:

⎧σ xx, x + τ xy , y = 0 ⎨ ⎩τ xy , x + σ yy , y = 0 Differentiating the first equation by x and the second one by y, we have

⎧σ xx, xx + τ xy , xy = 0 ⎨ ⎩τ xy , xy + σ yy , yy = 0

--- (2)

Adding the above two equations in (2) we obtain 1 2

τ xy , xy = − (σ xx , xx + σ yy , yy )

--- (3)

(e) Now substituting equation (3) into equation (1), we have

2.16.1

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1 (σ xx , yy − υσ yy , yy ) + ( −υσ xx , xx + σ yy , xx ) = 2(1 + υ )[− (σ xx , xx + σ yy , yy )] 2

=> σ xx , yy + σ yy , xx = −(σ xx , xx + σ yy , yy ) => σ xx , yy + σ yy , xx + σ xx , xx + σ yy , yy = 0 => ∇ 2 (σ xx + σ yy ) = 0 where ∇ 2 =

∂2 ∂2 + ∂x 2 ∂y 2 --- ANS

2.16.2

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2.17

Consider a thin rectangular panel loaded as shown in Fig. 2.20. Show that the Airy stress function

φ = c1 x 2 + c 2 xy + c3 y 2 solves the problem. Find the constants c1, c2, c3. y

σ0

σ0

b

x

a Figure 2.20

Thin rectangular panel subjected to uniform tension

Solution: (a) Since φ = c1 x 2 + c 2 xy + c3 y 2 is a second order equation in x and y it automatically satisfies the compatibility equation, ∇ 2 ∇ 2φ = 0 . So the given Airy stress function can be used to solve the problem if we can find c1, c2, c3 that satisfies the boundary conditions. (b) Stress matrix:

σ xx =

∂ 2φ , ∂y 2

=> σ xx = 2c3 ,

∂ 2φ , => σ yy = 2c1 , ∂x 2 ∂ 2φ =− => τ xy = −c 2 ∂x∂y

σ yy =

τ xy

So we have,

⎡ 2c [σ ] = ⎢ 3 ⎣− c 2

− c2 ⎤ 2c1 ⎥⎦

(c) Applying boundary conditions: (i)

a , => n x = 1 , n y = 0 , t x = σ 0 , t y = 0 2 we have [σ ]{n} = {t}

at x =

2.17.1

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⎡ 2c3 => ⎢ ⎣− c 2 (ii)

σ ⎧ − c 2 ⎤ ⎧1⎫ ⎧σ 0 ⎫ ⎪c3 = 0 ⎨ ⎬ = ⎨ ⎬ => ⎨ 2 2c1 ⎥⎦ ⎩0⎭ ⎩ 0 ⎭ ⎪⎩ c 2 = 0

b , => n x = 0 , n y = 1 , t x = 0 , t y = 0 2 we have [σ ]{n} = {t}

at y =

⎡ 2c3 => ⎢ ⎣− c 2

− c 2 ⎤ ⎧0⎫ ⎧0⎫ ⎧c 2 = 0 ⎨ ⎬ = ⎨ ⎬ => ⎨ ⎥ 2c1 ⎦ ⎩1⎭ ⎩0⎭ ⎩ c1 = 0

(d) The given Airy stress function is the solution to the problem and the values of the constants are c1 = 0 , c2 = 0 and c3 =

σo 2

. --- ANS

2.17.2

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Consider the a × b rectangular panel shown in Fig. 2.20. Find the problem that the Airy’s stress function φ = xy 3 solves. That is, find the tractions at the

2.18

boundary of the panel. y

b

x

a Figure 2.20

Thin rectangular panel subjected to uniform tension

Solution: (a) It is easy to verify that the stress function φ = xy 3 satisfies the compatibility equation, ∇ 2 ∇ 2φ = 0 . (b) The stresses are obtained from the stress function as

σ xx =

∂ 2φ , ∂y 2

=> σ xx = 6 xy ,

∂ 2φ σ yy = 2 , => σ yy = 0 , ∂x ∂ 2φ τ xy = − => τ xy = −3y 2 ∂x∂y So we have,

⎡ 6xy [σ ] = ⎢ 2 ⎣− 3 y

− 3y 2 ⎤ ⎥ 0 ⎦

(c) The tractions at the boundary (i)

On the vertical face at x =

a , => n x = 1 , n y = 0 , 2

we have {t} = [σ ]{n}

⎧t x ⎫ ⎡ 6 xy => ⎨ ⎬ = ⎢ 2 ⎩t y ⎭ ⎣− 3 y

− 3 y 2 ⎤ ⎧1⎫ ⎧ 6 xy ⎫ ⎧ 3ay ⎫ =⎨ ⎥⎨ ⎬ = ⎨ 2⎬ 2⎬ 0 ⎦ ⎩0⎭ ⎩− 3 y ⎭ ⎩− 3 y ⎭ --- ANS

2.18.1

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(ii)

a On the vertical face at x = − , => n x = −1 , n y = 0 , 2 we have {t} = [σ ]{n}

⎧t x ⎫ ⎡ 6 xy => ⎨ ⎬ = ⎢ 2 ⎩t y ⎭ ⎣− 3 y

− 3 y 2 ⎤ ⎧− 1⎫ ⎧− 6 xy ⎫ ⎧3ay ⎫ ⎥⎨ ⎬ = ⎨ 2 ⎬ = ⎨ 2 ⎬ 0 ⎦ ⎩ 0 ⎭ ⎩ 3 y ⎭ ⎩3 y ⎭ --- ANS

(iii) On the top face at y =

b , => n x = 0 , n y = 1 , 2

we have {t} = [σ ]{n} ⎧t x ⎫ ⎡ 6 xy => ⎨ ⎬ = ⎢ 2 ⎩t y ⎭ ⎣− 3 y

− 3 y 2 ⎤ ⎧0⎫ ⎧− 3 y 2 ⎫ ⎧⎪− 3 b 2 ⎫⎪ ⎬=⎨ 4 ⎬ ⎥⎨ ⎬ = ⎨ 0 ⎦ ⎩1⎭ ⎩ 0 ⎭ ⎪ 0 ⎪ ⎭ ⎩

--- ANS b (iv) On the bottom face at y = − , => n x = 0 , n y = −1 , 2 we have {t} = [σ ]{n} ⎧t x ⎫ ⎡ 6 xy => ⎨ ⎬ = ⎢ 2 ⎩t y ⎭ ⎣− 3 y

− 3 y 2 ⎤ ⎧ 0 ⎫ ⎧3 y 2 ⎫ ⎧⎪ 3 b 2 ⎫⎪ ⎬ = ⎨4 ⎬ ⎥⎨ ⎬ = ⎨ 0 ⎦ ⎩− 1⎭ ⎩ 0 ⎭ ⎪ 0 ⎪ ⎭ ⎩

--- ANS Note: The tractions on the edges can also be found directly from the stress components at the same locations.

2.18.2

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3.1

Show that there is no warping in a bar of circular cross-section.

Solution: (a) Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then, u = −θzy

v = θzx (3.1.1) w = θψ ( x, y ) where ψ ( x, y ) is some function of x and y, called warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small. (b) From the displacement field above, it is easy to obtain that

ε xx = ε yy = ε zz = γ xy = 0 So from the stress-strain relationship, we have

σ xx = σ yy = σ zz = τ xy = 0 Therefore the equilibrium equations reduce to ∂τ xz ∂τ yz + =0 ∂x ∂y This equation is identically satisfied if the stresses are derived from a stress function φ ( x, y ) , so that

τ xz =

∂φ , ∂y

τ yz = −

∂φ ∂x

(3.1.2)

(c) From the displacement field and stress-strain relationship, we can obtain

γ xz =

∂w ∂u ∂w + = − θy ∂x ∂z ∂x

(3.1.3)

γ yz =

∂w ∂v ∂w + = + θx ∂y ∂z ∂y

(3.1.4)

So it forms the compatibility equation or in terms of Prandtl stress function

∂γ yz ∂x

−

∂γ xz = 2θ , ∂y

∂ 2φ ∂ 2φ + = −2Gθ ∂x 2 ∂y 2

(3.1.5)

(d) Boundary conditions, dφ = 0 , or φ = const . But for a solid sections with a single contour boundary, ds

this constant can be chosen to be zero. Then we have the boundary condition φ = 0 on the lateral surface of the bar. (e) For a bar with circular cross-section, assume the Prandtl stress function as 3.1.1

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x2 y2 + − 1) which satisfies the boundary conditions stated above. a2 a2 1 Substitute φ into (3.1.5), we obtain C = − a 2 Gθ 2

φ = C(

Then φ = −

Gθ 2 (x + y 2 − a 2 ) 2

Using (3.1.2), we have 1 ∂φ γ xz = = −θy , G ∂y

and γ yz = −

1 ∂φ = θx G ∂x

Comparing with (3.1.3) and (3.1.4), we have

γ xz =

∂w ∂w − θy = −θy => = 0 . Thus, w = f ( y ) ∂x ∂x

γ yz =

∂w ∂w + θx = θx => = 0 , Thus, w = g (x ) ∂y ∂y

Hence we conclude w = const . This means that the cross-section remains plane after torsion. In other words, there is no warping. Therefore w( x, y ) = 0 can be verified, and it successfully expresses the statement. --- ANS

3.1.2

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3.2

Show that the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections as shown in Fig. 3.34. Find the torsion constant J in terms of the inner radius ai and outer radius a 0 , and compare with the torsion constant obtained using (3.59) for thin-walled sections. What is the condition on the wall thickness for the approximate J to be within 1 percent of the exact J ?

a0

ai

Figure 3.34

Bar of a hollow circular section

Solution: Recall: (a) Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then, u = −θzy

v = θzx (3.2.1) w = θψ ( x, y ) where ψ ( x, y ) is a function of x and y, called warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small. (b) From the displacement field above, it is easy to obtain that

ε xx = ε yy = ε zz = γ xy = 0 So from the stress-strain relationship, we have

σ xx = σ yy = σ zz = τ xy = 0 Therefore the equilibrium equations reduce to ∂τ xz ∂τ yz + =0 ∂x ∂y This equation is identically satisfied if the stresses are derived from a stress function φ ( x, y ) , so that

3.2.1

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τ xz =

∂φ , ∂y

τ yz = −

∂φ ∂x

(3.2.2)

(c) From the displacement field and stress-strain relationship, we can obtain

γ xz =

∂w ∂u ∂w + = − θy ∂x ∂z ∂x

(3.2.3)

γ yz =

∂w ∂v ∂w + = + θx ∂y ∂z ∂y

(3.2.4)

So it forms the compatibility equation

∂γ yz ∂x

−

∂γ xz = 2θ , ∂y

∂ φ ∂ 2φ + = −2Gθ ∂x 2 ∂y 2 2

or in terms of Prandtl stress function

(3.2.5)

(d) Boundary conditions, dφ = 0 , or φ = const . ds

--1.

To show that the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections, we have to show that the Prandtl stress function for hollow circular sections satifies equilibrium equations, compatibility equations as well as traction boundary conditions. (1) Equilibrium equations Prandtl stress functions by their definition must satify equilibrium equations.. (2) Compatibility equations Use the Prandtl stress function as it stated for bars of circular solid sections

φ = C(

x2 a0

2

+

y2 a0

2

− 1) (here we use a 0 . Assuming φ = C (

x2 ai

2

+

y2 ai

2

− 1)

would be fine too). 1 2 Then substitute φ into (3.2.5), we have C = − a 0 Gθ . Thus we have 2

Gθa 0 x 2 y2 φ=− ( 2 + 2 − 1) . 2 a0 a0 2

(3.2.6)

Therefore we have a stress function for bars of hollow circular sections satisfying the compatibility equation (3) Traction boundary conditions 3.2.2

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To satisfy the traction boundary conditions we must show

dφ = 0 on the ds

traction free surfaces.

φ (r = ai ) = −

Gθa0 ai ( 2 − 1) = const. 2 a0 2

that is

dφ | r = ai = 0 ds

that is

dφ | r = a0 = 0 ds

2

Gθa 0 a0 φ (r = a0 ) = − ( 2 − 1) = 0. 2 a0 2

2

It shows that the B.C.’s have been satisfied. (4) Since equilibrium equations, compatibility equations and traction boundary conditions are all satisfied, the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections. --- ANS 2.

Compare torsion constant (1) The torque produced by the stresses is ∂φ ∂φ T = ∫∫ (− x − y )dA ∂x ∂y A

(3.2.7)

Substituting (3.2.6) into (3.2.7) and use polar coordinates to perform integration, we have, 1 π 4 4 T = Gθ ∫∫ ( x 2 + y 2 ) dA =Gθ ∫∫ (r 2 )rdrdθ =Gθ ( 2π )( r 4 ) | aai0 = Gθ [ (a 0 − ai )] 4 2 A A Comparing with T = GθJ , we have the torsion constant J =

π 2

(a 0 − ai ) 4

4

(2) Using (3.59) in the textbook for thin-walled sections, we have the approximate torsion constant J app =

4A

2

∫ ds / t

where A is the area enclosed by the centerline of the wall section. a + ai 2 π A = π( 0 ) = (a 0 + ai ) 2 , and t = a0 − ai 2 4 Therefore J app =

3.

4A

2

∫ ds / t

π2 = 4

(a0 + ai ) 4

π (a 0 + ai )

=

π 4

(a 0 + ai ) 3 (a 0 − ai )

a 0 − ai

In order to have the approximate J to be within 1 percent of the exact J , one must have

J app − J J

≤ 0.01

3.2.3

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Substituting J app and J into the above error equation, we have

J app − J J

π = 4

((a 0 + ai ) 3 (a 0 − ai ) −

π 2

π 2

(a 0 − ai ) 4

4

=

(a 0 − ai ) 4

4

− (a 0 − ai ) 2 2(a 0 + ai ) 2

2

≤ 0.01

Because ai and a 0 are positive real number, we have (a0 − ai ) 2 2( a 0 + a i ) 2

2

≤ 0.01

=>

(

ai 2 a ) − 2.040816( i ) + 1 ≤ 0 a0 a0

We can obtain the solution of the above equation as a 0.8174 ≤ i ≤ 1.2235 a0 Since a0 > ai we have the solution

ai ≥ 0.8174 a0

Therefore the condition on the wall thickness t is t = a 0 − ai ≤ a 0 − 0.8174a 0 = 0.1826a0 (OR t = a 0 − ai ≤

1 ai − ai = 0.2235a i ) 0.8174

--- ANS

3.2.4

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3.3

Consider the straight bar of a uniform elliptical cross-section. The semimajor and semiminor axes are a and b, respectively. Show that the stress function of the form x2 y2 φ = C ( 2 + 2 − 1) a b provides the solution for torsion of the bar. Find the expression of C and show that J=

τ zx =

πa 3 b 3 a2 + b2

− 2Ty , πab 3

τ zy =

2Tx πa 3b

and the warping displacement w=

T (b 2 − a 2 ) xy πa 3b 3G

Solution: Recall: 1. Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then, u = −θzy

v = θzx (3.3.1) w = θψ ( x, y ) where ψ ( x, y ) is warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small. 2. From the displacement field above, it is easy to obtain that

ε xx = ε yy = ε zz = γ xy = 0 From the stress-strain relationship, we have

σ xx = σ yy = σ zz = τ xy = 0 Therefore the equilibrium equations reduce to ∂τ xz ∂τ yz + =0 ∂x ∂y which is identically satisfied if the stresses are derived from a stress function φ ( x, y ) , so that

τ xz = 3.

∂φ , ∂y

τ yz = −

∂φ ∂x

(3.3.2)

From the displacement field and stress-strain relationship, we can obtain 3.3.1

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γ xz =

∂w ∂u ∂w + = − θy ∂x ∂z ∂x

(3.3.3)

γ yz =

∂w ∂v ∂w + = + θx ∂y ∂z ∂y

(3.3.4)

The compatibility equation becomes or in terms of Prandtl stress function 4.

∂γ yz ∂x

−

∂γ xz = 2θ , ∂y

∂ 2φ ∂ 2φ + = −2Gθ ∂x 2 ∂y 2

(3.3.5)

The boundary condition along the bounding surface is dφ = 0 , or φ = const . ds

--x2 y2 + − 1) . In order to show this a2 b2 stress function provides the solution for torsion of the bar, we have to show that this stress function satisfies the equilibrium equations, compatibility equations and traction boundary conditions. (1) Equilibrium equations ∂φ 2y 2x ∂φ τ xz = = C( 2 ) , τ yz = − = C( 2 ) ∂y ∂x a b

(a) Let the stress function be of the form φ = C (

Substituting the above stress expressions into the equilibrium equations, we have ∂τ xz ∂τ yz =0+0= 0 + ∂y ∂x (2) Compatibility equations x2 y2 Substituting φ = C ( 2 + 2 − 1) into (3.3.5) we get a b a 2b 2 C = −Gθ 2 . (3.3.6) a + b2 Therefore we have a stress function satisfying compatibility equation (3) Traction boundary conditions To satisfy the traction boundary condition we must show

dφ = 0 on the ds

traction free lateral surface. Since the boundary of the cross section is given by the equation

3.3.2

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x2 y2 + −1 = 0 , a2 b2 x2 y2 + − 1) = 0 on the free surface and therefore a2 b2 dφ it satisfies the required condition =0 ds it is easy to see that φ = C (

Since equilibrium equations, compatibility equations and traction boundary conditions are all satisfied, the stated stress function provides the solution for torsion of the bar. --- ANS (b) Torsion constant J (1) We have the torque produced by the stresses is ∂φ ∂φ T = ∫∫ (− x − y )dA ∂x ∂y A

(3.3.7)

x2 y2 + − 1) into (3.3.7), then we have, a2 b2 2x 2y x2 y2 T = ∫∫ (− x(C 2 ) − y (C 2 ))dA = − C ∫∫ ( 2 + 2 )dA a b b A A a

Substituting φ = C (

Note that the integral part of the above equation is the area of the elliptical cross-section. It can be easily obtained that

∫∫ ( A

x2 y2 + )dA = πab a2 b2

So we have the torsion T = −Cπab By substituting C and utilizing T = GJθ , we have

J=

− Cπab = Gθ

− (−Gθ

a 2b 2 )πab πa 3b 3 a2 + b2 = 2 Gθ a + b2

(3.3.8) --- ANS

(2) τ xz =

∂φ 2y T 2y 2Ty = C( 2 ) = ( 2)=− , ∂y − πab b b πab 3

(3.3.9)

∂φ 2x 2Tx = −C ( 2 ) = 3 ∂x a πa b

(3.3.10)

and τ yz = −

--- ANS (c) The warping displacement can be derived from (3.3.3), (3.3.4), (3.3.9), (3.3.10) From (3.3.9) and (3.3.10), we have γ xz = −

2Ty 2Tx and γ yz = . 3 Gπab Gπa 3 b

T T (a 2 + b 2 ) Also we need to know θ = = GJ Gπa 3b 3 So from (3.3.3) and (3.3.4), we can rewrite in 3.3.3

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∂w 2Ty T (a 2 + b 2 ) y T (b 2 − a 2 ) y = γ xz + θy = − + = ∂x Gπab 3 Gπa 3b 3 Gπa 3 b 3

(3.3.11)

∂w 2Tx T (a 2 + b 2 ) x T (b 2 − a 2 ) x = γ yz − θx = − = ∂y Gπa 3 b Gπa 3 b 3 Gπa 3b 3

(3.3.12)

From (3.3.11), we can obtain T (b 2 − a 2 ) w( x, y ) = xy + f ( y ) Gπa 3 b 3 Then differentiating (3.3.13) with respect to y, we have

(3.3.13)

∂w( x, y ) T (b 2 − a 2 ) x + f ′( y ) . = ∂y Gπa 3b 3 Comparing this equation with (3.3.12) we have f ′( y ) = 0 , that is f ( y ) = const . For a symmetric cross-section w(0,0) = 0 , that is, f ( y ) = 0. Thus, the warping displacement is w( x, y ) =

T (b 2 − a 2 ) xy. Gπa 3 b 3 --- ANS

And it is easy to also find that the warping function T (b 2 − a 2 ) xy. 3 3 w( x, y ) a2 − b2 = Gπa2 b 2 =− 2 ψ ( x, y ) = xy θ a + b2 T (a + b ) Gπa 3 b 3

3.3.4

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3.4 A thin aluminum sheet is to be used to form a closed thin-walled section. If the total length of the wall contour is 100 cm, what is the shape that would achieve the highest torsional rigidity? Consider elliptical (including circular), rectangular, and equilateral triangular shapes. Solution: (a) We denote GJ as torsional rigidity, for the same material in comparison, only the torsion constant J needs to be taken into consideration. For the closed thin-walled section, the torsion constant J is J=

4A

2

(3.4.1)

∫ ds / t

where A is the area enclosed by the centerline of the wall section. We now have a thin aluminum sheet with its thickness t , all shapes of products made from this aluminum sheet will have the same thickness, t . Also, the total length of the wall contour is 100cm. Then

∫ ds / t

is the same for all shapes of the

cross-section. Consequently, only A needs to be taken into consideration in the evaluation of the torsional rigidity. (b) Comparison of A (1) Elliptical cross-section For the elliptical cross-section, the cross-sectional area is A ellp = πab ,

(3.4.2)

where a and b are the semimajor and semiminor axes, respectively. Unfortunately, the length of the perimeter of an elliptical cross-section is much more complicated to evaluate. The formula for the length of the perimeter can be found from many math handbook It is

L = 4a ∫

π /2

0

1 − k 2 sin 2 φ dφ ,

a2 − b2 = eccentricity a For the purpose to just comparing the area enclosed by the centerline of the wall section, We approximate the perimeter with where k =

a2 + b2 2 By changing the form of (3.4.3) se have L = 2π

(3.4.3)

3.4.1

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L 2 L ) − a 2 = C 2 − a 2 , where C = 2( ) 2 2π 2π Substituting (3.4.4) into (3.4.2) we have, b = 2(

(3.4.4)

A ellp = πa C 2 − a 2 We can find the optimum solution by

∂A = 0 , by some operations leads to ∂a

∂A C 2 − 2a 2 = 0 , therefore we have a = =π ∂a C 2 − a2

C2 for a, b > 0 2

C2 =a (3.4.5) 2 That means the optimum cross-section for elliptical shapes is a circle. Substitute it back to (3.4.4), we have b =

C2 2 L 2 L = ( ) = 2 2 2π 2π Finally, for a circle, the area enclosed by the centerline is Then from (3.4.5) we have a = b =

Acir = πa 2 = π (

L 2 ) = 0.0796 L2 2π

--- ANS (2) Rectangular section For rectangular section, the perimeter is L = 2( p + q ) ,

(3.4.6)

where p and q are length and width, respectively. The cross-sectional area of rectangular sections is simply, A rec = pq ,

(3.4.7)

Substituting (3.4.6) into (3.4.7), we have L A rec = pq = p ( − p ) 2

We use

∂A = 0 to find the optimal solution, ∂p

∂A L L = − 2 p = 0 , we have p = , and from (3.4.6), it is clear that ∂p 2 4 p=q=

L , i.e., the optimal cross-section for rectangular shapes is a square. 4

Finally, for a square thin-walled section, the area enclosed by the centerline is Asqu = pq = (

L 2 ) = 0.0625 L2 4

3.4.2

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--- ANS (3) Equilateral triangular section. For a equilateral triangle, the length of the lateral side is l =

L . 3

The area enclosed by the centerline of this triangular thin-walled section is Atri =

3 2 3 L 2 l = ( ) = 0.048 L2 4 4 3

--- ANS (c) Comparison From the results above we can easily tell A cir > A squ > A tri

Consequently we can conclude that the shape achieving the highest torsional rigidity is a CIRCLE. --- ANS

NOTE: It is interesting to compare in details with variables

q b and from 0~1. a p

(We here assume a>b and p>q)

For ellipse, A ellp = πab = πab

1 1 a2 + b2 ab L )(2( ) 2 ) = ( ) L2 ≈π( 2 2 2 2 2π 2π b a a +b a +b + a b

q ( p + q) pq L 1 p = pq = pq = ( )2 = L2 2 2 q 4 ( p + q) ( p + q) 2 (1 + ) 2 p 2

For rectangle, A rec

For equilateral triangle, A tri = We can illustrate

3 2 L 36

q A b in terms of and , and have the plot of torsional rigidity 2 p a L

of different shapes vs. variable aspect ratios.

3.4.3

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3.4.4

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3.5

The two-cell section in Fig.3.35 is obtained from the single-cell section of Fig.3.36 by adding a vertical web of the same thickness as the skin. Compare the torsional rigidity of the structures of Figs. 3.35 and 3.36 with L1 = L2 = 10cm and L1 = 5cm , L2 = 15cm , respectively.

t = 0.3cm Figure 3.35 Two-cell thin-walled section

Figure 3.36

Single-cell section

Solution: We denote GJ as torsional rigidity. For the same material in comparison, only the torsion constant J needs to be considered. (a) Single-cell thin-walled section The torsion constant J is J=

4A

2

(3.5.1)

∫ ds / t

where A is the area enclosed by the centerline of the wall section. We have A = ( L1 + L2 ) L3 = 20 × 10 = 200cm 2 . The torsion constant J can be simply derived as 3.5.1

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J 1cell =

4A

2

∫ ds / t

=

4[( L1 + L2 ) L3 ] 2 4(200) 2 = = 800cm 4 2( L1 + L2 + L3 ) / t 2(20 + 10) 0.3

(b) Two-cell thin-walled section (1) General Form We denote the shear flow on the left cell by q1 , and the shear flow on the right cell by q 2 . The shear flow in the vertical web is q12 = q1 − q 2 Also, we have the torque for two-cell section T = 2 A1 q1 + 2 A 2 q 2

(3.5.2)

where A1 = L1 L3 , A 2 = L2 L3 The twist angle of the section is obtained from eirher cell. For left cell we have 1 qds 1 (3.5.3) θ1 = = (q1 ( 2 L1 + L3 ) + (q1 − q 2 ) L3 ) ∫ cell 1 t 2GL1 L3 t 2G A1 and for the right cell 1 qds 1 θ2 = = (q 2 ( 2 L2 + L3 ) − ( q1 − q 2 ) L3 ) ∫ 2GL2 L3 t 2G A 2 cell 2 t

(3.5.4)

Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition θ1 = θ 2 = θ (3.5.5) From (3.5.3) to (3.5.5), we derive the relation between q1 and q 2 ,

L3 L3 + ) L1 L2 q2 = q L3 L3 1 (2 + 2 + ) L2 L1 (2 + 2

(3.5.6)

Substituting (3.5.6) into (3.5.2) and using J =

(2 A1 q1 + 2 A 2 q 2 )

J= G⋅

1 (q1 (2 L1 + L3 ) + (q1 − q 2 ) L3 ) 2GL1 L3 t

=

T and (3.5.3), we have Gθ

4 L1 L3 ( L1 L3 q1 + L2 L3 q 2 )t (2q1 L1 + 2q1 L3 − q 2 L3 ) (3.5.7)

(2) Case 1: L1 = L2 = 10cm and L3 = 10cm From (3.5.6), q 2 = q1 , then substituting into (3.5.7) we have 4 L L ( L L q + L2 L3 q 2 )t = 800cm 4 J 2 cell −1 = 1 3 1 3 1 (2q1 L1 + 2q1 L3 − q 2 L3 ) 3.5.2

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--- ANS (3) Case 2: L1 = 5cm , L2 = 15cm and L3 = 10cm

10 10 + ) 5 15 q = 1.25q From (3.5.6), q 2 = 1 10 10 1 (2 + 2 + ) 15 5 (2 + 2

Then substituting into (3.5.7) we have 4 ⋅ 5 ⋅ 10(5 ⋅ 10q1 + 15 ⋅ 10 ⋅ 1.25q1 ) ⋅ 0.3 = 814.2857cm 4 J 2 cell − 2 == (2q1 ⋅ 5 + 2q1 ⋅ 10 − 1.25q1 ⋅ 10) --- ANS (c) Comparison From the results above we have 814.2857 cm 4 = J 2 cell − 2 > J 2 cell −1 = J 1cell = 800cm 4

Adding a vertical web does not significantly improve the torsional rigidity. --- ANS

3.5.3

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3.6

Find the torsional rigidity if the side wall of one of the two cells in Fig. 3.35 (with L1 = L2 = 10cm ) is cut open. What is the reduction of torsional rigidity compared with the original intact structure?

t = 0.3cm Figure 3.35 Two-cell thin-walled section Solution: We denote torsional rigidity by GJ as. (a) Closed sidewall From the solution of Problem 3.5, we have the torsion constant J 2 cell −1 of the case with L1 = L2 = 10cm 4 L L ( L L q + L2 L3 q 2 )t = 800cm 4 J 2 cell −1 = 1 3 1 3 1 (2q1 L1 + 2q1 L3 − q 2 L3 ) So we have the original torsional rigidity GJ 2 cell −1 = 800G

(3.6.1)

(b) With one side wall cut open Assuming that the cell is cut open as shown in the figure, the torsional rigidity can be derived from GJ cut − open = GJ cell − not − cut + GJ cell − cut

(1)

(3.6.2)

+

(2)

Where 3.6.1

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J cell − not −cut =

4( A not −cut ) 2

∫ ds / t

=> J cell − not −cut =

, and A not −cut = L2 L3

(3.6.3)

4(10 × 10) 2 ⋅ 0.3 = 300cm 4 2(10 + 10)

and 1 3 J cell −cut = ∑ bi t i i 3

(3.6.4)

1 => J cell −cut = (10 + 10 + 10) ⋅ 0.33 = 0.27cm 4 3

So, from (3.6.2) we get GJ cut − open = GJ cell − not − cut + GJ cell − cut = 300.27G

---ANS (c) The reduction of torsional rigidity is obtained as GJ 2 cell −1 − GJ cut − open 800 − 300.27 R= = = 0.625 = 62.5% GJ 2 cell −1 800 --- ANS

3.6.2

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3.7

Find the torque capability of the thin-walled bar with the section shown in Fig. 3.36. Assume that the shear modulus G = 27GPa and the allowable shear stress of τ allow = 187 MPa .

t = 0.3cm Figure 3.36 Single thin-walled section Solution: Since the thickness of all walls are equal to t = 0.3cm , we can obtain the allowable shear flow from allowable shear stress, that is

qallow = τ allowt = 187 × 106 × 0.003 = 5.61 × 10 5

N/m

Then we have the torque capability as

Tallow = 2 Aqallow = 2( 0.1 × 0.2 ) × 5.61 × 10 5 = 22440 N − m --- ANS

3.7.1

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3.8

A two-cell thin-walled member with the cross-section shown in Fig. 3.37 is subjected to a torque T. The resulting twist angle θ is 3o / m . Find the shear flows of the applied torque, and the torsion constant. The material is aluminum alloy 2024-T3.

Figure 3.37

Two-cell section

Solution: (a) Assume the material is linearly elastic under the twist angle θ . For aluminum alloy 2024-T3, we have the shear modulus E 72 G= = = 27 GPa 2( 1 + υ ) 2( 1 + 0.33 ) (b) We denote the shear flow on the left cell q1 , and the shear flow on the right cell q 2 . The shear flow in the vertical web is q12 = q1 − q 2 , are the positive directions as shown in the figure above. Also, we have the torque for two-cell sections T = 2 A1 q1 + 2 A 2 q 2

where A1 = A2 =

πd 2

π ( 0.5 )2

= 0.098 m 2 , 8 8 The twist angle of the left cell is s s 1 1 qds θ1 = = ( 1 q1 + 12 (q1 − q 2 )) ∫ t12 2G A1 cell1 t 2G A1 t1 where s1 =

πd 2

=

(3.8.1)

(3.8.2)

= 0.785 m is the length of the left side wall, and s12 = 0.5m is

the length of the vertical web. The twist angle of the right cell is s s 1 1 qds θ2 = = ( 2 q 2 − 12 (q1 − q 2 )) ∫ t12 2G A 2 cell 2 t 2G A 2 t 2

(3.8.3)

3.8.1

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Again, we have s2 =

πd 2

= 0.785 m , the length of the right side wall.

Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition

θ1 = θ 2 = θ = 3o / m = 0.0524 rad / m

(3.8.4)

From (3.8.2) to (3.8.4) and noting that A1 = A 2 , we derive the relation between q1 and q 2 by substituting all the known quantities, 0.785398 0 .5 0.785398 0 .5 q1 + q12 = q2 − q12 0.001 0.002 0.003 0.002 Substituting q12 = q1 − q 2 , in the equation above, we obtain

q2 = 1.687 q1

(3.8.5)

Back substituting into (3.8.2) and (3.8.4), we have ( 785.398 − 250( 1.69732 − 1 )) θ1 = 0.0524 rad / m = q1 2( 27.0677 × 10 9 )( 0.098175 ) From which we obtain q1 = 453,500 N / m Subsequently from (3.8.5) we obtain q2 = 1.687 q1 = 765000 N / m --- ANS (c) The applied torque From (3.8.1), we compute the applied torque

T = 2 A1q1 + 2 A2 q2 = 2( 0.098 )( 453500 + 765000 ) = 239300 N − m = 2.393 × 10 5 N − m --- ANS (d) The torsion constant J From the fundamental relationship of torque and twist angle, we have T = GJθ So the torsion constant can be derived as T 239300 J= = = 1.69 × 10 − 4 m 4 Gθ ( 27 × 10 9 )( 0.0524 ) --- ANS

3.8.2

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3.9

For the bar of Fig. 3.37, find the maximum torque if the allowable shear stress is τ allow = 187 MPa . What is the corresponding maximum twist angle θ ?

Figure 3.37

Two-cell section

Solution: (a) Assume the material is linearly elastic under the twist angle θ . For aluminum alloy 2024-T3, we have the shear modulus E 72 G= = = 27 GPa 2( 1 + υ ) 2( 1 + 0.33 ) (b) We denote the shear flow on the left cell as q1 and that on the right cell as q 2 . The shear flow in the vertical web is q12 = q1 − q 2 . The positive directions for the shear flows are shown in the figure above. The torque for two-cell section is T = 2 A1 q1 + 2 A 2 q 2

where A1 = A2 =

πd 2

=

π ( 0.5 )2

(3.9.1) = 0.098 m 2 ,

8 8 The twist angle of the left cell is s s 1 1 qds θ1 = = ( 1 q1 + 12 (q1 − q 2 )) ∫ cell1 t t12 2G A1 2G A1 t1 where s1 =

πd 2

= 0.785 m is the length of the left side wall, and s12 = 0.5m is

the length of the vertical web. Also we have the twist angle of the right cell as s s 1 1 qds θ2 = = ( 2 q 2 − 12 (q1 − q 2 )) ∫ cell 2 t t 2G A 2 2G A 2 t 2 where s2 =

πd 2

(3.9.2)

(3.9.3)

= 0.785 m is the length of the right side wall.

