V & M Diagrams

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.

R A

R B R B

© F VERT ϭ 0:

V ϭ 1600 lb Ϫ 1400 lb ϭ 200 lb

⌺ M A ϭ 0: R B ϭ 140 1400 0 lb ⌺ M B ϭ 0: R A ϭ 100 1000 0 lb

© M D ϭ 0: M ϭ (1400 lb)(30 in.) ϭ 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.

C

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 A

1.0 m R A

1.0 m 4.0 m

B

Simple beam 6.0 kN

A

2.0 kN/m

1.0 m

⌺ M A ϭ 0: R B ϭ 4.5kN ⌺ M B ϭ 0: R A ϭ 5.5kN

2.0 m

C

1.0 m R B

V

M

1.0 m

R A

© F VERT ϭ 0:

V ϭ Ϫ0.5 kN

© M C ϭ 0: M ϭ 5.0 kN m

259

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260

CHAPTE CHA PTER R4

Shear She ar Forc Forces es and Ben Bendin ding g Momen Moments ts

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

¢

© M A ϭ 0: R B ϭ P 1 ϩ

B

2b L

≤

P A

b

L

b

R A

(downward)

b

R B

M

C L /2

R A

V

Free-body diagram (C is the midpoint)

© M B ϭ 0 R A ϭ

1 L

© F VERT ϭ 0:

¢

[ P ( L ϩ b ϩ b ) ]

¢

ϭP 1ϩ

2b L

≤

V ϭ R A Ϫ P ϭ P 1 ϩ

2b

≤ L

¢

(upward)

M ϭ P 1 ϩ PL

2

2b

L ≤ ¢ ≤ L 2

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.

ϩ Pb Ϫ Pb Ϫ

PL

2

L

L

2

≤

ϭ0

4.0 kN

1.5 kN/m

A

1.0 m

B

1.0 m

2.0 m

Cantilever Canti lever beam

4.0 kN

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

¢

ϪP bϩ

Problem 4.3-4

1.0 m

2bP

© M C ϭ 0:

M ϭ

Solution 4.3-4

ϪP ϭ

1.5 kN/m

© F VERT ϭ 0: m )(2.0 m) V ϭ 4.0 kN ϩ (1.5 kN ր m)(2.0 ϭ 4.0 kN ϩ 3.0 kN ϭ 7.0 kN © M D ϭ 0: M ϭ Ϫ (4.0 kN)(0.5 m) m )(2.0 m)(2.5 m) Ϫ (1.5 kN ր m)(2.0 ϭ Ϫ2.0 kN m Ϫ 7.5 kN m ϭ Ϫ9.5 kN m











SECTIO SEC TION N 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 She ar Forces Forces and and Bendin Bending g Moments Moments

10 ft

R A

Free-body diagram of segment AD C

6 ft

400 lb/ft D

A

6 ft

10 ft

6 ft V

R A

R B

M

⌺ M B ϭ 0: R A ϭ 246 2460 0 lb ⌺ M A ϭ 0: R B ϭ 274 2740 0 lb

Point D is 16 ft from support A.

© F VERT ϭ 0: V ϭ 2460 lb Ϫ (400 lb ր ft)(10 f t)(10 ft) ϭ Ϫ1540 lb

© M D ϭ 0: M ϭ (2460 lb)(16 ft) Ϫ (400 lb ր ft)(10 f t)(10 ft)(11 ft) ϭ Ϫ4640 lb-ft

Problem 4.3-6

The beam ABC shown in the figure is simply P1 = 4.0 kN supported at A and B and has an overhang from B to C . The loads consist of a horizontal force P1 ϭ 4.0 kN acting at the end of a vertical arm and a vertical force P2 ϭ 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 ( Note: use centerline dimensions when making calculations.)

Solution 4.3-6

P2 = 8.0 kN B

4.0 m

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

4.0 m

1.0 m

R A

M

D

3.0 m R

C

V











262

CHAPTE CHA PTER R4


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

C

b

Solution 4.3-7

L

b

Beam with overhangs q

Free-body diagram of left-hand half of beam:

A

D B b

Point E is at the midpoint of the beam.

