11.26 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. P11.26
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 midspan.] Relevant equation from Appendix C: 5 PL3 v A = − 48 EI Let L = 2 L
∴ v A = −
5 P(2 L)3
=−
48 EI
5 PL3 6 EI
Consider upward deflection of cantilever beam at A due to concentrated load R A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 v A = − 3 EI Let L = 2 L, P = − R A
∴ v A = −
(− R A )(2 L)3
=
3 EI
8 RA L3 3EI
Compatibility equation for deflection at A:
−
5 PL3 6 EI
+
8 R A L3 3EI
=0
∴ R A =
5 P ↑ 16
Ans.
Equilibrium equations for entire beam: Σ F y = RA + RB − P = 0
∴ R B = P −
5 P 16
=
11P
=
11P
16
↑
Ans.
16
Σ M B = M B − RA (2 L) + P( L) = 0 ∴ M B = RA (2 L) − P( L) =
5 PL 8
− PL = −
3PL 8
=
3PL 8
(cw)
Ans.
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11.27 For the beam and loading shown below, derive an expression for the reactions at supports A A and C . Assume that EI is constant for the beam.
Fig. P11.27
Solution Choose the reaction force at C as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due due to concentrated moment M 0. [Appendix C, Cantilever beam with concentrated moment.] Relevant equations from Appendix C: L2 ML v B = − and θ B = − 2 EI EI 2 M0L 3M 0 L2 0 L ∴ vC = − − ( L) = − 2 EI EI 2 EI Consider upward deflection of cantilever beam at C due due to concentrated load RC . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3 EI Let L = 2 L, P = − RC
∴ vC = −
(− RC )(2 L)3
=
3 EI
8 RC L3 3EI
Compatibility equation for deflection at 2
−
3 M 0 L 2 EI
3
+
8RC L 3EI
=0
∴ RC =
C :
9 M 0
↑
Ans.
16L
Equilibrium equations for entire beam: Σ F y = RA + RC = 0
∴ R A = −
9 M 0 16 L
=
9 M 0 16 L
↓
Ans.
Σ M A = − M A − M 0 + RC (2 L) = 0 ∴ M A = RC (2 L) − M 0 =
9 M 0 16 L
(2 L) − M 0 =
M0 8
=
M 0 8
(cw)
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.
11.28 For the beam and loading shown below, derive an expression for the reactions at supports A A and C . Assu Assume me that that EI is constant for the beam.
Fig. P11.28
Solution Choose the reaction force at C as as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 v B = − and θ B = − 8 EI 6 EI 4 3 wL wL 7 wL4 ∴ vC = − − ( L) = − 8 EI 6 EI 24 EI Consider upward deflection of cantilever beam at C due due to concentrated load RC . [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3 EI Let L = 2 L, P = − RC
∴ vC = −
(− RC )(2 L)3
=
8 RC L3
3 EI
3EI
Compatibility equation for deflection at 4
−
7 wL
24 EI
3
+
8 RC L 3EI
=0
C :
7 wL
∴ RC =
↑
Ans.
64
Equilibrium equations for entire beam: Σ F y = RA + RC − wL = 0
∴ R A = wL −
7 wL
=
57 wL
64
↑
Ans.
64
⎛ L ⎞ Σ M A = − M A − wL ⎜ ⎟ + RC (2 L) = 0 ⎝2⎠ ∴ M A = RC (2 L) −
wL2 2
=
7 wL 64
( 2 L) −
wL2 2
=−
18 wL2 64
=−
9 wL2 32
=
9 wL2 32
(ccw)
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.
11.29 For the beam and loading shown below, derive de rive an expression for the reaction forces at A, A, C , and D. D. Assume that EI is constant for the beam. ( Reminder: Reminder: The roller symbol implies that both upward and downward displacement is restrained.)
Fig. P11.29
Solution Choose the reaction force at C as as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b 2 − x2 ) (elastic curve) 6 LEI L EI Let L = 3L, b = L, x = L
∴ vC = −
due C due
to P .
