10.47 The simply supported beam shown in Fig. P10.47 consists of a W410 × 60 structural steel wide-flange shape [ E E = 200 6 4 GPa; I = 216 × 10 mm ]. For the loading shown, determine: (a) the beam deflection at point B point B.. (b) the beam deflection at point C . (c) the beam deflection at point D 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: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: = −180 kN-m, L kN-m, L = = 6 m, x m, x = = 1.5 m, M = 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−180 180 kNkN-m) m)(1 (1.5 .5 m)
⎡⎣ 2(6 m)2 − 3(6 m)(1.5 m) + (1.5 m)2 ⎤⎦ = 0.008203 m 6(6 m)(4.32 × 10 kN-m ) 4
2
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = 70 kN, L kN, L = = 6 m, a = 1.5 m, b = 4.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(70 kN)(1.5 m)(4.5 m)
⎡⎣(6 m) 2 − (1.5 m)2 − (4.5 m)2 ⎤⎦ = − 0.004102 m 6(6 m)(4.32 × 10 kN-m ) 4
2
Consider uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 v B = − (2 x3 − 6 Lx2 + a2 x + 4 L2 x − a2 L) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 6 m, a = 3 m, x m, x = = 4.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Computation: wa 2 v B = − (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a 2 L ) 24 LEI L EI
=−
(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)⎤⎦ 24(6.0 m)(4. (4.32 × 10 kN-m ) 4
2
= −0.010156 m Beam deflection at B v B = 0.008203 m − 0.004102 m − 0.010156 m = − 0.006055 m= 6.06 mm 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −180 kN-m, L kN-m, L = = 6 m, x m, x = = 3.0 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: M x (2 L2 − 3 Lx + x 2 ) vC = − 6 LEI L EI
=−
(−180 kNkN-m)(3 m)(3.0 .0 m)
⎡⎣ 2(6 m)2 − 3(6 m)(3.0 m) + (3.0 m)2 ⎤⎦ = 0.009375 m 6(6 m) m)(4.32 × 10 kN-m ) 4
2
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 L EI
Values: P = = 70 kN, L kN, L = = 6 m, x m, x = = 3.0 m, b = 1.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI L EI
=−
(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 × 10 kN-m ) 4
2
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Consider uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 vC = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 6 m, a = 3 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: wa 3 vC = − ( 4 L2 − 7aL + 3a 2 ) 24 LEI L EI
=−
(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. (4.32 × 10 kN-m ) 4
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: M x v D = − (2 L2 − 3 Lx + x2 ) (elastic curve) 6 LEI L EI
Values: M = = −180 kN-m, L kN-m, L = = 6 m, x m, x = = 4.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: M x v D = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−180 kNkN-m)(4 m)(4.5 .5 m)
⎡⎣ 2(6 m)2 − 3(6 m)(4.5 m) + (4.5 m)2 ⎤⎦ = 0.005859 m 6(6 m) m)(4.32 × 10 kN-m ) 4
2
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 v D = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI L EI
Values: P = = 70 kN, L kN, L = = 6 m, x m, x = = 1.5 m, b = 1.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m
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Computation: Pbx 2 v D = − ( L − b2 − x2 ) 6 LEI L EI
=−
(70 kN)(1.5 m)(1.5 m)
⎡⎣(6 m) 2 − (1.5 m)2 − (1.5 m)2 ⎤⎦ = − 0.003190 m 6(6 m)(4.32 × 10 kN-m ) 4
2
Consider uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span .] Relevant equation from Appendix C: wx ( Lx3 − 4 aLx2 + 2 a2 x2 + 4a2 L2 − 4 a3 L + a4 ) v D = − 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 6 m, a = 3 m, x m, x = = 1.5 m, 4 2 EI = 4.32 = 4.32 × 10 kN-m Computation: wx v D = − ( Lx3 − 4aLx 2 + 2a 2 x 2 + 4a 2 L2 − 4a 3L + a 4 ) 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 ⎤⎦ 24(6.0 m)(4. (4.32 × 10 kN-m ) 4
2
= −0.012109 m Beam deflection at D
v D = 0.005859 m − 0.003190 m − 0.012109 m = − 0.009440 m= 9.44 mm 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 E = = 200 GPa; I GPa; I = = 351 6 4 × 10 mm ]. For the loading shown, determine: (a) the beam deflection at point A point A.. (b) the beam deflection at point C . (c) the beam deflection at point E 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 v A = − (assuming fixed support at B at B)) 3 EI Values: 4 2 P = = 35 kN, L kN, L = = 4 m, EI m, EI = 7.02 = 7.02 × 10 kN-m
Computation: PL3 (35 kN)(4 m)3 v A = − =− = −0.0106363 m 3 EI 3(7.02 × 104 kN k N-m2 ) 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: L (slope magnitude) θ B = 3 EI Values: M = = (35 kN)(4 m) = 140 kN-m, L kN-m, L = = 8 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: ML (140 kN-m)(8 m) θ B = = = 0.0053181 rad 3 EI 3(7.02 × 104 kN-m2 )
v A = −(4 m)(0. (0.00531 53181 rad) ad) = − 0.02 .0212726 m Consider uniformly distributed loads between C and and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 θ B = (2 L2 − a 2 ) (slope magnitude) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 8 m, a = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m
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Computation: wa 2 (80 kN/m)(4 m) 2 2 2 ⎡2(8 m)2 − (4 m)2 ⎤⎦ = 0.0106363 rad θ B = (2 L − a ) = 4 2 ⎣ 24 LEI L EI 24(8 m)(7.02 × 10 kN kN-m ) v A = (4 m)( m)(0. 0.01 0106 0636 363 3 rad rad)) = 0.04 0.0425 2545 451 1m 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: L θ B = (slope magnitude) 6 EI Values: = (80 kN/m)(2 m)(1 m) = 160 kN-m, M = 4 2 L = L = 8 m, EI m, EI = 7.02 = 7.02 × 10 kN-m
Computation: ML (160 kN-m)(8 m) = = 0.0030389 rad θ B = 6 EI 6(7.02 × 104 kN-m2 ) v A = −(4 m)(0. (0.00303 30389 ra rad) = − 0.01215 21557 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 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = (35 kN)(4 m) = −140 kN-m, L kN-m, L = = 8 m, 4 2 x = x = 4 m, EI m, EI = 7.02 = 7.02 × 10 kN-m Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−140 140 kNkN-m)(4 )(4 m) m)
⎡⎣ 2(8 m)2 − 3(8 m)(4 m) + (4 m)2 ⎤⎦ = 0.0079772 m 6(8 m) m)(7.02 × 10 kN-m ) 4
2
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Consider uniformly distributed loads between C and and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 3 vC = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 8 m, a = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: wa 3 vC = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
=−
(80 kN/m)(4 m)3
⎡⎣ 4(8 m)2 − 7(4 m)(8 m) + 3(4 m)2 ⎤⎦ = − 0.0303894 m 24(8 m)(7.02 × 10 kN-m ) 4
2
