© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 1 CE 102 Statics Chapter 5 Forces

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

Vector Mechanics for Engineers: Statics

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7- 1

CE 102 Statics

Chapter 5

Forces in Beams and Cables



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Contents

IntroductionInternal Forces in MembersSample Problem 7.1Various Types of Beam Loading and

SupportShear and Bending Moment in a

BeamSample Problem 7.2Sample Problem 7.3Relations Among Load, Shear, and

Bending Moment

Sample Problem 7.4Sample Problem 7.5Cables With Concentrated LoadsCables With Distributed LoadsParabolic CableSample Problem 7.6

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 internal forces (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 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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Internal Forces in Members• 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.

• Multiforce member ABCD is 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-bodies JD and ABCJ.

• An internal force-couple system is required for equilibrium of two-force members which are not straight.



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Sample Problem 5.1

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

SOLUTION:

• Compute reactions and forces at connections for each member.

• Cut member ACF at J. The internal forces at J are represented by equivalent force-couple system which is determined by considering equilibrium of either part.

• Cut member BCD at K. Determine force-couple system equivalent to internal forces at K by applying equilibrium conditions to either part.



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Sample Problem 5.1

: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:



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Sample Problem 5.1Consider member BCD as free-body:

:0 BM

0m4.2m6.3N2400 yC N3600yC

:0 CM

0m4.2m2.1N2400 yB N1200yB

:0 xF 0 xx CB

Consider member ABE as free-body:

:0 AM 0m7.2 xB 0xB

:0xF 0 xx AB 0xA

:0yF 0N600 yy BA N1800yA

From member BCD,

:0 xF 0 xx CB 0xC



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Sample Problem 5.1• Cut member ACF at J. The internal forces at J are

represented by equivalent force-couple system.

Consider free-body AJ:

:0 JM

0m2.1N1800 M mN2160 M

:0 xF

07.41cosN1800 F N1344F

:0 yF

07.41sinN1800 V N1197V



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Sample Problem 5.1

• Cut member BCD at K. Determine a force-couple system equivalent to internal forces at K .

Consider free-body BK:

:0 KM

0m5.1N1200 M mN1800 M

:0 xF 0F

:0 yF

0N1200 V N1200V



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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 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

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



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Various Types of Beam Loading and Support

• 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.

• Determine reactions at supports by treating whole beam as free-body.

• 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.

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



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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 C and consider member AC,

22 PxMPV

• Cut beam at E and consider member EB,

22 xLPMPV

• For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly.



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Sample Problem 5.2

Draw the shear and bending moment diagrams for the beam and loading shown.

SOLUTION:

• 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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Sample Problem 5.2SOLUTION:

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

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

:0yF 0kN20 1 V kN201 V

:02 M 0m0kN20 1 M 01 M

mkN50kN26

mkN50kN26

mkN50kN26

mkN50kN26

66

55

44

33

MV

MV

MV

MV

Similarly,



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Sample Problem 5.2

• Plot results.

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



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Sample Problem 5.3

Draw the shear and bending moment diagrams for the beam AB. The distributed 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 within segments AC, CD, and DB.

• Plot results.



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

:0 AM

0in.22lb400in.6lb480in.32 yB

lb365yB

:0 BM

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

lb515A

:0 xF 0xB

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



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Sample Problem 5.3

:01 M 04051521 Mxxx

220515 xxM

:02 M 06480515 Mxx

in.lb 352880 xM

From C to D:

:0yF 0480515 V

lb 35V

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

From A to C:

:0yF 040515 Vx

xV 40515



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Sample Problem 5.3

:02 M

01840016006480515 Mxxx

in.lb 365680,11 xM

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

From D to B:

:0yF 0400480515 V

lb 365V



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Sample Problem 5.3

• Plot results.

