sm10 83 86

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________________________________________________________________________
PROBLEMS (10.83 through 10.86) A beam is loaded and supported as shown in Figs. P10.83
through P10.86. Determine:
(a) The reaction at support A.
(b) The deflection at the midspan.
SOLUTION (10.83)
(a)
0
0
'' A EIv R x M x a=−<−>
201
1
'2A
EIv R x M x a C=−<−>+
32
012
1
62
AM
EIv R x x a C x C
=−<−>++ (1)
Using boundary conditions:
2
(0) 0: 0vC==
201
1
'( ) 0: 0
2A
vL RL Mb C=−+= (2)
32
01
1
() 0: 0
62
AM
vL RL b CL
=−+= (3)
From Eqs.(2) and (3):
22
0
3
3()
2
AM
RLa
L
=−↑
and
10
(2)4CMbabL=−
(b) Making x=L/2 in Eq.(1)
22
0[3( ) 12 ( 2 )]
96
MM
vLabba
EI
=−+−
0()(511)
32
MLa L a
EI
=−−
SOLUTION (10.84)
Equivalent loading:
(a)
Continued on next slide
B
w
w
3a
a
R
A
A
C
D
y
x
RB
MB
b
a
R
A
A
C
B
y
x
RB
MB
M0
L

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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
22
'' 4
22
Aww
EIv R x x a x a
=−<−>−<−>
23 3
1
1
'4
26 6
Aww
EIv R x x a x a C
= − < − >+ < − >+
34 4
12
14
624 24
Aww
EIv R x x a x a C x C
= − < − >+ < − >+ + (1)
Using boundary conditions:
2
(0) 0: 0vC==
233
1
111
'(5 ) 0 : (5 ) (4 ) 0
266
A
va R a wa wa C=− − + += (2)
344
1
111
(5 ) 0: (5 ) (4 ) 5 0
62424
A
va R a wa wa Ca=−++= (3)
Multiply Eq.(2) by 5a & subtract Eq.(3) from it:
1.005
A
R
wa=↑
Then, Eq.(b) gives 3
12.0625Cwa=−
(b) Then, substitute x=2.5a into Eq.(1):
444
2.61719 0.21094 5.15625
M
EIv wa wa wa=−−
or
4
2.75
Mwa
vEI
=↓
SOLUTION (10.85)
(a)
2
'' 2
AA
w
EIv M R x x a
=− + − < − >
23
1
1
'26
AA
w
EIv M x R x x a C
=− + − < − > +
23 4
12
11
2624
AA
w
EIv M x R x x a C x C
=− + − < − > + + (1)
Continued on next slide
a
a
R
A
A
C
B
y
RB
MB
M
A
w

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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.
Using boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCvC== ==
3
2
'(2) 0: (2) (2) 0
26
A
ARwa
va M a a
=− + − = (2)
44 2
12
1111
() 0: 0
24 384 6 2
v L wL wL C L C L=− + + + = (3)
Multiply Eq.(2) by a & subtract Eq.(3) from it:
34
411 3
(2 ) ( ) 0,
3 6 24 16
AA
R
awaRwa−−− = =
Then Eq.(2) gives
2
5
48
A
M
wa=
(b) Make x=a in Eq.(1) and substitute the foregoing values:
4
48
Cwa
vEI
=↓
SOLUTION (10.86)
(a)
2
'' 22
AA
wL
EIv M R x x
=+−<−>
23
1
1
'262
AA
wL
EIv M x R x x C
=+ −<−>+
23 4
12
11
26242
AA
wL
EIv M x R x x C x C
=+−<−>++ (1)
Boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCvC== ==
4
23
11
() 0: 0
2 6 384
AA
wL
vL ML RL
=+−= (2)
Continued on next slide
L/
2
L/
2
R
A
A
C
B
y
RB
M
A
w
x

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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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Statics:
20
1
0: 0
8
BAA
MMRLwLM=−−+ −=
∑ (3)
Multiply Eq.(3) by 22&L add to Eq.(2):
34
0
11 1 1 1
() ( ) 0
6 2 16 384 2
A
RL wL ML−+− −=
0
3
23
128 2
A
M
RwL
L
=−
Equation (3) yields
20
71
128 2
A
M
wL M=−
(b) Then, making x=L/2 in Eq.(1):
4
22
400
723
1024 8 6144 32
C
M
LML
wL wL
EIv =− + + −
or
220
( 19 576 )
6144
CL
vwLM
EI
=−+