Mechanics of Materials Solutions Chapter10 Probs29 46

10.29a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.29a

Solution Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θA = (slope magnitude) 6 EI Values: M = 150 kN-m, L = 8 m, EI = 8 × 104 kN-m2

Computation: ML (150 kN-m)(8 m) θA = = = 0.00250 rad 6 EI 6(8 × 104 kN-m 2 ) Determine beam deflection at H. [Skill 1]

vH = (3 m)(0.00250 rad) = 0.00750 m = 7.50 mm ↑

Ans.

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10.29b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.29b

Solution Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2

Computation: wL4 (6 kN/m)(4 m)4 vA = − =− = −0.00240 m 8EI 8(8 × 104 kN-m 2 ) Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θA = (slope magnitude) 6 EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2

Computation: wL3 (6 kN/m)(4 m)3 θA = = = 0.00080 rad 6 EI 6(8 × 104 kN-m 2 ) Determine beam deflection at H. [Skill 2]

vH = −0.00240 m − (2 m)(0.00080 rad) = −0.00400 m = 4.00 mm ↓

Ans.

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10.29c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.29c

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vH = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI Values: P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m, EI = 8 × 104 kN-m2

Computation: Pbx 2 ( L − b2 − x2 ) vH = − 6 LEI (30 kN)(4 m)(4 m) ⎡ (12 m) 2 − (4 m) 2 − (4 m) 2 ⎤⎦ =− 4 2 ⎣ 6(12 m)(8 × 10 kN-m ) = 0.00933 m = 9.33 mm ↓

Ans.

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10.29c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.29d

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vH ,cant = − (assuming fixed support at B) 3EI Values: P = 15 kN, L = 4 m, EI = 8 × 104 kN-m2

Computation: vH ,cant

PL3 (15 kN)(4 m)3 =− =− = −0.004000 m 3EI 3(8 × 104 kN-m 2 )

Determine beam slope at B. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = (15 kN)(4 m) = 60 kN-m, L = 8 m, EI = 8 × 104 kN-m2

Computation: ML (60 kN-m)(8 m) θB = = = 0.002000 rad 3EI 3(8 × 104 kN-m 2 ) Determine beam deflection at H. [Skill 4]

vH = −0.00400 m − (4 m)(0.00200 rad) = −0.01200 m = 12.00 mm ↓

Ans.

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10.30a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.30a

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vB = − 2 EI Values: M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.2

Computation: ML2 (40 kip-ft)(9 ft) 2 (12 in./ft)3 vB = − =− = −0.23328 in. 2 EI 2(1.2 × 107 kip-in.2 ) Determine beam slope at B. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) EI Values: M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.2

Computation: ML (40 kip-ft)(9 ft)(12 in./ft) 2 θB = = = 0.004320 rad EI (1.2 × 107 kip-in.2 ) Determine beam deflection at H. [Skill 2]

vH = −0.23328 in. − (6 ft)(12 in./ft)(0.004320 rad) = −0.54432 in. = 0.544 in. ↓

Ans.

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10.30b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kipin.2 is constant for each beam.

Fig. P10.30b

Solution Determine beam slope at C. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 − a 2 ) θC = (slope magnitude) 6 LEI Values: P = 25 kips, L = 18 ft, a = 12 ft, EI = 1.2 × 107 kip-in.2

Computation: Pa ( L2 − a 2 ) θC = 6 LEI (25 kips)(12 ft) ⎡(18 ft) 2 − (12 ft) 2 ⎤⎦ (12 in./ft) 2 = 0.00600 rad = 7 2 ⎣ 6(18 ft)(1.2 × 10 kip-in. ) Determine beam deflection at H. [Skill 1]

vH = (7 ft)(12 in./ft)(0.00600 rad) = 0.5040 in. = 0.504 in. ↑

Ans.

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10.30c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.30c

Solution Determine beam deflection at H. [Skill 3] [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vH = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 2.5 kips/ft, L = 15 ft, x = 9 ft, EI = 1.2 × 107 kip-in.2

Computation: wx 2 (6 L2 − 4 Lx + x 2 ) vH = − 24 EI (2.5 kips/ft)(9 ft) 2 ⎡ 6(15 ft) 2 − 4(15 ft)(9 ft) + (9 ft) 2 ⎤⎦ (12 in./ft)3 =− 7 2 ⎣ 24(1.2 × 10 kip-in. ) = −1.082565 in. = 1.083 in. ↓

Ans.

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10.30d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.30d

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 (assuming fixed support at A) vH ,cant = − 8EI Values: w = 5 kips/ft, L = 8 ft, EI = 1.2 × 107 kip-in.2

Computation: vH ,cant

wL4 (5 kips/ft)(8 ft) 4 (12 in./ft)3 =− =− = −0.36864 in. 8EI 8(1.2 × 107 kip-in.2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θA = (slope magnitude) 3EI Values: M = (5 kips/ft)(8 ft)(4 ft) = 160 kip-ft, L = 22 ft, EI = 1.2 × 107 kip-in.2

Computation: ML (160 kip-ft)(22 ft)(12 in./ft) 2 θA = = = 0.014080 rad 3EI 3(1.2 × 107 kip-in.2 ) Determine beam deflection at H. [Skill 4]

vH = −0.36864 in. − (8 ft)(12 in./ft)(0.014080 rad) = −1.72032 in. = 1.720 in. ↓

Ans.

