Shear Forces and Bending Moments

4
Shear Forces and
Bending Moments
Shear Forces and Bending Moments
800 lb
1600 lb
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.
A
30 in. 60 in. 30 in.
120 in.
B
B
R B
Solution 4.3-1 Simple beam
Free-body diagram of segment DB
800 lb
1600 lb
1600 lb
A
D
V
B
M
D
R A ©F VERT 0:V 1600 lb 1400 lb
30 in.
30 in. 60 in. 30 in.
R B
200 lb
M A
0: R B
1400 lb
©M D 0:M (1400 lb)(30 in.)
M B
0: R A
1000 lb
42,000 lb-in.
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.
A
6.0 kN
C
2.0 kN/m
B
1.0 m 1.0 m
4.0 m
2.0 m
Solution 4.3-2
Simple beam
A
6.0 kN
C
2.0 kN/m
1.0 m 1.0 m 2.0 m
R A
B
R B
Free-body diagram of segment AC
6.0 kN
A
C
V M
1.0 m 1.0 m
R A
M A
0:
M B
0:
R B
4.5 kN
R A
5.5 kN
©F VERT 0:V 0.5 kN
©M C 0:M 5.0 kN m
259

260 CHAPTER 4 Shear Forces and Bending Moments
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.
P
P
b
L
b
Solution 4.3-3
Beam with overhangs
P
©M B 0
b
A
R A 1 [P(L b b)]
L
P
L
b
R A RB
P ¢1 2b L ≤(upward)
B
P
©M A 0:R B P ¢1 2b L ≤(downward)
Free-body diagram (C is the midpoint)
©F VERT 0:
V R A P P ¢1 2b L ≤ P
©M C 0:
M
A
C
b L/2
R A V
M P ¢1 2b L ≤ ¢L 2 ≤ P ¢b L 2 ≤
M PL
PL
Pb Pb
2 2 0
2bP
L
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.
A
4.0 kN
1.5 kN/m
B
1.0 m 1.0 m 2.0 m
Solution 4.3-4
A
Cantilever beam
4.0 kN
1.5 kN/m
1.0 m 1.0 m 2.0 m
Free-body diagram of segment DB
Point D is 0.5 m from support A.
B
©F VERT 0:
V 4.0 kN (1.5 kNm)(2.0 m)
4.0 kN 3.0 kN 7.0 kN
©M D 0:M (4.0 kN)(0.5 m)
(1.5 kNm)(2.0 m)(2.5 m)
2.0 kN m 7.5 kN m
9.5 kN m
M
V
D
4.0 kN
1.5 kN/m
B
0.5 m
1.0 m
2.0 m

SECTION 4.3 Shear Forces and Bending Moments 261
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.
A
400 lb/ft 200 lb/ft
B
C
10 ft 10 ft
6 ft
6 ft
Solution 4.3-5
A
Beam with an overhang
400 lb/ft 200 lb/ft
B
C
A
10 ft 10 ft 6 ft 6 ft
R A R B
Free-body diagram of segment AD
400 lb/ft
R A
10 ft
6 ft
D
V
M
M B
0:
M A
0:
R A
2460 lb
R B
2740 lb
Point D is 16 ft from support A.
©F VERT 0:
V 2460 lb (400 lbft)(10 ft)
1540 lb
©M D 0:M (2460 lb)(16 ft)
(400 lbft)(10 ft)(11 ft)
4640 lb-ft
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
4.0 kN acting at the
end of a vertical arm and a vertical force P 2
8.0 kN acting at
the end of the overhang.
Determine the shear force V and bending moment M at
a cross section located 3.0 m from the left-hand support.
(Note: Disregard the widths of the beam and vertical arm and
use centerline dimensions when making calculations.)
P 1 = 4.0 kN
1.0 m
A
P 2 = 8.0 kN
B
C
4.0 m 1.0 m
Solution 4.3-6 Beam with vertical arm
P 1 = 4.0 kN
P 2 = 8.0 kN
1.0 m
A
B
4.0 m 1.0 m
R A
R B
M B
0: R A
1.0 kN (downward)
M A
0: R B
9.0 kN (upward)
Free-body diagram of segment AD
Point D is 3.0 m from support A.
A
D
M
4.0 kN • m
3.0 m
R A
V
©F VERT 0:V R A 1.0 kN
©M D 0:M R A (3.0 m) 4.0 kN m
7.0 kN m

