sm10 159 160

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________________________________________________________________________
PROBLEMS (10.159 and *10.160) A two-span continuous beam ABC is loaded as shown in
Figs. P10.159 and P10.160. Find the reactions RA , RB , and RC for the beam.
SOLUTION (10.159)
Consider
B
R
as redundant.
2
111
(2 )
22 2
Pa
AaPa==
21(2 )
2BB
AMaMa==
/12
2
2()(2)
3
AB
EIt A a A a
=
−−⋅
/12
2
() ( 2)
3
BC
EIt A a A a
=
−−⋅
We have
233
2
// 4
2
:()
43 2 3
B
AB CB B
M
a
Pa Pa
tt Ma=− − − =− −
from which 1
8
B
M
Pa=
Segment BC:
(2 )
BC
M
RaPa
=
−
Continued on next slide
C
x
M
tan
g
ent at B
B
M
A
2
A
2E
I
2a a
x
EI
A
1
-MB
-tC/B= tA/B
A
P
A
a
-Pa/2
a
4a/3
C
A
2
R
A RC
RB
B
C
P
tC/B
A
1

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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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Thus,
(2 )
8C
Pa
R
aPa−= −, 7
16
C
R
P
=
↑
Segment AB:
(2 )
8A
Pa
R
aPa−=− +, 9
16
A
R
P
=
↓
Statics: 1
8
B
R
P=↑
Continued on next slide
B
A
C
tA/B
L/
2
w
L
tan
g
ent at B
tC/B
R
A RB
R
C
-RCL/2
wL2
/
8
x
A
1
A
2
L/
3
L/
2
M
2L/3
A
3

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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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SOLUTION (*10.160)
Assume
C
R
as redundant.
23
12()
38 12
wL wL
AL==
2
21()
22 4
CC
R
LRL
AL=− =−
2
31()
22 2 8
CC
R
LRL
L
A=− =−
3
4
/1 2
2
() ( )
23246
C
AB
R
L
LLwL
EIt A A=− =−
3
/3
()
324
C
CB
R
L
L
EIt A==−
We have 33
4
//
2:
12 24 6
CC
CB AB
R
LRL
wL
tt=− = − , 1
6
C
R
wL
=
↓
Statics:
3
4
B
R
wL=↑ 5
12
A
R
wL
=
↑