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10.31a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.31a

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vH = − 6 LEI Values: M = −60 kN-m, L = 12 m, x = 6 m, EI = 6 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vH = − 6 LEI (−60 kN-m)(6 m) ⎡ 2(12 m) 2 − 3(12 m)(6 m) + (6 m) 2 ⎤⎦ =− 6(12 m)(6 × 104 kN-m 2 ) ⎣ = 0.009000 m = 9.00 mm ↑

Ans.

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10.31b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.31b

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vH ,cant = − (assuming fixed support at A) 8EI Values: w = 7.5 kN/m, L = 3 m, EI = 6 × 104 kN-m2

Computation: vH ,cant = −

wL4 (7.5 kN/m)(3 m)4 =− = −0.00126563 m 8EI 8(6 × 104 kN-m 2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θA = (slope magnitude) 3EI Values: M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m, L = 6 m, EI = 6 × 104 kN-m2

Computation: ML (33.75 kN-m)(6 m) θA = = = 0.001125 rad 3EI 3(6 × 104 kN-m 2 ) Determine beam deflection at H. [Skill 4]

vH = −0.00126563 m − (3 m)(0.001125 rad) = −0.00464063 m = 4.64 mm ↓

Ans.

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10.31c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.31c

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2

Computation: PL3 (30 kN)(3 m)3 vB = − =− = −0.004500 m 3EI 3(6 × 104 kN-m 2 ) Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL2 θB = (slope magnitude) 2 EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2 Computation: PL2 (30 kN)(3 m)2 θB = = = 0.002250 rad 2 EI 2(6 × 104 kN-m 2 ) Determine beam deflection at H. [Skill 2]

vH = −0.004500 m − (3 m)(0.002250 rad) = −0.01125 m = 11.25 mm ↓

Ans.

Alternative solution for beam deflection at B. [Appendix C, Cantilever beam with concentrated load at midspan.] 5PL3 Relevant equation from Appendix C: vH = − 48EI 4 Values: P = 30 kN, L = 6 m, EI = 6 × 10 kN-m2 5PL3 5(30 kN)(6 m)3 Computation: vH = − =− = −0.011250 m = 11.25 mm ↓ 48 EI 48(6 × 104 kN-m 2 ) 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 Act without the permission of the copyright owner is unlawful.

10.31d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.31d

Solution Determine beam slope at C. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θC = (2 L − a) 2 (slope magnitude) 24 LEI Values: w = 5 kN/m, L = 9 m, a = 6 m, EI = 6 × 104 kN-m2

Computation: wa 2 (5 kN/m)(6 m) 2 2 θC = (2 L − a) 2 = 2(9 m) − (6 m)] = 0.00200 rad 4 2 [ 24 LEI 24(9 m)(6 × 10 kN-m ) Determine beam deflection at H. [Skill 1]

vH = (3 m)(0.00200 rad) = 0.00600 m = 6.00 mm ↑

Ans.

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10.32a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kipin.2 is constant for each beam.

Fig. P10.32a

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vH ,cant = − (assuming fixed support at A) 2 EI Values: M = 50 kip-ft, L = 6 ft, EI = 3.0 × 106 kip-in.2

Computation: vH ,cant = −

ML2 (50 kip-ft)(6 ft)2 (12 in./ft)3 =− = −0.51840 in. 2 EI 2(3.0 × 106 kip-in.2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θA = (slope magnitude) 3EI Values: M = 50 kip-ft, L = 18 ft, EI = 3.0 × 106 kip-in.2

Computation: ML (50 kip-ft)(18 ft)(12 in./ft)2 θA = = = 0.01440 rad 3EI 3(3.0 × 106 kip-in.2 ) Determine beam deflection at H. [Skill 4]

vH = −0.51840 in. − (6 ft)(12 in./ft)(0.01440 rad) = −1.5552 in. = 1.555 in. ↓

Ans.

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10.32b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.32b

Solution Determine beam deflection at H. [Skill 3] [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: Px 2 (3L − x) (elastic curve) vH = − 6 EI Values: P = 10 kips, L = 10 ft, x = 7 ft, EI = 3.0 × 106 kip-in.2

Computation: Px 2 vH = − (3L − x) 6 EI (10 kips)(7 ft) 2 (12 in./ft)3 =− [3(10 ft) − (7 ft)] = −1.081920 in. = 1.082 in. ↓ 6(3.0 × 106 kip-in.2 )

Ans.

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10.32c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.32c

Solution Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θA = (slope magnitude) 6 EI Values: M = (2 kips/ft)(8 ft)(4 ft) = 64 kip-ft, L = 18 ft, EI = 3.0 × 106 kip-in.2

Computation: ML (64 kip-ft)(18 ft)(12 in./ft) 2 θA = = = 0.009216 rad 6 EI 6(3.0 × 106 kip-in.2 ) Determine beam deflection at H. [Skill 1]

vH = (6 ft)(12 in./ft)(0.009216 rad) = 0.663552 in. = 0.664 in. ↑

Ans.