P ( L)( L)
⎡⎣(3L) 2 − ( L) 2 − ( L) 2 ⎤⎦ 6(3 L) EI
=−
PL
7 PL3
⎡7 L ⎤⎦ = − 18 EI ⎣ 18EI 2
Consider upward deflection of simply supported beam at C due due to concentrated load RC . [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI L EI
Let L = 3L, a = 2 L, b = L, P = − RC
∴ vC = − =
(− RC )(2 L)( L) 6(3 L) EI
2 RC L
⎡⎣(3L )2 − (2 L)2 − ( L)2 ⎤⎦
8 RC L3
⎡ 4 L ⎤⎦ = 18 EI ⎣ 18EI 2
Compatibility equation for deflection at
−
7 PL3 18 EI
+
8 RC L3 18EI
=0
∴ RC =
C :
7 P 8
↑
Ans.
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Equilibrium equations for entire beam: Σ M A = − PL + RC (2 L) + RD (3 L) = 0
R D (3 L) = PL − RC (2 L) = PL −
7 P 8
(2 L) = −
3PL 4
P P ∴ R D = − = ↓ 4 4
Ans.
Σ F y = RA + RC + RD − P = 0 ∴ R A = P − RC − RD = P −
7 P 8
5 P 3P 3P ⎛ P⎞ = P − = = ↑ ⎟ 8 8 8 ⎝ 4⎠
− ⎜−
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.
11.30 For the beam and loading shown below, derive an expression for the reaction force at B. B. Assume that EI is constant for the beam. ( Reminder: Reminder: The roller symbol implies that both upward and downward displacement is restrained.)
Fig. P11.30
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported. Consider upward deflection of simply supported beam at B due to M 0. [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
Let L = 2 L, x = L, M = − M 0 (−
)( L)
M L2
M
0 ⎡⎣ 2(2 L) 2 − 3(2 L)( L) + ( L) 2 ⎤⎦ = ⎡⎣3 L2 ⎤⎦ = 0 ∴ v B = − 6(2 L) EI 12 EI 4 EI 0
Consider upward deflection of simply supported beam at [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 v B = − 48 EI Let L = 2 L, P = − R B
∴ v B = −
(− R B )(2 L) 3 48 EI
=
B due
to concentrated load R B.
RB L3 6 EI
Compatibility equation for deflection at B:
M 0 L2 4 EI
+
R B L3 6 EI
=0
∴ R B = −
3M 0 2L
=
3M 0 2L
↓
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.
11.31 For the beam and loading shown below, derive de rive an expression for the reaction force at B. B. Assume that EI is constant for the beam.
Fig. P11.31
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Let L = 3L, a = 2 L
∴ v B = −
w(2 L)3
wL2
2 wL4
⎡ 4(3L ) − 7(2 L)( ⎡6 L ⎤⎦ = − )(3L ) + 3(2 L) ⎤⎦ = − 24(3 L ) EI ⎣ 9 EI ⎣ 3EI 2
2
2
Consider upward deflection of simply supported beam at B due to concentrated load R B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Let L = 3L, a = L, b = 2 L, P = − R B
∴ v B = −
(− R B )( L)(2 L) 6(3 L ) EI
RB L
4 RB L3
⎡⎣(3L) − ( L) − (2 L) ⎤⎦ = ⎡4 L ⎤⎦ = 9 EI ⎣ 9 EI 2
2
2
2
Compatibility equation for deflection at B:
−
2 wL4 3 EI
+
4 R B L3 9 EI
=0
∴ R B =
3w L 2
=
3w L 2
↑
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.
11.32 For the beam and loading shown below, derive de rive an expression for the reaction force at B. B. Assume that EI is constant for the beam. ( Reminder: Reminder: The roller symbol implies that both upward and downward displacement is restrained.)