262 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.3-7 The beam ABCD shown in the figure has overhangs
at each end and carries a uniform load of intensity q.
For what ratio b/L will the bending moment at the midpoint of the
beam be zero?
A
B
q
C
D
b
L
b
Solution 4.3-7 Beam with overhangs
q
Free-body diagram of left-hand half of beam:
Point E is at the midpoint of the beam.
A
D
B C
q
b
L
b
R B R A
M = 0 (Given)
C E
b L/2 V
From symmetry and equilibrium of vertical forces:
R B
R B R C q ¢b L 2 ≤
©M E 0
R B ¢ L 2 ≤ q ¢1 2 ≤ ¢b L 2 ≤ 2
0
q ¢b L 2 ≤ ¢L 2 ≤ q ¢1 2 ≤ ¢b L 2 ≤ 2
0
Solve for b/L :
b
L 1 2
Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the
bowstring of the bow shown in the figure. Determine the bending moment
at the midpoint of the bow.
70°
1400 mm
350 mm

SECTION 4.3 Shear Forces and Bending Moments 263
Solution 4.3-8
Archer’s bow
B
Free-body diagram of segment BC
B
P
A
C
H
T
H
2
b
C
M
©M C 0
b
P 130 N
70°
H 1400 mm
1.4 m
b 350 mm
0.35 m
Free-body diagram of point A
T(cos H b)¢ ≤ T(sin b)(b) M 0
2
M T H ¢ cosb b sin b≤
2
P 2 ¢H b tan b≤
2
Substitute numerical values:
M 130 N B 1.4 m (0.35 m)(tan 70)R
2 2
M 108 N m
T
P
A
T
T tensile force in the bowstring
F HORIZ
0: 2T cos P 0
T
P
2 cos b

264 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.3-9 A curved bar ABC is subjected to loads in the form
of two equal and opposite forces P, as shown in the figure. The axis of
the bar forms a semicircle of radius r.
Determine the axial force N, shear force V, and bending moment M
acting at a cross section defined by the angle .
P
A
B
O
r
C
P
P
A
M
V
N
Solution 4.3-9
Curved bar
P
A
B
O
r
C
P
P
P cos
P sin
A
B
M
V
N
O
©F N 0 Q b N P sin u 0
N P sin u
F V 0
R a V P cos u 0
V P cos u
©M O 0 M Nr 0
M Nr Pr sin u
Problem 4.3-10 Under cruising conditions the distributed load
acting on the wing of a small airplane has the idealized variation
shown in the figure.
Calculate the shear force V and bending moment M at the
inboard end of the wing.
1600 N/m
900 N/m
2.6 m
2.6 m
1.0 m
Solution 4.3-10 Airplane wing
1600 N/m
900 N/m
M
V
A
2.6 m
2.6 m
1.0 m
B
Loading (in three parts)
700 N/m 1
900 N/m
2
A
3
B
Shear Force
F VERT
0 c T
Bending
V 1 (700 Nm)(2.6 m) (900 Nm)(5.2 m)
2
1 (900 Nm)(1.0 m) 0
2
V 6040 N 6.04 kN
(Minus means the shear force acts opposite to the
direction shown in the figure.)
Moment
©M A 0
M 1 2
(900 Nm)(5.2 m)(2.6 m)
1 2
m
(700 Nm)(2.6 m)¢2.6 ≤
3
1.0 m
(900 Nm)(1.0 m)¢5.2 m ≤ 0
3
M 788.67 N • m 12,168 N • m 2490 N • m
15,450 N • m
15.45 kN m

