11.18a For the beams and loadings shown 4 2 below, assume that EI that EI = 3.0 = 3.0 × 10 kN-m is constant for each beam. (a) For the beam in Fig. P11.18a P11.18a, determine the concentrated upward force P force P required required to make the total beam deflection at B at B equal equal to zero (i.e., v B = 0).
Fig. P11.18a P11.18a
Solution Downward deflection at B due to 15 kN/m uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Values: w = 15 kN/m, L kN/m, L = = 7 m, a = 3.5 m, 4 2 EI = 3.0 = 3.0 × 10 kN-m Computation: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
=−
(15 kN/m)(3.5 m)3 24(7 m) EI
⎡⎣ 4(7 m) m) − 7(3.5 m)(7 m) + 3(3.5 m) ⎤⎦ = − 2
2
234.472656 kN-m3 EI
Upward deflection at B due to concentrated load P . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 v B = 48 EI Values: 4 2 L = L = 7 m, EI m, EI = 3.0 = 3.0 × 10 kN-m
Computation: v B =
PL3
=
48 EI
P (7 m)3
=
P (7.145833 m3 )
48EI
EI
Compatibility equation at B: 234.472656 kN kN-m3 P (7.145833 m3 ) − + =0 EI EI
∴ P =
234.472656 kN-m3 7.145833 m3
= 32.8125 kN = 32.8 kN
Ans.
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11.18b For the beams and loadings shown 4 2 below, assume that EI that EI = 3.0 = 3.0 × 10 kN-m is constant for each beam. (b) For the beam in Fig. P11.18b P11.18b, determine the concentrated moment M required to make the total beam slope at A at A equal equal to zero (i.e., θ A = 0).
Fig. P11.18b P11.18b
Solution Slope at A due to 32 kN concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL2 (slope magnitude) θ A = 2 EI Values: 4 2 P = = 32 kN, L kN, L = = 4 m, EI m, EI = 3.0 = 3.0 × 10 kN-m
Computation: θ A
=
PL2
=
(32 kN)(4 m)2
2 EI
= −
256 kN-m2
2 EI
EI
(negative slope by inspection)
Slope at A due to concentrated moment M . [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: L (slope magnitude) θ A = EI Values: 4 2 L = L = 4 m, EI m, EI = 3.0 = 3.0 × 10 kN-m
Computation: θ A
=
ML EI
=
M (4 m) EI
=
M (4 m) EI
(positive slope by inspection)
Compatibility equation at A: 256 kN-m2 M (4 m) − + =0 EI EI
∴ M =
256 kN-m2 4m
= 64.0 kN-m
Ans.
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11.19a For the beams and loadings shown 6 2 below, assume that EI that EI = 5.0 = 5.0 × 10 kip-in. is constant for each beam. (a) For the beam in Fig. P11.19a P11.19a, determine the concentrated upward force P force P required required to make the total beam deflection at B at B equal equal to zero (i.e., v B = 0).
Fig. P11.19a P11.19a
Solution Downward deflection at B due to 4 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 v B = − 8 EI Values: 6 2 w = 4 kips/ft, L kips/ft, L = = 13 ft, EI ft, EI = 5.0 = 5.0 × 10 kip-in.
Computation: v B = −
wL4
=−
(4 ki kips/ft)(13 ft ft) 4
8 EI
= −
14,28 ,280.5 ki kip-ft3
8EI
EI
Upward deflection at B due to concentrated load P . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = 3 EI Values: 6 2 L = L = 13 ft, EI ft, EI = 5.0 = 5.0 × 10 kip-in.
Computation: v B =
PL3 3 EI
=
P(13 ft)3
=
3EI
P (732.333333 ft3 ) EI
Compatibility equation at B: 14, 28 280.5 kip-ft3 P (732.333333 ft3 ) − + =0 EI EI
∴ P =
14,280.5 kip-ft kip-ft3 732.333333 ft3
= 19.50 kips
Ans.