From A to C: xV 40515

220515 xxM

From C to D:lb 35V

in.lb 352880 xM

From D to B:lb 365V

in.lb 365680,11 xM



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Relations Among Load, Shear, and Bending Moment• Relations between load and shear:

wx

V

dx

dV

xwVVV

x

0lim

0

curve loadunder area D

C

x

xCD dxwVV

• Relations between shear and bending moment:

VxwVx

M

dx

dM

xxwxVMMM

xx

21

00limlim

02

curveshear under area D

C

x

xCD dxVMM



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Relations Among Load, Shear, and Bending Moment

• Reactions at supports, 2

wLRR BA

• Shear curve,

xL

wwxwL

wxVV

wxdxwVV

A

x

A

22

0

• 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 5.4

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

SOLUTION:

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

• With uniform loading between D and E, 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 between D and E, the bending moment diagram is a parabola.



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Sample Problem 5.4

• 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.

:0AM

0ft 82kips 12

ft 14kips 12ft 6kips 20ft 24

D

kips 26D

:0 yF

0kips 12kips 26kips 12kips 20 yA

kips 18yA



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Sample Problem 5.4• Between concentrated load application

points, The change in moment between load application points is equal to area under the shear curve between points.

.constantVdxdM

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

048

ftkip 48140

ftkip 9216

ftkip 108108

EDE

DCD

CBC

BAB

MMM

MMM

MMM

MMM



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Sample Problem 5.5

Sketch the shear and bending-moment diagrams for the cantilever beam and loading shown.

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.

• 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.

• 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 5.5

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

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.

awVV AB 021 awVB 02

1

0,at wdx

dVB

0,0,at wwdx

dVVA A



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Sample Problem 5.5• The change in moment between A and B is equal

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

• 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.

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 force members

• Wish to determine shape of cable, i.e., vertical distance from support A to each load point.



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

cable at A and B are 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,

2 yields02

yMC yxyx TTFF , yield 0,0

• constantcos xx ATT

• Additional equation is obtained by considering equilibrium of portion of cable AD and assuming that coordinates of point D 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 curveb) internal force is a tension force directed along

tangent to curve.

• Consider free-body for portion of cable extending from lowest point C to given point D. Forces are horizontal force T0 at C and tangential force T at D.

• From force triangle:

0

220

0

tan

sincos

T

WWTT

WTTT

• 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.



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Parabolic Cable

• Consider a cable supporting a uniform, horizontally distributed load, e.g., support cables for a suspension 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

wxxwTT

• Summing moments about D,

02

:0 0 yTx

wxM D

0

2

2T

wxy

or

The cable forms a parabolic curve.



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Sample Problem 5.6

The cable AE supports three vertical loads 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 slope and maximum tension in the cable.

SOLUTION:

• Determine reaction force components at A from solution of two equations formed from taking entire cable as free-body and summing moments about E, and from 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 summing moments B. Similarly, calculate elevation of D using ABCD as a free-body.

• Evaluate maximum slope and maximum tension which occur in DE.



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• Determine two reaction force components at A from solution of two equations formed from taking entire cable as a free-body and summing moments about E,

06606020

041512306406020

:0

yx

yx

E

AA

AA

M

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

0610305

:0

yx

C

AA

M

Solving simultaneously,kips 5kips 18 yx AA



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Sample Problem 5.6• Calculate elevation of B by considering AB

as a free-body and summing moments B.

020518:0 BB yM

ft 56.5By

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

0121562554518

:0

Dy

M

ft83.5Dy



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Sample Problem 5.6

• Evaluate maximum slope and maximum tension which occur in DE.

15

7.14tan 4.43

cos

kips 18max T kips 8.24max 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 C to a point D given by the internal tension force magnitude is

wTcscwswTT 022222

0

,wsW

• To relate horizontal distance x to cable length s,

c

xcs

c

sc

csq

dsx

csq

ds

T

Tdsdx

ssinhandsinh

coscos

1

022

220



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Catenary• To relate x and y cable 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.

40

Problem 5.7

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G

A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b) the maximum tension in the cable, (c) the bending moment at F and G.