C

q

L

b RC

R B

M = 0 (Given)

A b

L /2

¢

≤

V

R B

From symmetry and equilibrium of vertical forces: L R B ϭ RC ϭ q b ϩ 2

E

© M E ϭ 0 2

¢ L2 ≤ q ¢ 12 ≤ ¢b L2 ≤ L L q ¢b ≤2 ¢ 2 ≤ q ¢ 12 ≤ ¢b L2 ≤ Ϫ R B

Ϫ

ϩ

ϩ

ϩ

ϩ

ϩ

ϭ0 2

ϭ0

Solve for b / L : b L

ϭ

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












Solution 4.3-8


Archer’s bow B

Free-body diagram of segment BC B ␤

␤

P

C

H 2

T

H

A b

C

M

© M C ϭ 0 b

P ϭ 130N

M ϭ

␤ ϭ 70° ϭ

H ϭ 140 1400 0 mm ϭ 1.4m

ϩ

Ϫ M ϭ 0

ϩ

ϩ

Substitute numerical value valuess:

b ϭ 35 350 0 mm M ϭ

ϭ 0.35m

T ␤

A

T

T ϭ tensile force in the bowstring ⌺F HORIZ ϭ 0:

2T cos ␤Ϫ P ϭ 0 T ϭ

P

2 cos b

B

130 N 1.4 m

2 M ϭ 108 N

Free-body diagram of point A

P

¢ H 2 ≤ T (sin (sin b ) ( b ) H T ¢ cos b b sin b≤ 2 P H ¢ b tan b≤ 2 2

T (cos (cos b )

m

2

R

ϩ (0.35 m)(tan 70Њ )

263











264

CHAPTE CHA PTER R4


Problem 4.3-9

A curve curved d 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 N P cos ␪ r

␪

A

P A

V

r

␪

O

P

P

␪

C

A

Curved bar

B P

M N

B

O

P C

B

N ϭ P sin u

V ␪

P

© F N ϭ 0 Qϩ b Ϫ N Ϫ P sin u ϭ 0

O

A P sin ␪

͚F V ϭ

0

ϩ

R a Ϫ

© M O ϭ 0

V Ϫ P cos u ϭ 0 V ϭ P cos u M Ϫ Nr ϭ 0 M ϭ Nr ϭ 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

Loading (in three parts)

900 N/m

700 N/m

V A

B

2.6 m

2.6 m

1.0 m

Shear Force ⌺F VERT ϭ 0

1.0 m

2.6 m

Airplane wing

1600 N/m M

900 N/m

1 2

900 N/m A

3 B

Bending Moment

cϩ TϪ

1 V ϩ (700 N ր m)(2.6 m )(2.6 m) ϩ (900 N ր m)(5.2 m )(5.2 m) 2 1 ϩ (900 N ր m)(1.0 m )(1.0 m) ϭ 0 2 V ϭ Ϫ6040 N ϭ Ϫ6.04 kN

© M A ϭ 0 Ϫ M ϩ

1 2

(700 N ր m)(2.6 m )(2.6 m)

¢ 2.63 m ≤

ϩ (900 N ր m)(5.2 m )(5.2 m)(2.6 m) ϩ

1

¢

(900 N ր m)(1.0 m )(1.0 m) 5.2 m ϩ

1.0 m

≤

ϭ0














Problem 4.3-11

A be beam am 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 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: Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

E

A

B

P

Cable A

B

6 ft 4P __ 9

UNITS: P in lb M in lb-ft

8 ft C

6 ft

D

6 ft

Beam with a cable E

P

P

Cable

6 ft

Solution 4.3-11

265


8 ft C

6 ft

Free-body diagram of section AC P

P

D

P 4 __ 5

A

M

C

N

6 ft

4P __ 9

6 ft 4 P __ 9

3P __ 5

B

6 ft V

© M C ϭ 0 M Ϫ

4P 5

M ϭ Ϫ

(6 ft) ϩ

4P 9

(12 ft) ϭ 0

8P

lb-ft 15 M equal Numerical value of M equalss 640 lb-f lb-ft. t.

