11.35 The beam shown in Fig. P11.35 consists of a W360 × 79 structural steel wide-flange shape [ E E = = 200 GPa; I GPa; I = = 225 × 6 4 10 mm ]. For the loading shown, determine: (a) the reactions at A at A,, B, B, and C . (b) the magnitude of the maximum bending stress in the beam.
Fig. P11.35
Solution (a) Reactions at A, B, and C . Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported between A between A and and C . Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 9 m, a = 6 m Calculation: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
=−
(80 kN/m)(6 m)3 24(9 m) EI
⎡⎣ 4(9 m) − 7(6 m)(9 m) + 3(6 m) ⎤⎦ = − 2
2
4, 320 kN-m3 EI
Consider downward deflection of simply supported beam at B due to concentrated moment. [Appendix C, SS beam with concentrated moment at one end of span.] Relevant equation from Appendix C: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = 240 kN-m, L kN-m, L = = 9 m, x m, x = = 3 m Calculation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(240 kN-m)(3 m) 6(9 m) EI
⎡⎣ 2(9 m) − 3(9 m)(3 m) + (3 m) ⎤⎦ = − 2
2
1, 200 kN-m3 EI
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Consider upward deflection of simply supported beam at B due to concentrated load R B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = − R B, L = L = 9 m, a = 3 m, b = 6 m
Calculation: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(− R B )(3 m)(6 m) 6(9 m) EI
⎡⎣(9 m) − (3 m) − (6 m) ⎤⎦ = 2
2
2
(12 m3 ) RB EI
Compatibility equation for deflection at B: 4, 320 kN-m3 1, 20 200 kN-m3 (12 m3 ) R B − − + =0 EI EI EI
∴ R B =
5,520 kN-m kN-m3 12 m 3
= 460 kN = 460 kN ↑
Ans.
Equilibrium equations for entire beam: Σ M A = RB (3 m) + RC (9 m) − 240 kN-m − (80 kN/m)(6 m)(6 m) = 0
∴ RC =
240 240 kNkN-m m + (80 kN/ kN/m)(6 m)(6 m) m) − (460 (460 kN) kN)(3 m) m) 9m
= 193.3333 kN kN = 193.3 kN kN ↑
Ans.
Σ F y = RA + RB + RC − (80 kN kN/m)(6 m) = 0
∴ R A = (80 kN/m)(6 m) − 460 kN − 193.3333 kN = −173.3333 kN = 173.3 kN ↓
Ans.
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Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 225 × 106 mm mm4 d = 353 mm
S = 1, 27 270 × 103 mm3 Maximum bending moment magnitude M max max = 280 kN-m Bending stresses at maximum moment (280 kN-m)(353 mm/2)(1,000) mm/2)(1,000)2 σ x = 225 × 106 mm4
= 220 MPa
Ans.
or using the tabulated section modulus value: (280 kN-m)(1,000)2 σ x = 1, 27 270 × 103 mm3
= 220 MPa
Ans.
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11.36 The beam shown in Fig. P11.36 consists of a W610 × 140 structural steel wide-flange shape [ E E = = 200 GPa; I GPa; I = = 1,120 6 4 × 10 mm ]. For the loading shown, determine: (a) the reactions at A at A,, B, B, and D and D.. (b) the magnitude of the maximum bending stress in the beam.
Fig. P11.36
Solution (a) Reactions at A, B, and D. Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported between A between A and and D D.. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wx v B = − ( L3 − 2 Lx 2 + x3 ) (elastic curve) 24 EI Values: w = 90 kN/m, L kN/m, L = = 7.5 m, x m, x = = 1.5 m
Calculation: wx v B = − ( L3 − 2 Lx 2 + x3 ) 24 EI
=−
(90 kN/m)(1.5 m) 24 EI
⎡⎣(7.5 m) − 2(7.5 m)(1.5 m) + (1.5 m) ⎤⎦ = − 3
2
3
2, 202.1875 kN-m3 EI
Consider downward deflection of simply supported beam at B due to concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 v B = − ( L − b2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 160 kN, L kN, L = = 7.5 m, b = 2.5 m, x m, x = = 1.5 m Calculation: Pbx 2 v B = − ( L − b2 − x2 ) 6 LEI L EI
=−
(160 kN)(2.5 m)(1.5 m) 6(7.5 m) EI
⎡⎣ (7.5 m) − (2.5 m) − (1.5 m) ⎤⎦ = − 2
2
2
636.6667 kN-m3 EI
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Consider upward deflection of simply supported beam at B due to concentrated load R B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = − R B, L = L = 7.5 m, a = 1.5 m, b = 6 m
Calculation: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(− R B )(1.5 m)(6 m) 6(7.5 m) EI
⎡⎣(7.5 m)2 − (1.5 m)2 − (6 m)2 ⎤⎦ =
(3.6 m3 ) RB EI
Compatibility equation for deflection at B: 2, 20 202.1875 kN-m3 636.6667 kN kN-m3 (3.6 m3 ) R B − − + =0 EI EI EI
∴ R B =
2,838.8542 kN-m3 3.6 m3
= 788.5706 kN kN = 789 kN kN ↑
Ans.
Equilibrium equations for entire beam: Σ M A = RB (1.5 m) m) + RD (7.5 m) − (90 kN kN/m)(7.5 m)(3.75 m) − (160 kN kN)(5 m) = 0
∴ R D =
(90 (90 kN/m kN/m)( )(7. 7.5 5 m)( m)(3.7 3.75 5 m) m) + (160 kN)( kN)(5 5 m) m) − (788. (788.57 5706 06 kN) kN)(1 (1.5 .5 m) 7.5 m
= 286.4525 kN kN = 286 kN kN ↑
Ans.
