STA CH07 Lecture

8/10/2019 STA CH07 Lecture

1/39

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.

7Forces in Beams and Cables


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Vector Mechanics for Engineers: StaticsEighth

Edition

7- 2

Contents

IntroductionInternal Forces in Members

Sample Problem 7.1

Various Types of Beam Loading and

Support

Shear and Bending Moment in aBeam

Sample Problem 7.2

Sample Problem 7.3

Relations Among Load, Shear, and

Bending Moment

Sample Problem 7.4Sample Problem 7.6

Cables With Concentrated Loads

Cables With Distributed Loads

Parabolic Cable

Sample Problem 7.8

Catenary


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Edition

7- 3

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.

The current chapter is concerned with determining the internalforces(i.e., tension/compression, shear, and bending) which hold

together the various parts of a given member.

Focus is on two important types of engineering structures:

a) Beams- usually long, straight, prismatic members designedto 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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Edition

7- 4

Internal Forces in Members Straight two-force memberABis in

equilibrium under application of Fand

-F.

Internal forcesequivalent toFand -Fare

required for equilibrium of free-bodiesAC

and CB.

Multiforce memberABCDis in equil-

ibrium under application of cable and

member contact forces.

Internal forces equivalent to a force-

couple system are necessary for equil-ibrium of free-bodiesJDandABCJ.

An internal force-couple system is

required for equilibrium of two-force

members which are not straight.EE


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Edition

7- 5

Sample Problem 7.1

Determine the internal forces (a)inmemberACFat pointJand (b)in

memberBCDatK.

SOLUTION:

Compute reactions and forces atconnections for each member.

Cut memberACFatJ. The internal

forces atJare represented by equivalent

force-couple system which is determinedby considering equilibrium of either part.

Cut memberBCDatK. Determine

force-couple system equivalent to

internal forces atKby applyingequilibrium conditions to either part.

EE


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Edition

7- 6

Sample Problem 7.1

:0 yF

0N1800N2400 yE NEy 600

:0 xF 0xE

SOLUTION:

Compute reactions and connection forces.

:0 EM

0m8.4m6.3N2400 F N1800F

Consider entire frame as a free-body:

EE


7/39 2007 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

7- 7

Sample Problem 7.1

Consider memberBCDas free-body:

:0 BM 0m4.2m6.3N2400 yC N3600yC

:0 CM

0m4.2m2.1N2400 yB N1200yB

:0 xF 0 xx CB

Consider memberABE as free-body:

:0 AM 0m4.2 xB 0xB

:0xF 0

xx AB 0

xA :0yF 0N600 yy BA N1800yA

From memberBCD,

:0 xF 0 xx CB 0xC

EE




Edition

7- 8

Sample Problem 7.1

Cut memberACFatJ. The internal forces atJare

represented by equivalent force-couple system.

Consider free-bodyAJ:

:0 JM

0m2.1N1800 M mN2160 M:0 xF

07.41cosN1800 F N1344F

:0 yF

07.41sinN1800 V N1197V

V M h i f E i S iEE




Edition

7- 9

Sample Problem 7.1

Cut memberBCDatK. Determine a force-couple

system equivalent to internal forces atK.

Consider free-bodyBK:

:0 KM

0m5.1N1200 M mN1800 M

:0 xF 0F

:0 yF

0N1200 V N1200V


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V t M h i f E i St tiEE




Edition

7- 11

Various Types of Beam Loading and Support

Beams are classified according to way in which they aresupported.

Reactions at beam supports are determinate if they

involve only three unknowns. Otherwise, they are

statically indeterminate.





Edition

7- 12

Shear and Bending Moment in a Beam

Wish to determine bending moment

and shearing force at any point in abeam subjected to concentrated and

distributed loads.

Determine reactions at supports by

treating whole beam as free-body.

Cut beam at Cand draw free-body

diagrams forACand CB. By

definition, positive sense for internal

force-couple systems are as shown.

From equilibrium considerations,

determineM and Vor Mand V.





Edition

7- 13

Shear and Bending Moment Diagrams

Variation of shear and bending

moment along beam may be

plotted.

Determine reactions at

supports.

Cut beam at Cand consider

memberAC,22 PxMPV

Cut beam atEand consider

memberEB,

22 xLPMPV

For a beam subjected to

concentrated loads, shear is

constant between loading points

and moment varies linearly.





Edition

7- 14

Sample Problem 7.2

Draw the shear and bending moment

diagrams for the beam and loading

shown.

SOLUTION:

Taking entire beam as a free-body,

calculate reactions atBandD.

Find equivalent internal force-couple

systems for free-bodies formed by

cutting beam on either side of loadapplication points.

Plot results.

V t M h i f E i St tiE

E




Edition

7- 15

Sample Problem 7.2SOLUTION:

Taking entire beam as a free-body, calculatereactions atBandD.

Find equivalent internal force-couple systems at

sections on either side of load application points.

:0yF 0kN20 1 V kN201 V

:02 M 0m0kN20 1M 01M

mkN50kN26

mkN50kN26

mkN50kN26

mkN50kN26

66

55

44

33

MV

MV

MV

MV

Similarly,

V t M h i f E i St tiEi

Ed




Edition

7- 16

Sample Problem 7.2

Plot results.

