sm10 25 26

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________________________________________________________________________
PROBLEM (10.25 and 10.26) For the beam and loading shown in Figs. P10.25 and 10.26, using
the double-integration method, determine:
(a) The equation of the deflection curve.
(b) The slope at the end A.
(c) The deflection at the midspan.
SOLUTION (10.25)
(a)
20
00sin
L
AB wL
x
RRw dx
L
π
π
== =
∫
We have
2
0
'' ( ) sin
Lx
EIv w L
π
π
=
3
01
'()cos
Lx
EIv w C
L
π
π
=− + (1)
4
012
()sin
Lx
EIv w C x C
L
π
π
=
−++ (2)
Boundary conditions:
2
(0) 0: 0vC==, 1 '( ) 0: 0
2
L
vC
=
=
Equations (1) and (2) become
3
0
3
'cos
wL
x
vEI L
π
π
= (3)
4
0
4sin
wL
x
vEI L
π
π
=↓ (4)
Continued on next slide
L/
2
L/
2
A
C
B
R
A
y
x
R
B
x
M
x
0cos
wL
x
VL
π
π
=
0sin
x
ww L
π
=
x
V
2
02sin
wL
x
ML
π
π
=

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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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Copyright Act without the permission of the copyright owner is unlawful.
(b) Make x=0 in Eq. (3):
3
0
3
AwL
EI
θπ =
(c) Make x=L/2 in Eq. (4):
4
0
4
CwL
vEI
π
=↓
SOLUTION (10.26)
(a)
Due to symmetry only part AC of the beam need be considered.
3
00
11
'' 34 EIv w x w Lx
L
=− +
42
001
11
'12 8 EIv w x w Lx C
L
=− + + (1)
53
0012
11
60 24
EIv w x w Lx C x C
L
=− + + + (2)
Boundary conditions:
3
210
5
(0) 0: 0, '( ) 0:
2 192
L
vCv CwL== ==−
Thus, Eqs.(1) and (2) become
342
0
' [ 80( ) 120( ) 25]
960
wL xx
vEI L L =−+ − (3)
453
0[ 16( ) 40( ) 25( )]
960
wL
x
xx
vEI L L L =−+− (4)
(b) Make x=0 in Eq. (3):
3
0
5
192
AwL
EI
θ =
Continued on next slide
O
3
00
11
34
w
M
xwLx
L
=− +
A
x
0
2
x
wL
y
20
x
wL
x/
3
R
A=w0L/4 V

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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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(c) Make x=L/2 in Eq. (4):
4
0
120
CwL
vEI =↓