Σ F y = RA + RB + RD − (90 kN kN/m)(7.5 m) − 160 kN kN = 0 kN/m)(7.5 m) + 160 kN kN − 788.5706 kN kN − 286.4525 kN kN ∴ R A = (90 kN
= −240.0231 kN = 240 kN ↓
Ans.
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Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 1,120 × 106 mm4 d = 617 mm
S = 3, 64 640 × 103 mm3 Maximum bending moment magnitude M max max = 461.28 kN-m Bending stresses at maximum moment (461.28 kN-m)(617 kN-m)(617 mm/2)(1,000)2 σ x = 1,120 × 106 mm4
= 127.1 MPa
Ans.
or using the tabulated section modulus value: (461.28 kN-m)(1,000)2 σ x = 3, 64 640 × 103 mm3
= 126.8 MPa
Ans.
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11.37 A propped cantilever beam is loaded as shown in Fig. P11.37. Determine the reactions at A at A and D and D for for the beam. Assume 6 2 EI = = 12.8 × 10 lb-in. .
Fig. P11.37
Solution Choose the reaction force at D at D as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at D due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly u niformly distributed load.] Relevant equations from Appendix C: wL4 wL3 θ C = − vC = − and 8 EI 6 EI Values: w = 20 lb/in., L lb/in., L = = 72 in.
Calculation: wL4 (20 (20 lb/ lb/iin.)(48 in.)4 13,271 ,271,040 ,040 lb lb-in.3 vC = − =− =− 8 EI 8EI EI θ C =
−
v D = −
wL3
=−
(20 (20 lb lb/in.)(48 in in.)3
6 EI
=−
368,64 ,640 lb lb-in.2
6 EI
13, 271, 040 lb-in.3 EI
EI
⎛ 368, 640 lb-in.2 ⎞ 22,118, 400 lb-in.3 − (24 in.) ⎜ ⎟= − EI EI ⎝ ⎠
Consider downward deflection of cantilever beam at D due to the 320-lb concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 θ B = − v B = − and 3 EI 2 EI Values: P = = 320 lb, L lb, L = = 24 in.
Calculation: PL3 (320 lb lb)(24 in.)3 1,474 ,474,56 ,560 lb lb-in.3 v B = − =− =− 3 EI 3EI EI θ B
=−
v D = −
PL2
=−
(320 lb lb)(24 in.)2
2 EI
=−
92,160 lb lb-in.2
2 EI
1, 474, 560 lb-in.3 EI
−
EI 92,160 lb-in.2 EI
(48 in.) = −
5, 898, 240 lb-in.3 EI
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Consider upward deflection of cantilever beam at D due to concentrated load R D. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v D = − 3 EI Values: P = = − R D, L = L = 72 in.
Calculation: v D = −
PL3
=−
3 EI
(− R D )(72 in.)3
=
(124, 416 in.3 ) RD
3EI
EI
Compatibility equation for deflection at C : 22,118,400 ,400 lb lb-in.3 5,89 ,898,240 ,240 lb lb-in.3 (124,416 ,416 in in.3 ) R D − − + =0 EI EI EI
∴ R D =
28,016,640 28,016, 640 lb-in. lb-in.3 124,416 in.3
Ans.
= 225.1852 lb lb = 225 lb ↑ Shear-force and bending-moment diagrams
Equilibrium equations for entire beam:
Σ F y = RA + RD − (20 lb lb/in.)(48 in.) − 320 lb lb = 0 ∴ R A = (20 lb lb/in.)(48 in.) + 320 lb lb − 225.1852 lb lb = 1,05 ,054.8148 lb = 1,05 ,055 lb ↑
Ans.
Σ M A = − M A − (20 lb lb/in.)(48 in.)(24 in.) − (320 lb lb)(24 in.) + RD (72 in in.) = 0 ∴ M A = (225.1 (225.185 852 2 lb)( lb)(72 72 in. in.)) − (20 lb/ lb/in in.) .)(4 (48 8 in.) in.)(2 (24 4 in.) in.) − (320 (320 lb) lb)(2 (24 4 in.) in.) = − 14,506. 14,506.66 6667 67 lblb-in. in. = 14,510 14,510 lb-i lb-in. n. (ccw (ccw))
Ans.
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11.38 A propped cantilever beam is loaded as 6 shown in Fig. P11.38. Assume EI = 24 × 10 2 kip-in. . Determine: (a) the reactions at B at B and and C for for the beam. (b) the beam deflection at A at A..
Fig. P11.38
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 v B = − 8 EI Values: w = 6 kips/ft, L kips/ft, L = = 24 ft
Calculation: v B = −
wL4 8 EI
=−
(6 ki kips/ft)(24 ft) 4
= −
248,83 ,832 ki kip-ft 3
8EI
EI
(a)
Consider downward deflection of cantilever beam at B due to the 30-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 v B = − (3L − x) (elastic curve) 6 EI Values: P = = 30 kips, L kips, L = = 36 ft, x ft, x = = 24 ft
Calculation: v B = −
Px 2 6 EI
(3L − x) = −
(30 kips)(24 ft)2 6 EI
[3(36 ft) − (24 ft)] = −
241, 920 kip-ft3 EI
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Consider upward deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: P = = − R B, L = L = 24 ft
Calculation: v B = −
PL3
=−
3 EI
(− R B )(24 ft)3
=
3EI
(4, 608 ft 3 ) RB EI
(b)
Compatibility equation for deflection at B: 248, 83 832 kip-ft3 241, 92 920 kip-ft3 (4, 60 608 ft3 ) R B − − + =0 EI EI EI
∴ R B =
490,752 kip-ft kip-ft3 4,608 ft 3
= 106 106.500 500 ki kips = 106.5 6.5 ki kips ↑
Ans.