SECTION 4.3 Shear Forces and Bending Moments 265
Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as
a simple beam at A and D (see figure). A cable passes over a small pulley
that is attached to the arm at E. One end of the cable is attached to the
beam at point B.
What is the force P in the cable if the bending moment in the
beam just to the left of point C is equal numerically to 640 lb-ft?
(Note: Disregard the widths of the beam and vertical arm and use
centerline dimensions when making calculations.)
A
B
Cable
C
E
D
8 ft
P
6 ft 6 ft 6 ft
Solution 4.3-11
Beam with a cable
E
P
Free-body diagram of section AC
P
P
__ 4P
9
A
B
Cable
C
6 ft 6 ft 6 ft
8 ft
D
__ 4P
9
P
__ 4P
9
A
©M C 0
4P __
5
__ 3P
5
6 ft B 6 ft
C
V
M
N
UNITS:
P in lb
M in lb-ft
M 4P 5 (6 ft) 4P 9
(12 ft) 0
M 8P
15 lb-ft
Numerical value of M equals 640 lb-ft.
∴ 640 lb-ft 8P
15 lb-ft
and P 1200 lb
Problem 4.3-12 A simply supported beam AB supports a trapezoidally
distributed load (see figure). The intensity of the load varies linearly
from 50 kN/m at support A to 30 kN/m at support B.
Calculate the shear force V and bending moment M at the midpoint
of the beam.
50 kN/m
A
30 kN/m
B
3 m

266 CHAPTER 4 Shear Forces and Bending Moments
Solution 4.3-12 Beam with trapezoidal load
50 kN/m
Free-body diagram of section CB
30 kN/m
Point C is at the midpoint of the beam.
40 kN/m
A
B
30 kN/m
V
3 m
M
C
B
R A RB
1.5 m
55 kN
Reactions
©M B 0 R A (3 m) (30 kNm)(3 m)(1.5 m)
(20 kNm)(3 m)( 1 2 )(2 m) 0
R A
65 kN
©F VERT 0 c
R A R B 1 2 (50 kNm 30 kNm)(3 m) 0
R B
55 kN
F VERT
0 c
T
V (30 kNm)(1.5 m) 1 2(10 kNm)(1.5 m)
55 kN 0
V 2.5 kN
©M C 0
M (30 kN/m)(1.5 m)(0.75 m)
1 2 (10 kNm)(1.5 m)(0.5 m)
(55 kN)(1.5 m) 0
M 45.0 kN m
Problem 4.3-13 Beam ABCD represents a reinforced-concrete
foundation beam that supports a uniform load of intensity q 1
3500 lb/ft
(see figure). Assume that the soil pressure on the underside of the beam is
uniformly distributed with intensity q 2
.
(a) Find the shear force V B
and bending moment M B
at point B.
(b) Find the shear force V m
and bending moment M m
at the midpoint
of the beam.
A
q 1 = 3500 lb/ft
B
C
q 2
3.0 ft 8.0 ft 3.0 ft
D
Solution 4.3-13 Foundation beam
q 1 = 3500 lb/ft
(b) V and M at midpoint E
A B C D
3500 lb/ft
A
B
M B M E
0:
F VERT
0:
M m
(2000 lb/ft)(7 ft)(3.5 ft)
2000 lb/ft V B
3 ft
V B 6000 lb
(3500 lb/ft)(4 ft)(2 ft)
A B E M m
q 2
3.0 ft 8.0 ft 3.0 ft
2000 lb/ft
V m
F VERT
0: q 2
(14 ft) q 1
(8 ft)
3 ft
4 ft
∴ q 2 8 14 q 1 2000 lbft
F VERT
0: V m
(2000 lb/ft)(7 ft) (3500 lb/ft)(4 ft)
(a) V and M at point B
V m 0
©M B 0:M B 9000 lb-ft
M m 21,000 lb-ft

SECTION 4.3 Shear Forces and Bending Moments 267
Problem 4.3-14 The simply-supported beam ABCD is loaded by
a weight W 27 kN through the arrangement shown in the figure.
The cable passes over a small frictionless pulley at B and is attached
at E to the end of the vertical arm.
Calculate the axial force N, shear force V, and bending moment
M at section C, which is just to the left of the vertical arm.
(Note: Disregard the widths of the beam and vertical arm and use
centerline dimensions when making calculations.)
A
E
Cable
1.5 m
B C
2.0 m 2.0 m 2.0 m
D
W = 27 kN
Solution 4.3-14
Beam with cable and weight
E
Free-body diagram of pulley at B
Cable
1.5 m
27 kN
A
B C
D
21.6 kN
2.0 m 2.0 m 2.0 m
10.8 kN
27 kN
27 kN
R A R D
R A
18 kN
R D
9kN
Free-body diagram of segment ABC of beam
10.8 kN
21.6 kN
A
B
C
2.0 m 2.0 m
V
18 kN
©F HORIZ 0:N 21.6 kN (compression)
©F VERT 0:V 7.2 kN
©M C 0:M 50.4 kN m
M
N