∴

640 lb-ft ϭ

8P

lb-ft 15 and P ϭ 1200 lb











266

CHAPTE CHA PTER R4

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 R A

C

R B

Reactions

⌺F VERT ϭ 0 Ϫ R A (3 m) ϩ (30 kN ր m)(3 m )(3 m)(1.5 m)

© M B ϭ 0

ϩ (20 kN ր m)(3 m )(3 m) ( 1΋2 ) (2 m) ϭ 0

R A ϭ 65kN

B

1.5 m

55 kN

cϩ TϪ

V Ϫ (30 kN ր m)(1.5 m )(1.5 m) Ϫ 12 (10 kN ր m)(1.5 m )(1.5 m) ϩ 55kNϭ 0

V ϭ Ϫ2.5 kN

© M C ϭ 0

ϩ

© F VERT ϭ 0 c m )(3 m) ϭ 0 R A ϩ R B Ϫ 1΋2 (50 kN ր m ϩ 30 kN ր m)(3 R B ϭ 55kN

Ϫ M Ϫ (30 kN/m) kN/m)(1.5 (1.5 m)(0 m)(0.75 .75 m) Ϫ 1΋2 (10 kN ր m)(1.5 m )(1.5 m)(0.5 m) ϩ (55kN)(1 (55kN)(1.5 .5 m) ϭ 0

M ϭ 45.0 kN

m q1 = 3500 lb/ft

Problem 4.3-13

Beam ABCD represents a reinforced-concret reinforced-concretee foundation beam that supports a uniform load of intensity q1 ϭ 3500 lb/ft (see figure). Assume that the soil pressure on the underside of the beam is uniformly distributed with intensity q2. (a) Fin Find d the shear shear forc forcee V B and bending moment M B at point B. (b) Fin Find d the shea shearr force force V m and bending moment M m at the midpoint of the beam.

Solution 4.3-13

B A

D q2

3.0 ft

Foundation beam

q1 = 3500 lb/ft

C

(b) V and M at midpoint E

8.0 ft

3.0 ft












E

Problem 4.3-14

The simply-supported beam ABCD is loaded by a weight W ϭ 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: 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

2.0 m

Free-body diagram of pulley at B 1.5 m

B

27 kN

C

2.0 m

D

21.6 kN

2.0 m 27 kN

27 kN

R A ϭ 18kN

10.8 kN

R D

R A

R D ϭ 9 kN

Free-body diagram of segment ABC of beam 10.8 kN 21.6 kN

A

2.0 m 18 kN

B

267


M

C

N

2.0 m V

© F HORIZ ϭ 0: N ϭ 21.6 kN (compression)

D

2.0 m











268

CHAPTE CHA PTER R4


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 ϭ 2.0 wL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b ϭ L /9 and c ϭ L /10.

y

c L

b

W

x ␣

W

Solution 4.3-15

Rotating centrifuge

c L b

W ( L __ L + b + c)␣ g

x

Tangential acceleration ϭ r ␣ Inertial force Mr ␣ ϭ

Substitute numerical dat data: a:

W r ␣ g

W ϭ 2.0 wL

Maximum V and M occur at x ϭ b. W V max ϭ ( L ϩ b ϩ c ) ␣ ϩ g W ␣ ( L ϩ b ϩ c ) ϭ g wL

__

w␣ x g

L ϩ b

Ύ

b

w␣ x dx x dx g

bϭ

2

V max ϭ M max ϭ

91wL ␣ 30g 229wL3␣ 75g

L

9

cϭ

L

10











SECTION SECT ION 4.5 4.5

269

Shear-Forc Shear -Force e and Bendin Bending-Mom g-Moment ent Diagra Diagrams ms

Shear-Force Shear-Fo rce 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

P V

0

a B

R A = P

a

R B = P













270

CHAPTE CHA PTER R4


Problem 4.5-2

A simp simple le beam beam AB is subjected to a counterclockwise M couple of moment 0 acting at distance a from the left-hand support (see figure). Draw the shear-force and bending-moment diagrams for this beam.