Equilibrium equations for entire beam: Σ F y = RB + RC − 30 ki kips − (6 ki kips/ft)(24 ft) = 0
∴ RC = 30 ki kips + (6 ki kips/ft)(24 ft ft) − 106.500 ki kips = 67.5 ki kips ↑
Ans.
Σ M C = M C + (30 kips)(36 ft) + (6 kips/ft)(24 ft)(12 ft) − RB (24 ft) = 0 ∴ M C = (106.5 106.500 00 kips kips)( )(24 24 ft) ft) − (30 (30 kips) kips)(3 (36 6 ft) ft) − (6 kips kips/f /ft) t)(2 (24 4 ft)( ft)(12 12 ft) ft) = − 252. 252.0 0 kipkip-fft = 252 252 kipkip-fft (cw) cw)
Ans.
(b) Beam deflection at A: Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 θ B = v B = − and (slope magnitude) 8 EI 6 EI Values: 6 2 w = 6 kips/ft, L kips/ft, L = = 24 ft, EI ft, EI = = 24 × 10 kip-in.
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Calculation: 248,832 kip-ft 3 v B = − EI θ B
=
wL3
=
(6 ki kips/f ps/ftt)(24 )(24 ft) ft)3
6 EI
v A = −
calculated previously in Eq. (a)
=
13,82 13,824 4 kipkip-ft ft2
6 EI
248, 832 kip-ft3 EI
EI
⎛ 13, 824 kip-ft2 ⎞ 414, 720 kip-ft3 − (12 ft) ⎜ ⎟= − EI EI ⎝ ⎠
Consider downward deflection of cantilever beam at A due to the 30-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v A = − 3 EI Values: 6 2 P = = 30 kips, L kips, L = = 36 ft, EI ft, EI = = 24 × 10 kip-in.
Calculation: v A = −
PL3
=−
(30 ki kips)(36 ft ft)3
3 EI
= −
3EI
466, 56 560 ki kip-ft3 EI
Consider upward deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 θ v B = − and (slope magnitude) B = 3 EI 2 EI Values: 6 2 P = = − R B = 106.5 kips, L kips, L = = 24 ft, EI ft, EI = = 24 × 10 kip-in.
Calculation: (4,60 (4,608 8 ft ft 3 )(106. 106.5 5 kips kips)) 490 490,752 ,752 kipkip-fft3 v B = = EI EI θ B
=
PL2
=
(106. 106.5 5 kip kipss)(24 )(24 ft)2
2 EI
v A =
2 EI
490, 752 kip-ft3 EI
=
using the results from Eq. (b)
30,67 30,672 2 kip kip--ft2 EI
⎛ 30, 672 kip-ft2 ⎞ 858, 816 kip-ft3 + (12 ft) ⎜ ⎟= EI EI ⎝ ⎠
Beam deflection at A. 414,72 ,720 kip-ft3 466,56 ,560 kip-ft3 858,81 ,816 kip-ft3 v A = − − + EI EI EI
=−
22,464 ,464 ki kip-ft3 EI
=−
(22 (22,464 ,464 ki kip-ft3 )(12 in in./ft)3 24 × 106 kip-in.2
= −1.6 1.61740 7408 in. in. = 1.617 in i n. ↓
Ans.
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11.39 A propped cantilever beam is loaded as shown in Fig. P11.39. Assume EI Assume EI = = 86.4 6 2 × 10 N-mm . Determine: (a) the reactions at A at A and and C for for the beam. (b) the beam deflection at B at B..
Fig. P11.39
Solution Choose the reaction force at C as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vC = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 25 N/mm, L N/mm, L = = 400 mm, x mm, x = = 300 mm, 6 2 EI = = 86.4 × 10 N-mm
Calculation: wx 2 vC = − (6 L2 − 4 Lx + x 2 ) 24 EI
=−
(25 N/mm)(300 mm)2 24 EI
= −
⎡⎣6(40 (400 mm mm)2 − 4(40 (400 mm mm)(300 mm) + (300 mm mm)2 ⎤⎦
53.43750 × 109 N-mm3 EI
Consider downward deflection of cantilever beam at C due due to the 4,000-N concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 θ B = v B = − and (slope magnitude) 3 EI 2 EI Values: 6 2 P = = 4,000 N, L N, L = = 120 mm, EI mm, EI = = 86.4 × 10 N-mm
Calculation: PL3 (4, 00 000 N)(120 mm)3 2.304 × 109 N-mm3 v B = − =− =− 3 EI 3EI EI θ B
=
PL2
=
(4, 00 000 N)(120 mm)2
2 EI
vC = −
2 EI
2.304 × 109 NN-mm3 EI
=
2.88 × 107 N-mm2 EI
⎛ 2.88 × 107 NN -mm2 ⎞ 7.488× 109 NN-mm3 − (180 mm) ⎜ ⎟= − EI EI ⎝ ⎠
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Consider upward deflection of cantilever beam at C due due to concentrated load RC . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3 EI Values: 6 2 P = = − RC , L = L = 300 mm, EI mm, EI = = 86.4 × 10 N-mm
Calculation: vC = −
PL3
=−
3 EI
(− RC )(300 mm)3
=
(9 × 106 mm3 ) RC
3EI
EI
Compatibility equation for deflection at C : 53.43750 × 109 N-mm3 7.488 × 109 N-mm3 (9× 106 mm3 ) RC − − + =0 EI EI EI
∴ RC =
60.9255 × 109 N-mm3 9 × 106 mm3
= 6, 76 769.5 N = 6, 77 770 N ↑
Ans.
