Mechanics of Materials Solutions Chapter10 Probs47 58

10.47 The simply supported beam shown in Fig. P10.47 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C. (c) the beam deflection at point D. Fig. P10.47

Solution (a) Beam deflection at point B Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2 L2 − 3Lx + x 2 ) (elastic curve) 6 LEI Values: M = −180 kN-m, L = 6 m, x = 1.5 m, EI = 4.32 × 104 kN-m2 Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−180 kN-m)(1.5 m) ⎡ 2(6 m) 2 − 3(6 m)(1.5 m) + (1.5 m) 2 ⎤⎦ = 0.008203 m =− 4 2 ⎣ 6(6 m)(4.32 × 10 kN-m ) Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a2 − b2 ) 6 LEI Values: P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m, EI = 4.32 × 104 kN-m2 Computation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI

=−

(70 kN)(1.5 m)(4.5 m) ⎡(6 m) 2 − (1.5 m)2 − (4.5 m) 2 ⎤⎦ = −0.004102 m 4 2 ⎣ 6(6 m)(4.32 × 10 kN-m )

Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vB = − (2 x 3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m, EI = 4.32 × 104 kN-m2 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.

Computation: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vB = − 24 LEI =−

(80 kN/m)(3 m) 2 ⎡ 2(4.5)3 − 6(6)(4.5) 2 + (3) 2 (4.5) + 4(6) 2 (4.5) − (3) 2 (6) ⎤⎦ 4 2 ⎣ 24(6.0 m)(4.32 × 10 kN-m )

= −0.010156 m Beam deflection at B

vB = 0.008203 m − 0.004102 m − 0.010156 m = −0.006055 m = 6.06 mm ↓

Ans.

(b) Beam deflection at point C 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) vC = − 6 LEI Values: M = −180 kN-m, L = 6 m, x = 3.0 m, EI = 4.32 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−180 kN-m)(3.0 m) ⎡ 2(6 m) 2 − 3(6 m)(3.0 m) + (3.0 m) 2 ⎤⎦ = 0.009375 m =− 4 2 ⎣ 6(6 m)(4.32 × 10 kN-m ) Consider concentrated 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 = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m, EI = 4.32 × 104 kN-m2

Computation: Pbx 2 ( L − b2 − x 2 ) vC = − 6 LEI (70 kN)(1.5 m)(3.0 m) ⎡(6 m) 2 − (1.5 m)2 − (3.0 m)2 ⎤⎦ = −0.005013 m =− 6(6 m)(4.32 × 104 kN-m 2 ) ⎣

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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 ) vC = − 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, EI = 4.32 × 104 kN-m2

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

(80 kN/m)(3 m)3 ⎡ 4(6 m) 2 − 7(3 m)(6 m) + 3(3 m) 2 ⎤⎦ = −0.015625 m 24(6.0 m)(4.32 × 104 kN-m 2 ) ⎣

Beam deflection at C

vC = 0.009375 m − 0.005013 m − 0.015625 m = −0.011263 m = 11.26 mm ↓

Ans.

(c) Beam deflection at point D Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vD = − (2 L2 − 3Lx + x 2 ) (elastic curve) 6 LEI Values: M = −180 kN-m, L = 6 m, x = 4.5 m, EI = 4.32 × 104 kN-m2

Computation: Mx vD = − (2 L2 − 3Lx + x 2 ) 6 LEI (−180 kN-m)(4.5 m) ⎡ 2(6 m) 2 − 3(6 m)(4.5 m) + (4.5 m) 2 ⎤⎦ = 0.005859 m =− 4 2 ⎣ 6(6 m)(4.32 × 10 kN-m ) Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vD = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI Values: P = 70 kN, L = 6 m, x = 1.5 m, b = 1.5 m, EI = 4.32 × 104 kN-m2

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Computation: Pbx 2 vD = − ( L − b2 − x2 ) 6 LEI (70 kN)(1.5 m)(1.5 m) ⎡(6 m) 2 − (1.5 m)2 − (1.5 m)2 ⎤⎦ = −0.003190 m =− 4 2 ⎣ 6(6 m)(4.32 × 10 kN-m ) Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wx vD = − ( Lx3 − 4aLx 2 + 2a 2 x 2 + 4a 2 L2 − 4a 3 L + a 4 ) 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 1.5 m, EI = 4.32 × 104 kN-m2

Computation: wx ( Lx3 − 4aLx 2 + 2a 2 x 2 + 4a 2 L2 − 4a 3 L + a 4 ) vD = − 24 LEI (80 kN/m)(1.5 m) ⎡(6)(1.5)3 − 4(3)(6)(1.5) 2 + 2(3) 2 (1.5) 2 + 4(3) 2 (6) 2 − 4(3)3 (6) + (3) 4 ⎤⎦ =− 4 2 ⎣ 24(6.0 m)(4.32 × 10 kN-m ) = −0.012109 m

Beam deflection at D

vD = 0.005859 m − 0.003190 m − 0.012109 m = −0.009440 m = 9.44 mm ↓

Ans.

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10.48 The simply supported beam shown in Fig. P10.48 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 A. (b) the beam deflection at point C. (c) the beam deflection at point E.

Fig. P10.48

Solution (a) Beam deflection at point A Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vA = − (assuming fixed support at B) 3EI Values: P = 35 kN, L = 4 m, EI = 7.02 × 104 kN-m2

Computation: PL3 (35 kN)(4 m)3 vA = − =− = −0.0106363 m 3EI 3(7.02 × 104 kN-m 2 ) Consider deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (140 kN-m)(8 m) θB = = = 0.0053181 rad 3EI 3(7.02 × 104 kN-m 2 )

v A = −(4 m)(0.0053181 rad) = −0.0212726 m Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 θB = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2

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Computation: wa 2 (80 kN/m)(4 m) 2 ⎡ 2(8 m) 2 − (4 m) 2 ⎤⎦ = 0.0106363 rad θB = (2 L2 − a 2 ) = 4 2 ⎣ 24 LEI 24(8 m)(7.02 × 10 kN-m ) v A = (4 m)(0.0106363 rad) = 0.0425451 m 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 = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2

Computation: ML (160 kN-m)(8 m) θB = = = 0.0030389 rad 6 EI 6(7.02 × 104 kN-m 2 ) v A = −(4 m)(0.0030389 rad) = −0.0121557 m Beam deflection at A v A = −0.0106363 m − 0.0212726 m + 0.0425451 m − 0.0121557 m

= −0.0015195 m = 1.520 mm ↓

Ans.

