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Shear Forces and Bending Moments
Electrical Engineering (University of Rizal System)
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4
Shear Forces and
Bending Moments
Shear Forces and Bending Moments
800 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
B
30 in.
Solution 4.3-1
1600 lb
60 in.
120 in.
30 in.
Simple beam
800 lb
Free-body diagram of segment DB
1600 lb
D
A
1600 lb
V
B
D
B
M
30 in.
30 in.
60 in.
30 in.
RA
RB
RB
⌺MA ⫽ 0: RB ⫽ 1400 lb
⌺MB ⫽ 0: RA ⫽ 1000 lb
©FVERT ⫽ 0:V ⫽ 1600 lb ⫺ 1400 lb
⫽ 200 lb
©MD ⫽ 0:M ⫽ (1400 lb)(30 in.)
⫽ 42,000 lb-in.
6.0 kN
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
1.0 m
Solution 4.3-2
2.0 m
Free-body diagram of segment AC
2.0 kN/m
C
6.0 kN
B
C V
A
1.0 m
RA
1.0 m
4.0 m
B
Simple beam
6.0 kN
A
2.0 kN/m
C
1.0 m
⌺MA ⫽ 0: RB ⫽ 4.5 kN
⌺MB ⫽ 0: RA ⫽ 5.5 kN
2.0 m
1.0 m
RB
M
1.0 m
RA
©FVERT ⫽ 0:V ⫽ ⫺0.5 kN
©MC ⫽ 0:M ⫽ 5.0 kN ⴢ m
259
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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
Solution 4.3-3
L
b
Beam with overhangs
P
P
A
B
©MA ⫽ 0:RB ⫽ P ¢ 1 ⫹
2b
≤(downward)
L
P
A
b
L
b
RA
b
RB
M
C
L/2
RA
V
Free-body diagram (C is the midpoint)
©MB ⫽ 0
©FVERT ⫽ 0:
1
RA ⫽ [P(L ⫹ b ⫹ b) ]
L
V ⫽ RA ⫺ P ⫽ P ¢ 1 ⫹
2b
2bP
≤⫺P ⫽
L
L
©MC ⫽ 0:
2b
⫽ P ¢ 1 ⫹ ≤(upward)
L
2b L
L
≤ ¢ ≤ ⫺ P ¢b ⫹ ≤
L 2
2
PL
PL
M⫽
⫹ Pb ⫺ Pb ⫺
⫽0
2
2
M ⫽ P ¢1 ⫹
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.
4.0 kN
1.0 m
Solution 4.3-4
1.5 kN/m
A
B
1.0 m
2.0 m
Free-body diagram of segment DB
Point D is 0.5 m from support A.
4.0 kN
V
1.0 m
2.0 m
©FVERT ⫽ 0:
V ⫽ 4.0 kN ⫹ (1.5 kNⲐm)(2.0 m)
⫽ 4.0 kN ⫹ 3.0 kN ⫽ 7.0 kN
©MD ⫽ 0:M ⫽ ⫺(4.0 kN)(0.5 m)
⫺ (1.5 kNⲐm)(2.0 m)(2.5 m)
⫽ ⫺2.0 kN ⴢ m ⫺ 7.5 kN ⴢ m
⫽ ⫺9.5 kN ⴢ m
1.5 kN/m
D
B
M
0.5 m
B
Cantilever beam
4.0 kN
1.0 m
1.5 kN/m
A
1.0 m
2.0 m
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SECTION 4.3
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.
400 lb/ft
200 lb/ft
B
A
10 ft
Solution 4.3-5
10 ft
C
6 ft
6 ft
Beam with an overhang
400 lb/ft
200 lb/ft
B
A
10 ft
261
Shear Forces and Bending Moments
10 ft
C
6 ft
RA
Free-body diagram of segment AD
400 lb/ft
D
A
6 ft
10 ft
V
RA
RB
M
6 ft
⌺MB ⫽ 0: RA ⫽ 2460 lb
⌺MA ⫽ 0: RB ⫽ 2740 lb
Point D is 16 ft from support A.