268 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal
plane (the xy plane) on a smooth surface about the z axis (which is vertical)
with an angular acceleration . Each of the two arms has weight w per unit
length and supports a weight W 2.0 wL at its end.
Derive formulas for the maximum shear force and maximum bending
moment in the arms, assuming b L/9 and c L/10.
y
b L c
W
x
W
Solution 4.3-15
Rotating centrifuge
L
b c
W
g
__ (L + b + c)
x
wx __
g
Tangential acceleration r
Substitute numerical data:
W
Inertial force Mr
g r
Maximum V and M occur at x b.
V max W g (L b c) Lb w
g x dx
W
g
(L b c)
wL (L 2b)
2g
M max W (L b c)(L c)
g
Lb
w
x(x b)dx
g
b
W (L b c)(L c)
g
w L2
(2L 3b)
6g
b
W 2.0 wLb L 9
91wL 2
V max
30g
M max
229wL 3
75g
c L 10

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 269
Shear-Force and Bending-Moment Diagrams
When solving the problems for Section 4.5, draw the shear-force and
bending-moment diagrams approximately to scale and label all critical
ordinates, including the maximum and minimum values.
Probs. 4.5-1 through 4.5-10 are symbolic problems and Probs. 4.5-11
through 4.5-24 are numerical problems. The remaining problems (4.5-25
through 4.5-30) involve specialized topics, such as optimization, beams
with hinges, and moving loads.
Problem 4.5-1 Draw the shear-force and bending-moment diagrams for
a simple beam AB supporting two equal concentrated loads P (see figure).
A
a
P
P
a
B
L
Solution 4.5-1
Simple beam
a
P
P
a
A
B
R A = P
L
R B = P
V
0
P
P
Pa
M
0

270 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-2 A simple beam AB is subjected to a counterclockwise
couple of moment M 0
acting at distance a from the left-hand support
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
a
L
M 0
B
Solution 4.5-2
Simple beam
A
B
R A = M 0
L
a
L
R B = M 0
L
V
0
M 0
M 0
L
M
0
M 0 (1 a L )
M 0 a
L
Problem 4.5-3 Draw the shear-force and bending-moment diagrams
for a cantilever beam AB carrying a uniform load of intensity q over
one-half of its length (see figure).
A
q
L
— 2
L —2
B
Solution 4.5-3
Cantilever beam
M A = 3qL2
8
A
q
B
R A = qL
2
L
— 2
L —2
qL —2
V
0
M
0
3qL 2
8
qL
2
8

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 271
Problem 4.5-4 The cantilever beam AB shown in the figure
is subjected to a concentrated load P at the midpoint and a
counterclockwise couple of moment M 1
PL/4 at the free end.
Draw the shear-force and bending-moment diagrams for
this beam.
A
P
M 1 = —–
PL
4
B
—
L
—2
L
2
Solution 4.5-4
Cantilever beam
P
M A
A
B
M 1 PL 4
R A P
R A
L/2 L/2
M A PL
4
V
M
0
0
P
PL
4
PL 4
Problem 4.5-5 The simple beam AB shown in the figure is subjected to
a concentrated load P and a clockwise couple M 1
PL/4 acting at the
third points.
Draw the shear-force and bending-moment diagrams for this beam.
A
P
M 1 = PL —–
4
L
— 3
L —3
L —3
B
Solution 4.5-5
Simple beam
A
P
M 1 = PL —–
4
B
R A = 5P —–
12
L
— 3
L —3
L —3
R B = 7P —–
12
V
0
5P/12
7P/12
5PL/36
7PL/36
M
0
PL/18

272 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-6 A simple beam AB subjected to clockwise couples M 1
and 2M 1
acting at the third points is shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
A
M 1 2M 1
B
L
— 3
L —3
L —3
Solution 4.5-6
Simple beam
M 1 2M 1
A
B
R A = 3M —– 1
L
—
L —3
L —3
L
R 3 B = 3M —– 1
L
0
V 3M —– 1
L
M 0 M 1
M 1 M 1
Problem 4.5-7 A simply supported beam ABC is loaded by a vertical
load P acting at the end of a bracket BDE (see figure).
Draw the shear-force and bending-moment diagrams for beam ABC.
A
B
D
E
C
P
—
L —4
L —2
L
4
L
Solution 4.5-7
Beam with bracket
A
P
B
PL
—–
4
C
R A = P —–
2
—
L
3L
— 4 4
R C = P —–
2
V
0
P
—–
2
—–
P
2
M
0
PL
—–
8
3PL
—–
8

