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Shear Forces and Bending Moments

Problem 4.3-1 Calculate the shear force Vand bending momentMat a cross section just to the left of the 1600-lb load acting on the simple

beamAB 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 segmentDB

42,000 lb-in.

MD 0:M (1400 lb)(30 in.) 200 lbFVERT 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 Vand bending momentMat the midpoint Cof the simple beamAB 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

MA

0: RB

4.5kN

MB

0: RA

5.5kN

Free-body diagram of segmentAC

MC 0:M 5.0 kN mFVERT 0:V 1.5 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 Vand bending momentMatthe 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

P1 2bL

(upward)RA

1

L[P(L b b) ]

MB 0

Free-body diagram (Cis the midpoint)

MPL

2 Pb Pb

PL

2 0

M P1 2bL

L2 Pb L

2

MC 0:

2bP

LVRA P P1 2b

L P

FVERT

0:

MA 0:RB P1 2bL

(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 Vand bending momentMat 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

A

B

1.5 kN/m4.0 kN

1.0 m1.0 m 2.0 m

Free-body diagram of segmentDB

PointD is 0.5m from supportA. 9.5

kN m

2.0 kN m 7.5 kN m

(1.5 kNm)(2.0 m)(2.5 m)MD 0:M (4.0 kN)(0.5 m) 4.0 kN 3.0 kN 7.0 kNV 4.0 kN (1.5 kNm)(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

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Problem 4.3-5 Determine the shear force Vand bending momentMat 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 segmentAD

PointD is 16 ft from supportA.

4640 lb-ft

(400 lbft)(10 ft)(11 ft)MD 0:M (2460 lb)(16 ft) 1540 lbV 2460 lb (400 lbft)(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 ft

RA V

M

Problem 4.3-6 The beamABCshown in the figure is simplysupported atA andB and has an overhang fromB 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

the end of the overhang.

Determine the shear force Vand bending momentMat

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 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

MB

0: RA

1.0kN (downward)

MA

0: RB

9.0kN (upward)

Free-body diagram of segmentAD

PointD is 3.0m from supportA.

7.0 kN m

MD 0:M RA(3.0 m) 4.0 kN m FVERT 0:V RA 1.0 kN3.0 m

A D

RA V

M

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Problem 4.3-7 The beamABCD shown in the figure has overhangsat 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 RC qb L2

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

PointEis at the midpoint of the beam.

Solve for b/L :

b

L

1

2

qb L2L

2 q1

2b L

2

2

0

RBL2 q1

2b 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 moment

at the midpoint of the bow.

350 mm

1400 mm

70

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Solution 4.3-8 Archers bow


P 130N

70

H 1400 mm

1.4m

b 350 mm

0.35m

Free-body diagram of pointA

T tensile force in the bowstring

FHORIZ

0: 2Tcos P 0

T

P

2 cosb

Free-body diagram of segmentBC

Substitute numerical values:

M 108 N m

M130 N

2B1.4 m

2 (0.35 m)(tan 70)R

P

2H

2 b tan b

M TH2

cosb b sin bT(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

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Problem 4.3-9 A curved barABCis subjected to loads in the formof 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 forceN, shear force V, and bending momentM

acting at a cross section defined by the angle .

Solution 4.3-9 Curved bar


PP P

C

B

A O

r

A

V

NM

MNr Prsin u

MO 0 MNr 0

V P cos u

FV 0 b ! V P cos u 0

N P sin u

FN 0 Q bN P sin u 0

PP P

C

B

A O

r

A

V

NM

O

Pcos

Psin

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 Vand bending momentMat 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 the

direction shown in the figure.)

V 6040 N 6.04 kN

1

2(900 Nm)(1.0 m) 0

V1

2(700 Nm)(2.6 m) (900 Nm)(5.2 m)

Bending Moment

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

15,450N m

15.45 kN m

1

2(900 Nm)(1.0 m)5.2 m 1.0 m

3 0

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

M1

2(700 Nm)(2.6 m)2.6 m

3

MA 0

A B

32

1700 N/m

900 N/m

Loading (in three parts)

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Problem 4.3-11 A beamABCD with a vertical arm CEis supported asa simple beam atA andD (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 pointB.

What is the force P in the cable if the bending moment in the

beam just to the left of point Cis equal numerically to 640 lb-ft?

(Note: Disregard the widths of the beam and vertical arm and usecenterline 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

Min lb-ft

Free-body diagram of sectionAC

Numerical value ofMequals 640 lb-ft.

and P 1200 lb

640 lb-ft 8P

15lb-ft

M 8P

15lb-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

__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 beamAB supports a trapezoidallydistributed load (see figure). The intensity of the load varies linearly

from 50 kN/m at supportA to 30 kN/m at supportB.

Calculate the shear force Vand bending momentMat 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

65kN

RB

55kN

30 kNm)(3 m) 0RA RB 12 (50 kNm

FVERT

0

c

(20 kNm)(3 m) (12) (2 m) 0

MB 0 RA(3 m) (30 kNm)(3 m)(1.5 m)

Free-body diagram of section CB

Point Cis at the midpoint of the beam.