Consider concentrated moment from overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = (80 kN/m)(2 m)(1 m) = 160 kN-m, 4 2 L = L = 8 m, x m, x = = 4 m, EI m, EI = 7.02 = 7.02 × 10 kN-m Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−160 160 kNkN-m)(4 )(4 m) m)
⎡⎣ 2(8 m)2 − 3(8 m)(4 m) + (4 m)2 ⎤⎦ = 0.0091168 m 6(8 m) m)(7.02 × 10 kN-m ) 4
2
Beam deflection at C
vC = 0.0079772 m − 0.0303894 m + 0.0091168 m = −0.0132954 m = 13.30 mm 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: L (slope magnitude) θ D = 6 EI Values: M = = (35 kN)(4 m) = 140 kN-m, L kN-m, L = = 8 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m
Computation:
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θ D
=
ML 6 EI
=
(140 kN-m)(8 m) 6(7.02 × 104 kN-m2 )
= 0.0026591 rad
v E = −(2 m) m)(0.002 0026591 rad) = − 0.0 0.0053181 181 m Consider uniformly distributed loads between C and and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 θ D = (2 L − a) 2 (slope magnitude) 24 LEI L EI
Values: w = 80 kN/m, L kN/m, L = = 8 m, a = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: wa 2 (80 kN/m)(4 m) 2 2 2 θ D = (2 L − a) = 2(8 m) − (4 m)] = 0.0136752 rad 4 2 [ 24 LEI L EI 24(8 m)(7.02 × 10 kN k N-m ) v E = (2 m)(0 m)(0.0 .013 1367 6752 52 rad) rad) = 0.02 0.0273 7350 504 4m Consider deflection at E resulting resulting from rotation at D caused by uniform load on overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: L θ D = (slope magnitude) 3 EI Values: M = = (80 kN/m)(2 m)(1 m) = 160 kN-m, 4 2 L = L = 8 m, EI m, EI = 7.02 = 7.02 × 10 kN-m Computation: ML (160 kN-m)(8 m) θ D = = = 0.0060779 rad 3 EI 3(7.02 × 104 kN-m2 )
v E = −(2 m)(0.006 0060779 rad) = − 0.0121 121557 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 v E = − (assuming fixed support at D at D)) 8 EI Values: 4 2 w = 80 kN/m, L kN/m, L = = 2 m, EI m, EI = 7.02 = 7.02 × 10 kN-m
Computation: v E = −
wL4 8 EI
=−
(80 kN/m)(2 m) 4 8(7.02 × 104 kN k N-m2 )
= −0.0022792 0.0022792 m
Beam deflection at E v E = −0.0053181 m + 0.0273504 m − 0.0121557 m − 0.0022792 m
= 0.0075974 m = 7.60 mm 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 E = 4 29,000 ksi; I = 518 in. ]. For the loading shown, determine: (a) the beam deflection at point B point B.. (b) the beam deflection at point C . (c) the beam deflection at point F 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 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = 40 kips, L kips, L = = 18 ft, a = 4 ft, b = 14 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in. Computation: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(40 kips)(4 ft)(14 ft)(12 in./ft)3
⎡⎣ (18 ft)2 − (4 ft)2 − (14 ft) 2 ⎤⎦ = −0.267213 in. 6(18 6(18 ft)( ft)(1. 1.50 5022 22 × 10 kipkip-iin. ) 7
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 v B = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 30 kips, L kips, L = = 18 ft, x ft, x = = 4 ft, b = 6 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in. Computation: Pbx 2 v B = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(30 kips)(6 ft)(4 ft)(12 in./ft) 3
⎡⎣(18 ft) 2 − (6 ft) 2 − (4 ft) 2 ⎤⎦ = −0.208590 in. 6(18 6(18 ft)( ft)(1. 1.50 5022 22 × 10 kipkip-iin. ) 7
2
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Consider 20-kip concentrated load at F . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: = −(20 kips)(6 ft) = −120 kip-ft, L kip-ft, L = = 18 ft, M = 7 2 x = x = 14 ft, EI ft, EI = 1.5022 = 1.5022 × 10 kip-in. Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−120 kip-f kip-ft)( t)(14 14 ft)( ft)(12 12 in./ft in./ft))3
⎡⎣ 2(18 ft)2 − 3(18 ft)(14 ft) + (14 ft)2 ⎤⎦ = 0.157465 in. 6(18 6(18 ft)( ft)(1. 1.50 5022 22 × 10 kipkip-in in.. ) 7
2
Beam deflection at B
v B = −0.267213 in. − 0.208590 in 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 vC = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI L EI
Values: P = = 40 kips, L kips, L = = 18 ft, b = 4 ft, x ft, x = = 10 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in. Computation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI L EI
=−
(40 kips)(4 ft)(10 ft)(12 in./ft)3
⎡⎣(18 ft)2 − (4 ft) 2 − (10 ft) 2 ⎤⎦ = −0.354467 in. 6(18 6(18 ft)( ft)(1. 1.50 5022 22 × 10 kipkip-in in.. ) 7
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 vC = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI L EI
Values: P = = 30 kips, L kips, L = = 18 ft, b = 6 ft, x ft, x = = 8 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in.
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Computation: Pbx 2 vC = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(30 kips)(6 ft)(8 ft)(12 in./ft) 3
⎡⎣(18 ft) 2 − (6 ft) 2 − (8 ft) 2 ⎤⎦ = −0.343560 in. 6(18 6(18 ft)( t)(1.50 1.5022 22 × 10 kipkip-iin. ) 7
2
Consider 20-kip concentrated load at F . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(20 kips)(6 ft) = −120 kip-ft, L kip-ft, L = = 18 ft, 7 2 x = x = 10 ft, EI ft, EI = 1.5022 = 1.5022 × 10 kip-in. Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−120 kip-f kip-ft) t)(10 (10 ft)( ft)(12 12 in./ft in./ft))3
⎡⎣ 2(18 ft)2 − 3(18 ft)(10 ft) + (10 ft)2 ⎤⎦ = 0.265850 in. 6(18 6(18 ft)( t)(1.50 1.5022 22 × 10 kipkip-in in.. ) 7
2
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 ) (slope magnitude) θ E = 6 LEI L EI
Values: P = = 40 kips, L kips, L = = 18 ft, a = 4 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in. Computation: Pa( L2 − a 2 ) (40 kips)(4 ft)(12 in./ft) 2 ⎡(18 ft)2 − (4 ft) 2 ⎤⎦ = 0.0043740 rad θ E = = 7 2 ⎣ 6 LEI L EI 6(18 ft)(1.5022 × 10 kip-in. ) v F = (6 ft)( ft)(12 12 in./ in./ft ft)( )(0. 0.004 00437 3740 40 rad rad)) = 0.314 0.314930 930 in. 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 ) (slope magnitude) θ E = 6 LEI L EI
Values: P = = 30 kips, L kips, L = = 18 ft, x ft, x = = 8 ft, a = 12 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in. Computation: Pa( L2 − a 2 ) (30 kips)(12 ft)(12 in./ft) 2 ⎡(18 ft) 2 − (12 ft) 2 ⎤⎦ = 0.0057516 rad θ E = = 7 2 ⎣ 6 LEI L EI 6(18 ft)(1.5022 × 10 kip-in. ) v F = (6 ft)( ft)(12 12 in. in./f /ft) t)(0 (0.00 .0057 57516 516 rad rad)) = 0.414 0.414113 113 in. in. Consider deflection at F resulting resulting from rotation at E caused caused by 20-kip load on overhang EF . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: L θ E = (slope magnitude) 3 EI Values: M = = (20 kips)(6 ft) = 120 kip-ft, k ip-ft, L L = = 18 ft, 7 2 EI = 1.5022 = 1.5022 × 10 kip-in.
