sm10 37

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PROBLEM (10.37) A cantilever of variable cross section supports a uniformly distributed load w
(Fig. P10.37). Verify, using the last of Eqs. (10.4), that the expression for the deflection curve is
32 3
3
0
(32)
wL
vxLxL
Eb h
=−+− (P10.37)
Here h represents the depth of the beam and b0 is the width at the fixed end.
SOLUTION
3
0
1()
12 x
I
bh
L
=
The last of Eq. (10.4):
3
222 2
0
222 2
()
()[ ]
12
bxLh
ddvd dv
EI E W
dx dx dx dx
==−
Integrating twice
32
01
2
()
[]
12
bxLh
ddv
EwxCV
dx dx =−+=−
322
012
2
() 1
12 2
bxLh dv
EwxCxCM
dx =− + + = (1)
Boundary conditions:
21
(0) 0: 0 (0) 0: 0MCVC== ==
Equation (1) becomes then
223
23 3
00
63dv wL dv wL
x
xC
dx Eb h dx Eb h
=− =− +
334
3
0
wL
vxCxC
Eb h
=− + + (2)
We have
34
34
33
00
32
'( ) 0: , ( ) 0 :
wL wL
vL C vL C
Eb h Eb h
== ==−
Substitution of these into Eq (2) results in Eq. (P10.37).