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11.19b For the beams and loadings shown 6 2 below, assume that EI that EI = 5.0 = 5.0 × 10 kip-in. is constant for each beam. (b) For the beam in Fig. P11.19b, determine the concentrated moment M required to make the total beam slope at C equal equal to zero (i.e., θ C 0). C =
Fig. P11.19b P11.19b
Solution Slope at C due due to 40-kip concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 θ C = (slope magnitude) 16 EI Values: 6 2 P = = 40 kips, L kips, L = = 18 ft, EI ft, EI = 5.0 = 5.0 × 10 kip-in.
Computation: θ C =
PL2
=
(40 (40 ki kips)(18 ft) 2
16 EI
=
810 ki kip-ft 2
16EI
EI
(negative slope by inspection)
Slope at C due due to concentrated moment M . [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: L θ C = (slope magnitude) 3 EI Values: 6 2 L = L = 18 ft, EI ft, EI = 5.0 = 5.0 × 10 kip-in.
Computation: θ C =
ML
=
M (18 ft)
3 EI
3EI
=
M (6 ft) EI
(positive slope by inspection)
Compatibility equation at C : 810 kip-ft 2 M (6 ft) − + =0 EI EI
∴ M =
810 kip-ft 2 6 ft
= 135.0 kip-ft
Ans.
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.
11.20a For the beams and loadings shown 4 2 below, assume that EI that EI = 5.0 × 10 kN-m is constant for each beam. (a) For the beam in Fig. P11.20a P11.20a, determine the concentrated downward force P required to make the total beam deflection at B at B equal equal to zero (i.e., v B = 0).
Fig. P11.20a P11.20a
Solution Upward deflection at B due to 105 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 = = −105 kN-m, L kN-m, L = = 8 m, x m, x = = 4 m, 4 2 EI = 5.0 = 5.0 × 10 kN-m Computation: M x v B = − (2 L2 − 3 Lx + x 2 ) 6 LEI L EI
=−
(−105 kN-m)(4 m) 6(8 m) EI
⎡⎣ 2(8 m) − 3(8 m)(4 m) + (4 m) ⎤⎦ = 2
2
420 kN-m3 EI
Downward deflection at B due to concentrated load P . [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 v B = − 48 EI Values: 4 2 L = L = 8 m, EI m, EI = 5.0 = 5.0 × 10 kN-m
Computation: v B = −
PL3 48 EI
=−
P(8 m)3
= −
P (10.666667 m3 )
48EI
EI
Compatibility equation at B: 420 kN kN-m3 P (10.666667 m3 ) − =0 EI EI
∴ P =
420 kN-m3 10.666667 10.666667 m3
= 39.375 kN = 39.4 kN kN
Ans.
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11.20b For the beams and loadings shown 4 2 below, assume that EI that EI = 5.0 = 5.0 × 10 kN-m is constant for each beam. (b) For the beam in Fig. P11.20b P11.20b, determine the concentrated moment M required to make the total beam slope at A at A equal equal to zero (i.e., θ A = 0).
Fig. P11.20b P11.20b
Solution Slope at A due to 6 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 θ A = (slope magnitude) 6 EI Values: 4 2 w = 6 kN/m, L kN/m, L = = 5 m, EI m, EI = 5.0 = 5.0 × 10 kN-m
Computation: θ A
=
wL3
=
(6 kN kN/m)(5 m)3
6 EI
=
125 kN kN-m2
6 EI
EI
(positive slope by inspection)
Slope at A due to concentrated moment M . [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: L (slope magnitude) θ A = EI Values: 4 2 L = L = 5 m, EI m, EI = 5.0 = 5.0 × 10 kN-m
Computation: θ A
=
ML
=
M (5 m)
EI
EI
(negative slope by inspection)
Compatibility equation at A: 125 kN-m2 M (5 m) − =0 EI EI
∴ M =
125 kN-m2 5m
= 25.0 kN-m
Ans.
