bending moment

Shear Forces and Bending Moments

Problem 4.3-1 Calculate the shear force V and bending moment Mat 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.

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

Free-body diagram of segment DB

� 42,000 lb-in.

©MD � 0:�M � (1400 lb)(30 in.)

� 200 lb

©FVERT � 0:�V � 1600 lb � 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 Mat 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 m4.0 m

2.0 m

AC

B

2.0 kN/m6.0 kN

1.0 m 1.0 m 2.0 mRA RB

�MA � 0: RB � 4.5 kN

�MB � 0: RA � 5.5 kN

Free-body diagram of segment AC

©MC � 0:�M � 5.0 kN � m

©FVERT � 0:�V � �0.5 kN

A C

6.0 kN

1.0 m 1.0 m

RA

V M

Problem 4.3-3 Determine the shear force V and bending moment M atthe midpoint of the beam with overhangs (see figure). Note that one loadacts downward and the other upward.

Solution 4.3-3 Beam with overhangs

260 CHAPTER 4 Shear Forces and Bending Moments

PP

bb L

� P ¢1 �2b

L≤�(upward)

RA �1

L[P(L � b � b) ]

©MB � 0

Free-body diagram (C is the midpoint)

M �PL

2� Pb � Pb �

PL

2� 0

M � P ¢1 �2b

L≤ ¢L

2≤� P ¢b �

L

2≤

©MC � 0:

�2bP

LV � RA � P � P ¢1 �

2b

L≤� P

©FVERT � 0:

©MA � 0:�RB � P ¢1 �2b

L≤�(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 across section located 0.5 m from the fixed support of the cantilever beamAB 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.� �9.5 kN � m� �2.0 kN � m � 7.5 kN � m

� (1.5 kN�m)(2.0 m)(2.5 m) ©MD � 0:�M � �(4.0 kN)(0.5 m) � 4.0 kN � 3.0 kN � 7.0 kN

V � 4.0 kN � (1.5 kN�m)(2.0 m) ©FVERT � 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

Problem 4.3-5 Determine the shear force V and bending moment Mat 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

�MB � 0: RA � 2460 lb

�MA � 0: RB � 2740 lb

Free-body diagram of segment AD

Point D is 16 ft from support A.

� �4640 lb-ft

� (400 lb�ft) (10 ft) (11 ft)

©MD � 0:�M � (2460 lb)(16 ft)

� �1540 lb

V � 2460 lb � (400 lb�ft) (10 ft)

©FVERT � 0:

A CB

400 lb/ft 200 lb/ft

6 ft6 ft10 ft 10 ft

RA RB

AD

400 lb/ft

6 ft10 ftRA V

M

Problem 4.3-6 The beam ABC shown in the figure is simplysupported at A and B and has an overhang from B to C. Theloads 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 atthe 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

BA C

P2 = 8.0 kNP1 = 4.0 kN

1.0 m

4.0 m 1.0 m

BA

P2 = 8.0 kNP1 = 4.0 kN

1.0 m

RA RB

�MB � 0: RA � 1.0 kN (downward)

�MA � 0: RB � 9.0 kN (upward)

Free-body diagram of segment AD

Point D is 3.0 m from support A.

� �7.0 kN � m

©MD � 0:�M � �RA(3.0 m) � 4.0 kN � m

©FVERT � 0:�V � �RA � � 1.0 kN

3.0 m

A D

RA V

M4.0 kN • m

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 thebeam be zero?



q

bb L

DAB C

From symmetry and equilibrium of vertical forces:

RB � RC � q ¢b �L

2≤

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

Point E is at the midpoint of the beam.

Solve for b /L :

b

L�

1

2

�q ¢b �L

2≤ ¢L

2≤� q ¢1

2≤ ¢b �

L

2≤

2

� 0

�RB ¢L2 ≤� q ¢12≤ ¢b �

L

2≤

2

� 0

©ME � 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 thebowstring of the bow shown in the figure. Determine the bending momentat the midpoint of the bow.

350 mm

1400 mm

70°

Solution 4.3-8 Archer’s bow


P � 130 N

� � 70°

H � 1400 mm

� 1.4 m

b � 350 mm

� 0.35 m

Free-body diagram of point A

T � tensile force in the bowstring

�FHORIZ � 0: 2T cos �� P � 0

T �P

2 cos b

Free-body diagram of segment BC

Substitute numerical values:

M � 108 N � m

M �130 N

2B1.4 m

2� (0.35 m)(tan 70�)R

�P

2 ¢H

2� b tan b≤

M � T ¢H2

cosb � b sin b≤T(cos b)¢H

2≤ � T(sin b) (b) � M � 0

©MC � 0 � �

b

HP

A

�C

B

PA

T

�

T

H2

Cb

B

T

�

M

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 ofthe bar forms a semicircle of radius r.