(c) Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition 3.9.1

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θ1 = θ 2 = θ

(3.9.4)

From (3.9.2) to (3.9.4) and note that A1 = A 2 , we derive the relation between q1 and q 2 by substituting all the known quantities, 0.785398 0 .5 0.785398 0 .5 q1 + q12 = q2 − q12 0.001 0.002 0.003 0.002 In view of the relation q12 = q1 − q 2 we obtain

q2 = 1.687 q1

(3.9.5)

Back substituting (3.9.5) into (3.9.2) and (3.9.4), we have (785.398 − 250(1.69732 − 1)) q1 , and then θ= 2( 27.0677 × 10 9 )(0.098175)

q1 = 8662000 θ Subsequently, q2 = 1.687 q1 = 14 ,600 ,000 θ q12 = q1 − q2 = −5 ,953,000 θ Note the units are θ

(3.9.6) (3.9.7) (3.9.8)

in rad / m , and q1 , q2 , q12 are in N/m.

(d) Stress in the wall From the above quantities of shear flow, we can then compute the shear stress in q . We have t q 8662028θ τ1 = 1 = = 8.66 × 10 9 θ t1 0.001

each wall by τ =

(3.9.9)

τ2 =

q2 14615612θ = = 4.87 × 10 9 θ t2 0.003

(3.9.10)

τ 12 =

q12 − 5953584θ = = −2.98 × 10 9 θ t12 0.002

(3.9.11)

(e) From the above stresses (3.9.9) to (3.9.11), because the negative value just denote the negative direction, the maximum absolute magnitude of shear stress is τ 1 = 8.66 × 10 9 θ ≤ τ allow = 187 × 106 Therefore the maximum twist angle is θ max = 0.0216 rad / m = 1.24o / m --- ANS (f) The maximum torque can be solved by using (3.9.1), (3.9.6), (3.9.7) and the maximum twist angle, that is T = 2 A1q1 + 2 A2 q2 = 2( 0.098 )( 8662000 + 14600000 )( 0.0216 )

= 98700 N − m --- ANS

3.9.2

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3.10

Find the shear flow and twist angle in the two-cell three-stringer thin-walled bar with the cross-section shown in Fig. 3.38. The material is Al2024-T3. The applied torque is 2 × 10 5 N ⋅ m .

Figure 3.38

Two-cell three-stringer thin-walled section

Solution: (a) Assume the material is linearly elastic under the applied torque. For aluminum alloy 2024-T3, we have the shear modulus E 72 G= = = 27GPa 2(1 + υ ) 2(1 + 0.33) (b) Denote the shear flow on the left cell as q1 , and the shear flow on the right cell as q 2 ; both are considered positive if counterclockwise. The shear flow in the vertical web is q12 = q1 − q 2 , which is positive if it is in the same direction as q1 . We have the torque for the two-cell section as T = 2 A1 q1 + 2 A 2 q 2

where A1 = and A 2 =

πd 2 8

=

π (1.2) 2 8

(3.10.1) = 0.565 m 2 ,

bh 2(1.2) = = 1 .2 m 2 2 2

The twist angle of the left. cell is s s qds 1 1 θ1 = = ( 1 q1 + 12 (q1 − q 2 )) ∫ t1 2G A1 cell1 t 2G A1 t1 where s1 =

πd 2

(3.10.2)

= 1.88 m is the length of the left half circular wall, and

s12 = 1.2m is the length of the vertical web. The twist angle of the right. cell is

3.10.1

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s s s 1 1 qds = ( 2 q 2 + 3 q 2 − 12 (q1 − q 2 )) ∫ cell 2 t t1 t3 2G A 2 2G A 2 t 2 Again, we have s2 = 2 m , the length of the lower wall,

θ2 =

(3.10.3)

and s3 = 2 2 + 1.2 2 = 2.33 m , the length of the inclined wall of thickness t3 in the right cell. Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition θ1 = θ 2 = θ (3.10.4) Solving the two equations, (3.10.2) and (3.10.4), we obtain 1 1.885 1 .2 1 2 2.332 1 .2 ( q1 + q12 ) = ( q2 + q2 − q12 ) 0.565 0.005 0.005 1.2 0.007 0.007 0.005

Eliminating q12 from the equation above using q12 = q1 − q 2 we obtain q2 = 1.132q1 (3.10.5) (c) To find the shear flow , we back substitute (3.10.5) into (3.10.1) and have T = 2 A1q1 + 2 A2 q2 = ( 2 A1 + 2.265 A2 ) q1

=> q1 =

T 2 × 105 = = 51966 N / m 2 A1 + 2.265 A2 (2)(0.565) + (2.265)(1.2) --- ANS

From (3.10.5), q2 = 1.132q1 = 58844 N / m --- ANS (d) For the twist angle, we can utilize the shear flows and equations (3.10.2) and (3.10.4) to get,

1.885 1.2 ( + (1 − 1.132)) × 51966 1 s1 s12 0 . 005 0 . 005 θ = θ1 = ( q1 + (q1 − q2 )) = --- ANS t1 2(27 × 109 )(0.565) 2G A1 t1 = 5.86 × 10− 4 rad / m = 0.0336o / m

3.10.2

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3.11

What is the maximum torque for the structure of Fig. 3.38 if the allowable twist angle θ is 2 o / m ?

Figure 3.38

Two-cell three-stringer thin-walled section

Solution: (a) Assume the material used is still Aluminum alloy 2024-T3. For aluminum alloy 2024-T3, we have the shear modulus E 72 G= = = 27 GPa 2( 1 + υ ) 2( 1 + 0.33 ) (b) Denote the shear flow on the left cell as q1 , and the shear flow on the right cell as q 2 . Both are assumed positive in the counterclockwise direction. The shear flow in the vertical web is q12 = q1 − q 2 , from bottom to top. The torque for two-cell section is T = 2 A1 q1 + 2 A 2 q 2

where A1 = and A 2 =

πd 2 8

=

π ( 1.2 )2 8

(3.11.1) = 0.56 m 2 ,

bh 2(1.2) = = 1 .2 m 2 2 2

The twist angle of the left cell is s s qds 1 1 θ1 = = ( 1 q1 + 12 (q1 − q 2 )) ∫ cell1 t t1 2G A1 2G A1 t1 where s1 =

πd 2

(3.11.2)

= 1.88 m is the length of the left half circular wall, and

s12 = 1.2m is the length of the vertical web. The twist angle of the right cell is s s s 1 1 qds θ2 = = ( 2 q 2 + 3 q 2 − 12 (q1 − q 2 )) ∫ cell 2 t t1 t3 2G A 2 2G A 2 t 2

(3.11.3)

3.11.1

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where s2 = 2 m is the length of the lower straight wall of thickness t2, and s3 = 2 2 + 1.2 2 = 2.33 m is the length of the inclined wall of thickness t3.

Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition θ1 = θ 2 = θ (3.11.4) From (3.10.2) to (3.10.4), we can derive the relation between q1 and q 2 by substituting all the known quantities, 1 1.88 1 .2 1 2 2.33 1 .2 ( q1 + q12 ) = ( q2 + q2 − q12 ) 0.566 0.005 0.005 1.2 0.007 0.007 0.005

Substituting q12 = q1 − q 2 into the above equation, we obtain q2 = 1.13q1

(3.11.5)

(c) Since we have the condition θ allowable = 2o / m = 0.035rad / m , and, thus,

θ1 = θ 2 = θ < θ allowable = 0.035rad / m . By equations (3.11.2), 1.88 1.2 + ( 1 − 1.13 )) × q1 1 s1 s12 0 . 005 0 . 005 θ = θ1 = ( q1 + ( q1 − q2 )) = t1 2G A1 t1 2( 27 × 10 9 )( 0.565 ) (

= 1.13 × 10 − 8 q1 ( rad / m ) < θ allowable = 0.035 ( rad / m ) We then have q1 ≤ 3,095,000 N / m (d) To find the maximum torque, we can use equation (3.11.1) T = 2 A1q1 + 2 A2 q2 = ( 2 A1 + 2.264708 A2 )q1 < [( 2 )( 0.56 ) + ( 2.26 )( 1.2 )] × 3 ,095 ,000 = 1.19 × 107 N − m

Therefore the maximum torque is

Tmax = 1.19 × 107 N − m --- ANS It should be noted that under this torque the shear stress has already exceeded the yield condition of Al 2024-T3. Consequently, this solution may not be of practical significance if allowable stress condition is to be satisfied too.

3.11.2

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3.12

The two shafts of thin-walled cross-sections shown in Fig. 3.39a and b, respectively. Contain the same amount of aluminum alloy. Compare the torsional rigidities of the two shafts without end constraints.

(a) Figure 3.39

(b) Cross-sections of two shafts

Solution: (a) Fig. 3.39a is a cross-section of an open thin-wall, its torsional rigidity is GJ a 1 1 GJ a = G ∑ biti 3 = 3( )( 200 )( 3 )3 G = 5400G mm4 3 i 3 --- ANS (b) Fig. 3.39b is a cross-section of a closed thin-wall, its torsional rigidity is GJ b 2

4A , GJ b = G ds ∫t

where A =

3b 2 , 4

2

4A 3b 4 t b 3 t GJ b = G =G = G = 6 × 10 6 Gmm 4 ds 4(3b) 4 ∫t

--- ANS (c) The ratio of the torsional rigidities is GJ b 6 × 106 G = = 1111 GJ a 5400G --- ANS

3.12.1

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3.13

Find the distributions of the primary warping displacement on the cross-sections shown in Fig. 3.39b. Due to symmetry, the center of twist coincides with the centroid of the section, and warp at the midpoint of each flat sheet section is zero. Sketch the warping displacement along the wall.

(b) Figure 3.39

Cross-sections of two shafts

Solution: (a) Observation. Because of the symmetry, the center of twist coincides with the centroid of the section, and warp at the midpoint of each flat sheet section is zero.

So from the figure above we set w = 0 at the midpoint of each flat sheet. First we assume the warp at point A is positive of z-direction. While going from A to B, we pass the midpoint and then the warp goes from positive into negative part, then end at point B with the maximum negative warping. Using the same concept on sheet BC will result in a maximum positive warping at point C. Now we consider the sheet CA by using the same conclusion, we will surprisingly find the warping at A is negative of z-direction. Hence it contradicts our assumption of A being 3.13.1

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positive. By applying the assumption of A is negative direction we will conclude in another contradiction. Therefore, we can confidently assure that there is no warping in this equilateral triangular thin-walled cross-section. In the following we will approve it by further derivatives. (b) For the closed thin-walled section, we have

∂w ∂u s ∂w + = + ρθ , (3.13.1) ∂s ∂z ∂s where ρ is the distance from the center of twist to the tangent line of point P of

γ sz =

interest, w is the warping that we are seeking. Also, we have

γ sz = where q s

τ sz

qs (3.13.2) G Gt is the shear flow along s-direction, t is the thickness of the wall and G =

is the shear modulus. Again, recall from the relationship between applied torque and shear flow, we

T 2A Combining (3.13.1) to (3.13.3) results in ∂w T T ∂w = − ρθ + ρθ = , or ∂s 2 AGt ∂s 2 AGt have q s =

s s T T ds − ∫ ρθ ds = ∫ ds − 2 Asθ 0 0 0 2 AGt 2 AGt Also the twist angle can be derived from 1 ds θ= ∫ 2G A t

=> w( s) − w(0) = ∫

s

(3.13.3)

(3.13.4)

(3.13.5)

(c) Assume the applied torque is uniformly applied to the cross-section. Also, the material is isotropic so that the shear modulus is constant. For the equilateral triangular section, we have 3b 2 (3.13.6) 4 And since the section is symmetric, we can just take the sheet CA into A=

consideration and applied to all other sheets. Assume the origin of s is on the midpoint of sheet CA, so w(0) = 0 , then we have 3bs 12 From (3.13.4) to (3.13.7), we obtain As =

w( s ) = ∫

s

0

T Ts ds − 2 Asθ = − 2 AGt 2G At

(3.13.7)

2(

3bs )(3b)T Ts 2 ( 3 / 8)b 2 12 = ( − ) Gt 3b 2 (3 / 16)b 4 4G ( A) 2 t

3.13.2

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=> w( s ) = 0 This approves our observation in part (a). --- ANS

3.13.3

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3.14

A shaft with a channel section shown in Fig. 3.40 is subjected to a torque T. Assume that neither end is constrained. Find the warping distribution on the cross-section, the maximum warp, and the location of the maximum warp.

Figure 3.40

Dimensions of a channel section

Solution: (a) For the open thin-walled section, we have

∂w ∂u s ∂w + = + ρθ = 0 , (3.14.1) ∂s ∂z ∂s where ρ is the distance from the center of twist to the tangent line of point P of

γ sz =

interest, w is the warping that we are seeking. So, we have ∂w = − ρθ ∂s s

=> w( s ) − w(0) = − ∫ ρθ ds = −2 Asθ 0

(3.14.2)

Also the twist angle can be derived from

θ=

T GJ

(3.14.3)

where the torsional constant is 1 3 J = ∑ bi t i i 3

(3.14.4)

for thin rectangular sections of thickness t and length b (see equation (3.38) in the textbook). (b) For the channel section in Fig. 3.40, assume the shear modulus is G. We can first derive some needed properties. The moment of inertia with respect to x axis is

3.14.1

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Ix =

1 t 1 t (b + )(2h + t ) 3 − (b − )(2h − t ) 3 = 1729080 mm 4 12 2 12 2

( Since the thickness of walls is relatively small, there are some approximated solutions such as I x = Ix =

t b (2h) 3 + 2 × [ t 3 + bth 2 ] = 1728270 mm 4 , or 12 12

t (2h) 3 + 2(bth 2 ) = 1728000 mm 4 are all the acceptable approximations) 12

The eccentric distance e =

tb 2 h 2 3( 60 )4 = = 22.49 mm Ix 1729080

(3.14.5)

The torsional constant is

1 3 60 × 33 120 × 33 J = ∑ bi t i = 2( )+ = 2160mm 4 3 3 3 i and the twist angle per unit length can be obtained from

θ=

T T = GJ 2160G

(3.14.6)

(c) Break up the contour s into two straight parts s1 and s2 , as shown below

For contour s1 , we have 1 (3.14.7) es1 2 and for contour s2 (the point s2 = 0 is at upper left corner of the section) we have A s1 =

1 (3.14.8) A s 2 = − hs 2 2 (d) On the contour s1 , the warping displacement w is calculated from equation

(3.14.2): T T = −0.0104 s1 2160G G In which the condition w( 0 ) = 0 has been used. This is obvious since the warping s

w ( s1 ) = − ∫0 ρθ ds = −2 As 1θ = −es1

at the middle point of the vertical web is zero because of anti-symmetry. Also note that 3.14.2

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w ( s1 = h ) = w ( h = 60 mm ) = −0.62

T G

--- ANS (e) On the contour s2 , the warping can be obtained from equation (3.14.2) by integrating from s1=0 to any point s2. Thus,

w( s2 ) = w( s1 = h ) − 2 As 2θ => w( s2 ) = −0.62

T T + 0.028s2 G G

---

ANS

So the warping displacement at the left upper corner is w( s1 = h) = −0.62

T G

and at the right upper edge is w( s2 = b) = 1.04

T G

--- ANS (f) Similar calculations show that the warping displacement is anti-symmetric with respect to the x-axis. From the above calculations, the maximum warp (absolute value) is wmax = w2 ( s2 = b) = 1.04

T G

and are located at both free edges. --- ANS

‧

w1 ( s1 ) = −0.0104 s1

T G

‧

w2 ( s 2 ) = 0.0278 s 2

T T − 0.62 G G

3.14.3

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3.15

Consider the shaft of the channel section shown in Fig. 3.40. If one end of the shaft is built in and the other end is free, find the effective torsional rigidity as a function of the distance from the built-in end. Assume that the length L of the shaft is sufficiently large so that near the free end the Saint-Venant torsion assumptions are valid. Compare the total twist angle with that for a free-free shaft for L = 2m .

Figure 3.40

Dimensions of a channel section

Solution: (a) The government equation for the twist angle under an applied torque T end constraints is d 2θ T = , 2 GJ dz EΓ where k 2 = , GJ

θ −k2

(3.15.1) s1 1 3 2 Γ = ∫ 4 A s tds and J = ∑ bi t i s0 i 3

(3.15.2)

The general solution is

θ = θh + θ p =

T (1 + C1e z / k + C 2 e − z / k ) GJ

(3.15.3)

(b) Applying boundary conditions We assume the shaft is built in at z = 0 and free at z = L where the torque T is applied. (1) First, assume the length L of the shaft is sufficiently large so that near the free end the Saint-Venant torsion assumptions are valid, so that T = GJθ (Saint-Venant torsion) when z → L . To satisfy this condition, we require that C1 = 0 , then (3.15.3) will converge to T = GJθ . (2) Second, at the built-in end ( z = 0 ), warping suppressed and w = 0 . From the equation w( s, z ) = − ws ( s )θ ( z ) , we conclude that θ = 0 . Thus, we have 3.15.1

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C 2 = −1 . From (1) and (2), we have the solution becoming T T θ= (1 − e − z / k ) = GJ GJ eff Then the effective torsion constant is J eff =

J 1 − e−z / k

(3.15.4)

(c) In Fig. 3.40, the channel cross-section has the properties from (3.15.2).

1 3 60 × 33 120 × 33 J = ∑ bi t i = 2( )+ = 2160mm 4 = 2.16 × 10 −9 m 4 3 3 3 i s1

Γ = ∫ 4 A s tds 2

s0

Because of the symmetric w.r.t x-axis, it is more convenient to measure distance from the middle point of the vertical web.

From the solution of Problem 3.14, we have ecot =

tb 2 h 2 3(60) 4 = = 22.49mm = 0.02249m Ix 1729080

Therefore, h h +b 1 1 1 2 4 A s tds = 2 × {∫ 4( ecot s1 ) 2 tds1 + ∫ 4[ ecot h − h( s1 − h)] 2 tds1 } 0 s0 h 2 2 2

∫

s1

~~~~~~A~~~~~~

~~~~~~~~~B~~~~~~~~~

Part A:

∫

h

0

1 1 2 2 1 3 4( ecot s1 ) 2 tds1 = ecot t ( ) s1 |0h = ecot th 3 = 1.09 × 10 −10 m 6 2 3 3

Part B:

3.15.2

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∫

h+b

h

=∫

1 1 4[ ecot h − h( s1 − h)]2 tds1 2 2

bb 2

0

1 2 (ecot h − hs2 ) 2 tds2 = t (ecot h 2b − ecot h 2b 2 + h 2b3 ) = 2.31 × 10−10 m6 3 s1

=> Γ = ∫ 4 As tds = 2( PartA + PartB) = 6.8 × 10−10 m6 2

s0

Taking E = 70GPa and G = 27GPa for aluminum, then

EΓ = 0.904 GJ Thus, the effective torsional rigidity is obtained as k=

GJ eff =

GJ 27 × 109 × 2.16 × 10−9 58.32 = = −z / k − z / 0.904 1− e 1− e 1 − e − z / 0.904 --- ANS

(d) The total twist angle. For the case with end constraints, L

α fixed = ∫ θdz = 0

=

2 T T (1 − e − z / 0.904 )dz = ( z + 0.904e − z / 0.904 ) |02 ∫ 58.32 0 58.32

T (2 − 0.805) = 0.0205T 58.32

For the case with free-free end,

α free = θL =

TL 2T = = 0.0343T GJ 58.32

The ratio of the two twist angles is

α fixed = 0.6 α free --- ANS It is clear that the end constraints reduce the twist angle. In other words, end constraints increase the torsional stiffness.

3.15.3

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3.16

Calculate the distributions of normal stress σ zz and shear flow distributions at the built-in end for Problem 3.15.

Figure 3.40

Dimensions of a channel section

Solution: (a) The solution for problem 3.15 is

θ=

T T (1 − e − z / k ) = (1 − e − z / 0.904 ) GJ 58.32

(3.16.1)

(b) The normal stress is

σ zz ( z , s ) = Eε zz ( z , s ) = E

dθ ∂w( z , s ) = − Ews ( s ) ∂z dz

(3.16.2)

where ws ( s ) = 2 A s Substituting (3.16.1) into (3.16.2) we have T e − z / 0.904 σ zz ( z, s ) = −(70)(2 As )( )( ) = −2.656T As e − z / 0.904 58.32 0.904 We can find the value from the solution of problem 3.15, On the vertical web s : 0 ~ s1 => A s =

1 ecot s1 2

On the horizontal sheet s : 0 ~ s 2 => A s = where ecot =

1 1 ecot h − hs 2 2 2

(3.16.3)

(3.16.4) (3.16.5)

tb 2 h 2 = 0.0225m Ix

(c) The distribution of normal stress σ zz at the built-in end ( z = 0 ) ‧

On the vertical web, we can calculate the normal stress with (3.16.3) and (3.16.4)

3.16.1

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1 2

σ zz ( z , s ) |z = 0 = −2.656T ( ecot s1 )e − z / 0.904 |z = 0 = −0.0299 s1T --- ANS At s1 = 0 , σ zz (0,0) = 0 At s1 = h = 0.06m , σ zz = −0.0299 × 0.06T = −1.79 × 10−3 T ( N / m 2 ) ‧

On the horizontal sheet, we can derive the normal stress with (3.16.3) and (3.16.5)

1 2 −3 = (−1.79 × 10 + 0.0797 s2 )T

1 2

σ zz ( z, s ) |z = 0 = −2.656T ( ecot h − hs2 )e − z / 0.904 |z = 0

--- ANS At s 2 = 0 , σ zz = −1.79 × 10 T At s 2 = b = 0.06m , −3

σ zz = (−1.79 × 10−3 + 0.0797 × 0.06)T = 2.99 × 10−3 T ( N / m 2 ) The distribution of normal stress on the cross-section is anti-symmetric. The distribution can be illustrated as the figure below.

--- ANS (d) The distribution of shear flow at the built-in end ( z = 0 ). From the equation (3.85) in the textbook, the shear flow at any location s at the built-in end ( z = 0 ) is q ( s) | z =0 = E

d 2θ | z =0 dz 2

∫

s

s0

ws tds

(3.16.6)

where ws ( s ) = 2 A s

3.16.2

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Substituting (3.16.1) in (3.16.6), we have q ( s ) | z = 0 = (70)(

s T − 1 2 − z / 0.904 )(−( ) e ) |z = 0 (2 × 0.003) ∫ As ds s0 58.32 0.904

(3.16.7)

s

= −8.82 × 10 T ∫ As ds −3

s0

Here it is important to emphasize that the s-direction is measured from the point where shear flow vanishes. Hence s begins from the free end of the horizontal sheet as shown in the figure below. Also, due to the symmetry w.r.t. x axis, we only need to consider the part above x-axis. This allows us to modify equation of

A s from (3.16.4) and (3.16.5).

For the horizontal sheet, we have As =

1 hs1 2

,

s1 : 0 → b

(3.16.8)

Moving along the vertical web, we have 1 A s = − ecot s 2 2

‧

s2 : 0 → h

(3.16.9)

On the horizontal sheet, we can derive the shear flow from (3.16.8) and (3.16.7) q1 ( s ) |z = 0 = −8.82 × 10 − 3 T ∫

s1

0

= −2.646 × 10− 4 T (

1 hxdx 2

x 2 s1 2 ) |0 = −1.323 × 10 − 4 s1 T 2

--- ANS At s1 = 0 , q (0) = 0 At s1 = b = 0.06m , q1 (0.06) = −1.323 × 10−4 (0.06) 2 T = −4.762 × 10−7 T ‧

N /m

On the vertical web, we can derive the shear flow with (3.16.9) and (3.16.7). Since the shear flow is continuous, we have s2 1 q2 ( s2 ) | z = 0 − q1 ( s1 = b) |z = 0 = −8.818341 × 10 − 3 T ∫ (− ecot x)dx 0 2 2 x s2 2 = 9.914 × 10 − 5 T |0 = 4.957 × 10− 5 s2 T 2

3.16.3

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=> q2 ( s2 ) |z = 0 = (4.957 × 10−5 s2 − 4.762 × 10−7 )T 2

--- ANS −7

At s 2 = 0 , q2 (0) = −4.762 × 10 T At s 2 = h = 0.06m , q2 ( s2 = 0.06) = (4.957 × 10 −5 (0.06) 2 − 4.762 × 10−7 )T = −2.977 × 10− 7 T The distribution of the shear flow at the fixed end is sketched in the figure below.

--- ANS

3.16.4

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3.17

Assume that the shaft of the channel section of Fig. 3.40 is built in at both ends. Find the torque that is necessary to produce a relative twist angle α = 5 o between two ends. Assume that L = 1m , Young’s modulus E = 70GPa , and shear modulus G = 27GPa . Compare this with the free-free case.

Figure 3.40

Dimensions of a channel section

Solution:

(a) Since both end of the channel are built-in, it allows us to set z = 0 at the middle of the channel as shown above. (b) The governing equation for the twist angle under an applied torque T end constraints is d 2θ T = , 2 GJ dz EΓ where k 2 = , GJ

θ −k2

(3.17.1) (3.17.2)

s1 1 3 2 Γ = ∫ 4 A s tds and J = ∑ bi t i for open thin-walled section. s0 i 3

(3.17.3)

The general solution of differential equation (3.17.1) is

3.17.1

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θ = θh + θ p =

T z z (1 + C1 cosh + C 2 sinh ) k k GJ

(3.17.4)

s1 1 3 2 (c) Compute J = ∑ bi ti and Γ = ∫ 4 A s tds . s 0 i 3

1 3 60 × 33 120 × 33 J = ∑ biti = 2( )+ = 2160mm4 = 2.16 ×10−9 m 4 3 3 3 i s1

Γ = ∫ 4 A s tds 2

s0

Because of symmetry w.r.t x axis, it is more convenient to set up the s contour with the origin at the middle point of the vertical web as shown in the figure below.

From the solution of Problem 3.14, we have tb 2 h 2 3(60) 4 = = 22.485946mm = 0.0225m Ix 1729000

ecot = Therefore,

∫

s1

s0

h h +b 1 1 1 2 4 A s tds = 2 × {∫ 4( ecot s1 ) 2 tds1 + ∫ 4[ ecot h − h( s1 − h)] 2 tds1 } 0 h 2 2 2

~~~~~~A~~~~~~

~~~~~~~~~B~~~~~~~~~

Part A:

∫

h

0

1 1 2 2 1 3 4( ecot s1 ) 2 tds1 = ecot t ( ) s1 |0h = ecot th 3 = 1.092 × 10 −10 m 6 2 3 3

Part B:

∫

h+b

h

=∫

bb 2

0

s1

1 1 4[ ecot h − h( s1 − h)]2 tds1 2 2 1 2 (ecot h − hs2 ) 2 tds2 = t (ecot h 2b − ecot h 2b 2 + h 2b3 ) = 2.31 × 10−10 m6 3

=> Γ = ∫ 4 A s tds = 2( PartA + PartB) = 6.804 × 10 −10 m 6 2

s0

3.17.2

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Taking E = 70GPa and G = 27GPa for aluminum, then

EΓ = 0.904 GJ (d) Applying boundary conditions (1) First, because of symmetry of θ with respect to z , the odd function sinh( z / k ) should be dropped. This is accomplished by setting C 2 = 0 . k=

(2) Second, at the built-in end ( z = L / 2 ), warping is suppressed and w = 0 . Since w( s, z ) = − ws ( s)θ ( z ) , we conclude that θ = 0 . Thus, we have T L (1 + C1 cosh ) = 0 2k GJ

1

Then C1 = −

cosh

L 2k

Since L = 1 m and k = 0.904 m , we have C1 = −

1 1 cosh( ) 2 × (0.904)

= −0.864

Thus the solution for the rate of twist angle is

θ=

T z T z (1 − 0.864 cosh ) = (1 − 0.864 cosh ) GJ k 58.32 0.904

The twist angle α related to distance z from the middle of the channel then is T u (1 − 0.864 cosh )du 0 0 58.32 0.904 T z = ( z − 0.78 sinh ) 58.32 0.904 z

α = ∫ θdz = ∫

z

(3.17.5)

This is the twist angle measured from the middle of the channel bar to the built-in end. (e) If we produce a relative twist angle α = 5 o , then the twist angle from the middle of the channel to the built-in end ( z = L / 2 = 0.5m ) is 5o α = = 2.5o = 0.0436 rad . 2 From equation (3.17.5) we can determine the required torque to produce such an angle. 0.0436 =

T 0 .5 (0.5 − 0.78 sinh ) = 7.795 × 10 − 4 T 58.32 0.904

Then T = 55.97 N − m --- ANS (f) For the free-free end case 3.17.3

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T free − end = GJθ = GJ

α z

= 58.32

0.0436 = 5.089 N − m . 0 .5

The ratio of built-in ends case and free-ends case is Tbuilt − in 55.974 = = 11 T free − end 5.0894 It is likely that the rigidity of the built-in ends case is enhanced eleven times more than the free-ends case. --- ANS

3.17.4

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4.1

A uniform beam of a thin-walled angle section as shown in Fig. 4.19 is subjected to the bending M y ( M z = 0 ). Find the neutral axis and bending stress distribution over the cross-section.

Figure 4.19

Thin-walled angle section

Solution: (a) For finding the location of the centroid, we select the corner of the thin-walled section as the origin of a Cartesian coordinate system with the horizontal and vertical distances between the centroid and the origin denoted by y c and z c , respectively. yc =

h ⋅ t ⋅ ( h / 2) h = 2ht 4

zc =

h ⋅ t ⋅ ( h / 2) h = 2ht 4

--- ANS (b) Set up a Cartesian coordinate system (y, z) in the pane of the section with the origin at the centroid. The moments of inertia with respect to this coordinate system are (assume t 15 z − 9 y = 0 2th 3 So the neutral plane is located at 15 z − 9 y = 0 in the y-z coordinate system (the centroid is the origin of this coordinate system). --- ANS

4.1.2

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4.2 Rotate the angle section of Fig. 4.19 counterclockwise for 45 o . Find the neutral axis and the maximum bending stress. Compare the load capacity with that of the original section given by Fig. 4.19.

Figure 4.19

Thin-walled angle section

Solution: Remove the primes in the coordinates

Set up a temporary Cartesian coordinate system with the origin at the corner of the thin-walled section to find the centroid. The horizontal and vertical distances from the centroid to the origin are denoted by yc and zc , respectively. Because of the symmetry, yc = 0 . Assuming t σ xx =

3 2M y th 2

Maximum negative stress is at

z=−

h 2 2

, => σ xx = −

3 2M y th 2

The absolute maximum stress is σ xx =

3 2M y

th 2 (c) The neutral axis (plane) is located along σ xx = 0 ,

σ xx = 12

My

z = 0 => z = 0 th 3 So the neutral axis coincides with the centroidal axis. Note that this section in this particular position is symmetric with respect to the y-z coordinate system. For symmetric sections the neutral axis always coincides with the location of the centroid. --- ANS (d) The load capacity with the original section For the same maximum bending stress in both beams,

σ xx = =>

3 2 M y ,rotate th 2

M y ,rotate M y ,origin

=

=

27 M y ,origin 4th 2

27 = 1.59 12 2

The load capacity of the rotated section is 1.59 times that of the original section. --- ANS

4.2.2

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4.3

The stringer-web sections shown in Figs. 4.20, 4.21, and 4.22 are subjected to the shear force Vz ≠ 0 , while V y = 0 . Find the bending stresses in the stringers for the same bending moment M y . Which section is most effective in bending?

Figure 4.20

Stringer-web section

Figure 4.21

Stringer-web section

Figure 4.22

Stringer-web section

Solution: The contribution of the thin sheets to bending is assumed to be negligible. Thus the neutral axis is only depends on the cross-sectional area of the stringers. Also, assume 4.3.1

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y and z are the horizontal axis and vertical axis, respectively. The origin of the system is located at the centroid. (a) Figure 4.20. (1) Because of symmetry, the centroid is located at the middle of the vertical web. (2) Moment of inertia I y = ∑ Ai z i = 2( 2 A × h 2 ) = 4 Ah 2 2

i

I z = ∑ Ai y i = 2( 2 A × 0 2 ) = 0 2

i

I yz = ∑ Ai y i z i = 0 i

(3) Bending stress. Considering M y ≠ 0 and M z = 0 My

σ xx =

Iy

z=

My 4 Ah 2

z

The stresses at the stringer are ①.

At z = h , σ xx =

②.

At z = −h , σ xx =

My 4 Ah

2

My 4 Ah

2

My

z=

4 Ah

z=−

My 4 Ah --- ANS

(b) Figure 4.21. (1) Because of symmetry (when neglecting the effects of webs), the centroid is located at the center of the section as shown in the figure. (2) Moment of inertia I y = ∑ Ai z i = 4( A × h 2 ) = 4 Ah 2 2

i

h 2 I z = ∑ Ai y i = 4( A × ( ) 2 ) = Ah 2 2 i I yz = ∑ Ai y i z i = 0 i

(3) Bending stress. Considering M y ≠ 0 and M z = 0

σ xx =

− I yz M y I y I z − I yz

2

y+

IzM y I y I z − I yz

2

z=

My Iy

z

The stresses at the stringers are (y position is not involved) 4.3.2

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My

①.

At z = h , σ xx =

②.

At z = −h , σ xx =

4 Ah 2

My

z=

My 4 Ah 2

4 Ah

z=−

My 4 Ah --- ANS

(c) Figure 4.22. (1) Again, when neglecting the effects of webs, the centroid is located at the middle of the vertical web. (2) Moment of inertia I y = ∑ Ai z i = 4( A × h 2 ) = 4 Ah 2 2

i

I z = ∑ Ai y i = 2( A × h 2 ) = 2 Ah 2 2

i

I yz = ∑ Ai y i z i = 2( A × h × (− h)) = −2 Ah 2 i

(3) Bending stress. Considering M y ≠ 0 and M z = 0

σ xx = =

− I yz M y I y I z − I yz

My 2 Ah 2

y+

2

My 2 Ah 2

y+

IzM y I y I z − I yz

2

z=

2M y [4 × 2 − (−2) 2 ] Ah 2

y+

2M y [4 × 2 − (−2) 2 ] Ah 2

z

z

The stresses at the stringer are At z = h , y = − h ,

σ xx = ①.

y+

My 2 Ah

2

z=

My 2 Ah 2

( − h + h) = 0

My 2 Ah 2

y+

My 2 Ah 2

z=

My 2 Ah 2

(0 + h ) =

My 2 Ah

At z = −h , y = 0 ,

σ xx = ③.