Note that shear is of constant value

between concentrated loads and

bending moment varies linearly.

V t M h i f E i St tiEi

Ed




Edition

7- 17

Sample Problem 7.3

Draw the shear and bending moment

diagrams for the beamAB. The

distributed load of 40 lb/in. extends

over 12 in. of the beam, fromAto C,and the 400 lb load is applied atE.

SOLUTION:

Taking entire beam as free-body,

calculate reactions atAandB.

Determine equivalent internal force-

couple systems at sections cut within

segmentsAC, CD, andDB.

Plot results.

Vector Mechanics for Engineers: StaticsEi

Ed



Vector Mechanics for Engineers: Staticsighth

dition

7- 18

Sample Problem 7.3

SOLUTION:

Taking entire beam as a free-body, calculatereactions atAandB.

:0 AM

0in.22lb400in.6lb480in.32 yB

lb365yB

:0 BM

0in.32in.10lb400in.26lb480 A

lb515A

:0 xF 0xB

Note: The 400 lb load atEmay be replaced by a

400 lb force and 1600 lb-in. couple atD.


Ed



Vector Mechanics for Engineers: Staticsghth

dition

7- 19

Sample Problem 7.3

:01M 040515 21 Mxxx

220515 xxM

:02 M 06480515 Mxx

in.lb352880 xM

From CtoD:

:0yF 0480515 Vlb35V

Evaluate equivalent internal force-couple systems

at sections cut within segmentsAC, CD, andDB.

From A to C:

:0yF 040515 VxxV 40515


Ed




dition

7- 20

Sample Problem 7.3

:02 M

01840016006480515 Mxxx

in.lb365680,11 xM

Evaluate equivalent internal force-couple

systems at sections cut within segmentsAC,

CD, andDB.

FromDtoB:

:0yF 0400480515 V

lb365V

Vector Mechanics for Engineers: StaticsEig

Ed




dition

7- 21

Sample Problem 7.3

Plot results.

From A to C:

xV 40515

220515 xxM

From CtoD:lb35V

in.lb352880 xM

FromDtoB:

lb365V

in.lb365680,11 xM


Ed




dition

7- 22

Relations Among Load, Shear, and Bending Moment

Relations between load and shear:

wx

V

dx

dV

xwVVV

x

0lim

0

curveloadunderarea D

C

x

xCD dxwVV

Relations between shear and bending moment:

VxwVx

M

dx

dM

xxwxVMMM

xx

21

00limlim

02

curveshearunderarea D

C

x

xCD dxVMM


Ed




dition

7- 23

Relations Among Load, Shear, and Bending Moment

Reactions at supports,

2

wLRR BA

Shear curve,

x

L

wwx

wL

wxVV

wxdxwVV

A

x

A

22

0

Moment curve,

0at8

22

2

max

2

0

0

Vdx

dMM

wLM

xxLw

dxxL

wM

VdxMM

x

x

A


Ed




dition

7- 24

Sample Problem 7.4

Draw the shear and bending-

moment diagrams for the beam

and loading shown.

SOLUTION:

Taking entire beam as a free-body, determinereactions at supports.

With uniform loading betweenDandE, the

shear variation is linear.

Between concentrated load application

points, and shear is

constant.

0 wdxdV


points, The change

in moment between load application points is

equal to area under shear curve between

points.

.constantVdxdM

With a linear shear variation betweenD

andE, the bending moment diagram is a

parabola.


Ed


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ition

7- 25

Sample Problem 7.4

Between concentrated load application points,

and shear is constant.0 wdxdV

With uniform loading betweenDandE, the shear

variation is linear.

SOLUTION:

Taking entire beam as a free-body,determine reactions at supports.

:0AM

0ft82kips12

ft14kips12ft6kips20ft24

D

kips26D

:0 yF

0kips12kips26kips12kips20 yA

kips18y

A


Ed


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ition

7- 26

Sample Problem 7.4


points, The changein moment between load application points is

equal to area under the shear curve between

points.

.constant

VdxdM

With a linear shear variation betweenDandE, the bending moment diagram is a

parabola.

048

ftkip48140

ftkip9216

ftkip108108

EDE

DCD

CBC

BAB

MMM

MMM

MMM

MMM


Edi


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ition

7- 27

Sample Problem 7.6

Sketch the shear and bending-

moment diagrams for the

cantilever beam and loading

shown.

SOLUTION:

The change in shear betweenAandBis equalto the negative of area under load curve

between points. The linear load curve results

in a parabolic shear curve.

With zero load, change in shear betweenB

and Cis zero.

The change in moment betweenAandBis


points. The parabolic shear curve results in

a cubic moment curve.

The change in moment betweenBand Cis


points. The constant shear curve results in a

linear moment curve.


Edi


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ition

7- 28

Sample Problem 7.6

With zero load, change in shear betweenBand Cis

zero.

SOLUTION:

The change in shear betweenAandBis equal tonegative of area under load curve between points.

