Shear Forces and Bending Moments - Vaftsy CAEvaftsycae.com/knowledge_files/SFD_BMD.pdfShear Forces...

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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.

Solution 4.3-1 Simple beam

4Shear Forces andBending Moments

259

A B

1600 lb800 lb

120 in.

30 in. 60 in. 30 in.

SMA

5 0: RB

5 1400 lb

SMB

5 0: RA

5 1000 lb

Free-body diagram of segment DB

5 42,000 lb-in.

©MD 5 0:ÊM 5 (1400 lb)(30 in.)

5 200 lb

©FVERT 5 0:ÊV 5 1600 lb 2 1400 lb

A B

1600 lb800 lb

30 in. 60 in. 30 in.

D

RA RB

B

1600 lb

30 in.

D

RB

V

M

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.


AC

B

2.0 kN/m6.0 kN

1.0 m 1.0 m

4.0 m

2.0 m

AC

B

2.0 kN/m6.0 kN

1.0 m 1.0 m 2.0 m

RA RB

SMA

5 0: RB

5 4.5 kN

SMB

5 0: RA

5 5.5 kN

Free-body diagram of segment AC

©MC 5 0:ÊM 5 5.0 kN ? m

©FVERT 5 0:ÊV 5 20.5 kN

A C

6.0 kN

1.0 m 1.0 m

RA

V M

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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.

Solution 4.3-3 Beam with overhangs

260 CHAPTER 4 Shear Forces and Bending Moments

PP

bb L

5 P ¢1 12b

L≤Ê(upward)

RA 51

L[P(L 1 b 1 b) ]

©MB 5 0

Free-body diagram (C is the midpoint)

M 5PL

21 Pb 2 Pb 2

PL

25 0

M 5 P ¢1 12b

L≤ ¢L

2≤2 P ¢b 1

L

2≤

©MC 5 0:

52bP

LV 5 RA 2 P 5 P ¢1 1

2b

L≤2 P

©FVERT 5 0:

©MA 5 0:ÊRB 5 P ¢1 12b

L≤Ê(downward)PP

bb L

A B

RA RB

P

b L/2

A C

RA V

M

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.

Solution 4.3-4 Cantilever beam

AB

1.5 kN/m4.0 kN

1.0 m1.0 m 2.0 m

Free-body diagram of segment DB

Point D is 0.5 m from support A.5 29.5 kN ? m

5 22.0 kN ? m 2 7.5 kN ? m

2 (1.5 kN/m)(2.0 m)(2.5 m)

©MD 5 0:ÊM 5 2(4.0 kN)(0.5 m)

5 4.0 kN 1 3.0 kN 5 7.0 kN

V 5 4.0 kN 1 (1.5 kN/m)(2.0 m)

©FVERT 5 0:A

B

1.5 kN/m4.0 kN

1.0 m1.0 m 2.0 m

DB

1.5 kN/m4.0 kN

1.0 m0.5 m

2.0 m

V

M

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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.

Solution 4.3-5 Beam with an overhang

SECTION 4.3 Shear Forces and Bending Moments 261

A CB

400 lb/ft 200 lb/ft

6 ft6 ft10 ft 10 ft

SMB

5 0: RA

5 2460 lb

SMA

5 0: RB

5 2740 lb

Free-body diagram of segment AD

Point D is 16 ft from support A.

5 24640 lb-ft

2 (400 lb/ft) (10 ft) (11 ft)

©MD 5 0:ÊM 5 (2460 lb)(16 ft)

5 21540 lb

V 5 2460 lb 2 (400 lb/ft) (10 ft)

©FVERT 5 0:

A CB

400 lb/ft 200 lb/ft

6 ft6 ft10 ft 10 ft

RA RB

AD

400 lb/ft

6 ft10 ft

RA V

M

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

loads consist of a horizontal force P1

5 4.0 kN acting at the

end of a vertical arm and a vertical force P2

5 8.0 kN acting at

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 Beam with vertical arm

4.0 m 1.0 m

BAC

P2 = 8.0 kN

P1 = 4.0 kN

1.0 m

4.0 m 1.0 m

BA

P2 = 8.0 kN

P1 = 4.0 kN

1.0 m

RA RB

SMB

5 0: RA

5 1.0 kN (downward)

SMA

5 0: RB

5 9.0 kN (upward)

Free-body diagram of segment AD

Point D is 3.0 m from support A.