Fig. P11.32
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported. Consider upward deflection of simply supported beam at B due to uniformly distributed load. [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
wL ⎛ L ⎞ ⎛ L ⎞ Let L = 2 L, x = L, M = − w ⎜ ⎟ ⎜ ⎟ = − 8 ⎝ 2 ⎠⎝ 4 ⎠ ⎛ wL2 ⎞ ⎜− ⎟ ( L) 8 ⎠ wL2 wL4 2 2 2 ⎝ ⎡ 2(2 L) − 3(2 L)( L) + ( L) ⎤⎦ = ⎡3 L ⎤⎦ = ∴ v B = − 6(2 L) EI ⎣ 96 EI ⎣ 32 EI 2
Consider upward deflection of simply supported beam at [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 v B = − 48 EI Let L = 2 L, P = − R B
∴ v B = −
(− R B )(2 L) 3 48 EI
=
B due
to concentrated load R B.
RB L3 6 EI
Compatibility equation for deflection at B:
wL4 32 EI
+
R B L3 6 EI
=0
∴ R B = −
3 wL 16
=
3 wL 16
↓
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.
11.33 For the beam and loading shown below, derive de rive an expression for the reaction force at B. B. Assume that EI is constant for the beam.
Fig. P11.33
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to one concentrated load P . [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 v B = − ( L − b2 − x2 ) (elastic curve) 6 LEI L EI
Let L = 4 L, b = L, x = 2 L P ( L)(2 L) PL 11PL3 2 2 2 2 ⎡(4 L) − ( L) − (2 L) ⎤⎦ = − ⎡11L ⎤⎦ = − ∴ v B = − 6(4 L) EI ⎣ 12 EI ⎣ 12 EI The second concentrated load will cause an additional deflection at B at B of of the same magnitude. Consider upward deflection of simply supported beam at [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 v B = − 48 EI Let L = 4 L, P = − R B
∴ v B = −
(− R B )(4 L)3
=
48 EI
64 RB L3
=
−
12 EI
−
11PL3 12 EI
+
16 R B L3
48EI
12 EI
=0
to concentrated load R B.
16 RB L3 12EI
Compatibility equation for deflection at
11 PL3
B due
C :
∴ R B =
11P ↑ 8
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.
11.34 For the beam and loading shown below, derive de rive an expression for the reaction force at B. B. Assume that EI is constant for the beam.
Fig. P11.34
Solution Choose the reaction force at B at B as as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam bea m with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 3 v B = − (4 L2 − 7 aL + 3a 2 ) 24 LEI L EI
Let L = 5 L, a = 3L
∴ v B = −
w(3L)3
27 wL2
99 wL4
⎡ 4(5 L ) − 7(3L )( ⎡22 L ⎤⎦ = − )(5 L ) + 3(3 L) ⎤⎦ = − 24(5 L ) EI ⎣ 120 EI ⎣ 20 EI 2
2
2
Consider downward deflection of simply supported beam at B due to concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 v B = − ( L − b2 − x2 ) (elastic curve) 6 LEI L EI
Let L = 5 L, b = L, x = 3L, P =
wL 3
⎛ wL ⎞ ⎜ 3 ⎟ ( L)(3L) wL2 wL4 2 2 2 2 ⎝ ⎠ ⎡⎣ (5 L) − ( L) − (3 L) ⎤⎦ = − ⎡15 L ⎤⎦ = − ∴ v B = − 6(5 L) EI 30 EI ⎣ 2 EI
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Consider upward deflection of simply supported beam at B due to concentrated load R B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 v B = − ( L − a 2 − b2 ) 6 LEI L EI
Let L = 5L, a = 3L, b = 2 L, P = − R B
∴ v B = −
(− R B )(3L)(2 L) 6(5 L) EI
RB L
12 RB L3
⎡⎣(5 L) − (3L) − (2 L) ⎤⎦ = ⎡12 L ⎤⎦ = 5 EI ⎣ 5 EI 2
2
2
2
Compatibility equation for deflection at B: 99wL4 wL4 12 R B L3 109wL − − + =0 ∴ R B = = 2.270833wL ↑ 20 EI 2 EI 5EI 48
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.