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10.32d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.32d

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8EI Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.2

Computation: wL4 (1.5 kips/ft)(10 ft)4 (12 in./ft)3 vB = − =− = −1.0800 in. 8EI 8(3.0 × 106 kip-in.2 ) Determine beam slope at B. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θB = (slope magnitude) 6 EI Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.2

Computation: wL3 (1.5 kips/ft)(10 ft)3 (12 in./ft) 2 θB = = = 0.01200 rad 6 EI 6(3.0 × 106 kip-in.2 ) Determine beam deflection at H. [Skill 2]

vH = −1.0800 in. − (4 ft)(12 in./ft)(0.0120 rad) = −1.6560 in. = 1.656 in. ↓

Ans.

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10.33 The simply supported beam shown in Fig. P10.33 consists of a W24 × 94 structural steel wide-flange shape [E = 29,000 ksi; I = 2,700 in.4]. For the loading shown, determine the beam deflection at point C.

Fig. P10.33

Solution Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 3.2 kips/ft, L = 28 ft, a = 21 ft, EI = 7.830 × 107 kip-in.2

Computation: wa 3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI =−

(3.2 kips/ft)(21 ft)3 (12 in./ft)3 ⎡ 4(28 ft) 2 − 7(21 ft)(28 ft) + 3(21 ft)2 ⎤⎦ = −0.333822 in. 7 2 ⎣ 24(28 ft)(7.830 × 10 kip-in. )

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: Px vC = − (3L2 − 4 x 2 ) (elastic curve) 48EI Values: P = 36 kips, L = 28 ft, x = 7 ft, EI = 7.830 × 107 kip-in.2

Computation: Px (3L2 − 4 x 2 ) vC = − 48 EI (36 kips)(7 ft)(12 in./ft)3 ⎡⎣3(28 ft) 2 − 4(7 ft) 2 ⎤⎦ = −0.249799 in. =− 7 2 48(7.830 × 10 kip-in. ) Beam deflection at C

vC = −0.333822 in. − 0.249799 in. = −0.583620 in. = 0.584 in. ↓

Ans.

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10.34 The simply supported beam shown in Fig. P10.34 consists of a W460 × 82 structural steel wide-flange shape [E = 200 GPa; I = 370 × 106 mm4]. For the loading shown, determine the beam deflection at point C.

Fig. P10.34

Solution Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 26 kN/m, L = 8 m, a = 6 m, EI = 7.4 × 104 kN-m2

Computation: wa 3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI =−

(26 kN/m)(6 m)3 ⎡ 4(8 m) 2 − 7(6 m)(8 m) + 3(6 m) 2 ⎤⎦ = −0.011068 m 4 2 ⎣ 24(8 m)(7.40 × 10 kN-m )

Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI Values: P = 60 kN, L = 8 m, b = 3 m, x = 2 m, EI = 7.4 × 104 kN-m2

Computation: Pbx 2 ( L − b2 − x 2 ) vC = − 6 LEI (60 kN)(3 m)(2 m) ⎡ (8 m) 2 − (3 m) 2 − (2 m) 2 ⎤⎦ = −0.005169 m =− 4 2 ⎣ 6(8 m)(7.40 × 10 kN-m ) Beam deflection at C

vC = −0.011068 m − 0.005169 m = −0.016237 m = 16.24 mm ↓

Ans.

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10.35 The simply supported beam shown in Fig. P10.35 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point B.

Fig. P10.35

Solution Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a2 − b2 ) vB = − 6 LEI Values: P = 60 kN, L = 9 m, a = 3 m, b = 6 m, EI = 4.32 × 104 kN-m2

Computation: Pab 2 ( L − a 2 − b2 ) vB = − 6 LEI (60 kN)(3 m)(6 m) ⎡(9 m)2 − (3 m)2 − (6 m)2 ⎤⎦ = −0.016667 m =− 4 2 ⎣ 6(9 m)(4.32 × 10 kN-m ) Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vB = − 6 LEI Values: M = −45 kN-m, L = 9 m, x = 6 m, EI = 4.32 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−45 kN-m)(6 m) ⎡ 2(9 m) 2 − 3(9 m)(6 m) + (6 m) 2 ⎤⎦ = 0.004167 m =− 4 2 ⎣ 6(9 m)(4.32 × 10 kN-m ) Beam deflection at B

vB = −0.016667 m + 0.004167 m = −0.012500 m = 12.50 mm ↓

Ans.

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10.36 The simply supported beam shown in Fig. P10.36 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine the beam deflection at point B.

