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These three relationships, plus three conditions of equilibrium, can be solved for the six
reactions
M
a
,R
Va
,R
Ha
,M
d
,R
Vd
,
and
R
Hd
of Fig. 3.5b.
Begin by computing the moment in each member using the free-body diagrams of
Fig. 3.5c, d, and e. From a summation of moments about the cut ends of these figures,
x
R
Hd
+
=
|
3
M
d
M
x
R
Vd
3
R
Hd
+
+
M
d
=
M
|
2
(2)
y
)
x
)
(
3
−
R
Hd
+
(
3
+
R
Vd
+
M
d
x
)
2
p
0
(
x
−
y
−
−
P
z
P
x
=
M
|
1
2
y
=
x
sin
x
=
x
cos
60
◦
,
0
x
≤
x
≤
x
≤
where
α
,
α
,
α
=
≤
3
,
0
≤
3
,
0
≤
3
.
464
,
and
.
To apply the first relationship of (1), form the strain energy for each of the three members,
sum the results to find
U,
and set
M
|
i
=
M
beam
i
,i
=
1
,
2
,
3
∂
U
/∂
R
Vd
=
0
.
Better still, utilize
beam 1
beam 2
beam 3
∂
∂
∂
1
EI
M
1
EI
M
1
EI
M
M
dx
+
M
dx
+
M
dx
=
0
(3)
∂
∂
∂
R
Vd
R
Vd
R
Vd
with
3
=
2
=
1
=
∂
M
∂
M
∂
M
x
,
x
0
,
3
+
(4)
∂
R
Vd
∂
R
Vd
∂
R
Vd
This leads to
32
.
088
R
Hd
+
61
.
638
R
Vd
+
17
.
89
M
d
−
524
.
694
=
0
(5)
For the second equation of (1), follow the above procedure using
3
=
2
=
1
=
∂
M
∂
M
∂
M
x
,
y
3
,
3
−
(6)
∂
R
Hd
∂
R
Hd
∂
R
Hd
to find
46
.
392
R
Hd
+
32
.
0884
R
Vd
+
18
.
696
M
d
−
122
.
35
=
0
(7)
For the third equation of (1), use
1
=
∂
2
=
∂
3
=
∂
M
M
M
1
(8)
∂
M
d
∂
M
d
∂
M
d
This gives
18
.
696
R
Hd
+
17
.
892
R
Vd
+
9
.
464
M
d
−
125
.
810
=
0
(9)
Solve (5), (7), and (9) to find
=−
.
=
.
=
.
R
Hd
10
205 kN
,
9
118 kN
,
16
217 kN
(10)
Vd
d
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