Computation: ML (120 kip-ft)(18 ft)(12 in./ft) 2 θ E = = = 0.0069019 rad 3 EI 3(1.5022 × 107 kip-in.2 ) v F = −(6 ft) ft)(1 (12 2 in. in./f /ft) t)(0 (0.0 .006 0690 9019 19 rad rad)) = − 0.49 0.4969 6935 35 in. in. Determine cantilever deflection due to concentrated load on overhang EF . [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 v F = − (assuming fixed support at E at E ) 3 EI Values: 7 2 P = = 20 kips, L kips, L = = 6 ft, EI ft, EI = 1.5022 = 1.5022 × 10 kip-in.
Computation: v F = −
PL3 3 EI
=−
(20 kips)( kips)(6 6 ft)3 (12 in./f in./ft) t)3 3(1.5022 × 107 kip-in.2 )
= −0.165645 in.
Beam deflection at F v F = 0.31493 14930 0 in. in. + 0.41411 14113 3 in. in. − 0.496 496935 935 in in. − 0.165 165645 in. in.
0.06646 66463 3 in in. = 0.0 0.0665 665 in. in. ↑ = 0.0
Ans.
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10.50 The cantilever beam shown in Fig. P10.50 consists of a rectangular structural steel tube shape [ E E = 200 GPa; I = 170 × 6 4 10 mm ]. For the loading shown, determine: (a) the beam deflection at point A point A.. (b) the beam deflection at point B 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 v A = − 8 EI Values: 4 2 w = −65 kN/m, L kN/m, L = = 6 m, EI m, EI = 3.4 = 3.4 × 10 kN-m
Computation: wL4 (−65 kN/m kN/m)( )(6 6 m) m)4 v A = − =− = 0.309706 m 8 EI 8(3.4 × 104 kN-m2 ) Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 v A = − 3 EI Values: 4 2 P = = 90 kN, L kN, L = = 6 m, EI m, EI = 3.4 = 3.4 × 10 kN-m
Computation: PL3 (90 kN)(6 m)3 v A = − =− = −0.190588 m 3 EI 3(3.4 × 104 kN-m2 ) Consider 30-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 θ B = v B = − and (magnitude) 3 EI 2 EI Values: 4 2 P = = 30 kN, L kN, L = = 3.5 m, EI m, EI = 3.4 = 3.4 × 10 kN-m
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Computation: PL3 (30 kN)(3.5 m)3 v B = − =− = − 0.012610 0.012610 m 3 EI 3(3.4 × 104 kN-m2 ) θ B
=
PL2
=
2 EI
(30 kN)(3.5 m)2 2(3.4 × 104 kN-m2 )
(a)
= 0.0054044 rad
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: L2 ML θ B = v B = − and (slope magnitude) 2 EI EI Values: 4 2 M = = 225 kN-m, L kN-m, L = = 3.5 m, EI m, EI = 3.4 = 3.4 × 10 kN-m
Computation: ML2 (225 kN-m)( k N-m)(3.5 3.5 m) 2 v B = − =− = −0.040533 m 2 EI 2(3.4 × 104 kN-m2 ) θ B
=
ML EI
=
(225 kN-m)(3.5 kN-m)(3.5 m) (3.4 × 104 kN-m2 )
(b)
= 0.0231618 rad
v A = −0.040533 m − (2. (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 v B = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −65 kN/m, L kN/m, L = = 6 m, x m, x = = 3.5 m, 4 2 EI = 3.4 = 3.4 × 10 kN-m
Computation: wx 2 v B = − (6 L2 − 4 Lx + x 2 ) 24 EI
=−
(−65 kN/ kN/m) m)(3 (3.5 .5 m)2
⎡⎣6(6 m)2 − 4(6 m)(3.5 m) + (3.5 m)2 ⎤⎦ = 0.140759 m 24(3.4 × 10 kN-m ) 4
2
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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 v B = − (3L − x) (elastic curve) 6 EI Values: P = = 90 kN, L kN, L = = 6 m, x m, x = = 3.5 m, 4 2 EI = 3.4 = 3.4 × 10 kN-m
Computation: Px 2 (90 kN)(3.5 m) 2 v B = − (3 L − x) = − [3(6 m) − (3.5 m)] = − 0.078364 m 6 EI 6(3.4 × 104 kN-m2 ) 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
v B = 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 E = = 200 GPa; I GPa; I = = 6 4 350 × 10 mm ]. For the loading shown, determine: (a) the beam deflection at point B point B.. (b) the beam deflection at point C . (c) the beam deflection at point E 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: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −315 kN-m, L kN-m, L = = 9 m, x m, x = = 3 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−315 315 kNkN-m)(3 m)(3 m)
⎡⎣ 2(9 m)2 − 3(9 m)(3 m) + (3 m)2 ⎤⎦ = 0.022500 m 6(9 m)(7.0 × 10 kN-m ) 4
2
Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 120 kN/m, L kN/m, L = = 9 m, a = 3 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: wa 3 v B = − (4 L2 − 7aL + 3a 2 ) 24 LEI L EI
=−
(120 kN/m)(3 m) 3
⎡⎣ 4(9 m)2 − 7(3 m)(9 m) + 3(3 m)2 ⎤⎦ = −0.034714 m 24(9 m) m)(7.0 × 10 kN-m ) 4
2
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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 v B = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 100 kN, L kN, L = = 9 m, b = 3 m, x m, x = = 3 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: Pbx 2 v B = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(100 kN)(3 m)(3 m)
⎡⎣(9 m)2 − (3 m) 2 − (3 m)2 ⎤⎦ = − 0.015000 m 6(9 m) m)(7.0 × 10 kN-m ) 4
2
Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, 4 2 L = L = 9 m, x m, x = = 6 m, EI m, EI = 7.0 = 7.0 × 10 kN-m Computation: v B = −
=−
M x 6 LEI L EI
(2 L2 − 3Lx + x2 )
(−270 270 kNkN-m)(6 m)(6 m)
⎡⎣ 2(9 m)2 − 3(9 m)(6 m) + (6 m)2 ⎤⎦ = 0.015429 m 6(9 m) m)(7.0 × 10 kN-m ) 4
2
Beam deflection at B
v B = 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −315 kN-m, L kN-m, L = = 9 m, x m, x = = 6 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m
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Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−315 315 kNkN-m)(6 )(6 m) m)
⎡⎣ 2(9 m)2 − 3(9 m)(6 m) + (6 m)2 ⎤⎦ = 0.018000 m 6(9 m) m)(7.0 × 10 kN-m ) 4
2
Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 x3 − 6 Lx2 + a2 x + 4 L2 x − a2 L) vC = − 24 LEI L EI
Values: w = 120 kN/m, L kN/m, L = = 9 m, a = 3 m, x m, x = = 6 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: wa 2 (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a 2 L ) vC = − 24 LEI L EI