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 273
Problem 4.5-8 A beam ABC is simply supported at A and B and
has an overhang BC (see figure). The beam is loaded by two forces
P and a clockwise couple of moment Pa that act through the
arrangement shown.
Draw the shear-force and bending-moment diagrams for
beam ABC.
A
P P Pa
C
B
a a a a
Solution 4.5-8
Beam with overhang
upper
beam:
P
P
C
a a a
Pa
P
P
lower
beam:
P
B
a a a
P
C
2P
V 0
P
P
M 0
Pa
Problem 4.5-9 Beam ABCD is simply supported at B and C and has
overhangs at each end (see figure). The span length is L and each
overhang has length L/3. A uniform load of intensity q acts along the
entire length of the beam.
Draw the shear-force and bending-moment diagrams for this beam.
A
L
3
B
q
L
C
L
3
D
Solution 4.5-9
Beam with overhangs
q
A
L/3
B
L
C
L/3
__ 5qL
R B = __ 5qL
R 6
C = 6
__ qL
qL/3
V 0
2
__ qL
– __ qL
– 3 __ 5qL 2 2
72
M
0
–qL 2 /18 X 1 –qL 2 /18
D
x 1 L 5
6
0.3727L

274 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-10 Draw the shear-force and bending-moment diagrams
for a cantilever beam AB supporting a linearly varying load of maximum
intensity q 0
(see figure).
A
L
B
q 0
Solution 4.5-10
Cantilever beam
__ x
q=q 0
L
q 0
q__
0 L
M 2
B = 6
A
x
L
B
__ q 0 L
R B = 2
V
0
q__
0 x
V = – 2
2L
__ q 0 L
– 2
M
0
q__
0 x
M = – 3
6L
__ q 0 L
– 2
6
Problem 4.5-11 The simple beam AB supports a uniform load of
intensity q 10 lb/in. acting over one-half of the span and a concentrated
load P 80 lb acting at midspan (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
P = 80 lb
q = 10 lb/in.
B
—
L
= 40 in. —
L
= 40 in.
2 2
Solution 4.5-11
Simple beam
P = 80 lb
10 lb/in.
A
B
R A =140 lb
40 in.
40 in.
R B = 340 lb
140
V
(lb)
0
60
6 in.
–340
M
(lb/in.)
0
5600
M max = 5780
46 in.

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 275
Problem 4.5-12 The beam AB shown in the figure supports a uniform
load of intensity 3000 N/m acting over half the length of the beam. The
beam rests on a foundation that produces a uniformly distributed load
over the entire length.
Draw the shear-force and bending-moment diagrams for this beam.
A
3000 N/m
B
0.8 m
1.6 m
0.8 m
Solution 4.5-12
Beam with distributed loads
3000 N/m
A
B
1500 N/m
0.8 m
1200
1.6 m
0.8 m
V
(N)
0
960
–1200
M
(N . m) 0
480
480
Problem 4.5-13 A cantilever beam AB supports a couple and a
concentrated load, as shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
A
400 lb-ft
200 lb
B
5 ft 5 ft
Solution 4.5-13
Cantilever beam
M A = 1600 lb-ft
A
400 lb-ft
200 lb
B
R A = 200 lb
5 ft 5 ft
+200
V
(lb)
0
0
M
(lb-ft)
–1600
–600
–1000

276 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-14 The cantilever beam AB shown in the figure is
subjected to a uniform load acting throughout one-half of its length and a
concentrated load acting at the free end.
Draw the shear-force and bending-moment diagrams for this beam.
A
2.0 kN/m
2 m 2 m
2.5 kN
B
Solution 4.5-14
Cantilever beam
2.0 kN/m
2.5 kN
M A = 14 kN . m
R A = 6.5 kN
V
(kN)
0
0
M
(kN . m)
A
2 m 2 m
6.5
2.5
–5.0
–14.0
B
Problem 4.5-15 The uniformly loaded beam ABC has simple supports at
A and B and an overhang BC (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
25 lb/in.
B
C
72 in.
48 in.
Solution 4.5-15
Beam with an overhang
25 lb/in.
A
B
C
72 in.
R A = 500 lb
500
V
0
(lb) 20 in.
5000
M 0
(lb-in.)
20 in.
40 in.
1200
48 in.
R B = 2500 lb
–1300
–28,800