©FVERT ⫽ 0:
V ⫽ 2460 lb ⫺ (400 lbⲐft)(10 ft)
⫽ ⫺1540 lb
©MD ⫽ 0:M ⫽ (2460 lb)(16 ft)
⫺ (400 lbⲐft)(10 ft)(11 ft)
⫽ ⫺4640 lb-ft
Problem 4.3-6 The beam ABC shown in the figure is simply P = 4.0 kN
1
supported at A and B and has an overhang from B to C. The
loads consist of a horizontal force P1 ⫽ 4.0 kN acting at the
end of a vertical arm and a vertical force P2 ⫽ 8.0 kN acting at 1.0 m
A
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.)
Solution 4.3-6
P2 = 8.0 kN
B
4.0 m
C
1.0 m
Beam with vertical arm
P1 = 4.0 kN
P2 = 8.0 kN
1.0 m
A
Free-body diagram of segment AD
Point D is 3.0 m from support A.
B
4.0 kN • m
A
3.0 m
4.0 m
RA
⌺MB ⫽ 0:
RA ⫽ 1.0 kN (downward)
⌺MA ⫽ 0:
RB ⫽ 9.0 kN (upward)
1.0 m
RB
RA
M
D
V
©FVERT ⫽ 0:V ⫽ ⫺RA ⫽ ⫺ 1.0 kN
©MD ⫽ 0:M ⫽ ⫺RA (3.0 m) ⫺ 4.0 kN ⴢ m
⫽ ⫺7.0 kN ⴢ m
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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?
q
A
D
B
b
Solution 4.3-7
C
L
b
Beam with overhangs
q
A
D
B
b
Free-body diagram of left-hand half of beam:
Point E is at the midpoint of the beam.
q
C
L
b
RC
RB
M = 0 (Given)
A
b
L
≤
2
V
RB
From symmetry and equilibrium of vertical forces:
RB ⫽ RC ⫽ q ¢ b ⫹
L/2
E
©ME ⫽ 0 哵 哴
L
1
L 2
⫺RB ¢ ≤ ⫹ q ¢ ≤ ¢ b ⫹ ≤ ⫽ 0
2
2
2
L L
1
L 2
⫺q ¢ b ⫹ ≤ ¢ ≤ ⫹ q ¢ ≤ ¢ b ⫹ ≤ ⫽ 0
2 2
2
2
Solve for b/L :
b 1
⫽
L 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
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SECTION 4.3
Solution 4.3-8
Shear Forces and Bending Moments
Archer’s bow
B
Free-body diagram of segment BC
B
␤
␤
P
C
H
2
T
H
A
b
C
M
©MC ⫽ 0 哵 哴
b
P ⫽ 130 N
␤ ⫽ 70°
H ⫽ 1400 mm
⫽ 1.4 m
H
≤ ⫹ T(sin b) (b) ⫺ M ⫽ 0
2
H
M ⫽ T ¢ cosb ⫹ b sin b≤
2
P H
⫽ ¢ ⫹ b tan b≤
2 2
T(cos b) ¢
Substitute numerical values:
b ⫽ 350 mm
130 N 1.4 m
B
⫹ (0.35 m)(tan 70⬚)R
2
2
M ⫽ 108 N ⴢ m
M⫽
⫽ 0.35 m
Free-body diagram of point A
T
␤
P
A
T
T ⫽ tensile force in the bowstring
⌺FHORIZ ⫽ 0:
2T cos ␤⫺ P ⫽ 0
T⫽
P
2 cos b
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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 ␪.