=−
(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) m)(7.0 × 10 kN-m ) 4
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 vC = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = 100 kN, L kN, L = = 9 m, a = 6 m, b = 3 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: Pab 2 ( L − a 2 − b2 ) vC = − 6 LEI L EI
=−
(100 kN)(6 m)(3 m)
⎡⎣(9 m) 2 − (6 m)2 − (3 m)2 ⎤⎦ = − 0.017143 m 6(9 m) m)(7.0 × 10 kN-m ) 4
2
Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, 4 2 L = L = 9 m, x m, x = = 3 m, EI m, EI = 7.0 = 7.0 × 10 kN-m
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Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−270 270 kNkN-m)(3 m)(3 m)
⎡⎣ 2(9 m)2 − 3(9 m)(3 m) + (3 m)2 ⎤⎦ = 0.019286 m 6(9 m) m)(7.0 × 10 kN-m ) 4
2
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: L θ D = (slope magnitude) 6 EI Values: M = = −315 kN-m, L kN-m, L = = 9 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m
Computation: ML (−315 315 kNkN-m)(9 m)(9 m) θ D = = = −0.0067500 rad 6 EI 6(7.0 × 104 kN-m2 ) v E = (3 m)( − 0.0067500 rad) = − 0.020250 m Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 L2 − a 2 ) (slope magnitude) θ D = 24 LEI L EI
Values: w = 120 kN/m, L kN/m, L = = 9 m, a = 3 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: wa 2 (120 kN/m)(3 m) 2 2 2 ⎡ 2(9 m) 2 − (3 m)2 ⎤⎦ = 0.0109286 rad θ D = (2 L − a ) = 4 2 ⎣ 24 LEI L EI 24(9 m)(7.0 × 10 kN kN-m ) v E = (3 m)(0 m)(0.0 .010 1092 9286 86 rad rad)) = 0.03 0.0327 2786 86 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 ) (slope magnitude) θ D = 6 LEI L EI
Values: P = = 100 kN, L kN, L = = 9 m, a = 6 m, 4 2 EI = 7.0 = 7.0 × 10 kN-m Computation: Pa( L2 − a 2 ) (100 kN)(6 m) ⎡ (9 m)2 − (6 m) 2 ⎤⎦ = 0.0071429 rad = θ D = 4 2 ⎣ 6 LEI L EI 6(9 m)(7.0 × 10 kN-m ) v E = (3 m)(0. )(0.00 0071 7142 429 9 rad rad)) = 0.02 0.0214 1429 29 m Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: L θ D = (slope magnitude) 3 EI Values: M = = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, 4 2 L = L = 9 m, EI m, EI = 7.0 = 7.0 × 10 kN-m
Computation: ML (270 kN-m)(9 m) θ D = = = 0.0115714 rad 3 EI 3(7.0 × 104 kN-m2 ) v E = −(3 m)(0.0 0.0115714 714 ra rad) = − 0.03471 4714 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 v E = − (assuming fixed support at D at D)) 8 EI Values: 4 2 w = 60 kN/m, L kN/m, L = = 3 m, EI m, EI = 7.0 = 7.0 × 10 kN-m
Computation: v E = −
wL4 8 EI
=−
(60 kN-m)(3 kN-m)(3 m)4 8(7.0 × 104 kN-m2 )
= −0.008679 m
Beam deflection at E v E = −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 6 steel tube shape [ E E = = 200 GPa; I GPa; I = = 95 × 10 4 mm ]. For the loading shown, determine: (a) the beam deflection at point B 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 v B = − 8 EI Values: 4 2 w = 50 kN/m, L kN/m, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: wL4 (50 kN/m)(2 m) 4 =− = −0.0052632 0.0052632 m v B = − 8 EI 8(1.9 × 104 kN k N-m2 )
(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 v B = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −25 kN/m, L kN/m, L = = 5 m, x m, x = = 2 m, 4 2 EI = 1.9 = 1.9 × 10 kN-m Computation: wx 2 v B = − (6 L2 − 4 Lx + x2 ) 24 EI
=−
(−25 kN/m kN/m)( )(2 2 m) m)2
⎡⎣ 6(5 m)2 − 4(5 m)(2 m) + (2 m)2 ⎤⎦ = 0.0250000 m 24(1.9 × 10 kN-m ) 4
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: wL4 v B = − 8 EI Values: 4 2 w = 25 kN/m, L kN/m, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
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Computation: wL4 (25 kN/m)(2 m) 4 v B = − =− = −0.0026316 0.0026316 m 8 EI 8(1.9 × 104 kN k N-m2 )
(b)
Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 v B = − 3 EI Values: 4 2 P = = −20 kN, L kN, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: PL3 (−20 kN kN)(2 m) m)3 v B = − =− = 0.0028070 0.0028070 m 3 EI 3(1.9 × 104 kN-m2 )
(c)
Consider 50-kN concentrated load at C . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 v B = − (3L − x) (elastic curve) 6 EI Values: 4 2 P = = 50 kN, L kN, L = = 5 m, x m, x = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: Px 2 (50 kN)(2 m)2 v B = − (3 L − x) = − [3(5 m) − (2 m)] = − 0.0228070 m 6 EI 6(1.9 × 104 kN-m2 ) Beam deflection at B v B = −0.0052632 m + 0.0250000 m − 0.0026316 m + 0.0028070 m− 0.0228070 m
= −0.0028947 m = 2.89 mm 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 θ B = (slope magnitude) 6 EI Values: 4 2 w = 50 kN/m, L kN/m, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: [v B previously calculated in Eq. (a)] wL3 (50 kN/m)(2 m)3 = = 0.0035088 rad θ B = 6 EI 6(1.9 × 104 kN-m2 ) 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 = − 8 EI Values: 4 2 w = −25 kN/m, L kN/m, L = = 5 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: wL4 (−25 kN/m kN/m)( )(5 5 m)4 vC = − =− = 0.1027961 m 8 EI 8(1.9 × 104 kN k N-m2 ) 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 θ B = (slope magnitude) 6 EI Values: 4 2 w = 25 kN/m, L kN/m, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: [v B previously calculated in Eq. (b)] wL3 (25 kN)(2 m)3 θ B = = = 0.0017544 rad 6 EI 6(1.9 × 104 kN-m2 ) 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 θ B = 2 EI Values: 4 2 P = = 20 kN, L kN, L = = 2 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: [v B previously calculated in Eq. (c)] PL2 (20 kN)(2 m) 2 θ B = = = 0.0021053 rad 2 EI 2(1.9 × 104 kN-m2 ) vC = 0.0 0.002807 8070 m + (3 m) m)(0. (0.002 0021053 rad) = 0.00 .0091228 228 m Consider 50-kN concentrated load at C . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vC = − 3 EI Values: 4 2 P = = 50 kN, L kN, L = = 5 m, EI m, EI = 1.9 = 1.9 × 10 kN-m
