eighth edition - apl100apl100.wdfiles.com/local--files/...beer_chapter_07mod-updated.pdf · vector...

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

7 Forces in Beams

and Cables


Vector Mechanics for Engineers: Statics

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Contents

Introduction

Internal Forces in Members

Sample Problem 7.1

Various Types of Beam Loading

and Support

Shear and Bending Moment in a

Beam

Sample Problem 7.2

Sample Problem 7.3

Relations Among Load, Shear,

and Bending Moment



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Contents

Sample Problem 7.4

Sample Problem 7.6

Cables With

Concentrated Loads

Cables With

Distributed Loads

Parabolic Cable

Sample Problem 7.8

Catenary



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Introduction

•Preceding chapters dealt with:

a) determining external forces acting on

a structure and

b)determining forces which hold together

the various members of a structure.



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Introduction

•The current chapter is concerned with

determining the internal forces (i.e.,

tension/compression, shear, and

bending) which hold together the

various parts of a given member.



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•Focus is on two important types of

engineering structures:

a)Beams - usually long, straight,

prismatic members designed to

support loads applied at various points

along the member.

b)Cables - flexible members capable of

withstanding only tension, designed to

support concentrated or distributed

loads.



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•Straight two-force

member AB is in

equilibrium under

application of F and

-F. •Internal forces

equivalent to F and -F

are required for

equilibrium of free-

bodies AC and CB.



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•Multiforce member

ABCD is in equili-

brium under

application of cable

and member contact

forces.



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•Internal forces

equivalent to a

force-couple

system are

necessary for

equili-brium of

free-bodies JD

and ABCJ.



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•An internal force-couple system is

required for equilibrium of two-

force members which are not

straight.



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Beams

7- 11



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4 - 12

Pure Bending

Pure Bending: Prismatic members subjected

to equal and opposite couples acting in the

same longitudinal plane



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4 - 13



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7- 14



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7- 15



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Post and Lintel

7- 16



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Parthenon, Athens

7- 17



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Corbelled domes and vaults

7- 18



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Various Types of Beam Loading and Support

•Beam - structural member

designed to support loads

applied at various points

along its length.

•Beam can be subjected

to concentrated loads or

distributed loads or

combination of both.



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• Beam design is two-step

process:

1)determine shearing forces

and bending moments

produced by applied loads

2)select cross-section best

suited to resist shearing

forces and bending

moments



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•Beams are classified according to way in

which they are supported.

•Reactions at beam supports are

determinate if they involve only three

unknowns. Otherwise, they are statically

indeterminate.



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Shear and Bending Moment in a Beam

• Wish to determine bending moment and

shearing force at any point in a beam

subjected to concentrated and distributed

loads.



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• Determine reactions at supports by

treating whole beam as free-body.



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• From equilibrium considerations,

determine M and V or M’ and V’.



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• Cut beam at C and draw free-body diagrams for AC

and CB. By definition, positive sense for internal

force-couple systems are as shown. This way we

don’t have to specify from which section ( AC or BC)

we are calculating V and M.



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Shear and Bending Moment Diagrams

• Variation of shear and bending moment along

beam may be plotted. • Determine reactions at supports.



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7- 30



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7- 31

22 PxMPV



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22 xLPMPV



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• Cut beam at E and consider member EB, 22 xLPMPV

• Cut beam at C and consider member AC,

22 PxMPV



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• For a beam subjected to concentrated loads,

shear is constant between loading points and

moment varies linearly.



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Sample Problem 7.2

Draw the shear and bending moment

diagrams for the beam and loading shown.



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• Taking entire beam as a free-body, calculate

reactions at B and D.

• Find equivalent internal force-couple systems for

free-bodies formed by cutting beam on either side

of load application points.

• Plot results.



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:0yF

0kN20 1V kN201V

:01M

0m0kN20 1M 01M



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Sample Problem 7.2



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Sample Problem 7.2

7- 42



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0kN14

mkN28kN14

mkN28kN26

mkN50kN26

66

55

44

33

MV

MV

MV

MV



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Sample Problem 7.2

•Plot results.

Note that shear is

of constant value

between

concentrated

loads and bending

moment varies

linearly.



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UDL (Uniformly distributed load)

7- 45



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Sample Problem 7.3

7- 53

Draw the shear and bending moment

diagrams for the beam AB. The

distributed load of 7200 N/m extends over

0.3 m of the beam, from A to C, and the

1800-N load is applied at E.



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Sample Problem 7.3

7- 54



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Sample Problem 7.3

7- 55

SOLUTION:

•Taking entire beam as free-body,

calculate reactions at A and B.

•Determine equivalent internal force-

couple systems at sections cut within

segments AC, CD, and DB.

•Plot results.



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Sample Problem 7.3

7- 56

SOLUTION:

• Taking entire beam as a free-body, calculate

reactions at A and B.

:0AM

0m550N1800m150N2160m80 ...By

N51642 .By

:0BM

0m8.0N25.0N1800m65.0N2160 A

N52317.A

:0xF 0xB

• Note: The 1800-N load at E may be replaced by

a 1800-N force and 180-N . m couple at D.



