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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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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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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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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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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
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Vector Mechanics for Engineers: StaticsEighth
Edition
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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
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Vector Mechanics for Engineers: StaticsEighth
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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
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Vector Mechanics for Engineers: StaticsEighth
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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
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Vector Mechanics for Engineers: StaticsEighth
Edition
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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.
V t M h i f E i St tiEE
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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.
V t M h i f E i St tiEE
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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.
V t M h i f E i St tiEE
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Vector Mechanics for Engineers: StaticsEighth
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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
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Vector Mechanics for Engineers: StaticsEighth
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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,
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Vector Mechanics for Engineers: StaticsEighth
Edition
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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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Vector Mechanics for Engineers: StaticsEighth
Edition
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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.
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Vector Mechanics for Engineers: Staticsighth
dition
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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.
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dition
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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
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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
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dition
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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
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dition
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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
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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
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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
Between concentrated load application
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.
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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
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ition
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Sample Problem 7.4
Between concentrated load application
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
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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
equal to area under shear curve between
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.
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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
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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
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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.
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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
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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.
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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.
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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.
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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
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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
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Sample Problem 7.8
Evaluate maximum slope and
maximum tension which occur inDE.
15
7.14tan 4.43
cos
kips18max T kips8.24max T
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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
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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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