Fig. P10.36

Solution Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 (4 L2 − 7aL + 3a 2 ) vB = − 24 LEI Values: w = 5 kips/ft, L = 24 ft, a = 16 ft, EI = 2.4447 × 107 kip-in.2

Computation: wa 3 (4 L2 − 7aL + 3a 2 ) vB = − 24 LEI =−

(5 kips/ft)(16 ft)3 (12 in./ft)3 ⎡ 4(24 ft) 2 − 7(16 ft)(24 ft) + 3(16 ft) 2 ⎤⎦ = −0.965066 in. 24(24 ft)(2.4447 × 107 kip-in.2 ) ⎣

Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vB = − 6 LEI Values: M = −200 kip-ft, L = 24 ft, x = 8 ft, EI = 2.4447 × 107 kip-in.2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−200 kip-ft)(8 ft)(12 in./ft)3 ⎡ 2(24 ft) 2 − 3(24 ft)(8 ft) + (8 ft) 2 ⎤⎦ = 0.502638 in. =− 7 2 ⎣ 6(24 ft)(2.4447 × 10 kip-in. ) Beam deflection at B

vB = −0.965066 in. + 0.502638 in. = −0.462428 in. = 0.462 in. ↓

Ans.

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10.37 The cantilever beam shown in Fig. P10.37 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 476 in.4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C.

Fig. P10.37

Solution (a) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2

Computation: wL4 (2 kips/ft)(6 ft)4 (12 in./ft)3 vB = − =− = −0.040559 in. 8EI 8(1.3804 × 107 kip-in.2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 (3L − x) (elastic curve) vB = − 6 EI Values: P = 12 kips, L = 10 ft, x = 6 ft, EI = 1.3804 × 107 kip-in.2

Computation: Px 2 (12 kips)(6 ft) 2 (12 in./ft)3 vB = − (3L − x) = − [3(10 ft) − (6 ft)] = −0.216313 in. 6 EI 6(1.3804 × 107 kip-in.2 ) Beam deflection at B

vB = −0.040559 in. − 0.216313 in. = −0.256872 in. = 0.257 in. ↓

Ans.

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(b) Beam deflection at point C Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θB = − 6 EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2

Computation: wL3 (2 kips/ft)(6 ft)3 (12 in./ft)3 θB = − =− = −0.0090130 rad 6 EI 6(1.3804 × 107 kip-in.2 ) vC = −0.040559 in. − (4 ft)(0.0090130 rad) = −0.076611 in.

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = 12 kips, L = 10 ft, EI = 1.3804 × 107 kip-in.2

Computation: PL3 (12 kips)(10 ft)3 (12 in./ft)3 vC = − =− = −0.500724 in. 3EI 3(1.3804 × 107 kip-in.2 ) Beam deflection at C

vC = −0.076611 in. − 0.500724 in. = −0.577336 in. = 0.577 in. ↓

Ans.

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10.38 The cantilever beam shown in Fig. P10.38 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 400 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B.

Fig. P10.38

Solution (a) Beam deflection at point A Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8EI Values: w = 25 kN/m, L = 4 m, EI = 8.0 × 104 kN-m2

Computation: wL4 (25 kN/m)(4 m) 4 vA = − =− = −0.010000 m 8EI 8(8.0 × 104 kN-m 2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 and θ B = − vB = − 3EI 2 EI Values: P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m2

Computation: PL3 (55 kN)(2.5 m)3 =− = −0.003581 m vB = − 3EI 3(8.0 × 104 kN-m 2 )

θB =

PL2 (55 kN)(2.5 m) 2 = = 0.002148 rad 2 EI 2(8.0 × 104 kN-m 2 )

v A = −0.003581 m − (1.5 m)(0.002148 rad) = −0.006803 m Beam deflection at A

v A = −0.010000 m − 0.006803 m = −0.016803 m = 16.80 mm ↓

Ans.

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(b) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 (6 L2 − 4 Lx + x 2 ) (elastic curve) vB = − 24 EI Values: w = 25 kN/m, L = 4 m, x = 2.5 m, EI = 8.0 × 104 kN-m2

Computation: wx 2 (6 L2 − 4 Lx + x 2 ) vB = − 24 EI (25 kN/m)(2.5 m) 2 ⎡6(4.0 m) 2 − 4(4.0 m)(2.5 m) + (2.5 m)2 ⎤⎦ = −0.005066 m =− 24(8.0 × 104 kN-m 2 ) ⎣ Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vB = − 3EI Values: P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m2

Computation: PL3 (55 kN)(2.5 m)3 vB = − =− = −0.003581 m 3EI 3(8.0 × 104 kN-m 2 ) Beam deflection at B

vB = −0.005066 m − 0.003581 m = −0.008647 m = 8.65 mm ↓

Ans.

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10.39 The solid 1.25-in.-diameter steel [E = 29,000 ksi] shaft shown in Fig. P10.39 supports two pulleys. For the loading shown, determine: (a) the shaft deflection at point B. (b) the shaft deflection at point C.

Fig. P10.39

Solution Section properties:

I=

π

64

(1.25 in.) 4 = 0.119842 in.4

(a) Shaft deflection at point B Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2

Computation: PL3 (200 lb)(10 in.)3 vB = − =− = −0.019182 in. 3EI 3(3.47543 × 106 lb-in.2 ) Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 (3L − x) (elastic curve) vB = − 6 EI Values: P = 120 lb, L = 25 in., x = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: Px 2 (120 lb)(10 in.) 2 vB = − (3L − x) = − [3(25 in.) − (10 in.)] = −0.037405 in. 6 EI 6(3.47543 × 106 lb-in.2 ) Shaft deflection at B

vB = −0.019182 in. − 0.037405 in. = −0.056588 in. = 0.0566 in. ↓

Ans.