2 Ah

2

At z = h , y = 0 ,

σ xx = ②.

My

My 2 Ah 2

y+

My 2 Ah 2

z=

My 2 Ah 2

(0 − h) = −

My 2 Ah

At z = −h , y = h ,

σ xx =

My 2 Ah

2

y+

My 2 Ah

2

z=

My 2 Ah 2

( h − h) = 0 --- ANS

4.3.3

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(d) Comparing the above results, sections in Figure 4.20 and Figure 4.21 are both more effective than the section in Figure 4.22 for this particular loading. --- ANS

4.3.4

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4.4

Compare the bending capabilities of the two sections of Figs. 4.21 and 4.22 if M y = 0, Mz ≠ 0.

Figure 4.21

Stringer-web section

Figure 4.22

Stringer-web section

Solution: The thin sheets are assumed to be negligible in bending. Thus, the location of the centroid of the cross-section only depends on stringers. The coordinates (y, z) are set up with the origin at the centroid with y and z designating the horizontal axis and vertical axis, respectively. (a) Figure 4.21. (1) The centroid is located at the center of of the space defined by the four stringers. (2) Moment of inertia I y = ∑ Ai z i = 4( A × h 2 ) = 4 Ah 2 2

i

4.4.1

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h 2 I z = ∑ Ai y i = 4( A × ( ) 2 ) = Ah 2 2 i I yz = ∑ Ai y i z i = 0 i

(3) Bending stress. Considering M y = 0 and M z ≠ 0 we have

IyM z

σ xx =

I y I z − I yz

2

y+

− I yz M z I y I z − I yz

2

z=

Mz y Iz

The stresses in stringers 1 and 4 are: At y =

h M , σ xx = z2 y 2 Ah

The stresses in stringers 2 and 3 are

M h At y = − , σ xx = z2 y 2 Ah

--- ANS (b) Figure 4.22. (1) The centroid is located at the middle of the vertical web. (2) Moment of inertia I y = ∑ Ai z i = 4( A × h 2 ) = 4 Ah 2 2

i

I z = ∑ Ai y i = 2( A × h 2 ) = 2 Ah 2 2

i

I yz = ∑ Ai y i z i = 2( A × h × (− h)) = −2 Ah 2 i

(3) Bending stress. For M y = 0 and M z ≠ 0

σ xx = =

I yM z I y I z − I yz

2

y+

− I yz M z I y I z − I yz

2

z=

4M z 2M z y+ z 2 2 [4 × 2 − (−2) ] Ah [4 × 2 − (−2) 2 ] Ah 2

Mz Mz y+ z 2 Ah 2 Ah 2

The stress in stringer 1 is At z = h , y = − h

4.4.2

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σ xx =

Mz Mz Mz M y+ z= (−2h + h) = − z 2 2 2 2 Ah 2 Ah 2 Ah Ah

Stringer 2: At z = h , y = 0 , σ xx =

Mz Mz Mz Mz y z + = (0 + h) = 2 2 2 2 Ah Ah 2 Ah 2 Ah

Stringer 3: At z = −h , y = 0 , σ xx =

Mz Mz Mz Mz y z h + = − = − ( 0 ) 2 Ah Ah 2 2 Ah 2 2 Ah 2

Stringer 4: At z = −h , y = h , σ xx =

Mz Mz Mz Mz y+ z= ( 2 h − h) = 2 2 2 2 Ah Ah 2 Ah 2 Ah --- ANS

(c) Comparing the above results, we see that sections in Figure 4.21 and Figure 4.22 have the same bending efficiency; they both reach the same maximum bending stress under the same moment. --- ANS

4.4.3

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4.5

Figure 4.23 shows the cross-section of a four-stringer box beam. Assume that the thin walls are ineffective in bending and the applied bending moments are M y = −500,000 N ⋅ cm

M z = 200,000N ⋅ cm . Find the bending stresses in all stringers.

Figure 4.23

Thin-walled section

Solution:

(a) Set up a temporary coordinate system with stringer 1 as the origin. The location of the centroid is yc =

∑ Ai yi i

∑ Ai

=

( 2 × 200 + 1 × 200 ) = 54.5cm (4 + 2 +1+ 4 )

=

( 1 × 50 + 4 × 100 ) = 40.9 cm (4 + 2 +1+ 4 )

i

zc =

∑ Ai zi i

∑ zi i

(b) The moment of inertia I y = ∑ Ai zi = ( 4 + 2 )( 40.909091 )2 + 1 × ( 50 − 40.909091 )2 + 4 × ( 100 − 40.909091 )2 2

i

= 240901cm 4

Similarly,

I z = ∑ Ai yi 2 = 87273cm4 i

4.5.1

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I yz = ∑ Ai yi zi = −14545 cm 4 i

It is more convenient to put in the chart, for instance: Ai Stringer No. (cm 2 )

yi (cm )

zi (cm )

2

Ai y i z i

(cm 4 )

(cm 4 )

(cm 4 )

Ai z i

2

Ai yi

1

4

-54.5

-40.9

6694

11901

89256

2

2

145.5

-40.9

3347

42314

-11901

3

1

145.5

9.1

82.6

21157

1322

4

4

-54.5

59.1

13967

11901

-12893

∑=

24091

87273

-14545

(c) Bending stress in the stringers. By using the equation: σ xx =

I y M z − I yz M y I y I z − I yz

2

y+

I z M y − I yz M z I y I z − I yz

2

z , and

M y = −500,000 N ⋅ cm

M z = 200,000N ⋅ cm . I y = 24090.909cm 4 I z = 87272.727cm 4

I yz = −14545.455cm 4 We obtain σ xx = −1.298 y − 21.54 z Therefore the bending stresses in the stringers are:

σ xx

No.

yi (cm )

zi (cm )

( N / cm 2 )

1

-54.54

-40.91

951.92

2

145.45

-40.91

692.31

3

145.45

9.09

-384.62

4

-54.54

59.09

-1201.92

Stringer

--- ANS

4.5.2

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4.6

Find the neutral axis in the tin-walled section of Fig. 4.23 for the loading given in Problem 4.5. M y = −500,000 N ⋅ cm

M z = 200,000N ⋅ cm . Find the bending stresses in all stringers.

Figure 4.23

Thin-walled section

Solution: (a) From Problem 4.5 we get the centroid position as follows. yc = 54.5 cm , zc = 40.9 cm These are the horizontal and vertical distances, respectively, from stringer 1. (b) Set up the coordinate system (y,z) with the origin located at the centroid. Neutral plane is located at the position that centroid is the origin. From the bending stress formulas we find the neutral plane by setting the bending stress to zero, i.e.,

σ xx = −1.298 y − 21.538 z = 0 On the cross-section, this equation represents the line passing through the centroid with y = −16.59 z and an angle z y

α = tan −1( − ) = tan −1(

1 ) = 3.45o 16.59

--- ANS 4.6.1

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4.7

Find the bending stresses in the stringers at the fixed end of the box beam loaded as shown in Fig. 4.24. Assume that the thin sheets are negligible in bending. Find the neutral axis.

Figure 4.24

Loaded box beam

Solution: (a) Name the stringers from top to bottom and left to right as stringer 1, stringer 2, and stringer 3, respectively. Relative to string 2 the centroid position is given by yc =

∑ Ai yi i

∑ Ai

=

4 × 80 = 26.67 cm 3×4

=

4 × 40 = 13.33cm 3× 4

i

zc =

∑ Ai zi i

∑ Ai i

(b) The bending moments at the fixed end of the box beam produced by the loads are M y = −2 PL = −2( 200)(500) = −200000 N ⋅ cm ( M y is positive in positive y)

M z = −2 PL = −2(200)(500) = −200000N ⋅ cm ( M z is positive in negative z) (c) Set up the coordinate system (x,y,z) with the origin at the centroid. Moment of inertia (see table below for details):

I y = ∑ Ai zc 2 = 4( 2 × 13.332 + 26.67 2 ) = 4266 cm4 i

I z = ∑ Ai yc 2 = 4( 2 × 26.67 2 + 53.332 ) = 17067 cm4 i

4.7.1

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I yz = ∑ Ai yc zc = −4266 cm 4 i

Ai Stringer No. (cm 2 )

yi (cm )

zi (cm )

2

Ai y i z i

(cm 4 )

(cm 4 )

(cm 4 )

Ai z i

2

Ai yi

1

4

-26.67

26.67

2844

2844

-2844

2

4

-26.67

-13.33

711

2844

1422

3

4

53.33

-13.33

711

11377

-2844

∑=

4266

17067

-4267

(d) Bending stress in the stringers. Using the equation σ xx =

I y M z − I yz M y I y I z − I yz

2

y+

I z M y − I yz M z I y I z − I yz

2

z,

we obtain σ xx = −31.25 y − 78.125 z and the bending stresses in the stringers are:

σ xx

No.

yi (cm )

zi (cm )

( N / cm 2 )

1

-26.67

26.67

-1250

2

-26.67

-13.33

1875

3

53.33

-13.33

-625

Stringer

--- ANS (e) Neutral plane by angle α . Neutral plane is located at the position where bending stresses vanish under this particular loading. We have

σ xx = −31.25 y − 78.125 z = 0 It is the line passing through the centroid with y = −2.5 z z y

α = tan −1( − ) = tan −1(

1 ) = 21.8 o 2.5

--- ANS

4.7.2

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4.8

Find the deflection of the box beam of Fig. 4.24 using the simple beam theory.

Figure 4.24

Loaded box beam

Solution: (a) Name the stringers from top to bottom and then left to right as stringer 1, stringer 2, and stringer 3, respectively. From the solution of problem 4.7, we have the following moments of inertia: I y = 4266 cm 4 I z = 17066 cm 4

I yz = −4266 cm 4 Let the origin ( x = 0 ) of the coordinate system be located at the fixed end. The bending moments produced by the forces applied at the free end are M y = −400 × (500 − x ) N ⋅ cm

M z = −400 × (500 − x) N ⋅ cm

(b) The governing equations (see p. 122 in the book) for the bidirectional bending are

I y M z − I yz M y d 2v E 2 =− = 0.063(500 − x) ( N / cm 3 ) , 2 dx I y I z − I yz E

I z M y − I yz M z d 2w =− = 0.156(500 − x) ( N / cm 3 ) 2 2 dx I y I z − I yz

Integrating twice the above differential equations, we obtain

4.8.1

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x3 Ev = 0.063(250 x − ) + C1 x + C 2 6 2

x3 Ew = 0.156(250 x − ) + C 3 x + C 4 6 2

By applying the boundary conditions, the integration constants are solved as v ( x = 0) = 0 ,

dv ( x = 0) = 0 => C1 = C 2 = 0 dx

w( x = 0) = 0 ,

dw ( x = 0) = 0 => C 3 = C 4 = 0 dx

Then the lateral (in y-direction) and vertical (in z-direction) deflections are, respectively, v( x) =

0.063 x3 (250 x 2 − ) E 6

w( x) =

0.156 x3 (250 x 2 − ) E 6

In the expressions above, distance x is measured in cm, and the units of Young’s modulus and deflection are N / cm 2 and cm , respectively. --- ANS As an example, consider Aluminum 2024-T3, E = 72GPa = 72 × 10 5 ( N / cm 2 ) . The deflections in y and z directions at the free end are: 0.063 5003 2 v( x = 500) = × (250 × 500 − ) = 0.36cm 72 × 105 6 0.156 5003 2 w( x = 500) = × (250 × 500 − ) = 0.90cm 72 × 105 6

4.8.2

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4.9

Find the bending stresses in the stringers of the box beam in Fig. 4.24 for the bending moments given in Problem 4.5. M y = −500,000 N ⋅ cm

M z = 200,000N ⋅ cm .

Figure 4.24

Loaded box beam

Solution: (a) Name the stringers from top to bottom and then left to right as stringer 1, stringer 2, and stringer 3, respectively. The centroid position is given by yc =

∑ Ai yi i

∑ Ai

=

4 × 80 = 26.67 cm 3×4

=

4 × 40 = 13.33cm 3× 4

i

zc =

∑ Ai zi i

∑ Ai i

relative stringer 2. (b) Moment of inertia (see the table below for details)

I y = ∑ Ai zc 2 = 4( 2 × 13.332 + 26.67 2 ) = 4267 cm4 i

I z = ∑ Ai yc 2 = 4( 2 × 26.666667 2 + 53.3333332 ) = 17067 cm4 i

I yz = ∑ Ai yc zc = −4267 cm 4 i

4.9.1

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Ai Stringer No. (cm 2 )

yi (cm )

zi (cm )

2

Ai y i z i

(cm )

(cm )

(cm 4 )

Ai z i

2

4

Ai yi 4

1

4

-26.67

26.67

2844

2844

-2844

2

4

-26.67

-13.33

711

2844

1422

3

4

53.33

-13.33

711

11378

-2844

∑=

4267

17067

-4267

(c) Bending stress in the stringers. Subsituting the moments and moments of inertia in the bending stress formula

σ xx =

I y M z − I yz M y I y I z − I yz

2

y+

I z M y − I yz M z I y I z − I yz

2

z,

we obtain σ xx = −23.44 y − 140.62 z Therefore the bending stresses in the stringers are:

σ xx

No.

yi (cm )

zi (cm )

( N / cm 2 )

1

-26.67

26.67

-3125

2

-26.67

-13.33

2500

3

53.33

-13.33

625

Stringer

--- ANS

4.9.2

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4.10

A cantilever beam of a solid rectangular cross-section is loaded as shown in Fig. 4.25. Assume that the material is isotropic. Find the deflections of the beam using the simple beam theory and Timoshenko beam theory, respectively. Plot the ratio of the maximum deflections of the two solutions (at the free end) versus L/h. Use the shear correction factor k =

5 . 6

P x

h

t

L Figure 4.25

Cantilever beam subjected to a shear force P

Solution: (a) Simple beam theory The displacement equilibrium equations for the simple beam theory is:

EI y

d 4 w0 dx 4

= pz

(4.10.1)

In this particular problem, we have I y = EI y

1 3 th , p z = 0 . Thus, 12

d 4 w0 =0 dx 4

(4.10.2)

Integrating the equation (4.10.2) and applying boundary conditions, EI y

d 3 w0 = C 0 = − P (shear force) dx 3

Integrating again, we obtain EI y

d 2 w0 = − Px + C1 . dx 2

At x = L , M = EI y

(4.10.3)

d 2 w0 ( x = L) = 0 = − PL + C1 dx 2

=> C1 = PL From (4.10.3), 4.10.1

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EI y

dw0 1 = − Px 2 + PLx + C 2 2 dx

dw0 = 0 => C 2 = 0 dx

At x = 0 ,

1 1 Finally, EI y w0 ( x) = − Px 3 + PLx 2 + C 3 , and C 3 = 0 because w0 = 0 6 2

at x = 0 .

Therefore, the deflection curve is

w0 ( x) =

1 1 1 P x L x (− Px 3 + PLx 2 ) = [−2( ) 3 + 6 ( ) 2 ] 2 EI y 6 Et h h h --- ANS

The maximum deflection occurs at x = L : wmax,S =

P L L L 4P L 3 [ − 2( ) 3 + 6 ( ) 2 ] = ( ) Et h h h Et h

--- ANS (b) Timoshenko beam theory The displacement equilibrium equations for Timoshenko beam theory are:

EI y

d 2ψ y dx

2

− kGA(

dw0 +ψ y ) = 0 dx

(4.10.4)

d 2 w0 dψ y kGA( 2 + ) + pz = 0 dx dx

(4.10.5)

and can be combined into the following equation, EI y

EI y d 2 p z d 4 w0 = p − z GA dx 2 dx 4

(4.10.6)

In this particular problem, we have I y = EI y

1 3 th , and p z = 0 . Hence we have 12

d 4 w0 = 0 as the governing equation. dx 4

The concentrated loading at the free end produces a constant shear force along the beam, so we have

kGA(

dw0 + ψ y ) = shear force = P dx

(4.10.7)

Substituting (4.10.7) in (4.10.4) yields

4.10.2

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EI y

d 2ψ y dx 2

=P

(4.10.8)

Integrating equation (4.10.8) twice, we obtain EI yψ y =

1 2 Px + B0 x + B1 2

(4.10.9)

Using (4.10.7) and (4.10.9), we obtain dw0 P P 1 1 2 = −ψ y = − ( Px + B0 x + B1 ) dx kGA kGA EI y 2 Integrating the equation above,

w0 ( x) =

1 1 3 1 P x− ( Px + B0 x 2 + B1 x) + B2 kGA EI y 6 2

(4.10.10)

The following boundary conditions are used to determine the arbitrary constants in (4.10.10): M ( x = L) = EI y

dψ y dx

( x = L) = 0

=> B0 = − PL

ψ y ( x = 0) = 0 (no rotation of the cross-section)

=> B1 = 0

w0 ( x = 0) = 0

=> B2 = 0

Then the deflection equation (4.10.10) becomes

w0 ( x) = With k =

1 1 3 1 P x− ( Px − PLx 2 ) kGA EI y 6 2

E 5 1 , A = th , I y = th 3 , and G = , we obtain 2(1 + υ ) 6 12

w0 ( x) =

P x L x 12(1 + υ ) P x ( ) − [ 2( ) 3 − 6 ( ) 2 ] h Et h h h 5 Et

--- ANS The maximum deflection occurs at x = L : wmax,T =

12(1 + υ ) P L 4 P L 3 ( )+ ( ) h Et h 5 Et

--- ANS (c) The ratio of the maximum deflections of the two solutions versus L/h Assume the material to be Aluminum 2024-T3 with E = 72GPa , υ = 0.33 . For convenience, we let

L =β. h

The maximum deflection according to the simple beam theory: 4.10.3

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wmax, S =

P 4P ( β ) 3 = 0.055556 β 3 t 72t

The maximum deflection according to the Timoshenko beam theory: wmax,T =

12(1 + 0.33) P 4P P P (β ) + ( β ) 3 = 0.044333 β + 0.055556 β 3 5(72)t 72t t t Maximum deflections vs. L/h

14 12

α (w = αP/t)

10 8 Simple Timoshenko

6 4 2 0 0

1

2

3

4

5

6

7

L/h

Define Error (%) =

wmax,T − wmax, S wmax,T

× 100%

Error (%) vs. L/h 100 90 80

Error (%)

70 60 50 40 30 20 10 0 0

1

2

3

4

5

6

7

L/h

4.10.4

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4.11 A thin-walled beam of length 2 m long with one end built into a rigid wall and the other end is subjected to a shear force Vz = 5000 N . The cross-section is given by Fig. 4.21 with h = 0.2m and the wall thickness = 0.002m . The material is aluminum 2024-T3 with E = 70GPa , G = 27GPa , and the cross-sectional area of each stringer is 25cm 2 . Assume that thin walls carry only shear stresses. Find the deflections at the free end using the simple beam theory and the Timoshenko beam theory, respectively. Compare the transverse shear stress in the vertical web obtained from the two theories.

Figure 4.21

Stringer-web section

Solution: (a) Simple beam theory (1) The displacement equilibrium equation for the simple beam theory is: EI y

d 4 w0 =0 dx 4

(4.11.1)

Integrate the equation (4.11.1) and apply shear force boundary condition to yield, EI y

d 3 w0 = C 0 = −V z (shear force) dx 3

Integrate again to obtain EI y

d 2 w0 = −V z x + C1 , dx 2

At the free end, x = L , M = EI y

d 2 w0 ( x = L) = 0 = −V z L + C1 dx 2

=> C1 = V z L 4.11.1

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Again by integration, we have

EI y

dw0 1 = − V z x 2 + V z Lx + C 2 . 2 dx

At the fixed end, x = 0 , the rotation of the cross-section vanishes, i.e.,

ψy =−

dw0 = 0 => C 2 = 0 dx

Thus, we have, after integration, 1 1 EI y w0 ( x) = − V z x 3 + V z Lx 2 + C 3 6 2

In which the integration constant C3 is determined by the boundary condition w0 = 0 at x = 0 => C 3 = 0 . The deflections curve is

w0 ( x) =

1 1 1 (− V z x 3 + V z Lx 2 ) EI y 6 2

(4.11.2)

(2) Properties of the cross-section I y = ∑ Ai z i = 4 Ah 2 = 4( 25 × 10 −4 )(0.2) 2 = 4 × 10 −4 m 4 2

i

E = 70GPa L = 2m Vz = 5000 N (3) Deflections Compute deflection curve (4.11.2): w0 ( x) =

1 1 1 (− (5000) x 3 + (5000)(2) x 2 ) −4 2 (70 × 10 )(4 × 10 ) 6 9

= −2.9762 × 10 −5 x 3 + 1.7857 × 10 − 4 x 2

( m)

Deflection at the free end:

w0 ( x = 2m) = −2.9762 × 10 −5 (2) 3 + 1.7857 × 10 −4 (2) 2 = 4.762 × 10 − 4 ≅ 0.48 mm

m

--- ANS (b) Timoshenko beam theory (1) The displacement equilibrium equations for the Timoshenko beam theory are:

4.11.2

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EI y

GA(

d 2ψ y dx

2

− GA(

dw0 +ψ y ) = 0 dx

(4.11.3)

d 2 w0 dψ y + ) + pz = 0 dx 2 dx

(4.11.4)

which can be combined into the following equation, EI y

EI y d 2 p z d 4 w0 = p − z GA dx 2 dx 4

(4.11.5)

In the equations above, the area A in the GA term is the effective area of the thin-walled section that carries shear stress and should not be confused with the stringer cross-sectional area. Since p z = 0 we have d 4 w0 EI y = 0 as the governing equation. dx 4 The concentrated shear loading at the free end produces a constant shear force along the beam; so we have GA(

dw0 + ψ y ) = Vz dx

(4.11.6)

Substitution of (4.11.6) in (4.11.3) yields

EI y

d 2ψ y dx 2

= Vz

(4.11.7)

Integrating (4.11.7), we obtain 1 EI yψ y = V z x 2 + B0 x + B1 2

(4.11.8)

Using (4.11.6) and (4.11.8), we have

dw0 Vz V 1 1 = −ψ y = z − ( Vz x 2 + B0 x + B1 ) dx GA GA EI y 2 Integrating the above equation with the result,

w0 ( x) =

Vz 1 1 1 x− ( Vz x3 + B0 x 2 + B1 x) + B2 GA EI y 6 2

(4.11.9)

Applying boundary conditions to equation (4.11.8) and (4.11.9), we have M ( x = L) = EI y

ψ y ( x = 0) = 0

dψ y dx

( x = L) = 0 => B0 = −V z L => B1 = 0

4.11.3

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=> B2 = 0

w0 ( x = 0) = 0 Then equation (4.11.9) becomes

Vz V 1 1 x − z ( x3 − Lx 2 ) GA EI y 6 2

w0 ( x) =

(4.11.10)

(2) Properties of the cross-section In the Timoshenko beam theory the area A (in the GA term) of the thin-walled cross-section is Ashear = 2ht = 2(0.2)(0.002) = 8 × 10 −4 m 2

I y = ∑ Ai z i = 4 Ah 2 = 4( 25 × 10 −4 )(0.2) 2 = 4 × 10 −4 m 4 2

i

E = 70GPa , G = 27GPa L = 2m Vz = 5000 N (3) Deflection Compute the deflection curve (4.11.10) using the above properties: w0 ( x) =

5000 x − (2.9762 × 10 −5 x 3 − 1.7857 × 10 − 4 x 2 ) (27 × 10 9 )(8 × 10 − 4 )

= 2.3148 × 10 − 4 x + 1.7857 × 10 − 4 x 2 − 2.9762 × 10 −5 x 3

( m)

Deflection at the free end is

w0 ( x = 2m) = 2.3148 × 10 −4 (2) + 1.7857 × 10 −4 (2) 2 − 2.9762 × 10 −5 (2) 3 = 9.391 × 10 − 4 ≅ 0.94 mm

m

--- ANS

The difference between the two theories is Error(%) =

w0 ,Tim − w0 ,Sim 9.391 − 4.762 = × 100% = 49.3% w0 ,Tim 9.391

(c) Transverse Shear Stress (1) Simple beam theory From the derivation of the simple beam theory, we assume γ xy = 0 as an approximation. As a result, the transverse shear stress can not be directly obtained from the stress-strain relations. It is obtained usually from the 4.11.4

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equilibrium equation. We have

τ xz =

Vz 5000 = = 6.25 × 106 N / m 2 = 6.25MPa −4 Ashear 8 × 10

(2) Timoshenko beam theory

τ xz = Gγ xz =

Vz = 6.25MPa Ashear

--- ANS

4.11.5

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4.12

A 2024-T3 aluminum box beam with a thin-walled section is shown in Fig. 4.26. Assume that thin walls (thickness t = 0.3 cm) are ineffective in bending. The cross-sectional area of each stringer is 20 cm2. Find the deflections at the free end using the simple beam theory for shear loads Vz = 5000 N and V y = 5000 N separately. Solve the same problem using Timoshenko beam

theory. In which loading case is the simple beam theory more accurate in predicting the deflection? Explain.

Figure 4.26

Box beam with a triangular thin-walled section

Solution: (a) First, we need to know the centroid of this section. Take stringer 2 as the origin of a coordinate system. Then the centroid is located at

∑A y = ∑A i

yc

i

i

=

20 × 60 = 20 cm 3 × 20

=

20 × (40 + 20) = 20 cm 3 × 20

i

i

∑Az = ∑A i

zc

i

i

i

i

The moments of inertia with respect to the coordinate system with the origin at the centroid are I y = ∑ Ai z c = 20( 20 2 + ( −20) 2 ) = 16000 cm 4 2

i

I z = ∑ Ai y c = 20( 40 2 + 2 × ( −20) 2 ) = 48000 cm 4 2

i

4.12.1

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I yz = ∑ Ai y c z c = 4[40 × 0 + (−20) × 20 + (−20) × (−20)] = 0 (This should be i

obvious because the section is symmetric with respect to y-axis) For 2024-T3, E = 72GPa = 72 × 10 5 N / cm 2 , υ = 0.33 => G = 27.068 × 10 5 N / cm 2

(b) Simple beam theory The displacement equilibrium equations for the simple beam theory are: d 4 w0 EI y = 0, dx 4

for V z loading

(4.12.1)

d 4 v0 EI z = 0, dx 4

for V y loading

(4.12.2)

Integrating the above equations, we get EI y

d 3 w0 = −V z dx 3

(4.12.3)

EI z

d 3 v0 = −V y dx 3

(4.12.4)

Thus, d 3 w0 − V z − 5000 = = = −4.3403 × 10 −8 3 EI y 72 × 10 5 × 16000 dx

(1 / cm 2 )

d 3 v0 − V y − 5000 = = = −1.4468 × 10 −8 3 EI z dx 72 × 10 5 × 48000

(1 / cm 2 )

Integrating the above equations, we have 1 w0 ( x ) = −7.234 × 10 −9 x 3 + C1 x 2 + C 2 x + C 3 2

1 v0 ( x) = −2.411 × 10 −9 x 3 + C 4 x 2 + C 5 x + C 6 2

The arbitrary constants are determined by the boundary conditions,

For w0 ( x) w0 ( x = 0) = 0 ,

=> C 3 = 0

dw0 ( x = 0) = 0 dx

=> C 2 = 0

4.12.2

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d 2 w0 EI y ( x = L) = M ( x = L) = 0 dx 2 d 2 w0 => EI y ( x = 200cm) = −5000(200) + (72 × 10 5 × 16000)C1 = 0 2 dx => C1 = 8.681 × 10 −6 So, w0 ( x ) = −7.234 × 10 −9 x 3 + 4.340 × 10 −6 x 2

(4.12.5)

For v 0 ( x ) v0 ( x = 0) = 0 ,

=> C 6 = 0

dv0 ( x = 0) = 0 dx

=> C 5 = 0

d 2 v0 EI z ( x = L) = M ( x = L) = 0 dx 2 d 2 v0 => EI z ( x = 200cm) = −5000(200) + (72 × 10 5 × 48000)C 4 = 0 2 dx => C 4 = 2.894 × 10 −6 So, v 0 ( x ) = −2.411 × 10 −9 x 3 + 1.447 × 10 −6 x 2

(4.12.6)

--Therefore deflections at the free end can be obtained from (4.12.5) and (4.12.6) by setting x = 200cm : w0 ( x = 200cm) = −7.234 × 10 −9 (200) 3 + 4.340 × 10 −6 (200) 2 = 0.116 cm v 0 ( x = 200cm) = −2.411 × 10 −9 (200) 3 + 1.447 × 10 −6 (200) 2 = 0.039 cm --- ANS (c) Timoshenko beam theory The displacement equilibrium equations for Timoshenko beam theory for V z loading are:

EI y

d 2ψ y dx

2

− GAz (

dw0 +ψ y ) = 0 dx

(4.12.7)

d 2 w0 dψ y GAz ( 2 + ) + pz = 0 dx dx

(4.12.8)

4.12.3

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which can be combined into the following equation, EI y

EI y d 2 p z d 4 w0 = p − z GAz dx 2 dx 4

(4.12.9)

Note that Az is the projection of the cross-sectional area of the thin sheets onto z-axis. In this case, Az = 2 × 40 × 0.3 = 24cm 2 . In this particular problem, we have p z = 0 . Hence EI y

d 4 w0 =0 dx 4

is the governing equation. The concentrated shear loading at the free end produces a constant shear force along the beam, so we have

GAz (

dw0 + ψ y ) = Vz dx

(4.12.10)

Substituting the above in equation (4.12.7) yields

EI y

d 2ψ y dx 2

= Vz

(4.12.11)

Integrating equation (4.12.11), we obtain 1 EI yψ y = V z x 2 + B0 x + B1 2

(4.12.12)

Using equation (4.11.10) and (4.11.12), we have dw0 V V 1 1 = z −ψ y = z − ( V z x 2 + B0 x + B1 ) dx GAz GAz EI y 2 Integrating the above equation, w0 ( x) =

Vz 1 1 1 x− ( V z x 3 + B0 x 2 + B1 x) + B2 GAz EI y 6 2

(4.12.13)

Applying boundary conditions to equation (4.12.13) M ( x = L) = EI y

dψ y dx

( x = L) = 0

=> B0 = −V z L

ψ y ( x = 0) = 0

=> B1 = 0

w0 ( x = 0) = 0

=> B2 = 0

Then equation (4.12.13) becomes w0 ( x) =

Vz V 1 1 x − z ( x 3 − Lx 2 ) GAz EI y 6 2

(4.12.14)

4.12.4

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Similarly, the deflection in y-direction due to V y is v0 ( x) =

Vy GAy

x−

Vy

1 1 ( x 3 − Lx 2 ) EI z 6 2

(4.12.15)

where Ay is the projection of the cross-sectional area of the thin sheets onto y-axis. We have Ay = 2 × 60 × 0.3 = 36cm 2 .

The deflection due to V z is 5000 5000 1 1 x− ( x 3 − (200) x 2 ) 5 5 2 (27.068 × 10 )(24) (72 × 10 )(16000) 6

w0 ( x) =

= 7.697 × 10 −5 x − (7.234 × 10 −9 x 3 − 4.340 × 10 −6 x 2 ) And for V y : v0 ( x) ==

5000 5000 1 1 x− ( x 3 − (200) x 2 ) 5 5 2 (27.068 × 10 )(36) (72 × 10 )(48000) 6

= 5.131 × 10 −5 x − (2.411 × 10 −9 x 3 − 1.447 × 10 −6 x 2 ) --At the free end the respective deflection can be obtained from (4.12.5) and (4.12.6) by substituting in x = 200cm w0 ( x = 200cm) = 7.697 × 10 −5 (200) − [7.234 × 10 −9 (200) 3 − 4.340 × 10 −6 (200) 2 ] = 0.131 cm v 0 ( x = 200cm) = 5.131 × 10 −5 (200) − [2.411 × 10 −9 (200) 3 − 1.447 × 10 −6 (200) 2 ] = 0.049cm

--- ANS (d) Summary (1) Deflections at the free end Simple Beam Theory

Timoshenko Beam Theory

Error (%)

Vz = 5000 N

0.116 cm

0.131 cm

11.5

(2) V y = 5000 N

0.039 cm

0.049 cm

20.4

(1)

Error (%) =

d Timoshenko − d Simple d Timoshenko

,

where d = w0 or v0

4.12.5

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(2) Case 1 ( V z = 5000 N ) of the above results is more accurate. It is mainly because of the reason that I y is smaller than I z , and, as a result, the bending behavior for z-direction is more likely to resemble a slender beam than that for y-direction. --- ANS

4.12.6

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4.13

Consider the structure with a cutout as shown in Fig. 4.17. Find the axial force distribution in stringers 3-4 and 5-6. Assume that both stringers and webs have the same material properties of E = 70GPa and G = 27GPa . Also assume that b = 200mm , the thickness of the web t = 2mm , and the cross-sectional area of the stringer A = 64mm 2 . Hint: The zero-stress condition in the web at the cutout cannot be enforced because of the simplified assumption that shear stress and strain are uniform across the width of the web. Use the known condition that the force in the side stringers is 1.5P at the cutout.

1

2

3

4

5

P

6

P

P L1

Figure 4.17

L2

Cutout in a stringer sheet panel

Solution: (a) First, we consider the part left hand side of the cutout.