5 27.0 kN ? m

©MD 5 0:ÊM 5 2RA(3.0 m) 2 4.0 kN ? m

©FVERT 5 0:ÊV 5 2RA 5 2 1.0 kN

3.0 m

A D

RAV

M

4.0 kN • m

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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

bb L

DAB C

From symmetry and equilibrium of vertical forces:

RB 5 RC 5 q ¢b 1L

2≤

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

Point E is at the midpoint of the beam.

Solve for b /L :

b

L5

1

2

2q ¢b 1L

2≤ ¢L

2≤1 q ¢1

2≤ ¢b 1

L

2≤

2

5 0

2RB ¢L2≤1 q ¢1

2≤ ¢b 1

L

2≤

2

5 0

©ME 5 0 ` ~

q

bb L

DAB C

RB RC

q

b L/2

A

RB

V

M = 0 (Given)

E

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.

350 mm

1400 mm

70°

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Solution 4.3-8 Archer’s bow


P 5 130 N

b 5 70°

H 5 1400 mm

5 1.4 m

b 5 350 mm

5 0.35 m

Free-body diagram of point A

T 5 tensile force in the bowstring

SFHORIZ

5 0: 2T cos b2 P 5 0

T 5P

2 cos b

Free-body diagram of segment BC

Substitute numerical values:

M 5 108 N ? m

M 5130 N

2B1.4 m

21 (0.35 m)(tan 708)R

5P

2 ¢H

21 b tan b≤

M 5 T ¢H2

cosb 1 b sin b≤

T(cos b)¢H2≤ 1 T(sin b) (b) 2 M 5 0

©MC 5 0 ` ~

b

H

PA

bC

B

PA

T

b

T

H2

Cb

B

T

b

M

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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 u.

Solution 4.3-9 Curved bar


PP P

C

B

A O

r

A

V

NM

uu

M 5 Nr 5 Pr sin u

©MO 5 0 ` ~ M 2 Nr 5 0

V 5 P cos u

oFV 5 0 1R a2 V 2 P cos u 5 0

N 5 P sin u

©FN 5 0 Q1 b2ÊN 2 P sin u5 0

PP P

C

B

A O

r

A

V

NM

uuO

P cos u

P sin u

B

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.

Solution 4.3-10 Airplane wing

1.0 m

1600 N/m 900 N/m

2.6 m2.6 m

1.0 m

1600 N/m 900 N/m

2.6 m2.6 m

A B

VM

Shear Force

SFVERT

5 0 c1 T2

(Minus means the shear force acts opposite to the

direction shown in the figure.)

V 5 26040 N 5 26.04 kN

11

2 (900 N/m)(1.0 m) 5 0

V 11

2(700 N/m)(2.6 m) 1 (900 N/m)(5.2 m)

Bending Moment

M 5 788.67 N • m 1 12,168 N • m 1 2490 N • m

5 15,450 N • m

5 15.45 kN ? m

11

2(900 N/m)(1.0 m)¢5.2 m 1

1.0 m

3≤5 0

1 (900 N/m)(5.2 m)(2.6 m)

2M 11

2 (700 N/m)(2.6 m)¢2.6 m

3≤

©MA 5 0 `~

A B

32

1700 N/m

900 N/m

Loading (in three parts)

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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.)

Solution 4.3-11 Beam with a cable


A

E P

C DB

Cable8 ft

6 ft 6 ft 6 ft

UNITS:

P in lb

M in lb-ft

Free-body diagram of section AC

Numerical value of M equals 640 lb-ft.

and P 5 1200 lb

∴ 640 lb-ft 58P

15 lb-ft

M 5 28P

15 lb-ft

M 24P

5(6 ft) 1

4P

9 (12 ft) 5 0

©MC 5 0 `~

A

E P

C DB

Cable8 ft

6 ft 6 ft 6 ft

P

__9

__9

4P 4P

A

P

C

B6 ft 6 ft

P

__5

__5

N

M

V__9

4P

4P

3P

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.BA

50 kN/m

30 kN/m

3 m

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Solution 4.3-12 Beam with trapezoidal load


Reactions

RA

5 65 kN

RB

5 55 kN

1 30 kN/m)(3 m) 5 0RA 1 RB 2 1@2 (50 kN/m

©FVERT 5 01

c

1 (20 kN/m)(3 m)(1@2)(2 m) 5 0

©MB 5 0 ` 2 RA(3 m) 1 (30 kN/m)(3 m)(1.5 m)

Free-body diagram of section CB

Point C is at the midpoint of the beam.