Equilibrium equations for entire beam: Σ F y = RA + RC − 4, 00 000 N − (25 N/ N/mm)(400 mm mm) = 0
∴ R A = 4, 00 000 N + (25 N/mm)(400 mm) − 6, 76 769.5 N = 7, 23 230.5 N = 7, 23 230 N ↑
Ans.
Σ M A = − M A − (4,00 ,000 N) N)(120 mm) − (25 N/mm)(400 mm)(200 mm) + RC (300 mm mm) = 0 ∴ M A = (6,769. (6,769.5 5 N)( N)(300 300 mm) mm) − (4,000 (4,000 N)(1 N)(120 20 mm) mm) − (25 N/mm N/mm)( )(400 400 mm)( mm)(20 200 0 mm) mm) = − 449, 449,15 150 0 N-mm N-mm = 449,00 449,000 0 N-m N-mm (ccw (ccw))
Ans.
(b) Beam deflection at B: Consider downward deflection of cantilever beam at B due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 25 N/mm, L N/mm, L = = 400 mm, x mm, x = = 120 mm, 6 2 EI = = 86.4 × 10 N-mm
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Calculation: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) 24 EI
=−
(25 N/mm)(120 mm)2 24 EI
= −
⎡⎣ 6(40 (400 mm mm)2 − 4(40 (400 mm mm)(120 mm mm) + (120 mm mm)2 ⎤⎦
11.736 × 109 N-mm3 EI
Consider downward deflection of cantilever beam at B due to the 4,000-N concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: 6 2 P = = 4,000 N, L N, L = = 120 mm, EI mm, EI = = 86.4 × 10 N-mm
Calculation: v B = −
PL3
=−
(4, 00 000 N)(120 mm)3
3 EI
3EI
= −
2.304 × 109 N-mm3 EI
Consider upward deflection of cantilever beam at B due to concentrated load RC . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 v B = − (3L − x) 6 EI Values: P = = − RC = = −6,769.5 N, L N, L = = 300 mm, x mm, x = = 120 mm, 6 2 EI = = 86.4 × 10 N-mm
Calculation: Px 2 (−6,769.5 6,769.5 N)(12 N)(120 0 mm) mm)2 v B = − (3 L − x) = − (120 mm)] [3(300 mm) − (1 6 EI 6 EI
=
12.6725 × 109 N-mm3 EI
Beam deflection at B. 11.736 × 109 N-mm3 2.304 × 109 N-mm3 12.6725× 109 N-mm3 v B = − − + EI EI EI
=−
1.367496 × 109 N-mm3 EI
=−
1.367496 × 109 N-mm3 86.4 × 106 N-mm2
= −15.8275 mm mm = 15.83 mm mm ↓
Ans.
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11.40 The beam shown in Fig. P11.40 consists of a W610 × 82 structural steel wide-flange 6 4 shape [ E E = = 200 GPa; I GPa; I = = 562 × 10 mm ]. For the loading shown, determine: (a) the reaction force at C . (b) the beam deflection at A at A..
Fig. P11.40
Solution (a) Reaction force at C . Choose the reaction force at C as as the redundant; therefore, the released beam is simply supported between B between B and and D D.. Consider upward deflection of simply supported beam at C due due to uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(80 kN/m)(3 m)(1.5 m) = −360 kN-m, L = L = 14 m, x m, x = = 7 m Calculation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−360 kN-m)(7 m) 6(14 m) EI
⎡⎣ 2(14 m) − 3(14 m)(7 m) m) + (7 m) m) ⎤⎦ = 2
2
3
4, 410 kN-m EI
Consider downward deflection of simply supported beam at C due due to uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC = − 384 EI Values: w = 80 kN/m, L kN/m, L = = 14 m
Calculation: vC = −
5wL4 384 EI
=−
5(80 kN/m)(14 m)4 384EI
= −
40, 016.6667 kN-m3 EI
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Consider upward deflection of simply supported beam at C due due to concentrated load RC . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = = − RC , L = L = 14 m
Calculation: vC = −
PL3
=−
48 EI
(− RC )(14 m)3
=
(57.1667 m3 ) RC
48EI
EI
Compatibility equation for deflection at B: 4, 41 410 kN kN-m3 40, 01 016.6667 kN kN-m3 (57.1667 m3 ) RC − + =0 EI EI EI
∴ RC =
35,606.6667 kN-m kN-m3 57.1667 m3
= 622.8571 kN = 623 kN kN ↑
Ans.