1.5P

F1

F2

3

4

b

1.5P

F1 L1 x

The balance of forces in the x-direction yields

2 F1 + F2 = 3P

(4.13.1)

Also we have the differential equation 4.13.1

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τ=

1 dF1 t dx

(4.13.2)

F F dτ G = ( 1 − 2) dx E (b / 2) A1 A2

(4.13.3)

Combining equations (4.13.2) and (4.13.3), we have F F 1 d 2 F1 G 6G = ( 1 − 2)= ( F1 − P) 2 t dx E (b / 2) A1 A2 EAb =>

d 2 F1 6Gt = ( F1 − P) , EAb dx 2

=>

d 2 F1 − λ2 F1 = −λ2 P dx 2

let λ2 =

6Gt EAb

(4.13.4)

The general solution of this second-order differential equation is

F1 ( x) = C1 cosh λx + C 2 sinh λx + P (where cosh λx =

e λx + e − λx e λx − e − λ x and sinh λx = ) 2 2

Applying the boundary conditions, => τ = 0 , that is

At x = 0 (fixed end) =>

dF1 ( x = 0) = λC 2 = 0 dx

At x = L1

dF1 ( x = 0) = 0 dx

=> C 2 = 0

=> F1 ( x = L1 ) = 1.5P

=> F1 ( x = L1 ) = C1 cosh λL1 + P = 1.5P => C1 =

P 2 cosh λL1

Therefore the solution of the differential equation is F1 ( x ) = P (

cosh λx + 1) 2 cosh λL1

(4.13.5)

--The axial force distribution in stringers 3-4 can be obtained from (4.13.1) and (4.13.5), that is F2 ( x) = 3P − 2 F1 ( x ) = P(1 −

with λ =

6Gt = EAb

cosh λx ) cosh λL1

6(27 × 10 9 )(2 × 10 −3 ) = 19.016 (70 × 10 9 )(64 × 10 −6 )(0.2)

(1 ) m

4.13.2

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--- ANS

(b) Next, we consider the part to the right of the cutout: F1

1.5P

5 F2

1.5P

F1 x

From the balance of forces in the x-direction we have

2 F1 + F2 = 3P

(4.13.6)

Also we have the differential equation

τ=

1 dF1 t dx

(4.13.7)

F F dτ G ( 1 − 2) = dx E (b / 2) A1 A2

(4.13.8)

Combining equations (4.13.7) and (4.13.8), we have F F 1 d 2 F1 G 6G = ( 1 − 2)= ( F1 − P) 2 t dx E (b / 2) A1 A2 EAb =>

d 2 F1 6Gt = ( F1 − P) , EAb dx 2

let λ2 =

6Gt EAb

d 2 F1 => − λ2 F1 = −λ2 P 2 dx

(4.13.9)

The general solution of this second-order differential equation is

F1 ( x) = C1 cosh λx + C 2 sinh λx + P Applying the boundary conditions, at x = 0

=> F1 ( x = 0) = 1.5P

=> F1 ( x = 0) = C1 + P = 1.5P at x = L2

=> C1 = 0.5P

=> F1 ( x = L2 ) = P

=> F1 ( x) = 0.5P cosh λL2 + C 2 sinh λL2 + P = P

4.13.3

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=> C 2 = −

P 2 tanh λL2

The solution of the differential equation is F1 ( x ) =

P sinh λx (cosh λx − + 2) 2 tanh λL2

(4.13.10)

--The axial force distribution in stringer 5-6 can be obtained from (4.13.6) and (4.13.10), that is F2 ( x) = 3P − 2 F1 ( x) = P (1 − cosh λx +

where λ =

6Gt = EAb

sinh λx ) tanh λL2

6(27 × 10 9 )(2 × 10 −3 ) = 19.016 ( 1 ) m (70 × 10 9 )(64 × 10 −6 )(0.2) --- ANS

4.13.4

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5.1

Find the flexural shear flow produced by the transverse shear force Vz = 1000N in the beam with the thin-walled section given by Fig. 5.30.

Figure 5.30

Thin-walled section with a side cut

Solution: (a) Assume that the transverse shear force acts through the shear center, and, thus, no torsional effect exists. Assume that the loss of material at the cut is negligible. Hence the centroid of the cross-sectional area is obviously at the center as shown in Fig. 5.30. The shear flow is obtained by VQ qs = − z (5.1.1) Iy where Q = ∫∫ zdA = As z c is the first moment of area As (the area measured As

along the wall from an free edge to the current position of interest), and z c is the vertical distance from the centroid of As to the y-axis. We have Iy =

1 [(0.1 + 0.002)(0.2 + 0.002) 3 − (0.1 − 0.002)(0.2 − 0.002) 3 ] = 6.6676 × 10 −6 m 4 12

(b) Setup the shear flow contour as shown in the figure below.

5.1.1

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(1) On s1 : 0 ~ 0.1m s 2 Q = ∫∫ zdA = As z c = ( s1t )( 1 ) = 0.001s1 2 As

VQ 1000 × 0.001s1 2 qs = − z = − = −1.5 × 10 5 s1 −6 Iy 6.6676 × 10 2

At s1 = 0 , q s = −1.5 × 10 5 × 0 2 = 0 At s1 = 0.1 , qs = −1.5 × 105 × 0.12 = −1500 N / m

(The negative sign

means that the actual direction of this shear flow is opposite to contour direction of s1 . --- ANS (2) On s 2 : 0 ~ 0.1m The first moment Q for this segment must include the entire first moment of the segment covered by contour s1. Thus, Q = 0.001(0.1) 2 + 0.002 s 2 (0.1) = 10 −5 + 2 × 10 −4 s 2 qs = −

Vz Q 1000 × (10 −5 + 2 × 10 −4 s 2 ) =− = −1500 − 3 × 10 4 s 2 −6 Iy 6.6676 × 10

Note that the distribution of the shear flow is linear along the contour. At s2 = 0 , qs = −1500 N / m At s 2 = 0.1m , qs = −1500 − 3 × 10 4 × 0.1 = −4500 N / m --- ANS 5.1.2

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(3) On s 3 : 0 ~ 0.1m In a similar manner, the first moments for the previous two segments must be added to that for the additional area along contour s3. We have Q = 0.001(0.1) 2 + 0.002(0.1) 2 + 0.002 s 3 (0.1 − 0.5s3 ) = 3 × 10 −5 + 2 × 10 − 4 ( s 3 − 5s3 ) 2

1000 × [3 × 10 −5 + 2 × 10 −4 ( s 3 − 5s3 )] Vz Q =− Iy 6.6676 × 10 −6 2

qs = −

= −4500 − 30000 s 3 + 150000 s3

2

At s3 = 0 , qs = −4500 − 30000 (0) + 150000 (0) 2 = −4500 N / m At s3 = 0.1m , qs = −4500 − 30000 (0.1) + 150000 (0.1) 2 = −6000 N / m --- ANS

(4) On s 4 : 0 ~ 0.1m For convenience, we start another contour s4 from the other free edge at the cut. Q = ∫∫ zdA = As z c = ( s 4 t )( As

s4 2 ) = −0.001s 4 2

V Q 1000 × 0.001s 4 2 qs = − z = = 1.5 × 10 5 s 4 −6 Iy 6.6676 × 10 2

At s4 = 0 , q s = 1.5 × 10 5 × 0 2 = 0 At s 4 = 0.1m , q s = 1.5 × 10 5 × 0.12 = 1500 N / m --- ANS (5) On s 5 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002 s 5 (0.1) = −10 −5 − 2 × 10 −4 s 5

qs = −

V z Q 1000 × (10 −5 + 2 × 10 −4 s5 ) = 1500 + 3 × 10 4 s 5 = −6 Iy 6.6676 × 10

5.1.3

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s 5 = 0m , q s = 1500 N / m s5 = 0.1m , q s = 15003 × 10 4 × 0.1 = 4500 N / m --- ANS (6) On s 6 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002(0.1) 2 − 0.002 s 6 (0.1 − 0.5s 6 ) = −3 × 10 −5 − 2 × 10 − 4 ( s 6 − 5s 6 ) 2

V z Q 1000 × [3 × 10 −5 + 2 × 10 −4 ( s 6 − 5s 6 )] = Iy 6.6676 × 10 −6 2

qs = −

= 4500 + 30000 s 6 − 150000 s 6

2

s 6 = 0m , q s = 4500 + 30000 (0) − 150000 (0) 2 = 4500 N / m s 6 = 0.1m , q s = 4500 + 30000 (0.1) − 150000 (0.1) 2 = 6000 N / m --- ANS (c) The complete shear flow along the entire section is shown below.

5.1.4

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5.2

Find the shear flow of the wide-flange beam (Fig. 5.31) subjected to Vz = 1000N .

Figure 5.31

Section of an I-beam

Solution: (a) Assume that the transverse shear force acts through the shear center and produces no torsion. From symmetry, it is obvious that the centroid of the section is located at the mid point of the vertical web. The shear flow is obtained by VQ qs = − z (5.2.1) Iy where Q = ∫∫ zdA = As z c is the first moment of area As , and z c is the vertical As

distance from the centroid of As to the y-axis. The moment of inertia of the cross-section is Iy =

1 [(0.1)(0.1 + 0.003) 3 − (0.1 − 0.003)(0.1 − 0.003) 3 ] = 1.7286 × 10 −6 m 4 12

(b) Set up the shear flow contours as in the following figure.

5.2.1

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(1) On s1 : 0 ~ 0.05m h 0.1 Q = ∫∫ zdA = As z c = ( s1t )( ) = (0.003s1 )( ) = 1.5 × 10 − 4 s1 2 2 As qs = −

Vz Q 1000 × 1.5 × 10 −4 s1 =− = −8.678 × 10 4 s1 −6 Iy 1.7286 × 10

At s1 = 0 , q s = −8.678 × 10 4 × 0 = 0 At s1 = 0.05m , q s = −8.678 × 10 4 × 0.05 = −4339 N / m --- ANS (2) On s 2 : 0 ~ 0.05m This is similar to s1 q s = −8.678 × 10 4 s 2

At s2 = 0 , q s = 0 At s 2 = 0.2m , q s = −8.678 × 10 4 × 0.05 = −4339 N / m --- ANS (3) On s3 : 0 ~ 0.05m

Q = 0.003(0.1)(0.05) + 0.003s3 (0.05 − 0.5s3 ) = 1.5 × 10 −5 + 1.5 × 10 − 4 ( s3 − 10s3 ) 2

1000 × [1.5 × 10 −5 + 1.5 × 10 −4 ( s3 − 10 s3 )] Vz Q qs = − =− Iy 1.7286 × 10 −6 2

= −8678 − 86775s3 + 867754 s3

2

At s3 = 0 , qs = −8678 − 86775(0) + 867754 (0) 2 = −8678 N / m At s 3 = 0.05m , qs = −8678 − 86775(0.05) + 867754 (0.05) 2 = −10847 N / m --- ANS (4) On s 4 : 0 ~ 0.05m & s5 : 0 ~ 0.05m The equation is the same as that of s1 and s2 with opposite sign. --- ANS (5) On s 6 : 0 ~ 0.05m 5.2.2

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The equation is the same as that of s 3 with opposite sign. --- ANS (c) Shear flows in the cross-section

5.2.3

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5.3

Find the shear center y sc for the sections of Figs. 5.30 and 5.32. For the four-stringer section (Fig. 5.32), assume that the thin sheets are ineffective in bending.

Figure 5.30

Thin-walled section with a side cut

Figure 5.32

Open four-stringer section

Solution: (a) Figure 5.30 (1) Since this cross-section is symmetric with respect to y axis, shear center is located on the y axis. So it is only necessary to determine the y position of the shear center. (2) Also, the centroid is at the center of the section as shown in Fig.5.30. Assume that the transverse shear force V z ( and V y = 0 ) acts through the shear center, at the distance y sc to the right-top corner. The shear flow can be obtained by VQ qs = − z Iy

(5.3.1)

where Q = ∫∫ zdA = As z c is the first moment of area As , and z c is the As

vertical distance from the centroid of As to the y-axis. 5.3.1

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1 [(0.1 + 0.002)(0.2 + 0.002) 3 − (0.1 − 0.002)(0.2 − 0.002) 3 ] 12 = 6.6676 × 10 −6 m 4 Iy =

(3) Setup the shear flow contour as the following figure,

y sc

i.

Vz

On s1 : 0 ~ 0.1m s 2 Q = ∫∫ zdA = As z c = ( s1t )( 1 ) = 0.001s1 2 As

VQ V × 0.001s1 2 qs = − z = − z = −150V z s1 −6 Iy 6.6676 × 10 2

ii.

On s 2 : 0 ~ 0.1m Q = 0.001(0.1) 2 + 0.002 s 2 (0.1) = 10 −5 + 2 × 10 −4 s 2 qs = −

iii.

Vz Q V × (10 −5 + 2 × 10 −4 s 2 ) =− z = −1.5V z − 30V z s 2 Iy 6.6676 × 10 −6

On s 3 : 0 ~ 0.1m Q = 0.001(0.1) 2 + 0.002(0.1) 2 + 0.002 s 3 (0.1 − 0.5s3 ) = 3 × 10 −5 + 2 × 10 − 4 ( s 3 − 5s3 ) 2

V z × [3 × 10 −5 + 2 × 10 −4 ( s3 − 5s3 )] Vz Q qs = − =− Iy 6.6676 × 10 −6 2

= −4.5V z − 30V z s 3 + 150V z s3 iv.

2

On s 4 : 0 ~ 0.1m Q = ∫∫ zdA = As z c = ( s 4 t )( As

s4 2 ) = −0.001s 4 2

5.3.2

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V Q V × 0.001s 4 2 qs = − z = z = 150V z s 4 −6 Iy 6.6676 × 10 2

v.

On s 5 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002 s 5 (0.1) = −10 −5 − 2 × 10 −4 s 5

qs = −

vi.

V z Q V z × (10 −5 + 2 × 10 −4 s5 ) = 1.5V z + 30V z s5 = Iy 6.6676 × 10 −6

On s 6 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002(0.1) 2 − 0.002 s 6 (0.1 − 0.5s 6 ) = −3 × 10 −5 − 2 × 10 − 4 ( s 6 − 5s 6 ) 2

V z Q V z × [3 × 10 −5 + 2 × 10 −4 ( s 6 − 5s 6 )] = Iy 6.6676 × 10 −6 2

qs = −

= 4.5V z + 30V z s 6 − 150V z s 6

2

(4) Shear center The moment produced by Vz and the shear flow about the right-top corner must be equal. And we can see that only q s 3 , q s 5 , q s 6 produce the moment with respect to the right-top corner. i. moment produced by Vz M Vz = V z y sc (counterclockwise) ii. moment produced by q s 3 0.1

0.1

0

0

M q 3 = ∫ ( q s 3 )(0.1) ds 3 = ∫ ( −4.5V z − 30V z s 3 + 150V z s 3 )(0.1)ds 3 2

= ( −0.45V z s 3 − 1.5V z s 3 + 5V z s3 ) | 00.1 = −0.055V z 2

iii.

3

(counterclockwise) moment produced by q s 5 0.1

0.1

M q 5 = ∫ ( q s 5 )(0.2) ds 5 = ∫ (1.5V z + 30V z s5 )(0.2)ds 5 0

0

= (0.3V z s5 + 3V z s3 ) | = 0.06V z 2

iv.

0.1 0

(clockwise) moment produced by q s 6 0.1

0.1

0

0

M q 6 = ∫ ( q s 6 )(0.1)ds 6 = ∫ (4.5V z + 30V z s 6 − 150V z s 6 )(0.1)ds 6 2

= (0.45V z s 6 + 1.5V z s 6 − 5V z s 6 ) | 00.1 = 0.055V z 2

3

(clockwise) 5.3.3

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So we can determine the shear center by M V = ∑ M q => V z y sc = (−0.055V z ) − 0.06V z − 0.055V z => y sc = −0.17 m The negative sign indicates the opposite side as we assumed. --- ANS (b) Figure 5.32 (1) Since this cross-section is symmetric with respect to y axis, shear center is located in the y axis. So it is only necessary to determine the y position of the shear center. (2) Assume that the transverse shear force Vz ( and V y = 0 ) acts through the shear center and produces no torsion to the cross-section. Also assume that the thin sheets are ineffective in bending. Let the left-bottom stringer be the center of origin. We can now determine the centroid of this four-stringer section.

∑A y = ∑A i

yc

i

i

=

2(2 A0 )(2h) = 1.3333h 2( A0 ) + 2(2 A0 )

=

(3 A0 )(h) = 0.5h 2( A0 ) + 2(2 A0 )

i

i

∑A y = ∑A i

yc

i

i

i

i

Now set the y and z axis to match the centroid. The moment of inertia then can be obtained I y = ∑ Ai z i = 2(3 A0 )(0.5h) 2 = 1.5 A0 h 2 2

i

I z = ∑ Ai y i = 2( 2 A0 )(0.6667 h) 2 + 2( A0 )(1.3333h) 2 = 5.3333 A0 h 2 2

i

(3) The shear flow of the stringer section is VQ qi = − z i Iy

(5.3.2)

i

where Qi = ∑ Ak z k is the first moment of stringer area Ak , and z k is k =1

the vertical distance from the centroid of Ak to the y-axis.

5.3.4

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i.

On s1 Q1 = (2 A0 )(0.5h) = A0 h q1 = −

ii.

On s2 Q2 = (3 A0 )(0.5h) = 1.5 A0 h q2 = −

iii.

V × A0 h V z Q1 0.6667V z =− z =− 2 Iy h 1.5 A0 h

V × 1.5 A0 h V z Q2 V =− z =− z 2 Iy h 1.5 A0 h

On s 3 Q3 = (3 A0 )(0.5h) − ( A0 )(0.5h) = 1A0 h q3 = −

V z Q3 V × A0 h 0.6667V z =− z =− 2 Iy h 1.5 A0 h

(4) Shear center The moment produced by Vz and the shear flow about the centroid must be equal. i. moment produced by Vz M Vz = V z y sc (counterclockwise) ii.

moment produced by q1

M q1 = (q1 )(2h)(0.5h) = (−

0.6667V z )(2h)(0.5h) = −0.6667V z h h

iii.

(counterclockwise) moment produced by q 2

iv.

V 1 M q 2 = 2 A(q 2 ) = 2( πh 2 )(− z ) = −0.7854V z h 8 h (counterclockwise) moment produced by q 3 M q 3 = (q3 )(2h)(0.5h) = (−

0.6667V z )(2h)(0.5h) = −0.6667V z h h

(counterclockwise) 5.3.5

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So we can determine the shear center by M V = ∑ M q => V z y sc = −0.6667V z h − 0.7854V z − 0.6667V z h = −2.1188V z h => y sc = −2.1188h The negative sign indicates the opposite side as we assumed. --- ANS

5.3.6

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5.4 Find the flexural shear flow in the section of Fig. 5.32 for Vz = 5000 N .

Figure 5.32

Open four-stringer section

Solution: (a) Assume that the transverse shear force acts through the shear center and produces no torsion to the cross-section. Also assume that the thin sheets are ineffective in bending. Let the left-bottom stringer be the origin of a coordinate system with respect to which we now determine the centroid of this four-stringer section. We obtain

∑Ay = ∑A i

yc

i

i

=

2(2 A0 )(2h) = 1.33h 2( A0 ) + 2(2 A0 )

=

(3 A0 )(h) = 0 .5 h 2( A0 ) + 2(2 A0 )

i

i

∑Az = ∑A i

zc

i

i

i

i

Now we set up the (y, z) coordinate system with the origin located at the centroid. The moments of inertia with respect to y-axis and z-axis are obtained as I y = ∑ Ai z i = 2(3 A0 )(0.5h) 2 = 1.5 A0 h 2 2

i

I z = ∑ Ai yi = 2( 2 A0 )(0.67 h) 2 + 2( A0 )(1.33h) 2 = 5.33 A0 h 2 2

i

(b) The shear flow calculation The shear flow formula for symmetric and open sections VQ qi = − z i Iy

(5.4.1)

Is used with the contour direction shown in Fig.5.32. (1) On s1 5.4.1

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Q1 = (2 A0 )(0.5h) = A0 h q1 = −

5000 × A0 h V z Q1 3333.33 =− =− 2 Iy h 1.5 A0 h --- ANS

(2) On s2 Q2 = (3 A0 )(0.5h) = 1.5 A0 h q2 = −

5000 × 1.5 A0 h V z Q2 5000 =− =− 2 Iy h 1.5 A0 h --- ANS

(3) On s 3 Q3 = (3 A0 )(0.5h) − ( A0 )(0.5h) = 1A0 h q3 = −

V z Q3 5000 × A0 h 3333.33 =− =− 2 Iy h 1.5 A0 h --- ANS

The negative signs indicates that the actual shear flow direction is opposite to the assumed direction of the contour.

5.4.2

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5.5

Find the shear flow for the three-stringer section shown in Fig. 5.33 for

Vz = 5000 N and V y = 0 . Given shear modulus G = 27GPa , find the twist angle per unit length. Also determine the shear center. Is the shear flow statically determinate?

Figure 5.33

Single-cell closed section

Solution: (a) Assume that the thin sheets are ineffective in bending. Let stringer 2 be the reference point for the location of the centroid of this four-stringer section. We have the horizontal and vertical distances of the centroid from stringer 2 as

∑A y = ∑A i

yc

i

i

=

(10)(80) = 26.6667cm = 0.2667 m 3(10)

=

(10)(20) + (10)(10) = 10cm = 0.1m 3(10)

i

i

∑Az = ∑A i

zc

i

i

i

i

Now we set up the y and z axes with the origin at the centroid. The moments of inertia are I y = ∑ Ai z i = 2(10 × 10 −4 )(0.1) 2 = 2 × 10 −5 m 4 2

i

I z = ∑ Ai y i = 2(10 × 10 −4 )(0.2667) 2 + (10 × 10 −4 )(0.8 − 0.2667) 2 2

i

= 4.2667 × 10 − 4 m 4

(b) Shear flows Since this cross-section is symmetric with respect to y axis, the shear center is located on the y axis. Hence only the y position of the shear center needs to be found. We first 5.5.1

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make a fictitious cut between stringers 1 and 3 and consider the shear flow as the superposition of two shear flow systems as shown in the figure below.

+

(1) First, calculate the shear flows by assuming a cut in the wall between stringers 1 and 3. Then q '31 = 0 , and the shear on the cut section is calculated by using q'i = −

V z Qi Iy

(5.5.1)

i

where Qi = ∑ Ak z k is the first moment of stringer area Ak , and z k is the k =1

vertical distance from the centroid of Ak to the y-axis. We obtain q '12 = −

V z Q1 5000(10 × 10 −4 × 0.1) =− = −25000 N / m Iy 2 × 10 −5

q ' 23 = −

V z Q2 5000(10 × 10 −4 × 0.1 − 10 × 10 −4 × 0.1) =− =0 Iy 2 × 10 −5

(2) Adding the shear flow q0 from the second part, we have the total shear flow as q12 = q '12 + q 0 = −25000 + q 0 q 23 = q ' 23 + q0 = q 0 q31 = q '31 + q 0 = q 0 The resulting moment of the total shear flow must be equal to the moment produced by V z . Taking moment about stringer 1, we have V z × 0 = 2 A12 q12 + 2 A123 q 23

π

= 2( (0.2) 2 )(−25000 + q 0 ) + 2 8 = −785.4 + 0.1914q 0 = 0

(0.8)(0.2) q0 2

=> q0 = 4103N / m Note: A12 is the area enclosed by the curved sheet and the line connecting stringers 1 and 2; A123 is the area enclosed by the lines connecting stringers 1, 2, and 3. Also note that the shear flow passes stringer 1 and, thus, does not produce any moment.

5.5.2

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The final shear flows are q12 = −25000 + q0 = −20897 N / m q23 = q0 = 4103 N / m q31 = q0 = 4103N / m --- ANS (c) Twist angle per unit length The equation of the twist angle per unit length is

1 q ds ∫ 2G A t We can obtain the twist angle per unit length as

θ=

(5.5.2)

0.2π ) + 2 × 4103 × (0.806) 1 q 51.54 2 θ= ds = = 6 ∫ π 2G A t 2(27 × 109 )( 0.22 + 0.2 × 0.8 / 2)(0.001) 5.17 × 10 8 −6 −4 = 9.97 × 10 rad / m = 5.7 × 10 deg/ m − 20896 × (

--- ANS (d) Shear center To determine the horizontal location of the shear center, we assume that the shear force V z acts through the shear center which is assumed to be located at a horizontal distance e to the right of stringer 1. We now rewrite the moment equation as (V z )(e) = 2 A12 q12 + 2 A123 q 23

π

= 2( (0.2) 2 )(−25000 + q 0 ) + 2 8 = −785.40 + 0.19q 0 = 5000e

(0.8)(0.2) q0 2

We can solve q 0 in terms of e , => q 0 = 26123.30e + 4103.44 So that the shear flows within each sheet are q12 = q '12 + q 0 = −25000 + q 0 = 26123.30e − 20896.56 q 23 = q' 23 + q 0 = q 0 = 26123.30e + 4103.44 q31 = q '31 + q 0 = q 0 = 26123.30e + 4103.44 Since the shear force V z passes through the shear center, the twist angle is equal to

1 q ds = 0 ∫ 2G A t q q 1 q12 => ( s12 + 23 s 23 + 31 s31 ) = 0 t t 2G A t

zero, i.e. θ =

5.5.3

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=>

1 0.2π [(26123.30e − 20896.56)( ) + 2 × (26123.30e + 4103.44)(0.806)] = 0 2 2G At => 50328.09e + 51.54 = 0 => e = −1 × 10 −3 m The negative sign means that the shear center is to the left of stringer 1. --- ANS

5.5.4

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5.6

Do Problem 5.5 for Vz = 5000 N and V y = 10,000 N .

Figure 5.33

Single-cell closed section

Solution: (a) Assume that the thin sheets are ineffective in bending. The centroid of this four-stringer section is

∑A y = ∑A i

yc

i

i

=

(10)(80) = 26.6667cm = 0.2667 m 3(10)

=

(10)(20) + (10)(10) = 10cm = 0.1m 3(10)

i

i

∑Az = ∑A i

zc

i

i

i

i

The distance is measured from stringer 2. Now we set up the (y, z) coordinate with the origin at the centroid. The moments of inertia are obtained as I y = ∑ Ai z i = 2(10 × 10 −4 )(0.1) 2 = 2 × 10 −5 m 4 2

i

I z = ∑ Ai y i = 2(10 × 10 −4 )(0.2667) 2 + (10 × 10 −4 )(0.8 − 0.2667) 2 2

i

= 4.2667 × 10 − 4 m 4

(b) Shear flows We can solve the problem by considering the two applied forces separately and then superposed the two solutions. Alternatively, we can calculate the shear flows by taking advantage of the fact that they are statically determinate shear flows.

5.6.1

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Denote the shear flows by q12 , q 23 , q 31 , respectively, in the sheet between stringers 1 and 2, stringers 2 and 3, and stringers 3 and 1, as shown in the figure below.

From the condition that the resultants of the shear flows must be equal to the applied shear forces, we have

∑F

= V y , => q 23 (0.8) − q31 (0.8) = 10000

(5.6.1)

= V z , => q 23 (0.1) + q31 (0.1) − q12 (0.2) = 5000

(5.6.2)

y

∑F

z

∑M

o

= moment about x-axis passing through o

π ⋅ 0.22

) + (q23 + q31 )(0.8)(0.1) = 5000 × 0 + 10000 × 0.1 4 Solving the equations (5.6.1) to (5.6.3), we obtain q12 = −15672 N / m q23 = 15577 N / m q31 = 3077 N / m => q12 (

(5.6.3)

--- ANS (c) Twist angle per unit length The equation for twist angle per unit length is

θ=

=

1 q ds ∫ 2G A t

− 15672 × (

(5.6.4)

0.2π ) + (15577 + 3077) × (0.806) 2

π

2(27 × 109 )( 0.22 + 0.2 × 0.8 / 2)(0.001) 8 10115 = = 1.97 × 10− 3 rad / m = 0.11o / m 5.168 × 106 --- ANS 5.6.2

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(d) Shear center Since shear center is independent of applied forces, it should be the same as in the solution for Problem 5.5. --- ANS

5.6.3

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5.7

Find the shear flow on the four-stringer section (Fig. 5.34) subjected to Vz = 5000 N . Assume that the thin sheets are ineffective in bending.

Figure 5.34

Unsymmetrical open section

Solution: (a) The centroid of this four-stringer section is located at

∑A y = ∑A i

yc

i

i

=

2(10)(80) = 32cm 2(15) + 2(10)

=

(15)(40) + (10)(20) = 16cm 2(15) + 2(10)

i

i

∑Az = ∑A i

zc

i

i

i

i

from stringer 2. Now set up the (y,z) coordinate system with the origin at the centroid.. The moments of inertia are obtained as I y = ∑ Ai z i = (15)(24 2 + 16 2 ) + (10)(4 2 + 16 2 ) 2

i

= 15200cm 4 I z = ∑ Ai y i = 2(15)(32) 2 + 2(10)(80 − 32) 2 2

i

= 76800cm 4 I yz = ∑ Ai y i z i = (15)[(−32)(24) + (−32)(−16)] i

+ (10)[(48)(−16) + (48)(4)] = −9600cm 4

(b) Shear flows Assume the shear force acts through the shear center and no torsion is produced.. The shear flow in the unsymmetrical thin-walled section is calculated using 5.7.1

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q s = −(k yV y − k yzV z )Q z − ( k zV z − k yzV y )Q y

where k y =

Iy I y I z − I yz

2

, kz =

Iz I y I z − I yz

2

(5.7.1) , k yz =

I yz I y I z − I yz

2

and Q z = ∫∫ ydA , Q y = ∫∫ zdA As

As

In this problem, we have ky =

kz =

k yz =

Iy I y I z − I yz

2

Iz I y I z − I yz

2

I yz I y I z − I yz

2

=

15200 = 1.413 × 10 −5 2 ( 15200 )( 76800 ) − ( −9600 )

=

76800 = 7.143 × 10 −5 2 ( 15200 )( 76800 ) − ( −9600 )

=

− 9600 = −0.893 × 10 −5 2 ( 15200 )( 76800 ) − ( −9600 )

Given Vz = 5000 N , V y = 0 , we have from (5.7.1) q s = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y = −4.4645 × 10 − 2 Q z − 35.7145 × 10 − 2 Q y

(5.7.2)

The shear flow contour direction is indicated in the figure below.

(1) For shear flow q1 Q z = (15)(−32) = −480cm 3 Q y = ( 15 )( 24 ) = 360cm 3 From equation (5.7.2) we have

q1 = −4.4645 × 10 −2 ( −480 ) − 35.7145 × 10 −2 ( 360 ) = −107.14 N / cm

5.7.2

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(Negative sign indicates the opposite direction) --- ANS (2) For shear flow q 2 Q z = 2(15)(−32) = −960cm 3

Q y = (15)(24 − 16) = 120cm 3 From equation (5.7.2) we have q 2 = −4.4645 × 10 −2 ( −960 ) − 35.7145 × 10 −2 ( 120 ) = 0

--- ANS (3) For shear flow q 3 Q z = 2(15)(−32) + (10)(48) = −480cm 3

Q y = (15)(24 − 16) + (10)(−16) = −40m 3 From equation (5.7.2) we have

q3 = −4.4645 × 10 −2 ( −480 ) − 35.7145 × 10 −2 ( −40 ) = 35.71 N / cm --- ANS

5.7.3

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5.8

Find the shear center ( y sc , z sc ) for the open section in Fig. 5.34.

Figure 5.34

Unsymmetrical open section

Solution: (a) Assume that the thin sheets are ineffective in bending. The centroid of this four-stringer section is located at

∑A y = ∑A i

yc

i

i

=

2(10)(80) = 32cm = 0.32m 2(15) + 2(10)

=

(15)(40) + (10)(20) = 16cm = 0.16m 2(15) + 2(10)

i

i

∑Az = ∑A i

zc

i

i

i

i

from stringer 2. Now set up the (y , z) coordinate system with the origin at the centroid as shown in Fig. 5.34. The moments of inertia are I y = ∑ Ai z i = (15 × 10 −4 )(0.24 2 + 0.16 2 ) + (10 × 10 −4 )(0.04 2 + 0.16 2 ) 2

i

= 1.52 × 10 − 4 m 4 I z = ∑ Ai y i = 2(15 × 10 −4 )(0.32) 2 + 2(10 × 10 −4 )(0.8 − 0.32) 2 2

i

= 7.68 × 10 − 4 m 4 I yz = ∑ Ai y i z i = (15 × 10 −4 )[(−0.32)(0.24) + ( −0.32)(−0.16)] i

+ (10 × 10 − 4 )[(0.48)(−0.16) + (0.48)(0.04)] = −0.96 × 10 − 4 m 4

(b) Shear flows Assume the shear force acts through the shear center and no torsion is produced. The shear flow in an unsymmetrical thin-walled section is calculated using the 5.8.1

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equation q s = −(k yV y − k yzV z )Q z − ( k zV z − k yzV y )Q y

where k y =

Iy I y I z − I yz

2

, kz =

Iz I y I z − I yz

2

(5.8.1) , k yz =

I yz I y I z − I yz

2

and Q z = ∫∫ ydA , Q y = ∫∫ zdA As

As

In this problem, we have

ky =

kz =

k yz =

Iy I y I z − I yz

2

=

1.52 × 10 4 = 1.4137 × 10 3 2 (1.52)(7.68) − (−0.96)

2

=

7.68 × 10 4 = 7.1429 × 10 3 2 (1.52)(7.68) − ( −0.96)

=

− 0.96 × 10 4 = −0.8929 × 10 3 2 (1.52)(7.68) − (−0.96)

Iz I y I z − I yz

I yz I y I z − I yz

2

Then equation (5.8.1) becomes q s = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y = −(1.4137V y + 0.8929V z ) × 10 3 Q z − (7.1429V z + 0.8929V y ) × 10 3 Q y

(5.8.2)

Assume that the shear center is located at ( y sc , z sc ) with respect to stringer 2 as shown in the figure above (1) For shear flow q1 Q z = (15 × 10 −4 )(−0.32) = −4.8 × 10 −4 m 3

Q y = (15 × 10 −4 )(0.24) = 3.6 × 10 −4 m 3 From equation (5.8.2) we have

5.8.2

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q1 = −( 1.4137V y + 0.8929V z ) × 10 3 Q z − ( 7.1429V z + 0.8929V y ) × 10 3 Q y = −( 1.4137V y + 0.8929V z )( −4.8 ) × 10 −1 − ( 7.1429V z + 0.8929V y )( 3.6 ) × 10 −1 = 0.357V y − 2.143V z (2) For shear flow q 2 Q z = 2(15 × 10 −4 )(−0.32) = −9.6 × 10 −4 m 3

Q y = (15 × 10 −4 )(0.24 − 0.16) = 1.2 × 10 −4 m 3 From equation (5.8.2) we have q 2 = −(1.4137V y + 0.8929V z ) × 10 3 Q z − (7.1429V z + 0.8929V y ) × 10 3 Q y = −(1.4137V y + 0.8929V z )(−9.6) × 10 −1 − (7.1429V z + 0.8929V y )(1.2) × 10 −1 = 1.25V y

(3) For shear flow q 3 Q z = 2(15 × 10 −4 )(−0.32) + (10 × 10 −4 )(0.48) = −4.8 × 10 −4 m 3

Q y = (15 × 10 −4 )(0.24 − 0.16) + (10 × 10 −4 )(−0.16) = −0.4 × 10 −4 m 3 From equation (5.8.2) we have q 3 = −( 1.4137V y + 0.8929V z ) × 10 3 Q z − ( 7.1429V z + 0.8929V y ) × 10 3 Q y = −( 1.4137V y + 0.8929V z )( −4.8 ) × 10 −1 − ( 7.1429V z + 0.8929V y )( −0.4 ) × 10 −1 = 0.714V y + 0.714V z Check balance of forces:

∑F

y

∑ Fz

= q 2 s 2 = (1.25V y )(0.8) = V y = − q1 s 1 + q 3 s 3

= −( 0.357V y − 2.143V z )( 0.4 ) + ( 0.714V y + 0.714V z )( 0.2 ) = V z

Satisfied!

