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Chapter 5 Solution

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PROBLEM 5.1

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION   Reactions: 0: 0 C Pb M LA bP A L Σ = − = = 0: 0 A Pa M LC aP C L Σ = − = = From A to B: 0< < x a 0: 0 y Pb F V L Σ = − = Pb V L =  0: 0 J Pb M M x L Σ = − = Pbx M L =  From B to C: a < < x L 0: 0 y Pa F V L Σ = + = Pa V L = −  0: ( ) 0 K Pa M M L x L Σ = − + − = ( ) Pa L x M L − = At section B: 2 Pab M L = 

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PROBLEM 5.2

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION Reactions: B 0: 2 0 2 L wL M AL wL A Σ = − + ⋅ = = A 0: 2 0 2 L wL M BL wL B Σ = − ⋅ = =

Free body diagram for determining reactions. Over whole beam, 0 < < x L

Place section at x.

Replace distributed load by equivalent concentrated load.

0: 0 2 y wL F wx V Σ = − − = 2 L V = wx    0: 0 2 2 J wL x M x wx M Σ = − + + = 2 ( ) 2 w M = Lxx ( ) 2 w M = x L−  x Maximum bending moment occurs at .

2 L x = 2 max 8 wL M = 

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PROBLEM 5.15

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION Reaction at A:

0:MB = −4.5A+(3.0)(3)+(1.5)(3) +(1.8)(4.5)(2.25) = 0 A=7.05 kN ↑ Use AC as free body.

3 0: (7.05)(1.5) (1.8)(1.5)(0.75) 0 8.55 kN m 8.55 10 N m C C C M M M Σ = − + = = ⋅ = × ⋅ 3 3 6 4 6 4 1 1 (80)(300) 180 10 mm 12 12 180 10 m 1 (300) 150 mm 0.150 m 2 I bh c − = = = × = × = = = 3 6 6 (8.55 10 )(0.150) 7.125 10 Pa 180 10 Mc I σ = = × = × × 7.13 σ = MPa 

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F n SOLUTION U U

PROBLEM 5.20

For the beam and loading shown, determine the max normal stress due to bending on a transverse section at C

Use entire beam as free body. 0 : B M= 4.8A (3.6)(216) (1.6)(150) (0.8)(150) 0 − + + + = 237 kN = A

Use portion AC as free body. 0 : C M= (2.4)(237) (1.2)(216) 0 309.6 kN m M M − + = = ⋅ For W460×113, S = 2390×10 mm6 3 Normal stress: 3 6 3 6 309.6 10 N m 2390 10 m 129.5 10 Pa M S σ = = × ⋅ × = × 129.5 σ = M ximum C. MPa 

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P

D an d SOLUTION

P

ROBLEM 5.22

Draw the shear and bending-moment diagrams for the nd loading shown and determine the maximum normal ue to bending. Reactions: 0 : 4 64 (24)(2)(1) 0 28 D M A= = A= 0 : 28 (24)(2) 0 76 k y F D  = − + − = D = A to C: 0 < <x 2m 0 : 28 0 28 kN y F V V= − − = = − 0 : 28 0 ( 28 ) kN m J M M x M x  = + = = − ⋅ C to D: 2m < <x 4m 0 : 28 0 2 y F V V= − − = = − 0 : 28 64 ( 28 64) kN m J M M x M x= + = − + ⋅ D to B: 4m < <x 6m 0 : 24(6 ) 0 ( 24 144) kN y F V x V x  = − − = = − + 2 0 : 6 24(6 ) 2 12(6 ) kN m J M x M x M x  = −   − − −    = − − ⋅ 3 maxM =56 kN m ⋅ =56×10 N m⋅ For S250×52 section, S = 482×10 mm3 3 Normal Stress: 3 6 3 56 10 N m 116.2 10 482 10 m M S σ = = × ⋅ = × × 116.2 MPσ = beam stress 8 kN kN N 28 kN 4 0 m = 0 m  =   6 0 Pa Pa 

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PROBLEM 5.42

Using the method of Sec. 5.3, Solve Prob. 5.9

PROBLEM 5.9 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION Reactions: 0 : 2 (12)(2)(1) (40)(1) 0 8 kN C M A  = + − = = ↓ A 0 : 2 (12)(2)(1) (40)(3) 0 72 kN A M C= = = ↑ C Shear diagram: VA = −8 kN A to C: 0< <x 2 m w=12kN/m 2 2 0 012 24 kN 24 8 32 kN C A C V V wdx dx V − = − = − = − = − − = −

C to B: 32VB = − +72 = 40 kN Areas of shear diagram:

A to C: 1( 8 32)(2) 40 kN m 2 Vdx= − − = − C to B: Vdx=(1)(40)=40 kN m⋅ Bending moments: 0 0 40 40 kN m 40 40 0 A C A B C M M M Vdx M M Vdx = = +  = − = − ⋅ = +  = − + = (a) Maximum V = 40.0 kN  (b) Maximum M = 40.0 kN m⋅ 

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PROBLEM 5.93

Beams AB, BC, and CD have the cross section shown and are pin-connected at B and C. Knowing that the allowable normal stress is +110 MPa in tension and −150 MPa in compression, determine (a) the largest permissible value of P if beam BC is not to be overstressed, (b) the corresponding maximum distance a for which the cantilever beams AB and CD are not overstressed.