SFVERT

5 0 c1 T2

1 55 kN5 0

2 M 2 (30 kN/m)(1.5 m)(0.75 m)

1 (55 kN)(1.5 m) 5 0

M 5 45.0 kN ? m

2 1@2(10 kN/m)(1.5 m)(0.5 m)

©MC 5 0 `~

V 5 22.5 kN

V 2 (30 kN/m)(1.5 m) 212(10 kN/m)(1.5 m)

BA

50 kN/m

30 kN/m

3 m

RA RB

B

V

40 kN/m

30 kN/m

1.5 m55 kN

CM

Problem 4.3-13 Beam ABCD represents a reinforced-concrete

foundation beam that supports a uniform load of intensity q1

5 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 Foundation beam

A

B C

D

3.0 ft 3.0 ft

q2

q1 = 3500 lb/ft

8.0 ft

SFVERT

5 0: q2(14 ft) 5 q

1(8 ft)

(a) V and M at point B

SFVERT

5 0:

©MB 5 0:ÊMB 5 9000 lb-ft

VB 5 6000 lb

∴ q2 58

14 q1 5 2000 lb/ft

(b) V and M at midpoint E

SFVERT

5 0: Vm

5 (2000 lb/ft)(7 ft) 2 (3500 lb/ft)(4 ft)

SME

5 0:

Mm

5 (2000 lb/ft)(7 ft)(3.5 ft)

2 (3500 lb/ft)(4 ft)(2 ft)

Mm 5 21,000 lb-ft

Vm 5 0

A B C D

3.0 ft 3.0 ft

q2

q1 = 3500 lb/ft

8.0 ft

A B

3 ft2000 lb/ft VB

MB

A B E

4 ft

2000 lb/ft

3500 lb/ft

3 ft

Mm

Vm

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Problem 4.3-14 The simply-supported beam ABCD is loaded by

a weight W 5 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.)

Solution 4.3-14 Beam with cable and weight


A

E

DCB

W = 27 kN

2.0 m 2.0 m 2.0 m

Cable1.5 m

RA

5 18 kN RD

5 9 kN

Free-body diagram of pulley at B

A

E

DCB

27 kN

2.0 m 2.0 m 2.0 m

Cable1.5 m

RA RD

27 kN

21.6 kN

10.8 kN

27 kN

Free-body diagram of segment ABC of beam

©MC 5 0:ÊM 5 50.4 kN ? m

©FVERT 5 0:ÊV 5 7.2 kN

©FHORIZ 5 0:ÊN 5 21.6 kN (compression)

A

N

MCB21.6 kN

2.0 m 2.0 m

V

10.8 kN

18 kN

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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 5 2.0 wL at its end.

Derive formulas for the maximum shear force and maximum bending

moment in the arms, assuming b 5 L/9 and c 5 L/10.

Solution 4.3-15 Rotating centrifuge


b

c

L

W

x

W

y

a

b

c

L

x

Wg__ (L + b + c)a

waxg

__

Tangential acceleration 5 ra

Maximum V and M occur at x 5 b.

1w L2a

6g (2L 1 3b)

5Wa

g (L 1 b 1 c)(L 1 c)

1 #L1b

b

wa

g x(x 2 b)dx

Mmax 5Wa

g (L 1 b 1 c)(L 1 c)

1wLa

2g (L 1 2b)

5Wa

g (L 1 b 1 c)

Vmax 5W

g(L 1 b 1 c)a 1 #

L1b

b

wa

g x dx

Inertial force Mr a 5

Wg

ra

Substitute numerical data:

Mmax 5

229wL3a

75g

Vmax 5

91wL2a

30g

W 5 2.0 wLÊb 5L

9 c 5

L

10

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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).


SECTION 4.5 Shear-Force and Bending-Moment Diagrams 269

A B

L

P Pa a

A B

L

P Pa a

RA = P RB = P

P

]P

V

Pa

M

0

0

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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.



