sm10 81 82

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________________________________________________________________________
PROBLEMS (10.81 and *10.82) A beam ABC loaded and supported as shown in Figs. P10.81
and P10.82.Determine :
(a) The reactions at supports A and B.
(b) The deflection at the middle of span AB.
SOLUTION (10.81)
(a)
2
2
'' 2
AA B PL
EIv M R x R x L x L
=− + + < − >− < − >
223
1
1
'()
22 62
AB
ARx R P
EIv M x x L x L C
L
=− + + <−>− <−>+
23 34
12
1()
266 242
AAB
Mx Rx R P
EIv x L x L C x C
L
=− + + <−>− <−>+ + (1)
Using boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCvC== ==
23
11
() 0: 0, 3
26 A
AA A
M
vL ML RL R L =− + = = (2)
Statics:
3
8
A
M
PL= 913
88
AB
R
PRP
=
↑=↓
(b) Then make x=L/2 in Eq.(1):
3
23
3193
()
16 4 6 8 8 128
MPL L P L
EIv PL
=− + =−
or
3
3
128
MPL
vEI
=↓
SOLUTION (*10.82)
By inspection, the beam is statically indeterminate to the first degree. Due to
symmetry AC
R
R= and 0
B
θ
=
.
(a) 2
() 2
AB
wx
M
xRx RxL=−+<−>
Continued on next slide
L
L
R
A
A
C
B
y
RB
M
A
P/
2L P

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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.
So
22
22
AB
dv wx
EI R x R x L
dx =−+<−>
23 2
1
26 2
AB
dv x wx x L
EI R R C
dx
<−>
=−+ +
34 3
12
624 6
AB
xwx xL
EIv R R C x C
<−>
=−+ ++ (1)
Boundary conditions:
2
(0) 0: 0EIv C==
23
1
'( ) 0: 0
26
ALwL
EIv L R C
=−+=
34
1
() 0: 0
624
ALwL
EIv L R C L
=−+=
Solving
3
1
3
848
ACwL
RwLRC
== =−
From equilibrium
35
244
B
R
wL wL wL=− =
(b) Equation (1) becomes
3343
6 6 24 48
AB
Rx R x L wx wL
vEI EI EI EI
<−>
=− −−
At the middle of span AB:
344
5
()
2 48 384 192
A
RL
LwLwL
vEI EI EI
=− = ↓