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In case of using method of joints the supporting
reactions are not required.
Joint B:
3
0;
cos45
100
0
5
xB
C
B
A
FF
F
⎛⎞
=°
−
−
⎜⎟
⎝⎠
∑
=
①
4
0;
sin45
500
0
5
yB
C
B
A
F
+
−
∑
=
=
=
b
②
Solving Eqs.
① and ②
yields
286
BA
Fl
b
=
(T)
384
BC
b
=
(T)
Joint C:
0;
384cos45
0
xC
A
=−
°
∑
271
CA
b
=
(C)
B
C
0;
384sin 45
0
yy
FC
°
∑
271.43
y
Cl
=
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View Full Document First, find reaction forces at pin A from the entire structure.
0;
0
xx
FA
==
∑
0;
(12)
4(8)
8(6)
0
Gy
MA
=−
+
+
=
∑
6.667
y
A
kN
=
Then, consider equilibrium equations from the sectioned part of the structure.
(Cutting line goes through the members which we are interested in.)
0;
6.667(4)
(2)
0
CK
MF
+
=
∑
J
13.3
KJ
Fk
N
=
(T)
To find
BC
F
,
move
BC
F
to point C and resolve
to x and y components.
2
()
5
BC x
BC
F
F
⎛⎞
=
⎜⎟
⎝⎠
,
1
5
BC y
BC
F
F
=
0;
6.667(4)
(
) (2)
0
KB
+
=
∑
C
x
∑
14.9
BC
N
=
(C)
0
AC
K
First, find the reaction forces at point E. (
)
0
x
E
=
0;
1000(10) 1000(20) 1000(30)
(40)
0
Ay
ME
=−
−
−
+
=
∑
lb
1500
y
E
=
Next, find the force in GJ by taking moment about point C.
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.
 Fall '08
 Thouless

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