Solution 4.3-9
M
P cos ␪
r
␪
A
P
A
V
r
␪
O
N
P
P
C
␪
A
Curved bar
B
P
M
B
P
O
©FN ⫽ 0 Q⫹ b⫺ N ⫺ P sin u⫽ 0
B
N ⫽ P sin u
V
␪
P
C
N
O
A
P sin ␪
兺FV ⫽ 0
⫹
R a⫺
©MO ⫽ 0 哵 哴
V ⫺ P cos u ⫽ 0
V ⫽ P cos u
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
2.6 m
Solution 4.3-10
1.0 m
2.6 m
Airplane wing
1600 N/m
M
900 N/m
Loading (in three parts)
900 N/m
700 N/m 1
V
2
900 N/m
A
B
2.6 m
2.6 m
1.0 m
Shear Force
⌺FVERT ⫽ 0
A
3
B
Bending Moment
⫺
c⫹ T
©MA ⫽ 0 哵哴
1
V ⫹ (700 NⲐm)(2.6 m) ⫹ (900 NⲐm)(5.2 m)
2
1
⫹ (900 NⲐm)(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.)
⫺M ⫹
1
2.6 m
(700 NⲐm)(2.6 m) ¢
≤
2
3
⫹ (900 NⲐm)(5.2 m)(2.6 m)
1
1.0 m
⫹ (900 NⲐm)(1.0 m) ¢ 5.2 m ⫹
≤⫽0
2
3
M ⫽ 788.67 N • m ⫹ 12,168 N • m ⫹ 2490 N • m
⫽ 15,450 N • m
⫽ 15.45 kN ⴢ m
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SECTION 4.3
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.)
E
P
Cable
A
8 ft
B
6 ft
Solution 4.3-11
265
Shear Forces and Bending Moments
C
D
6 ft
6 ft
Beam with a cable
E
P
Free-body diagram of section AC
P
Cable
A
P
B
6 ft
4P
__
9
UNITS:
P in lb
M in lb-ft
8 ft
C
6 ft
P
D
4P
__
9
3P
__
5
M
C
N
6 ft
4P
__
9
6 ft
4P
__
5
A
B
6 ft
V
©MC ⫽ 0 哵哴
4P
4P
(6 ft) ⫹
(12 ft) ⫽ 0
5
9
8P
M⫽ ⫺
lb-ft
15
Numerical value of M equals 640 lb-ft.
M ⫺
8P
lb-ft
15
and P ⫽ 1200 lb
∴ 640 lb-ft ⫽
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
30 kN/m
A
B
3m
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266
CHAPTER 4 Shear Forces and Bending Moments
Solution 4.3-12
Beam with trapezoidal load
Free-body diagram of section CB
50 kN/m
30 kN/m
Point C is at the midpoint of the beam.
40 kN/m
A
B
30 kN/m
V
M
3m
RA
B
C
1.5 m
RB
55 kN
c⫹ T⫺
Reactions
⌺FVERT ⫽ 0
©MB ⫽ 0 哵 ⫺ RA (3 m) ⫹ (30 kNⲐm)(3 m)(1.5 m)
V ⫺ (30 kNⲐm)(1.5 m) ⫺ 21 (10 kNⲐm)(1.5 m)
⫹ (20 kNⲐm)(3 m)( 1冫2 )(2 m) ⫽ 0
RA ⫽ 65 kN
⫹ 55 kN⫽ 0
V ⫽ ⫺2.5 kN
⫹
©FVERT ⫽ 0 c
RA ⫹ RB ⫺ 1冫2 (50 kNⲐm ⫹ 30 kNⲐm)(3 m) ⫽ 0
RB ⫽ 55 kN
©MC ⫽ 0 哵哴
⫺ M ⫺ (30 kN/m)(1.5 m)(0.75 m)
⫺ 1冫2 (10 kNⲐm)(1.5 m)(0.5 m)
⫹ (55 kN)(1.5 m) ⫽ 0
M ⫽ 45.0 kN ⴢ m
q1 = 3500 lb/ft
Problem 4.3-13 Beam ABCD represents a reinforced-concrete
foundation beam that supports a uniform load of intensity q1 ⫽ 3500 lb/ft
(see figure). Assume that the soil pressure on the underside of the beam is
uniformly distributed with intensity q2.