A B

L

a

M0

A B

L

a

M0

M0

L0

V

M

M0a

L0

] M0 (1]aL

)

RA =M0

LRB =

M0

L

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).


AB

q

L—2

L—2

AB

q

L—2

L—2

qL—2

V

MqL2

0

0

MA =3qL2

8

RA =qL

2

3qL2

8

82

2

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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

5 PL/4 at the free end.

Draw the shear-force and bending-moment diagrams for

this beam.



A B

P

L—2

L—2

M1 =PL—–4

MA

P

RAL/2 L/2

A BM1 5

PL4

MA 5 PL4

RA 5 P

V

M

0

0

2 PL4

PL4

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

5 PL /4 acting at the

third points.



A B

P

L—3

L—3

L—3

M1 =PL—–4

A B

P

L—3

L—3

L—3

M1 =PL—–

4

RA =5P—–12

RB =7P—–12

5P/12V

M

0

0

5PL/367PL/36

2PL/18

27P/12

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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.




A B

M1 2M1

L—3

L—3

L—3

A B

M1 2M1

L—3

L—3

L—3

RB =3M1—–

LRA =

3M1—–L

V 23M1—–

L

0

M0

M1

2M1 2M1

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.

Solution 4.5-7 Beam with bracket

A C

L

DE

P

B

L—4

L—4

L—2

A C

P

B

L—4

—4

3L

RA =P

—–2

RC =P

—–2

V

M

0

0

P—–2

PL—–

8

PL—–4

3PL—–8

P—–2

2

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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 Beam with overhang


A CB

a a a a

P P Pa

C

P P

Pa

a a a

P P

upperbeam:

B

P P

a a a

2P

lowerbeam:

C

V 0

M 0

P

2Pa

2P

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.



q

LL3

DAB C

L3

q

LL/3

–qL2/18 –qL2/18

qL/3

L/3DA

B C

__5qLRB =

6

__qL–

3__qL

–2

__5qLRC =

6

V

M

X1

__5qL2

72

0

0

__qL

2

x1 5 L Ï5

6 5 0.3727L

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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).



AB

L

q0

A

V

M

B

L

q0

x

__xq=q0L __q0L2

MB =6

__q0x3M = –

6L

__q0x2V = –

2L

__q0LRB = 2

__q0 L–

2

__q0L2–

6

0

0

Problem 4.5-11 The simple beam AB supports a uniform load of

intensity q 5 10 lb/in. acting over one-half of the span and a concentrated

load P 5 80 lb acting at midspan (see figure).



A B

q = 10 lb/in.

P = 80 lb

= 40 in.L—2

= 40 in.L—2

A B

10 lb/in.

P = 80 lb

40 in.

46 in.

6 in.

40 in.

60

RB = 340 lbRA =140 lb

140

–340

V

M

Mmax = 57805600

(lb)

(lb/in.)

0

0

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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.


Solution 4.5-12 Beam with distributed loads


0.8 m

3000 N/m

A B

0.8 m1.6 m

0.8 m

3000 N/m

A

V

M

B

0.8 m1.6 m

1500 N/m

1200

–1200960

480480

(N)

(N . m)

0

0

Problem 4.5-13 A cantilever beam AB supports a couple and a

concentrated load, as shown in the figure.



AB

5 ft 5 ft

200 lb

400 lb-ft

AB

5 ft 5 ft

200 lb

400 lb-ft

MA = 1600 lb-ft

RA = 200 lb

V

M

(lb)

+200

–600–1600

–1000

0

0

(lb-ft)

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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.




AB

2 m 2 m

2.5 kN2.0 kN/m

AB

2 m 2 m

2.5 kN2.0 kN/m

RA = 6.5 kN

MA = 14 kN . m

6.5

–14.0

–5.0

2.5V

M

(kN)

(kN . m)

0

0

Problem 4.5-15 The uniformly loaded beam ABC has simple supports at

A and B and an overhang BC (see figure).



A

V

M

CB

72 in.

25 lb/in.

48 in.

RA = 500 lb RB = 2500 lb

1200500

20 in.

–1300

–28,800

20 in.

40 in.

(lb)

(lb-in.)

0

0

5000

A CB

72 in.

25 lb/in.

48 in.

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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).




