ch07-forces in beams and cables

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VECTOR MECHANICS FOR ENGINEERS: STATICS

Ninth Edition

Ferdinand P. BeerE. Russell Johnston, Jr.

Lecture Notes:J. Walt OlerTexas Tech University

CHAPTER

© 2010 The McGraw-Hill Companies, Inc. All rights reserved.

7Forces in Beams andCables




Vector Mechanics for Engineers: StaticsNi n

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Contents

7- 2

Introduction Internal 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 7.4 Sample Problem 7.6 Cables With ConcentratedLoads Cables With Distributed Loads

Parabolic Cable Sample Problem 7.8 Catenary





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Introduction

7- 3

• Preceding chapters dealt with:

a) determining external forces acting on a structure and

b) determining forces which hold together the various membersof a structure.

• The current chapter is concerned with determining the internal forces (i.e., tension/compression, shear, and bending) which holdtogether 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 onlytension, designed to support concentrated or distributed loads.





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Sample Problem 7.1

7- 5

Determine the internal forces (a) inmember ACF at point J and (b) inmember BCD at K .

SOLUTION:

• Compute reactions and forces atconnections for each member.

• Cut member ACF at J . The internalforces at J are represented by equivalentforce-couple system which is determined

by considering equilibrium of either part.

• Cut member BCD at K . Determineforce-couple system equivalent tointernal forces at K by applyingequilibrium conditions to either part.

NE





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Sample Problem 7.1

7- 6

:0 y F

0 N1800 N2400 y E N E y 600

:0 x F 0 x E

SOLUTION:

• Compute reactions and connection forces.

:0 E M

0m8.4m6.3 N2400 F N1800 F

Consider entire frame as a free-body:

NE


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Sample Problem 7.1

7- 7

Consider member BCD as free-body:

:0 B M 0m4.2m6.3 N2400 yC N3600 yC

:0 C M

0m4.2m2.1 N2400 y B N1200 y B

:0 x F 0 x x C B

Consider member ABE as free-body:

:0 A M 0m4.2 x B 0 x B

:0 x F 0 x x A B 0 x A

:0 y F 0 N600 y y B A N1800 y A

From member BCD ,

:0 x F 0 x x C B 0 xC

NE




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Sample Problem 7.1

7- 8

• Cut member ACF at J . The internal forces at J are

represented by equivalent force-couple system.

Consider free-body AJ :

:0 J M

0m2.1 N1800 M m N2160 M :0 x F

07.41cos N1800 F N1344 F

:0 y F

07.41sin N1800 V N1197V

NE




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Sample Problem 7.1

7- 9

• Cut member BCD at K . Determine a force-couple

system equivalent to internal forces at K .

Consider free-body BK :

:0 K M

0m5.1 N1200 M m N1800 M

:0 x F 0 F

:0 y F

0 N1200 V N1200V

NE




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Various Types of Beam Loading and Support

7- 10

• Beam - structural member designed to support

loads applied at various points along its length.

• Beam design is two-step process:

1) determine shearing forces and bendingmoments produced by applied loads

2) select cross-section best suited to resistshearing forces and bending moments

• Beam can be subjected to concentrated loads ordistributed loads or combination of both.

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Various Types of Beam Loading and Support

7- 11

• Beams are classified according to way in which they aresupported.

• Reactions at beam supports are determinate if theyinvolve only three unknowns. Otherwise, they arestatically indeterminate.

NE




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Shear and Bending Moment in a Beam

7- 12

• Wish to determine bending moment

and shearing force at any point in a beam subjected to concentrated anddistributed loads.

• Determine reactions at supports by

treating whole beam as free-body.

• Cut beam at C and draw free-bodydiagrams for AC and CB. Bydefinition, positive sense for internal

force-couple systems are as shown.

• From equilibrium considerations,determine M and V or M’ and V’ .

NE




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Shear and Bending Moment Diagrams

7- 13

• Variation of shear and bendingmoment along beam may be

plotted.• Determine reactions at

supports.• Cut beam at C and consider

member AC ,22 Px M P V

• Cut beam at E and considermember EB,

22 x L P M P V

• For a beam subjected toconcentrated loads, shear isconstant between loading pointsand moment varies linearly.

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Sample Problem 7.2

7- 14

Draw the shear and bending momentdiagrams for the beam and loadingshown.

SOLUTION:

• Taking entire beam as a free-body,calculate reactions at B and D.

• Find equivalent internal force-couplesystems for free-bodies formed bycutting beam on either side of loadapplication points.

• Plot results.

h fNE




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Sample Problem 7.2

7- 15

SOLUTION:

• Taking entire beam as a free-body, calculatereactions at B and D.

• Find equivalent internal force-couple systems atsections on either side of load application points.

