Problem 1 (Reduction of a distributed load):

Urban and Civil

Nov 15, 2013 (4 years and 5 months ago)

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Problem
1 (Reduction of a distributed load
):

equivalent resultant force and
specify its location on the beam,
measured from point B.

Solution:

ft

479
0
5
4
4500
3
1350
4
4800
10650
kip

7
10
lb

10650
4500
1350
4800
F
lb

4500
9
500
F
lb

1350
9
300
2
1
F
lb

4800
12
800
2
1
F
R
3
2
1
.
)
.
(
)
(
)
(
.
)
)(
(
)
)(
(
)
)(
(

d
d
M
M
F
F
F
B
R
R
R
y
B

ft

479
0
5
4
4500
3
1350
4
4800
10650
kip

7
10
lb

10650
4500
1350
4800
F
lb

4500
9
500
F
lb

1350
9
300
2
1
F
lb

4800
12
800
2
1
F
R
3
2
1
.
)
.
(
)
(
)
(
.
)
)(
(
)
)(
(
)
)(
(

d
d
M
M
F
F
F
B
R
R
R
y
B

Problem
2 (Reduction of a distributed load
):

R
equivalent resultant force and
couple moment at point A.

Solution:

lb

425
200
60
300
60
150
F
lb

71
389
60
300
60
150
lb

200
50
4
F
lb

300
50
6
F
lb

150
50
6
2
1
F
y
R
3
2
1

cos
cos
.
sin
sin
)
)(
(
)
)(
(
)
)(
(
y
x
x
R
y
R
x
R
F
F
F
F
F

)
kip.ft

2.2

lb.ft

2200
2
60
6
200
3
300
2
150
5
47
71
389
425
θ
lb

577
425
71
389
lb

425
;
lb

71
389
lb

200
F
;
lb

300
F
;
lb

150
F
1
2
2
3
2
1

A
A
A
y
x
R
R
A
R
R
R
R
M
M
M
M
F
F
F
)
cos
(
)
(
)
(
.
.
tan
)
(
)
.
(
.

Problem
3 (Reduction of a distributed load
):

Wet concrete exerts a pressure distribution
along the wall of the form. Determine the
resultan
t force of this distribution and specify
the height h where the bracing

strut should be
placed so that it lies through the line of action
of the resultant force. The wall has a width of
5 m.

2 ft
F
1
F
2
F
3
A
60
º
1 ft
2 ft
3 ft
2 ft
F
1
F
2
F
3
A
60
º
1 ft
2 ft
3 ft
Solution:

Free Body Diagram:

h
z
F
R
A
4 m
h
z
z
F
R
A
4 m

N/m
10
20z
10
4z
5
5
N/m
10
4z
kPa

4z
m

5

width
Wall
3
1/2
3
1/2
2
3
1/2
1/2
)
(
)
(
)
(
)
(
)
(

p
w
p

h
z
dF=dA
A
4 m

3
2
1
3
2
1
10
20
5
10
4
/
/
)
(
z
w
z
w

dA=
w
dz

kPa

4
2
1
/
z
p

dz
h
z
z
dF=dA
A
4 m

3
2
1
3
2
1
10
20
5
10
4
/
/
)
(
z
w
z
w

dA=
w
dz

kPa

4
2
1
/
z
p

dz

kN
107
10
67
106
10
4
3
40
10
3
2
20
10
20
:
Force
Resultant

Equivalent
3
3
2
3
4
0
3
2
3
4
0
3
2
1
0
N
F
F
z
F
dz
z
F
wdz
dA
-F
F
F
R
R
m
R
m
R
z
A
R
x
R
)
(
.
)
(
)
(
)
(

R
m
m
m
z
z
A
A
F
dz
z
z
z
dz
z
dz
z
z
z
wdz
zwdz
dA
zdA
z

4
0
3
2
1
4
0
3
2
1
4
0
3
2
1
0
0
10
20
10
20
10
20
:
Force
Resultant

Equivalent

of
Location
)
(
)
(
)
(

m

40
2
10
67
106
10
256
10
67
106
10
4
5
40
10
67
106
10
5
2
20
10
67
106
10
20
:
Force
Resultant

Equivalent

of
Location
3
3
3
3
2
5
3
4
0
3
2
5
3
4
0
3
2
3
.
)
(
.
.
)
(
)
(
.
)
(
)
(
.
)
(
)
(
.
)
(

z
N
m
N
N
z
N
z
z
N
dz
z
z
m
m

m

6
1
4
2
4
4
m

40
2
:
Force
Resultant

Equivalent

of
Location
.
.
.

h
z
h
z

Problem
4

(
):

The lifting force along the wing of
a jet aircraft consists of a uniform
distribution along AB, and a semi
parabolic distribution along BC
with origin at B.

resultant force and specify it
s
location measured from point A.

Solution:

FBD:

6
'
A
2880(12)=34560 lb
6
'
x
dx
dF=dA
w
=(2880
-
5x
2
) lb/ft
24
'
6
'
A
2880(12)=34560 lb
6
'
x
dx
dF=dA
w
=(2880
-
5x
2
) lb/ft
24
'

kip

80.6

lb

80640
24
3
5
24
2880
34560
3
5
2880
34560
5
2880
34560
34560
:
Force
Resultant

Equivalent
3
24
0
3
24
0
2
0
R
R
ft
R
ft
R
x
R
y
R
F
F
x
x
F
dx
x
F
wdx
F
F
F
)
(
)
(

ft
ft
ft
x
A
R
x
x
x
x
x
dx
x
x
x
x
dx
x
x
x
wdx
x
x
M
M
A
24
0
3
4
2
24
0
2
3
24
0
2
0
3
60
34560
4
5
2
2880
207360
80640
60
34560
5
2880
207360
80640
5
2880
12
207360
80640
12
6
34560
80640
:
Force
Resultant

Equivalent

of
Location

)
(
)
)(
(
)
(
)
(

ft
ft
ft
x
A
R
x
x
x
x
x
dx
x
x
x
x
dx
x
x
x
wdx
x
x
M
M
A
24
0
3
4
2
24
0
2
3
24
0
2
0
3
60
34560
4
5
2
2880
207360
80640
60
34560
5
2880
207360
80640
5
2880
12
207360
80640
12
6
34560
80640
:
Force
Resultant

Equivalent

of
Location

)
(
)
)(
(
)
(
)
(

ft

6
14
967680
207360
80640
24
20
24
34560
4
24
5
24
1440
207360
80640
3
60
34560
4
5
2
2880
207360
80640
:
Force
Resultant

Equivalent

of
Location
3
4
2
24
0
3
4
2
.
)
(
)
(
)
(
)
(

x
x
x
x
x
x
x
x
ft