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(b) Shaft deflection at point C Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 and θ B = (magnitude) vB = − 3EI 2 EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2

Computation: PL3 (200 lb)(10 in.)3 vB = − =− = −0.019182 in. 3EI 3(3.47543 × 106 lb-in.2 )

θB =

PL2 (200 lb)(10 in.) 2 = = 0.0028773 rad 2 EI 2(3.47543 × 106 lb-in.2 )

vC = −0.019182 in. − (15 in.)(0.0028773 rad) = −0.062342 in.

Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = 120 lb, L = 25 in., EI = 3.47543 × 106 lb-in.2

Computation: PL3 (120 lb)(25 in.)3 vC = − =− = −0.179834 in. 3EI 3(3.47543 × 106 lb-in.2 ) Shaft deflection at C

vC = −0.062342 in. − 0.179834 in. = −0.242176 in. = 0.242 in. ↓

Ans.

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10.40 The cantilever beam shown in Fig. P10.40 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 1,710 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B.

Fig. P10.40

Solution (a) Beam deflection at point A Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vA = − 2 EI Values: M = −200 kip-ft, L = 15 ft, EI = 4.959 × 107 kip-in.2

Computation: ML2 (−200 kip-ft)(15 ft) 2 (12 in./ft)3 vA = − =− = 0.784029 in. 2 EI 2(4.959 × 107 kip-in.2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 and θ B = (slope magnitude) vB = − 3EI 2 EI Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2

Computation: PL3 (−18 kips)(9 ft)3 (12 in./ft)3 vB = − =− = 0.152415 in. 3EI 3(4.959 × 107 kip-in.2 ) PL2 (18 kips)(9 ft) 2 (12 in./ft) 2 = = 0.0021169 rad θB = 2 EI 2(4.959 × 107 kip-in.2 ) v A = 0.152415 in. + (6 ft)(12 in./ft)(0.0021169 rad) = 0.304830 in. Beam deflection at A

v A = 0.784029 in. + 0.304830 in. = 1.088860 in. = 1.089 in. ↑

Ans.

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(b) Beam deflection at point B Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: Mx 2 vB = − (elastic curve) 2 EI Values: M = −200 kip-ft, L = 15 ft, x = 9 ft, EI = 4.959 × 107 kip-in.2

Computation: Mx 2 (−200 kip-ft)(9 ft) 2 (12 in./ft)3 vB = − =− = 0.282250 in. 2 EI 2(4.959 × 107 kip-in.2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2

Computation: PL3 (−18 kips)(9 ft)3 (12 in./ft)3 vB = − =− = 0.152415 in. 3EI 3(4.959 × 107 kip-in.2 ) Beam deflection at B

vB = 0.282250 in. + 0.152415 in. = 0.434665 in. = 0.435 in. ↑

Ans.

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10.41 The simply supported beam shown in Fig. P10.41 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C.

Fig. P10.41

Solution (a) Beam deflection at point A Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vA = − (assuming fixed support at B) 8EI Values: w = 4 kips/ft, L = 8 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL4 (4 kips/ft)(8 ft) 4 (12 in./ft)3 vA = − =− = −0.144760 in. 8EI 8(2.4447 × 107 kip-in.2 ) Consider deflection at A resulting from rotation at B caused by distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = (4 kips/ft)(8 ft)(4 ft) = 128 kip-ft, L = 22 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (128 kip-ft)(22 ft)(12 in./ft) 2 θB = = = 0.0055290 rad 3EI 3(2.4447 × 107 kip-in.2 ) v A = (8 ft)(12 in./ft)(0.0055290 rad) = −0.530786 in.

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Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 θB = (slope magnitude) 16 EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.2

Computation: PL2 (45 kips)(22 ft) 2 (12 in./ft) 2 θB = = = 0.0080182 rad 16 EI 16(2.4447 × 107 kip-in.2 ) v A = (8 ft)(12 in./ft)(0.0080182 rad) = 0.769744 in. Beam deflection at A

v A = −0.144760 in. − 0.530786 in. + 0.769744 in. = 0.094198 in. = 0.0942 in. ↑

Ans.

(b) Beam deflection at point C Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft, L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.2

Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI

(−128 kip-ft)(11 ft)(12 in./ft)3 ⎡ 2(22 ft) 2 − 3(22 ft)(11 ft) + (11 ft) 2 ⎤⎦ = 0.273687 in. =− 7 2 ⎣ 6(22 ft)(2.4447 × 10 kip-in. ) Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.2

Computation: PL3 (45 kips)(22 ft)3 (12 in./ft)3 vC = − =− = −0.705598 in. 48EI 48(2.4447 × 107 kip-in.2 ) Beam deflection at C

vC = 0.273687 in. − 0.705598 in. = −0.431912 in. = 0.432 in. ↓

Ans.

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10.42 The simply supported beam shown in Fig. P10.42 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point D.