(b) Beam deflection at point C Consider concentrated moment from overhang AB. [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 = (35 kN)(4 m) = −140 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−140 kN-m)(4 m) ⎡ 2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2 ⎤⎦ = 0.0079772 m =− 6(8 m)(7.02 × 104 kN-m 2 ) ⎣

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Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 3 (4 L2 − 7aL + 3a 2 ) vC = − 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2

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

(80 kN/m)(4 m)3 ⎡ 4(8 m) 2 − 7(4 m)(8 m) + 3(4 m) 2 ⎤⎦ = −0.0303894 m 4 2 ⎣ 24(8 m)(7.02 × 10 kN-m )

Consider concentrated moment from 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 = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−160 kN-m)(4 m) ⎡ 2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2 ⎤⎦ = 0.0091168 m =− 4 2 ⎣ 6(8 m)(7.02 × 10 kN-m ) Beam deflection at C

vC = 0.0079772 m − 0.0303894 m + 0.0091168 m = −0.0132954 m = 13.30 mm ↓

Ans.

(c) Beam deflection at point E Consider deflection at E resulting from rotation at D caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θD = (slope magnitude) 6 EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2

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θD =

ML (140 kN-m)(8 m) = = 0.0026591 rad 6 EI 6(7.02 × 104 kN-m 2 )

vE = −(2 m)(0.0026591 rad) = −0.0053181 m Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 θD = (2 L − a ) 2 (slope magnitude) 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 2 (80 kN/m)(4 m)2 2 θD = (2 L − a ) 2 = 2(8 m) − (4 m) ] = 0.0136752 rad 4 2 [ 24 LEI 24(8 m)(7.02 × 10 kN-m )

vE = (2 m)(0.0136752 rad) = 0.0273504 m 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 = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (160 kN-m)(8 m) θD = = = 0.0060779 rad 3EI 3(7.02 × 104 kN-m 2 )

vE = −(2 m)(0.0060779 rad) = −0.0121557 m Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with distributed load.] Relevant equation from Appendix C: wL4 vE = − (assuming fixed support at D) 8EI Values: w = 80 kN/m, L = 2 m, EI = 7.02 × 104 kN-m2 wL4 (80 kN/m)(2 m) 4 Computation: vE = − =− = −0.0022792 m 8EI 8(7.02 × 104 kN-m 2 ) Beam deflection at E vE = −0.0053181 m + 0.0273504 m − 0.0121557 m − 0.0022792 m

= 0.0075974 m = 7.60 mm ↑

Ans.

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10.49 The simply supported beam shown in Fig. P10.49 consists of a W16 × 40 structural steel wide-flange shape [E = 29,000 ksi; I = 518 in.4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C. (c) the beam deflection at point F.

Fig. P10.49

Solution (a) Beam deflection at point B Consider 40-kip concentrated load at B. [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 = 40 kips, L = 18 ft, a = 4 ft, b = 14 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pab 2 ( L − a 2 − b2 ) vB = − 6 LEI =−

(40 kips)(4 ft)(14 ft)(12 in./ft)3 ⎡⎣(18 ft) 2 − (4 ft)2 − (14 ft)2 ⎤⎦ = −0.267213 in. 7 2 6(18 ft)(1.5022 × 10 kip-in. )

Consider 30-kip concentrated load at D. [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 = 30 kips, L = 18 ft, x = 4 ft, b = 6 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pbx 2 ( L − b2 − x2 ) vB = − 6 LEI (30 kips)(6 ft)(4 ft)(12 in./ft)3 ⎡ (18 ft) 2 − (6 ft) 2 − (4 ft) 2 ⎤⎦ = −0.208590 in. =− 7 2 ⎣ 6(18 ft)(1.5022 × 10 kip-in. )

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Consider 20-kip concentrated load at F. [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 = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft, x = 14 ft, EI = 1.5022 × 107 kip-in.2

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

(−120 kip-ft)(14 ft)(12 in./ft)3 ⎡ 2(18 ft) 2 − 3(18 ft)(14 ft) + (14 ft) 2 ⎤⎦ = 0.157465 in. 6(18 ft)(1.5022 × 107 kip-in.2 ) ⎣

Beam deflection at B

vB = −0.267213 in. − 0.208590 in. + 0.157465 in. = −0.318338 in. = 0.318 in. ↓

Ans.

(b) Beam deflection at point C Consider 40-kip concentrated load at B. [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 = 40 kips, L = 18 ft, b = 4 ft, x = 10 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pbx 2 ( L − b2 − x 2 ) vC = − 6 LEI =−

(40 kips)(4 ft)(10 ft)(12 in./ft)3 ⎡(18 ft) 2 − (4 ft) 2 − (10 ft) 2 ⎤⎦ = −0.354467 in. 6(18 ft)(1.5022 × 107 kip-in.2 ) ⎣

Consider 30-kip concentrated load at D. [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 = 30 kips, L = 18 ft, b = 6 ft, x = 8 ft, EI = 1.5022 × 107 kip-in.2

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Computation: Pbx 2 ( L − b2 − x2 ) vC = − 6 LEI =−

(30 kips)(6 ft)(8 ft)(12 in./ft)3 ⎡(18 ft) 2 − (6 ft) 2 − (8 ft) 2 ⎤⎦ = −0.343560 in. 6(18 ft)(1.5022 × 107 kip-in.2 ) ⎣

Consider 20-kip concentrated load at F. [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 = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft, x = 10 ft, EI = 1.5022 × 107 kip-in.2

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

(−120 kip-ft)(10 ft)(12 in./ft)3 ⎡ 2(18 ft) 2 − 3(18 ft)(10 ft) + (10 ft) 2 ⎤⎦ = 0.265850 in. 7 2 ⎣ 6(18 ft)(1.5022 × 10 kip-in. )

Beam deflection at C

vC = −0.354467 in. − 0.343560 in. + 0.265850 in. = −0.432177 in. = 0.432 in. ↓

Ans.