Computation: PL3 (50 kN)(5 m)3 vC = − =− = −0.1096491 m 3 EI 3(1.9 × 104 kN k N-m2 ) 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 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 E = 4 29,000 ksi; I = 170 in. ]. For the loading shown, determine: (a) the beam deflection at point A point A.. (b) the beam deflection at point C . (c) the beam deflection at point D 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: L2 v A = − 2 EI Values: 6 2 M = = 85 kip-ft, L kip-ft, L = = 3 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: ML2 (85 kip-ft kip-ft)(3 )(3 ft) ft)2 (12 in./ft) in./ft)3 v A = − =− = −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: L (slope magnitude) θ B = 3 EI Values: 6 2 M = = 85 kip-ft, L kip-ft, L = = 15 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: ML (85 kip-ft)(15 ft)(12 in./ft) 2 θ B = = = 0.0124138 rad 3 EI 3(4.93 × 106 kip-in.2 ) v A = −(3 ft) ft)((12 in. in./f /ftt)(0. )(0.01 0124 2413 138 8 rad rad)) = − 0.44 0.4468 6897 97 in. 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 v A = − 8 EI Values: 6 2 w = 5 kips/ft, L kips/ft, L = = 3 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-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 Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Computation: wL4 (5 kips/ft) kips/ft)(3 (3 ft) ft) 4 (12 in./ft) in./ft)3 v A = − =− = −0.017744 in. 8 EI 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: L (slope magnitude) θ B = 3 EI Values: M = = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft, 6 2 L = L = 15 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: ML (22.5 kip-ft)(15 ft)(12 ft)(12 in./ft)2 = = 0.0032860 rad θ B = 3 EI 3(4.93 × 106 kip-in.2 ) v A = −(3 ft)( ft)(12 12 in./ in./ft ft)( )(0. 0.00 0032 3286 860 0 rad rad)) = − 0.11 0.1182 8296 96 in. in. Consider 5 kips/ft uniformly distributed load on segment BC . [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 L − a) 2 (slope magnitude) θ B = 24 LEI L EI
Values: w = 5 kips/ft, L kips/ft, L = = 15 ft, a = 5 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: θ B
=
wa 2 24 LEI
(2 L − a ) = 2
(5 kips/ft)( kips/ft)(5 5 ft)2 (12 in./ft) in./ft)2 24(15 ft)(4.93 × 10 ki k ip-in. ) 6
2
2
[ 2(15 ft) − (5 ft)] = 0.0063387 rad
ft)(1 (12 2 in./f in./ft) t)(0 (0.0 .006 0633 3387 87 rad) rad) = 0.22 0.2281 8195 95 in. in. v A = (3 ft)
with concentrated load not at midspan.] Consider 25-kip concentrated load. [Appendix C, SS beam with Relevant equation from Appendix C: Pb( L2 − b2 ) θ B = (slope magnitude) 6 LEI L EI Values: P = = 25 kips, L kips, L = = 15 ft, b = 5 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: Pb( L2 − b2 ) (25 kips)(5 ft)(12 in./ft) 2 ⎡ (15 ft)2 − (5 ft) 2 ⎤⎦ = 0.0081136 rad θ B = = 6 2 ⎣ 6 LEI L EI 6(15 ft)(4.93 × 10 kip-in. ) v A = (3 ft) ft)(1 (12 2 in./ in./ft ft)( )(0. 0.008 00811 1136 36 rad rad)) = 0.29 0.29208 2089 9 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 Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Beam deflection at A v A = −0.13 .134069 069 in. − 0.44689 6897 in. − 0.01774 7744 in.− 0.118 118296 in.+ 0.228 228195 in. in.+ 0.29208 2089 in.
= −0.19 .196722 722 in in. = 0.19 .1967 in 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −85 kip-ft, L kip-ft, L = = 15 ft, x ft, x = = 5 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−85 kip-ft) kip-ft)(5 (5 ft)( ft)(12 12 in./ft) in./ft)3
⎡⎣ 2(15 ft) 2 − 3(15 ft)(5 ft) + (5 ft) 2 ⎤⎦ = 0.413793 in. 6(15 ft ft)(4. )(4.9 93 × 10 kip-i p-in. ) 6
2
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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, 6 2 L = L = 15 ft, x ft, x = = 5 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in. Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−22.5 kip-f kip-ft)( t)(5 5 ft)(12 ft)(12 in./ft) in./ft)3
⎡⎣ 2(15 ft)2 − 3(15 ft)(5 ft) + (5 ft) 2 ⎤⎦ = 0.109533 in. 6(15 ft ft)(4 )(4.93 × 10 kip-in. ) 6
2
Consider 5 kips/ft uniformly distributed load on segment BC . [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 vC = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 5 kips/ft, L kips/ft, L = = 15 ft, a = 5 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in.
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Computation: wa 3 vC = − ( 4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
=−
(5 kips/ft)( kips/ft)(5 5 ft)3 (12 in./ft) in./ft)3
⎡⎣ 4(15 ft)2 − 7(5 ft)(15 ft) + 3(5 ft)2 ⎤⎦ = −0.273834 in. 24(15 ft ft)(4 )(4.93 × 10 kip-in. ) 6
2
with concentrated load not at midspan.] Consider 25-kip concentrated load. [Appendix C, SS beam with Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI L EI Values: P = = 25 kips, L kips, L = = 15 ft, b = 5 ft, x ft, x = = 5 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: Pbx 2 vC = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(25 kips)(5 ft)(5 ft)(12 ft)(12 in./ft) 3
⎡⎣(15 ft)2 − (5 ft) 2 − (5 ft) 2 ⎤⎦ = −0.425963 in. 6(1 6(15 ft ft)(4. (4.93 × 10 kip-in. ) 6
2
Beam deflection at C vC = 0.4 0.41379 3793 in in. + 0.1095 09533 in. in.− 0.273 273834 in in.− 0.4 0.425963 963 in in.
