TAM 210 / 211, Fall, 2004 Homework solutions for 09/20
3.143 Given:
The distributed load shown.
Required:
Find the equipollent force/couple system at the wall.
Solution:
By inspection, we can easily see that the area under the
curve is zero, so the resultant force should be zero.
However, we could test this to get a bit of
practice:
()
(
)
()
()
(
)
[]
0
0
cos
2
cos
2
/
ˆ
/
2
cos
2
/
ˆ
/
2
sin
ˆ
ˆ
ˆ
0
0
0
=
−
=
=
=
=
=
=
=
=
∫
∫
π
L
q
L
x
L
q
dx
L
x
q
qdx
F
o
L
x
o
L
x
o
L
x
y
j
j
j
j
j
F
()
()
∫
∫
=
=
=
×
=
L
x
o
L
x
o
O
r
dx
L
x
x
q
dx
L
x
q
x
0
0
,
/
2
sin
ˆ
ˆ
/
2
sin
ˆ
k
j
i
M
To solve this, it helps to integrate by parts:
Let
u = x
, so
du = dx
;
()
dx
L
x
dv
/
2
sin
=
, so
()
(
)
L
x
L
v
/
2
cos
2
/
=
.
Then
()
(
)
()
[]
()(
)
()(
)
[]
()
(
)
[]
{}
()
2
/
ˆ
/
2
sin
2
/
0
cos
0
2
cos
2
/
ˆ
/
2
cos
2
/
/
2
cos
2
/
ˆ
/
2
sin
ˆ
2
0
2
0
0
0
,
o
L
x
o
L
x
L
x
o
L
x
o
O
r
q
L
L
x
L
L
L
q
dx
L
x
L
L
x
x
L
q
dx
L
x
x
q
k
k
k
k
M
=
−
−
=
−
=
=
=
=
=
=
∫
∫
3.147 Given:
The concrete ramp shown, width = 1m. Density of concrete = 22,600 N/m
3
Find:
The weight of the ramp and its line of action
Solution:
Weight = density x volume
( )
()
()
(
)
(
)
()
N
m
m
m
m
N
W
900
,
33
1
5
.
0
6
2
/
1
/
600
,
22
3
=
=
The weight is distributed in proportion to the local volume distribution. Assume y is “up” and
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 Fall '05
 Downing
 Force, 1 ft, 0 l, 0.5m, Single Force

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