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The distributed load 75lb/ft can be considered as a resultant force 75lb/ft×12ft=900lb at 6ft
away from the point A.
From the entire structure, normal forces at the point of A and B can be calculated.
0;
(14)
2500(20)
900(8)
3000(2)
0
By
MA
=−
+
+
+
=
∑
,
4514
y
A
lb
=
0
xx
FB
==
∑
4514
2500
900
3000
0
yy
−
−
+
∑
=
1886
y
B
lb
=
Section at the point C, and consider the F.B.D. of the left part.
2500(6)
0
CC
MM
=+
∑
=
15000
15.0
C
M
lb ft
kip ft
⋅
⋅
0
xC
FN
∑
k
i
p
=
2500
4514
0
yC
FV
+
−
=
∑
,
Vl
2014
2.01
C
b
Section at the point D, and consider the F.B.D. of the right part.
1886(2)
0
DD
∑
3771
3.77
D
M
lb ft
kip ft
=⋅
=
⋅
0
xD
∑
k
i
p
3000 1886
0
yD
+
+
=
∑
,
1114
1.11
D
b
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View Full Document The distributed load can be considered as one resultant force,
1
(2
)
2
abw
+
, acting at the
centroid of the triangle.
From the entire structure, find reaction forces at the point C
11
0;
(2
)
(
)
( )
0
23
By
Ma
b
w
b
a
A
⎡⎤
=+
−
−
⎢⎥
⎣⎦
∑
b
=
)(
)
6
y
w
A
abba
b
−
This problem requires
0
C
V
=
From the F.B.D. of the left part of the sectioned structure at C
1
(2
)(
)
0
62
2
y
wb
Fa
b
b
a
a
b
⎛⎞
⎛
⎞
−
−
+
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
∑
2
w
=
)(
)
)
68
ww
ab
b
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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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