= −0.176 176471 in. = 0.17 .1765 in. 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: M x v D = − (2 L2 − 3 Lx + x2 ) (elastic curve) 6 LEI L EI
Values: M = = −85 kip-ft, L kip-ft, L = = 15 ft, x ft, x = = 10 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: M x v D = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−85 kip-ft) kip-ft)(10 (10 ft)( ft)(12 12 in./ft) in./ft)3
⎡⎣ 2(15 ft) 2 − 3(15 ft)(10 ft) + (10 ft)2 ⎤⎦ = 0.331034 in. 6(15 ft ft)(4. )(4.9 93 × 10 kip-i p-in. ) 6
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: M x v D = − (2 L2 − 3 Lx + x2 ) (elastic curve) 6 LEI L EI
Values: M = = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, 6 2 L = L = 15 ft, x ft, x = = 10 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in. Computation: M x v D = − ( 2 L2 − 3Lx + x 2 ) 6 LEI L EI
=−
(−22.5 22.5 kip-ft)( kip-ft)(10 10 ft)( ft)(12 in./ft) in./ft)3
⎡⎣ 2(15 ft)2 − 3(15 ft)(10 ft) + (10 ft)2 ⎤⎦ = 0.087627 in. 6(15 ft) ft)(4 (4..93 × 10 kip-in. in. ) 6
2
Consider 5 kips/ft uniformly distributed load on segment BC . [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 v D = − (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a2 L) 24 LEI L EI
Values: w = 5 kips/ft, L kips/ft, L = = 15 ft, a = 5 ft, x ft, x = = 10 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: wa 2 v D = − (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a 2 L ) 24 LEI
=−
(5 kips/ft)( kips/ft)(5 5 ft)2 (12 in./ft) in./ft)3
⎡⎣ 2(10 ft ft)3 − 6(15 ft)(10 ft)2 + (5 ft ft)2 (10 ft ft) + 4(15 ft)2 (10 ft ft) − (5 ft)2 (15 ft)⎤⎦ 24(15 ft) ft)(4 (4..93 × 10 kip-in. in. ) 6
2
= −0.228195 in. with concentrated load not at midspan.] Consider 25-kip concentrated load. [Appendix C, SS beam with Relevant equation from Appendix C: Pab 2 v D = − ( L − a 2 − b2 ) 6 LEI L EI Values: P = = 25 kips, L kips, L = = 15 ft, a = 10 ft, b = 5 ft, EI = 4.93 = 4.93 × 106 kip-in.2
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Computation: Pab 2 ( L − a 2 − b2 ) v D = − 6 LEI L EI
=−
(25 kips)(10 ft)(5 ft)(12 ft)(12 in./ft) 3
⎡⎣ (15 ft)2 − (10 ft)2 − (5 ft) 2 ⎤⎦ = −0.486815 in. 6(1 6(15 ft) ft)(4 (4..93 × 10 kip-in. ) 6
2
Beam deflection at D v D = 0.3 0.33103 1034 in in. + 0.08762 7627 in. in.− 0.22 .228195 195 in in.− 0.4868 86815 in in.
= −0.296 296349 in 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 E = 29,000 ksi; I = 4 170 in. ]. For the loading shown, determine: (a) the beam deflection at point A 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 v A = − 30 EI Values: 6 2 w0 = 8 kips/ft, L kips/ft, L = = 9 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: w0 L4 (8 kips/ft kips/ft)(9 )(9 ft) ft) 4 (12 in./ft) in./ft) 3 v A = − =− = −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: L (slope magnitude) θ B = 3 EI Values: M = = ½(8 kips/ft)(9 ft)(3 ft) = 108 kip-ft, 6 2 L = L = 18 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 = = 0.0189274 rad θ B = 3 EI 3(4.93 × 106 kip-in.2 ) v A = −(9 ft) ft)(12 in./f n./ft) t)(0 (0.0 .018 1892 9274 74 rad rad)) = − 2.04 2.0441 4157 57 in. 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: w0 L3 θ B = (slope magnitude) 45 EI Values: 6 2 w0 = 8 kips/ft, L kips/ft, L = = 18 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
Computation: w0 L3 (8 kips/ft kips/ft)(18 )(18 ft)3 (12 in./ft in./ft)) 2 θ B = = = 0.0302838 rad 45 EI 45(4.93 × 106 kip-in.2 ) v A = (9 ft) ft)(1 (12 2 in./ in./ft ft)( )(0.0 0.030 30283 2838 8 rad) rad) = 3.270 3.27065 652 2 in. Consider 4 kips/ft uniformly distributed load on segment CD. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θ B = (2 L2 − a 2 ) (slope magnitude) 24 LEI L EI
Values: w = 4 kips/ft, L kips/ft, L = = 18 ft, a = 9 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: wa 2 θ B = (2 L2 − a 2 ) 24 LEI L EI
=
(4 kips/ft)( kips/ft)(9 9 ft)2 (12 in./ft) in./ft)2
⎡⎣ 2(18 ft)2 − (9 ft)2 ⎤⎦ = 0.0124211 rad 24(18 ft) ft)(4 (4..93 × 10 kip-in. in. ) 6
2
v A = (9 ft)( ft)(12 12 in./ft) in./ft)(0. (0.012 012421 4211 1 rad) rad) = 1.3414 1.341478 78 in. Beam deflection at A
v A = −0.613247 in in. − 2.044157 in in.+ 3.270652 in in.+ 1.341478 in in.= 1.954726 in in.= 1.955 in 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: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
Values: M = = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft, 6 2 L = L = 18 ft, x ft, x = = 9 ft, EI ft, EI = 4.93 = 4.93 × 10 kip-in.
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Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−108 kip-f kip-ft)( t)(9 9 ft)(12 ft)(12 in./ft) in./ft)3
⎡⎣ 2(18 ft)2 − 3(18 ft)(9 ft) + (9 ft) 2 ⎤⎦ = 0.766559 in. 6(18 ft ft)(4 )(4.93 × 10 kip-in. ) 6
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 vC = − (7 L4 − 10 L2 x2 + 3 x4 ) 360 LEI L EI
Values: kips/ft, L = = 18 ft, x ft, x = = 9 ft, w0 = 8 kips/ft, L 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: w0 x vC = − (7 L4 − 10 L2 x2 + 3 x4 ) 360 LEI L EI
=−
(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. 360(1 0(18 ft) ft)(4 (4..93 × 10 kip-in. ) 6
2
Consider 4 kips/ft uniformly distributed load on segment CD. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 vC = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 4 kips/ft, L kips/ft, L = = 18 ft, a = 9 ft, 6 2 EI = 4.93 = 4.93 × 10 kip-in. Computation: wa 3 vC = − ( 4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
=−
(4 kips/ft)(9 kips/ft)(9 ft)3 (12 in./ft) in./ft)3
⎡⎣ 4(18 ft)2 − 7(9 ft)(18 ft) + 3(9 ft)2 ⎤⎦ = −0.958199 in. 24(18 ft) ft)(4 (4..93 × 10 kip-in. in. ) 6
2
Beam deflection at C
vC = 0.766559 in in. − 1.916398 in. − 0.958199 in in.= − 2.108037 in 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 E = 4 29,000 ksi; I = 843 in. ]. For the loading shown, determine: (a) the beam deflection at point A 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 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 v A = − 8 EI Values: 7 2 w = 4 kips/ft, L kips/ft, L = = 12 ft, EI ft, EI = 2.4447 = 2.4447 × 10 kip-in.