(c) Shear center ( y sc , z sc ) (1) To determine y sc , we consider the case V y = 0 and Vz ≠ 0 , and consider moment about stringer 2.

∑ M = (q s ) s 3 3

2

= V z y sc

2

5.8.3

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=> y sc =

( 0.714308V z )( 0.2 )( 0.8 ) = 0.114 m Vz

(2) For z sc we consider the case V y ≠ 0 and Vz = 0 . We have

∑ M = (q s ) s 3 3

2

= −V y z sc

2

=> z sc = −

( 0.714292V y )( 0.2 )( 0.8 ) Vy

= −0.114 m

So the shear center is located at ( 0.114 m ,−0.114 m ) with respect to stringer 2. This says that the shear center is located below the sheet between stringers 2 and 3. --- ANS

5.8.4

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5.9 Find the shear center of the z section given by Fig. 4.8.

Figure 4.8

Thin-walled Z-section

Solution: (a) The centroid of this Z-section is obviously located at the midpoint of the vertical web. The moments of inertia are 2

2bt 3 th 3 tbh 2 t 3 b th 3 ⎛h⎞ + = + + I y = 2bt ⎜ ⎟ + 12 12 2 6 12 ⎝2⎠ 2

2tb 3 ht 3 2tb 3 t 3 h ⎛b⎞ + = + I z = 2bt ⎜ ⎟ + 12 12 3 12 ⎝2⎠

I yz = bt

2 bh ⎛ b ⎞⎛ h ⎞ tb h + bt ⎜ − ⎟⎜ − ⎟ = 22 2 ⎝ 2 ⎠⎝ 2 ⎠

(b) Shear flows Assume the shear force acts through the shear center and results without producing torsion. The shear flow in the unsymmetrical thin-walled section is calculated using q s = −(k yV y − k yzV z )Q z − ( k zV z − k yzV y )Q y

where k y =

Iy I y I z − I yz

2

, kz =

Iz I y I z − I yz

2

(5.9.1) , k yz =

I yz I y I z − I yz

2

and Q z = ∫∫ ydA , Q y = ∫∫ zdA

(5.9.2)

(5.9.3)

(1) For shear flow q1 Calculate shear flow q1 from the lower left end of the Z-section. 5.9.1

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Q z = ∫∫ ydA = −(b −

1 1 s1 )ts1 = ( s1 − b)ts1 2 2

h 1 Q y = ∫∫ zdA = (− )ts1 = − ths1 2 2 From equation (5.9.1) we have q1 ( s1 ) = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y

1 2 1 = −(k yV y − k yzV z )( ts1 − tbs1 ) − (k zV z − k yzV y )(− ths1 ) 2 2 Then the resultant force of shear flow q1 is b b 1 2 1 V1 = ∫ q1 ds1 = ∫ [−(k yV y − k yzV z )( ts1 − tbs1 ) − (k zV z − k yzV y )(− ths1 )]ds1 0 0 2 2 1 3 1 1 2 2 = [−(k yV y − k yzV z )( ts1 − tbs1 ) − (k zV z − k yzV y )(− ths1 )] |b0 6 2 4 1 1 = tb 3 (k yV y − k yzV z ) + tb 2 h(k zV z − k yzV y ) 3 4

(2) For shear flow q 2 1 Q z = ∫∫ ydA = − tb 2 2 1 1 Q y = ∫∫ zdA = − tbh + ts 2 ( s 2 − h) 2 2 From equation (5.9.1) we have q 2 ( s 2 ) = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y

1 1 1 1 2 = −(k yV y − k yzV z )(− tb 2 ) − (k zV z − k yzV y )(− tbh − ths 2 + ts 2 ) 2 2 2 2 Then the resultant force of shear flow q 2 is h

V2 = ∫ q 2 ds 2 0

h 1 1 1 1 2 = ∫ [−(k yV y − k yzV z )(− tb 2 ) − (k zV z − k yzV y )(− tbh − ths 2 + ts 2 )]ds 2 0 2 2 2 2 1 1 1 1 3 2 = [−(k yV y − k yzV z )(− tb 2 s 2 ) − (k zV z − k yzV y )(− tbhs 2 − ths 2 + ts 2 )] |0h 2 2 4 6 1 1 1 = tb 2 h(k yV y − k yzV z ) + ( tbh 2 + th 3 )(k zV z − k yzV y ) 2 2 12 (3) For shear flow q 3 1 1 2 Q z = ∫∫ ydA = − tb 2 + ts3 2 2

1 1 Q y = ∫∫ zdA = − tbh + ths3 2 2 From equation (5.9.1) we have 5.9.2

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q3 ( s3 ) = −(k yV y − k yzV z )Qz − (k zV z − k yzV y )Q y 1 1 2 1 1 = −(k yV y − k yzV z )(− tb 2 + ts3 ) − (k zV z − k yzV y )(− tbh + ths3 ) 2 2 2 2 Then the resultant force of shear flow q 3 is b

V3 = ∫ q 3 ds3 0

1 1 2 1 1 = ∫ [−(k yV y − k yzV z )(− tb 2 + ts 3 ) − (k zV z − k yzV y )(− tbh + ths3 )]ds3 0 2 2 2 2 1 1 3 1 1 2 = [−(k yV y − k yzV z )(− tb 2 s 3 + ts 3 ) − (k zV z − k yzV y )(− tbhs3 + ths 3 )] |b0 2 6 2 4 1 1 = tb 3 (k yV y − k yzV z ) + tb 2 h(k zV z − k yzV y ) 3 4 We find that V3 = V1 (c) Shear center ( y sc , z sc ) b

(1) To determine ysc , we let V y = 0 and Vz ≠ 0 The moment about the centroid by the shear force must be equal to that produced by the shear flow: ⎛h⎞

⎛h⎞

∑ M = V ⎜⎝ 2 ⎟⎠ − V ⎜⎝ 2 ⎟⎠ = V 1

3

z

y sc

c

Since V3 = V1 , we conclude y sc = 0 (2) To determine zsc , we consider the case V y ≠ 0 and Vz = 0 Thus, ⎛h⎞

⎛h⎞

∑ M = V ⎜⎝ 2 ⎟⎠ − V ⎜⎝ 2 ⎟⎠ = −V 1

3

y

z sc

c

=> z sc = 0 The shear center is at the centroid of the Z-section. --- ANS

5.9.3

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5.10

Find the shear flow in the two-cell section loaded as shown in Fig. 5.35 for Vz = 5000 N . Given G = 27GPa , find the twist angle θ .

Figure 5.35

Two-cell closed section

Solution: (a) Assume that the thin sheets are ineffective in bending. Because of symmetry, the centroid of this two-cell section is obviously located at the mid point of the vertical web. Set up the (y, z) coordinate system with the origin at the centroid. The moment of inertia with respect to y axis is I y = ∑ Ai z i = 2(10 × 10 −4 )(0.2 2 ) = 8 × 10 −5 m 4 2

i

(b) Shear flows

Cuts are made on the curved webs as shown in the above figure. Note that the cut section is basically reduced to a single vertical web and, thus, the shear flows are simply: ′ r =0 q'12l = q12 q'12 v =

Vz 5000 − = −12 ,500 N / m h 0 .4

So the shear flows in the original section are q12 v = q'12 v −q1 + q 2 = −12500 − q1 + q 2 q12l = q1 5.10.1

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q12 r = q 2 where q1 and q2 are the constant shear flows in the left and the right cells, respectively. Note that q1 and q2 are both assumed to be positive if counterclockwise. (1) Moment equation Take the moment about the centroid of this cross-section, we have Vz × 0 = 2 A1q1 + 2 A2 q2

1 where A1 = A2 = πh 2 = 0.0628 m 2 8 => q1 + q 2 = 0

(5.10.2)

(2) Compatibility equation

1 q ds ∫ 2G A t We have θ = θ1 = θ 2 , 1 1 πh πh [(q12l )( ) + (−q12v )(h)] = [(q12v )(h) + (q12 r )( )] 2 2 2G A1t 2G A2 t

θ=

=> 0.6283q1 − 0.6283q 2 = −10000 − 0.8q1 + 0.8q 2 => q1 − q2 = −7001( N / m )

(5.10.3)

Solving equations (5.10.2) and (5.10.3), we have q1 = −3500( N / m ) q2 = 3500( N / m ) Then the final shear flows are q12 v = −12500 − q1 + q2 = −5498 q12l = q1 = −3500 N / m

N/m

q12 r = q2 = 3500 N / m Negative sign means the actual direction is opposite to the assumed. --- ANS And the twist angle is q 1 [(−3500.67)(0.6283) + (5498.66)(0.4)] θ = θ1 = ds = ∫ 2(27 × 10 9 )(0.06283)(0.001) 2G A1 t = 0(rad / m) = 0o / m As expected, there is no twist angle produced since the vertical load is applied

through the shear center. --- ANS

5.10.2

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5.11

Find the shear flow of the structure with the cross-section given in Fig. 5.35 if the vertical force V z is applied at 20cm to the right of the stringers. Also find the corresponding angle of twist θ .

Figure 5.35

Two-cell closed section

Solution: (a) Assume that the thin sheets are ineffective in bending. Because of the symmetry to both y and z axis, the centroid of this two-cell section is obviously located at the center of the vertical web. So we can shift the y and z axis to match the centroid as a new origin of the new coordinate system. The moment of inertia with respect to y axis is I y = ∑ Ai z i = 2(10 × 10 −4 )(0.2 2 ) = 8 × 10 −5 m 4 2

i

(b) Shear flows

Assume both a cut on the circular webs, as shown in the above figure. So the shear flow in the vertical web can be obtained by VQ q'i = − z i (5.11.1) Iy i

where Qi = ∑ Ak z k is the first moment of stringer area Ak , and z k is the k =1

vertical distance from the centroid of Ak to the y-axis, and V z = 5000N . We obtain 5.11.1

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q '12 v = −

V z Q1 5000(10 × 10 −4 × 0.2) =− = −12500 N / m Iy 8 × 10 −5

So the shear flows on the thin-wall segments are q12 v = q'12 v −q1 + q 2 = −12500 − q1 + q 2 q12l = q1

q12 r = q 2 where q12l and q12 r are shear flows, respectively, in the left and right curved walls. q1 and q 2 are constant shear flows, in the left and right cells, respectively. Both are assumed to be counterclockwise. (1) Moment equation Take moment about the centroid of this cross-section, we have 1 V z (0.2) = 2 A1 q1 + 2 A2 q 2 , where A1 = A2 = πh 2 = 0.063m 2 8 => q1 + q2 = 7957.98( N / m )

(5.11.2)

(2) Compatibility equation

1 q ds ∫ 2G A t We have θ = θ1 = θ 2 , 1 1 πh πh [(q12l )( ) + (−q12v )(h)] = [(q12v )(h) + (q12 r )( )] 2 2 2G A1t 2G A2 t

θ=

=> 0.6283q1 − 0.6283q 2 = −10000 − 0.8q1 + 0.8q 2 => q1 − q 2 = −7001.33( N / m)

(5.11.3)

Solving equations (5.11.2) and (5.11.3), we have q1 = 478.33( N / m) q 2 = 7479.65( N / m) Then the complete solution for the shear flows is q12 v = −12500 − q1 + q 2 = −5498.66( N / m) q12l = q1 = 478.33( N / m)

q12 r = q 2 = 7479.65( N / m) Negative sign means the actual direction is opposite to the assumed. --- ANS The twist angle is 1 q [( 478.33 )( 0.6283 ) + ( 5498.66 )( 0.4 )] ds = θ = θ1 = ∫ 2G A1 t 2( 27 × 10 9 )( 0.06283 )( 0.001 ) = 7.37 × 10 − 4 rad / m = 0.042 o / m

--- ANS 5.11.2

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An alternative approach in solving the problem is to consider the shift of the vertical force to the center position as in Problem 5.10 in conjunction with a torque of 0.2Vz. Thus the shear flow produced by the off-center load can be considered as the superposition of the shear flows produced by the vertical load (applied at the center location) and the torque. Since the shear flows of the first problem have been solved by in Problem 5.10, we need only to solve the second torsion problem with a torque of 0.2Vz. Because of symmetry, the shear flow along the vertical web is zero and the constant shear flow along the circular wall is q=

T 2A

(5.11.4) 2

⎛h⎞ in which A = π ⎜ ⎟ = 0.1257 m2, and T = 0.2Vz = 1000 Nm ⎝2⎠

Thus, q = 3978

N /m

(counterclockwise)

and the complete shear flows are, after adding q to the shear flow solution for Problem 4.10. q12 v = −5498 + 0 = −5498 N / m q12l = −3500 + q = −3500 + 3978 = 478 N / m

q12 r = 3500 + q = 3500 + 3978 = 7478 N / m

5.11.3

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5.12

Solve Example 5.9 by assuming cuts on the webs between stringers 1 and 2 and stringers 6 and 3.

Figure 5.27

Assumed cuts and shear flow contours

Solution: (a) Because the section is symmetric with respect to y axis, the centroid is located on y axis. The moment of inertia with respect to y axis is I y = ∑ Ai z i = 2(1 + 2 + 3)(20 2 ) = 4800cm 4 2

i

(b) Shear flows

The shear flows in the cut section as shown in the left figure above are calculated first. We have VQ 4800(1)(−20) q' 23 = − z 1 = − = 20 N / cm Iy 4800 q '34 = −

4800(1 + 2)(−20) = 60 N / cm 4800

q ' 45 = −

4800(1 + 2 + 3)(−20) = 120 N / cm 4800

q '56 = −

4800[(1 + 2 + 3)(−20) + 3( 20)] = 60 N / cm 4800

5.12.1

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q ' 61 = −

4800[(1 + 2 + 3)( −20) + (3 + 2)(20)] = 20 N / cm 4800

Thus, the shear flows are given by q12 = q1 q 23 = q' 23 + q1 = 20 + q1 q36 = q1 − q 2 q34 = q '34 + q 2 = 60 + q 2 q 45 = q' 45 + q1 = 120 + q 2 q56 = q '56 + q 2 = 60 + q 2 q 61 = q ' 61 + q1 = 20 + q1 in which q1 and q2 are the two assumed constant shear flows in the left and right cells of the original closed section.. (1) Moment equation The moment produced by the shear force Vz must be equal to that produced by the shear flows. Taking moment about stringer 1, we have Vz × 0 = q23 (40)(40) + q36 (40)(40) + q34 (40)(40) + q45 (40)(80) => q1 + q2 = −160 N / cm

(5.12.2)

(2) Compatibility equation

1 q ds ∫ 2G A t We have θ = θ1 = θ 2 ,

θ=

1 (q12 )(40) (q23 )(40) (q36 )(40) (q61 )(40) [ ] + + + 0 .1 0 .1 0 .1 0 .1 2G A1 =

1 (q63 )(40) (q34 )(40) (q45 )(40) (q56 )(40) [ ] + + + 0 .1 0 .2 0 .2 0 .2 2G A2

=> 10q1 − 7q2 = 160 N / cm

(5.12.3)

Solving equations (5.12.2) and (5.12.3), we have q1 = −56.47 N / cm q2 = −103.53 N / m Then the shear flows are q12 = q1 = −56.47 N / cm q23 = 20 + q1 = −36.47 N / cm q36 = q1 − q2 = 47.06 N / cm q34 = 60 + q2 = −43.53 N / cm q 45 = 120 + q 2 = 16.47( N / cm) q56 = 60 + q2 = −43.53 N / cm q61 = 20 + q1 = −36.47 N / cm 5.12.2

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The negative sign means the actual shear flow is opposite to the assumed direction. --- ANS

5.12.3

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5.13

A thin-walled box beam is obtained by welding the cut of the section shown in Fig. 5.30. Find the shear flow produced by a vertical shear force Vz = 1000N applied at 100mm to the right of the vertical wall that contains the original cut.

Figure 5.30

Thin-walled section with a side cut

Solution: The shear flow in the welded section loaded by the off center shear force Vz is of interest. To solve the problem, we first make a fictitious cut at the original gap of the wall as shown in Fig. 5.30. The centroid is at the center of the section for both welded and the cut sections The moment of inertia about y axis is 1 [(0.1 + 0.002)(0.2 + 0.002) 3 − (0.1 − 0.002)(0.2 − 0.002) 3 ] 12 = 6.6676 × 10 −6 m 4 Iy =

(a) Consider the shear flow in the cut section first. Setup the shear flow contours in the thin wall segments as shown in the figure below. Since this is an open section, the shear flow can be calculated using the formula VQ q'i = − z i Iy (1) On s1 : 0 ~ 0.1m s 2 Q = ∫∫ zdA = As z c = ( s1t )( 1 ) = 0.001s1 2 As

VQ 1000 × 0.001s1 2 q1 = − z = − = −1.5 × 10 5 s1 −6 Iy 6.6676 × 10 2

(2) On s 2 : 0 ~ 0.1m 5.13.1

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Q = 0.001(0.1) 2 + 0.002 s 2 (0.1) = 10 −5 + 2 × 10 −4 s 2 q2 = −

Vz Q 1000 × (10 −5 + 2 × 10 −4 s 2 ) =− = −1500 − 3 × 10 4 s 2 −6 Iy 6.6676 × 10

(3) On s 3 : 0 ~ 0.1m Q = 0.001(0.1) 2 + 0.002(0.1) 2 + 0.002 s 3 (0.1 − 0.5s3 ) = 3 × 10 −5 + 2 × 10 − 4 ( s 3 − 5s3 ) 2

1000 × [3 × 10 −5 + 2 × 10 −4 ( s 3 − 5s 3 )] Vz Q q3 = − =− Iy 6.6676 × 10 − 6 2

= −4500 − 30000 s 3 + 150000 s 3

2

(4) On s 4 : 0 ~ 0.1m Q = ∫∫ zdA = As z c = ( s 4 t )( As

s4 2 ) = −0.001s 4 2

V Q 1000 × 0.001s 4 2 q4 = − z = = 1.5 × 10 5 s 4 −6 Iy 6.6676 × 10 2

(5) On s 5 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002 s 5 (0.1) = −10 −5 − 2 × 10 −4 s 5

q5 = −

V z Q 1000 × (10 −5 + 2 × 10 −4 s5 ) = 1500 + 3 × 10 4 s 5 = Iy 6.6676 × 10 −6

(6) On s 6 : 0 ~ 0.1m Q = −0.001(0.1) 2 − 0.002(0.1) 2 − 0.002 s 6 (0.1 − 0.5s 6 ) = −3 × 10 −5 − 2 × 10 − 4 ( s 6 − 5s 6 ) 2

V z Q 1000 × [3 × 10 −5 + 2 × 10 −4 ( s 6 − 5s 6 )] = Iy 6.6676 × 10 −6 2

q6 = −

= 4500 + 30000 s 6 − 150000 s 6

2

(b) The shear flow in the welded section is that in the cut section plus a constant shear flow q 0 along the contour of the wall. This unknown constant shear flow is needed to produce the same moment as produced by the vertical shear force Vz = 1000N applied at 100mm to the right of the vertical wall that contains the original cut. 5.13.2

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Take moment about the out-of-plane axis that passes through the upper right corner of the thin-walled section. The moment produced by the shear flow must be equal to the moment produced by the applied shear force about the same axis, i.e.,

∑M =V y z

1000 × 0.1 = [ ∫ q3 ds3 − ∫ q6 ds 6 ] × 0.1 − [ ∫ q5 ds5 ] × 0.2 0.1

= [ ∫ (−4500 − 30000s3 + 150000s3 + q 0 )ds3 2

0

0.1

− ∫ (4500 + 30000s 6 − 150000s 6 − q 0 )ds 6 ] × 0.1 2

0

0.1

− [ ∫ (1500 + 3 × 10 4 s5 − q 0 )ds5 ] × 0.2 0

=> 100 = −170 + 0.04q0 => q 0 = 6750 N / m (c) Shear flows in the cross-section are shown q1 = 6750 − 1.5 × 10 5 s1

2

q 2 = 5250 − 3 × 10 4 s 2

q3 = 2250 − 30000s3 + 150000s3 q 4 = −6750 + 1.5 × 10 5 s 4

2

2

q 5 = −5250 + 3 × 10 4 s 5

q6 = −2250 + 30000s 6 − 150000s 6

2

--- ANS

5.13.3

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5.14

Show that the shear center for the section of Fig. 5.36 is at a distance a ( a + αb ) e= (a + b)(1 + α ) to the left of stringer 1.

Figure 5.36

Four-stringer thin-walled section

Solution: (a) Assume that the thin sheets are ineffective in bending. The centroid of this four-stringer section is located at

∑A y = ∑A i

yc

i

i

=

2(αA)(a ) αa = 2( A + αA) 1 + α

=

( A + αA)(b) b = 2( A + αA) 2

i

i

∑Az = ∑A i

zc

i

i

i

i

relative to stringer 3. We set up the (y,z) coordinate system with the origin placed at the centroid as shown in Fig. 5.36. The moments of inertia are b 1+α 2 2 I y = ∑ Ai z i = 2( A + αA)( ) 2 = b A 2 2 i

αa 2 αa 2 2αa 2 A 2 ) + 2(αA)(a − ) = I z = ∑ Ai y i = 2( A)( 1+α 1+α 1+α i (b) Shear flows Since this cross-section is symmetric with respect to y axis, the shear center is located on the y axis. Hence it is only necessary to determine the y position of the shear center. We can consider a fictitious cut section with shear flow q ' plus the existing constant shear flow q 0 . (1) First, calculate the shear flows by assuming a cut in the wall between stringers 1 and 4. Then 5.14.1

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q' 41 = 0 b V z (αA)( ) VQ 2 = − αV z q'12 = − z 1 = − 1+ α 2 Iy (1 + α )b b A 2 b V z (1 + α ) A( ) VQ 2 = − Vz q' 23 = − z 2 = − 1+α 2 Iy b b A 2 b b V z [(1 + α ) A( ) + A(− )] VQ 2 2 = − αV z q '34 = − z 3 = − 1+α 2 Iy (1 + α )b b A 2 (2) The total shear flow and their resultant forces are,

αaV z + q0 a (1 + α )b q 23 = q ' 23 + q0 => V2 = q 23 b = −V z + q 0 b q12 = q '12 + q 0 => V1 = q12 a = −

αaV z + q0 a (1 + α )b q 41 = q' 41 + q 0 = q 0 => V4 = q 41b = q 0 b Assume that the force V z is acting through the shear center and, thus, twist angle is produced. Consequently, we require 1 q θ= ds = 0 ∫ 2G A t αaV z αaV z => (− + q 0 a ) + ( −V z + q 0 b) + (− + q 0 a ) + ( q 0 b) = 0 (1 + α )b (1 + α )b 2αa + (1 + α )b => q 0 = Vz 2(1 + α )(a + b)b q34 = q '34 + q0 => V3 = q 34 a = −

5.14.2

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αaV z (1 − α ) a + q0 a = Vz (1 + α )b 2(1 + α )(a + b) − 2 a − b − αb V2 = q 23 b = −V z + q 0 b = Vz 2(1 + α )(a + b)

V1 = q12 a = −

αaV z (1 − α ) a + q0 a = Vz (1 + α )b 2(1 + α )(a + b) 2αa + (1 + α )b V4 = q 41b = q 0 b == Vz 2(1 + α )(a + b) V3 = q34 a = −

Take moment about stringer 1.The moment equivalence condition gives Vz × ysc = V2 × a + V3 × b = =−

− 2a 2 − ab − αab + ab − αab Vz 2(1 + α )(a + b)

a ( a + αb ) Vz (1 + α )(a + b)

=> y sc = −

a ( a + αb ) (1 + α )(a + b)

The negative sign indicates that the shear center is to the left of stringer 1.It is noted that if α = 1 , then the shear center is at the center of the square section as expected. --- ANS

5.14.3

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5.15

Find the shear flow in the two-cell thin-walled section for Vz = 5000 N shown in Fig. 5.37. Also determine the shear center. Assume thin sheets to be ineffective in bending.

Figure 5.37

Assumed cuts and shear flow contours

Solution: (a) The centroid of this four-stringer section is

∑A y = ∑A i

yc

i

i

=

2 A(40) = 26.67cm 3A

=

A(20) = 6.67cm 3A

i

i

∑Az = ∑A i

zc

i

i

i

i

relative to stringer 1. Set up the (y,z) coordinate system with the origin place at the centroid. The moments of inertia are 20 40 2 I y = ∑ Ai z i = 2(10)(− ) 2 + (10)( ) 2 = 2666.67cm 4 3 3 i I z = ∑ Ai y i = 2(10)( 2

i

40 2 80 ) + (10)(− ) 2 = 10666.67cm 4 3 3

I yz = ∑ Ai y i z i = 2666.67cm 4 i

(b) Shear flows First we make fictitious cuts in the sheet between stringers 1 and 2 and the vertical sheet between stringers 2 and 3. The shear flow in the unsymmetrical thin-walled section is calculated using the formula below. q s = −(k yV y − k yzV z )Q z − ( k zV z − k yzV y )Q y

(5.15.1)

5.15.1

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where k y =

Iy I y I z − I yz

2

, kz =

Iz I y I z − I yz

2

, k yz =

I yz I y I z − I yz

2

and Q z = ∫∫ ydA , Q y = ∫∫ zdA As

(5.15.2)

(5.15.3)

As

In this problem, we have

ky =

k yz =

Iy I y I z − I yz

2

I yz I y I z − I yz

2

Iz

= 1.25 × 10 − 4 cm − 4 ,

kz =

= 1.25 × 10 − 4 cm − 4 ,

Vz = 5000 N ,

I y I z − I yz

2

= 5 × 10 − 4 cm − 4

Vy = 0

Then equation (5.15.1) becomes

q s = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y = 0.625Q z − 2.5Q y

(5.15.4)

(1) For shear flow q'12 Q z = (10)( −

80 ) = −266.67cm 3 3

Q y = (10)(−

20 ) = −66.67cm 3 3

From equation (5.15.4) we have q'12 = 0.625(−266.67) − 2.5(−66.67) = 0 (2) For shear flow q ' 23c (in the curved sheet) Q z = −133.33cm 3

Q y = −133.33cm 3 From equation (5.15.4) we have q ' 23c = 0.625(−133.33) − 2.5(−133.33) = 250( N / cm) The total shear flow is obtained by adding the constant counterclockwise shear flows q1 and q 2 in the left and right cells, respectively. Thus, q12 = q'12 + q1 = q1 q 23v = q1 − q 2 (vertical sheet) q 23c = q ' 23c + q 2 = 250 + q 2 q31 = q1

5.15.2

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(1) Moment equation Take moment about stringer 1, We have 1 1 π V z × 0 = 2q 23c [( 20 2 ) + × 20 × 40] + 2q 23v × × 20 × 40 8 2 2 => 800q1 + 314.16q2 = −278540

(5.15.5)

(2) Compatibility equation

1 q ds ∫ 2G A t We have θ = θ1 = θ 2 , 1 1 [40q12 + 20q 23v + 44.7214q31 ] = [20(−q 23v ) + 10πq 23c ] 2G A1t 2G A2 t

θ=

=> 155.65q1 − 155.93q 2 = 20000

(5.15.6)

Solve equations (5.15.5) and (5.15.6), we have q1 = −210.78( N / cm) q 2 = −349.88( N / cm) Then the shear flows in each sheet are q12 = −210.78( N / cm) q 23v 139.10( N / cm) q 23c = −99.88( N / cm) q31 − 210.78( N / cm) Negative sign means the actual direction of the shear flow is opposite to the assumed. --- ANS Check:

∑F

y

= 40(−210.78) + 40(210.78) = 0

(OK)

z

= 20(139.10) + 20(−99.88) + 20(210.78) = 5000 N

(OK)

∑F

(c) Shear center Assume the shear force is acting through the shear center which is assumed to be at a distance e y to the right of stringer 3. (1) Moment equation Take moment about stringer 3. We have 1 π 5000(e y ) = 2q12 (40)(20) + 2q23c ( 20 2 ) 2 8

5.15.3

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=> 800 q1 + 314.16q 2 = 5000e y − 25000π

(5.15.7)

(2) Compatibility equation

1 q ds ∫ 2G A t We have θ = θ1 = θ 2 = 0 when the force acts through the shear center, then 1 [40q12 + 20q 23v + 44.7214q31 ] = 0 2G A1t

θ=

=> q 2 = 5.236q1 1 [20(−q 23v ) + 10πq 23c ] = 0 and 2G A2 t

(5.15.8)

=> 51.416q 2 − 20q1 + 7853.98 = 0

(5.15.9)

Solving equations (5.15.8) and (5.15.9), we have q1 = −31.52 ( N / cm) q 2 = −165.01 ( N / cm)

Substituting back to equation (5.15.7), we obtain e y = 0.3cm (to the right of stringers 2 and 3)

--- ANS

5.15.4

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5.16

Find the shear flow in the five-stringer thin-walled section produced by the loads shown in Fig. 5.38.

Figure 5.38

Three-cell closed section

Solution: (a) The centroid of this four-stringer section is at

∑A y = ∑A i

yc

i

i

=

2 A( h) + A(2h) = 0.8h = 32cm 5A

=

2 A(h) = 16cm 5A

i

i

∑Az = ∑A

i i

zc

i

i

i

relative to stringer 1. The origin of the coordinate system (y, z) is at the centroid. The moments of inertia are I y = ∑ Ai z i = 3(10)( −16) 2 + 2(10)( 24) 2 = 19200cm 4 2

i

I z = ∑ Ai y i = 2(10)(−32) 2 + 2(10)(8) 2 + (10)( 48) 2 = 44800cm 4 2

i

I yz = ∑ Ai yi z i = 10[(−32)(−16) + (8)(−16) + (48)(−16) + (8)(24) + (−32)(24)] i

= −9600cm 4 (b) Shear flows First we make three fictitious cuts as shown in Fig.5.38. The shear flow in the unsymmetrical thin-walled section is calculated with the formulas: q s = −(k yV y − k yzV z )Q z − ( k zV z − k yzV y )Q y

(5.16.1)

5.16.1

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where k y =

Iy I y I z − I yz

2

, kz =

Iz I y I z − I yz

2

, k yz =

I yz I y I z − I yz

2

and Q z = ∫∫ ydA , Q y = ∫∫ zdA As

As

In this problem, we have

ky =

k yz =

Iy I y I z − I yz

2

I yz I y I z − I yz

2

= 2.5 × 10 −5 cm −4 , k z =

Iz I y I z − I yz

2

= 5.8333 × 10 −5 cm − 4

= −1.25 × 10 −5 cm − 4 , Vz = 5000 + 3000 = 8000 N , V y = 0

Then equation (5.16.1) becomes

q s = −(k yV y − k yzV z )Q z − (k zV z − k yzV y )Q y = −0.1Q z − 0.4667Q y

(5.16.2)

(1) For shear flow q'12 Q z = (10)(−32) = −320cm 3

Q y = (10)(−16) = −160cm 3 From equation (5.16.2) we have q'12 = −0.1(−320) − 0.4667(−160) = 106.67( N / cm) (2) For shear flow q ' 23 Q z = (10)(−32 + 8) = −240cm 3

Q y = (10)(−16 − 16) = −320cm 3 From equation (5.16.2) we have q'23 = −0.1(−240) − 0.4667(−320) = 173.33 N / cm (3) For shear flow q ' 34 Q z = (10)(−32 + 8 + 48) = 240cm 3

Q y = (10)(−16 − 16 − 16) = −480cm 3 From equation (5.16.2) we have q '34 = −0.1(240) − 0.4667(−480) = 200 N / cm (4) For shear flow q ' 45 Q z = 240 + (10)(8) = 320cm 3 5.16.2

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Q y = −480 + (10)(24) = −240cm 3 From equation (5.16.2) we have q '45 = −0.1(320) − 0.4667(−240) = 80 N / cm

Now three counterclockwise constant shear flows q1 , q 2 , and q 3 must be added to the three cells from left to right, respectively. of the thin walls are q12 = q'12 + q 2 = 106.67 + q 2 q 23 = q 23 + q3 = 173.33 + q3 q34 = q '34 + q 3 = 200 + q3 q 45 = q ' 45 + q 2 = 80 + q 2 q51c = q1 (curved sheet) q15v = q1 − q 2 (vertical sheet) q 24 = q 2 − q 3

Thus, the shear flows on the segments

(1) Moment equation Take the moment about stringer 1. We have

π

3000(40) = q51c (2)( 40 2 ) + q34 ( 40)(60) + q 24 ( 40)(40) + q 45 ( 40)(40) 8 => 10πq1 + 80q 2 + 40q3 = −16200 (5.16.5)

(2) Compatibility equation

1 q ds ∫ 2G A t We require θ1 = θ 2 , 1 1 [20πq51c + 40q15v ] = [40q12 + 40q 24 + 40q 45 − 40q15v ] 2G A1t 2G A2 t => 7.5465q1 − 6.546q 2 + q3 = 186.67 (5.16.6) and θ 2 = θ 3 , 1 1 [40q12 + 40q 24 + 40q 45 − 40q15v ] = [40q 23 + 40 2q34 − 40q 24 ] 2G A2 t 2G A3t => q1 − 6q 2 + 7.828q3 = −725.69 (5.16.7)

θ=

Solving equations (5.15.5) to (5.15.7), we have q1 = −42.437 N / cm q2 = −102.801 N / cm 5.16.3

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q3 = −166.068 N / cm Then the complete shear flows are q12 = 106.67 + q 2 = 3.869( N / cm) q 23 = 173.33 + q3 = 7.262( N / cm) q34 = 200 + q3 = 33.932( N / cm) q 45 = 80 + q 2 = −22.801( N / cm) q 51c = q1 = −42.437( N / cm) (curved sheet) q15v = q1 − q 2 = 60.364( N / cm) (vertical sheet) q 24 = q 2 − q3 = 63.267( N / cm) The negative sign means that the actual shear flow direction is opposite to the assumed direction. The assumed directions are shown in Fig.5.38 . --- ANS

5.16.4

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6.1

Derive the distortional energy expression for plane stress.

