sm10 53 54

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PROBLEMS (*10.53 and *10.54) An overhanging beam is supported and loaded as shown in
Fig. P10.53 and P10.54. Determine the equation of the elastic curve and deflection at point A.
SOLUTION (*10.53)
y
wa2
w
C
a
B
A
x
Equivalent loading
w
RA=wa/2
RB
Refer to the above figure:
w
w
wa
a
EIv '' = − ( x + a) 2 + < x > 2 +
< x > + a2 w < x − >0
2
2
2
2
w
w
wa
a
EIv ' = − ( x + a)3 + < x >3 +
< x > 2 + a 2 w < x − > +C1
6
6
4
2
2
w
w
wa
a w
a
EIv = − ( x + a ) 4 +
< x >4 +
< x >3 +
< x − > 2 +C1 x + C2
24
24
12
2
2
Using boundary conditions:
wa 4
wa 4
v(0) = 0 : −
+ C2 = 0,
C2 =
24
24
4
4
2
wa
wa
wa 4
wa 4
3
v(a) = 0 : − wa 4 +
+
+
+ C1a +
= 0,
C1 = wa 3
3
24
12
8
24
8
Equation (1) becomes
w
a
v=
[−( x + a) 4 + < x > 4 +2a < x >3 +12a 2 < x − > 2 +9a 3 x + a 4 ]
24 EI
2
Let x=-a:
wa 4
vC =
↓
3EI
(1)
SOLUTION (*10.54)
From conditions of equilibrium:
RA = 2 P ↑
RB = 0
Using singularity functions:
M = − Px + 2 P < x − a > − P < x − 2a >
Using singularity functions:
d 2v
EI 2 = − Px + 2 P < x − a > − P < x − 2a >
dx
(1)
Continued on next slide
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dv
1
1
= − Px 2 + P < x − a > 2 − P < x − 2a > 2 +C1
dx
2
2
1
1
1
EIv = − Px 3 + P < x − a >3 − P < x − 2a >3 +C1 x + C2
6
3
6
EI
(2)
(3)
Boundary conditions:
1 3
Pa
6
EIv(3a) = 0 : 3aC1 + C2 = 2 Pa 2
11
3
C1 = Pa 2 , then C2 = − Pa 3
Eq.(5)-Eq.(4):
12
4
Equation (3) is therefore
P
x3 < x − a > < x − 2a >3 11a 2 x 3a 3
v=
[− +
−
+
−
]
6
3
6
12
4
EI
At the free end (x=0), we have
3Pa 3 3Pa 3
vC = −
=
↓
4 EI
4 EI
EIv(a) = 0 : C1a + C2 =
(4)
(5)
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.