This

**preview**has**blurred**sections. Sign up to view the full version! View Full DocumentShear Forces and Bending Moments
Problem 4.3-1
Calculate the shear force
V
and bending moment
M
at a cross section just to the left of the 1600-lb load acting on the simple
beam
AB
shown in the figure.
Solution 4.3-1
Simple beam
4
Shear Forces and
Bending Moments
259
AB
1600 lb
800 lb
120 in.
30 in.
60 in.
30 in.
S
M
A
5
0:
R
B
5
1400 lb
S
M
B
5
0:
R
A
5
1000 lb
Free-body diagram of segment
DB
5
42,000
lb-in.
©
M
D
5
0:
Ê
M
5
(1400
lb)(30
in.)
5
200
lb
©
F
VERT
5
0:
Ê
V
5
1600
lb
2
1400
lb
1600 lb
800 lb
30 in.
60 in.
30 in.
D
R
A
R
B
B
1600 lb
30 in.
D
R
B
V
M
Problem 4.3-2
Determine the shear force
V
and bending moment
M
at the midpoint
C
of the simple beam
AB
shown in the figure.
Solution 4.3-2
Simple beam
A
C
B
2.0 kN/m
6.0 kN
1.0 m
1.0 m
4.0 m
2.0 m
A
C
B
2.0 kN/m
6.0 kN
1.0 m
1.0 m
2.0 m
R
A
R
B
S
M
A
5
0:
R
B
5
4.5 kN
S
M
B
5
0:
R
A
5
5.5 kN
Free-body diagram of segment
AC
©
M
C
5
0:
Ê
M
5
5.0
kN
?
m
©
F
VERT
5
0:
Ê
V
52
1.5
kN
A
C
6.0 kN
1.0 m
1.0 m
R
A
V
M

Problem 4.3-3
Determine the shear force
V
and bending moment
M
at
the midpoint of the beam with overhangs (see figure). Note that one load
acts downward and the other upward.
Solution 4.3-3
Beam with overhangs
260
CHAPTER 4 Shear Forces and Bending Moments
P
P
b
b
L
5
P
¢
1
1
2
b
L
≤
Ê
(upward)
R
A
5
1
L
[
P
(
L
1
b
1
b
)]
©
M
B
5
0
Free-body diagram
(
C
is the midpoint
)
M
5
PL
2
1
Pb
2
Pb
2
PL
2
5
0
M
5
P
¢
1
1
2
b
L
≤
¢
L
2
≤
2
P
¢
b
1
L
2
≤
©
M
C
5
0:
5
2
bP
L
V
5
R
A
2
P
5
P
¢
1
1
2
b
L
≤
2
P
©
F
VERT
5
0:
©
M
A
5
0:
Ê
R
B
5
P
¢
1
1
2
b
L
≤
Ê
(downward)
P
P
b
b
L
AB
R
A
R
B
P
b
L/2
AC
R
A
V
M
Problem 4.3-4
Calculate the shear force
V
and bending moment
M
at a
cross section located 0.5 m from the fixed support of the cantilever beam
AB
shown in the figure.
Solution 4.3-4
Cantilever beam
A
B
1.5 kN/m
4.0 kN
1.0 m
1.0 m
2.0 m
Free-body diagram of segment
DB
Point
D
is 0.5 m from support
A
.
52
9.5
kN
?
m
2.0
kN
?
m
2
7.5
kN
?
m
2
(1.5
kN
/
m)(2.0
m)(2.5
m)
©
M
D
5
0:
Ê
M
(4.0
kN)(0.5
5
4.0
kN
1
3.0
kN
5
7.0
kN
V
5
4.0
kN
1
(1.5
kN
/
©
F
VERT
5
0:
A
B
1.5 kN/m
4.0 kN
1.0 m
1.0 m
2.0 m
D
B
1.5 kN/m
4.0 kN
1.0 m
0.5 m
2.0 m
V
M

Problem 4.3-5
Determine the shear force
V
and bending moment
M
at a cross section located 16 ft from the left-hand end
A
of the beam
with an overhang shown in the figure.
Solution 4.3-5
Beam with an overhang
SECTION 4.3 Shear Forces and Bending Moments
261
A
C
B
400 lb/ft
200 lb/ft
6 ft
6 ft
10 ft
10 ft
S
M
B
5
0:
R
A
5
2460 lb
S
M
A
5
0:
R
B
5
2740 lb
Free-body diagram of segment
AD
Point
D
is 16 ft from support
A
.
52
4640
lb-ft
2
(400
lb
/
ft)(10
ft)(11
ft)
©
M
D
5
0:
Ê
M
5
(2460
lb)(16
ft)
1540
lb
V
5
2460
lb
2
(400
lb
/
ft)
©
F
VERT
5
0:
A
C
B
400 lb/ft
200 lb/ft
6 ft
6 ft
10 ft
10 ft
R
A
R
B
A
D
400 lb/ft
6 ft
10 ft
R
A
V
M
Problem 4.3-6
The beam
ABC
shown in the figure is simply
supported at
A
and
B
and has an overhang from
B
to
C
. The
loads consist of a horizontal force
P
1
5
4.0 kN acting at the
end of a vertical arm and a vertical force
P
2
5
8.0 kN acting at
the end of the overhang.

This is the end of the preview. Sign up to
access the rest of the document.