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11.21a For the beams and loadings shown 6 2 below, assume that EI that EI = 8.0 = 8.0 × 10 kip-in. is constant for each beam. (a) For the beam in Fig. P11.21a P11.21a, determine the concentrated downward force P required to make the total beam deflection at B at B equal equal to zero (i.e., v B = 0).
Fig. P11.21a P11.21a
Solution Upward deflection at B due to 125 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: L2 v B = − 2 EI Values: 6 2 M = = −125 kip-ft, L kip-ft, L = = 15 ft, EI ft, EI = 8.0 = 8.0 × 10 kip-in.
Computation: ML2 (−125 ki kip-ft)(15 ft ft)2 14,062 ,062.5 ki kip-ft3 v B = − =− = 2 EI 2 EI EI Downward deflection at B due to concentrated load P . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v B = − 3 EI Values: 6 2 L = L = 15 ft, EI ft, EI = 8.0 = 8.0 × 10 kip-in.
Computation: v B = −
PL3 3 EI
=−
P(15 ft)3 3EI
= −
P (1,125 ft 3 ) EI
Compatibility equation at B: 14, 06 062.5 kip-ft3 P (1,125 ft3 ) − =0 EI EI
∴ P =
14,062.5 kip-ft 3 1,125 ft 3
= 12.50 kips
Ans.
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11.21b For the beams and loadings shown 6 2 below, assume that EI that EI = 8.0 = 8.0 × 10 kip-in. is constant for each beam. (b) For the beam in Fig. P11.21b P11.21b, determine the concentrated moment required to make the total beam slope at A at A equal to zero (i.e., θ A = 0).
Fig. P11.21b P11.21b
Solution Slope at A due to 7 kips/ft uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 2 θ A = (2 L2 − a 2 ) (slope magnitude) 24 LEI L EI
Values: w = 7 kips/ft, L kips/ft, L = = 23 ft, a = 15 ft, 6 2 EI = 8.0 = 8.0 × 10 kip-in. Computation: wa 2 (2 L2 − a 2 ) θ A = 24 LEI L EI
=
(7 kips/ft)(15 ft) 2 24(23 ft) EI
= −
⎡⎣ 2(23 ft)2 − (15 ft) 2 ⎤⎦
2,376.766304 kip-ft 2
(negative slope by inspection)
EI
Slope at A due to concentrated moment M . [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: L θ A = (slope magnitude) 3 EI Values: 6 2 L = L = 23 ft, EI ft, EI = 8.0 = 8.0 × 10 kip-in.
Computation: θ A
=
ML 3 EI
=
M (23 ft)
=
M (7.666667 ft)
3EI
EI
(positive slope by inspection)
Compatibility equation at A: 2,37 ,376.766304 kip-ft2 M (7.666667 ft) − + =0 EI EI
∴ M =
2,376.766304 kip-ft 2 7.666667 7.666667 ft
= 310 kip-ft
Ans.
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11.22 For the beam and loading shown below, derive an expression for the reactions at supports A A and B. B. Assume that EI is constant for the beam.
Fig. P11.22
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to concentrated moment M 0. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: L2 M 0 L2 v B = − =− 2 EI 2 EI
Consider upward deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 R B L3 v B = = 3 EI 3EI
Compatibility equation for deflection at B:
M 0 L2 R B L3 − + =0 2 EI 3EI
∴ R B =
3M 0
↑
Ans.
2L
Equilibrium equations for entire beam:
Σ F y = RA + RB = 0
∴ RA = − RB = −
3 M 0
=
3M 0
2 L
↓
Ans.
(cw)
Ans.
2L
Σ M A = − M A − M 0 + RB L = 0 ∴ M A = RB L − M 0 =
3 M 0 2 L
( L) − M 0 =
3M 0 2
− M 0 =
M 0 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
11.23 For the beam and loading shown below, derive an expression for the reactions at supports A A and B. B. Assume that EI is constant for the beam.