(c) Beam deflection at point F Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 − a 2 ) θE = (slope magnitude) 6 LEI Values: P = 40 kips, L = 18 ft, a = 4 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pa ( L2 − a 2 ) (40 kips)(4 ft)(12 in./ft) 2 ⎡(18 ft) 2 − (4 ft) 2 ⎤⎦ = 0.0043740 rad = θE = 6 LEI 6(18 ft)(1.5022 × 107 kip-in.2 ) ⎣ vF = (6 ft)(12 in./ft)(0.0043740 rad) = 0.314930 in.

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Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 − a 2 ) θE = (slope magnitude) 6 LEI Values: P = 30 kips, L = 18 ft, x = 8 ft, a = 12 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pa ( L2 − a 2 ) (30 kips)(12 ft)(12 in./ft) 2 ⎡(18 ft) 2 − (12 ft) 2 ⎤⎦ = 0.0057516 rad = θE = 7 2 ⎣ 6 LEI 6(18 ft)(1.5022 × 10 kip-in. ) vF = (6 ft)(12 in./ft)(0.0057516 rad) = 0.414113 in.

Consider deflection at F resulting from rotation at E caused by 20-kip load on overhang EF. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θE = (slope magnitude) 3EI Values: M = (20 kips)(6 ft) = 120 kip-ft, L = 18 ft, EI = 1.5022 × 107 kip-in.2

Computation: ML (120 kip-ft)(18 ft)(12 in./ft) 2 θE = = = 0.0069019 rad 3EI 3(1.5022 × 107 kip-in.2 ) vF = −(6 ft)(12 in./ft)(0.0069019 rad) = −0.496935 in.

Determine cantilever deflection due to concentrated load on overhang EF. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vF = − (assuming fixed support at E) 3EI Values: P = 20 kips, L = 6 ft, EI = 1.5022 × 107 kip-in.2

Computation: vF = −

PL3 (20 kips)(6 ft)3 (12 in./ft)3 =− = −0.165645 in. 3EI 3(1.5022 × 107 kip-in.2 )

Beam deflection at F vF = 0.314930 in. + 0.414113 in. − 0.496935 in. − 0.165645 in.

= 0.066463 in. = 0.0665 in. ↑

Ans.

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10.50 The cantilever beam shown in Fig. P10.50 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 170 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B.

Fig. P10.50

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 = −65 kN/m, L = 6 m, EI = 3.4 × 104 kN-m2

Computation: wL4 (−65 kN/m)(6 m) 4 vA = − =− = 0.309706 m 8EI 8(3.4 × 104 kN-m 2 ) Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vA = − 3EI Values: P = 90 kN, L = 6 m, EI = 3.4 × 104 kN-m2

Computation: PL3 (90 kN)(6 m)3 vA = − =− = −0.190588 m 3EI 3(3.4 × 104 kN-m 2 ) Consider 30-kN concentrated load at 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 = 30 kN, L = 3.5 m, EI = 3.4 × 104 kN-m2

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Computation: PL3 (30 kN)(3.5 m)3 vB = − =− = −0.012610 m 3EI 3(3.4 × 104 kN-m 2 )

θB =

(a)

PL2 (30 kN)(3.5 m) 2 = = 0.0054044 rad 2 EI 2(3.4 × 104 kN-m 2 )

v A = −0.012610 m − (2.5 m)(0.0054044 rad) = −0.026121 m Consider 225 kN-m concentrated moment at B. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equations from Appendix C: ML2 ML and θ B = (slope magnitude) vB = − 2 EI EI Values: M = 225 kN-m, L = 3.5 m, EI = 3.4 × 104 kN-m2

Computation: ML2 (225 kN-m)(3.5 m) 2 vB = − =− = −0.040533 m 2 EI 2(3.4 × 104 kN-m 2 ) ML (225 kN-m)(3.5 m) θB = = = 0.0231618 rad EI (3.4 × 104 kN-m 2 )

(b)

v A = −0.040533 m − (2.5 m)(0.0231618 rad) = −0.098438 m Beam deflection at A

v A = 0.309706 m − 0.190588 m − 0.026121 m − 0.098438 m = −0.005441 m = 5.44 mm ↓

Ans.

(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 = −65 kN/m, L = 6 m, x = 3.5 m, EI = 3.4 × 104 kN-m2

Computation: wx 2 (6 L2 − 4 Lx + x 2 ) vB = − 24 EI (−65 kN/m)(3.5 m) 2 ⎡6(6 m) 2 − 4(6 m)(3.5 m) + (3.5 m)2 ⎤⎦ = 0.140759 m =− 4 2 ⎣ 24(3.4 × 10 kN-m )

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Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 (3L − x) (elastic curve) vB = − 6 EI Values: P = 90 kN, L = 6 m, x = 3.5 m, EI = 3.4 × 104 kN-m2

Computation: Px 2 (90 kN)(3.5 m) 2 vB = − (3L − x) = − [3(6 m) − (3.5 m)] = −0.078364 m 6 EI 6(3.4 × 104 kN-m 2 ) Consider 30-kN concentrated load at B. Previously calculated in Eq. (a). Consider 225 kN-m concentrated moment at B. Previously calculated in Eq. (b). Beam deflection at B

vB = 0.140759 m − 0.078364 m − 0.012610 m − 0.040533 m = 0.009252 m = 9.25 mm ↑

Ans.

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10.51 The simply supported beam shown in Fig. P10.51 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 350 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C. (c) the beam deflection at point E.