M 0 A

B a L

Solution 4.5-2

Simple beam M 0 A

M R A = 0 L

B a

M R B = 0 L

L

M0 L

V

0

M

Problem 4.5-3

0

M 0a L ᎐ M 0 (1 ᎐

a ) L

Draw the shear-force and bending-moment diagrams

q













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 ϭ PL /4 at the free end. Draw the shear-force and bending-moment diagrams for this beam.

Solution 4.5-4

271

PL M 1 = —– 4

P

A

B L — 2

L — 2

Cantileve Cant ileverr beam P A

B M 1 ϭ

M A L/2

R A

V

M

PL 4

R A ϭ P

L/2 M A ϭ

PL 4

P

0

PL 4

0 Ϫ

PL 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 M 1 ϭ PL /4 acting at the third points. Draw the shear-force and bending-moment diagrams for this beam.

A

PL M 1 = —– 4 B











272

CHAPTE CHA PTER R4


Problem 4.5-6

A simp simple le beam beam AB subjected to clockwise couples M 1 M and 2 1 acting at the third points is shown in the figure. Draw the shear-force and bending-moment diagrams for this beam.

M 1 A

B L — 3

Solution 4.5-6

2 M 1

L — 3

L — 3

Simple beam M 1

2 M 1

A

B

3 M R A = —–1 L

L — 3

L — 3

3 M R B = —–1 L

L — 3

0

3 M 1 L

Ϫ —–

V

M 1 M

0 Ϫ M 1

Problem 4.5-7

Ϫ M 1

A simp simply ly supported supported beam ABC is loaded by a vertical P load acting at the end of a bracket BDE (see figure).

B A

C












Problem 4.5-8

A be beam am ABC is simply supported at A and B and BC has an overhang (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

a

a

P

a

Pa

a

a

P

P P

P B

lower beam:

C a

a

a

2P P V

0

P

Pa

C

B

C

upper beam:

P

A

Beam with overhang P

273


a

a











274

CHAPTE CHA PTER R4


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 Canti lever beam x q=q0 __ L

q0 2 q__ 0 L M B = 6

B

A x V 0

M 0

Problem 4.5-11 intensity

L

R B =

q0 L __

2

q0 x 2 V = – __ 2 L

q0 L – __ 2

q0 x 3 M = – __ 6 L

q0 L2 – __ 6

The simple beam AB supports a uniform load of 10 lb/in. acting over one-half of the span and a concentrated

P = 80 lb














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

Beam with distributed loads 3000 N/m A

B

1500 N/m 0.8 m

1.6 m

275


0.8 m

1.6 m

0.8 m











276

CHAPTE CHA PTER R4


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

B

A

2m

Solution 4.5-14

Cantilever Canti lever 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

2.5 kN

2.5 2. 5

2m












Problem 4.5-16

A be beam am 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 R A = 13.2 kN

13.2

1.6 m

C

B

Beam with an overhang 12 kN/m

2.4 kN

A

1.6 m

Solution 4.5-16

277


1.6 m R B = 8.4 kN

1.6 m

1.6 m











278

CHAPTE CHA PTER R4


Problem 4.5-18

A simp simple le 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 4 kN/m

16 kN . m

2m

6.0

4 kN/m B

2m

2m

Simple beam

A

R A = 6 kN

2m

2m

2m

V (kN)

0

1.5 m

Ϫ 2.0

2m R B = 10 kN

16.0

Ϫ 10.0












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

5.1 kN/m

279


V 0 (kN)

Ϫ6.36













280

CHAPTE CHA PTER R4


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

1.0 kN/m

A

0.8 m

B

0.8 m

1.6 m












Solution 4.5-24


Simple beam 2.0

3.0 kN/m 1.0 kN/m

V (kN)

281











282

CHAPTE CHA PTER R4


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

4 kN

1m 2 kN

A

B

C

D

1m E












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.


283

12 kN V

0 –12 kN











284

CHAPTE CHA PTER R4


Problem 4.5-30

A simp simple le 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 Determine the dista distance nce x that will produce the maximum shear force in the beam, and also determine the maximum shear force V .

P x

2P d

V & M Diagrams

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