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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
MA 0: MB 0:
RB RB
RB 1400 lb 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
6.0 kN
B
C V
A
MA 0: MB 0:
2.0 m
Free-body diagram of segment AC
2.0 kN/m
C
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
RB 4.5 kN RA 5.5 kN
2.0 m
1.0 m RB
M
1.0 m
RA
©FVERT 0:V 1.5 kN ©MC 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
Solution 4.3-3
L
b
Beam with overhangs
P
P A
B
©MA 0:RB P ¢ 1 P A
b
L
b
RA
2b ≤(downward) L
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
1.5 kN/m
D
B
M 0.5 m
B 1.0 m
2.0 m
Cantilever beam
4.0 kN
1.0 m
1.5 kN/m
A
1.0 m
2.0 m
©FVERT 0: V 4.0 kN (1.5 kNm)(2.0 m) 4.0 kN 3.0 kN 7.0 kN ©MD 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
261
SECTION 4.3 Shear Forces and Bending Moments
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
6 ft
6 ft
Beam with an overhang
400 lb/ft
200 lb/ft B
A 10 ft
10 ft
C
10 ft
RA
Free-body diagram of segment AD C
6 ft
400 lb/ft 10 ft
RA 2460 lb
MA 0:
RB 2740 lb
M
6 ft V
RA
RB
MB 0:
D
A
6 ft
Point D is 16 ft from support A. ©FVERT 0: V 2460 lb (400 lbft)(10 ft) 1540 lb ©MD 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 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
B
Free-body diagram of segment AD Point D is 3.0 m from support A. 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 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?
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 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
SECTION 4.3 Shear Forces and Bending Moments
Solution 4.3-8
Archer’s bow B
Free-body diagram of segment BC B
P
A
H 2
T
H C
b
C M
©MC 0 b
P 130 N 70° H 1400 mm 1.4 m
Substitute numerical values:
b 350 mm
130 N 1.4 m B (0.35 m)(tan 70)R 2 2 M 108 N m M
0.35 m Free-body diagram of point A T P
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) ¢
A
T
T tensile force in the bowstring FHORIZ 0:
2T cos P 0 T
P 2 cos b
263
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 Pcos r
A
A
V
r
P
O
N
P
P
C
A
Curved bar
B P
M B
P
O
©FN 0 Q b N P sin u 0 N P sin u
V
P
C
N
A
O
FV 0
b !
V P cos u 0 V P cos u
Psin
©MO 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
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
3 B
Bending Moment
Shear Force FVERT 0
A
c T
1 V (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.)
©MA 0 M
1 2.6 m (700 Nm)(2.6 m) ¢ ≤ 2 3
(900 Nm)(5.2 m)(2.6 m) 1 1.0 m (900 Nm)(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 m
265
SECTION 4.3 Shear Forces and Bending Moments
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
Cable A
8 ft
B
C
6 ft
Solution 4.3-11
P
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 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
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
A
Point C is at the midpoint of the beam. 40 kN/m
B
30 kN/m V M
3m RA
B
C
1.5 m
RB
FVERT 0
Reactions ©MB 0 RA (3 m) (30 kNm)(3 m)(1.5 m)
c T
V (30 kNm)(1.5 m) 12 (10 kNm)(1.5 m)
(20 kNm)(3 m)( 12 )(2 m) 0 RA 65 kN
55 kN
55 kN 0 V 2.5 kN
©FVERT 0 c RA RB 12 (50 kNm 30 kNm)(3 m) 0 RB 55 kN
©MC 0 M (30 kN/m)(1.5 m)(0.75 m) 12 (10 kNm)(1.5 m)(0.5 m) (55 kN)(1.5 m) 0 M 45.0 kN 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 lbft 14 1 (a) V and M at point B B
A
MB
FVERT 0: 2000 lb/ft 3 ft
VB
VB 6000 lb
©MB 0:MB 9000 lb-ft
3 ft
Mm
4 ft
FVERT 0: Vm (2000 lb/ft)(7 ft) (3500 lb/ft)(4 ft) Vm 0 ME 0: Mm (2000 lb/ft)(7 ft)(3.5 ft) (3500 lb/ft)(4 ft)(2 ft) Mm 21,000 lb-ft
267
SECTION 4.3 Shear Forces and Bending Moments
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 RD
RA
RA 18 kN
10.8 kN
27 kN
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
©FHORIZ 0:N 21.6 kN (compression) ©FVERT 0:V 7.2 kN ©MC 0:M 50.4 kN m
D
2.0 m
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 a. 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.