Computation: wL4 (4 kips/ft)(12 kips/ft)(12 ft)4 (12 in./ft) in./ft)3 v A = − =− = −0.732847 0.732847 in. 8 EI 8(2.4447 × 107 kip-in.2 ) Consider cantilever beam deflection of upward 4 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 θ = v =− and (slope magnitude) 8 EI 6 EI Values: 7 2 w = −4 kips/ft, L kips/ft, L = = 6 ft, EI ft, EI = 2.4447 = 2.4447 × 10 kip-in.
Computation: wL4 (−4 kips kips/f /ft) t)(6 (6 ft) ft) 4 (12 in. in./f /ft) t)3 v=− =− = 0.045803 in. 8 EI 8(2.4447 × 107 kip-in.2 ) θ
=
wL3 6 EI
=
(4 kips/ft)(6 kips/ft)(6 ft) ft)3 (12 in./ft) in./ft) 2 6(2.4447 × 107 kip-in.2 )
= 0.0008482 rad
v A = 0.04580 0.045803 3 in. in. + (6 ft)( ft)(12 12 in./ in./ft) ft)(0.0 (0.00084 008482 82 rad) 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: L (slope magnitude) θ B = 3 EI Values: M = = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft, 7 2 L = L = 24 ft, EI ft, EI = 2.4447 = 2.4447 × 10 kip-in.
Computation: ML (216 kip-ft)(24 ft)(12 ft)(12 in./ft) 2 θ B = = = 0.0101784 rad 3 EI 3(2.4447 × 107 kip-in.2 ) v A = −(12 ft) ft)(1 (12 2 in./ in./ft ft))(0.0 (0.010 1017 1784 84 rad rad)) = − 1.46 1.4656 5693 93 in. in. with concentrated load not at midspan.] Consider 42-kip concentrated load. [Appendix C, SS beam with Relevant equation from Appendix C: Pb( L2 − b2 ) θ B = (slope magnitude) 6 LEI L EI Values: P = = 42 kips, L kips, L = = 24 ft, b = 18 ft, 7 2 EI = 2.4447 = 2.4447 × 10 kip-in. Computation: Pb( L2 − b2 ) (42 kips)(18 ft)(12 in./ft) 2 ⎡(24 ft) 2 − (18 ft)2 ⎤⎦ = 0.0077929 rad θ B = = 7 2 ⎣ 6 LEI L EI 6(24 ft)(2.4447 × 10 kip-in. ) v A = (12 ft)(1 ft)(12 2 in./f in./ft) t)(0. (0.00 0077 77929 929 rad) rad) = 1.12 1.12217 2172 2 in. Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 θ B = (2 L2 − a 2 ) (slope magnitude) 24 LEI L EI
Values: w = 4 kips/ft, L kips/ft, L = = 24 ft, a = 6 ft, 7 2 EI = 2.4447 = 2.4447 × 10 kip-in. Computation: wa 2 (4 kips/ft)( kips/ft)(6 6 ft)2 (12 in./ft) in./ft)2 2 2 ⎡ 2(24 ft)2 − (6 ft) 2 ⎤⎦ = 0.0016434 rad ( 2 L − a ) = θ B = 7 2 ⎣ 24 LEI L EI 24(24 ft)(2.4447 × 10 kip-in. ) v A = (12 ft)(12 ft)(12 in./ft in./ft)(0 )(0.00 .00164 16434 34 rad) rad) = 0.2366 0.236648 48 in. Beam deflection at A v A = −0.73 .732847 in in. + 0.10 .106873 in.− 1.4656 65693 in in.+ 1.122172 172 in in.+ 0.236 236648 in.
= −0.73 .732847 in. = 0.7 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft, 7 2 L = L = 24 ft, x ft, x = = 12 ft, EI ft, EI = 2.4447 = 2.4447 × 10 kip-in. Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−216 kip-ft) kip-ft)(12 (12 ft)( ft)(12 12 in./ft) in./ft)3
⎡⎣ 2(24 ft)2 − 3(24 ft)(12 ft) + (12 ft)2 ⎤⎦ = 0.549635 in in. 6(24 6(24 ft)(2 t)(2..4447 × 10 kip kip-in. ) 7
2
with concentrated load not at midspan.] Consider 42-kip concentrated load. [Appendix C, SS beam with Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x 2 ) (elastic curve) 6 LEI L EI Values: P = = 42 kips, L kips, L = = 24 ft, b = 6 ft, 7 2 x = x = 12 ft, EI ft, EI = 2.4447 = 2.4447 × 10 kip-in. Computation: Pbx 2 ( L − b2 − x2 ) vC = − 6 LEI L EI
=−
(42 kips)(6 ft)(12 ft)(12 ft)(12 in./ft) i n./ft)3
⎡⎣(24 ft) 2 − (6 ft) 2 − (12 ft) 2 ⎤⎦ = −0.587804 in. 6(24 (24 ft)(2. (2.4447 × 10 kip-in. ) 7
2
Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC = − (2 x3 − 6 Lx2 + a2 x + 4 L2 x − a2 L) 24 LEI L EI
(elastic curve) Values: w = 4 kips/ft, L kips/ft, L = = 24 ft, a = 6 ft, x ft, x = = 12 ft, 7 2 EI = 2.4447 = 2.4447 × 10 kip-in.
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Computation: wa 2 vC = − (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a 2 L ) 24 LEI L EI
=−
(4 kips/ft)(6 kips/ft)(6 ft)2 (12 in./ft) 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 (24 ft) ft)(2 (2..4447 × 10 kip-in. in. ) 7
2
= −0.175578 in.