Fig. P10.51

Solution (a) Beam deflection at point B Consider 315 kN-m 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 = −315 kN-m, L = 9 m, x = 3 m, EI = 7.0 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−315 kN-m)(3 m) ⎡ 2(9 m) 2 − 3(9 m)(3 m) + (3 m) 2 ⎤⎦ = 0.022500 m =− 6(9 m)(7.0 × 104 kN-m 2 ) ⎣ Consider 120 kN/m 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 = 120 kN/m, L = 9 m, a = 3 m, EI = 7.0 × 104 kN-m2

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

(120 kN/m)(3 m)3 ⎡ 4(9 m) 2 − 7(3 m)(9 m) + 3(3 m) 2 ⎤⎦ = −0.034714 m 4 2 ⎣ 24(9 m)(7.0 × 10 kN-m )

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Consider 100-kN concentrated 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 = 100 kN, L = 9 m, b = 3 m, x = 3 m, EI = 7.0 × 104 kN-m2

Computation: Pbx 2 ( L − b2 − x2 ) vB = − 6 LEI (100 kN)(3 m)(3 m) ⎡(9 m) 2 − (3 m) 2 − (3 m) 2 ⎤⎦ = −0.015000 m =− 4 2 ⎣ 6(9 m)(7.0 × 10 kN-m ) Consider 60 kN/m uniformly distributed load on overhang DE. [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 = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, x = 6 m, EI = 7.0 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) 6 LEI (−270 kN-m)(6 m) ⎡ 2(9 m)2 − 3(9 m)(6 m) + (6 m) 2 ⎤⎦ = 0.015429 m =− 6(9 m)(7.0 × 104 kN-m 2 ) ⎣

vB = −

Beam deflection at B

vB = 0.022500 m − 0.034714 m − 0.015000 m + 0.015429 m = −0.011785 m = 11.79 mm ↓

Ans.

(b) Beam deflection at point C Consider 315 kN-m 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) vC = − 6 LEI Values: M = −315 kN-m, L = 9 m, x = 6 m, EI = 7.0 × 104 kN-m2

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Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−315 kN-m)(6 m) ⎡ 2(9 m) 2 − 3(9 m)(6 m) + (6 m) 2 ⎤⎦ = 0.018000 m =− 4 2 ⎣ 6(9 m)(7.0 × 10 kN-m ) Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 x 3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI Values: w = 120 kN/m, L = 9 m, a = 3 m, x = 6 m, EI = 7.0 × 104 kN-m2 Computation: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI =−

(120 kN/m)(3 m) 2 ⎡ 2(6 m)3 − 6(9 m)(6 m) 2 + (3 m) 2 (6 m) + 4(9 m) 2 (6 m) − (3 m) 2 (9 m) ⎤⎦ 24(9 m)(7.0 × 104 kN-m 2 ) ⎣

= −0.028929 m

Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a 2 − b2 ) vC = − 6 LEI Values: P = 100 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.0 × 104 kN-m2 Computation: Pab 2 ( L − a 2 − b2 ) vC = − 6 LEI (100 kN)(6 m)(3 m) ⎡(9 m) 2 − (6 m) 2 − (3 m)2 ⎤⎦ = −0.017143 m =− 4 2 ⎣ 6(9 m)(7.0 × 10 kN-m ) Consider 60 kN/m uniformly distributed 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 = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, x = 3 m, EI = 7.0 × 104 kN-m2

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Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−270 kN-m)(3 m) ⎡ 2(9 m) 2 − 3(9 m)(3 m) + (3 m) 2 ⎤⎦ = 0.019286 m =− 4 2 ⎣ 6(9 m)(7.0 × 10 kN-m ) Beam deflection at C

vC = 0.018000 m − 0.028929 m − 0.017143 m + 0.019286 m = −0.008786 m = 8.79 mm ↓

Ans.

(c) Beam deflection at point E Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML θD = (slope magnitude) 6 EI Values: M = −315 kN-m, L = 9 m, EI = 7.0 × 104 kN-m2

Computation: ML (−315 kN-m)(9 m) θD = = = −0.0067500 rad 6 EI 6(7.0 × 104 kN-m 2 ) vE = (3 m)( − 0.0067500 rad) = −0.020250 m Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θD = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 120 kN/m, L = 9 m, a = 3 m, EI = 7.0 × 104 kN-m2

Computation: wa 2 (120 kN/m)(3 m) 2 ⎡ 2(9 m) 2 − (3 m) 2 ⎤⎦ = 0.0109286 rad (2 L2 − a 2 ) = θD = 4 2 ⎣ 24 LEI 24(9 m)(7.0 × 10 kN-m ) vE = (3 m)(0.0109286 rad) = 0.032786 m

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Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 − a 2 ) θD = (slope magnitude) 6 LEI Values: P = 100 kN, L = 9 m, a = 6 m, EI = 7.0 × 104 kN-m2

Computation: Pa ( L2 − a 2 ) (100 kN)(6 m) ⎡ (9 m) 2 − (6 m) 2 ⎤⎦ = 0.0071429 rad = θD = 4 2 ⎣ 6 LEI 6(9 m)(7.0 × 10 kN-m ) vE = (3 m)(0.0071429 rad) = 0.021429 m

Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML θD = (slope magnitude) 3EI Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, EI = 7.0 × 104 kN-m2

Computation: ML (270 kN-m)(9 m) θD = = = 0.0115714 rad 3EI 3(7.0 × 104 kN-m 2 ) vE = −(3 m)(0.0115714 rad) = −0.034714 m Determine cantilever deflection due to 60 kN/m uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: wL4 vE = − (assuming fixed support at D) 8EI Values: w = 60 kN/m, L = 3 m, EI = 7.0 × 104 kN-m2

Computation: vE = −

wL4 (60 kN-m)(3 m)4 =− = −0.008679 m 8EI 8(7.0 × 104 kN-m 2 )

Beam deflection at E vE = −0.020250 m + 0.032786 m + 0.021429 m − 0.034714 m − 0.008679 m

= −0.009429 m = 9.43 mm ↓

Ans.

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10.52 The cantilever beam shown in Fig. P10.52 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 95 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C.