(b) Beam deflection at A. Consider downward cantilever beam deflection caused by uniformly distributed load on overhang AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 v A = − 8 EI Values: w = 80 kN/m, L kN/m, L = = 3 m
Calculation: v A = −
wL4 8 EI
=−
(80 kN/m)(3 m)4 8EI
= −
810 kN-m3 EI
Consider downward deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: L (slope magnitude) θ B = 3 EI Values: M = = (80 kN/m)(3 m)(1.5 m) = 360 kN-m, k N-m, L = L = 14 m
Computation: ML (360 (360 kNkN-m)(14 )(14 m) 1,680 kNkN-m2 θ B = = = 3 EI 3EI EI
⎛ 1, 680 kN-m2 ⎞ 5, 040 kN-m3 v A = −(3 m) ⎜ ⎟= − EI EI ⎝ ⎠ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Consider upward deflection at A due to uniformly distributed load between B and D. [Appendix C, SS beam bea m with uniformly distributed load.] Relevant equation from Appendix C: wL3 θ B = (slope magnitude) 24 EI Values: w = 80 kN/m, L kN/m, L = = 14 m
Calculation: wL3 (80 kN/ kN/m)(14 )(14 m)3 9,146 ,146.6 .666 667 7 kNkN-m m2 = = θ B = 24 EI 24EI EI
⎛ 9,146.6667 kN-m2 ⎞ 27, 44 440 kN-m3 v A = (3 m) ⎜ ⎟= EI EI ⎝ ⎠ Consider downward deflection at A due to concentrated load RC . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 θ B = (slope magnitude) 16 EI Values: P = = − RC = −622.8571 kN, L kN, L = = 14 m
Calculation: PL2 (622 622.8571 571 kN)( N)(14 m) m)2 7,630 ,630 kN kN-m2 θ B = = = 16 EI 16EI EI
⎛ 7, 630 kN-m2 ⎞ 22, 890 kN-m3 v A = −(3 m) ⎜ ⎟= − EI EI ⎝ ⎠ Beam deflection at A. 810 kN-m3 5, 04 040 kN-m3 27, 44 440 kN-m3 22, 890 kN-m3 v A = − − + − EI EI EI EI
=
−1, 300 kN-m3 EI
=
−1, 300 kN-m3 112,400 kN-m2
= −0.011566 m = 11.57 mm mm ↓
Ans.
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11.41 The beam shown in Fig. P11.41 consists of a W8 × 15 structural steel wide4 flange shape [ E E = 29,000 ksi; I = 48 in. ]. For the loading shown, determine: (a) the reactions at A at A and and B B.. (b) the magnitude of the maximum bending stress in the beam. beam. ( Reminder: Reminder: The roller symbol implies that both upward and downward displacement is restrained.)
Fig. P11.41
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −80 lb/in., L lb/in., L = = 150 in., x in., x = = 100 in.
Calculation: wx 2 v B = − (6 L2 − 4 Lx + x2 ) 24 EI
=−
(−80 lb/ lb/in. in.)( )(10 100 0 in.) in.)2 24 EI
=
⎡⎣ 6(150 in in.)2 − 4(150 in in.)(100 in in.) + (100 in in.)2 ⎤⎦
2.8333 × 109 lb-in.3 EI
Consider deflection of cantilever beam at B due to the force caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: P = = −½(50 in.)(60 lb/in.) = −1,500 lb, L lb, L = = 100 in.
Calculation: v B = −
PL3 3 EI
=−
(−1, 50 500 lb)(100 in.)3 3EI
=
500 × 106 lb-in.3 EI
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Consider deflection of cantilever beam at B due to the moment caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: L2 v B = − 2 EI Values: M = = −½(50 in.)(60 lb/in.)[⅔(50 in.)] = −50,000 lb-in., L lb-in., L = = 100 in.
Calculation: ML2 (−50, 00 000 lb-in.)(100 in.)2 250 × 106 lb-in.3 v B = − =− = 2 EI 2 EI EI Consider deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: P = = − R B, L = L = 100 in.
Calculation: v B = −
PL3
=−
(− R B )(100 in.)3
3 EI
=
(333.3333 in.3 )RB
3EI
EI
Compatibility equation for deflection at B: 2.8333 × 109 lb-in.3 500 × 106 lb-in.3 250 × 106 lb-in.3 (333.3333× 103 in.3 R ) B + + + =0 EI EI EI EI
∴ R B = −
3.5833 × 109 lb-in.3 333.3333 × 103 in.3
= −10,75 ,750 lb lb = 10,75 ,750 lb lb ↓
Ans.
Equilibrium equations for entire beam:
Σ F y = RA + RB + (80 lb lb/in.)(150 in.) +
1 2
∴ R A = −(80 lb/in.)(150 in.) −
(140 lb lb/in. − 80 lb lb/in.)(50 in.) = 0 (60 lb/in.)(50 in.)
= −2, 75 750 lb = 2, 75 750 lb ↓
2
− (− 10, 750 lb) Ans.
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Σ M A = − M A + (80 lb/in.)(150 in.)(75 in.) 2 ⎡ (140 lb/in. − 80 lb/in.)(50 in.) ⎤ ⎡ ⎤ 100 in. + (50 in.)⎥ + R B (100 in.) = 0 ⎥ ⎢ 2 3 ⎣ ⎦⎣ ⎦
+⎢
∴ M A = (80 lb/in.)(150 in.)(75 in.) +
(60 lb/in.)(50 in.) 2
(133.3333 in.) + (− 10, 750 in.)(100 in.)
= 25,000 25,000 lb-in b-in.. = 25,000 25,000 lb-in b-in.. (cw) cw)
Ans.
(b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 48 in.4 d = 8.11 in. S = 11.8 in.3 Maximum bending moment magnitude M max (at B)) max = 150,000 lb-in. (at B Bending stresses at maximum moment (150,000 lb-in.)(8.11 in./2) σ x = = 12,571 ,571.875 875 ps psi = 12,57 ,570 psi psi 48 in.4 or, using the tabulated value for the section modulus: 150,000 lb-in. lb-in. = 12,711 ,711.864 864 ps psi = 12,710 ,710 ps psi σ x = 11.8 in.3
Ans.
Ans.
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11.42 The beam shown in Fig. P11.42 consists of a W24 × 94 structural steel wide4 flange shape [ E E = = 29,000 ksi; I ksi; I = = 2,700 in. ]. For the loading shown, determine: (a) the reactions at A at A and and D D.. (b) the magnitude of the maximum bending stress in the beam. beam.