A CB

1.6 m 1.6 m 1.6 m

2.4 kN12 kN/m

A CB

1.6 m 1.6 m 1.6 m

2.4 kN

2.4

13.2

5.76

–3.84

–6.0

12 kN/m

V

M

Mmax = 7.26

RA = 13.2 kN RB = 8.4 kN

(kN . m)

0

0

1.1m

1.1m

0.64 m

Mmax

(kN)

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

5 400 lb acting at the end

of the vertical arm and a vertical force P2

5 900 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 Beam with vertical arm

4.0 ft 1.0 ft

BAC

P2 = 900 lb

P1 = 400 lb

1.0 ft

V(lb)

M(lb)

900

0

0

24002900

4.0 ft 1.0 ft

BAC

P2 = 900 lbP1 = 400 lb

1.0 ft

RA = 125 lb RB = 1025 lb

A400 lb-ft

125 lb

B900 lb

C

1025 lb

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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).




A B

4 kN/m8 kN4 kN/m

2 m2 m2 m 2 m

1 m

1 m

8 kN

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 center-

line dimensions when making calculations.)

Solution 4.5-19 Beam with a cable

A

E

C DB

Cable8 ft

1800 lb

6 ft 6 ft 6 ft

A B

4 kN/m4 kN/m

2 m2 m2 m 2 m

RA = 6 kN RB = 10 kN

V(kN)

M

(kN . m)

0

0

22.0

16 kN . m

1.5 m

1.5 m

6.0

210.0

4.54.0

16.012.0

Note: All forces have units of pounds.

A

E

C DB

Cable8 ft

1800 lb

6 ft 6 ft 6 ft

1800 lb

RD = 800 lb RD = 800 lb

Free-body diagram of beam ABCD

A C DB1800

1440 1800

1440

5760 lb-ft

8001080 720

800

V(lb)

M

(lb-ft)

640

00

24800

4800

28002800

2960

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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.


overhanging beam.



A D

1.2 m

4.2 m 4.2 m

5.1 kN/m5.1 kN/m

10.6 kN/m

B C

A D

1.2 m4.2 m 4.2 m

5.1 kN/m5.1 kN/m

10.6 kN/m

B C

RB = 39.33 kN RC = 39.33 kN

V(kN)

232.97

6.36

0

32.97

26.36

M0

(kN . m)

261.15 261.15

259.24

Problem 4.5-21 The simple beam AB shown in the figure supports a

concentrated load and a segment of uniform load.



AC

B

2.0 k/ft4.0 k

20 ft

10 ft5 ft

AC

B

2.0 k/ft4.0 k

10 ft5 ft 5 ftRA = 8 kRB = 16 k

Mmax = 64 k-ft

V(k)

216

M(k-ft)

0

0

84

8 ft

12 ft

8 ft12 ft

40

60 64

C

C

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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.


Problem 4.5-22 The cantilever beam shown in the figure supports

a concentrated load and a segment of uniform load.


cantilever beam.



AB

1.0 kN/m3 kN

1.6 m0.8 m 0.8 m

AB

1.0 kN/m3 kN

1.6 m0.8 m 0.8 m

RA = 4.6 kN

26.24

M(kN . m)

V(kN)

0

0

4.6

1.6

22.56

21.28

MA =

6.24 kN . m

Problem 4.5-23 The simple beam ACB shown in the figure is subjected

to a triangular load of maximum intensity 180 lb/ft.



BC

A

180 lb/ft

7.0 ft

6.0 ft

B

CA

180 lb/ft

1.0 ft6.0 ft

RA = 240 lb RB = 390 lb

Mmax = 640

V(lb)

2300

M(lb-ft)

0

0

240

x1 = 4.0 ft

2390

360

BA

3.0 kN/m

1.0 kN/m

2.4 m

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BA

3.0 kN/m

1.0 kN/m

2.4 m

RA = 2.0 kNRB = 2.8 kN

Set V 5 0: x1

5 1.2980 m

V 5 2.0 2 x 2x2

2.4Ê(x 5 meters; V 5 kN)

M

(kN . m)

22.8

2.0

0

0

Mmax = 1.450

x1 = 1.2980 m

x

V

(kN)

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.


condition.



A B

L

a

q

A B

a

q

RA = qL/2 RB = qL/2

(L 2 a)/2 (L 2 a)/2

M2

M1 M1

0M

The maximum bending moment is smallest when

M15 M

2 (numerically).