Fig. P10.42

Solution (a) Beam deflection at point B Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vB = − 384 EI Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m2

Computation: 5wL4 5(55 kN/m)(7.2 m) 4 vB = − =− = −0.027415 m 384 EI 384(7.02 × 104 kN-m 2 ) Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vB = − 6 LEI Values: M = (55 kN/m)(2.8 m)(1.4 m) = −215.6 kN-m, L = 7.2 m, x = 3.6 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−215.6 kN-m)(3.6 m) ⎡ 2(7.2 m) 2 − 3(7.2 m)(3.6 m) + (3.6 m)2 ⎤⎦ = 0.009951 m =− 6(7.2 m)(7.02 × 107 kN-m 2 ) ⎣ Beam deflection at B

vB = −0.027415 m + 0.009951 m = −0.017464 m = 17.46 mm ↓

Ans.

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(b) Beam deflection at point D Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θC = (slope magnitude) 24 EI Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m2

Computation: wL3 (55 kN/m)(7.2 m)3 θC = = = 0.0121846 rad 24 EI 24(7.02 × 104 kN-m 2 ) vD = (2.8 m)(0.0121846 rad) = 0.034117 m Consider deflection at D resulting from rotation at C caused by distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θC = (slope magnitude) 3EI Values: M = (55 kN/m)(2.8 m)(1.4 m) = 215.6 kN-m, L = 7.2 m, EI = 7.02 × 104 kN-m2

Computation: ML (215.6 kN-m)(7.2 m) θC = = = 0.0073709 rad 3EI 3(7.02 × 107 kN-m 2 ) vD = −(2.8 m)(0.0073709 rad) = −0.020639 m Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 (assuming fixed support at C) vD = − 8EI Values: w = 55 kN/m, L = 2.8 m, EI = 7.02 × 104 kN-m2

Computation: wL4 (55 kN/m)(2.8 m)4 vD = − =− = −0.006020 m 8EI 8(7.02 × 104 kN-m 2 ) Beam deflection at D

vD = 0.034117 m − 0.020639 m − 0.006020 m = 0.007459 m = 7.46 mm ↑

Ans.

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10.43 The simply supported beam shown in Fig. P10.43 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. (c) the beam deflection at point E.

Fig. P10.43

Solution (a) Beam deflection at point A Determine cantilever deflection due to linearly distributed load on overhang. [Appendix C, Cantilever beam with linear load.] Relevant equation from Appendix C: w0 L4 (assuming fixed support at B) vA = − 30 EI Values: w0 = −6 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.2

Computation: w L4 (−6 kips/ft)(12 ft) 4 (12 in./ft)3 vA = − 0 = − = 0.293139 in. 30 EI 30(2.4447 × 107 kip-in.2 ) Consider deflection at A resulting from rotation at B caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (144 kip-ft)(18 ft)(12 in./ft) 2 θB = = = 0.0050892 rad 3EI 3(2.4447 × 107 kip-in.2 ) v A = (12 ft)(12 in./ft)(0.0050892 rad) = 0.732847 in.

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Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θB = (slope magnitude) 24 EI Values: w = 6 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL3 (6 kips/ft)(18 ft)3 (12 in./ft) 2 θB = = = 0.0085880 rad 24 EI 24(2.4447 × 107 kip-in.2 ) v A = −(12 ft)(12 in./ft)(0.0085880 rad) = −1.236679 in. Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) 6 EI Values: M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 θB = = = 0.0019085 rad 6 EI 6(2.4447 × 107 kip-in.2 ) v A = (12 ft)(12 in./ft)(0.0019085 rad) = 0.274818 in.

Beam deflection at A

v A = 0.293139 in. + 0.732847 in. − 1.236679 in. + 0.274818 in. = 0.064124 in. = 0.0641 in. ↑

Ans.

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(b) Beam deflection at point C Consider deflection at C from moment caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (144 kip-ft)(9 ft)(12 in./ft)3 ⎡ 2(18 ft) 2 − 3(18 ft)(9 ft) + (9 ft)2 ⎤⎦ = −0.206112 in. =− 7 2 ⎣ 6(18 ft)(2.4447 × 10 kip-in. )

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC = − 384 EI Values: w = −6 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: 5wL4 5(−6 kips/ft)(18 ft)4 (12 in./ft) 2 vC = − =− = 0.579693 in. 384 EI 384(2.4447 × 107 kip-in.2 ) Consider deflection at C resulting from moment caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

(108 kip-ft)(9 ft)(12 in./ft)3 ⎡ 2(18 ft) 2 − 3(18 ft)(9 ft) + (9 ft)2 ⎤⎦ = −0.154585 in. 7 2 ⎣ 6(18 ft)(2.4447 × 10 kip-in. )

Beam deflection at C

vC = −0.206112 in. + 0.579693 in. − 0.154585 in. = 0.218995 in. = 0.219 in. ↑

Ans.

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(c) Beam deflection at point E Consider deflection at E resulting from rotation at D caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θD = (slope magnitude) 6 EI Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (144 kip-ft)(18 ft)(12 in./ft) 2 θD = = = 0.0025446 rad 6 EI 6(2.4447 × 107 kip-in.2 ) vE = (6 ft)(12 in./ft)(0.0025446 rad) = 0.183212 in.