Fig. P11.42
Solution Choose the reaction force at A at A as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to the 50-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 v B = − and θ B = (slope magnitude) 3 EI 2 EI Values: P = = 50 kips, L kips, L = = 20 ft
Calculation: PL3 (50 ki kips)(20 ft)3 133,33 ,333.33 .333 kip-ft3 v B = − =− =− 3 EI 3EI EI θ B
=
PL2
=
(50 kip kips)(20 ft) 2
2 EI
v A = −
=
10,000 ,000 kip kip-ft3
2 EI
133, 333.333 kip-ft3 EI
EI
⎛ 10, 000 kip-ft3 ⎞ 193, 333.333 kip-ft3 − (6 ft) ⎜ ⎟= − EI EI ⎝ ⎠
Consider downward deflection of cantilever beam at A due to the uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vC = − and (slope magnitude) θ C = 8 EI 6 EI Values: w = 4 kips/ft, L kips/ft, L = = 14 ft
Calculation: wL4 (4 kips/ft)(14 ft) 4 19,208 ,208 kip-ft 3 vC = − =− =− 8 EI 8EI EI θ C =
wL3
=
(4 kip kips/ s/ft ft)( )(14 14 ft) ft)3
6 EI
v A = −
6 EI
19, 208 kip-ft 3 EI
=
1,829. ,829.33 333 3 kip kip--ft 3 EI
⎛ 1, 829.333 kip-ft3 ⎞ 41,160 kip-ft3 − (12 ft) ⎜ ⎟= − EI EI ⎝ ⎠
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Consider upward deflection of cantilever beam at A due to concentrated load R A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v A = − 3 EI Values: P = = − R A, L = L = 26 ft
Calculation: v A = −
PL3
=−
3 EI
(− R A )(26 ft)3
=
(5, 858.667 ft 3 ) RA
3EI
EI
(b)
Compatibility equation for deflection at A: 193,33 ,333.333 kip-ft3 41,160 kip-ft3 (5,85 ,858.667 ft3 ) R A − − + =0 EI EI EI
∴ R A =
234,493.333 kip-ft 3 5,858.667 ft 3
= 40.025 ki kips = 40.0 ki kips ↑
Ans.
Equilibrium equations for entire beam: Σ F y = RA + RD − 50 ki kips − (4 ki kips/ft)(14 ft) = 0
∴ R D = 50 ki kips + (4 ki kips/ft)(14 ft) − 40.025 ki kips = 65.975 ki kips = 66.0 ki kips ↑
Ans.
Σ M D = M D + (50 ki kips)(20 ft ft) + (4 ki kips/ft)(14 ft ft)(7 ft) − RA (26 ft ft) = 0 ∴ M D = (40.0 (40.025 25 kip kips) s)(2 (26 6 ft) ft) − (50 (50 kips kips)( )(20 20 ft) ft) − (4 kips kips/f /ft) t)(1 (14 4 ft) ft)(7 (7 ft) ft) = − 351. 351.35 350 0 kipkip-ft ft = 351 351 kipkip-ft ft (cw) (cw)
Ans.
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Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 2,700 in.4
d = 24.3 in. S = 222 in.3 Maximum bending moment magnitude M max max = 351.350 kip-ft Bending stresses at maximum moment (351.350 kip-ft)(24.3 in./2)(12 in./ft) σ x = 2,700 in.4
= 18.97 ksi
Ans.
or, using the tabulated section modulus: (351.350 kip-ft)(12 in./ft) σ x = 222 in.3
= 18.99 ksi
Ans.
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11.43 The solid 20-mm-diameter steel [ E = E = 200 GPa] shaft shown in Fig. P11.43 supports two belt pulleys. Assume that the bearing at A at A can be idealized as a pin support and that the bearings at C and E can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings A bearings A,, C , and E and E . (b) the magnitude of the maximum bending stress in the shaft.
Fig. P11.43
Solution (a) Reaction forces at A, C, and E . Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between A between A and and E E . Consider downward deflection of simply supported beam at C due due to pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 750 N, L N, L = = 2,000 mm, b = 600 mm, x = x = 1,000 mm Calculation: vC = −
=−
Pbx 6 LEI L EI
( L2 − b 2 − x 2 )
(750 N)(600 mm)(1,000 mm) 6(2,000 mm) EI
= −
⎡⎣(2, 00 000 mm)2 − (600 mm)2 − (1, 00 000 mm)2 ⎤⎦
99.0 × 109 N-mm3 EI
Consider downward deflection of simply supported beam at C due due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 500 N, L N, L = = 2,000 mm, b = 400 mm, x = x = 1,000 mm
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Calculation: Pbx 2 vC = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(500 N)(400 mm)(1,000 mm) 6(2,000 mm) EI
= −
⎡⎣ (2, 00 000 mm)2 − (400 mm)2 − (1, 00 000 mm)2 ⎤⎦
47.3333 × 109 N-mm3 EI
Consider upward deflection of simply supported beam at C due due to concentrated load RC . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = = − RC , L = L = 2,000 mm
Calculation: vC = −
PL3
=−
48 EI
(− RC )(2, 000 mm)3
=
(166.6667 × 106 mm3 )RC
48EI
EI
Compatibility equation for deflection at C : 99.0 × 109 N-mm3 47.3333 × 109 N-mm3 (166.6667× 106 mm3 ) RC − − + =0 EI EI EI
∴ RC =
146.3333 × 109 N-mm3 166.6667 × 106 mm3
= 878 N = 878 N ↑
Ans.