FVERT

0 c

T

55kN 0

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

(55kN)(1.5 m) 0

M 45.0 kN m

12(10 kNm)(1.5 m)(0.5 m)

MC 0

V 2.5 kN

V (30 kNm)(1.5 m) 12(10 kNm)(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 BeamABCD represents a reinforced-concretefoundation beam that supports a uniform load of intensity q

1 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 momentMB

at pointB.

(b) Find the shear force Vm

and bending momentMm

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(14ft) q

1(8ft)

(a) VandMat pointB

FVERT

0:

MB 0:MB 9000 lb-ftVB 6000 lb

q2 8

14 q1 2000 lbft

(b) VandMat midpointE

FVERT

0: Vm

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

ME

0:

Mm

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

(3500 lb/ft)(4ft)(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

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Problem 4.3-14 The simply-supported beamABCD is loaded bya weight W 27 kN through the arrangement shown in the figure.

The cable passes over a small frictionless pulley atB and is attached

atEto the end of the vertical arm.

Calculate the axial forceN, 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

18kN RD

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 segmentABCof beam

MC 0:M 50.4 kN mFVERT 0:V 7.2 kNFHORIZ 0:N 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 horizontalplane (thexy 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 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.

Solution 4.3-15 Rotating centrifuge


b

c

L

W

x

W

y

b

c

L

x

Wg__ (L + b + c)

wxg__

Tangential acceleration r

Maximum VandMoccur atx b.

wL2

6g(2L 3b)

W

g (L b c) (L c)

(L c) Lb

b

w

gx(x b)dx

Mmax W

g(L b c)

wL

2g(L 2b)

W

g(L b c)

Vmax W

g(L b c)

Lb

b

w

gxdx

Inertial forceMr Wg

r

Substitute numerical data:

Mmax 229wL3

75g

Vmax 91wL2

30g

W 2.0 wLb L9

c 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 fora simple beamAB 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 beamAB is subjected to a counterclockwisecouple of momentM

0acting 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

0V

M

M0a

L0

M0(1aL

)

RA =M0L

RB =M0L

Problem 4.5-3 Draw the shear-force and bending-moment diagrams

for a cantilever beamAB carrying a uniform load of intensity q over

one-half of its length (see figure).


AB

q

L2

L2

AB

q

L2

L2

qL2

V

M qL2

0

0

MA =3qL2

8

RA =qL

2

3qL2

8

8

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Problem 4.5-4 The cantilever beamAB shown in the figureis subjected to a concentrated load P at the midpoint and a

counterclockwise couple of momentM1

PL/4 at the free end.

Draw the shear-force and bending-moment diagrams for

this beam.



A B

P

L2

L2

M1=PL4

MA

P

RAL/2 L/2

A BM1

PL4

MAPL4

RA P

V

M

0

0

PL4

PL4

P

Problem 4.5-5 The simple beamAB shown in the figure is subjected toa concentrated load P and a clockwise coupleM

1 PL/4 acting at the

third points.



A B

P

L3

L3

L3

M1=PL4

A B

P

L3

L3

L3

M1=PL4

RA=5P12

RB=7P12

5P/12V

M

0

0

5PL/367PL/36

PL/18

7P/12

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Problem 4.5-6 A simple beamAB subjected to clockwise couplesM1

and 2M1

acting at the third points is shown in the figure.




A B

M1 2M1

L3

L3

L3

A B

M1 2M1

L3

L3

L3

RB=3M1L

RA=3M1L

V 3M1L

0

M0

M1

M1 M1

Problem 4.5-7 A simply supported beamABCis loaded by a vertical

load P acting at the end of a bracketBDE(see figure).

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

Solution 4.5-7 Beam with bracket

A C

L

DE

P

B

L

4

L

4

L

2

A C

P

B

L4

2

3L

RA=P

2

RC=P2

V

M

0

0

P2

PL8

3PL8

P2

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Problem 4.5-8 A beamABCis simply supported atA andB andhas an overhangBC(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

beamABC.

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

P 2P

lowerbeam:

C

V 0

M 0

P

Pa

P

Problem 4.5-9 BeamABCD is simply supported atB and Cand hasoverhangs at each end (see figure). The span length is L and each

overhang has lengthL/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

6__

qL

2

__

5qLRC=6

V

MX1

__

5qL2

72

0

0

__

qL

2

__

qL

2

x1 L 5

6 0.3727L

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Problem 4.5-10 Draw the shear-force and bending-moment diagramsfor a cantilever beamAB supporting a linearly varying load of maximum

intensity q0

(see figure).



AB

L

q0

A

V

M

B

L

q0

x

__

xq=q0

L__

q0L2

MB =6

__

q0x3M =6L

__

q0x2

V =2L

__

q0LRB = 2

__

q0L2

__

q0L26

0

0

Problem 4.5-11 The simple beamAB supports a uniform load ofintensity q 10 lb/in. acting over one-half of the span and a concentrated

load 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 = 340lbRA =140lb

140

340

V

M

Mmax = 57805600

(lb)

(lb/in.)