(a) Find the shear force VB and bending moment MB at point B.
(b) Find the shear force Vm and bending moment Mm at the midpoint
of the beam.
Solution 4.3-13
C
D
3.0 ft
q2
8.0 ft
3.0 ft
Foundation beam
q1 = 3500 lb/ft
A
B
A
B
(b) V and M at midpoint E
C
D
3500 lb/ft
B
A
3.0 ft
q2
8.0 ft
E
3.0 ft
Vm
2000 lb/ft
⌺FVERT ⫽ 0:
q2(14 ft) ⫽ q1(8 ft)
8
∴ q2 ⫽
q ⫽ 2000 lbⲐft
14 1
(a) V and M at point B
B
A
VB
4 ft
⌺FVERT ⫽ 0: Vm ⫽ (2000 lb/ft)(7 ft) ⫺ (3500 lb/ft)(4 ft)
Vm ⫽ 0
MB
⌺FVERT ⫽ 0:
2000 lb/ft
3 ft
3 ft
Mm
VB ⫽ 6000 lb
©MB ⫽ 0:MB ⫽ 9000 lb-ft
⌺ME ⫽ 0:
Mm ⫽ (2000 lb/ft)(7 ft)(3.5 ft)
⫺ (3500 lb/ft)(4 ft)(2 ft)
Mm ⫽ 21,000 lb-ft
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SECTION 4.3
E
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.)
Cable
1.5 m
A
B
C
2.0 m
2.0 m
W = 27 kN
Solution 4.3-14
Beam with cable and weight
E
Cable
A
B
2.0 m
Free-body diagram of pulley at B
1.5 m
27 kN
C
2.0 m
D
21.6 kN
2.0 m
27 kN
RA ⫽ 18 kN
10.8 kN
27 kN
RD
RA
RD ⫽ 9 kN
Free-body diagram of segment ABC of beam
10.8 kN
21.6 kN
A
2.0 m
B
M
C
N
2.0 m
V
18 kN
267
Shear Forces and Bending Moments
©FHORIZ ⫽ 0:N ⫽ 21.6 kN (compression)
©FVERT ⫽ 0:V ⫽ 7.2 kN
©MC ⫽ 0:M ⫽ 50.4 kN ⴢ m
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D
2.0 m
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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
c
L
b
W
x
␣
W
Solution 4.3-15
Rotating centrifuge
c
L
b
W (L + b + c)␣
__
g
x
Tangential acceleration ⫽ r␣
Substitute numerical data:
W
Inertial force Mr ␣ ⫽ g r␣
Maximum V and M occur at x ⫽ b.
冮
W ⫽ 2.0 wLb ⫽
L
9
91wL2␣
30g
229wL3␣
⫽
75g
Vmax ⫽
L⫹b
w␣
W
(L ⫹ b ⫹ c)␣ ⫹
x dx
g
g
b
W␣
⫽
(L ⫹ b ⫹ c)
g
wL␣
⫹
(L ⫹ 2b)
2g
W␣
Mmax ⫽
(L ⫹ b ⫹ c)(L ⫹ c)
g
L⫹b
w␣
⫹
x(x ⫺ b)dx
g
b
W␣
⫽
(L ⫹ b ⫹ c)(L ⫹ c)
g
w L2␣
⫹
(2L ⫹ 3b)
6g
Vmax ⫽
w␣x
__
g
Mmax
冮
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c⫽
L
10
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SECTION 4.5
269
Shear-Force and Bending-Moment Diagrams
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
P
P
A
B
L
Solution 4.5-1
Simple beam
a
P
P
A
a
B
L
RA = P
RB = P
P
V
0
᎐P
Pa
M
a
0
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270
CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-2 A simple beam AB is subjected to a counterclockwise
couple of moment M0 acting at distance a from the left-hand support
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
M0
A
B
a
L
Solution 4.5-2
Simple beam
M0
A
RA =
B
a
M0
L
RB =
L
M0
L
V
0
M
M0
L
M0a
L
0
᎐ M0 (1 ᎐ a )
L
q
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
B
L
—
2
Solution 4.5-3
Cantilever beam
MA =
3qL2
8
q
A
B
RA =
qL
2
L
—
2
L
—
2
qL
—
2
V
M
0
0
⫺
3qL2
qL2
⫺ 8
8
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L
—
2
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SECTION 4.5
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 M1 ⫽ PL/4 at the free end.
