Shear Forces and Bending Moments PDF

Preview

Summary

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 A B at a cross section just to the left of the 1600-lb load acting on the simple beam AB shown in the figure. 30 in. 60 in. 30 in. 120 in. Solution 4.3-1 ...

Description

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

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 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

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

RA

Free-body diagram of segment AD C

6 ft

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

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

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

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

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 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

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.)

©MA ⫽ 0 哵哴 ⫺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

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

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

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

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

©FHORIZ ⫽ 0:
N ⫽ 21.6 kN (compression) ©FVERT ⫽ 0:
V ⫽ 7.2 kN ©MC ⫽ 0:
M ⫽ 50.4 kN ⴢ m

267

Shear Forces and Bending Moments

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 ␣. 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

冮

W ⫽ 2.0 wL
b ⫽ 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

L 9

c⫽

L 10

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

L

RB = P

P V 0 ᎐P

Pa M

0

a B

RA = P

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

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 8

qL2 ⫺ 8

L — 2

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

7PL/36

0 ⫺PL/18

PL M1 = —– 4

P

L — 3

Solution 4.5-5

PL 4

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.

2M1

M1 A

B L — 3

Solution 4.5-6

L — 3

L — 3

Simple beam 2M1

M1 A

B