sm10 80

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________________________________________________________________________
PROBLEM (10.80) A propped cantilever beam AB constructed of a S10 x 35 shape (see Table B.9)
is subjected to a linearly varying load at the half-span as shown in Fig. P10.80. Calculate:
(a) The reaction RA at support A.
(b) The deflection at C.
Given: E = 29 x 106 psi I = 147 in.4
SOLUTION
Note: forces in kips and lengths in ft.
Slop of load : 2
61.5
4
kip ft
k kip ft
ft
==
Refer to superposed load diagram:
0
() 1.5 6 4 1.5 4
dV wx x x x
dx
=
=− +<−>+ <−>
22
0.75 6 4 0.75 4
A
dM VR x x x
dx = = − +<−>+ <−>
(a) 232 3
2( ) 0.25 3 4 0.25 4
A
dv
EI M x R x x x x
dx ==− +<−>+<−>
243 4
1
10.0625 4 0.0625 4
2A
dv
EI R x x x x C
dx = − +<−>+ <−>+
35 4 5
12
10.0125 0.25 4 0.0125 4
6A
EIv R x x x x C x C= − + <−>+ <−>+ +
Continued on next slide
R
A
A
C
A
B
B
k=1.5
x
4
ft
6 kip/ft
6 kip/ft
C
k=1.5
y
4 ft

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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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Boundary conditions:
2
(0) 0: 0EIv C==
243 4
1
1
'(8) 0: (8) 0.625(8) (4) 0.0625(4) 0
2A
EIv R C=−+++=
1176 32 A
CR
=
−
3545
1
(8) 0: (8) 0.0125(8) 0.25(4) 0.0125(4)
6A
EIv R=−++
176(8) 32 (8) 0, 6.3
AA
R
Rkips
+
−= =↑
Then
1176 32(6.3) 25.6C=− =−
(b) Deflection at midspan (x=4):
35
1(6.3)(4) 0.0125(4) 0 0 25.6(4)
6
C
EIv = − ++− 2 48 kip ft
=
−⋅
We have
6622
(29 10 )(147) 4.263 10 . 29,604EI kip in kip ft=× = × ⋅= ⋅
It follows that
3
48 1.621 10
29,604
C
vft
−
=− =− × 0.019 .in
=
↓