Solution: The stress-strain relation can be expressed as σ 0 = Kε 0

(6.1.1)

1 where σ 0 = (σ xx + σ yy + σ zz ) is the average stress, ε 0 = ε xx + ε yy + ε zz is the 3

dilatation for small strain, and K =

E is the bulk modulus. 3(1 − 2υ )

For the state of plane stress, we have σ zz = 0 and ε zz = −υ (ε xx + ε yy ) 1 1 Thus, σ 0 = (σ xx + σ yy + σ zz ) = (σ xx + σ yy ) 3 3

and ε 0 = ε xx + ε yy + ε zz = (1 − υ )(ε xx + ε yy )

The strain energy density associated with the volume dilatation is given by, 1 1 1 Wv = σ 0 ε 0 = σ 02 = (σ xx + σ yy ) 2 2 2K 18 K

(6.1.2)

The total strain energy density for plane stress is W =

1 (σ xx ε xx + σ yy ε yy + τ xy γ xy ) 2

(6.1.3)

1 1 2 2 = τ xy 2 (σ xx + σ yy − 2υσ xxσ yy ) + 2E 2G

where ε xx =

1 1 1 (σ xx − υσ yy ) , ε yy = ( −υσ xx + σ yy ) , and γ xy = τ xy E E G

The strain energy associated with distortional deformation is obtained as Wd = W − Wv Substituting (6.1.2) and (6.1.3) in W and Wv, and using G = Wd =

E we have 2(1 + υ )

1 1 1 2 2 (σ xx + σ yy − 2υσ xxσ yy ) + τ xy 2 − (σ xx + σ yy ) 2 2E 2G 18 K

=

1 1 + υ 2 1 − 2υ 2 2 2 2 (σ xx + σ yy − 2υσ xxσ yy ) + τ xy − (σ xx + σ yy + 2σ xxσ yy ) 2E 6E E

6.1.1

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=

1 1+υ 2 2 2 2 2 τ xy [3σ xx + 3σ yy − 6υσ xxσ yy − (1 − 2υ )(σ xx + σ yy + 2σ xxσ yy )] + E 6E

=

1 1+υ 2 2 2 τ xy [( 2 + 2υ )σ xx + (2 + 2υ )σ yy − (2 + 2υ )σ xxσ yy ] + E 6E

=

1+υ 2 2 2 [σ xx + σ yy − σ xxσ yy + 3τ xy ] 3E

=

1 2 2 2 [(σ xx − σ yy ) 2 + σ xx + σ yy + 6τ xy ] 12G

We can rewrite this equation in Wd =

1 J2 2G

1 2 2 2 where J 2 = [(σ xx − σ yy ) 2 + σ xx + σ yy + 6τ xy ] 6

6.1.2

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6.2

A thin-walled hollow sphere 2 m in diameter is subjected to internal pressure p 0 . The wall thickness is 5 mm and the yield stress of the material is 250 MPa. Use both Tresca and von Mises yield criteria to determine the maximum internal pressure p 0 that does not cause yielding.

Solution: The stresses caused by the internal pressure p 0 is

σ xx = σ yy

1 p 0 ( πD 2 ) p D 2 p0 4 = = 0 = = 100 p 0 , τ xy = 0 , σ zz = 0 (πD)t 4t 4(0.005)

where x and y are orthogonal axes in the directions tangential to the surface at the point of interest, and z is perpendicular to the surface at the same point. There are only normal stresses presnt. Thus,

σ 1 = σ xx = 100 p 0 , σ 2 = σ yy = 100 p 0 , σ 3 = 0 (1) Tresca yield criterion Yielding occurs when

σ1 − σ 2 ≥ σY

(6.2.1a)

σ 2 − σ 3 ≥ σY

(6.2.1b)

σ 3 − σ1 ≥ σY

(6.2.1c)

After substituting the stress values, all three inequalities become one: 100 p0 ≥ σ Y = 250 MPa . Thus, the maximum internal pressure p 0 that does not cause yielding is p 0 = 2.5MPa --- ANS (2) von Mises yield criterion von Mises yiled criterion is 1 1 [( σ 1 − σ 2 )2 + ( σ 2 − σ 3 )2 + ( σ 3 − σ 1 )2 ] = σ Y 2 6 3 or in plane stress form ( σ 3 = 0 ) J2 =

J2 =

1 1 1 [( σ 1 − σ 2 )2 + σ 12 + σ 2 2 ] = ( σ 12 − σ 1σ 2 + σ 2 2 ) = σ Y 2 6 3 3

(6.2.2)

(6.2.2a)

Substituting the values to equation (6.2.2) (or (6.2.2a)), we have 2( 100 p0 )2 = 2σ Y 2 = 2( 250 )2 ,

6.2.1

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so the maximum internal pressure p 0 that does not cause yielding is p 0 = 2.5MPa --- ANS Both criteria give the same maximum pressure of p 0 = 2.5MPa .

6.2.2

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6.3

Consider the problem of Example 6.2. Find the maximum p 0 without causing yielding if N = −5 × 10 6 N (compression).

Solution: From example 6.2, we have yield stress σ Y = 280 MPa , the radius of the thin-walled hollow cylinder is a = 1m , and its thickness is t = 5 × 10 −3 m . N −5 Axial stress: σ xx = = = −159.16 MPa (2πa )t ( 2π )(1)(5 × 10 −3 ) Shear stress: τ xy = 0

p 0 ( 2a ) L p 0 a p0 = = = 200 p0 2tL t 0.005 By von Mises yielded criterion for plane stress, Hoop stress: σ yy =

1 1 2 2 2 2 J 2 = [(σ xx − σ yy ) 2 + σ xx + σ yy + 6τ xy ] = σ y 6 3

(6.3.1)

Substituting the values, we have (200 p 0 + 159.16) 2 + (200 p 0 ) 2 + (159.16) 2 = 2σ y = 2(280) 2 2

80000 p0 + 63664 p0 − 106136.2 = 0 2

=> p 0 = 0.82Mpa , p 0 = −1.62 Mpa Since inner pressure p 0 should be positive, the negative solution is neglected. Thus, the maximum p 0 without causing yielding is p 0 = 0.82Mpa . --- ANS

6.3.1

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6.4

An aluminum alloy 2024-T651 (see Table 6.1) panel is subjected to biaxial loading as shown in Fig. 6.24. Assume that σ 1 = 300MPa and σ 2 can be either tension or compression. Find the maximum values of σ 2 in tension and compression that the panel can withstand before yielding according to von Mises yield criterion.

σ2

σ1

σ1

σ2 Figure 6.24

Material under biaxial stress

Solution: From Table 6.1, we have the yield stress σ Y = 415MPa for aluminum 2024-T651. By von Mises yield criterion for plane stress, 1 1 ( σ 1 2 − σ 1σ 2 + σ 2 2 ) = σ Y 2 3 3 Substituting σ 1 = 300MPa and σ Y = 415MPa , we have

(6.4.1)

300 2 − 300σ 2 + σ 2 = 415 2 2

=> σ 2 − 300σ 2 − 82225 = 0 , solve for σ 2 2

=> σ 2 = 473.61MPa or σ 2 = −173.61MPa The maximum values of σ 2 is In tension: σ 2 = 473.61MPa In compression: σ 2 = 173.61MPa This solution indicates that if loads in both directions are tensile (or compressive), it would be more difficult to yield the material than if one is tensile and the other is compressive. --- ANS 6.4.1

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6.5

Find the total strain energy release rate for the split beam loaded as shown in Figs. 6.25 and 6.26.

2

3

1

Figure 6.25

Split beam subjected to shear force

2

3

1

Figure 6.26

Split beam subjected to extension and bending

Solution: (1) Figure 6.25 The strain energy stored in the beam due to a bending moment M is M2 dx 0 2 EI For segment 1, the strain energy is U =∫

L

U1 = ∫

a

0

( Px) 2 P 2a3 2P 2a3 dx = = , 2 EI1 6 EI1 Eth 3

(6.5.1)

where I 1 =

th 3 12

For segment 2, U2 = 0 For segment 3, U3 = ∫

L

a

( Px) 2 P 2 ( L3 − a 3 ) P 2 ( L3 − a 3 ) t (2h) 3 2th 3 dx = = , where I = = 3 2 EI 3 6 EI 3 12 3 4 Eth 3

The total strain energy stored in the entire split beam is 2 P 2 a 3 P 2 ( L3 − a 3 ) U = U1 + U 2 + U 3 = + Eth 3 4 Eth 3 The strain energy release rate is 6.5.1

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1 dU 1 6 P 2 a 2 3P 2 a 2 21 P 2 a 2 = ( − ) = t da t Eth 3 4 Et 2 h 3 4 Eth 3

G=

--- ANS (2) Figure 6.26 Assume the axial load acting on segment 1 is located at the distance of

h from the 2

center of the entire split beam. The strain energy stored in the beam due to an axial force P is P2L 2 EA For segment 1, the strain energy is U=

P2a P2a = , 2 EA1 2 Eth

U1 =

(6.5.2)

where A1 = th

For segment 2, U2 = ∫

a

0

2

2

(M 0 ) 2 M a 6M 0 a th 3 , where I = dx = 0 = 2 12 2 EI 2 2 EI 2 Eth 3

For segment 3, the load consists of an axial force of P and bending moment of M0 + P ×

h . Strain energy stored in segment 3 is 2

U3 = =

P ( L − a) +∫ a 2 EA3 2

L

h (M 0 + P ) 2 2 dx 2 EI 3

,

P ( L − a ) 3M 0 ( L − a) 3M 0 Ph( L − a ) 3P h ( L − a ) + + + 4 Eth 4 Eth 3 4 Eth 3 16 Eth 3 2

2

2

2

t (2h) 3 2th 3 where A3 = 2th , and I 3 = = 12 3 The total strain energy stored in the entire split beam is U = U1 + U 2 + U 3 P 2 a 6 M 0 a P ( L − a ) 3M 0 ( L − a ) 3M 0 Ph( L − a ) 3P 2 h 2 ( L − a ) = + + + + + 2 Eth 4 Eth Eth 3 4 Eth 3 4 Eth 3 16 Eth 3 The strain energy release rate is 2

2

2

2

6M 0 3M 0 3M 0 Ph 3P 2 1 dU 1 P 2 P2 ) = ( + − − − − t da t 2 Eth Eth 3 4 Eth 4 Eth 3 4 Eth 3 16 Eth 2 3M 0 P 1 P2 21 M 0 = + − 2 3 2 16 Et h 4 Et h 4 Et 2 h 2

G=

--- ANS

6.5.2

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6.6

Consider the split beam with loading shown in Fig. 6.27. Loadings in both Fig.6.11 and 6.27 are antisymmetric, and both are mode II fracture problems. For the same value of P, which loading is more efficient in cracking the beam? Assume that the beam dimensions and the elastic properties are E = 70GPa , υ = 0.3 a = 10 × 10 −2 m ,

t = 2 × 10 −2 m

L = 15 × 10 −2 m ,

h = 1 × 10 −2 m

2

3

1

Figure 6.27

Split beam subjected to shear forces

2

3

1

Figure 6.11

Split beam subjected to horizontal forces

Solution: (1) Figure 6.11 The strain energy stored in the beam due to an axial load P is P2L 2 EA The strain energy stored in the beam due to a bending moment M is U=

M2 U =∫ dx 0 2 EI For segments 1 and 2, the strain energy is L

U1 = U 2 =

P2a P2a = , 2 EA1 2 Eth

(6.6.1)

(6.6.2)

where A1 = th

6.6.1

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For segment 3, the two axial forces are assumed to be completely cancelled out by themselves and only the unbalanced couple Ph is taken up by this segment of beam. U3 = ∫

L

a

( Ph) 2 P 2 h 2 ( L − a ) 3P 2 ( L − a ) dx = = , 2 EI 3 2 EI 3 4 Eth

where I 3 =

t (2h) 3 2th 3 = 12 3

The total strain energy stored in the entire split beam is P2a 3P 2 ( L − a ) U = U 1 + U 2 + U 3 = 2( )+ 2 Eth 4 Eth The strain energy release rate is 1 dU 1 P 2 3P 2 1 P2 = ( − )= t da t Eth 4 Eth 4 Et 2 h (2) Figure 6.27 For segment 1 and 2, the strain energy is G6.11 =

U1 = U 2 = ∫

a

0

( Px) 2 P 2 a 3 2P 2 a 3 dx = = , 2 EI 1 6 EI 1 Eth 3

where I 1 =

th 3 = I2 12

For segment 3,

U3 = ∫

L

a

2 2 L 2P x (2 Px) 2 P2 dx = ∫ dx = ( L3 − a 3 ) , 3 3 a 2 EI 3 Eth 2th E( ) 3

t (2h) 3 2th 3 = 12 3 The total strain energy stored in the entire split beam is where I 3 =

2P 2 a 3 P2 U = U 1 + U 2 + U 3 = 2( )+ ( L3 − a 3 ) 3 3 Eth Eth The strain energy release rate is 1 dU 1 12 P 2 a 2 P2 9P 2 a 2 2 = [ + ( − 3 a )] = t da t Eth 3 Eth 3 Et 2 h 3 (3) Comparison G6.27 =

G6.27 G6.11

9P 2 a 2 2 3 a2 10 = Et 2h = 36 2 = 36( ) 2 = 3600 1 h P 2 4 Et h

It is obvious to see that the loading in Fig. 6.27 produces more energy release and is much more efficient in cracking the beam. --- ANS

6.6.2

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6.7

To further split the beam of Fig. 6.27, a rigid pin of diameter d = 0.5cm is inserted as shown in Fig. 6.28. How far does one have to drive the cylinder in order to split the beam? Assume a plane strain fracture condition with

K Ic = 50MPa m .

Figure 6.28

Split beam opened by a cylinder

Solution: Use the beam dimensions and the material properties in Problem 6.6.

E = 70GPa , L = 15 × 10 −2 m ,

υ = 0.3

a = 10 × 10 −2 m ,

t = 2 × 10 −2 m

h = 1 × 10 −2 m

The vertical displacement of each split beam at the position of the cylinder is

δ =

d = 0.25cm . This is an approximate value because the exact contact points may 2

not be diametric. Consider the upper leg as a cantilever beam subjected to a vertical load P at a distance a - c from the crack tip. The load-deflection relation can be found in any mechanics of solid book: 3EI 1 3(70 × 10 9 )(1.6667 × 10 −9 ) 0.875 P= δ = × 0.0025 = (6.7.1) 3 3 (a − c) (0.1 − c) (0.1 − c) 3 th 3 (2 × 10 −2 )(1 × 10 −2 ) 3 = = 1.6667 × 10 −9 m 4 12 12 Let x measures the distance from the load to a location to the right. The total strain energy stored in the upper and lower beams due to a bending is where I =

U = 2∫

a −c

0

( Px) 2 P 2 (a − c) 3 dx = 2 EI 1 3EI 1

The strain energy release rate is GI =

1 dU P 2 (a − c) 2 = t da tEI 1

6.7.1

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=(

0.875 2 ( 0.1 − c )2 0.3281 ) = 3 9 −9 ( 0.1 − c ) ( 0.02 )( 70 × 10 )( 1.6667 × 10 ) ( 0.1 − c )4

Relation between K and G For plane strain fracture condition, 1 −υ2 1 − 0.32 2 GIc = K Ic = ( 50 × 106 )2 = 32500 N − m / m 2 9 E 70 × 10 Crack grows when G I ≥ G Ic 0.3281 ≥ 32500 ( 0 .1 − c ) 4

=> 0.1 − c ≤ 0.05637 => c ≥ 0.0436 m When the cylinder is driven to c = 0.0436 m , the beam would start to split. --- ANS

6.7.2

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6.8

Consider a long thin-walled cylinder of a brittle material subjected to an internal pressure p 0 . The diameter of the cylinder is 2 m, the wall thickness is 5 mm, and the mode I fracture toughness of the material (of the same thickness of the wall) is K Ic = 5MPa m (here, K Ic may not be the plane strain fracture toughness). If there is a through-the-thickness longitudinal crack of 5 cm in length on the cylinder, estimate the maximum internal pressure that the cracked cylinder can withstand. If the cracked cylinder is subjected to a torque and the mode II toughness of the material is the same as that of mode I, estimate the maximum torque. Provide justifications of the approach employed in the estimation.

Solution: (1) Under the internal pressure p 0 . The only nonvanishing stress is the hoop stress which is given by

p 0 ( 2r ) L p 0 r p0 = = = 200 p 0 2tL t 0.005 For a large cylinder with a small crack that is perpendicular to the uniform hoop stress it can be approximated as a flat plate subjected to a remotely applied uniform tension as shown in the figure below. Then the stress intensity factor can be approximately taken as

σ yy =

K I = σ yy πa = 200 p0

0.05 π( ) = 56.05 p0 2

σ yy

σ yy The mode I fracture toughness of the material is K Ic = 5 MPa m . Therefore, the maximum internal pressure that the cracked cylinder can withstand is estimated as K I = K Ic => 56.05 p0 = 5 Thus, p0 = 0.0892 MPa = 89.2 KPa --- ANS (2) Under a torque T (without internal pressure p 0 ) 6.8.1

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The state of stress in the uncracked thin-walled cylinder is

σ xx = σ yy = 0 ,

τ=

q T T = = = 31.83 T 2 t 2 At 2π ( 1 ) ( 0.005 )

N / m2

where τ is the shesr stress associated with the shear flow q along the wall produced by the torque T. If there is a small longitudinal crack on the cylinder of a larger diameter, then in region of the crack may be approximated by a flat plat subjected to antisymmetric shear loading as shown in the figure below.

τ τ

τ τ

So this is a mode II fracture problem and the large plate solution for stress intensity factor may be used. We have

K II = τ 0 πa = 31.83T π (

0.05 ) = 8.92T 2

The mode II fracture toughness of the material is K IIc = 5MPa m . Therefore, the maximum torque that the cracked cylinder can withstand is estimated as K II = K IIc => 8.92T = 5 Thus, the maximum torque is T = 0.56

MN − m = 560

KN − m

--- ANS

6.8.2

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6.9

Consider the thin-walled box beam in Fig. 6.17. The top wall contains a crack parallel to the x-axis. The crack length is 0.02 m (i.e., a = 0.01m ). Assume that the material is brittle and that modes I and II have the same toughness value of 5MPa m . If the box beam has already been subjected to a torque T = 100kN ⋅ m , estimate the maximum additional axial force N by using the mixed mode fracture criterion.

Figure 6.17

Box of a rectangular thin-walled section (w/ crack parallel to x-axis)

Solution: (1) Under the torque T Without the crack, a constant shear flow in the thin wall is produced by the torque and in the top panel the state of uniform stress would be T 100 σ xx = σ zz = 0 , τ xz = = = 20000kPa = 20 MPa , 2 At 2(0.5)(0.005) Since the crack is small as compared with the panel, the top panel with the crack can be approximated as a large panel subjected to a remote shear stress condition as depicted in the figure below.

τ z

τ

x

τ

τ 6.9.1

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This is a mode II fracture problem and the stress intensity factor can be approximated by the expression for a crack in an infinite plate:

K II = τ xz πa = 20 π (0.01) = 3.545 MPa m The mode II fracture toughness of the material is K IIc = 5 MPa m .

(2) Under the axial force N σ xx = 0 , τ xz = 0

σ zz =

N N = = 66.67 N A 2(1 + 0.5)(0.005)

Pa

This is a mode I fracture problem. Again, using the large panel (relative to the crack size) argument, we approximate the mode I stress intensity factor as

K I = σ zz πa = 66.67 N π (0.01) = 11.82 N The mode I fracture toughness of the material is K Ic = 5 MPa m .

(3) Mix mode fracture criterion ⎛ KI ⎜⎜ ⎝ K Ic

2

⎞ ⎛ K II ⎟⎟ + ⎜⎜ ⎠ ⎝ K IIc

2

⎞ ⎟⎟ = 1 ⎠ 2

6 ⎛ 11.82 N max ⎞ ⎛ 3.545 × 10 ⎞ ⎜ ⎟⎟ = 1 => ⎜ + ⎟ ⎜ 6 6 ⎝ 5 × 10 ⎠ ⎝ 5 × 10 ⎠ 2

=> N max = 298487 Newton --- ANS

6.9.2

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6.10

Derive the strain energy (bending and shear together) per unit length of a Timoshenko beam with a solid rectangular cross-section. The counterpart of the simple beam theory is given by (6.26). Use this expression to derive the mode I strain energy release rate for the split beam of Fig. 6.10. Compare the Timoshenko beam solution with the simple beam solution. How long (in terms of a / h ) does the crack length have to be for the simple beam solution to be within 5 percent of the Timoshenko beam solution.

2

3

1

Figure 6.10

Loaded split beam

Solution: The strain energy in a Timoshenko beam consists of two parts; one part is associated with bending deformation and the other part is transverse shear deformation. (1) Strain energy associated with bending deformation M2 UM = ∫ dx (6.10.1) 0 2 EI For this particular problem, M = Px in beams 1 and 2, and M = 0 in beam 3, and L

I1 = I 2 =

th 3 12

U M1 = U M 2 = ∫

a

0

( Px) 2 2P 2 a 3 dx = , UM3 = 0 th 3 Eth 3 2E( ) 12

Thus, U M = ∑ U Mi = i

4P 2 a 3 Eth 3

(2) Strain energy associated with transverse deformation The strain energy density for the transverse shear deformation is 1 G W = τγ = γ 2 2 2 where τ is the transverse shear stress and γ is the transverse shear strain. By the

6.10.1

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Timoshenko beam theory, the transverse shear strain produced by the shear force V is

γ=

dw0 V +ψ y = dx kGA

where k =

5 for the rectangular solid cross-section. 6

Then the total stain energy associated with the transverse shear deformation in a Timoshenko beam is U V = ∫ WdV = ∫

L

0

V

L G V 2 V2 dA dx = ( ) ∫∫A 2 kGA ∫0 2k 2 GA dx

(6.10.2)

For this particular problem, V = P in beams 1 and 2, and V = 0 in beam 3, and A1 = A2 = th UV1 = UV 2 = ∫

a

0

( P) 2 18 P 2 a dx = , UV 3 = 0 25Gth 2k 2 G (th)

Thus the total strain energy associated with the transverse shear deformation is 36 P 2 a UV = UV 1 + UV 2 = 25Gth (3) Total strain energy Assume Poisson’s ratio υ = 0.3 , G =

E = 0.385E 2(1 + υ )

(a) Simple beam theory 4P 2 a 3 U Sim = U M = ( ) Et h (b) Timoshenko beam theory U Tim = U M + U V

U Tim =

(6.10.3)

a P2 a P2 a 4P2 a 3 36 P 2 ( ) + ( ) = 4 ( )3 + 3.74 ( ) Et h Et h Et h 25(0.385E )t h

(6.10.4) --- ANS

(4) Energy release rate G for the Timoshenko beam theory 1 dU P 2a 2 P2 G= = 12 2 3 + 3.74 2 t da Et h Et h --- ANS (5) The rate of Let

a within 5% error h

a = q , from (6.10.3) and (6.10.4), we have h

Err =

U Tim − U Sim ≤ 0.05 , U Tim

=>

3.744q ≤ 0.05 4q + 3.744q 3

6.10.2

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=> q ≥ 4.22 or − 4.22 ≤ q ≤ 0 Since

a a = q is positive, we have = q ≥ 4.22 h h

--- ANS

6.10.3

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6.11

Compare the plastic zone sizes for plane strain mode I fracture at failure in Al 2024-T651 and Al 7075-T651.

Solution: The plastic zone size under plane strain is rp = (1 − 2υ )

2

KI

2

2πσ Y

(6.11.1)

2

(1) For Al 2024-T651 ( υ = 0.33 ) The plane strain toughness is K Ic = 24MPa m , and the yield stress is σ Y = 415MPa . At failure, the plastic zone is rp = (1 − 2υ ) 2

2

K Ic (24) 2 2 = ( 1 − 2 × 0 . 33 ) = 6.153 × 10 − 5 m = 0.062 mm 2 2 π 2 ( 415 ) 2πσ Y

--- ANS (2) For Al 7075-T651 ( υ = 0.33 ) The plane strain toughness is K Ic = 29MPa m , and the yield stress is σ Y = 505MPa . The plastic zone is 2

K Ic ( 29) 2 2 rp = (1 − 2υ ) = (1 − 2 × 0.33) = 6.067 × 10 − 5 m = 0.061 mm 2 2 2π (505) 2πσ Y 2

--- ANS The plastic zone sizes for plane strain mode I fracture at failure in Al 2024-T651 and Al 7075-T651 are almost identical.

6.11.1

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6.12

A center-cracked thin Al 2024-T651 flat panel with a very large width-to-crack length ratio is subjected to uniform remote tensile stress. The initial crack length is 50 mm and it grows to 55 mm when the applied load reaches the maximum value of 136 MPa. Determine the fracture toughness using Irwin’s plastic zone adjustment method. Is the crack length valid for this method?

Solution: By using Irwin’s plastic zone adjustment method, the fracture toughness can be characterized by KI evaluated at a eff which is aeff = a0 + rp

where a0 = 27.5 mm. Since rp depends on K I , which in turn depends on a eff , the determination of a eff requires a few iterations. We start with K I = σ 0 πa 0 = 136 π (0.0275) = 39.97 MPa m The plastic zone under plane stress is

rp1 =

KI

2

2πσ Y

=

2

(136) 2 π 0.0275 = 1.477 × 10 −3 m 2 2π (415)

The effective crack length is obtained as a eff 1 = a 0 + rp1 = 0.02898m

Then we can compute the new stress intensity factor by using the new effective crack length, that is K I ( eff )1 = σ 0 πa eff 1 = 136 π (0.02898) = 41.04 MPa m

Thus, the new plastic zone is rp 2 =

K I ( eff )1 2πσ y

2

2

=

(39.123866) 2 = 1.556 × 10 −3 m 2π (415) 2

It produces the new effective crack length a eff 2 = a 0 + rp 2 = 0.02906 m

In the same manner, after 5 iterations we obtain the converged values of a eff = 0.029060425 m = 0.291 m

K I ( eff ) = 41.1 MPa m This is the fracture toughness obtained using Irwin’s plastic zone adjustment

6.12.1

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method. r p = 0.001560466 m = 0.00156 m

Since

rp 0.00156 = = 0.057 < 0.1 , this crack is valid for Irwin’s method. a0 0.0275 --- ANS

6.12.2

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6.13

The split beam of Fig. 6.10 is subjected to a pair of cyclic opening forces P with

Pmin = 0

Pmax = 2000 N ,

The initial crack length a 0 is 40 mm. The material is 2024-T651 Al, and t = 2 × 10 −2 m , h = 1 × 10 −2 m . The crack growth rate is given by

da = 1.6 × 10 −11 (ΔK I ) 3.59 m / cycle dN

in which K I is in MPa m . Find the number of cycles to failure (at which the crack becomes unstable under the load Pmax ). Assume that the plane strain condition exists.

Figure 6.10

Loaded split beam

Solution: Assume a plane strain fracture condition so that GI =

1−υ 2 2 KI E

(6.13.1)

Since the strain energy release rate for the split beam is GI =

P2a2 tEI

(6.13.2)

Comparing (6.13.1) and (6.13.2), we have KI = 2

P2a2 (1 − υ 2 )tI

(6.13.3)

For this problem, we have Pmax = 2000 N , Pmin = 0 ,

6.13.1

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th 3 = 1.6667 × 10 −9 m 4 t = 2 × 10 m , h = 1 × 10 m => I = 12 −2

−2

For the material of 2024-T651 Al, Plane strain toughness K Ic = 24MPa m , and E = 72GPa , υ = 0.33

When K I = K Ic = 24MPa m , the crack becomes unstable under the load Pmax . Thus, using all the numerical values in (6.13.3) and setting K I = K Ic , we obtain the crack length ac at which fracture occurs. We have a c = (1 − υ 2 )tI

K Ic 24 × 10 6 = (1 − 0.33 2 )(0.02)(1.6667 × 10 −9 ) Pmax 2000

= 0.0654m = 65.4mm

For the given cyclic loading ΔK I =

Pmax a (1 − υ )tI 2

=

2000 a (1 − 0.33 )(0.02)(1.6667 × 10 − 9 ) 2

= 366.964a MPa m

The crack growth rate is given by da = 1.6 × 10 −11 ( ΔK I ) 3.59 = 1.6 × 10 −11 (366.964a ) 3.59 = 0.0257702 a 3.59 dN

=>

da = 0.0257702 dN a 3.59

Integrating the above differential equation, we have a 1−3.59 ac | a = 0.0257702 N 1 − 3.59 0 Substituting a 0 = 0.04m and ac = 0.0654m in the equation above, we find the number of cycles to failure, N = 45045 cycles . --- ANS

6.13.2

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6.14

Consider Example 6.6. Instead of a static torque, a cyclic torque with

Tmin = 0

Tmax = 0.1 MN ⋅ m ,

is applied. The Paris law for the material is

da = 5 × 10 −11 ( ΔK I ) 3 m/cycle. dN

Find the number of cycles for the crack of initial length a 0 = 0.01m to grow to a length a = 0.02m .

Solution: The maximum crack opening stress is σ 1(max) .

Since the member is under pure

torsion, σ 1(max) = τ max in which τ max can be determined by

Tmax = 2 Atτ max = 0.1 MN ⋅ m => τ max = and

Tmax

=

0 .1 = 20 MPa 2 ⋅ (0.5 × 1) ⋅ 0.005

2 At τ min = 0 MPa

Therefore, Δσ 1 = 20 − 0 = 20 MPa

Now, the Paris fatigue model for the material is

da = 5 × 10 −11 (ΔK I ) 3 m/cycle. dN

=>

da = 5 × 10 −11 (ΔK I ) 3 = 5 × 10 −11 ( Δσ 1 πa ) 3 dN

=>

da = 5 × 10 −11 ( 20 ⋅ π ) 3 dN 3/ 2 a

(6.14.1)

The number of cycles (N) required for a crack to grow from ao to a is then obtained from integrating (6.14.1), a

N

da 3 −11 ∫a a 3 / 2 = ∫0 5 × 10 (20 ⋅ π ) dN 0 => N =

a 0−1 / 2 − a −1 / 2 1 × 5 × 10 −11 (20 ⋅ π ) 3 2

= 897936 × (a0−1 / 2 − a −1 / 2 )

For a crack grown from ao = 0.01m to a = 0.02m, the number of cycle required is

6.14.1

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N = 897936 × (a 0−1 / 2 − a −1 / 2 ) = 2630000 = 2.63 × 10 6

cycles --- ANS

6.14.2

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7.1

The truss structure consists of two bars connected by a pin-joint (which allows free rotation of the bars). The other ends of the bars are hinged as shown in Fig. 7.26. A weight W is hung at the joint. Find the maximum weight the truss can sustain before buckling occurs.

Figure 7.26

Two-bar truss

Solution: From equilibrium, axial forces of the bars can be easily determined as W (Tension) sin θ

N 12 =

N 13 = − N 12 cos θ = −

W cos θ (Compression) sin θ

Only the compressed bar 13 may suffer buckling when the weight W increases. Since bar 13 is connected with pin at both ends, its buckling load is Pcr =

π 2 EI

L2 When the axial force N 13 reaches the critical load Pcr , buckling occurs. That is N13,max

Wmax cos θ π 2 EI = = 2 = Pcr sin θ L

=> Wmax =

π 2 EI L2

tan θ --- ANS

7.1.1

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7.2

A bar is built-in at the left end and supported at the tight end by a linear spring with spring constant α . Find the equation for buckling loads. Hint: The boundary conditions are w = 0 and dw / dx = 0 at the left end; and M = 0 and V = −αw at the right end.

Figure 7.27

Bar with a built-in end and an elastically supported end

Solution: The equilibrium equation in terms of deflection is 2 d 4w 2 d w +k =0 dx 4 dx 2

(7.2.1)

P EI The general solution is w = C1 sin kx + C 2 cos kx + C 3 x + C 4 where k =

Its first, second and third derivatives, respectively, are dw = C1 k cos kx − C 2 k sin kx + C 3 dx

d 2w = −C1 k 2 sin kx − C 2 k 2 cos kx 2 dx d 3w = −C1 k 3 cos kx + C 2 k 3 sin kx 3 dx Boundary conditions: At the left end, x = 0 , w=0 => C 2 + C 4 = 0 dw / dx = 0 => C1 k + C 3 = 0

(7.2.2a) (7.2.2b)

At the right end, x = L , M = − EI V = − EI

d 2w =0 dx 2

=>

d 3w dw −P = −αw => 3 dx dx

k 2 (C1 sin kL + C 2 cos kL) = 0

(7.2.2c)

d 3w dw αw + k2 = 3 dx EI dx 7.2.1

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− C1 k 3 cos kL + C 2 k 3 sin kL + Ck 3 cos kL − C 2 k 3 sin kL + C3 k 2 =>

=

α EI

(C1 sin kL + C 2 cos kL + C3 L + C 4 )

=> C 3 k 2 =

α EI

(C1 sin kL + C 2 cos kL + C 3 L + C 4 )

From (7.2.2c), From (7.2.2b),

C 2 = −C1 tan kL C 3 = −C1 k

From (7.2.2a),

C 4 = −C 2 = C1 tan kL

(7.2.2d)

Plug into (7.2.2d), we have − C1 k 3 =

α EI

=> C1 [ k 3 +

[C1 sin kL + (−C1 tan kL) cos kL + ( −C1 k ) L + C1 tan kL]

α EI

( − kL + tan kL)] = 0

Since C1 = 0 leads to trivial solution, the value inside the bracket must vanish, that is, k3 +

α EI

( − kL + tan kL) = 0

tan kL − kL + k 3

EI

= 0, α Letting β = kL , then the equation for buckling loads is

=>

tan β − β + β 3

EI =0 αL3

--- ANS

7.2.2

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7.3

Two steel bars ( E = 210GPa ) are connected by a hinge as shown in Fig. 7.28. The square cross-section of the bar is 5cm × 5cm . Find the buckling load for the bar with a built-in end. Hint: Treat the simply support beam as an elastic spring and find the effective elastic constant α first.

Figure 7.28

Two-bar structure

Solution: In this problem, we have Lab = 3m , Lcd = 2m , I ab = I cd =

0.05 4 = 5.21 × 10 −7 m 4 , E ab = E cd = 210GPa 12

By treating the simply support beam as an elastic spring, the effective elastic constant α can be found by the following procedure,

The deflection at the midpoint of the simply support beam caused by a vertical concentrated force F at the mid span of the beam can be found from books on mechanics of solids as 3

FLcd δm = 48 E cd I cd

The above relation can be expressed in the form F = αδ m with the effective elastic constant given by 48Ecd I cd α= 3 Lcd

(7.3.1)

And this structure can be reconsidered equivalent to the structure loaded as shown in the figure below. This is exactly the problem 7.2. The solution procedure is given as 7.3.1

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follows.