Draw the shear-force and bending-moment diagrams for
this beam.
Solution 4.5-4
271
Shear-Force and Bending-Moment Diagrams
PL
M1 = —–
4
P
B
A
L
—
2
L
—
2
Cantilever beam
P
A
B
MA
L/2
RA
V
M
PL
M1 ⫽ 4
RA ⫽ P
L/2
MA ⫽
P
0
PL
4
0
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 M1 ⫽ PL/4 acting at the
third points.
Draw the shear-force and bending-moment diagrams for this beam.
A
B
Simple beam
PL
M1 = —–
4
P
A
5P
RA = —–
12
B
L
—
3
L
—
3
L
—
3
7P
RB = —–
12
5P/12
V
0
⫺7P/12
5PL/36
M
PL
M1 = —–
4
P
L
—
3
Solution 4.5-5
PL
4
7PL/36
0
⫺PL/18
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L
—
3
L
—
3
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272
CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-6 A simple beam AB subjected to clockwise couples M1
and 2M1 acting at the third points is shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
2M1
M1
A
B
L
—
3
Solution 4.5-6
L
—
3
L
—
3
Simple beam
2M1
M1
A
B
3M
RA = —–1
L
3M
RB = —–1
L
L
—
3
L
—
3
L
—
3
0
3M
⫺ —–1
L
V
M1
M 0
⫺M1
⫺M1
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.
B
A
C
D
E
P
L
—
2
L
—
4
L
—
4
L
Solution 4.5-7
Beam with bracket
P
A
PL
—–
4
C
B
P
RA = —–
2
V
P
RC = —–
2
P
—–
2
P
⫺ —–
2
0
PL
—–
8
M
3L
—
4
L
—
4
3PL
—–
8
0
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lOMoARcPSD|36318479
SECTION 4.5
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.
Solution 4.5-8
273
Shear-Force and Bending-Moment Diagrams
P
P
A
Pa
C
B
a
a
a
a
Beam with overhang
P
P
C
upper
beam:
a
Pa
a
a
P
P
P
P
B
lower
beam:
C
a
a
a
2P
P
V
0
M
0
⫺P
⫺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.
Solution 4.5-9
q
A
D
B
L
3
C
L
L
3
Beam with overhangs
q
A
D
L /3
B
C
L
L/3
5qL
RC = __
6
5qL
RB = __
6
qL
__
2
V 0
qL/3
qL
– __
3
5qL2
__
72
qL
– __
2
M 0
–qL2/18
X1
–qL2/18
x1 ⫽ L
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
兹5
⫽ 0.3727L
6
lOMoARcPSD|36318479
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 q0 (see figure).
q0
A
B
L
Solution 4.5-10
Cantilever beam
q0
x
q=q0 __
L
q0 L 2
MB = __
6
B
A
q0 L
RB = __
2
L
x
V 0
q0 x2
V = – __
2L
q0 L
– __
2
q0 x3
M = – __
6L
q0 L2
– __
6
M 0
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.
P = 80 lb
q = 10 lb/in.
A
B
L =
—
40 in.
2
Solution 4.5-11
Simple beam
P = 80 lb
10 lb/in.
A
RA =140 lb
B
40 in.
40 in.
RB = 340 lb
140
V
(lb)
60
0
6 in.
–340
5600
M
(lb/in.)
Mmax = 5780
0
46 in.
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
L =
—
40 in.