41

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G

A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b)

Solving Problems on Your Own

1. Identify points of the cable where useful information (position, slope,etc.) exists. Cut the cable at these points and draw a free-body diagram of one of the two portions of the cable.2. Use M = 0 if you know the position, or Fx = 0 and Fy = 0 if you know the slope, to generate needed equations of equilibrium.

Problem 7.158

the maximum tension in the cable, (c) the bending moment at F and G.

42

4. For a cable supporting vertical loads only, the horizontal component of the tension force is the same at any point. For such a cable, the maximum tension occurs in the steepest portion of the cable.

3. The tension in each section can be determined from the equations of equilibrium.

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G

A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b)


Problem 7.158

the maximum tension in the cable, (c) the bending moment at F and G.

43

+

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G

Problem 7.158 Solution

Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use M = 0 to generate the equations of equilibrium.

A

B

2 ft

5 ft

FBF

TBC

T0

Ay

Free Body: Portion AB

MB = 0: T0(2 ft) - Ay(5 ft) = 0

Ay = 0.4 T0

44


+

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G


A

B3 ft

5 ft

FBF

TCDT0

Free Body: Portion ABC

MC = 0: T0(3 ft) - 0.4 T0 (10 ft) + FBF (5 ft) = 0

0.4T0

C

5 ft

FCH

FBF = 0.2 T0

45

+

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G


3 ft TBCT0

Free Body: Portion CD

MC = 0: Dy(5 ft) - T0 (3 ft) = 0

C

5 ft

FCH

FBF = 0.2 T0

Dy

Dy= 0.6 T0D


46

+

A

BC

D

E

2 ft 3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G


2 ft T0

Free Body: Portion BCD

MB = 0: 0.6T0 (10 ft) - T0 (2 ft) - FCH (5ft) = 0

C

5 ft

FCH

FBF = 0.2 T0

0.6 T0

5 ft

TAB

0.2 T0

BD

FCH = 0.8 T0


47

A

BC

D

E

3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G FCH = 0.8 T0

2 ftThe tension in each section can be determined from the equations of equilibrium.

Free Body: Beam EFH

E7.5 ft 200 lb

F

H

G

Ey

Ex

0.8 T00.2 T0

5 ft 5 ft

FBF = 0.2T0

+

0.2T0 (5 ft) + 0.8 T0 (10 ft) - (200 lb)(7.5 ft) = 0

ME = 0:

T0 = 166.67 lb

FCH = 0.8(166.67) FCH = 133.33 lb T

FBF = 0.2(166.67) FBF = 33.33 lb T


48

A

BC

D

E

3 ft

5 ft 5 ft 5 ft

7.5 ft 200 lb

F H

G

2 ftThe maximum tension occurs in the steepest portion of the cable.

T0

TmDy= 0.6 T0

T0 = 166.67 lb

Tm = T02 + (0.6T0)2 = 1.1662T0

= 1.1662(166.67)

Tm = 194.4 lb

D


49

+

E7.5 ft

200 lb

F H

G

Ey

Ex

5 ft 5 ft

FCH = 133.33 lbFBF = 33.33 lb

Beam EFH

133.33 lb

2.5 ft

VMG

GMG = 0: (133.33 lb)(2.5 ft) -MG = 0 MG = + 333 lb-ft

133.33 lb

2.5 ft

VMF

G

2.5 ft

200 lb

+ MF = 0: (133.33 lb)(5 ft) - (200 lb)(2.5 ft) -MF = 0 MF = +166. 7 lb-ft

The moments at points G and F are determined by using free-body diagrams for two sections of the beam.


50

Problem 5.8

For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.x

y

L

AB

w = w0 cos x2L

51


For beams supporting a distributed load expressed as a function w(x), the shear V can be obtained by integrating the function -w(x) , and the moment M can be obtained by integrating V (x).