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 277
Problem 4.5-16 A beam ABC with an overhang at one end supports a
uniform load of intensity 12 kN/m and a concentrated load of magnitude
2.4 kN (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
12 kN/m
B
2.4 kN
C
1.6 m 1.6 m 1.6 m
Solution 4.5-16
Beam with an overhang
12 kN/m
2.4 kN
A
B
C
V
(kN)
M
(kN . m)
0
0
1.6 m 1.6 m 1.6 m
R A = 13.2 kN
R B = 8.4 kN
13.2
2.4
1.1m
M max
–6.0
5.76
0.64 m
M max = 7.26
1.1m
–3.84
Problem 4.5-17 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
400 lb acting at the end
of the vertical arm and a vertical force P 2
900 lb acting at the
end of the overhang.
Draw the shear-force and bending-moment diagrams for this
beam. (Note: Disregard the widths of the beam and vertical arm
and use centerline dimensions when making calculations.)
P 1 = 400 lb
1.0 ft
A
P 2 = 900 lb
B
C
4.0 ft 1.0 ft
Solution 4.5-17
Beam with vertical arm
1.0 ft
A
P 1 = 400 lb
A
R A = 125 lb
400 lb-ft
125 lb
P 2 = 900 lb
B
C
4.0 ft 1.0 ft
R B = 1025 lb
900 lb
B
C
1025 lb
V
(lb)
M
(lb)
0
0
400
125
900
900

278 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-18 A simple beam AB is loaded by two segments of
uniform load and two horizontal forces acting at the ends of a vertical
arm (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
4 kN/m
1 m
8 kN
1 m
4 kN/m
B
8 kN
2 m 2 m
2 m
2 m
Solution 4.5-18 Simple beam
4 kN/m
16 kN . m
A
2 m 2 m 2 m
R A = 6 kN
4 kN/m
2 m
B
R B = 10 kN
V
(kN)
M
(kN . m)
0
0
6.0
1.5 m
1.5 m
2.0
4.5 4.0
16.0
12.0
10.0
Problem 4.5-19 A beam ABCD with a vertical arm CE is supported as a
simple beam at A and D (see figure). A cable passes over a small pulley
that is attached to the arm at E. One end of the cable is attached to the
beam at point B. The tensile force in the cable is 1800 lb.
Draw the shear-force and bending-moment diagrams for beam ABCD.
(Note: Disregard the widths of the beam and vertical arm and use centerline
dimensions when making calculations.)
A
B
Cable
C
E
D
1800 lb
8 ft
6 ft 6 ft 6 ft
Solution 4.5-19
Beam with a cable
E
1800 lb
Free-body diagram of beam ABCD
1800 lb
A
Cable
B
C
D
8 ft
1800
1440
1800
1440
5760 lb-ft
A C D
B
800
1080 720
800
6 ft 6 ft 6 ft
R D = 800 lb
R D = 800 lb
640
Note: All forces have units of pounds.
4800
V
(lb)
0
M
(lb-ft)
0
960
800
800
4800

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 279
Problem 4.5-20 The beam ABCD shown in the figure has
overhangs that extend in both directions for a distance of 4.2 m
from the supports at B and C, which are 1.2 m apart.
Draw the shear-force and bending-moment diagrams for this
overhanging beam.
5.1 kN/m
A
B
10.6 kN/m
C
5.1 kN/m
D
4.2 m 4.2 m
1.2 m
Solution 4.5-20
Beam with overhangs
32.97
5.1 kN/m
10.6 kN/m
5.1 kN/m
V
(kN)
0
6.36
6.36
A
B
C
D
32.97
4.2 m 4.2 m
R
1.2 m
B = 39.33 kN R C = 39.33 kN
M 0
(kN . m)
61.15 61.15
59.24
Problem 4.5-21 The simple beam AB shown in the figure supports a
concentrated load and a segment of uniform load.
Draw the shear-force and bending-moment diagrams for this beam.
A
4.0 k
C
2.0 k/ft
B
5 ft
20 ft
10 ft
Solution 4.5-21
Simple beam
A
4.0 k
C
2.0 k/ft
B
R A = 8 k
5 ft 5 ft
10 ft
R B = 16 k
V
(k)
0
8
4
12 ft
40
C
60 64
8 ft
16
M max = 64 k-ft
M
(k-ft)
0
12 ft
C
8 ft