Fig. P11.23
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 L4 v B = − 30 EI
Consider upward deflection of cantilever beam at B due to concentrated load R B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 R B L3 v B = = 3 EI 3EI
Compatibility equation for deflection at B:
−
w0 L4 30 EI
+
R B L3 3EI
=0
∴ R B =
w0 L
↑
Ans.
10
Equilibrium equations for entire beam:
Σ F y = RA + RB − Σ M A = − M A − ∴ M A = RB L −
w0 L 2
=0
∴ RA =
w0 L 2
−
w0 L 10
=
4 w0 L 10
=
2 w0 L 5
↑
Ans.
w0 L ⎛ L ⎞
⎜ ⎟ + RB L = 0
2 ⎝3⎠ w0 L2 6
=
w0 L 10
( L) −
w0 L2 6
=−
w0 L2 15
=
w0 L2 15
(ccw)
Ans.
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11.24 For the beam and loading shown below, derive an expression for the reactions at supports A A and B. B. Assume that EI is constant for the beam.
Fig. P11.24
Solution Choose the reaction force at A at A as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to concentrated load P . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 v A = − (3L − x) (elastic curve) 6 EI 3 L Let x = L , L = 2
P ( L) 2 ∴ v A = − 6 EI
7 PL3 ⎡ ⎛ 3L ⎞ ⎤ ⎢3 ⎜ 2 ⎟ − L ⎥ = − 12EI ⎣ ⎝ ⎠ ⎦
Consider upward deflection of cantilever beam at A due to concentrated load R A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 R A L3 v A = = 3 EI 3EI
Compatibility equation for deflection at A:
−
7 PL3 12 EI
+
R A L3 3EI
=0
∴ R A =
7 P ↑ 4
Ans.
Equilibrium equations for entire beam: Σ F y = RA + RB − P = 0
∴ R B = P −
7 P
=−
4
3P 4
=
3P 4
↓
Ans.
⎛ 3 L ⎞ ⎟=0 2 ⎝ ⎠ 3PL PL PL ⎛ 3 L ⎞ 7 P ∴ M B = RA L − P ⎜ = ( L) − = = (ccw) ⎟ 2 4 2 4 4 ⎝ ⎠ Σ M B = M B − RA L + P ⎜
Ans.
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11.25 For the beam and loading shown below, derive an expression for the reactions at supports A A and B. B. Assume that EI is constant for the beam.
Fig. P11.25
Solution Choose the reaction force at A at A as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 v A = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI
Let x = L , L =
3 L 2
2 ⎤ w( L) 2 ⎡ ⎛ 3 L ⎞ 17 wL4 ⎛ 3L ⎞ 2 ∴ v A = − ⎢ 6 ⎜ ⎟ − 4 ⎜ ⎟ ( L ) + ( L) ⎥ = − 24 EI ⎣⎢ ⎝ 2 ⎠ 48EI ⎝ 2 ⎠ ⎦⎥
Consider upward deflection of cantilever beam at A due to concentrated load R A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 R A L3 v A = = 3 EI 3EI
Compatibility equation for deflection at A:
−
17 wL4 48 EI
+
R A L3 3EI
=0
∴ R A =
17 wL 16
↑
Ans.
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Equilibrium equations for entire beam:
Σ F y = RA + RB −
3wL 2
=0
∴ RB =
3wL
−
2
17 wL
=
16
7 wL wL
=
16
7 wL 16
↑
Ans.
⎛ 3wL ⎞ ⎛ 3L ⎞ ⎟⎜ ⎟ = 0 ⎝ 2 ⎠⎝ 4 ⎠
Σ M B = M B − RA L + ⎜ ∴ M B = RA L −
9 wL2 8
=
17 wL 16
( L) −
9 wL2 8
=−
wL2 16
=
wL2 16
(cw)
Ans.
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