Fig. P10.52

Solution (a) Beam deflection at point B Consider the downward 50 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8EI Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: wL4 (50 kN/m)(2 m) 4 vB = − =− = −0.0052632 m 8EI 8(1.9 × 104 kN-m 2 )

(a)

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-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 = −25 kN/m, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m2 Computation: wx 2 (6 L2 − 4 Lx + x 2 ) vB = − 24 EI (−25 kN/m)(2 m) 2 ⎡ 6(5 m) 2 − 4(5 m)(2 m) + (2 m)2 ⎤⎦ = 0.0250000 m =− 4 2 ⎣ 24(1.9 × 10 kN-m ) Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

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Computation: wL4 (25 kN/m)(2 m)4 vB = − =− = −0.0026316 m 8EI 8(1.9 × 104 kN-m 2 )

(b)

Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vB = − 3EI Values: P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: PL3 (−20 kN)(2 m)3 vB = − =− = 0.0028070 m 3EI 3(1.9 × 104 kN-m 2 )

(c)

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 (3L − x) (elastic curve) vB = − 6 EI Values: P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m2

Computation: Px 2 (50 kN)(2 m) 2 vB = − (3L − x) = − [3(5 m) − (2 m)] = −0.0228070 m 6 EI 6(1.9 × 104 kN-m 2 ) Beam deflection at B vB = −0.0052632 m + 0.0250000 m − 0.0026316 m + 0.0028070 m − 0.0228070 m

= −0.0028947 m = 2.89 mm ↓

Ans.

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(b) Beam deflection at point C Consider the downward 50 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θB = (slope magnitude) 6 EI Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: [vB previously calculated in Eq. (a)] wL3 (50 kN/m)(2 m)3 θB = = = 0.0035088 rad 6 EI 6(1.9 × 104 kN-m 2 ) vC = −0.0052632 m − (3 m)(0.0035088 rad) = −0.0157895 m

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC = − 8EI Values: w = −25 kN/m, L = 5 m, EI = 1.9 × 104 kN-m2

Computation: wL4 (−25 kN/m)(5 m) 4 vC = − =− = 0.1027961 m 8EI 8(1.9 × 104 kN-m 2 ) Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θB = (slope magnitude) 6 EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: [vB previously calculated in Eq. (b)] wL3 (25 kN)(2 m)3 θB = = = 0.0017544 rad 6 EI 6(1.9 × 104 kN-m 2 ) vC = −0.0026316 m − (3 m)(0.0017544 rad) = −0.0078948 m

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Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL2 θB = 2 EI Values: P = 20 kN, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: [vB previously calculated in Eq. (c)] PL2 (20 kN)(2 m) 2 θB = = = 0.0021053 rad 2 EI 2(1.9 × 104 kN-m 2 ) vC = 0.0028070 m + (3 m)(0.0021053 rad) = 0.0091228 m

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vC = − 3EI Values: P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m2

Computation: PL3 (50 kN)(5 m)3 vC = − =− = −0.1096491 m 3EI 3(1.9 × 104 kN-m 2 ) Beam deflection at C vC = −0.0157895 m + 0.1027961 m − 0.0078948 m + 0.0091228 m − 0.1096491 m

= −0.0214145 m = 21.4 mm ↓

Ans.

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10.53 The simply supported beam shown in Fig. P10.53 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 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 D.

Fig. P10.53

Solution (a) Beam deflection at point A Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.] Relevant equation from Appendix C: ML2 vA = − 2 EI Values: M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.2

Computation: ML2 (85 kip-ft)(3 ft)2 (12 in./ft)3 vA = − =− = −0.134069 in. 2 EI 2(4.93 × 106 kip-in.2 ) Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) θB = 3EI Values: M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2

Computation: ML (85 kip-ft)(15 ft)(12 in./ft) 2 θB = = = 0.0124138 rad 3EI 3(4.93 × 106 kip-in.2 ) v A = −(3 ft)(12 in./ft)(0.0124138 rad) = −0.446897 in.

Consider cantilever beam deflection of 5 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8EI Values: w = 5 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.2

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Computation: wL4 (5 kips/ft)(3 ft)4 (12 in./ft)3 vA = − =− = −0.017744 in. 8EI 8(4.93 × 106 kip-in.2 ) Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2

Computation: ML (22.5 kip-ft)(15 ft)(12 in./ft) 2 θB = = = 0.0032860 rad 3EI 3(4.93 × 106 kip-in.2 ) v A = −(3 ft)(12 in./ft)(0.0032860 rad) = −0.118296 in.

Consider 5 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θB = (2 L − a) 2 (slope magnitude) 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2

Computation:

θB =

wa 2 (5 kips/ft)(5 ft) 2 (12 in./ft) 2 2 (2 L − a ) 2 = 2(15 ft) − (5 ft) ] = 0.0063387 rad 6 2 [ 24 LEI 24(15 ft)(4.93 × 10 kip-in. )

v A = (3 ft)(12 in./ft)(0.0063387 rad) = 0.228195 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 − b 2 ) θB = (slope magnitude) 6 LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2

Computation: Pb( L2 − b 2 ) (25 kips)(5 ft)(12 in./ft) 2 ⎡ (15 ft) 2 − (5 ft) 2 ⎤⎦ = 0.0081136 rad θB = = 6 2 ⎣ 6 LEI 6(15 ft)(4.93 × 10 kip-in. ) v A = (3 ft)(12 in./ft)(0.0081136 rad) = 0.292089 in. 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.

Beam deflection at A vA = −0.134069 in. − 0.446897 in. − 0.017744 in. − 0.118296 in. + 0.228195 in. + 0.292089 in.

= −0.196722 in. = 0.1967 in. ↓

Ans.