The equilibrium equation in terms of deflection is 2 d 4w 2 d w + k =0 dx 4 dx 2

where k =

(7.3.2)

P E ab I ab

Foe convenience, we let E ab = E , Lab = L and I ab = I . The general solution for equation (7.3.2) is w = C1 sin kx + C 2 cos kx + C 3 x + C 4 Its first, second and third derivatives, respectively, are dw = C1 k cos kx − C 2 k sin kx + C 3 dx

d 2w = −C1 k 2 sin kx − C 2 k 2 cos kx 2 dx d 3w = −C1 k 3 cos kx + C 2 k 3 sin kx 3 dx Boundary conditions: At the left end, x = 0 , w=0 => C 2 + C 4 = 0 dw / dx = 0 => C1 k + C 3 = 0

(7.3.3a) (7.3.3b)

At the right end, x = L d 2w M = − EI 2 = 0 dx

=>

d 3w dw V = − EI 3 − P = −αw => dx dx

k 2 (C1 sin kL + C 2 cos kL) = 0

(7.3.3c)

d 3w dw αw + k2 = 3 dx EI dx

− C1 k 3 cos kL + C 2 k 3 sin kL + Ck 3 cos kL − C 2 k 3 sin kL + C3 k 2 =>

=

α EI

(C1 sin kL + C 2 cos kL + C3 L + C 4 )

=> C 3 k 2 =

α EI

(C1 sin kL + C 2 cos kL + C 3 L + C 4 )

(7.3.3d)

7.3.2

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From (7.3.3c), From (7.3.3b),

C 2 = −C1 tan kL C 3 = −C1 k

From (7.3.3a),

C 4 = −C 2 = C1 tan kL

Substituting the above into (7.3.3d), we have − C1 k 3 =

α EI

=> C1 [ k 3 +

[C1 sin kL + (−C1 tan kL) cos kL + ( −C1 k ) L + C1 tan kL]

α EI

( − kL + tan kL)] = 0

Since C1 = 0 will lead to trivial solution, the value inside the bracket must vanish, that is k3 +

=>

α EI

( − kL + tan kL) = 0

tan kL − kL + k 3

EI

=0

α This is the buckling equation Letting β = kL = kLab , E = E ab and I = I ab , then the equation for buckling loads becomes E I tan β − β + β 3 ab ab3 = 0 αLab

(7.3.4)

Now plug in all the values to equation (7.3.4)

EI ab

αLab 3

3

Lcd E ab I ab 1 2 3 1 ( ) = = = 3 48Ecd I cd Lab 48 3 162

=> tan β − β +

1 3 β =0 162

(7.3.5)

The buckling equation (7.3.5) can be solved by numerical methods. The easiest way to find the numerical solution for β is to plot the value of the quantity on the left hand side of (7.3.5) vs incremental value of β . The lowest value of β that makes the left hand side quantity in (7.3.5) equal to zero is the solution. We have

β ≈ 4.462 = kLab = Lab

Pcr Eab I ab

=> Pcr = 242 kN where Pcr is the buckling load. --- ANS 7.3.3

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7.4

Find the buckling load equation for the bar with the left end built-in and the right end simply supported but constrained by a rotational spring (see Fig. 7.29). The spring constant β relates the bending moment M and the rotation θ = dw / dx by M = βθ .

Figure 7.29

Bar with a built-in end and a rotationally constrained end

Solution: The equilibrium equation in terms of deflection is 2 d 4w 2 d w +k =0 dx 4 dx 2

(7.4.1)

P EI The general solution is w = C1 sin kx + C 2 cos kx + C 3 x + C 4 where k =

Its first, second and third derivatives, respectively, are dw = C1 k cos kx − C 2 k sin kx + C 3 dx

d 2w = −C1 k 2 sin kx − C 2 k 2 cos kx 2 dx d 3w = −C1 k 3 cos kx + C 2 k 3 sin kx 3 dx Boundary conditions: At the left end, x = 0 , w=0 => C 2 + C 4 = 0 dw / dx = 0 => C1 k + C 3 = 0

(7.4.2a) (7.4.2b)

At the right end, x = L , w=0 => C1 sin kL + C 2 cos kL + C 3 L + C 4 = 0

(7.4.2c)

M = − EI

2

d w dw =β => 2 dx dx

− EI (−C1 k 2 sin kL − C 2 k 2 cos kL ) = β (C1 k cos kL − C 2 k sin kL + C 3 )

(7.4.2d)

7.4.1

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C 4 = −C 2 C 3 = −C1 k

From (7.4.2a), From (7.4.2b),

Plug into (7.4.2c) and (7.4.2d), let γ =

β

EI C1 (sin kL − kL) + C 2 (cos kL − 1) = 0

, we have

C1 k sin kL + C 2 k cos kL = γ (C1 k cos kL − C 2 k sin kL − C1 k ) => C1 (k sin kL − γ cos kL + γ ) + C 2 (k cos kL + γ sin kL) = 0 2

(7.4.3a)

2

(7.4.3b)

For a non-trivial solution of equations (7.4.3a) and (7.4.3b), we need sin kL − kL cos kL − 1 =0 k sin kL − γ cos kL + γ k cos kL + γ sin kL Expanding the above equation leads to the buckling load equation: (2γ + k 2 L) cos kL + (γkL − k ) sin kL − 2γ = 0 where k =

Pcr β and γ = EI EI --- ANS

7.4.2

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7.5

Two steel bars of a 4-cm circular cross-section are rigidly connected into a T-shaped structure. The diameter of the bars is 4 cm. Three ends are built-in as shown in Fig. 7.30. At the joint, a roller support is provided to prevent vertical deflection of the joint. Compression is applied as shown in the figure. Find the lowest buckling load.

Figure 7.30

Bar with a built-in end and a rotationally constrained end

Solution: In this problem, we have Lab = 3m , Lcd = 2m , Cross-sectional area: Aab = Acd = Moment of inertia: I ab = I cd =

Shear modulus: Gab = Gcd =

4

π (0.04) 4

Torsional constant: J ab = J cd = Young's modulus: E ab = E cd

π (0.04) 2

64

π (0.04) 4

32 = 210GPa ,

= 1.256637 × 10 −3 m 2

= 1.256637 × 10 −7 m 4 = 2.513274 × 10 −7 m 4 Poisson’s ratio: υ = 0.32

210 = 79.5455GPa 2(1.32)

Two key procedures for this problem: (1) Force applied to bar AB Considering the free body diagram as shown,

The force applied to bar AB is denoted Pab . By compatibility of displacement, the 7.5.1

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deflection at the middle of bar CD, named δ cd ,m , is equal to the axial displacement of bar AB, δ ab . That is,

δ cd ,m

( P − Pab ) Lcd = 192 E cd I cd

3

=

δ ab =

Pab Lab E ab Aab

from which we obtain Pab = 0.992852 P

(7.5.1)

(2) Rotational constraint By treating bar CD as an elastic rotational spring, the effective rotational spring constant β can be found by determining the torque needed when a unit angle of twist is produced at the mid span of the rod CD. Because of symmetry, the total torque is two times the torque taken by each half of the rod of length Lcd/2. For each half rod the torque is α 2Gcd J cd α T = GJθ = Gcd J cd = Lcd / 2 Lcd where α is the rotation angle at the mid span of rod CD. Thus the effective rotational spring constant provided by rod CD is given by 4G J β = cd cd Lcd

(7.5.2)

Now the loading condition and boundary conditions on rod AB are depicted by the figure below.

In this problem, β =

4Gcd J cd = 39984 N − m Lcd

Derive the buckling load: In the following derivation, simply let Lab = L , ( EI ) ab = EI , Pab = P The displacement equilibrium equation for rod AB is 2 d 4w 2 d w + k =0 dx 4 dx 2

where k =

(7.5.3)

P EI 7.5.2

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Its general solution is w = C1 sin kx + C 2 cos kx + C 3 x + C 4 And its first, second and third derivatives, respectively, are dw = C1 k cos kx − C 2 k sin kx + C 3 dx

d 2w = −C1 k 2 sin kx − C 2 k 2 cos kx 2 dx d 3w = −C1 k 3 cos kx + C 2 k 3 sin kx 3 dx Boundary conditions: At the left end, x = 0 , w=0 => C 2 + C 4 = 0 dw / dx = 0 => C1 k + C 3 = 0

(7.5.4a) (7.5.4b)

At the right end, x = L , w=0 => C1 sin kL + C 2 cos kL + C 3 L + C 4 = 0

(7.5.4c)

M = − EI

d 2w dw =β => 2 dx dx

− EI (−C1 k 2 sin kL − C 2 k 2 cos kL ) = β (C1 k cos kL − C 2 k sin kL + C 3 )

(7.5.4d)

C 4 = −C 2 C 3 = −C1 k

From (7.5.4a), From (7.5.4b),

Using the above relations in (7.5.4c) and (7.5.4d), and letting γ =

β EI

, we obtain

C1 (sin kL − kL) + C 2 (cos kL − 1) = 0

(7.5.5a)

C1 k sin kL + C 2 k cos kL = γ (C1 k cos kL − C 2 k sin kL − C1 k ) => C1 (k sin kL − γ cos kL + γ ) + C 2 (k cos kL + γ sin kL) = 0

(7.5.5b)

2

2

For a non-trivial solution for equations (7.5.5a) and (7.5.5b), the determinant of the coefficient matrix must vanish, i.e., sin kL − kL cos kL − 1 =0 k sin kL − γ cos kL + γ k cos kL + γ sin kL Expanding the above equation leads to the buckling load equation: (2γ + k 2 L) cos kL + (γkL − k ) sin kL − 2γ = 0 where k =

Pab ,cr E ab I ab

, γ =

β E ab I ab

and L = Lab 7.5.3

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Substituting all the known numerical values into the buckling load equation, we have

γ =

β E ab I ab

= 1.515474 and L = Lab = 3 m

The buckling load equation becomes (3.030948 + 3k 2 ) cos 3k + (3.546422k ) sin 3k − 3.030948 = 0

(7.5.6)

Solve (7.5.6) by any numerical method. The result is k = 1.797181807 ≈ 1.7972 =

Pab ,cr E ab I ab

=> Pab ,cr = 85234 N The buckling load in terms of the total force P is obtained from (7.5.1) as Pcr =

Pab ,cr 0.992852

= 85847 N --- ANS

7.5.4

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7.6

For the structure of Problem 7.5, find the buckling load if the roller support at the joint is removed.

Figure 7.30

Bar with a built-in end and a rotationally constrained end (With the roller support removed)

Solution: In this problem, we have Lab = 3m , Lcd = 2m , Cross-sectional area: Aab = Acd = Moment of inertia: I ab = I cd =

Shear modulus: Gab = Gcd =

4

π (0.04) 4

Torsional constant: J ab = J cd = Young's modulus: E ab = E cd

π (0.04) 2

64

π (0.04) 4

32 = 210GPa ,

= 1.256637 × 10 −3 m 2

= 1.256637 × 10 −7 m 4 = 2.513274 × 10 −7 m 4 Poisson’s ratio: υ = 0.32

210 = 79.5455GPa 2(1.32)

Two key procedures for this problem: (1) Force applied to bar AB Considering the free body diagram as shown,

The force applied to bar AB is denoted Pab . By compatibility of the displacement, the deflection at the middle of bar CD, named δ cd ,m , is equal to the axial displacement of

7.6.1

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bar AB, δ ab . That is,

δ cd ,m

( P − Pab ) Lcd = 192 E cd I cd

3

=

δ ab =

Pab Lab E ab Aab

Plugging in the numbers gets Pab = 0.992852 P

(7.6.1)

(2) Lateral and torsional constraints This structure can be reconsidered as the following figure.

By treating bar CD as an elastic torsional spring, the effective elastic torsional constant β can be found by determining the torque needed when a unit twist angle is produced. Using the torsion of a rod of a circular cross-section, we have 4G J β = cd cd (7.6.2) Lcd In this problem, β =

4Gcd J cd = 39983.93 N − m Lcd

In addition, the rod CD also provides vertical support to rod AB. This support to rod AB can be viewed as an effective elastic spring with an elastic constant α . This elastic spring constant can be obtained by considering a vertical force applied at the mid span of rod CD and the resulting vertical deflection. The result gives 192 E cd I cd α= (7.6.3) 3 Lcd In this problem, α =

192 Ecd I cd = 633345 N / m 3 Lcd

Derive the equation of buckling load for rod AB under Pab: In the following derivation, we let Lab = L , ( EI ) ab = EI , Pab = P The equilibrium equation in terms of deflection is 2 d 4w 2 d w +k =0 dx 4 dx 2

(7.6.4)

7.6.2

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P EI Its general solution is w = C1 sin kx + C 2 cos kx + C 3 x + C 4 where k =

And its first, second and third derivatives, respectively, are dw = C1 k cos kx − C 2 k sin kx + C 3 dx

d 2w = −C1 k 2 sin kx − C 2 k 2 cos kx 2 dx d 3w = −C1 k 3 cos kx + C 2 k 3 sin kx 3 dx Boundary conditions: At the left end, x = 0 , w=0 => C 2 + C 4 = 0 dw / dx = 0 => C1 k + C 3 = 0

(7.6.5a) (7.6.5b)

At the right end, x = L , d 3w dw V = − EI 3 − P = −αw => dx dx

d 3w dw αw + k2 = 3 dx EI dx

− C1 k 3 cos kL + C 2 k 3 sin kL + Ck 3 cos kL − C 2 k 3 sin kL + C3 k 2 =>

=

α EI

(C1 sin kL + C 2 cos kL + C3 L + C 4 )

=> C 3 k 2 = M = − EI

α EI

(C1 sin kL + C 2 cos kL + C 3 L + C 4 )

(7.6.5c)

d 2w dw =β => 2 dx dx

− EI (−C1 k 2 sin kL − C 2 k 2 cos kL ) = β (C1 k cos kL − C 2 k sin kL + C 3 )

(7.6.5d)

C 4 = −C 2 C 3 = −C1 k

From (7.6.5a), From (7.6.5b),

Eliminating C3 and C4

from

(7.6.5c) and (7.6.5d), and letting λ =

obtain C1 (λ sin kL − λkL + k 3 ) + C 2 (λ cos kL − λ ) = 0 C1 (k sin kL − γ cos kL + γ ) + C 2 (k cos kL + γ sin kL) = 0

α EI

γ =

β EI

, we

(7.6.6a) (7.6.6b)

For a non-trivial solution for equations (7.6.6a) and (7.6.6b), the determinant of the 7.6.3

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coefficient matrix muse vanish, i.e.,

λ sin kL − λkL + k 3 k sin kL − γ cos kL + γ

λ cos kL − λ =0 k cos kL + γ sin kL

Expanding the above equation leads to the buckling load equation: (k 4 − λk 2 L − 2λγ ) cos kL + (γk 3 − λγkL + λk ) sin kL + 2λγ = 0 Pab ,cr

where k =

E ab I ab

, λ=

α E ab I ab

β

, γ =

E ab I ab

and L = Lab

By substituting all the known numerical values

λ=

α E ab I ab

= 24 , γ =

β E ab I ab

= 1.515152 and L = Lab = 3 m

in the buckling load equation, the buckling load equation becomes (k 4 − 72k 2 − 72.727296) cos 3k + (1.515152 k 3 − 85.090944 k ) sin 3k + 72.727296 = 0

(7.6.7)

Solve (7.6.7) by any numerical method to find the minimum value for k. k = 1.796869538 ≈ 1.7969 =

Pab ,cr E ab I ab

=> Pab , cr = 85204 N Thus, the buckling load of total force P, from (7.6.1), is Pcr =

Pab ,cr 0.992852

= 85818 N --- ANS

7.6.4

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7.7

A simply supported bar has doubly symmetrical cross-section consisting of a thin web and thin flanges as shown in Fig. 7.31. Find the length of the bar at which the flexural buckling load is equal to the torsional buckling load.

Figure 7.31

Cross-section of a simply supported bar

Solution: In this problem, the (y, z) coordinate system is set up with the origin at the centroid (the mid pont of the vertical web). We have (assume b, h >> t ) Cross-sectional area:

A = 2bt + ht

Moment of inertia:

Iy =

th 3 tbh 2 + 12 2

I0 = I y + I z =

Iz =

tb 3 6

th 3 tbh 2 tb 3 + + 12 2 6

(2b + h)t 3 3

Torsional constant:

J=

Warping constant:

Cw =

Young's modulus:

E

Shear modulus:

G=

th 2 b 3 24 Poisson’s ratio:

υ

E 2(1 + υ )

Flexural buckling load: For simply supported bar, Pcr , z =

Pcr , y =

π 2 EI z L2 π 2 EI y

L2

=

=

π 2 Etb 3 6L2

π 2 Eth 2 (h + 6b) 12 L2

It should be noticed that for typical I-cross-sectional bar, I y > I z , hence Pcr , y > Pcr , z 7.7.1

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and Pcr , z is the flexural buckling load.

Torsional buckling load: For simply supported bar, PcrT

=

A π2 12(2b + h) (2b + h)t 3 π 2th 2b3 (GJ + 2 ECw ) = 3 [ + ]E I0 L (h + 6bh 2 + 2b3 ) 6(1 + υ ) 24 L2

If the flexural buckling load is to equal the torsional buckling load, the length of the bar can be calculated from the condition Pcr , z = PcrT . We have

π 2 Etb 3 6 L2

=

=> L2 =

=> L =

12(2b + h) (2b + h)t 3 π 2 th 2 b 3 [ + ]E (h 3 + 6bh 2 + 2b 3 ) 6(1 + υ ) 24 L2

π 2 (1 + υ )b 4 ( 2h 2 + b 2 ) 6t 2 (2b + h) 2

πb 2

(1 + υ )(2h 2 + b 2 ) 6 t (2b + h)

--- ANS

7.7.2

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7.8

A simply supported aluminum ( E = 70GPa , G = 27GPa ) bar 2 m in length has the cross-section shown in Fig. 7.32. Find the lowest three buckling loads.

z y

Figure 7.32

Cross-section of a thin-walled bar

Solution: In this problem, the cross-sectional area is A = 2(0.02)(0.002) + (0.04)(0.002) = 1.6 × 10 −4 m 4 It is easy to verify that the centroid is located at mid point of the vertical web, and that the shear center coincides with the centroid. The moments of inertia with respect to the centroid and the coordinate system shown in the figure are (0.002)(0.04) 3 (0.02)(0.002) 3 + 2 ⋅[ + (0.02)(0.002)(0.02) 2 ] 12 12 −8 4 = 4.269 × 10 m

Iy =

Iz =

(0.04)(0.002) 3 (0.002)(0.04) 3 + = 1.069 × 10 −8 m 4 12 12

I yz = 2(0.02)(0.002)(0.01)(0.02) = 1.6 × 10 −8 m 4 To determine the buckling load, the principal centroidal axes are used and its corresponding moments of inertia are utilized in the equilibrium equations. The moments of inertia about these principal axes can be found by solving the eigenvalue problem, that is,

Iy −λ I yz

I yz 4.269 − λ 1.6 = =0 Iz − λ 1.6 1.069 − λ

=> λ = 4.932

or

λ = 0.407

It gives the moments of inertia about the principal axes as I 1 = 4.932 × 10 −8 m 4 7.8.1

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I 2 = 0.407 × 10 −8 m 4 Because the shear center coincides with the centroid, the polar moment of inertia is I 0 = I 1 + I 2 = 5.339 × 10 −8 m 4

The torsion constant is [2(0.02) + 0.04](0.002) 3 J= = 2.1339 × 10 −10 m 4 3 The warping constant can be found in Table 7.2 of the textbook. For the current section, it is Cw =

b3h 2 [2t f (b 2 + bh + h 2 ) + 3t w bh] = 2.6697 × 10 −12 m 6 2 12(2b + h)

For the material of aluminum, Young's modulus is E = 70GPa and the shear modulus is G = 27GPa . Flexural and torsional buckling loads: For simply supported bar, the general solutions are given by (7.119) through (7.121) in the text book. They are reproduced here: ( P − Pcr ,1 )C1 + Pz 0 C 3 = 0

(7.8.1)

( P − Pcr , 2 )C 2 + Py 0 C 3 = 0

(7.8.2)

I0 ( P − Pα )C3 = 0 (7.8.3) A where ( y0 , z0 ) is the position of the shear center relative to the centroid. Since the shear center coincides with the centroid, thus, y 0 = z 0 = 0 and Equations (7.8.1) – Pz 0 C1 − Py 0 C 2 +

(7.8.3) reduce to ( P − Pcr ,1 )C 1 = 0 ( P − Pcr , 2 )C 2 = 0

I0 ( P − Pα )C 3 = 0 A A nontrivial solution for C1 , C2 , and C 3 exists only if the determinant of the coefficient matrix vanishes, i.e.,

( P − Pcr ,1 )( P − Pcr , 2 )

I0 ( P − Pα ) = 0 A 7.8.2

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where in this problem Pcr ,1 = Pcr , 2 =

π 2 EI 1 L2

π 2 (70 × 10 9 )(4.9329 × 10 −8 )

=

π 2 EI 2 L2

22

=

π 2 (70 × 10 9 )(0.407 × 10 −8 ) 22

= 8518.6 N = 702.3 N ,

A 1.6 × 10 −4 π2 Pα = (GJ + EC w 2 ) = [(27 × 10 9 )(2.133 × 10 −10 ) −8 I0 L 5.339 × 10 + (70 × 10 )(2.667 × 10 9

−12

)

π2 22

] = 18643.1 N

So the three possible roots for P are Pcr = 702.3 N Pcr = 8518.6 N Pcr = 18643 .1 N

And these are the three lowest buckling loads. --- ANS

7.8.3

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7.9

Find the buckling load of a 1 m long and simply supported bar having a thin-walled circular cross-section 50 mm in diameter and 2 mm wall thickness. If the closed section is made into an open one by cutting a longitudinal slit over the entire length of the bar, what is the buckling load? Assume that E = 70GPa and G = 27GPa .

Solution: The cross-sectional area can be obtained by assuming small thickness, A ≈ 2πat = 3.142 × 10 −4 m 4 The centroid is located at center of the circular section, and that the shear center coincides with the centroid. The moments of inertia with respect to the centroid are

I y = I z ≈ πa 3t = 9.817 × 10 −8 m 4

Closed thin-walled circular cross-section Because the shear center coincides with the centroid, the polar moment of inertia is

I 0 = I y + I z = 19.635 × 10 −8 m 4 The warping constant for closed circular section is zero, C w = 0 For simply supported bar, ( P − Pcr , y )C1 = 0

(7.9.1)

( P − Pcr , z )C 2 = 0

(7.9.2)

I0 ( P − Pα )C 3 = 0 (7.9.3) A Since the above three equations are not coupled, they can be solved individually. Thus the possible buckling loads are P = Pcr , y = Pcr , z = P = Pα =

π 2 EI y L2

= 67826 N

or

A (GI 0 ) = 8482301 N I0

Therefore the global buckling load is the flexural buckling load given by Pcr = 67826.2 N .

--- ANS

7.9.1

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Opened thin-walled circular cross-section When the cross-section is cut into an open one, shear center then is no longer coincided with the centroid. Assume that the cut is at the wall to the left of the centroid (y = -0.025m, z = 0) as shown in the figure below.

z y C

The position of the shear center can be found in Table 7.2 in the textbook for the case of α = π . We have e = 2a

sin α − α cos α sin π − π cos π = 2(0.025) = 0.05 m α − sin α cos α π − sin π cos π

where e is the distance from the centroid to the shear center. => y 0 = 0.05 m ,

z0 = 0

The polar moment of inertia is still I 0 = I 1 + I 2 = 5.339 × 10 −8 m 4 The torsion constant is bt 3 J= = 4.189 × 10 −10 m 4 3 The warping constant can be found in Table 7.2 of the textbook. For the current section, it is 2ta 5 3 6(sin α − α cos α ) 2 2(0.002)(0.025) 5 3 6(π ) 2 [α − ]= [π − ] 3 α − sin α cos α 3 π = 1.583 × 10 −10 m 6

Cw =

For simply supported bar, ( P − Pcr , y )C1 + Pz 0 C 3 = 0

(7.9.4)

( P − Pcr , z )C 2 − Py 0 C 3 = 0

(7.9.5)

7.9.2

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Pz 0 C1 − Py 0 C 2 +

I0 ( P − Pα )C3 = 0 A

(7.9.6)

where

Pcr , y = Pcr , z = Pα =

π 2 EI y L2

= 67826.2 N

or

A π2 (GJ + EC w 2 ) = 193065.6 N I0 L

Plugging in the numbers and the equations (7.9.4) to (7.9.6) become ( P − 67826.2)C1 = 0 ( P − 67826.2)C 2 − 0.05 PC 3 = 0 − 0.05PC 2 + 0.000625( P − 193065.6)C 3 = 0

(7.9.7) (7.9.8) (7.9.9)

Since (7.9.7) is not coupled with the other two equations, it can be solved individually and the solution for nontrivial coefficients is P = 67826 N

This is the buckling load for buckling about the y axis in which the cut lies in the neutral plane of bending. In order to have a nontrivial solution for (7.9.8) and (7.9.9), we require that P − 67826.2 − 0.05 P =0 − 0.05P 0.000625( P − 193065.6) => 0.001875 P 2 + 163.051375 P − 8184316 = 0 => P = 35611 N

or

P = −122572 N

The possible lowest buckling load is Pcr = 35611 N . This buckling mode involves coupled torsion and bending. Apparently, the thin-walled tube with a longitudinal cut is weaker in buckling strength than the tube with a closed section. --- ANS

7.9.3

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8.1

Given a carbon/epoxy composite panel under uniaxial loading, i.e., σ xx = σ 0 ,

σ yy = σ xy = 0 , plot γ xy as a function of the fiber orientation θ . The composite properties are

E1 = 140GPa ,

E 2 = 10GPa ,

G12 = 7GPa ,

υ12 = 0.3

Solution: The strain-stress relation can be written as, ⎧σ xx ⎫ ⎧ε xx ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ε yy ⎬ = S ⎨σ yy ⎬ ⎪σ ⎪ ⎪γ ⎪ ⎩ xy ⎭ ⎩ xy ⎭

[]

(8.1.1)

where

S 11 = S11 cos 4 θ + (2S12 + S 66 ) sin 2 θ cos 2 θ + S 22 sin 4 θ S 12 = S 21 = S12 (sin 4 θ + cos 4 θ ) + ( S11 + S 22 − S 66 ) sin 2 θ cos 2 θ S 22 = S11 sin 4 θ + (2S12 + S 66 ) sin 2 θ cos 2 θ + S 22 cos 4 θ

(8.1.2)

S 16 = S 61 = (2S11 − 2S12 − S 66 ) sin θ cos 3 θ + (2S12 − 2S 22 + S 66 ) sin 3 θ cosθ S 26 = S 62 = (2S11 − 2S12 − S 66 ) sin 3 θ cosθ + (2S12 − 2S 22 + S 66 ) sin θ cos 3 θ S 66 = 2(2S11 + 2S 22 − 4S12 − S 66 ) sin 2 θ cos 2 θ + S 66 (sin 4 θ + cos 4 θ ) and S11 =

1 1 = = 7.143 × 10 −12 m 2 / N E1 140 × 109

S 21 = − S12 = −

υ12 E1

υ 21 E2

=−

0 .3 = 2.143 × 10 −12 m 2 / N 140 × 109

= S 21 = 2.143 × 10 −12 m 2 / N

S 22 =

1 1 = = 100 × 10 −12 m 2 / N E 2 10 × 10 9

S 66 =

1 1 = = 142.857 × 10 −12 m 2 / N G12 7 × 109

S16 = S 61 = S 26 = S 62 = 0 8.1.1

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From (8.1.1),

γ xy = S 61σ xx + S 62σ yy + S 66σ xy = S 61σ 0 and from (8.1.2) S 61 = (2 S11 − 2 S12 − S66 ) sin θ cos3 θ + (2 S12 − 2 S 22 + S66 ) sin 3 θ cosθ = [−132.857 sin θ cos3 θ − 52.857 sin 3 θ cosθ ] × 10 −12 So

γ xy = S 61σ 0 = [−132.857 sin θ cos3 θ − 52.857 sin 3 θ cosθ ] × 10−12 σ 0

(Unit: 10 −12 σ 0 ) 60

40

20

0 0

45

90

135

180

-20 Gamma(xy) -40

-60

--- ANS

8.1.2

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8.2

Consider a rectangular composite panel with θ = 45 o (material properties are given in Problem 8.1) subjected to σ xx = 10 MPa , σ yy = 0 , σ xy = τ . Find τ that is necessary to keep the deformed shape rectangular.

Solution: The composite properties are E1 = 140GPa , E 2 = 10GPa ,

G12 = 7GPa ,

υ12 = 0.3

The strain-stress relation can be written as, ⎧σ xx ⎫ ⎧ε xx ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ε yy ⎬ = S ⎨σ yy ⎬ ⎪σ ⎪ ⎪γ ⎪ ⎩ xy ⎭ ⎩ xy ⎭

[]

(8.2.1)

where

S 11 = S11 cos 4 θ + (2S12 + S 66 ) sin 2 θ cos 2 θ + S 22 sin 4 θ S 12 = S 21 = S12 (sin 4 θ + cos 4 θ ) + ( S11 + S 22 − S 66 ) sin 2 θ cos 2 θ S 22 = S11 sin 4 θ + (2S12 + S 66 ) sin 2 θ cos 2 θ + S 22 cos 4 θ

(8.2.2)

S 16 = S 61 = (2S11 − 2S12 − S 66 ) sin θ cos 3 θ + (2S12 − 2S 22 + S 66 ) sin 3 θ cosθ S 26 = S 62 = (2S11 − 2S12 − S 66 ) sin 3 θ cosθ + (2S12 − 2S 22 + S 66 ) sin θ cos 3 θ S 66 = 2(2S11 + 2S 22 − 4S12 − S 66 ) sin 2 θ cos 2 θ + S 66 (sin 4 θ + cos 4 θ ) and S11 =

1 1 = = 7.142857 × 10 −12 m 2 / N 9 E1 140 × 10

S 21 = − S12 = −

υ12 E1

υ 21 E2

=−

0.3 = −2.142857 × 10 −12 m 2 / N 9 140 × 10

= S 21 = −2.142857 × 10 −12 m 2 / N

S 22 =

1 1 = = 100 × 10 −12 m 2 / N 9 E 2 10 × 10

S 66 =

1 1 = = 142.857142 × 10 −12 m 2 / N 9 G12 7 × 10 8.2.1

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S16 = S 61 = S 26 = S 62 = 0 To keep the deformed shape rectangular implies γ xy = 0 in this problem. From (8.2.1),

γ xy = S 61σ xx + S 62σ yy + S 66σ xy = 0 and from (8.1.2) S 61 = (2 S11 − 2 S12 − S 66 ) sin θ cos3 θ + (2 S12 − 2 S 22 + S66 ) sin 3 θ cosθ = [−124.285714(

2 2 2 2 )( )3 − 61.428571( )3 ( )] × 10 −12 2 2 2 2

= −46.429 × 10 −12 m 2 / N S 66 = 2(2 S11 + 2S 22 − 4S12 − S66 ) sin 2 θ cos 2 θ + S66 (sin 4 θ + cos 4 θ ) = [160(

2 2 2 2 2 2 ) ( ) + 142.857142(( ) 4 + ( ) 4 )] × 10−12 2 2 2 2

= 111.286 × 10−12 m 2 / N Thus,

γ xy = S 61σ xx + S 62σ yy + S 66σ xy = (−46.429 × 10 −12 )(10 × 106 ) + (111.286 × 10 −12 )τ = 0 => τ = 4.172 MPa --- ANS Alternative solution: From the strain-stress relation, we have

γ xy =

η x , xy Ex

σ xx +

η y , xy Ey

σ yy +

1 σ xy G xy

For the specific loading case, it becomes 0=

η x , xy Ex

× 10 +

1 τ G xy

=>

τ =−

10η x , xy Ex

G xy ( MPa)

(8.2.3)

From (8.17) in the text book, we have

η x ,xy Ex

⎡⎛ 2 2ν ⎤ ⎛ 2 2ν 12 1 ⎞ 1 ⎞ 3 ⎟⎟ sinθ cos 3 θ − ⎜⎜ ⎟⎟ sin θ cos θ ⎥ = ⎢⎜⎜ + 12 − + − E1 G12 ⎠ E1 G12 ⎠ ⎝ E2 ⎣⎝ E1 ⎦ 8.2.2

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and

⎛ 1 1 1 1 2ν 12 1 ⎞ 2 ⎟⎟ sin θ cos 2 θ = + 4⎜⎜ + + − G xy G12 E1 G12 ⎠ ⎝ E1 E 2 Let θ = 45 0 , then the relation becomes

η x , xy Ex and

=

1⎡ 1 1 ⎤ E 2 − E1 ⎢ − ⎥= 2 ⎣ E1 E 2 ⎦ 2 E1 E 2

(8.2.4)

1 1 1 2ν 12 E1 + E 2 (1 + 2ν 12 ) = + + = G xy E1 E 2 E1 E1 E 2

(8.2.5)

Combining (8.2.4) and (8.2.5) into (8.2.3) and plugging in the numbers, we have 5(E1 − E 2 ) = 4.167 MPa τ= E1 + E 2 (1 + 2ν 12 ) --- ANS

8.2.3

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8.3

Plot the extension-shear coupling coefficients η x, xy and η xy , x versus θ for the composite given in Problem 8.1. Find the θ ’s that correspond to the maximum values of η x, xy and η xy , x , respectively.