2
lOMoARcPSD|36318479
SECTION 4.5
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.
3000 N/m
A
B
0.8 m
Solution 4.5-12
275
Shear-Force and Bending-Moment Diagrams
1.6 m
0.8 m
Beam with distributed loads
3000 N/m
B
A
1500 N/m
0.8 m
1.6 m
0.8 m
1200
V
(N) 0
960
480
M
–1200
480
(N . m) 0
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.
200 lb
400 lb-ft
A
B
5 ft
Solution 4.5-13
5 ft
Cantilever beam
200 lb
400 lb-ft
A
B
MA = 1600 lb-ft
5 ft
RA = 200 lb
5 ft
+200
V
(lb)
0
0
M
(lb-ft)
–1600
–600
–1000
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lOMoARcPSD|36318479
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.
2.0 kN/m
2.5 kN
B
A
2m
Solution 4.5-14
2m
Cantilever beam
2.0 kN/m
M A = 14 kN . m
2.5 kN
B
A
2m
R A = 6.5 kN
2m
6.5
V
(kN)
2.5
0
0
M
(kN . m)
–5.0
–14.0
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.
25 lb/in.
A
C
B
72 in.
Solution 4.5-15
Beam with an overhang
25 lb/in.
A
C
B
72 in.
RA = 500 lb
48 in.
RB = 2500 lb
1200
500
V 0
(lb)
20 in.
–1300
5000
M 0
(lb-in.) 20 in.
40 in.
–28,800
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
48 in.
lOMoARcPSD|36318479
SECTION 4.5
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.
12 kN/m
2.4 kN
A
C
B
1.6 m
Solution 4.5-16
277
Shear-Force and Bending-Moment Diagrams
1.6 m
1.6 m
Beam with an overhang
2.4 kN
12 kN/m
A
C
B
1.6 m
RA = 13.2 kN
1.6 m
1.6 m
RB = 8.4 kN
13.2
V
(kN)
2.4
0
1.1m
Mmax
–6.0
5.76
M
(kN . m) 0
Mmax = 7.26
0.64 m
1.1m
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 P1 ⫽ 400 lb acting at the end
of the vertical arm and a vertical force P2 ⫽ 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.)
Solution 4.5-17
–3.84
P1 = 400 lb
P2 = 900 lb
1.0 ft
A
B
4.0 ft
C
1.0 ft
Beam with vertical arm
P1 = 400 lb
P2 = 900 lb
1.0 ft
A
B
900
C
V
(lb) 0
4.0 ft
RA = 125 lb
A
400 lb-ft
RB = 1025 lb
900 lb
B
M
(lb)
0
⫺400
C
125 lb
⫺125
1.0 ft
1025 lb
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
⫺900
lOMoARcPSD|36318479
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.
8 kN
4 kN/m
4 kN/m
1m
A
B
1m
8 kN
2m
Solution 4.5-18
2m
2m
2m
Simple beam
4 kN/m
6.0
4 kN/m
16 kN . m
A
V
(kN)
B
2m
2m
2m
0
1.5 m
⫺2.0
2m
RA = 6 kN
⫺10.0
RB = 10 kN
16.0
12.0
4.5
M
(kN . m) 0
4.0
1.5 m
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.)
E
Cable
A
B
1800 lb
Cable
6 ft
6 ft
B
Free-body diagram of beam ABCD
1440
1800
B
C
1440
5760 lb-ft
8 ft
C
D
1800 A
1080
D
720
800
6 ft
RD = 800 lb
6 ft
6 ft
RD = 800 lb
Note: All forces have units of pounds.
640
V
(lb)
D
Beam with a cable
E
1800 lb A
8 ft
C
6 ft
Solution 4.5-19
1800 lb
4800
0
M 0
(lb-ft)
⫺800
⫺960
⫺800
⫺4800
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800
lOMoARcPSD|36318479
SECTION 4.5
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.