280 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-22 The cantilever beam shown in the figure supports
a concentrated load and a segment of uniform load.
Draw the shear-force and bending-moment diagrams for this
cantilever beam.
A
3 kN
0.8 m 0.8 m
1.0 kN/m
1.6 m
B
Solution 4.5-22
Cantilever beam
4.6
M A =
6.24 kN . m
A
3 kN
1.0 kN/m
B
V
(kN)
0
1.6
0.8 m 0.8 m
R A = 4.6 kN
1.6 m
M
(kN . m)
0
2.56
1.28
6.24
Problem 4.5-23 The simple beam ACB shown in the figure is subjected
to a triangular load of maximum intensity 180 lb/ft.
Draw the shear-force and bending-moment diagrams for this beam.
180 lb/ft
A
C
B
6.0 ft
7.0 ft
Solution 4.5-23
Simple beam
A
180 lb/ft
C
B
V
(lb)
0
240
x 1 = 4.0 ft
300
390
M max = 640
R A = 240 lb
6.0 ft
1.0 ft
R B = 390 lb
M
(lb-ft)
0
360
Problem 4.5-24 A beam with simple supports is subjected to a
trapezoidally distributed load (see figure). The intensity of the load varies
from 1.0 kN/m at support A to 3.0 kN/m at support B.
Draw the shear-force and bending-moment diagrams for this beam.
A
1.0 kN/m
3.0 kN/m
B
2.4 m

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 281
Solution 4.5-24
A
1.0 kN/m
Simple beam
3.0 kN/m
B
V
(kN)
0
2.0
x 1 = 1.2980 m
x
R A = 2.0 kN
2.4 m
R B = 2.8 kN
Set V 0:
V 2.0 x x2
2.4 (x meters; V kN) M
2.8
M max = 1.450
x 1
1.2980 m
(kN . m)
0
Problem 4.5-25 A beam of length L is being designed to support a uniform load
of intensity q (see figure). If the supports of the beam are placed at the ends,
creating a simple beam, the maximum bending moment in the beam is qL 2 /8.
However, if the supports of the beam are moved symmetrically toward the middle
of the beam (as pictured), the maximum bending moment is reduced.
Determine the distance a between the supports so that the maximum bending
moment in the beam has the smallest possible numerical value.
Draw the shear-force and bending-moment diagrams for this
condition.
A
q
a
L
B
Solution 4.5-25
Beam with overhangs
q
Solve for a: a (2 2)L 0.5858L
M
0
A
(L a)/2 (L a)/2
a
R A = qL/2
R B = qL/2
M 2
M 1 M 1
The maximum bending moment is smallest when
M 1
M 2
(numerically).
q(L a)2
M 1
8
M 2 R a A ¢
2 ≤ qL2
8 qL (2a L)
8
M 1 M 2 (L a) 2 L(2a L)
B
M 1 M 2 q (L a)2
8
qL2
8
V 0
M 0
(3 22) 0.02145qL2
0.2071L
0.2929 qL
0.2929L
0.2071 qL
0.02145 qL 2
x 1 x 1
0.2071 qL
0.2929 qL
0.02145 qL 2 0.02145 qL 2
x 1 = 0.3536 a
= 0.2071 L

