sm10 102 103

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________________________________________________________________________
PROBLEMS (10.102 and 10.103) The compound beam AHB shown in Figs. P10.102 and
10.103 has fixed supports at ends A and B and consists of two members joined by pin connection at H.
Determine all reactions of the beam due to the concentrated load P.
SOLUTION (10.102)
Use Table B.14 (case 1):
(1)
22
(3)
218
CPL PL
EI EI
θ ==
332
(2 3) (3) ()
33183
H
HRL PL PL L
vEI EI EI =−−
33
85
81 162
H
RL PL
EI EI
=− (2)
From Eqs.(1) and (2):
5
18
H
R
P=
Then
5
18
A
R
P
=
↑
5
54
A
M
PL=
13
18
B
R
P
=
↑
8
54
B
M
PL=
Continued on next slide
C
L/
3
A
H
2L/3
H
C
θ
B
H
B
+
R
H
R
H
33
(3)
381
HH
H
R
LRL
vEI EI =− =−
RB
5
18 P
P
RA
C
M
B
MA
L/
3
A
H
B
5
18 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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SOLUTION (10.103)
Use Table B.14 (case 1):
3
()
() 3H
HL PRL
vEI
−
=
3
() 3H
HR
R
a
vEI =
Since () ():
H
LHR
vv=
33 33 3
() ,
33
HH HH
PRL Ra PL R L R a
EI EI
−=−=
or
3
33
HPL
RLa =+
Statics:
Loading I:
3
33
0:
yA
Pa
FR
La
==↑
+
∑
3
33
0: ( )
AAH
Pa L
MMPP
La
==−=
+
∑
Loading II:
3
33
0:
yB
Pa
FR
La
==↑
+
∑
3
33
0:
BB
PL a
MM
La
==
+
∑
RB
Loading I
RA
H
M
B
MA
L
A
H
B
a Loading II
P
R
H
RH