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θD = (slope magnitude) 24 EI Values: w = 6 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL3 (6 kips/ft)(18 ft)3 (12 in./ft) 2 θD = = = 0.0085880 rad 24 EI 24(2.4447 × 107 kip-in.2 ) vE = −(6 ft)(12 in./ft)(0.0085880 rad) = −0.618336 in. Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θD = (slope magnitude) 3EI Values: M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 θD = = = 0.0038169 rad 3EI 3(2.4447 × 107 kip-in.2 ) vE = (6 ft)(12 in./ft)(0.0038169 rad) = 0.274818 in.

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Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 (assuming fixed support at D) vE = − 8EI Values: w = −6 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL4 (−6 kips/ft)(6 ft) 4 (12 in./ft)3 vE = − =− = 0.068704 in. 8 EI 8(2.4447 × 107 kip-in.2 ) Beam deflection at E vE = 0.183212 in. − 0.618336 in. + 0.274818 in. + 0.068704 in.

= −0.091602 in. = 0.0916 in. ↓

Ans.

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10.44 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.44 supports two belt pulleys. Assume that the bearing at B can be idealized as a roller support and that the bearing at D can be idealized as a pin support. For the loading shown, determine: (a) the shaft deflection at pulley A. (b) the shaft deflection at pulley C.

Fig. P10.44

Solution Section properties:

I=

π

64

(30 mm) 4 = 39,760.78 mm 4

(a) Shaft deflection at pulley A Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 (assuming fixed support at B) vA = − 3EI Values: P = 700 N, L = 500 mm, EI = 7.95216 × 109 N-mm2

Computation: PL3 (700 N)(500 mm)3 vA = − =− = −3.6678 mm 3EI 3(7.95216 × 109 N-mm 2 ) Consider deflection at A resulting from rotation at B caused by pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) θB = 3EI Values: M = (700 N)(500 mm) = 350,000 N-mm, L = 1,800 mm, EI = 7.95216 × 109 N-mm2

Computation: ML (350,000 N-mm)(1,800 mm) θB = = = 0.0264079 rad 3EI 3(7.95216 × 109 N-mm 2 ) v A = −(500 mm)(0.0264079 rad) = −13.2040 mm Consider deflection at A resulting from rotation at B caused by pulley C load. [Appendix C, SS beam with concentrated load.] 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 Act without the permission of the copyright owner is unlawful.

Relevant equation from Appendix C: PL2 θB = (slope magnitude) 16 EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: PL2 (1,000 N)(1,800 mm) 2 θB = = = 0.0254648 rad 16 EI 16(7.95216 × 109 N-mm 2 ) v A = (500 mm)(0.0254648 rad) = 12.7324 mm

Shaft deflection at A

v A = −3.6678 mm − 13.2040 mm + 12.7324 mm = −4.1393 mm = 4.14 mm ↓

Ans.

(b) Shaft deflection at pulley C Consider pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = (700 N)(500 mm) = −350,000 N-mm, L = 1,800 mm, x = 900 mm, EI = 7.95216 × 109 N-mm2 Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−350,000 N-mm)(900 mm) ⎡ 2(1,800 mm) 2 − 3(1,800 mm)(900 mm) + (900 mm)2 ⎤⎦ =− 9 2 ⎣ 6(1,800 mm)(7.95216 × 10 N-mm ) = 8.9127 mm

Consider pulley C load. [Appendix C, SS beam with concentrated load.] Relevant equation from Appendix C: PL3 vC = − 48EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2

Computation: PL3 (1,000 N)(1,800 mm)3 vC = − =− = −15.2789 mm 48 EI 48(7.95216 × 109 N-mm 2 ) Shaft deflection at C

vC = 8.9127 mm − 15.2789 mm = −6.3662 mm = 6.37 mm ↓

Ans.

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10.45 The cantilever beam shown in Fig. P10.45 consists of a W530 × 92 structural steel wide-flange shape [E = 200 GPa; I = 552 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B.

Fig. P10.45

Solution (a) Beam deflection at point A Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8EI Values: w = −85 kN/m, L = 4 m, EI = 1.104 × 105 kN-m2

Computation: wL4 (−85 kN/m)(4 m) 4 vA = − =− = 0.024638 m 8EI 8(1.104 × 105 kN-m 2 ) Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 and θ B = (magnitude) vB = − 8EI 6 EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2

Computation: wL4 (85 kN)(2.5 m) 4 =− = −0.003759 m vB = − 8EI 8(1.104 × 105 kN-m 2 )

θB =

wL3 (85 kN)(2.5 m)3 = = 0.0020050 rad 6 EI 6(1.104 × 105 kN-m 2 )

v A = −0.003759 m − (1.5 m)(0.0020050 rad) = −0.006767 m Beam deflection at A

v A = 0.024638 m − 0.006767 m = 0.017871 m = 17.87 mm ↑

Ans.