279
Shear-Force and Bending-Moment Diagrams
10.6 kN/m
5.1 kN/m
5.1 kN/m
A
D
B
C
4.2 m
4.2 m
1.2 m
Solution 4.5-20
Beam with overhangs
32.97
6.36
10.6 kN/m
5.1 kN/m
V
0
(kN)
5.1 kN/m
A
⫺6.36
D
B
4.2 m
RB = 39.33 kN
⫺32.97
C
M
0
(kN . m)
4.2 m
RC = 39.33 kN
1.2 m
⫺59.24
⫺61.15
⫺61.15
4.0 k
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.
2.0 k/ft
C
A
5 ft
10 ft
20 ft
Solution 4.5-21
Simple beam
4.0 k
A
RA = 8 k
V
(k)
5 ft
B
5 ft
10 ft
RB = 16 k
8
0
2.0 k/ft
C
4
12 ft
8 ft
C
⫺16
60 64
Mmax = 64 k-ft
40
M
(k-ft) 0
12 ft
C
8 ft
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B
lOMoARcPSD|36318479
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.
3 kN
B
0.8 m
Solution 4.5-22
1.0 kN/m
A
0.8 m
1.6 m
Cantilever beam
4.6
3 kN
MA =
6.24 kN . m
V
(kN)
1.0 kN/m
A
1.6
0
B
0.8 m
0.8 m
1.6 m
M
(kN . m)
RA = 4.6 kN
0
⫺1.28
⫺2.56
⫺6.24
180 lb/ft
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.
A
B
C
6.0 ft
7.0 ft
Solution 4.5-23
Simple beam
240
180 lb/ft
V
(lb)
0
x1 = 4.0 ft
B
⫺300
⫺390
A
C
Mmax = 640
6.0 ft
RA = 240 lb
1.0 ft
RB = 390 lb
360
M
(lb-ft) 0
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.
3.0 kN/m
1.0 kN/m
A
B
2.4 m
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lOMoARcPSD|36318479
SECTION 4.5
Solution 4.5-24
281
Shear-Force and Bending-Moment Diagrams
Simple beam
2.0
3.0 kN/m
V
(kN)
1.0 kN/m
0
A
x1 = 1.2980 m
x
B
⫺2.8
2.4 m
RA = 2.0 kN
RB = 2.8 kN
x2
V ⫽ 2.0 ⫺ x ⫺
(x ⫽ meters; V ⫽ kN)
2.4
Mmax = 1.450
M
(kN . m)
Set V ⫽ 0: x1 ⫽ 1.2980 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 qL2/8.
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.
Solution 4.5-25
q
A
B
a
L
Beam with overhangs
q
A
B
(L ⫺ a)/ 2
(L ⫺ a)/ 2
a
RB = qL/2
RA = qL/2
Solve for a: a ⫽ (2 ⫺ 兹2)L ⫽ 0.5858L
q
M1 ⫽ M2 ⫽ (L ⫺ a) 2
8
2
qL
⫽
(3 ⫺ 2兹2) ⫽ 0.02145qL2
8
0.2929 qL
M2
0.2071 qL
0.2071L
V
0
M 0
0.2929L
⫺ 0.2071 qL
M1
M1
The maximum bending moment is smallest when
M1⫽ M2 (numerically).
q(L ⫺ a) 2
M1 ⫽
8
qL2
qL
a
M2 ⫽ RA ¢ ≤ ⫺
⫽ (2a ⫺ L)
2
8
8
M1 ⫽ M2
(L ⫺ a) 2 ⫽ L(2a ⫺ L)
⫺ 0.2929 qL
0.02145 qL2
M
0
x1
⫺ 0.02145 qL2
x1 = 0.3536 a
= 0.2071 L
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
x1
⫺ 0.02145 qL2
lOMoARcPSD|36318479
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
1m
2 kN
B
C
D
A
E
2m
Solution 4.5-26
1m
2m
2m
2m
Compound beam
4 kN
Hinge
4 kN . m
B
2 kN
C
D
A
2m
2m
2m
RA = 2.5 kN
1m 1m
RC = 2.5 kN
2.5
V
(kN)
E
RE = 1 kN
1.0
0
D
⫺1.5
⫺1.0
5.0
M
0
(kN . m)
1.0
D
1.0
2.67 m
⫺2.0
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.