SOLUTION 0 B C B C M M V V P = = = − =

Area B to E of shear diagram: 2.4 P

0 2.4 P 2.4 P

E F

M = + = =M

Centroid and moment of inertia:

Part A(mm )2 y(mm) Ay(mm )3 d(mm) Ad2(mm )4 I (mm )4  2500 156.25 390625 34.82 6 3.031 10× 0.0326 10× 6  1875 75 140625 46.43 4.042 10× 6 3.516 10× 6 Σ 4375 531250 7.073 10× 6 3.548 10× 6 2 6 4 531250 121.43 mm 4375 10.621 10 mm Y I Ad I = = = Σ + Σ = × Location y(mm) I y/ (10 mm )3 3 ← also (10−6m )3 Top 41.07 258.6 Bottom −121.43 −87.47

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PROBLEM 5.93 (Continued)

Bending moment limits: M = −σI y/

Tension at E and F: −(110 10 ) ( 87.47 10 )× 6 − × −6 =9.622 10 N m× 3 ⋅ Compression at E and F: − −( 150 10 )(258.6 10 )× 6 × −6 =38.8 10 N m× 3 ⋅ Tension at A and D: −(110 10 ) (258.6 10 )× 6 × −6 = −28.45 10 N m× 3 ⋅

Compression at A and D: − −( 150 10 )( 87.47 10 )× 6 − × −6 = −13.121 10 N m× 3 ⋅

(a) Allowable load P: 2.4 P=9.622 10× 3 P=4.01 10 N× 3 P=4.01 kN  Shear at A: VA=P

Area A to B of shear diagram: aVA=aP

Bending moment at A: MA = −aP = −4.01 10× 3a

(b) Distance a: −4.01 10× 3a= −13.121 10× 3 3.27 a= m 

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PROBLEM 5.120

Using a computer and step functions, calculate the shear and bending moment for the beam and loading shown. Use the specified increment ΔL, starting at point A and ending at the right-hand support.

SOLUTION 1 0: 6 (4)(120) (1) (3)(36) 0 2 D A M R   Σ = − + +    = 89 kN A R = 1 1 36 3 12 3 3 w=  −  =  −  x x 0 2 89 120 2 6 3 kN V = −  −  −  − x x  1 3 89 120 2 2 3 kN m M = x−  −  −  − x x ⋅  x V M m kN kN ⋅ m 0.0 89.0 0.0 0.3 89.0 22.3 0.5 89.0 44.5 0.8 89.0 66.8 1.0 89.0 89.0 1.3 89.0 111.3 1.5 89.0 133.5 1.8 89.0 155.8 2.0 −31.0 178.0 2.3 −31.0 170.3 2.5 −31.0 162.5 2.8 −31.0 154.8 3.0 −31.0 147.0 3.3 −31.4 139.2 3.5 −32.5 131.3 3.8 −34.4 122.9 4.0 −37.0 114.0 4.3 −40.4 104.3 4.5 −44.5 93.8 4.8 −49.4 82.0 x V M m kN kN ⋅ m 5.0 −55.0 69.0 5.3 −61.4 54.5 5.5 −68.5 38.3 5.8 −76.4 20.2 6.0 −85.0 −0.0

(10)

PROBLEM 5.121

Using a computer and step functions, calculate the shear and bending moment for the beam and loading shown. Use the specified increment ΔL, starting at point A and ending at the right-hand support.

SOLUTION 0: (5.2)(12) 4 (2)(4)(16) 0 C M B Σ = − + = 47.6 kN = ↑ B 0: (1.2)(12) (2)(4)(16) 4 0 B M C Σ = − + = 28.4 kN = ↑ C 0 16 1.2 dV w x dx =  −  = − 1 0 16 1.2 12 47.6 1.2 V= −  −  − +x  − x  2 1 8 1.2 12 47.6 1.2 M = −  −  −x x+  − xx V M m kN kN ⋅ m 0.0 −12.0 0.00 0.4 −12.0 −4.80 0.8 −12.0 −9.60 1.2 35.6 −14.40 1.6 29.2 −1.44 2.0 22.8 8.96 2.4 16.4 16.80 2.8 10.0 22.08 3.2 3.6 24.80 3.6 −2.8 24.96 4.0 −9.2 22.56 4.4 −15.6 17.60 4.8 −22.0 10.08 5.2 −28.4 −0.00 

References

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