(b) Beam deflection at point C Consider 85 kip-ft 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) vC = − 6 LEI Values: M = −85 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2

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

(−85 kip-ft)(5 ft)(12 in./ft)3 ⎡ 2(15 ft) 2 − 3(15 ft)(5 ft) + (5 ft) 2 ⎤⎦ = 0.413793 in. 6 2 ⎣ 6(15 ft)(4.93 × 10 kip-in. )

Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2

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

(−22.5 kip-ft)(5 ft)(12 in./ft)3 ⎡ 2(15 ft) 2 − 3(15 ft)(5 ft) + (5 ft) 2 ⎤⎦ = 0.109533 in. 6(15 ft)(4.93 × 106 kip-in.2 ) ⎣

Consider 5 kips/ft uniformly distributed load on segment BC. [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 ) vC = − 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2

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

(5 kips/ft)(5 ft)3 (12 in./ft)3 ⎡ 4(15 ft) 2 − 7(5 ft)(15 ft) + 3(5 ft) 2 ⎤⎦ = −0.273834 in. 24(15 ft)(4.93 × 106 kip-in.2 ) ⎣

Consider 25-kip concentrated 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 = 25 kips, L = 15 ft, b = 5 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2

Computation: Pbx 2 ( L − b2 − x2 ) vC = − 6 LEI =−

(25 kips)(5 ft)(5 ft)(12 in./ft)3 ⎡⎣(15 ft) 2 − (5 ft) 2 − (5 ft) 2 ⎤⎦ = −0.425963 in. 6 2 6(15 ft)(4.93 × 10 kip-in. )

Beam deflection at C vC = 0.413793 in. + 0.109533 in. − 0.273834 in. − 0.425963 in.

= −0.176471 in. = 0.1765 in. ↓

Ans.

(c) Beam deflection at point D Consider 85 kip-ft 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) vD = − 6 LEI Values: M = −85 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2

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

(−85 kip-ft)(10 ft)(12 in./ft)3 ⎡ 2(15 ft) 2 − 3(15 ft)(10 ft) + (10 ft)2 ⎤⎦ = 0.331034 in. 6(15 ft)(4.93 × 106 kip-in.2 ) ⎣

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Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vD = − 6 LEI Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2

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

(−22.5 kip-ft)(10 ft)(12 in./ft)3 ⎡ 2(15 ft)2 − 3(15 ft)(10 ft) + (10 ft)2 ⎤⎦ = 0.087627 in. 6(15 ft)(4.93 × 106 kip-in.2 ) ⎣

Consider 5 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vD = − 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2

Computation: wa 2 vD = − (2 x 3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI =−

(5 kips/ft)(5 ft) 2 (12 in./ft)3 ⎡ 2(10 ft)3 − 6(15 ft)(10 ft) 2 + (5 ft) 2 (10 ft) + 4(15 ft) 2 (10 ft) − (5 ft) 2 (15 ft) ⎤⎦ 24(15 ft)(4.93 × 106 kip-in.2 ) ⎣

= −0.228195 in.

Consider 25-kip concentrated 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 = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2

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Computation: Pab 2 ( L − a 2 − b2 ) vD = − 6 LEI =−

(25 kips)(10 ft)(5 ft)(12 in./ft)3 ⎡ (15 ft) 2 − (10 ft) 2 − (5 ft) 2 ⎤⎦ = −0.486815 in. 6(15 ft)(4.93 × 106 kip-in.2 ) ⎣

Beam deflection at D vD = 0.331034 in. + 0.087627 in. − 0.228195 in. − 0.486815 in.

= −0.296349 in. = 0.296 in. ↓

Ans.

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10.54 The simply supported beam shown in Fig. P10.54 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C.

Fig. P10.54

Solution (a) Beam deflection at point A Consider cantilever beam deflection of linearly distributed load on overhang AB. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 L4 vA = − 30 EI Values: w0 = 8 kips/ft, L = 9 ft, EI = 4.93 × 106 kip-in.2

Computation: w0 L4 (8 kips/ft)(9 ft)4 (12 in./ft)3 vA = − =− = −0.613247 in. 30 EI 30(4.93 × 106 kip-in.2 ) Consider rotation at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = ½(8 kips/ft)(9 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 4.93 × 106 kip-in.2

Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 θB = = = 0.0189274 rad 3EI 3(4.93 × 106 kip-in.2 ) v A = −(9 ft)(12 in./ft)(0.0189274 rad) = −2.044157 in.

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Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C: w L3 θB = 0 (slope magnitude) 45EI Values: w0 = 8 kips/ft, L = 18 ft, EI = 4.93 × 106 kip-in.2

Computation: w L3 (8 kips/ft)(18 ft)3 (12 in./ft) 2 θB = 0 = = 0.0302838 rad 45 EI 45(4.93 × 106 kip-in.2 ) v A = (9 ft)(12 in./ft)(0.0302838 rad) = 3.270652 in.

Consider 4 kips/ft uniformly distributed load on segment CD. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θB = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 4 kips/ft, L = 18 ft, a = 9 ft, EI = 4.93 × 106 kip-in.2

Computation: wa 2 θB = (2 L2 − a 2 ) 24 LEI (4 kips/ft)(9 ft) 2 (12 in./ft) 2 ⎡ 2(18 ft) 2 − (9 ft) 2 ⎤⎦ = 0.0124211 rad = 6 2 ⎣ 24(18 ft)(4.93 × 10 kip-in. ) v A = (9 ft)(12 in./ft)(0.0124211 rad) = 1.341478 in.

Beam deflection at A

vA = −0.613247 in. − 2.044157 in. + 3.270652 in. + 1.341478 in. = 1.954726 in. = 1.955 in. ↑

Ans.