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(b) Beam deflection at point B Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 (6 L2 − 4 Lx + x 2 ) (elastic curve) vB = − 24 EI Values: w = −85 kN/m, L = 4 m, x = 2.5 m, EI = 1.104 × 105 kN-m2

Computation: wx 2 (6 L2 − 4 Lx + x 2 ) vB = − 24 EI (−85 kN/m)(2.5 m) 2 ⎡ 6(4 m) 2 − 4(4 m)(2.5 m) + (2.5 m) 2 ⎤⎦ = 0.012481 m =− 5 2 ⎣ 24(1.104 × 10 kN-m ) Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2

Computation: wL4 (85 kN)(2.5 m) 4 vB = − =− = −0.003759 m 8EI 8(1.104 × 105 kN-m 2 ) Beam deflection at B

vB = 0.012481 m − 0.003759 m = 0.008722 m = 8.72 mm ↑

Ans.

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10.46 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.46 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearing at E can be idealized as a roller support. For the loading shown, determine: (a) the shaft deflection at pulley B. (b) the shaft deflection at point C. (c) the shaft deflection at pulley D.

Fig. P10.46

Solution Section properties:

I=

π

(30 mm) 4 = 39,760.78 mm 4

64 (a) Shaft deflection at pulley B Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a2 − b2 ) vB = − 6 LEI Values: P = 750 N, L = 1,000 mm, a = 300 mm, b = 700 mm, EI = 7.95216 × 109 N-mm2 Computation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI =−

(750 N)(300 mm)(700 mm) ⎡(1,000 mm) 2 − (300 mm) 2 − (700 mm) 2 ⎤⎦ 9 2 ⎣ 6(1,000 mm)(7.95216 × 10 N-mm )

= −1.38642 mm

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 ( L − b 2 − x 2 ) (elastic curve) vB = − 6 LEI Values: P = 500 N, L = 1,000 mm, x = 300 mm, b = 200 mm, EI = 7.95216 × 109 N-mm2 Computation: Pbx 2 ( L − b2 − x 2 ) vB = − 6 LEI =−

(500 N)(200 mm)(300 mm) ⎡(1,000 mm) 2 − (200 mm) 2 − (300 mm) 2 ⎤⎦ 9 2 ⎣ 6(1,000 mm)(7.95216 × 10 N-mm )

= −0.54702 mm 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 Act without the permission of the copyright owner is unlawful.

Shaft deflection at B

vB = −1.38642 mm − 0.54702 mm = −1.93344 mm = 1.933 mm ↓

Ans.

(b) Shaft deflection at point C Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 ( L − b 2 − x 2 ) (elastic curve) vC = − 6 LEI Values: P = 750 N, L = 1,000 mm, x = 500 mm, b = 300 mm, EI = 7.95216 × 109 N-mm2

Computation: Pbx 2 ( L − b2 − x2 ) vC = − 6 LEI (750 N)(300 mm)(500 mm) ⎡(1,000 mm) 2 − (300 mm) 2 − (500 mm) 2 ⎤⎦ =− 9 2 ⎣ 6(1,000 mm)(7.95216 × 10 N-mm ) = −1.55618 mm

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 ( L − b 2 − x 2 ) (elastic curve) vC = − 6 LEI Values: P = 500 N, L = 1,000 mm, x = 500 mm, b = 200 mm, EI = 7.95216 × 109 N-mm2

Computation: Pbx 2 ( L − b2 − x2 ) vC = − 6 LEI (500 N)(200 mm)(500 mm) ⎡(1,000 mm) 2 − (200 mm) 2 − (500 mm) 2 ⎤⎦ =− 6(1,000 mm)(7.95216 × 109 N-mm 2 ) ⎣ = −0.74403 mm

Shaft deflection at C

vC = −1.55618 mm − 0.74403 mm = −2.30021 mm = 2.30 mm ↓

Ans.

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(c) Shaft deflection at pulley D Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 ( L − b 2 − x 2 ) (elastic curve) vD = − 6 LEI Values: P = 750 N, L = 1,000 mm, x = 200 mm, b = 300 mm, EI = 7.95216 × 109 N-mm2

Computation: Pbx 2 ( L − b2 − x 2 ) vD = − 6 LEI (750 N)(300 mm)(200 mm) ⎡(1,000 mm) 2 − (300 mm) 2 − (200 mm) 2 ⎤⎦ =− 6(1,000 mm)(7.95216 × 109 N-mm 2 ) ⎣ = −0.82053 mm

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a 2 − b2 ) vD = − 6 LEI Values: P = 500 N, L = 1,000 mm, a = 800 mm, b = 200 mm, EI = 7.95216 × 109 N-mm2

Computation: Pab 2 ( L − a 2 − b2 ) vD = − 6 LEI (500 N)(800 mm)(200 mm) ⎡(1,000 mm) 2 − (800 mm) 2 − (200 mm) 2 ⎤⎦ =− 9 2 ⎣ 6(1,000 mm)(7.95216 × 10 N-mm ) = −0.53654 mm

Shaft deflection at D

vD = −0.82053 mm − 0.53654 mm = −1.35707 mm = 1.357 mm ↓

Ans.

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