Equilibrium equations for entire beam: Σ M A = −(750 N) N)(600 mm) − (500 N) N)(1,600 mm) + RC (1,000 mm) + RE (2,000 mm) = 0
∴ R E =
(750 (750 N)( N)(60 600 0 mm) mm) + (500 500 N)(1 N)(1,6 ,600 00 mm) mm) − (878 878 N)(1 N)(1,0 ,000 00 mm) mm) 2,000 mm
= 186 N ↑
Ans.
Σ F y = RA + RC + RE − 750 N − 500 N = 0 ∴ R A = 750 N + 500 N − 878 N − 186 N = 186 N ↑
Ans.
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Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties:
I =
π
64
(20 mm mm)4 = 7,85 ,853.9816 mm mm4
Maximum bending moment magnitude M max max = 114,000 N-mm Bending stresses at maximum moment (114,000 114, 000 N-mm)(20 N-mm)(20 mm/2) σ x = 7,853.9816 mm4
= 145.1 MPa
Ans.
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11.44 The solid 1.00-in.-diameter steel [ E E = 29,000 ksi] shaft shown in Fig. P11.44 supports three belt pulleys. Assume that the bearing at A A can be idealized as a pin support and that the bearings at C and E can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings A, A, C , and E and E . (b) the magnitude of the maximum bending stress in the shaft.
Fig. P11.44
Solution (a) Reaction forces at A, C, and E . Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between A between A and and E E . Consider downward deflection of simply supported beam at C due due to pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 200 lb, L lb, L = = 60 in., b = 15 in., x in., x = = 30 in. Calculation: Pbx 2 vC = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(200 lb)(15 in.)(30 in.) 6(60 in.) EI
= −
⎡⎣(60 in.)2 − (15 in.)2 − (30 in.)2 ⎤⎦
618,750 lb-in. lb-in.3 EI
Consider downward deflection of simply supported beam at C due due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 200 lb, L lb, L = = 60 in., b = 15 in., x in., x = = 30 in. Calculation: Pbx 2 vC = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(200 lb)(15 in.)(30 in.) 6(60 in.) EI
= −
⎡⎣(60 in.)2 − (15 in.)2 − (30 in.)2 ⎤⎦
618,750 lb-in. lb-in.3 EI
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Consider upward deflection of simply supported beam at C due due to pulley F load. load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(120 lb)(10 in.) = −1,200 lb-in., L = L = 60 in., x in., x = = 30 in. Calculation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−1, 200 200 lb-i lb-in.) n.)(3 (30 0 in.) in.) 6(60 in.) EI
=
⎡⎣ 2(60 in.)2 − 3(60 in.)(30 in in.) + (30 in.)2 ⎤⎦
270,000 lb-in. lb-in.3 EI
Consider upward deflection of simply supported beam at C due due to concentrated load RC . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = = − RC , L = L = 60 in.
Calculation: vC = −
PL3
=−
48 EI
(− RC )(60 in.)3 48EI
=
(4, 500 in.3 ) RC EI
Compatibility equation for deflection at C : 618, 75 750 lb-in.3 618, 75 750 lb-in.3 270, 00 000 lb-in.3 (4, 50 500 in.3 ) RC − − + + =0 EI EI EI EI
∴ RC =
967,500 lb-in. lb-in.3 4,500 in.3
= 215 lb = 215 lb ↑
Ans.
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Equilibrium equations for entire beam: Σ M A = −(200 lb)(15 in.) − (200 lb)(45 in.) − (120 lb)(70 in.) + RC (30 in.) + RE (60 in.) = 0
∴ R E =
(200 (200 lb)( lb)(15 15 in.) in.) + (200 (200 lb)(4 b)(45 5 in.) in.) + (120 120 lb) lb)(7 (70 0 in.) in.) − (215 (215 lb)( lb)(30 30 in.) in.) 60 in.
= 232.50 lb = 233 lb ↑
Ans.
Σ F y = RA + RC + RE − 200 lb − 200 lb − 120 lb = 0 ∴ R A = 200 lb + 200 lb + 120 lb − 215 lb − 232.50 lb = 72.5 lb ↑
Ans.
Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties:
I =
π
64
(1.00 in.)4 = 0.04 .0490874 874 in.4
Maximum bending moment magnitude M max max = 1,200 lb-in. Bending stresses at maximum moment (1, 200 lb-in.)(1.00 in./2) σ x = 0.0490874 in.4 = 12,223.1 psi
= 12,220 psi
Ans.
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11.45 The solid 1.00-in.-diameter steel [ E = E = 29,000 ksi] shaft shown in Fig. P11.45 supports two belt pulleys. Assume that the bearing at E can be idealized as a pin support and that the bearings at B at B and and C can can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings B bearings B,, C , and E . (b) the magnitude of the maximum bending stress in the shaft.
Fig. P11.45
Solution (a) Reaction forces at B, C, and E . Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between B between B and and E E . Consider upward deflection of simply supported beam at C due due to pulley A load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(110 lb)(7 in.) = −770 lb-in., L = L = 45 in., x in., x = = 15 in. Calculation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−770 lblb-in in.)( .)(15 15 in.) in.) 6(45 in.) EI
=
⎡⎣ 2(45 (45 in in.)2 − 3(45 (45 in in.)(15 in in.) + (15 in in.)2 ⎤⎦
96,250 96, 250 lb-in. lb-in.3 EI
Consider downward deflection of simply supported beam at C due due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 260 lb, L lb, L = = 45 in., b = 15 in., x in., x = = 15 in.