(b) Beam deflection at point C Consider moment at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI Values: M = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft, L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.2

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

(−108 kip-ft)(9 ft)(12 in./ft)3 ⎡ 2(18 ft) 2 − 3(18 ft)(9 ft) + (9 ft) 2 ⎤⎦ = 0.766559 in. 6(18 ft)(4.93 × 106 kip-in.2 ) ⎣

Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C: w0 x (7 L4 − 10 L2 x 2 + 3 x 4 ) vC = − 360 LEI Values: w0 = 8 kips/ft, L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.2

Computation: w0 x (7 L4 − 10 L2 x 2 + 3 x 4 ) vC = − 360 LEI =−

(8 kips/ft)(9 ft)(12 in./ft)3 ⎡7(18 ft) 4 − 10(18 ft) 2 (9 ft) 2 + 3(9 ft)4 ⎤⎦ = −1.916398 in. 6 2 ⎣ 360(18 ft)(4.93 × 10 kip-in. )

Consider 4 kips/ft uniformly distributed load on segment CD. [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 ) vC = − 24 LEI Values: w = 4 kips/ft, L = 18 ft, a = 9 ft, EI = 4.93 × 106 kip-in.2

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

(4 kips/ft)(9 ft)3 (12 in./ft)3 ⎡ 4(18 ft) 2 − 7(9 ft)(18 ft) + 3(9 ft) 2 ⎤⎦ = −0.958199 in. 6 2 ⎣ 24(18 ft)(4.93 × 10 kip-in. )

Beam deflection at C

vC = 0.766559 in. − 1.916398 in. − 0.958199 in. = −2.108037 in. = 2.11 in. ↓

Ans.

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10.55 The simply supported beam shown in Fig. P10.55 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. P10.55

Solution (a) Beam deflection at point A Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8EI Values: w = 4 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL4 (4 kips/ft)(12 ft)4 (12 in./ft)3 vA = − =− = −0.732847 in. 8EI 8(2.4447 × 107 kip-in.2 ) Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 and θ = (slope magnitude) v=− 8EI 6 EI Values: w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.2

Computation: wL4 (−4 kips/ft)(6 ft)4 (12 in./ft)3 v=− =− = 0.045803 in. 8EI 8(2.4447 × 107 kip-in.2 )

θ=

wL3 (4 kips/ft)(6 ft)3 (12 in./ft) 2 = = 0.0008482 rad 6 EI 6(2.4447 × 107 kip-in.2 )

vA = 0.045803 in. + (6 ft)(12 in./ft)(0.0008482 rad) = 0.106873 in.

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Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML θB = (slope magnitude) 3EI Values: M = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft, L = 24 ft, EI = 2.4447 × 107 kip-in.2

Computation: ML (216 kip-ft)(24 ft)(12 in./ft) 2 θB = = = 0.0101784 rad 3EI 3(2.4447 × 107 kip-in.2 ) v A = −(12 ft)(12 in./ft)(0.0101784 rad) = −1.465693 in.

Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 − b 2 ) θB = (slope magnitude) 6 LEI Values: P = 42 kips, L = 24 ft, b = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation: Pb( L2 − b 2 ) (42 kips)(18 ft)(12 in./ft) 2 ⎡(24 ft) 2 − (18 ft) 2 ⎤⎦ = 0.0077929 rad θB = = 7 2 ⎣ 6 LEI 6(24 ft)(2.4447 × 10 kip-in. ) v A = (12 ft)(12 in./ft)(0.0077929 rad) = 1.122172 in.

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θB = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 4 kips/ft, L = 24 ft, a = 6 ft, EI = 2.4447 × 107 kip-in.2

Computation: wa 2 (4 kips/ft)(6 ft)2 (12 in./ft) 2 2 2 ⎡ 2(24 ft) 2 − (6 ft) 2 ⎤⎦ = 0.0016434 rad θB = (2 L − a ) = 7 2 ⎣ 24 LEI 24(24 ft)(2.4447 × 10 kip-in. ) v A = (12 ft)(12 in./ft)(0.0016434 rad) = 0.236648 in.

Beam deflection at A vA = −0.732847 in. + 0.106873 in. − 1.465693 in. + 1.122172 in. + 0.236648 in.

= −0.732847 in. = 0.733 in. ↓

Ans.

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(b) Beam deflection at point C Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) (elastic curve) vC = − 6 LEI Values: M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft, L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2

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

(−216 kip-ft)(12 ft)(12 in./ft)3 ⎡ 2(24 ft)2 − 3(24 ft)(12 ft) + (12 ft) 2 ⎤⎦ = 0.549635 in. 6(24 ft)(2.4447 × 107 kip-in.2 ) ⎣

Consider 42-kip concentrated 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 = 42 kips, L = 24 ft, b = 6 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2

Computation: Pbx 2 ( L − b2 − x 2 ) vC = − 6 LEI =−

(42 kips)(6 ft)(12 ft)(12 in./ft)3 ⎡ (24 ft) 2 − (6 ft) 2 − (12 ft) 2 ⎤⎦ = −0.587804 in. 7 2 ⎣ 6(24 ft)(2.4447 × 10 kip-in. )

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI (elastic curve) Values: w = 4 kips/ft, L = 24 ft, a = 6 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2

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Computation: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI =−

(4 kips/ft)(6 ft) 2 (12 in./ft)3 ⎡ 2(12 ft)3 − 6(24 ft)(12 ft) 2 + (6 ft) 2 (12 ft) + 4(24 ft) 2 (12 ft) − (6 ft) 2 (24 ft) ⎤⎦ 24(24 ft)(2.4447 × 107 kip-in.2 ) ⎣

= −0.175578 in.

θB =

wa 2 (4 kips/ft)(6 ft)2 (12 in./ft) 2 ⎡ 2(24 ft) 2 − (6 ft) 2 ⎤⎦ = 0.0016434 rad (2 L2 − a 2 ) = 24 LEI 24(24 ft)(2.4447 × 107 kip-in.2 ) ⎣

v A = (12 ft)(12 in./ft)(0.0016434 rad) = 0.236648 in.

Beam deflection at C

vC = 0.549635 in. − 0.587804 in. − 0.175578 in. = −0.213747 in. = 0.214 in. ↓

Ans.

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10.56 The simply supported beam shown in Fig. P10.56 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 C.