W 2.0 wLb 91wL2 30g 229wL3 75g
Vmax
Lb
W w (L b c) x dx g g b W (L b c) g wL (L 2b) 2g W Mmax (L b c) g Lb w (L c) x(x b)dx g b W (L b c)(L c) g w L2 (2L 3b) 6g Vmax
wx __ g
Mmax
L 9
c
L 10
269
SECTION 4.5 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
P
a
P
a
A
B
L
RA = P
RB = P
P V 0 P
Pa M
0
a
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
0
M0 (1 a ) L
M0a 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 =
q
3qL2 8
A B
RA =
qL 2
L — 2
L — 2
qL — 2 V
0 0
M
3qL2 8
qL2 8
L — 2
271
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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
PL M1 = —– 4
P
B
A L — 2
L — 2
Cantilever beam P A
B
PL M1 4
MA RA
V
M
L/2 RA P
L/2 PL MA 4 P
0
PL 4
0 PL 4
A
B L — 3
Solution 4.5-5
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
7PL/36
0 PL/18
PL M1 = —– 4
P
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.
L — 3
L — 3
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.
M1
2M1
A
B L — 3
Solution 4.5-6
L — 3
L — 3
Simple beam M1
2M1
A
B
3M RA = —–1 L
L — 3
L — 3
3M RB = —–1 L
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 — 4
L — 4
L — 2 L
Solution 4.5-7
Beam with bracket P A
C B
P RA = —– 2
V
3L — 2
L — 4
P RC = —– 2
P —– 2 P —– 2
0
PL —– 8 M 0
3PL —– 8
273
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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
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
P
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
5qL RB = __ 6 qL __ 2 V 0
5qL RC = __ 6 qL/3
qL – __ 6 M 0
L/3
qL __ 2 –qL2/18
5qL2 __ 72 X1
qL – __ 2
–qL2/18
x1 L
5 0.3727L 6
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.)
0 46 in.
Mmax = 5780
L = — 40 in. 2
275
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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
1.6 m
0.8 m
Beam with distributed loads 3000 N/m A
B 1500 N/m 0.8 m
1.6 m
0.8 m
1200 V (N) 0 –1200
960 480
M
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 = 200lb
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.
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 = 6.5kN
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 M 0 (lb-in.) 20 in. 40 in. –28,800
48 in.
277
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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 A
C
B 1.6 m
Solution 4.5-16
2.4 kN
1.6 m
1.6 m
Beam with an overhang 2.4 kN
12 kN/m A
C
B 1.6 m RA = 13.2kN
1.6 m
1.6 m RB = 8.4kN
13.2 V (kN)
2.4 0 1.1m Mmax
–6.0
Mmax = 7.26
5.76 .64m
M kN . m 0 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 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
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
1.0 ft
RA = 125 lb
A
400 lb-ft
RB = 1025 lb 900 lb
B
C 125 lb
1025 lb
M (lb)
125
0 400
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.
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 RA = 800 lb
6 ft
6 ft RB = 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
800
960 4800
800
279
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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.
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 V 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
B
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
V (kN)
1.0 kN/m
1.6 0
A
B
0.8 m
0.8 m
1.6 m M (kN . m)
RA = 4.6 kN
0 2.56
1.28
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
A
390
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
281
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Solution 4.5-24
Simple beam 2.0
3.0 kN/m V (kN)
1.0 kN/m
0 A
B
x1 = 1.2980 m x 2.8
2.4 m RA = 2.0 kN
RB = 2.8 kN
x2 (x meters; V kN) V 2.0 x 2.4 Set V 0:
Mmax = 1.450
M (kN . m)
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 2)L 0.5858L q M1 M2 (L a) 2 8 2 qL (3 22) 0.02145qL2 8 0.2071 qL
0.2071L
M2 V
0
0.2929L 0.2071 qL
M 0 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 (L a) 2 L(2a L) M1 M2
0.2929 qL 2
0.02145 qL
M
0
x1 0.02145 qL2
x1 0.02145 qL2
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
2L
283
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
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
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
–448
4 ft 1028 lb 500 lb
V (lb/ft) 0 14.50 ft –2160
500
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: MD RB (L x d) P (2d 3x)(L x d) L P [3x2 (3L 5d)x 2d(L d) ] L RB
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
PL d 2 ¢ 3 ≤ 78.4 kN m 12 L
M (kN . m) 4.0m
D
Mmax = 78.4
64
2.4m
5.6m
B
P d ¢ 3 ≤ 16 kN 2 L P d RB ¢ 3 ≤ 14 kN 2 L
Note:RA L
L 5d ≤ ¢3 ≤ 2d(L d)R 6 L
0
2P
P x
2P
RB
Eq.(1)
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