0

0

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Problem 4.5-12 The beamAB shown in the figure supports a uniformload 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 beamAB 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 = 200lb

V

M

(lb)

+200

6001600

1000

0

0

(lb-ft.)

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Problem 4.5-14 The cantilever beamAB shown in the figure issubjected 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

MA = 14 kN . m

kN

6.5

14.0

5.0

2.5V

M

(kN)

(kN . m)

0

0

Problem 4.5-15 The uniformly loaded beamABChas simple supports atA andB and an overhangBC(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

A CB

72 in.

25 lb/in.

48 in.

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Problem 4.5-16 A beamABCwith an overhang at one end supports auniform 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 = 13.2kN RB = 8.4kN

kN . m

0

0

1.1m

1.1m

.64m

Mmax

(kN)

Problem 4.5-17 The beamABCshown in the figure is simplysupported atA andB and has an overhang fromB 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

BAC

P2 = 900 lb

P1 = 400 lb

1.0 ft

V(lb)

M(lb)

900

0

0

400900

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

125

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Problem 4.5-18 A simple beamAB is loaded by two segments ofuniform 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 beamABCD with a vertical arm CEis supported as asimple beam atA andD (see figure). A cable passes over a small pulley

that is attached to the arm atE. One end of the cable is attached to the

beam at pointB. The tensile force in the cable is 1800 lb.

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

(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

4.54.0

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

RA = 800 lb RB = 800 lb

Free-body diagram of beamABCD

A C DB1800

1440 1800

1440

5760 lb-ft

8001080 720

800

V(lb)

M

(lb-ft)

640

00

4800

4800

800800

960

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Problem 4.5-20 The beamABCD shown in the figure hasoverhangs that extend in both directions for a distance of 4.2 m

from the supports atB and C, which are 1.2 m apart.


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

V0

(kN . m)

61.15 61.15

59.24

Problem 4.5-21 The simple beamAB 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)

16

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 supportA to 3.0 kN/m at supportB.


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

6.24

M(kN . m)

V(kN)

0

0

4.6

1.6

2.561.28

Problem 4.5-23 The simple beamACB shown in the figure is subjectedto 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/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 0: x1

1.2980m

V 2.0 xx2

2.4(x meters; V kN)

M

(kN . m)

2.8

2.0

0

0

Mmax= 1.450

x1= 1.2980 m

x

V

(kN)

Problem 4.5-25 A beam of lengthL 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 a)/2 (L a)/2

M2

M1 M1

0M

The maximum bending moment is smallest when

M1M

2(numerically).

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

M2 RAa2 qL

2

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

qL2

8(3 2 2) 0.02145qL2

M1 M2 q

8

(L a)2

Solve for a: a (2 2)L 0.5858L

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Problem 4.5-26 The compound beamABCDEshown in the figureconsists of two beams (AD andDE) joined by a hinged connection atD.

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 pointB and a 2-kN force at the midpoint of beamDE.


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

1.0

M(kN . m)

4 kN . m Hinge

RC= 2.5 kN RE= 1 kN

V

(kN) 0

2.5 1.0

1.5 D

D1.0

5.0

2.0

02.67 m

1.0

Problem 4.5-27 The compound beamABCDEshown in the figure

consists of two beams (AD andDE) joined by a hinged connection atD.

The hinge can transmit a shear force but not a bending moment. A force Pacts upward atA and a uniform load of intensity q acts downward on

beamDE.


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

PqL

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 beamis 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 (Vis 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 (Vis 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

V

652

500580

448

(lb)

(lb/ft)

4 ft4 ft 16 ft

20 lb/ft

652 lb 700 lb 1028 lb 500 lb

Force diagram

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Problem 4.5-30 A simple beamAB supports two connected wheel loads Pand 2P that are distance dapart (see figure). The wheels may be placed at

any distancexfrom the left-hand support of the beam.

(a) Determine the distancexthat will produce the maximum shear force

in the beam, and also determine the maximum shear force Vmax

.

(b) Determine the distancexthat will produce the maximum bending

moment in the beam, and also draw the corresponding bending-moment diagram. (Assume P 10 kN, d 2.4 m, andL 12 m.)

Solution 4.5-30 Moving loads on a beam

(a) Maximum shear force

By inspection, the maximum shear force occurs at

supportB 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 pointD, under the larger load 2P.

Vmax RB P3 dL

28 kNxL d 9.6 m

Reaction at supportB:

Bending moment atD:

Eq.(1)

Substitutexinto Eq (1):

RB P

23 d

L 14 kN

Note:RA P23 d

L 16 kN

PL

123 d

L

2

78.4 kN m

L63 5d

L 2d(L d)R

Mmax P

LB 3L

6

2

3 5dL

2

(3L 5d)

Solve for x: xL

63 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

Lx

2P

L(x d)

P

L(2d 3x)

L

BA

x d

P 2P

L

BA

x d

P 2P

BA

x = L d

P 2P

RB = P(3

)

d

LRA = L

Pd

d

BA

L

P 2P

x dD

RB

64 Mmax = 78.4

2.4m4.0m 5.6m

0

M

(kN . m)

P 10kN

d 2.4m

L 12 m

bascis

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