For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.

x

y

L

AB

w = w0 cos x2L

Problem 7.159

52


x

y

L

AB

w = w0 cos x2L

The shear V can be obtained by integrating the function -w(x)and the moment M can be obtained by integrating V (x).x

= -w = -w0 cosdVdx

x2L

V = - wdx = - w0 sin + C1( )2L

x2L

dMdx

= V = - w0 sin + C1( )2L

x2L

M = Vdx = w0 cos + C1x + C2( )2L

2 x2L

53

x

y

L

AB

w = w0 cos x2L


V = - w0 sin + C1( )2L

x2L

M = w0 cos + C1x + C2( )2L

2 x2L

Boundary conditions

At x = 0 : V = C1 = 0 C1 = 0

At x = 0 : M = w0 (2L/)2 cos (0) + C2 = 0

C2 = -w0 (2L/)2


54

x

y

L

AB

w = w0 cos x2L


V = - w0 sin( )2L

x2L

M = w0 cos - w0 ( )2L

2 x2L ( )2L

2

M = w0 ( -1 + cos )( )2L

2 x2L

Mmax at x = L: Mmax = w0 [-1 + 0]( )2L

2

42


Mmax = w0 L2

55

Problem 5.9

It has been experimentally determined that the bending moment at point K of the frame shown is 300 N-m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at point J.

AB

D

EJ

k

F

100 mm

100 mm

100 mm

120 mm

280 mm

C

56


1. Cut the member at a point, and draw the free-body diagram of each of the two portions.

2. Select one of the two free-body diagrams and use it to write the equations of equilibrium.

It has been experimentally determined that the bending moment at point K of the frame shown is 300 N-m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at point J.

AB

D

EJ

k

F

100 mm

100 mm

100 mm

120 mm

280 mm

C

Problem 7.161

57


AB

D

Tx

J T100 mm

100 mm

120 mm

Cut the member at a point, and draw the free-body diagram of each of the two portions.

D100 mm

280 mmC

TykV

FMK = 300 N-m

k

MK = 300 N-m

V

F

Tx

T Ty

300

280 - 120 = 160

3002 + 1602 = 340

A

D

A

D

8

1517

Ty = T1517

Tx = T817

58

+

AB

D

J200 mm

120 mm

kV

FMK = 300 N-m

Ty = T1517

Tx = T8

17 Select one of the two free-body diagrams and use it to writethe equations of equilibrium.

Mk = 0: 300 N-m - T(0.2 m) - T(0.12 m) = 0 1517

817

T = 1500 N

Free Body: ABK


59

AB

D

J100 mm

120 mm

Ty = (1500) = 1323.53 N1517

Tx = (1500) = 705.88 N8

17

Free Body: ABJ

VF

MJ

+ MJ = 0: MJ - (705.88 N)(0.1 m) - (1323.53 N)(0.12 m) = 0

MJ = 229 N-m

Fx = 0: 705.88 N - V = 0 V = 706 N +

Fy = 0: -F - 1323.53 N = 0 F = 1324 N+


60

Problem 5.10

Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest point C is located 9 m to the right of A. Determine (a) the vertical

B2.25 m

C

A

a

9 m 6 m

distance a, (b) the length of the cable, (c) the components of the reaction at A.

60 kg/m

61


Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest point C is located 9 m to the right of A.

B2.25 m

C

A

a

9 m 6 m

Determine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A.

1. Identify points of the cable where useful information (position, slope,etc.) exists. Cut the cable at these points and draw a free-body diagram of the two portions of the cable.

2. Use M = 0 if you know the position, or Fx = 0 and Fy = 0 if you know the slope, to generate needed equations of equilibrium.

3. The length of the cable can be determined from (7.10).

60 kg/m

Problem 7.162

62

+


C

Aa

9 m

Ay

T0

T0

Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or Fx = 0 and Fy = 0 if you know the slope to generate needed equations ofequilibrium.