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Calculation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI L EI
=−
(260 lb)(15 in.)(15 in.) 6(45 in.) EI
= −
⎡⎣(45 in.)2 − (15 in.)2 − (15 in.)2 ⎤⎦
341,250 lb-in.3 EI
Consider upward deflection of simply supported beam at C due due to concentrated load RC . [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = − RC , L = L = 45 in., a = 15 in., b = 30 in. Calculation: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(− RC )(15 in in.)(30 in in.) 6(45 in.) EI
=
⎡⎣(45 in.)2 − (15 in.)2 − (30 in.)2 ⎤⎦
(1,500 in.3 ) RC EI
Compatibility equation for deflection at C : 96,25 ,250 lb-in.3 341,25 ,250 lb-in.3 (1,50 ,500 in.3 ) RC − + =0 EI EI EI
∴ RC =
245,000 lb-in. lb-in.3 1,500 in.3
= 163.333 3333 lb = 163. 63.3 lb ↑
Ans.
Equilibrium equations for entire beam: Σ M E = (260 lb)(15 in.) + (110 lb)(52 in.) − RB (45 in.) − RC (30 in.) = 0
∴ R B =
(260 (260 lb) lb)(1 (15 5 in. in.)) + (110 110 lb) lb)(5 (52 2 in. in.)) − (163.3 163.333 333 3 lb) lb)(3 (30 0 in. in.))
= 104.8889 lb lb = 104.9 lb lb ↑
45 in. Ans.
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Σ F y = RB + RC + RE − 110 lb − 260 lb = 0 ∴ R E = 110 lb + 260 lb − 163.3333 lb − 104.8889 lb = 101.7778 lb = 101.8 lb↑
Ans.
Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties:
I =
π
64
(1.00 in.)4 = 0.04908 90874 in. in.4
Maximum bending moment magnitude M max max = 1,526.7 lb-in. Bending stresses at maximum moment (1, 526.7 lb-in.)(1.00 lb-in.)(1.00 in./2) in./2 ) σ x = 4 0.0490874 in. = 15,550.8 psi
= 15,550 psi
Ans.
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11.46 The beam shown in Fig. P11.46 consists of a W360 × 101 structural steel wide-flange 6 4 shape [ E E = = 200 GPa; I GPa; I = = 301 × 10 mm ]. For the loading shown, determine: (a) the reactions at A at A and and B B.. (b) the magnitude of the maximum bending stress in the beam.
Fig. P11.46
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly un iformly distributed load.] Relevant equations from Appendix C: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 30 kN/m, L kN/m, L = = 8 m, x m, x = = 5.5 m
Calculation: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) 24 EI
=−
(30 kN/m)(5.5 m) 2 24 EI
= −
⎡⎣6(8 m)2 − 4(8 m)(5.5 m) + (5.5 m)2 ⎤⎦
9,008.828125 kN-m3 EI
Consider deflection of cantilever beam at B due a linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 x 2 v B = − (10 L3 − 10 L2 x + 5 Lx2 − x3 ) (elastic curve) 120 LEI L EI
Values: w0 = 60 kN/m, L kN/m, L = = 8 m, x m, x = = 5.5 m Calculation: w0 x 2 v B = − (10 L3 − 10 L2 x + 5Lx2 − x3 ) 120 LEI L EI
=−
(60 kN/m)(5.5 m)2 120(8 m) EI
= −
⎡⎣10(8 m)3 − 10(8 m)2 (5.5 m) + 5(8 m)(5.5 m)2 − (5.5 m)3 ⎤⎦
4,998.103516 kN-m3 EI
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Consider deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: P = = − R B, L = L = 5.5 m
Calculation: v B = −
PL3
=−
3 EI
(− R B )(5.5 m)3
=
(55.458333 m3 ) RB
3EI
EI
Compatibility equation for deflection at B: 9,008 ,008.828125 kN-m3 4,99 ,998.103516 kN-m3 (55.458333 m3 R ) B − − + =0 EI EI EI
∴ R B =
14,006.93164 kN-m3 55.458333 m3
= 252.56676 kN kN = 253 kN ↑
Ans.
Equilibrium equations for entire beam: 1 Σ F y = RA + RB − (30 kN kN/m)(8 m) − (90 kN kN/m − 30 kN kN/m)(8 m) = 0 2 1 ∴ R A = (30 kN kN/m)(8 m) + (90 kN kN/m − 30 kN kN/m)(8 m) − 252.56676 kN kN 2
= 227.43324 kN = 227 kN kN ↑
Ans.
⎛ (90 kN/m − 30 kN/m)(8 m) ⎞ ⎛ 8 m ⎞ ⎟ ⎜ 3 ⎟ + RB (5.5 m) = 0 2 ⎝ ⎠⎝ ⎠
Σ M A = − M A − (30 kN/m)(8 m)(4 m) − ⎜
∴ M A = −960 960 kN kN-m − 640 640 kNkN-m + (252 (252.5 .566 6676 76 kN)( kN)(5. 5.5 5 m) = − 210. 210.88 8828 281 1 kNkN-m = 211 211 kNkN-m (ccw (ccw))
Ans.
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Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 301 × 106 mm4
d = 356 mm S = 1, 69 690 × 103 mm3 Maximum bending moment magnitude M max max = 210.88281 kN-m Bending stresses at maximum moment (210.88281 kN-m)(356 mm/2)(1,000)2 σ x = 301 × 106 mm4
= 124.7 MPa or, using the tabulated value for the section modulus: (210.88281 kN-m)(1,000)2 σ x = 1, 69 690 × 103 mm3 = 124.8 MPa
Ans.
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