Shear Forces and Bending Moments

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Shear Forces and Bending Moments

Shear 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.

800 lb A

30 in.

60 in. 120 in.

Solution 4.3-1 Simple beam

800 lb A

1600 lb D

B

30 in. RA

60 in.

MA 0: RB 1400 lb MB 0: RA 1000 lb

30 in. RB

Free-body diagram of segment DB

M ?FVERT 0:

?MD 0:

1600 lb VD

B

30 in. RB

V 1600 lb 1400 lb 200 lb

M (1400 lb)(30 in.) 42,000 lb-in.

1600 lb B

30 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

2.0 kN/m

C B

Solution 4.3-2 Simple beam

6.0 kN

2.0 kN/m

C

A

B

1.0 m 1.0 m RA

2.0 m RB

MA 0: RB 4.5 kN MB 0: RA 5.5 kN

1.0 m 1.0 m

2.0 m

4.0 m

Free-body diagram of segment AC

6.0 kN

A

CV M

1.0 m 1.0 m RA

?FVERT 0: V 0.5 kN ?MC 0: M 5.0 kN m

259

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CHAPTER 4 Shear Forces and Bending Moments

Problem 4.3-3 Determine the shear force V and bending moment M at

P

P

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

P A

P B

b

L

RA

b RB

?MB 0

RA

1 L

[

P(L

b

b)

]

P ?1 2b (upward) L

b

L

b

?MA 0: P

RB

P

?1

2b L

(downward)

A

C

M

b

L /2

RA

V

Free-body diagram (C is the midpoint)

?FVERT 0:

V

RA

P

P

?1

2b L

P

2bP L

?MC 0:

M P ?1 2b ? L P ?b L

L2

2

M PL Pb Pb PL 0

2

2

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 A

1.0 m 1.0 m

1.5 kN/m B

2.0 m

Solution 4.3-4 Cantilever beam

4.0 kN

1.5 kN/m

A B

1.0 m 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

D M

1.5 kN/m B

0.5 m 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

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.

400 lb/ft A

200 lb/ft B

C

10 ft

10 ft 6 ft 6 ft

Solution 4.3-5 Beam with an overhang

400 lb/ft A

200 lb/ft B

C

10 ft

10 ft 6 ft 6 ft

RA

RB

MB 0: RA 2460 lb MA 0: RB 2740 lb

Free-body diagram of segment AD

400 lb/ft

A 10 ft

RA

D M

6 ft V

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 supported at A and B and has an overhang from B to C. The

P1 = 4.0 kN

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

the end of the overhang.

A

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.)

4.0 m

P2 = 8.0 kN

B

C

1.0 m

Solution 4.3-6 Beam with vertical arm

P1 = 4.0 kN

P2 = 8.0 kN

1.0 m

A

B

4.0 m RA

MB 0: RA 1.0 kN (downward) MA 0: RB 9.0 kN (upward)

1.0 m RB

Free-body diagram of segment AD Point D is 3.0 m from support A.

A 4.0 kN ? m

RA

3.0 m

DM V

?FVERT 0: ?MD 0:

V RA 1.0 kN 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?

A b

q

B

C

L

D b

Solution 4.3-7 Beam with overhangs q

A

D

B

C

b

L

RB

b RC

From symmetry and equilibrium of vertical forces:

L

RB

RC

q ?b 2

Free-body diagram of left-hand half of beam: Point E is at the midpoint of the beam.

q

A b

E

L /2

V

RB

M = 0 (Given)

?ME 0

L

1

L2

RB

?

2

q

?

2

?b

2

0

q

?b

L

?

L

q

?

1

?b

L

2

0

22

2

2

Solve for b/L :

b1 L2

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

Solution 4.3-8 Archer's bow B

P

A

H C

b P 130 N 70? H 1400 mm

1.4 m b 350 mm

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

SECTION 4.3 Shear Forces and Bending Moments

263

Free-body diagram of segment BC

B

T

H 2

bC M

?MC 0

H T(cos b)?

T(sin b) (b) M 0

2

M

H T ? cos b

b sin b

2

P

H

?

b

tan

b

2 2

Substitute numerical values:

M 130 N B1.4 m (0.35 m)(tan 70)R

2

2

M 108 N m

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 .

MN B

P

r

V

PP

A

O

C

A

Solution 4.3-9 Curved bar

B

P

r

P

A

O

C

MN

B P cos P

A

V

O

P sin

?FN 0 Q b

FV 0

R

a

?MO 0

N P sin u 0 N P sin u

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

900 N/m

2.6 m

2.6 m

1.0 m

Solution 4.3-10 Airplane wing

MV

1600 N/m

900 N/m

A 2.6 m

2.6 m

B 1.0 m

Shear Force

FVERT 0

c T

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.)

Loading (in three parts)

700 N/m 1

900 N/m

2

A

3 B

Bending Moment

?MA 0

M

1

(700

N m) (2.6

2.6 m)?

m

2

3

(900 Nm)(5.2 m)(2.6 m)

1

(900

Nm) (1.0

m)?5.2

m

1.0

m

0

2

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.)

E

Cable

A

B

C

P

8 ft D

6 ft

6 ft

6 ft

Solution 4.3-11 Beam with a cable

E

P

Cable

P

A

B

C

6 ft

6 ft

_4_P 9

UNITS: P in lb M in lb-ft

8 ft D

6 ft

_4_P 9

Free-body diagram of section AC P

4_P_

P

A

5

_4_P

6 ft

B

9

?MC 0

_3_P

M

5C

N

6 ft

V

M

4P (6

ft)

4P

(12

ft)

0

5

9

M 8P lb-ft 15

Numerical value of M equals 640 lb-ft.

640 lb-ft 8P lb-ft 15

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

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CHAPTER 4 Shear Forces and Bending Moments

Solution 4.3-12 Beam with trapezoidal load

50 kN/m

30 kN/m

A

B

3 m

RA

RB

Free-body diagram of section CB

Point C is at the midpoint of the beam.

40 kN/m

30 kN/m V

M

C

B

1.5 m

55 kN

Reactions

?MB 0 RA(3 m) (30 kNm) (3 m) (1.5 m) (20 kNm) (3 m) (12) (2 m) 0

RA 65 kN

?FVERT 0 c RA RB 12 (50 kNm 30 kNm) (3 m) 0 RB 55 kN

FVERT 0 c T

V

(30

kNm)

(1.5

m)

1 2

(10

kNm)

(1.5

m)

55 kN 0 V 2.5 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

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.

q1 = 3500 lb/ft

B

C

A

D

3.0 ft

q2 8.0 ft

3.0 ft

Solution 4.3-13 Foundation beam

q1 = 3500 lb/ft

AB

CD

3.0 ft

q2 8.0 ft

3.0 ft

FVERT 0: q2(14 ft) q1(8 ft)

8 q2 14 q1 2000 lbft (a) V and M at point B

A

B

MB

FVERT 0:

2000 lb/ft VB 3 ft

VB 6000 lb

?MB 0: MB 9000 lb-ft

(b) V and M at midpoint E

3500 lb/ft

A

B

E Mm

2000 lb/ft

Vm

3 ft

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

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