:0 y F 0kN20 1 V kN201 V

:02 M 0m0kN20 1 M 01 M

mkN50kN26

mkN50kN26

mkN50kN26

mkN50kN26

66

55

4433

M V

M V

M V

M V

Similarly,

h fNE



Vector Mechanics for Engineers: Staticsi n t h

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Sample Problem 7.2

7- 16

• Plot results.

Note that shear is of constant value between concentrated loads and bending moment varies linearly.

V M h i f E i S iNE

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Sample Problem 7.3

7- 17

Draw the shear and bending momentdiagrams for the beam AB. Thedistributed load of 40 lb/in. extends

over 12 in. of the beam, from A to C ,and the 400 lb load is applied at E .

SOLUTION:

• Taking entire beam as free-body,calculate reactions at A and B.

• Determine equivalent internal force-couple systems at sections cut withinsegments AC , CD, and DB.

• Plot results.

V M h i f E i S iNi

E d




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Sample Problem 7.3

7- 18

SOLUTION:

• Taking entire beam as a free-body, calculatereactions at A and B.

:0 A M

0in.22lb400in.6lb480in.32 y B

lb365 y B

:0 B M

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

lb515 A

:0 x F 0 x B

• Note: The 400 lb load at E may be replaced by a400 lb force and 1600 lb-in. couple at D.

V M h i f E i S iNi

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Sample Problem 7.3

7- 19

:01 M 040515 21 M x x x

220515 x x M

:02 M 06480515 M x x

in.lb352880 x M

From C to D:

:0 y F 0480515 V lb35V

• Evaluate equivalent internal force-couple systemsat sections cut within segments AC , CD, and DB.

From A to C : :0 y F 040515 V x

xV 40515

V M h i f E i S iNi

E d




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Sample Problem 7.3

7- 20

:02 M

01840016006480515 M x x x

in.lb365680,11 x M

• Evaluate equivalent internal force-couplesystems at sections cut within segments AC ,CD, and DB.

From D to B:

:0 y F 0400480515 V

lb365V

V t M h i f E i St tiNi

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Sample Problem 7.3

7- 21

• Plot results.

From A to C : xV 40515

220515 x x M

From C to D:lb35V

in.lb352880 x M

From D to B:lb365V

in.lb365680,11 x M

V t M h i f E i St tiNi

E d




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Relations Among Load, Shear, and Bending Moment

7- 22

• Relations between load and shear:

w

xV

dxdV

xwV V V

x 0lim

0

curveloadunderarea D

C

x

xC D dxwV V

• Relations between shear and bending moment:

V xwV x

M dx

dM

x xw xV M M M

x x 21

00limlim

02

curveshearunderarea D

C

x

xC D dxV M M

V t M h i f E gi St tiNi

E d




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Relations Among Load, Shear, and Bending Moment

7- 23

• Reactions at supports,

2

wL R R B A

• Shear curve,

x L

wwxwL

wxV V

wxdxwV V

A

x

A

22

0

• Moment curve,

0at8

22

2

max

2

0

0

V dx

dM M

wL M

x x Lw

dx x L

w M

Vdx M M

x

x

A

V t M h i f E gi St tiNi

E d




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Sample Problem 7.4

7- 24

Draw the shear and bending-moment diagrams for the beamand loading shown.

SOLUTION:

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

• With uniform loading between D and E , theshear variation is linear.

• Between concentrated load application points, and shear isconstant.

0 wdxdV

• Between concentrated load application points, The changein moment between load application points isequal to area under shear curve between

points.

.constantV dxdM

• With a linear shear variation between D and E , the bending moment diagram is a

parabola.

Vector Mechanics for Engineers: StaticsNi nE d



Vector Mechanics for Engineers: Staticsn t h

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Sample Problem 7.4

7- 25

• Between concentrated load application points,and shear is constant.0 wdxdV

• With uniform loading between D and E , the shear

variation is linear.

SOLUTION:

•

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

:0 A M

0ft82kips12

ft14kips12ft6kips20ft24 D

kips26 D

:0 y F

0kips12kips26kips12kips20 y A

kips18 y

A





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Sample Problem 7.4

7- 26

• Between concentrated load application

points, The changein moment between load application points isequal to area under the shear curve between

points.

.constantV dxdM

•

With a linear shear variation between D and E , the bending moment diagram is a parabola.

048

ftkip48140ftkip9216

ftkip108108

E D E

DC D

C BC

B A B

M M M

M M M M M M

M M M






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Sample Problem 7.6

7- 27

Sketch the shear and bending-moment diagrams for thecantilever beam and loading

shown.

SOLUTION:

• The change in shear between A and B is equalto the negative of area under load curve

between points. The linear load curve resultsin a parabolic shear curve.