θ B
=
wa 2 24 LEI L EI
( 2 L2 − a 2 ) =
(4 kips/ft)( kips/ft)(6 6 ft)2 (12 in./ft) in./ft)2
⎡⎣ 2(24 ft)2 − (6 ft) 2 ⎤⎦ = 0.0016434 rad 24(24 ft)(2.4447 × 10 kip-in. ) 7
2
ft)(12 in./ft in./ft)(0 )(0.00 .00164 16434 34 rad) rad) = 0.2366 0.236648 48 in. v A = (12 ft)(12 Beam deflection at C
vC = 0.549635 in. − 0.587804 in.− 0.175578 in.= − 0.213747 in 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 E = = 200 6 4 GPa; I = 351 × 10 mm ]. For the loading shown, determine: (a) the beam deflection at point B 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: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −300 kN-m, L kN-m, L = = 9 m, x m, x = = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−300 300 kNkN-m)(4 m)(4 m)
⎡⎣ 2(9 m)2 − 3(9 m)(4 m) + (4 m)2 ⎤⎦ = 0.022159 m 6(9 m) m)(7.02 × 10 kN-m ) 4
2
Consider 85 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 85 kN/m, L kN/m, L = = 9 m, a = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: wa 3 v B = − (4 L2 − 7aL + 3a 2 ) 24 LEI L EI
=−
(85 kN/m)(4 m)3
⎡⎣ 4(9 m)2 − 7( 4 m)(9 m) + 3( 4 m)2 ⎤⎦ = −0.043052 m 24(9 m)(7.02 × 10 kN-m ) 4
2
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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 v B = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI
Values: P = = 140 kN, L kN, L = = 9 m, b = 3 m, x m, x = = 4 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: Pbx 2 v B = − ( L − b 2 − x2 ) 6 LEI L EI
=−
(140 kN)(3 m)(4 m)
⎡⎣ (9 m) 2 − (3 m)2 − (4 m)2 ⎤⎦ = −0.024818 m 6(9 m) m)(7.02 × 10 kN-m ) 4
2
Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: M x v B = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −175 kN-m, L kN-m, L = = 9 m, x m, x = = 5 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−175 175 kNkN-m) m)(5 (5 m)
⎡⎣ 2(9 m)2 − 3(9 m)(5 m) + (5 m)2 ⎤⎦ = 0.012003 m 6(9 m)(7.02 × 10 kN-m ) 4
2
Beam deflection at B
v B = 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: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −300 kN-m, L kN-m, L = = 9 m, x m, x = = 6 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−300 300 kNkN-m)(6 m)(6 m)
⎡⎣ 2(9 m)2 − 3(9 m)(6 m) + (6 m)2 ⎤⎦ = 0.017094 m 6(9 m) m)(7.02 × 10 kN-m ) 4
2
Consider 85 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC = − (2 x3 − 6 Lx2 + a2 x + 4 L2 x − a2 L) 24 LEI L EI
Values: w = 85 kN/m, L kN/m, L = = 9 m, a = 4 m, x m, x = = 6 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: wa 2 vC = − (2 x3 − 6 Lx 2 + a2 x + 4 L2 x − a 2 L ) 24 LEI L EI
=−
(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)⎤⎦ 24(9 m)(7.02 × 10 kN-m ) 4
2
= −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 vC = − ( L − a 2 − b2 ) 6 LEI L EI
Values: P = = 140 kN, L kN, L = = 9 m, a = 6 m, b = 3 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m
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Computation: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI L EI
=−
(140 kN)(6 m)(3 m)
⎡⎣(9 m)2 − (6 m)2 − (3 m)2 ⎤⎦ = −0.023932 m 6(9 m) m)(7.02 × 10 kN-m ) 4
2
Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: M x vC = − (2 L2 − 3 Lx + x 2 ) (elastic curve) 6 LEI L EI
Values: M = = −175 kN-m, L kN-m, L = = 9 m, x m, x = = 3 m, 4 2 EI = 7.02 = 7.02 × 10 kN-m Computation: M x vC = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−175 175 kNkN-m)( m)(3 m) m)
⎡⎣ 2(9 m)2 − 3(9 m)(3 m) + (3 m)2 ⎤⎦ = 0.012464 m 6(9 m) m)(7.02 × 10 kN-m ) 4
2
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. A. Added support is supplied by a tieback anchor at B at B,, which exerts a force of 4,000 lb on the soldier beam. Determine the horizontal deflection of the soldier 8 2 beam at point C . Assume EI Assume EI = = 5 × 10 lb-in. .
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 = − 8 EI Values: 8 2 w = 260 lb/ft, L lb/ft, L = = 25 ft, EI ft, EI = 5.0 = 5.0 × 10 lb-in.
Computation: wL4 (260 lb/ft)(25 lb/ft)(25 ft)4 (12 in./ft) in./ft)3 =− = −43.875 in. vC = − 8 EI 8(5.0 × 108 lb lb-in.2 )
Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 θ B = v B = and (slope magnitude) 3 EI 2 EI Values: 8 2 P = = 4,000 lb, L lb, L = = 18 ft, EI ft, EI = 5.0 = 5.0 × 10 lb-in. Computation: PL3 (4,000 lb)(18 lb)(18 ft)3 (12 in./ft) in./ft)3 = = 26.873856 in. v B = 3 EI 3(5.0 × 108 lb-in.2 ) θ B
=
PL2 2 EI
=
(4,000 lb)(18 lb)(18 ft)2 (12 in./ft) in./ft)2 2(5.0 × 108 lb lb-in.2 )
= 0.1866240 rad
vC = 26.8 26.873 7385 856 6 in. in. + (7 ft) ft)(1 (12 2 in./ in./ft ft)( )(0. 0.18 1866 6624 240 0 rad) rad) = 42.5 42.550 50272 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. A. Added support is supplied by a tieback anchor at B, B, which exerts a force of 5,000 lb on the soldier beam. Determine the horizontal deflection of the 8 soldier beam at point C . Assume EI Assume EI = = 5 × 10 lb2 in. .
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 = − 8 EI Values: 8 2 w = 260 lb/ft, L lb/ft, L = = 25 ft, EI ft, EI = 5.0 = 5.0 × 10 lb-in.
Computation: wL4 (260 lb/ft)(25 lb/ft)(25 ft)4 (12 in./ft) in./ft)3 vC = − =− = −43.875 in. 8 EI 8(5.0 × 108 lb 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: w0 L4 vC = − 30 EI Values: w0 = 260 lb/ft, L lb/ft, L = = 25 ft, EI ft, EI = 5.0 = 5.0 × 108 lb-in.2
Computation: w0 L4 (260 lb/ft)(25 lb/ft)(25 ft)4 (12 in./ft) in./ft)3 vC = − =− = −11.700 in. 30 EI 30(5.0 × 108 lb 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 θ B = v B = and (slope magnitude) 3 EI 2 EI Values: 8 2 P = = 5,000 lb, L lb, L = = 18 ft, EI ft, EI = 5.0 = 5.0 × 10 lb-in.
Computation: PL3 (5,000 lb)(18 lb)(18 ft) ft)3 (12 in./ft) in./ft)3 v B = = = 33.592320 in. 3 EI 3(5.0 × 108 lb lb-in.2 ) θ B
=
PL2 2 EI
=
(5, (5, 000 lb)(18 lb)(18 ft) ft)2 (12 in./ft) in./ft)2 2(5.0 × 108 lb lb-in.2 )
= 0.2332800 rad
vC = 33.5 33.592 9232 320 0 in. in. + (7 ft)( ft)(12 12 in./ in./ft ft)( )(0. 0.23 2332 3280 800 0 rad rad)) = 53.1 53.187 87840 in. Beam deflection at C
vC = −43.875 in in. − 11.700 in in. + 53.187840 in in. = −2.387160 in in. = 2.39 in in. →
Ans.
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