sm10 72

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________________________________________________________________________
PROBLEM (*10.72) Figure P10.72 shows a nonprismatic propped cantilever beam AB with
flexural rigidity EI from A to C and 2EI from C to B. The beam supports a uniform load of intensity w.
Using a direct-integration approach, find :
(a) The support reactions (RA, MA, and RB).
(b) The mid span deflection if w = 60 lb/ft, L = 16 ft, and EI = 200 x 106 lb
⋅
in.2
*SOLUTION
(a)
2
11
2
AA
M
MRxwx=− + −
2[( )]
22
AA
LL
MMR x=− + + −
2
1
[( )] ( )
24 2 2 2
wL L L L
xwx−+−−−
Segment AC
2
11
'' 2
AA
EIv M R x wx=− + −
23
11
11
'26
AA
EIv M x R x wx C=− + − +
234
112
111
2624
AA
EIv M x R x wx C x C=− + − + + (1)
Segment CB
22
2
21
'' ( )()()
224162 2242
AA A
MMRL
wL wL L w L
EIv R x x==−+ −+ − −− −
223
2 3
1
'()()()
24164 22122
AA A
Mx RLx wL x wL L w L
EIv R x x C=− + − + − − − − +
2222 34
234
1( )() ()
4 8 32 12 2 2 48 2
AA A
Mx RLx wL x wL L w L
EIv R x x C x C=− + − + − − − − + +
Continued on next slide
L/
2
R
A
A
C
B
L/
2
M
A
R
B
w
R
A
A
L/
2
M
A
M
w
x
C V

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
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Copyright Act without the permission of the copyright owner is unlawful.
Boundary & continuity conditions:
111 2
'(0) 0: 0 (0) 0: 0vCv C== = =
2233
12 3
'( ) '( ) :
2 2 2848 4832
AA AA
ML RL ML RL
LL wL wL
vv C= −+−=−+−+
3
3496
A
ML wL
C=− +
23 2344
12
() ():
2 2 8 48 384 16 32 128
AA AA
ML RL ML RL
LL wL wL
vv= − +−=− +−
24
4
8 192
A
ML wL C
−
++
23
416 96
AA
M
LRL
C=− −
23 4
2717
() 0: 0
16 8 256
AA
vL ML RL wL=− + − =, 77
232
A
AM
R
wL
L
=+
Statics:
2
9
0: 80
BA
M
MwL==
∑
and
49
80
A
R
wL=↑
31
0: 80
yB
FRwL==↑
∑
(b) Equation (1) becomes
22 3 4
0
1(27 49 20 )
480
w
vLxxx
EI
=−+−
Letting x=L/2:
4
256
CwL
vEI =↓
Substitute the given data:
4
3
60(16)
256(200 10 )
C
v=×0.077 .in
=