Determine the axial force N, shear force V, and bending moment Macting at a cross section defined by the angle �.

Solution 4.3-9 Curved bar


PP P

C

B

A O

r

A

V

NM

��

M � Nr � Pr sin u

©MO � 0 � � M � Nr � 0

V � P cos u �FV � 0 �R a� V � P cos u � 0

N � P sin u

©FN � 0 Q� b��N � P sin u� 0

PP P

C

B

A O

r

A

V

NM

��O

P cos �

P sin �

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

�FVERT � 0 c� T�

(Minus means the shear force acts opposite to thedirection shown in the figure.)

V � �6040 N � �6.04 kN

�1

2 (900 N�m)(1.0 m) � 0

V �1

2(700 N�m)(2.6 m) � (900 N�m)(5.2 m)

Bending Moment

M � 788.67 N • m � 12,168 N • m � 2490 N • m

� 15,450 N • m

� 15.45 kN � m

�1

2(900 N�m)(1.0 m)¢5.2 m �

1.0 m

3≤� 0

� (900 N�m)(5.2 m)(2.6 m)

�M �1

2 (700 N�m)(2.6 m)¢2.6 m

3≤

©MA � 0 ��

A B

32

1700 N/m

900 N/m

Loading (in three parts)

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 thebeam 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 � 1200 lb

∴ 640 lb-ft �8P

15 lb-ft

M � �8P

15 lb-ft

M �4P

5(6 ft) �

4P

9 (12 ft) � 0

©MC � 0 ��

A

E P

C DB

Cable8 ft

6 ft 6 ft 6 ft

P

__9

__94P 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 trapezoidallydistributed 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 midpointof the beam.

BA

50 kN/m30 kN/m

3 m

Solution 4.3-12 Beam with trapezoidal load


Reactions

RA � 65 kN

RB � 55 kN

� 30 kN�m)(3 m) � 0RA � RB � 1�2 (50 kN�m

©FVERT � 0�

c

� (20 kN�m)(3 m)(1�2) (2 m) � 0

©MB � 0 � � RA(3 m) � (30 kN�m)(3 m)(1.5 m)

Free-body diagram of section CB

Point C is at the midpoint of the beam.

�FVERT � 0 c� T�

� 55 kN� 0

� M � (30 kN/m)(1.5 m)(0.75 m)

� (55 kN)(1.5 m) � 0

M � 45.0 kN � m

� 1�2(10 kN�m)(1.5 m)(0.5 m)

©MC � 0 ��

V � �2.5 kN

V � (30 kN�m)(1.5 m) � 12(10 kN�m)(1.5 m)

BA

50 kN/m30 kN/m

3 mRA RB

B

V

40 kN/m

30 kN/m

1.5 m55 kN

CM

Problem 4.3-13 Beam ABCD represents a reinforced-concretefoundation 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 isuniformly 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

�FVERT � 0: q2(14 ft) � q1(8 ft)

(a) V and M at point B

�FVERT � 0:

©MB � 0:�MB � 9000 lb-ft

VB � 6000 lb

∴ q2 �8

14 q1 � 2000 lb�ft

(b) V and M at midpoint E

�FVERT � 0: Vm � (2000 lb/ft)(7 ft) � (3500 lb/ft)(4 ft)

�ME � 0:

Mm � (2000 lb/ft)(7 ft)(3.5 ft)

� (3500 lb/ft)(4 ft)(2 ft)

Mm � 21,000 lb-ft

Vm � 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

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 usecenterline 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 � 18 kN RD � 9 kN

Free-body diagram of pulley at B

A

E

DCB

27 kN

2.0 m 2.0 m 2.0 m

Cable 1.5 m

RA RD

27 kN

21.6 kN

10.8 kN

27 kN

Free-body diagram of segment ABC of beam

©MC � 0:�M � 50.4 kN � m

©FVERT � 0:�V � 7.2 kN

©FHORIZ � 0:�N � 21.6 kN (compression)

A

N

MCB21.6 kN

2.0 m 2.0 m

V

10.8 kN

18 kN

Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontalplane (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 unitlength and supports a weight W � 2.0 wL at its end.

Derive formulas for the maximum shear force and maximum bendingmoment in the arms, assuming b � L/9 and c � L/10.

Solution 4.3-15 Rotating centrifuge


b

c

L

W

x

W

y

�

b

c

L

x

Wg__ (L + b + c)�

w�xg

__

Tangential acceleration � r�

Maximum V and M occur at x � b.