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Sample Problem 7.3

7- 57

• Evaluate equivalent internal force-

couple systems at sections cut

within segments AC, CD, and DB.



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:01M 072005231721 Mxxx.

m.N )360052317( 2xx.M

From A to C:

:0yF 0720052317 Vx.

N)720052317( x.V



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Sample Problem 7.3

7- 61

:02M 0150216052317 M.xx.

mN 5157324 x.M

From C to D:

:0yF 0216052317 V.

N 5157.V



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Sample Problem 7.3

7- 63

:03M

04501800

180150216052317

M.x

.xx.

mN 516421314 x.M

From D to B:

:0yF

01800216052317 V.

N 51642.V



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Sample Problem 7.3

7- 64



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Sample Problem 7.3

7- 65

• Plot results.

From A to C:

From C to D:

From D to B:

N)720052317( x.V

m.N )360052317( 2xx.M

N 5157.V

mN 5157324 x.M

N 51642.V

mN 516421314 x.M



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Relations Among Load, Shear, and Bending Moment



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Relation between load and shear

wx

V

dx

dV

xwVVV

x 0lim

0

curve loadunder areaD

C

x

xCD dxwVV



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Relation between Shear, and Bending Moment

VxwVx

M

dx

dM

xxwxVMMM

xx 21

00limlim

02

curveshear under areaD

C

x

xCD dxVMM



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•Reactions at supports,

2

wLRR BA



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•Reactions at supports,

2

wLRR BA

•Shear curve,

xL

wwxwL

wxVV

wxdxwVV

A

x

A

22

0



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•Moment curve,

0at 8

22

2

max

2

0

0

Vdx

dMM

wLM

xxLw

dxxL

wM

VdxMM

x

x

A



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Sample Problem 7.4

Draw the shear and bending-

moment diagrams for the beam

and loading shown.



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Sample Problem 7.4

SOLUTION:

•Taking entire beam as a

free-body, determine

reactions at supports.

•Between concentrated load

application points,

and shear is constant.

0wdxdV



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Sample Problem 7.4

•With uniform loading

between D and E, the

shear variation is

linear.



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Sample Problem 7.4

•Between concentrated load application

points, The

change in moment between load

application points is equal to area under

shear curve between points.

const.VdxdM



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Sample Problem 7.4

•With a linear shear variation

between D and E, the bending

moment diagram is a

parabola.



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SOLUTION:

Taking entire beam as a free-body,

determine reactions at supports.



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Sample Problem 7.4

:0AM

0m .48kN 12

m .24kN 12m .81kN 20m .27D

kN 26D

:0yF

0kN 12kN 26kN 12kN 20yA

kN 18yA



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Sample Problem 7.4

•Between concentrated

load application points,

and shear is constant.

0wdxdV

•With uniform loading

between D and E, the

shear variation is linear.



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Sample Problem 7.4

•Between concentrated load application

points,

•The change in moment between load

application points is equal to area under

the shear curve between points.

.constVdxdM



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048

mkN 48140

mkN 9216

mkN 108108

EDE

DCD

CBC

BAB

MMM

MMM

MMM

MMM



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Sample Problem 7.4

•With a linear shear variation between D

and E, the bending moment diagram is a

parabola.



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Sample Problem 7.4

•With a linear shear variation between

D and E, the bending moment diagram

is a parabola.



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Sample Problem 7.6

Sketch the shear

and bending-

moment

diagrams for the

cantilever beam

and loading

shown.



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Sample Problem 7.6

SOLUTION:

The change in shear

between A and B is equal

to the negative of area

under load curve between

points. The linear load

curve results in a

parabolic shear curve.



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Sample Problem 7.6

•With zero load, change

in shear between B

and C is zero.

•The change in moment between A and B

is equal to area under shear curve

between points. The parabolic shear

curve results in a cubic moment curve.



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Sample Problem 7.6

•The change in moment

between B and C is

equal to area under

shear curve between

points. The constant

shear curve results in a

linear moment curve.



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Sample Problem 7.6

SOLUTION:

•The change in shear

between A and B is equal

to negative of area under

load curve between

points. The linear load

curve results in a

parabolic shear curve.



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Sample Problem 7.6

•With zero load, change in shear between

B and C is zero.

awVV AB 021 awVB 02

1

0,at wdx

dVB

0,0,at wwdx

dVVA A



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Sample Problem 7.6


between A and B is

equal to area under

shear curve between

the points. The

parabolic shear curve

results in a cubic

moment curve.



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Sample Problem 7.6

aLawMaLawMM

awMawMM

CBC

BAB

3061

021

203

1203

1

0,0,at Vdx

dMMA A



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Sample Problem 7.6


between B and C is

equal to area under

shear curve between

points. The constant

shear curve results in a

linear moment curve.

eighth edition - apl100apl100.wdfiles.com/local--files/...beer_chapter_07mod-updated.pdf · vector...

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