The linear load curve results in a parabolic shear

curve.

awVV AB 021 awVB 02

1

0,at w

dx

dVB

0,0,at wwdx

dV

VA A


Edi


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tion

7- 29

Sample Problem 7.6

The change in moment betweenAandBis equal

to area under shear curve between the points.The parabolic shear curve results in a cubic

moment curve.

The change in moment betweenBand Cis equal

to area under shear curve between points. The

constant shear curve results in a linear moment

curve.

aLawMaLawMM

awMawMM

CBC

BAB

3061

021

203

1203

1

0,0,at Vdx

dMMA A


Edi


30/39


Vector Mechanics for Engineers: Staticshthtion

7- 30

Cables With Concentrated Loads

Cables are applied as structural elements

in suspension bridges, transmission lines,aerial tramways, guy wires for high

towers, etc.

For analysis, assume:

a) concentrated vertical loads on given

vertical lines,b) weight of cable is negligible,

c) cable is flexible, i.e., resistance to

bending is small,

d) portions of cable between successive

loads may be treated as two forcemembers

Wish to determine shape of cable, i.e.,

vertical distance from supportAto each

load point.


Edit


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7- 31

Cables With Concentrated Loads Consider entire cable as free-body. Slopes of

cable atAandBare not known - two reaction

components required at each support.

Four unknowns are involved and three

equations of equilibrium are not sufficient to

determine the reactions.

For other points on cable,

2yields02 yMC

yxyx TTFF ,yield0,0

constantcos xx ATT

Additional equation is obtained byconsidering equilibrium of portion of cable

ADand assuming that coordinates of pointD

on the cable are known. The additional

equation is .0 DM

Vector Mechanics for Engineers: StaticsEigh

Edit


32/39



7- 32

Cables With Distributed Loads For cable carrying a distributed load:

a) cable hangs in shape of a curve

b) internal force is a tension force directed along

tangent to curve.

Consider free-body for portion of cable extending

from lowest point Cto given pointD. Forces are

horizontal force T0at C and tangential force TatD.

From force triangle:

0

220

0

tan

sincos

T

WWTT

WTTT

Horizontal component of Tis uniform over cable.

Vertical component of Tis equal to magnitude of W

measured from lowest point.

Tension is minimum at lowest point and maximum

atAandB.


Edit


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7- 33

Parabolic Cable

Consider a cable supporting a uniform, horizontally

distributed load, e.g., support cables for asuspension bridge.

With loading on cable from lowest point Cto a

pointDgiven by internal tension force

magnitude and direction are

,wxW

0

2220 tan

T

wxxwTT

Summing moments aboutD,

02:00 yT

xwxMD

0

2

2T

wxy

or

The cable forms a parabolic curve.


Edit


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7- 34

Sample Problem 7.8

The cableAEsupports three vertical

loads from the points indicated. If

point Cis 5 ft below the left support,

determine (a)the elevation of points

BandD, and (b)the maximum slope

and maximum tension in the cable.

SOLUTION:

Determine reaction force components at

Afrom solution of two equations formed

from taking entire cable as free-body

and summing moments aboutE, and

from taking cable portionABCas a free-

body and summing moments about C.

Calculate elevation ofBby considering

ABas a free-body and summing

momentsB. Similarly, calculate

elevation ofDusingABCDas a free-body.

Evaluate maximum slope and

maximum tension which occur inDE.


Edit


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7- 35

Sample Problem 7.8

SOLUTION:

Determine two reaction force components atAfrom solution of two equations formed from

taking entire cable as a free-body and summing

moments aboutE,

06606020

041512306406020

:0

yx

yx

E

AAAA

M

and from taking cable portionABCas a

free-body and summing moments about C.

0610305:0

yx

C

AA

M

Solving simultaneously,

kips5kips18 yx AA


Edit


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Vector Mechanics for Engineers: Staticshthion

7- 36

Sample Problem 7.8

Calculate elevation ofBby consideringABas

a free-body and summing momentsB.

020518:0 BB yM

ft56.5By

Similarly, calculate elevation ofDusingABCDas a free-body.

0121562554518

:0

Dy

M

ft83.5Dy


Edit


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7- 37

Sample Problem 7.8

Evaluate maximum slope and

maximum tension which occur inDE.

15

7.14tan 4.43

cos

kips18max T kips8.24max T


Editi


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7- 38

Catenary

Consider a cable uniformly loaded along the cable

itself, e.g., cables hanging under their own weight.

With loading on the cable from lowest point Cto a

pointDgiven by the internal tension force

magnitude is

wTcscwswTT 0222220

,wsW

To relate horizontal distancexto cable lengths,

c

xcs

c

sc

csq

dsx

csq

ds

T

Tdsdx

s

sinhandsinh

coscos

1

0 22

22

0


Editi


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Vector Mechanics for Engineers: StaticshthionCatenary

To relatexandycable coordinates,

c

xcy

cc

xcdx

c

xcy

dxc

xdx

c

sdx

T

Wdxdy

x

cosh

coshsinh

sinhtan

0

0

which is the equation of a catenary.

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