Fig. P10.56

Solution (a) Beam deflection at point B Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2 L2 − 3Lx + x 2 ) (elastic curve) 6 LEI Values: M = −300 kN-m, L = 9 m, x = 4 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−300 kN-m)(4 m) ⎡ 2(9 m)2 − 3(9 m)(4 m) + (4 m)2 ⎤⎦ = 0.022159 m =− 6(9 m)(7.02 × 104 kN-m 2 ) ⎣ Consider 85 kN/m uniformly distributed load on segment AB. [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 = 85 kN/m, L = 9 m, a = 4 m, EI = 7.02 × 104 kN-m2

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

(85 kN/m)(4 m)3 ⎡ 4(9 m) 2 − 7(4 m)(9 m) + 3(4 m) 2 ⎤⎦ = −0.043052 m 4 2 ⎣ 24(9 m)(7.02 × 10 kN-m )

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Consider 140-kN concentrated 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 = 140 kN, L = 9 m, b = 3 m, x = 4 m, EI = 7.02 × 104 kN-m2

Computation: Pbx 2 ( L − b2 − x2 ) vB = − 6 LEI (140 kN)(3 m)(4 m) ⎡(9 m) 2 − (3 m) 2 − (4 m) 2 ⎤⎦ = −0.024818 m =− 6(9 m)(7.02 × 104 kN-m 2 ) ⎣ Consider 175 kN-m 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 = −175 kN-m, L = 9 m, x = 5 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−175 kN-m)(5 m) ⎡ 2(9 m) 2 − 3(9 m)(5 m) + (5 m) 2 ⎤⎦ = 0.012003 m =− 6(9 m)(7.02 × 104 kN-m 2 ) ⎣ Beam deflection at B

vB = 0.022159 m − 0.043052 m − 0.024818 m + 0.012003 m = −0.033708 m = 33.7 mm ↓

Ans.

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(b) Beam deflection at point C Consider 300 kN-m 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) vC = − 6 LEI Values: M = −300 kN-m, L = 9 m, x = 6 m, EI = 7.02 × 104 kN-m2

Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−300 kN-m)(6 m) ⎡ 2(9 m) 2 − 3(9 m)(6 m) + (6 m) 2 ⎤⎦ = 0.017094 m =− 4 2 ⎣ 6(9 m)(7.02 × 10 kN-m ) Consider 85 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI Values: w = 85 kN/m, L = 9 m, a = 4 m, x = 6 m, EI = 7.02 × 104 kN-m2

Computation: wa 2 (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) vC = − 24 LEI (85 kN/m)(4 m)2 ⎡ 2(6 m)3 − 6(9 m)(6 m) 2 + (4 m) 2 (6 m) + 4(9 m) 2 (6 m) − (4 m) 2 (9 m) ⎤⎦ =− 4 2 ⎣ 24(9 m)(7.02 × 10 kN-m ) = −0.034441 m

Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a 2 − b2 ) vC = − 6 LEI Values: P = 140 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.02 × 104 kN-m2

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Computation: Pab 2 ( L − a 2 − b2 ) vC = − 6 LEI (140 kN)(6 m)(3 m) ⎡(9 m)2 − (6 m)2 − (3 m)2 ⎤⎦ = −0.023932 m =− 4 2 ⎣ 6(9 m)(7.02 × 10 kN-m ) Consider 175 kN-m 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) vC = − 6 LEI Values: M = −175 kN-m, L = 9 m, x = 3 m, EI = 7.02 × 104 kN-m2

Computation: Mx (2 L2 − 3Lx + x 2 ) vC = − 6 LEI (−175 kN-m)(3 m) ⎡ 2(9 m) 2 − 3(9 m)(3 m) + (3 m) 2 ⎤⎦ = 0.012464 m =− 4 2 ⎣ 6(9 m)(7.02 × 10 kN-m ) Beam deflection at C

vC = 0.017094 m − 0.034441 m − 0.023932 m + 0.012464 m = −0.028814 m = 28.8 mm ↓

Ans.

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10.57 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a uniformly distributed soil loading of 260 lb/ft, as shown in Fig. P10.57. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 4,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lb-in.2.

Fig. P10.57

Solution Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC = − 8EI Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2

Computation: wL4 (260 lb/ft)(25 ft)4 (12 in./ft)3 vC = − =− = −43.875 in. 8EI 8(5.0 × 108 lb-in.2 )

Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 and θ B = (slope magnitude) vB = 3EI 2 EI Values: P = 4,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2 Computation: PL3 (4,000 lb)(18 ft)3 (12 in./ft)3 = = 26.873856 in. vB = 3EI 3(5.0 × 108 lb-in.2 )

θB =

PL2 (4,000 lb)(18 ft)2 (12 in./ft) 2 = = 0.1866240 rad 2 EI 2(5.0 × 108 lb-in.2 )

vC = 26.873856 in. + (7 ft)(12 in./ft)(0.1866240 rad) = 42.550272 in. Beam deflection at C vC = −43.875 in. + 42.550272 in. = −1.324728 in. = 1.325 in. →

Ans.

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10.58 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a soil loading that is linearly distributed from 520 lb/ft to 260 lb/ft, as shown in Fig. P10.58. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 5,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lbin.2.

Fig. P10.58

Solution Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC = − 8EI Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2

Computation: wL4 (260 lb/ft)(25 ft)4 (12 in./ft)3 vC = − =− = −43.875 in. 8EI 8(5.0 × 108 lb-in.2 )

Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at B. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w L4 vC = − 0 30 EI Values: w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2

Computation: w L4 (260 lb/ft)(25 ft)4 (12 in./ft)3 vC = − 0 = − = −11.700 in. 30 EI 30(5.0 × 108 lb-in.2 )

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Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 vB = and θ B = (slope magnitude) 3EI 2 EI Values: P = 5,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2

Computation: PL3 (5,000 lb)(18 ft)3 (12 in./ft)3 = = 33.592320 in. vB = 3EI 3(5.0 × 108 lb-in.2 ) PL2 (5,000 lb)(18 ft) 2 (12 in./ft) 2 θB = = = 0.2332800 rad 2 EI 2(5.0 × 108 lb-in.2 ) vC = 33.592320 in. + (7 ft)(12 in./ft)(0.2332800 rad) = 53.187840 in. Beam deflection at C vC = −43.875 in. − 11.700 in. + 53.187840 in. = −2.387160 in. = 2.39 in. →

Ans.

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