�w L2�

6g (2L � 3b)

�W�

g (L � b � c)(L � c)

� �L�b

b

w�

g x(x � b)dx

Mmax �W�

g (L � b � c)(L � c)

�wL�

2g (L � 2b)

�W�

g (L � b � c)

Vmax �Wg

(L � b � c)� � �L�b

b

w�

g x dx

Inertial force Mr � �

Wg r�

Substitute numerical data:

Mmax �

229wL3�

75g

Vmax �

91wL2�

30g

W � 2.0 wL�b �L

9 c �

L

10

Shear-Force and Bending-Moment Diagrams

When solving the problems for Section 4.5, draw the shear-force andbending-moment diagrams approximately to scale and label all criticalordinates, including the maximum and minimum values.

Probs. 4.5-1 through 4.5-10 are symbolic problems and Probs. 4.5-11through 4.5-24 are numerical problems. The remaining problems (4.5-25through 4.5-30) involve specialized topics, such as optimization, beamswith 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

Problem 4.5-2 A simple beam AB is subjected to a counterclockwisecouple 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 L

0

V

M

M0a L

0� M0 (1� a

L)

RA =M0L RB =

M0L

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 =qL2

3qL2

8

8�

�

Problem 4.5-4 The cantilever beam AB shown in the figure is subjected to a concentrated load P at the midpoint and acounterclockwise couple of moment M1 � PL/4 at the free end.




A B

P

L—2

L—2

M1 =PL—–4

MA

P

RAL/2 L/2

A BM1 �

PL4

MA � PL4

RA � P

V

M

0

0

� 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 � PL /4 acting at thethird 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

�PL/18

�7P/12

Problem 4.5-6 A simple beam AB subjected to clockwise couples M1and 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 �3M1—–

L

0

M 0

M1

�M1 �M1

Problem 4.5-7 A simply supported beam ABC is loaded by a verticalload 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

�

Problem 4.5-8 A beam ABC is simply supported at A and B andhas an overhang BC (see figure). The beam is loaded by two forcesP and a clockwise couple of moment Pa that act through thearrangement 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

CP P

Paa a a

P P

upperbeam:

BP P

a a a

2P

lowerbeam:

C

V 0

M 0

P

�Pa

�P

Problem 4.5-9 Beam ABCD is simply supported at B and C and hasoverhangs at each end (see figure). The span length is L and eachoverhang has length L /3. A uniform load of intensity q acts along theentire 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

MX1

__5qL2

72

0

0

__qL2

x1 � L �5

6 � 0.3727L

Problem 4.5-10 Draw the shear-force and bending-moment diagramsfor a cantilever beam AB supporting a linearly varying load of maximumintensity q0 (see figure).



AB

L

q0

A

V

M

B

L

q0

x

__xq=q0 L __q0L2MB = 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 ofintensity q � 10 lb/in. acting over one-half of the span and a concentratedload P � 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

Problem 4.5-12 The beam AB shown in the figure supports a uniformload of intensity 3000 N/m acting over half the length of the beam. Thebeam rests on a foundation that produces a uniformly distributed loadover 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 aconcentrated 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)

Problem 4.5-14 The cantilever beam AB shown in the figure issubjected to a uniform load acting throughout one-half of its length and aconcentrated 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

05000

A CB

72 in.

25 lb/in.

48 in.

Problem 4.5-16 A beam ABC with an overhang at one end supports auniform load of intensity 12 kN/m and a concentrated load of magnitude2.4 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 � 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 Beam with vertical arm

4.0 ft 1.0 ft

BA C

P2 = 900 lbP1 = 400 lb

1.0 ft

V(lb)

M(lb)

900

0

0

�400�900

4.0 ft 1.0 ft

BA C

P2 = 900 lbP1 = 400 lb

1.0 ft

RA = 125 lb RB = 1025 lb

A400 lb-ft

125 lb

B900 lb

C

1025 lb

�125

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 verticalarm (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 asimple beam at A and D (see figure). A cable passes over a small pulleythat is attached to the arm at E. One end of the cable is attached to thebeam 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 mRA = 6 kN RB = 10 kN

V(kN)

M

(kN . m)

0

0

�2.0

16 kN . m

1.5 m

1.5 m

6.0

�10.0

4.5 4.0

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 18001440

5760 lb-ft

8001080 720

800

V(lb)

M(lb-ft)

640

0 0

�4800

4800

�800�800

�960

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.



A D

1.2 m4.2 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)

�32.97

6.36

0

32.97

�6.36

M 0(kN . m)

�61.15 �61.15

�59.24

Problem 4.5-21 The simple beam AB shown in the figure supports aconcentrated load and a segment of uniform load.



AC

B

2.0 k/ft4.0 k

20 ft10 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)

�16

M(k-ft)

0

0

84

8 ft

12 ft

8 ft12 ft

40

60 64

C

C

Problem 4.5-24 A beam with simple supports is subjected to atrapezoidally distributed load (see figure). The intensity of the load variesfrom 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.