M1 5 M2 (L 2 a)2 5 L(2a 2 L)

M2 5 RA¢a2≤2

qL2

8 5

qL

8(2a 2 L)

M1 5q(L 2 a)2

8

0.2071L0.2071 qL

0.02145 qL2

0.2929L

2 0.2071 qL 2 0.2929 qL

V 0

M 0

2 0.02145 qL22 0.02145 qL2

x1 x1

0.2929 qL

x1 = 0.3536 a

= 0.2071 L

5qL2

8 (3 2 2Ï2) 5 0.02145qL2

M1 5 M2 5q

8 (L 2 a)2

Solve for a: a 5 (2 2 Ï2)L 5 0.5858L

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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.


compound beam.

Solution 4.5-26 Compound beam


A EB C D

4 kN

2 m2 m2 m2 m

1 m

2 kN1 m

A EB C D

4 kN

1 m1 m 1 m1 m2 m2 m2 m

2 kN

RA = 2.5 kN

21.0

M(kN . m)

4 kN . m Hinge

RC = 2.5 kN RE = 1 kN

V

(kN) 0

2.51.0

21.5 D

D1.0

5.0

22.0

02.67 m

1.0

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.


compound beam.

Solution 4.5-27 Compound beam

A EB

P

C D

2LL L L

q

A

V

M

EB

P

C D

2LL L L

q

PL

D

D

P

−P−qL

−qL2

–qL

L L

RC = P + 2qL RE = qLRB = 2P + qL

0

0

Hinge

qL

qL

2

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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 bending-

moment diagram, assuming that no couples act as loads on

the beam.

Solution 4.5-28 Simple beam (V is given)


1.0 m1.0 m2.0 m

12 kN

–12 kN

0

V

12

−1212

0

0

V

M

6.0 kN/m 12 kN

A B

2 m 1 m 1 m

(kN . m)

(kN)

RA = 12kN RB = 12kN

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 bending-

moment diagram.

Solution 4.5-29 Forces on a beam (V is given)

4 ft4 ft 16 ft

572 lb

–128 lb

0

V

652 lb

500 lb580 lb

–448 lb

14.50 ft

572

2448

–2160

–128

0

0

V

M

652

500580

–448

(lb)

(lb-ft)

4 ft4 ft 16 ft

20 lb/ft

652 lb 700 lb 1028 lb 500 lb

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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.

(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 bending-

moment diagram. (Assume P 5 10 kN, d 5 2.4 m, and L 5 12 m.)

Solution 4.5-30 Moving loads on a beam

(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.

(b) Maximum bending moment

By inspection, the maximum bending moment occurs

at point D, under the larger load 2P.

Vmax 5 RB 5 P ¢3 2d

L≤5 28 kN

x 5 L 2 d 5 9.6 m

Reaction at support B:

Bending moment at D:

Eq.(1)

Substitute x into Eq (1):

RB 5

P

2¢3 2

d

L≤ 5 14 kN

Note:ÊRA 5

P

2¢3 1

d

L≤ 5 16 kN

5

PL

12 ¢3 2

d

L≤

2

5 78.4 kN ? m

3 ¢L6≤ ¢3 2

5d

L≤1 2d(L 2 d)R

Mmax 5

P

LB2 3¢L

6≤

2

¢3 25d

L≤

2

1 (3L 2 5d)

Solve for x: x 5L

6 ¢3 2

5d

L≤5 4.0 m

dMD

dx 5

P

L (26x 1 3L 2 5d) 5 0

5P

L[23x2 1 (3L 2 5d)x 1 2d(L 2 d) ]

5

P

L (2d 1 3x)(L 2 x 2 d)

MD 5 RB(L 2 x 2 d)

RB 5P

L x 1

2P

L (x 1 d) 5

P

L (2d 1 3x)

L

BA

x d

P 2P

L

BA

x d

P 2P

BA

x = L − d

P 2P

RB = P(3 − )dL

RA = LPd

d

BA

L

P 2P

x dD

RB

64 Mmax = 78.4

2.4 m4.0 m 5.6 m

0

M

(kN . m)

P 5 10 kN

d 5 2.4 m

L 5 12 m

Shear Forces and Bending Moments - Vaftsy CAEvaftsycae.com/knowledge_files/SFD_BMD.pdfShear Forces...

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