PROBLEM 9.2
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
ΣM J = 0: (wx)
x
+M =0
2
1
M = − wx 2
2
d2y
1
= M = − wx 2
2
dx 2
1
dy
= − wx3 + C1
EI
dx
6
EI
x = L,
EI
dy
1
1
= 0 : 0 = − wL3 + C1 C1 = wL3
6
6
dx
dy
1
1
= − wx3 + wL3
dx
6
6
1
1
wx 4 + wL3 x + C2
24
6
EIy = −
[ x = L, y = 0] 0 = −
C2 =
(a)
Elastic curve.
(b)
y at x = 0.
(c)
dy
at x = 0.
dx
1
1
wL4 + wL4 + C2 = 0
24
6
1
1
3
−
wL4 = − wL4
24 6
24
y =−
yA = −
dy
dx
=
A
3wL4
wL4
=−
24EI
8EI
wL3
6EI
w
( x 4 − 4 L3 x + 3L4 )
24EI
yA =
θA =
wL4
↓
8EI
wL3
6 EI
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PROBLEM 9.3
For the loading shown, determine (a) the equation of the elastic
curve for the cantilever beam AB, (b) the deflection at the free end,
(c) the slope at the free end.
SOLUTION
ΣM J = 0: − M − P( L − x) = 0
M = − P ( L − x)
EI
d2y
= − P( L − x) = − PL + Px
dx 2
EI
dy
1
= − PLx + Px 2 + C1
dx
2
x = 0,
dy
=0 :
dx
0 = −0 + 0 + C1
C1 = 0
1
1
EIy = − PLx 2 + Px3 + C1x + C2
2
6
[ x = 0, y = 0] :
(a)
0 = −0 + 0 + 0 + C2
C2 = 0
y =−
Elastic curve.
Px 2
(3L − x)
6 EI
dy
Px
=−
(2 L − x)
dx
2 EI
(b)
y at x = L.
(c)
dy
at x = L.
dx
yB = −
dy
dx
=−
B
PL2
PL3
(3L − L) = −
6EI
3EI
PL
PL2
(2 L − L) = −
2EI
2EI
yB =
θB =
PL3
↓
3EI
PL2
2EI
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PROBLEM 9.7
For the beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the slope at A, (c) the slope at B.
SOLUTION
ΣM B = 0:
− RA L +
RA =
3
wL
2
1
L =0
4
3
wL
8
For portion AB only, (0 ≤ x < L)
3
x
wLx + ( wx) + M = 0
8
2
3
1 2
M = wLx − wx
8
2
ΣM J = 0: −
d2y 3
= wLx −
8
dx 2
dy
3
EI
wLx 2
=
dx 16
1
EIy =
wLx3
16
EI
(a)
1 2
wx
2
1
− wx3 + C1
6
1
−
wx 4 + C1x + C2
24
[ x = 0, y = 0]
0 = 0 − 0 + 0 + C2
[ x = L, y = 0]
0=
C2 = 0
1
1
wL3 −
wL4 + C1L
16
24
y =
Elastic curve.
C1 = −
1
wL3
48
w 1 3
1 4
1 3
Lx −
x −
Lx
24
48
EI 16
dy
w 3 2 1 3
1 3
=
Lx − x −
L
6
48
dx
EI 16
(b)
(c)
dy
at x = 0.
dx
dy
at x = L.
dx
dy
dx
dy
dx
=
w
1 3
wL3
0−0−
L =−
EI
48
48EI
=
w 3 3 1 3
1 3
L − L −
L =0
EI 16
6
48
A
B
θA =
wL3
48EI
θB = 0
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PROBLEM 9.65
For the cantilever beam and loading shown, determine the slope and deflection
at the free end.
SOLUTION
Counterclockwise couple PL at B.
Loading I:
Case 3 of Appendix D applied to portion BC.
θ B′ = −
y′B =
( PL)( L / 2) 1 PL2
=
2 EI
EI
( PL)( L / 2) 2
1 PL3
=
2EI
8 EI
AB remains straight.
θ ′A = θ B′ =
y′A = y′B −
Loading II:
1 PL2
2 EI
L
1 PL3 1 PL3
3 PL3
θ B′ = −
−
=−
2
8 EI
4 EI
8 EI
Case 1 of Appendix D.
θ ′′A =
PL2
,
2 EL
y′′A −
PL3
3EI
By superposition,
θ A = θ ′A + θ ′′A =
1 PL2 1 PL2
PL2
+
=
EI
2 EI
2 EI
y A = y′A + y′′A = −
3 PL3 1 PL3
17 PL3
−
=−
8 EI
3 EI
24 EI
θA =
yA =
PL2
EI
17 PL3
↓
24 EI
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PROBLEM 9.67
For the cantilever beam and loading shown, determine the slope and
deflection at the end.
SOLUTION
Loading I:
w = P/L.
Uniformly distributed downward loading with
Case 2 of Appendix D.
θC′ = −
( P / L) L3
1 PL2
=−
6 EI
6 EI
yC′ = −
( P / L) L4
8EI
Loading II:
=−
1 PL3
8 EI
Upward concentrated load at P at point B.
Case 1 of Appendix D applied to portion AB.
θ B′′ =
P( L / 2) 2
1 PL2
=
2EI
8 EI
y′′B =
P( L / 2)3
1 PL3
=
3EI
24 EI
.
Portion BC remains straight.
θC′′ = θ B′′ =
1 PL2
8 EI
yC′′ = y′′B +
L
1 PL3
1 PL3
5 PL3
θ B′′ =
+
=
2
24 EI
16 EI
48 EI
By superposition,
θC = θC′ + θC′′ = −
1 PL2 1 PL2
1 PL2
+
=−
6 EI
8 EI
24 EI
θC =
PL2
24EI
1 PL3
5 PL2
1 PL3
+
=−
8 EI
48 EI
48 EI
yC =
PL3
↓
48EI
yC = yC′ + yC′′ = −
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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PROBLEM 9.75
For the cantilever beam and loading shown, determine the
slope and deflection at end A. Use E = 29 × 106 psi.
SOLUTION
Units:
Forces in kips; lengths in ft.
Loading I:
Concentrated load at A.
Case 1 of Appendix D.
θ ′A =
PL2
(1)(5)2 12.5
=
=
2EI
2 EI
EI
y′A = −
Loading II:
PL3
(1)(5)3
41.667
=−
=−
EI
3EI
3EI
Uniformly distributed load over portion BC.
Case 2 of Appendix D applied to portion BC.
θ B′′ =
wL3
(1)(3)3
4.5
=
=
6 EI
6EI
EI
y′′B = −
wL4
(1)(3)4
10.125
=−
=−
EI
8EI
8EI
Portion AB remains straight. θ ′′A = θ B′′ =
4.5
EI
y′′A = y′′B − aθ B′′ = −
10.125
4.5
19.125
− (2)
=−
EI
EI
EI
By superposition,
12.5 4.5 17
+
=
EI
EI
EI
41.667 19.125
60.792
−
=−
y A = y′A + y′′A = −
EI
EI
EI
θ A = θ ′A + θ A′′ =
Data:
E = 29 × 106 psi = 29 × 103 ksi
I =
1
(2.0)(4.0)3 = 10.667 in 4
12
EI = (29 × 103 ) (10.667) = 309.33 × 103 kip ⋅ in 2 = 2148 kip ⋅ ft 2
17
2148
Slope at A.
θA =
Deflection at A.
yA = −
60.792
= −28.30 × 10−3 ft
2148
θ A = 7.91 × 10−3 rad
y A = 0.340 in. ↓
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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