Solution 4.5-27
Compound beam
P
q
B
C
D
A
E
L
P
L
L
q
B
C
D
A
E
Hinge
L
RB = 2P + qL
P
V
L
L
RC = P + 2qL
2L
RE = qL
qL
0
D
PL
–qL
−P−qL
M
qL
2
D
0
−qL2
L
L
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
2L
lOMoARcPSD|36318479
SECTION 4.5
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.
12 kN
V
0
–12 kN
2.0 m
Solution 4.5-28
283
Shear-Force and Bending-Moment Diagrams
Simple beam (V is given)
1.0 m
6.0 kN/m
1.0 m
12 kN
B
A
2m
RA = 12kN
1m
1m
12
−12
RB = 12kN
12
V
(kN)
0
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.
652 lb
580 lb
572 lb
500 lb
V
0
–128 lb
–448 lb
4 ft
Solution 4.5-29
Forces on a beam (V is given)
16 ft
4 ft
652
580
572
Force diagram
V
(lb)
20 lb/ft
0
–128
2448
4 ft
652 lb
700 lb
16 ft
4 ft
1028 lb 500 lb
–448
M
(lb-ft)
0
14.50 ft
–2160
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
500
lOMoARcPSD|36318479
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.
P
x
(a) Determine the distance x that will produce the maximum shear force
in the beam, and also determine the maximum shear force Vmax.
(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.)
Solution 4.5-30
d
A
B
L
Moving loads on a beam
P
2P
x
d
P ⫽ 10 kN
d ⫽ 2.4 m
L ⫽ 12 m
A
B
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.
2P
P
x=L−d
d
A
B
Reaction at support B:
P
2P
P
x⫹
(x ⫹ d) ⫽ (2d ⫹ 3x)
L
L
L
Bending moment at D:
RB ⫽
MD ⫽ RB (L ⫺ x ⫺ d)
P
⫽ (2d ⫹ 3x)(L ⫺ x ⫺ d)
L
P
⫽ [⫺3x2 ⫹ (3L ⫺ 5d)x ⫹ 2d(L ⫺ d) ]
L
dMD
P
⫽ (⫺6x ⫹ 3L ⫺ 5d) ⫽ 0
dx
L
L
5d
Solve for x: x ⫽ ¢ 3 ⫺ ≤ ⫽ 4.0 m
6
L
Substitute x into Eq (1):
Mmax ⫽
RA = Pd
L
P
L 2
5d 2
B ⫺ 3¢ ≤ ¢ 3 ⫺ ≤ ⫹ (3L ⫺ 5d)
L
6
L
d)
RB = P(3 − L
⫻¢
x ⫽ L ⫺ d ⫽ 9.6 m
d
Vmax ⫽ RB ⫽ P ¢ 3 ⫺ ≤ ⫽ 28 kN
L
⫽
(b) Maximum bending moment
By inspection, the maximum bending moment occurs
at point D, under the larger load 2P.
d
A
L
5d
≤ ¢3 ⫺
≤ ⫹ 2d(L ⫺ d)R
6
L
PL
d 2
¢ 3 ⫺ ≤ ⫽ 78.4 kN ⴢ m
12
L
M
(kN . m)
4.0 m
D
Mmax = 78.4
64
0
2P
P
x
2P
2.4 m
B
P
d
¢ 3 ⫹ ≤ ⫽ 16 kN
2
L
P
d
RB ⫽ ¢ 3 ⫺ ≤ ⫽ 14 kN
2
L
Note:RA ⫽
L
5.6 m
RB
Downloaded by manuel adame (manu_xulo_2000@hotmail.com)
Eq.(1)