Free Body : Portion AC

Fy = 0: Ay - 9w =0, Ay = 9w

4.5 m9w

MA = 0: To a - (9w)(4.5 m) = 0, To a = 40.5w (1)

+

63

BC

6 mIdentify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or Fx = 0 and Fy = 0 if you know the slope to generate needed equations ofequilibrium.

Free Body : Portion CB

T0T0

By

yB

3 m6w

+

Fy = 0: By - 6w = 0, By = 6w

MB = 0: (6w)(3 m) - To yB = 0, To yB = 18w

+


64

B2.25 m

C

A

yB = a - 2.25

9 m 6 m

Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or Fx = 0 and Fy = 0 if you know the slope, to generate needed equations of equilibrium.

T0

6w

T0

9w

15w7.5 m

+ MA = 0: 6w (15 m) - 15w(7.5 m) + To (2.25 m) = 0

To a = 40.5w (1)

Free Body : Entire Cable

To = 10w

Using (1) (10w) a = 40.5w a = 4.05 m


65

B2.25 m

C

A

4.05

9 m 6 m

6w9w

15w7.5 m

(b) Length of AC & CB

The length of the cable can be determined from (7.10).

10w

10w

Portion AC xA = 9 m, yA = a = 4.05 m; = = 0.45yAxA

4.059

SAC = xA 1 + - + ...yAxA

23 ( )

4[ ]

yAxA

( )2

25

SAC = 9 m [ 1 + (2/3)(0.45)2 - (2/5)(0.45)2 + ... ]= 10.067 m

Portion CB xB = 6 m, yB = 4.05 - 2.25 = 1.8 m; = 0.3yBxB

SCB = 6 m [ 1 + (2/3)(0.3)2 - (2/5)(0.3)2 + ... ]= 6.341 m

SABC = SAC + SCB = 10.067 + 6.341 SABC = 16.41 m


66

B2.25 m

C

A

4.05

9 m 6 m

6w9w

15w7.5 m

(c) Components of Reaction at A

10w

10w

Ay = 9w = 9(60 kg/m)(9.81 m/s2) = 5297.4 N

Ax = 10w = 10(60 kg/m)(9.81 m/s2) = 5886 N

Ay = 5300 NAx = 5890 N


67

Problem 5.13

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

4 ft 6 ft

6000 lb

1500 lb/ft

A BC

68


1. Draw a free-body diagram for the entire beam, and use it to determine the reactions at the beam supports.

2. Draw the shear diagram.

3. Draw the bending-moment diagram by computing the area under each portion of the shear curve.

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

4 ft 6 ft

6000 lb

1500 lb/ft

A BC

Problem 7.163

69

+


4 ft 6 ft 6 kips

9 kips

A BC

Draw a free-body diagram for the entire beam, and use it to determine the reactions at the beam supports.

3 ft

MA = 0 : (6 kips)(10 ft) - (9 kips)(7 ft) + B (4 ft) = 0

B = 0.75 kips

Fy = 0 : A + 0.75 kips + 6 kips - 9 kips = 0

A = 2.25 kips

+

70

6 kips

1.5 kips/ft

A B C

2.25 kips 0.75 kips

4 ft 6 ft

Draw the shear diagram.

x+ 2.25 kip

+ 3 kip

- 6 kip

A B C

D

2 ft 4 ft

V (kips)

BD = =VB

w3 kips

1.5 kips/ft


71

6 kips

1.5 kips/ft

A B C

2.25 kips 0.75 kips

4 ft 6 ft

Draw the bending-moment diagram.

x+ 2.25 kip

+ 3 kip

- 6 kip

A B C

D

2 ft 4 ft

V (kips)

BD = =VB

w3 kips

1.5 kips/ft

xA B CD

M (kip-ft)

(+9 kip-ft)

(3 kips)(2 ft) = (+3 kip-ft)12

(- 12 kip-ft)

(+9 kip-ft)(+12 kip-ft) Mmax = 12 kip-ft


6 ft from A

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 1 CE 102 Statics Chapter 5 Forces