• With zero load, change in shear between B and C is zero.

• The change in moment between A and B isequal to area under shear curve between

points. The parabolic shear curve results ina cubic moment curve.

• The change in moment between B and C isequal to area under shear curve between

points. The constant shear curve results in alinear moment curve.

Vector Mechanics for Engineers: StaticsNi nE d i





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Sample Problem 7.6

7- 28

• With zero load, change in shear between B and C iszero.

SOLUTION:

• The change in shear between A and B is equal tonegative of area under load curve between points.The linear load curve results in a parabolic shearcurve.

awV V A B 021 awV B 02

1

0,at wdx

dV B

0,0,at wwdxdV

V A A




Vector Mechanics for Engineers: Statics t h i t i on

Cables With Concentrated Loads

7- 30

• Cables are applied as structural elements

in suspension bridges, transmission lines,aerial tramways, guy wires for hightowers, 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 support A to eachload point.






Cables With Concentrated Loads

7- 31

• Consider entire cable as free-body. Slopes ofcable at A and B are not known - two reactioncomponents required at each support.

• Four unknowns are involved and threeequations of equilibrium are not sufficient todetermine the reactions.

• For other points on cable,

2yields02

y M C

y x y x T T F F ,yield 0,0 • constantcos x x AT T

• Additional equation is obtained byconsidering equilibrium of portion of cable

AD and assuming that coordinates of point D on the cable are known. The additionalequation is .0 D M






Cables With Distributed Loads

7- 32

• For cable carrying a distributed load:a) cable hangs in shape of a curve

b) internal force is a tension force directed alongtangent to curve.

• Consider free-body for portion of cable extendingfrom lowest point C to given point D. Forces arehorizontal force T 0 at C and tangential force T at D.

• From force triangle:

0

220

0

tan

sincos

T W

W T T

W T T T

• Horizontal component of T is uniform over cable.• Vertical component of T is equal to magnitude of W

measured from lowest point.• Tension is minimum at lowest point and maximum

at A and B.






Parabolic Cable

7- 33

• Consider a cable supporting a uniform, horizontally

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

• With loading on cable from lowest point C to a point D given by internal tension force

magnitude and direction are

,wxW

0

2220 tan

T wx

xwT T

• Summing moments about D,

02:0 0 yT x

wx M D

0

2

2T wx

y or

The cable forms a parabolic curve.

Vector Mechanics for Engineers: StaticsNi nE d it





Sample Problem 7.8

7- 34

The cable AE supports three verticalloads from the points indicated. If

point C is 5 ft below the left support,determine (a) the elevation of points

B and D, and (b) the maximum slopeand maximum tension in the cable.

SOLUTION:

• Determine reaction force components at A from solution of two equations formedfrom taking entire cable as free-bodyand summing moments about E , andfrom taking cable portion ABC as a free-

body and summing moments about C .

• Calculate elevation of B by considering AB as a free-body and summingmoments B. Similarly, calculate

elevation of D using ABCD as a free- body.

• Evaluate maximum slope andmaximum tension which occur in DE .

Vector Mechanics for Engineers: StaticsNi nt

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Vector Mechanics for Engineers: Statics t h t i on

Sample Problem 7.8

7- 35

SOLUTION:

•

Determine two reaction force components at A from solution of two equations formed fromtaking entire cable as a free-body and summingmoments about E ,

06606020

041512306406020

:0

y x

y x

E

A A A A

M

and from taking cable portion ABC as afree-body and summing moments about C .

0610305:0

y x

C

A A M

Solving simultaneously,kips5kips18 y x A A


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Sample Problem 7.8

7- 36

• Calculate elevation of B by considering AB asa free-body and summing moments B.

020518:0 B B y M

ft56.5 B y

Similarly, calculate elevation of D using ABCD as a free-body.

0121562554518

:0

D y

M

ft83.5 D y


E d i t





Sample Problem 7.8

7- 37

• Evaluate maximum slope and

maximum tension which occur in DE .

157.14

tan 4.43

coskips18

max T kips8.24max T


E d i ti





Catenary

7- 38

• 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 C to a

point D given by the internal tension forcemagnitude is

wT c scw swT T 0222220

,wsW

• To relate horizontal distance x to cable length s,

c x

c sc s

cc sq

ds x

c sq

dsT T

dsdx

ssinhandsinh

coscos

1

0 22

220

Vector Mechanics for Engineers: StaticsNi nth

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Catenary• To relate x and y cable coordinates,

c x

c y

cc x

cdxc x

c y

dxc x

dxc s

dxT W

dxdy

x

cosh

coshsinh

sinhtan

0

0

which is the equation of a catenary.

ch07-forces in beams and cables

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