Solution: The composite properties are E1 = 140GPa , E 2 = 10GPa ,

υ12 = 0.3

G12 = 7GPa ,

The strain-stress relation in an arbitrary coordinate system (x,y) can be written as,

⎡ 1 ⎢ E ⎧ε xx ⎫ ⎢ x ⎪ ⎪ ⎢ υ xy ⎨ε yy ⎬ = ⎢− ⎪γ ⎪ ⎢ E x ⎩ xy ⎭ ⎢ η x , xy ⎢ ⎢⎣ E x

−

υ yx

Ey 1 Ey

η y , xy Ey

η xy , x ⎤

⎥ G xy ⎥ ⎧σ ⎫ η xy , y ⎥ ⎪ xx ⎪ ⎥ ⎨σ yy ⎬ G xy ⎥ ⎪ ⎪ σ 1 ⎥ ⎩ xy ⎭ ⎥ G xy ⎥⎦

(8.3.1)

where ⎡1 ⎤ 2υ 1 1 − 12 ) sin 2 θ cos 2 θ + sin 4 θ ⎥ E x = ⎢ cos 4 θ + ( G12 E1 E2 ⎣ E1 ⎦

−1

⎡υ ⎤ 1 1 2υ12 1 υ xy = E x ⎢ 12 − ( + + − ) sin 2 θ cos 2 θ ⎥ ⎣ E1

E1

E2

E1

G12

⎦

⎡1 ⎤ 2υ 1 1 − 12 ) sin 2 θ cos 2 θ + cos 4 θ ⎥ E y = ⎢ sin 4 θ + ( G12 E1 E2 ⎣ E1 ⎦

G xy

⎡ 1 ⎤ 1 1 2υ12 1 =⎢ + 4( + + − ) sin 2 θ cos 2 θ ⎥ E1 E 2 E1 G12 ⎣ G12 ⎦

−1

(8.3.2)

−1

⎡ 2 2υ12 ⎤ 1 2 2υ 1 + − ) sin θ cos 3 θ − ( + 12 − ) sin 3 θ cos θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η x , xy = E x ⎢(

⎡ 2 2υ12 ⎤ 1 2 2υ 1 + − ) sin 3 θ cos θ − ( + 12 − ) sin θ cos 3 θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η y , xy = E y ⎢(

8.3.1

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Plugging in the numbers and then some properties are obtained. ⎡1 ⎤ 2υ 1 1 − 12 ) sin 2 θ cos 2 θ + sin 4 θ ⎥ E x = ⎢ cos 4 θ + ( G12 E1 E2 ⎣ E1 ⎦

−1

[

= 7.142857 cos 4 θ + 138.571429 sin 2 θ cos 2 θ + 100 sin 4 θ

]

−1

× 1012

⎡ 2 2υ12 ⎤ 1 2 2υ12 1 + − ) sin θ cos 3 θ − ( + − ) sin 3 θ cos θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η x , xy = E x ⎢(

[

]

= E x − 124.285714 sin θ cos 3 θ − 61.428571sin 3 θ cos θ × 10 −12 − 124.285714 sin θ cos 3 θ − 61.428571sin 3 θ cos θ = 7.142857 cos 4 θ + 138.571429 sin 2 θ cos 2 θ + 100 sin 4 θ

ηx,xy 2.5 2 1.5 1 0.5 0 0

45

90

135

180

-0.5 -1 ηx,xy

-1.5 -2 -2.5

From the plot above, the maximum values of η x, xy are obtained as follows.

θ = 12.73 o => η x , xy = −1.989 θ = 167 .27 o => η x , xy = 1.989 --- ANS

8.3.2

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From the relation of S 16 = S 61 , we have η xy , x =

G xy Ex

η x , xy

⎡ 1 ⎤ 1 1 2υ12 1 + 4( + + − G xy = ⎢ ) sin 2 θ cos 2 θ ⎥ E1 E 2 E1 G12 ⎣ G12 ⎦

where

[

= 142.857142 − 108.571428 sin 2 θ cos 2 θ

]

−1

−1

× 1012

so ⎡ 2 2υ12 ⎤ 1 2 2υ12 1 + − ) sin θ cos 3 θ − ( + − ) sin 3 θ cos θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η xy , x = G xy ⎢(

[

]

= G xy − 124.285714 sin θ cos 3 θ − 61.428571sin 3 θ cos θ × 10 −12 − 124.285714 sin θ cos 3 θ − 61.428571sin 3 θ cos θ = 142.857142 − 108.571428 sin 2 θ cos 2 θ

ηxy,x 0.5 0.4 0.3 0.2 0.1 0 0

45

90

135

180

-0.1 ηxy,x

-0.2 -0.3 -0.4 -0.5

By numerical analysis, the maximum values of η xy , x can be obtained as follows.

θ = 38.96 o => η xy , x = −0.416 θ = 141 .04 o => η xy , x = 0.416 --- ANS 8.3.3

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8.4 If the carbon/epoxy composite panel is subjected to a shear stress τ xy , find (a) the fiber orientation at which σ 11 is maximum. (b) the fiber orientation at which γ xy is minimum. Compare the result with that of Problem 8.3. Solution: (a) The two sets of stress components with respect to these two coordinates systems are related by the transformation matrix [Tσ ] , ⎧σ xx ⎫ ⎧σ 11 ⎫ ⎪ ⎪ ⎪ ⎪ ⎨σ 22 ⎬ = [Tσ ]⎨σ yy ⎬ ⎪σ ⎪ ⎪σ ⎪ ⎩ 12 ⎭ ⎩ xy ⎭

(8.4.1)

where ⎡ cos 2 θ [Tσ ] = ⎢⎢ sin 2 θ ⎢− sin θ cos θ ⎣

sin 2 θ cos θ sin θ cos θ 2

2 sin θ cos θ ⎤ ⎥ − 2 sin θ cos θ ⎥ cos 2 θ − sin 2 θ ⎥⎦

(8.4.2)

If the carbon/epoxy composite panel is subjected to a shear stress τ xy , then

σ 11 = 2 sin θ cos θτ xy

(8.4.3)

The maximum value of (8.4.3) can be determined by

∂σ 11 =0 ∂θ

θ=

=>

π 4

= 45 o

--- ANS

(b) The strain-stress relation in an arbitrary coordinate system (x,y) can be written as,

⎡ 1 ⎢ E ⎧ε xx ⎫ ⎢ x ⎪ ⎪ ⎢ υ xy ⎨ε yy ⎬ = ⎢− ⎪γ ⎪ ⎢ E x ⎩ xy ⎭ ⎢ η x , xy ⎢ ⎣⎢ E x

−

υ yx

Ey 1 Ey

η y , xy Ey

η xy , x ⎤

⎥ G xy ⎥ ⎧σ ⎫ η xy , y ⎥ ⎪ xx ⎪ ⎥ ⎨σ yy ⎬ G xy ⎥ ⎪ ⎪ σ 1 ⎥ ⎩ xy ⎭ ⎥ G xy ⎦⎥

(8.4.4)

where 8.4.1

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⎡1 ⎤ 2υ 1 1 − 12 ) sin 2 θ cos 2 θ + sin 4 θ ⎥ E x = ⎢ cos 4 θ + ( G12 E1 E2 ⎣ E1 ⎦

−1

⎡υ ⎤ 1 1 2υ12 1 υ xy = E x ⎢ 12 − ( + + − ) sin 2 θ cos 2 θ ⎥ ⎣ E1

E1

E2

E1

⎦

G12

⎡1 ⎤ 2υ 1 1 − 12 ) sin 2 θ cos 2 θ + cos 4 θ ⎥ E y = ⎢ sin 4 θ + ( G12 E1 E2 ⎣ E1 ⎦

G xy

⎡ 1 ⎤ 1 1 2υ12 1 =⎢ + 4( + + − ) sin 2 θ cos 2 θ ⎥ E1 E 2 E1 G12 ⎣ G12 ⎦

−1

(8.4.5)

−1

⎡ 2 2υ12 ⎤ 1 2 2υ 1 + − ) sin θ cos 3 θ − ( + 12 − ) sin 3 θ cos θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η x , xy = E x ⎢(

⎡ 2 2υ12 ⎤ 1 2 2υ 1 + − ) sin 3 θ cos θ − ( + 12 − ) sin θ cos 3 θ ⎥ E1 G12 E2 E1 G12 ⎣ E1 ⎦

η y , xy = E y ⎢(

If the carbon/epoxy composite panel is subjected to a shear stress τ xy only, then

γ xy =

σ xy

(8.4.6)

Gxy

Assume the composite properties as E1 = 140GPa , E 2 = 10GPa , G12 = 7GPa , υ12 = 0.3 , and then

⎡ 1 ⎤ 1 1 2υ12 1 + 4( + + − ) sin 2 θ cos 2 θ ⎥ G xy = ⎢ E1 E 2 E1 G12 ⎣ G12 ⎦

[

= 142.857 − 125.714 sin 2 θ cos 2 θ

]

−1

−1

× 1012

From (8.4.6) the shear strain becomes

γ xy =

σ xy Gxy

[

]

= 142.857 − 125.714 sin 2 θ cos 2 θ × 10−12 × σ xy

(8.4.7)

The minimum value of (8.4.7) can be determined by ∂γ xy π =0 => θ = = 45 o ∂θ 4 --- ANS

8.4.2

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The maximum value of σ 12 and the minimum value of γ xy both appear at θ = 45 o , whereas in Problem 8.3, the minimum values of η x, xy and η xy , x appear at

θ = 13.22 o and θ = 43.996 o , respectively. --- ANS

8.4.3

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8.5

Consider a [± 45]s laminate. If the constituent composite material is highly anisotropic, i.e.,

E1 >> E 2 and E1 >> G12 show that the effective engineering moduli for the laminate can be expressed approximately as E x ≈ 4Q66 ≈ 4G12

Q11 E1 ≈ 4 4 Q − 4Q66 E1 − 4G12 ≈ 11 ≈ Q11 + 4Q66 E1 + 4G12

G xy ≈

υ xy

Compare these approximate values with the exact values for AS4/3501-6 carbon/epoxy composite. Solution: For the [± 45]s laminate, the effective moduli can be expressed explicitly as A A − A12 E x = 11 22 hA22

2

(8.5.1)

A66 h A = 12 A22

G xy =

(8.5.2)

υ xy

(8.5.3)

where in this case the extensional stiffnesses are ( 45)

Aij = 2t (Q ij

( −45 )

+ Q ij

)

(8.5.4)

and the thickness is h = 4t From equation (8.11) of the textbook, we have

Q11 = Q11 cos 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 sin 4 θ Q12 = Q 21 = Q12 (sin 4 θ + cos 4 θ ) + (Q11 + Q22 − 4Q66 ) sin 2 θ cos 2 θ Q 22 = Q11 sin 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 cos 4 θ

(8.5.5)

Q16 = Q 61 = (Q11 − Q12 − 2Q66 ) sin θ cos 3 θ + (Q12 − Q22 + 2Q66 ) sin 3 θ cosθ Q 26 = Q 62 = (Q11 − Q12 − 2Q66 ) sin 3 θ cosθ + (Q12 − Q22 + 2Q66 ) sin θ cos 3 θ

8.5.1

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Q 66 = (Q11 + Q22 − 2Q12 − 2Q66 ) sin 2 θ cos 2 θ + Q66 (sin 4 θ + cos 4 θ ) For θ = 45 o and θ = −45 o 1 1 1 Q11 = Q11 + (Q12 + 2Q66 ) + Q22 4 2 4 Q 12 = Q 21 =

1 1 Q12 + (Q11 + Q22 − 4Q66 ) 2 4

Q 22 =

1 1 1 Q11 + (Q12 + 2Q66 ) + Q22 4 2 4

Q 66 =

1 1 1 (Q11 + Q22 − 2Q12 − 2Q66 ) + Q66 = (Q11 + Q22 − 2Q12 ) 4 2 4

Plug in (8.5.4), we have ( 45 )

( −45 )

) = t[Q11 + 2(Q12 + 2Q66 ) + Q22 ]

( 45 )

( −45 )

) = t[ 2Q12 + (Q11 + Q22 − 4Q66 )]

( 45 )

( −45 )

( 45 )

( −45 )

A11 = 2t (Q 11 + Q 11

A12 = 2t (Q 12 + Q 12

A22 = 2t (Q 22 + Q 22 ) = t[Q11 + 2(Q12 + 2Q66 ) + Q22 ] A66 = 2t (Q 66 + Q 66 ) = t[(Q11 + Q22 − 2Q12 )]

(1) Effective modulus E x

Ex =

A11 A22 − A12 hA22

2

t 2 [Q11 + 2(Q12 + 2Q66 ) + Q22 ] 2 − t 2 [2Q12 + (Q11 + Q22 − 4Q66 )]2 = 4t ⋅ t[Q11 + 2(Q12 + 2Q66 ) + Q22 ] where

Q11 = Q22 =

E1 1 − υ12υ 21 E2 1 − υ12υ 21

,

Q12 = Q21 =

,

Q66 = G12

υ12 E 2 1 − υ12υ 21

Expand the above equation and use the highly anisotropic assumption in the statement, i.e., Q Q12 Q22 ≈ 0, ≈ 0, and 66 ≈ 0 Q11 Q11 Q11

8.5.2

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Also Q11 =

E1 1 − υ12υ 21

≈ E1 because υ 21 =

E2 υ12 ≈ 0 E1

The above equation therefore can be simplified as

Ex ≈

4(Q12 + 2Q66 ) − 2(Q22 − 4Q66 ) 1 = Q12 − Q22 + 4Q66 ≈ 4Q66 = 4G12 4 2

(2) G xy

G xy =

A66 t[(Q11 + Q22 − 2Q12 )] Q11 E1 = ≈ ≈ h 4t 4 4

(3) υ xy

υ xy = ≈

A12 t[2Q12 + (Q11 + Q22 − 4Q66 )] Q11 − 4Q66 + 2Q12 + Q22 = = A22 t[Q11 + 2(Q12 + 2Q66 ) + Q22 ] Q11 + 4Q66 + 2Q12 + Q22 E1 − 4G12 E1 + 4G12

--- ANS The material properties for AS4/3501-6 carbon/epoxy composite are E1 = 140 GPa E 2 = 10 GPa G12 = 7 GPa υ12 = 0.3 E 10 (0.3) = 0.021 => υ 21 = 2 υ12 = E1 140 therefore, Q11 =

140 = 140.89 GPa 1 − (0.3)(0.021)

Q12 = Q21 = E2

υ12 E 2 (0.3)(10) = = 3.02 GPa 1 − υ12υ 21 1 − (0.3)(0.021)

10 = 10.06 GPa 1 − υ12υ 21 1 − (0.3)(0.021) Q66 = G12 = 7 GPa Q22 =

=

(a) Exact effective engineering moduli [Q + 2(Q12 + 2Q66 ) + Q22 ] 2 − [2Q12 + (Q11 + Q22 − 4Q66 )]2 E x = 11 4[Q11 + 2(Q12 + 2Q66 ) + Q22 ] [140.89 + 2(3.02 + 2 × 7) + 10.06] 2 − [2 × 3.02 + (140.89 + 10.06 − 4 × 7)]2 = 4[140.89 + 2(3.02 + 2 × 7) + 10.06] = 23.76 GPa 8.5.3

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[(Q11 + Q22 − 2Q12 )] (140.89 + 10.06 − 2 × 3.02) = = 36.23 GPa 4 4 [2Q12 + (Q11 + Q22 − 4Q66 )] = [Q11 + 2(Q12 + 2Q66 ) + Q22 ]

G xy =

υ xy

=

(2 × 3.02 + 140.89 + 10.06 − 4 × 7) = 0.70 GPa [140.89 + 2 × (3.02 + 2 × 7) + 10.06]

(b) Approximate effective engineering moduli E x ≈ 4Q66 = 4G12 = 4 × 7 = 28 GPa

Err (%) = 17.85%

Q11 E1 140 ≈ = = 35 GPa 4 4 4 E − 4G12 140 − 4 × 7 = = 0.67 GPa ≈ 1 E1 + 4G12 140 + 4 × 7

G xy ≈

Err (%) = 3.4%

υ xy

Err (%) = 4.3%

8.5.4

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8.6

Compare the in-plane longitudinal stiffnesses in the x-direction for [± 30 / 0]s and [30 2 / 0]s laminates of AS4/3501-6 carbon/epoxy composite. Which is stiffer?

Solution: The effective engineering moduli in the x-direction for the laminate is 1 Ex = hA11 '

(8.6.1)

where

[A'] = [A]−1

(8.6.2)

and in this case ( 30 )

Aij = 2t (Q ij

( −30 )

+ Q ij

( 0)

+ Q ij )

(8.6.3)

also the thickness is h = 6t From equation (8.11) of the textbook, we have

Q11 = Q11 cos 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 sin 4 θ Q12 = Q 21 = Q12 (sin 4 θ + cos 4 θ ) + (Q11 + Q22 − 4Q66 ) sin 2 θ cos 2 θ Q 22 = Q11 sin 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 cos 4 θ

(8.6.4)

Q16 = Q 61 = (Q11 − Q12 − 2Q66 ) sin θ cos 3 θ + (Q12 − Q22 + 2Q66 ) sin 3 θ cosθ Q 26 = Q 62 = (Q11 − Q12 − 2Q66 ) sin 3 θ cosθ + (Q12 − Q22 + 2Q66 ) sin θ cos 3 θ Q 66 = (Q11 + Q22 − 2Q12 − 2Q66 ) sin 2 θ cos 2 θ + Q66 (sin 4 θ + cos 4 θ ) For θ = 30 o Q 11 =

9 3 1 Q11 + (Q12 + 2Q66 ) + Q22 16 8 16

5 3 Q 12 = Q 21 = Q12 + (Q11 + Q22 − 4Q66 ) 8 16 Q 22 =

1 3 9 Q11 + (Q12 + 2Q66 ) + Q22 16 8 16

8.6.1

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Q 16 = Q 61 =

3 3 3 (Q11 − Q12 − 2Q66 ) + (Q12 − Q22 + 2Q66 ) 16 16

3 3 3 (Q11 − Q12 − 2Q66 ) + (Q12 − Q22 + 2Q66 ) 16 16 3 5 3 1 = (Q11 + Q22 − 2Q12 − 2Q66 ) + Q66 = (Q11 + Q22 − 2Q12 ) − Q66 16 16 16 16

Q 26 = Q 62 = Q 66

It can be written in the following form, Q11 Q12

Q22

Q16

Q26

Q66 3 4

Q11

9 16

3 8

1 16

0

0

Q 12

3 16

5 8

3 16

0

0

Q 22

1 16

3 8

9 16

0

0

Q 16

3 3 16

3 8

−

3 16

0

0

Q 26

3 16

3 8

−

3 3 16

0

0

Q 66

3 16

−

3 8

3 16

0

0

Q11

Q12

Q22

Q16

Q26

Q66

Q11

9 16

3 8

1 16

0

0

3 4

Q 12

3 16

5 8

3 16

0

0

Q 22

1 16

3 8

9 16

0

0

3 4 3 4

−

3 4

3 4

3 4

−

3 4 1 16

−

For θ = −30 o

−

3 4

Q 16

−

3 3 16

−

3 8

3 16

0

0

Q 26

−

3 16

−

3 8

3 3 16

0

0

−

3 4

3 8

3 16

0

0

−

1 16

Q 66

3 16

−

8.6.2

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For θ = 0 o

Q11

Q12

Q22

Q16

Q26

Q66

Q11

1

0

0

0

0

0

Q 12

0

1

0

0

0

0

Q 22

0

0

1

0

0

0

Q 16

0

0

0

0

0

0

Q 26

0

0

0

0

0

0

Q 66

0

0

0

0

0

1

For AS4/3501-6 carbon/epoxy composite material, the elastic moduli are E1 = 140GPa , E 2 = 10GPa , G12 = 6.9GPa , υ12 = 0.3 E 10 0.3 = 0.02143 => υ 21 = 2 υ12 = 140 E1 And the reduced stiffnesses can be obtained by E1 υ E Q11 = = 140.906 GPa , Q12 = Q21 = 12 2 = 3.0194 GPa 1 − υ12υ 21 1 − υ12υ 21

Q22 =

E2

= 10.0647 GPa , 1 − υ12υ 21 Q16 = Q61 = Q26 = Q62 = 0

Q66 = G12 = 6.9 GPa

For [± 30 / 0]s laminates of AS4/3501-6 carbon/epoxy composite The extensional stiffnesses are

9 3 1 3 × 2 + 1)(140.906) + ( × 2)(3.0194) + ( × 2)(10.0647) + ( × 2)(6.9)] 16 8 16 4 = 626.5958t

A11 = 2t[(

3 5 3 3 A12 = 2t[( × 2)(140.906) + ( × 2 + 1)(3.0194) + ( × 2)(10.0647) + (− × 2)(6.9)] 16 8 16 4 = 106.1153t

8.6.3

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1 3 9 3 × 2)(140.906) + ( × 2)(3.0194) + ( × 2 + 1)(10.0647) + ( × 2)(6.9)] 16 8 16 4 = 103.2306t

A22 = 2t[(

A16 = 0 A26 = 0 3 3 3 1 × 2)(140.906) + (− × 2)(3.0194) + ( × 2)(10.0647) + (− × 2 + 1)(6.9)] 16 8 16 16 = 120.7739t

A66 = 2t[(

That is

0 ⎤ ⎡626.596 106.115 ⎢ [A] = ⎢106.115 103.231 0 ⎥⎥t ⎢⎣ 0 0 120.774⎥⎦

(10 9 N / m)

Hence,

[A]

−1

0 ⎤ ⎡ 1.921 − 1.975 1⎢ = ⎢− 1.975 11.717 0 ⎥⎥ t ⎢⎣ 0 0 8.280⎥⎦

(10 −12 m / N )

From (8.6.1) and (8.6.2), the effective engineering moduli in the x-direction for the laminate is Ex =

1 = hA11 '

1 = 86.8 GPa 1.921 × 10 −12 (6t )( ) t

(*)

For [30 2 / 0]s laminates of AS4/3501-6 carbon/epoxy composite

The extensional stiffnesses are

9 3 1 3 × 2 + 1)(140.906) + ( × 2)(3.0194) + ( × 2)(10.0647) + ( × 2)(6.9)] 16 8 16 4 = 626.5958t

A11 = 2t[(

3 5 3 3 A12 = 2t[( × 2)(140.906) + ( × 2 + 1)(3.0194) + ( × 2)(10.0647) + (− × 2)(6.9)] 16 8 16 4 = 106.1153t

8.6.4

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1 3 9 3 × 2)(140.906) + ( × 2)(3.0194) + ( × 2 + 1)(10.0647) + ( × 2)(6.9)] 16 8 16 4 = 103.2306t

A22 = 2t[(

3 3 3 3 3 × 2)(140.906) + ( × 2)(3.0194) + (− × 2)(10.0647) + (− × 2)(6.9)] 16 8 16 4 = 169.3478t

A16 = 2t[(

3 3 3 3 3 × 2)(140.906) + ( × 2)(3.0194) + (− × 2)(10.0647) + ( × 2)(6.9)] 16 8 16 4 = 62.5057t

A22 = 2t[(

3 3 3 1 × 2)(140.906) + (− × 2)(3.0194) + ( × 2)(10.0647) + (− × 2 + 1)(6.9)] 16 8 16 16 = 120.7739t

A66 = 2t[(

Thus,

⎡626.596 106.115 169.348⎤ [A] = ⎢⎢106.115 103.231 62.506 ⎥⎥t ⎢⎣169.348 62.506 120.774⎥⎦

(10 9 N / m)

Hence,

[A]

−1

⎡ 2.602 − 0.678 3.298 ⎤ 1⎢ = ⎢− 0.678 14.285 − 6.442⎥⎥ t ⎢⎣ 3.298 − 6.442 16.238 ⎥⎦

(10 −12 m / N )

From (8.6.1) and (8.6.2), the effective engineering moduli in the x-direction for the laminate is Ex =

1 = hA11 '

1 = 64.1 GPa 2.602 × 10 −12 (6t )( ) t

(**)

By comparing (*) and (**), [± 30 / 0]s laminates is stiffer in the x-direction. --- ANS

8.6.5

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8.7

Plot the effective moduli E x , G xy , and υ xy versus θ for the angle-ply

laminate [± θ ]s of AS4/3501-6 carbon/epoxy composite.

Solution: The effective engineering moduli for the symmetric laminate are A A − A12 E x = 11 22 hA22

2

(8.7.1)

A66 h A = 12 A22

G xy =

(8.7.2)

υ xy

(8.7.3)

and in this case (θ )

( −θ )

Aij = 2t (Q ij + Q ij )

(8.7.4)

Also, the thickness is h = 4t From equation (8.11) of the textbook, we have (θ )

Q 11 = Q11 cos 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 sin 4 θ (θ )

(θ )

Q 12 = Q 21 = Q12 (sin 4 θ + cos 4 θ ) + (Q11 + Q22 − 4Q66 ) sin 2 θ cos 2 θ (θ )

Q 22 = Q11 sin 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 cos 4 θ (θ )

(θ )

(θ )

(θ )

Q 16 = Q 61 = (Q11 − Q12 − 2Q66 ) sin θ cos 3 θ + (Q12 − Q22 + 2Q66 ) sin 3 θ cos θ Q 26 = Q 62 = (Q11 − Q12 − 2Q66 ) sin 3 θ cos θ + (Q12 − Q22 + 2Q66 ) sin θ cos 3 θ (θ )

Q 66 = (Q11 + Q22 − 2Q12 − 2Q66 ) sin 2 θ cos 2 θ + Q66 (sin 4 θ + cos 4 θ )

For the case of − θ ( −θ )

= Q11 cos 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 sin 4 θ

( −θ )

= Q 21 = Q12 (sin 4 θ + cos 4 θ ) + (Q11 + Q22 − 4Q66 ) sin 2 θ cos 2 θ

Q 11 Q 12

( −θ )

( −θ )

Q 22 = Q11 sin 4 θ + 2(Q12 + 2Q66 ) sin 2 θ cos 2 θ + Q22 cos 4 θ 8.7.1

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Q 16

( −θ )

( −θ )

= −(Q11 − Q12 − 2Q66 ) sin θ cos 3 θ − (Q12 − Q22 + 2Q66 ) sin 3 θ cos θ

( −θ )

= −(Q11 − Q12 − 2Q66 ) sin 3 θ cos θ − (Q12 − Q22 + 2Q66 ) sin θ cos 3 θ

= Q 61

Q 26 = Q 62 ( −θ )

Q 66

= (Q11 + Q22 − 2Q12 − 2Q66 ) sin 2 θ cos 2 θ + Q66 (sin 4 θ + cos 4 θ )

For AS4/3501-6 carbon/epoxy composite material, the elastic moduli are E1 = 140GPa , E 2 = 10GPa , G12 = 6.9GPa , υ12 = 0.3 E 10 0.3 = 0.02143 => υ 21 = 2 υ12 = E1 140 and the reduced stiffnesses are obtained from E1 υ E Q11 = = 140.906 GPa , Q12 = Q21 = 12 2 = 3.0194 GPa 1 − υ12υ 21 1 − υ12υ 21

Q22 =

E2

= 10.0647 GPa , 1 − υ12υ 21 Q16 = Q61 = Q26 = Q62 = 0

Q66 = G12 = 6.9 GPa

For [± θ ]s laminates of AS4/3501-6 carbon/epoxy composite The extensional stiffnesses are A11 = 2t[2(140.906) cos 4 θ + 4(3.0194 + 2 × 6.9) sin 2 θ cos 2 θ + 2(10.0647 ) sin 4 θ ] = t[563.624 cos 4 θ + 134.5552 sin 2 θ cos 2 θ + 40.2588 sin 4 θ ] A12 = 2t[2(140.906 + 10.0647 − 4 × 6.9) sin 2 θ cos 2 θ + 2(3.0194)(sin 4 θ + cos 4 θ )] = t[493.4828 sin 2 θ cos 2 θ + 12.0776(sin 4 θ + cos 4 θ )] A22 = 2t[2(140.906) sin 4 θ + 4(3.0194 + 2 × 6.9) sin 2 θ cos 2 θ + 2(10.0647) cos 4 θ ] = t[563.624 sin 4 θ + 134.5552 sin 2 θ cos 2 θ + 40.2588 cos 4 θ ]

A16 = 0 A26 = 0 A66 = 2t[2((140.906 + 10.0647 − 2 × 3.0194 − 2 × 6.9) sin 2 θ cos 2 θ + 2(6.9)(sin 4 θ + cos 4 θ )] = t[524.5276 sin 2 θ cos 2 θ + 27.6(sin 4 θ + cos 4 θ )]

Effective engineering moduli From (8.7.1) to (8.7.3), we have

8.7.2

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A11 A22 − A12 , hA22 2

Ex =

G xy =

A66 , h

υ xy =

A12 A22

Where h = 4t After plugging in the values of extensional stiffness, the effective engineering moduli can be plotted as follows. Ex : Ex 160 140 120 100 80 60 40

Ex

20 0 0

45

90

135

180

G xy : Gxy 40 35 30 25 20 15 10

Gxy

5 0 0

45

90

135

180

8.7.3

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υ xy : υxy 1.4

1.2

1

0.8

0.6

0.4 υxy 0.2

0 0

45

90

135

180

--- ANS

8.7.4

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Mechanics of Aircraft structures C.T. Sun

8.8

Find the shear strains ( γ xy ) in the AS4/3501-6 carbon/epoxy composite [± 45]s

and [0 / 90]s laminates subjected to the shear loading N xy = 1000 N / m . Also find the lamina stresses σ 11 , σ 22 , and σ 12 . If the maximum shear strength of the composite is σ 12 = 100MPa , what are the shear loads ( N xy ) the two laminates can carry?

Solution: For AS4/3501-6 carbon/epoxy composite material, the elastic moduli are E1 = 140GPa , E 2 = 10GPa , G12 = 7GPa , υ12 = 0.3 => υ 21 =

E2 10 υ12 = 0.3 = 0.02143 140 E1

The ply thickness = 0.127 mm. And the reduced stiffnesses are E1 Q11 = = 140.9 GPa , 1 − υ12υ 21

Q22 =

Q12 = Q21 =

E2

= 10.06 GPa , 1 − υ12υ 21 Q16 = Q61 = Q26 = Q62 = 0

υ12 E 2 = 3.02 GPa 1 − υ12υ 21

Q66 = G12 = 7 GPa

The following stiffness matrices are readily calculated:

[Q]

0⎤ ⎡140.9 3.02 ⎢ = ⎢ 3.02 10.06 0 ⎥⎥ × 10 9 Pa ⎢⎣ 0 0 7.0⎥⎦

[Q]

0⎤ ⎡10.6 3.02 ⎢ = ⎢3.02 140.9 0 ⎥⎥ × 10 9 Pa ⎢⎣ 0 0 7.0⎥⎦

0o

90o

[Q]

± 45o

32.25 ± 32.71⎤ ⎡ 46.25 ⎢ 46.25 ± 32.71⎥⎥ × 10 9 Pa = ⎢ 32.25 ⎢⎣± 32.71 ± 32.71 36.23 ⎥⎦

(1) For [± 45]s laminate, 0 ⎤ ⎡23.495 16.383 ⎢ [A] = ⎢16.383 23.495 0 ⎥⎥ × 10 6 N / m ⎢⎣ 0 0 18.405⎥⎦ 8.8.1

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[A]

−1

0 ⎤ ⎡ 82.842 − 57.765 ⎢ 0 ⎥⎥ × 10 −9 m / N = ⎢− 57.765 82.842 ⎢⎣ 0 0 54.333⎥⎦

The strains are ⎧ 0 ⎫ ⎧ ⎧ε xx ⎫ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ 0 ⎬ ⎨ε yy ⎬ = [A] ⎨ 0 ⎬ = ⎨ ⎪ N ⎪ ⎪54.333 × 10 −6 ⎪ ⎪γ ⎪ ⎭ ⎩ xy ⎭ ⎩ ⎩ xy ⎭

--- ANS Note:

Since only the load N xy = 1000 N / m exists, and in addition, A16 = A26 = 0 , we can just consider the 66-component, which is γ xy =

1 N xy . Therefore the resulting shear A66

strain can be also easily found as 1 1000 γ xy = N xy = = 54.333 × 10 − 6 6 A66 18.405 × 10 The laminar stresses are

⎧σ xx ⎫ ⎧ 0 ⎫ ⎧± 1777.23⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = Q ± 45o ⎨ 0 ⎬ = ⎨± 1777.23⎬ × 103 Pa ⎨σ yy ⎬ ⎪σ ⎪ ⎪γ ⎪ ⎪ 1968.48 ⎪ ⎭ ⎩ xy ⎭± 45o ⎩ xy ⎭ ⎩

[]

and ⎧σ xx ⎫ ⎡ cos 2 θ ⎧σ 11 ⎫ ⎪ ⎪ ⎢ ⎪ ⎪ 2 ⎨σ 22 ⎬ = [Tσ ]⎨σ yy ⎬ = ⎢ sin θ ⎪σ ⎪ ⎢− sin θ cos θ ⎪σ ⎪ ⎩ 12 ⎭ ⎩ xy ⎭ ⎣

sin 2 θ cos θ sin θ cos θ 2

2 sin θ cos θ ⎤ ⎧σ xx ⎫ ⎥⎪ ⎪ − 2 sin θ cos θ ⎥ ⎨σ yy ⎬ cos 2 θ − sin 2 θ ⎥⎦ ⎪⎩σ xy ⎪⎭

Therefore 0.5 ± 1⎤ ⎧σ xx ⎫ ⎧± 3745.71⎫ ⎧σ 11 ⎫ ⎡ 0.5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ = ⎢ 0.5 0.5 m 1⎥⎥ ⎨σ yy ⎬ = ⎨ m 191.25 ⎬ KPa ⎨σ 22 ⎬ ⎪σ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎩ 12 ⎭ ± 45o ⎢⎣m 0.5 ± 0.5 0 ⎥⎦ ⎩σ xy ⎭ ⎩ ⎭ --- ANS 8.8.2

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(2) For [0 / 90]s laminate, Similarly,

0 ⎤ ⎡38.481 1.534 ⎢ [A] = ⎢ 1.534 38.481 0 ⎥⎥ × 10 6 N / m ⎢⎣ 0 0 3.556⎥⎦

[A]

−1

0 ⎤ ⎡ 23.323 − 1.038 ⎢ 0 ⎥⎥ × 10 −9 m / N = ⎢− 1.038 23.323 ⎢⎣ 0 0 281.215⎥⎦

Hence, the strains are ⎧ 0 ⎫ ⎧ ⎧ε xx ⎫ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ 0 ⎬ ⎨ε yy ⎬ = [A] ⎨ 0 ⎬ = ⎨ ⎪ N ⎪ ⎪281.215 × 10 −6 ⎪ ⎪γ ⎪ ⎭ ⎩ xy ⎭ ⎩ ⎩ xy ⎭

--- ANS The laminar stresses are ⎧σ xx ⎫ ⎧σ 11 ⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3 ⎨σ yy ⎬ = ⎨σ 22 ⎬ = Q 0 o {ε } = ⎨ 0 ⎬ × 10 Pa ⎪σ ⎪ ⎪ ⎪ ⎪1968.48⎪ ⎩ ⎭ ⎩ xy ⎭0 o ⎩σ 12 ⎭

[]

⎧σ xx ⎫ ⎧σ 22 ⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3 ⎨σ yy ⎬ = ⎨σ 22 ⎬ Q 90 o {ε } = ⎨ 0 ⎬ × 10 Pa ⎪σ ⎪ ⎪ ⎪ ⎪1968.48⎪ ⎩ ⎭ ⎩ xy ⎭90 o ⎩σ 12 ⎭

[]

--- ANS If the maximum shear strength σ 12 = 100MPa of the composite is the only failure condition, then the maximum shear load Nxy can be obtained as follows. For [± 45]s laminate, σ 12 = 0 , then theoretically the laminate can carry unbounded shear loads. In reality, of course, other failure mechanisms would take over, and result in a finite maximum shear load. 8.8.3

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For [0 / 90]s laminate, 1.96848x ≤ 100 , => x ≤ 50.8 . The maximum shear load is N xy , max = 50.8 kN / m --- ANS

8.8.4

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