Draw the shear-force and bending-moment diagrams for thiscantilever 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

�6.24

M(kN . m)

V(kN)

0

0

4.6

1.6

�2.56�1.28

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)

�300

M(lb-ft)

0

0

240

x1 = 4.0 ft

�390

360

BA

3.0 kN/m1.0 kN/m

2.4 m


BA

3.0 kN/m1.0 kN/m

2.4 mRA = 2.0 kN

RB = 2.8 kN

Set V � 0: x1 � 1.2980 m

V � 2.0 � x �x2

2.4�(x � meters; V � kN)

M

(kN . m)

�2.8

2.0

0

0

Mmax = 1.450

x1 = 1.2980 mx

V(kN)

Problem 4.5-25 A beam of length L is being designed to support a uniform loadof 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 middleof the beam (as pictured), the maximum bending moment is reduced.

Determine the distance a between the supports so that the maximum bendingmoment in the beam has the smallest possible numerical value.

Draw the shear-force and bending-moment diagrams for thiscondition.



A B

L

a

q

A B

a

q

RA = qL/2 RB = qL/2

(L � a)/2 (L � a)/2

M2

M1 M1

0M

The maximum bending moment is smallest whenM1� M2 (numerically).

M1 � M2 (L � a)2 � L(2a � L)

M2 � RA¢a2≤�qL2

8 �

qL

8(2a � L)

M1 �q(L � a)2

8

0.2071L0.2071 qL

0.02145 qL2

0.2929L

� 0.2071 qL � 0.2929 qL

V 0

M 0

� 0.02145 qL2 � 0.02145 qL2

x1 x1

0.2929 qL

x1 = 0.3536 a

= 0.2071 L

�qL2

8 (3 � 2�2) � 0.02145qL2

M1 � M2 �q

8 (L � a)2

Solve for a: a � (2 � �2)L � 0.5858L

Problem 4.5-26 The compound beam ABCDE shown in the figureconsists 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. Theloads on the beam consist of a 4-kN force at the end of a bracket attachedat point B and a 2-kN force at the midpoint of beam DE.

Draw the shear-force and bending-moment diagrams for thiscompound 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

�1.0

M(kN . m)

4 kN . m Hinge

RC = 2.5 kN RE = 1 kN

V(kN) 0

2.51.0

�1.5 D

D1.0

5.0

�2.0

02.67 m

1.0

Problem 4.5-27 The compound beam ABCDE shown in the figureconsists 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 Pacts upward at A and a uniform load of intensity q acts downward onbeam DE.

Draw the shear-force and bending-moment diagrams for thiscompound 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

qL2

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

Force diagram


Problem 4.5-30 A simple beam AB supports two connected wheel loads Pand 2P that are distance d apart (see figure). The wheels may be placed atany distance x from the left-hand support of the beam.

(a) Determine the distance x that will produce the maximum shear forcein the beam, and also determine the maximum shear force Vmax.

(b) Determine the distance x that will produce the maximum bendingmoment in the beam, and also draw the corresponding bending-moment diagram. (Assume P � 10 kN, d � 2.4 m, and L � 12 m.)

Solution 4.5-30 Moving loads on a beam

(a) Maximum shear forceBy inspection, the maximum shear force occurs atsupport B when the larger load is placed close to, butnot directly over, that support.

(b) Maximum bending moment

By inspection, the maximum bending moment occursat point D, under the larger load 2P.

Vmax � RB � P ¢3 �d

L≤� 28 kN

x � L � d � 9.6 m

Reaction at support B:

Bending moment at D:

Eq.(1)

Substitute x into Eq (1):

RB �

P

2¢3 �

d

L≤ � 14 kN

Note:�RA �

P

2¢3 �

d

L≤ � 16 kN

�

PL

12 ¢3 �

d

L≤

2

� 78.4 kN � m

� ¢L6≤ ¢3 �

5d

L≤� 2d(L � d)R

Mmax �

P

LB� 3¢L

6≤

2

¢3 �5d

L≤

2

� (3L � 5d)

Solve for x: x �L

6 ¢3 �

5d

L≤� 4.0 m

dMD

dx �

P

L (�6x � 3L � 5d) � 0

�P

L[�3x2 � (3L � 5d)x � 2d(L � d) ]

�

P

L (2d � 3x)(L � x � d)

MD � RB(L � x � d)

RB �P

L x �

2P

L (x � d) �

P

L (2d � 3x)

L

BA

x d

P 2P

L

BA

x d

P 2P

BA

x = L − dP 2P

RB = P(3 − )dLRA = L

Pd

d

BA

L

P 2P

x dD

RB

64 Mmax = 78.4

2.4 m4.0 m 5.6 m0

M(kN . m)

P � 10 kNd � 2.4 mL � 12 m

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bending moment