282 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-26 The compound beam ABCDE shown in the figure
consists of two beams (AD and DE) joined by a hinged connection at D.
The hinge can transmit a shear force but not a bending moment. The
loads on the beam consist of a 4-kN force at the end of a bracket attached
at point B and a 2-kN force at the midpoint of beam DE.
Draw the shear-force and bending-moment diagrams for this
compound beam.
4 kN
A
1 m
2 kN 1 m
B C D
E
2 m
2 m
2 m
2 m
Solution 4.5-26
Compound beam
4 kN
4 kN . m Hinge
2 kN
A
B C D
E
2 m
2 m
2 m
1 m 1 m
R A = 2.5 kN
R C = 2.5 kN
R E = 1 kN
V
(kN) 0
2.5
1.5
1.0
D
1.0
5.0
M
(kN . m)
0
2.67 m
1.0
2.0
D
1.0
Problem 4.5-27 The compound beam ABCDE shown in the figure
consists of two beams (AD and DE) joined by a hinged connection at D.
The hinge can transmit a shear force but not a bending moment. A force P
acts upward at A and a uniform load of intensity q acts downward on
beam DE.
Draw the shear-force and bending-moment diagrams for this
compound beam.
A
P
B C
L L L
D
q
2L
E
Solution 4.5-27
Compound beam
A
P
B
C
D
q
E
Hinge
L L L
R B = 2P + qL R C = P + 2qL
V
P
qL
0
D
2L
R E = qL
–qL
M
0
PL
−P−qL
D
−qL 2
L
L
qL
2

SECTION 4.5 Shear-Force and Bending-Moment Diagrams 283
Problem 4.5-28 The shear-force diagram for a simple beam
is shown in the figure.
Determine the loading on the beam and draw the bendingmoment
diagram, assuming that no couples act as loads on
the beam.
V
0
12 kN
–12 kN
2.0 m
1.0 m
1.0 m
Solution 4.5-28
Simple beam (V is given)
6.0 kN/m 12 kN
A
B
R A = 12kN
2 m 1 m 1 m
R B = 12kN
V
(kN)
0
12
12
−12
M
(kN . m)
0
Problem 4.5-29 The shear-force diagram for a beam is shown
in the figure. Assuming that no couples act as loads on the beam,
determine the forces acting on the beam and draw the bendingmoment
diagram.
V
652 lb
0
572 lb
580 lb
500 lb
–128 lb
–448 lb
4 ft 16 ft
4 ft
Solution 4.5-29
Force diagram
Forces on a beam (V is given)
20 lb/ft
V
(lb)
652
0
572
580
500
–128
4 ft 16 ft
4 ft
652 lb 700 lb 1028 lb 500 lb
M
(lb-ft)
2448
–448
0
14.50 ft
–2160

284 CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-30 A simple beam AB supports two connected wheel loads P
and 2P that are distance d apart (see figure). The wheels may be placed at
any distance x from the left-hand support of the beam.
(a) Determine the distance x that will produce the maximum shear force
in the beam, and also determine the maximum shear force V max
.
(b) Determine the distance x that will produce the maximum bending
moment in the beam, and also draw the corresponding bendingmoment
diagram. (Assume P 10 kN, d 2.4 m, and L 12 m.)
A
x
P
d
L
2P
B
Solution 4.5-30
A
x
Moving loads on a beam
P
d
L
2P
P 10 kN
d 2.4 m
L 12 m
B
Reaction at support B:
R B P L x 2P L (x d) P (2d 3x)
L
Bending moment at D:
M D R B (L x d)
P
(2d 3x)(L x d)
L
(a) Maximum shear force
By inspection, the maximum shear force occurs at
support B when the larger load is placed close to, but
not directly over, that support.
dM D
dx
P L [3x2 (3L 5d)x 2d(L d)]
P
(6x 3L 5d) 0
L
Eq.(1)
A
R A = L
Pd
x L d 9.6 m
x = L − d
V max R B P ¢3 d ≤ 28 kN
L
P
d
R B = P(3 −
d
L)
2P
B
Solve for x: x L 6 ¢3 5d L ≤ 4.0 m
Substitute x into Eq (1):
M max
(b) Maximum bending moment
64
M max = 78.4
By inspection, the maximum bending moment occurs
M
at point D, under the larger load 2P.
(kN . m)
P 2P
0
4.0 m 2.4 m 5.6 m
x d
A
D
B
P
Note:R A
2 ¢3 d
≤ 16 kN
L
L
R
P
R B
2 ¢3 d
B ≤ 14 kN
L
P
L B 3¢L 6 ≤ 2
¢3 5d L ≤ 2
(3L 5d)
L ¢
6 ≤ ¢3 5d ≤ 2d(L d)R
L
PL
12 ¢3 d L ≤ 2
78.4 kN m