Vector Mechanics for Engineers: Statics

CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
Eighth Edition
Forces in Beams and Cables
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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dition
Vector Mechanics for Engineers: Statics
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
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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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Vector Mechanics for Engineers: Statics
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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Vector Mechanics for Engineers: Statics
Internal Forces in Members
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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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Vector Mechanics for Engineers: Statics
Sample Problem 7.1
Determine the internal forces (a) in
member ACF at point J and (b) in
member BCD at K.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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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Vector Mechanics for Engineers: Statics
Sample Problem 7.1
SOLUTION:
Fy
x
0 :
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Compute reactions and connection forces.
Consider entire frame as a free-body:
E M
0 :
2400 N3.
6m
4. 8m
0
F F 1800 N
y
2400 N 1800
N E 0 E y 600 N
F 0 :
E 0
x
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Vector Mechanics for Engineers: Statics
Sample Problem 7.1
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Consider member BCD as free-body:
B 0 : M
N 3.
6 m y
2400 2. 4 m
0
M C 0 :
N 1.
2 m y
2400 2. 4 m
0
x
F 0 : B 0
C C 3600 N
B B 1200 N
x Cx
Consider member ABE as free-body:
A
0 : M B 2. 4 m
0 B 0
x
F 0 : B 0
A 0
x
x Ax
F 0 : A 600 N 0 A 1800 N
y
From member BCD,
x
y By
F 0 : B 0 C
0
x Cx
y
y
x
x
y
x
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Vector Mechanics for Engineers: Statics
Sample Problem 7.1
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Cut member ACF at J. The internal forces at J are
represented by equivalent force-couple system.
Consider free-body AJ:
M J
0 :
1800 N1.
2m
0
M M 2160 N m
Fx
0 :
1800 Ncos
41.
7
0
F
F 1344 N
Fy
V
0 :
1800 Nsin
41.
7
0
V
1197 N
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Vector Mechanics for Engineers: Statics
Sample Problem 7.1
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Cut member BCD at K. Determine a force-couple
system equivalent to internal forces at K .
Consider free-body BK:
K M
0 :
1200 N1.
5m
0
x
M M 1800
N m
F 0 :
F 0
Fy
0 :
1200 N V
0
V
1200
N
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Vector Mechanics for Engineers: Statics
Various Types of Beam Loading and Support
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
• 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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Vector Mechanics for Engineers: Statics
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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Vector Mechanics for Engineers: Statics
Shear and Bending Moment in a Beam
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• 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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Vector Mechanics for Engineers: Statics
Shear and Bending Moment Diagrams
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• Variation of shear and bending
moment along beam may be
plotted.
• Determine reactions at
supports.
• Cut beam at C and consider
member AC,
V P 2 M Px 2
• Cut beam at E and consider
member EB,
V P 2 M P L x
2
• For a beam subjected to
concentrated loads, shear is
constant between loading points
and moment varies linearly.
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Vector Mechanics for Engineers: Statics
Sample Problem 7.2
Draw the shear and bending moment
diagrams for the beam and loading
shown.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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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Vector Mechanics for Engineers: Statics
Sample Problem 7.2
SOLUTION:
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
F 0 : 20kN 1 0 V kN 20 V
y
1
M 0 : 20kN0 m
1 0 M 0 M
2
Similarly,
V
V
V
V
3
4
5
6
26kN
26kN
26kN
26kN
M
M
M
M
3
4
5
6
50
kN m
50
kN m
50
kN m
50
kN m
1
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Vector Mechanics for Engineers: Statics
Sample Problem 7.2
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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: Statics
Sample Problem 7.3
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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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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Vector Mechanics for Engineers: Statics
Sample Problem 7.3
SOLUTION:
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Taking entire beam as a free-body, calculate
reactions at A and B.
A M
By
0 :
32in. 480lb6in. 400lb22in. 0
B M
0 :
By
365lb
480lb26in. 400lb10in. A32in.
0
x
A 515lb
F 0 :
B 0
• Note: The 400 lb load at E may be replaced by a
400 lb force and 1600 lb-in. couple at D.
x
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Vector Mechanics for Engineers: Statics
Sample Problem 7.3
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From A to C:
Fy 0 : 515 40x
V
0
V
515 40x
M 0 : 515x
40x
x
M 0
1
From C to D:
1
2
M 515x 20x
Fy 0 : 515 480 V
0
V 35 lb
M 0 : 515x 480x
6
M 0
2
M
2
2880 35 lb in.
x
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Vector Mechanics for Engineers: Statics
Sample Problem 7.3
M
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Evaluate equivalent internal force-couple
systems at sections cut within segments AC,
CD, and DB.
From D to B:
F 0 : 515 480 400 V
0
y
2
0 :
V
x 6
1600 400x
18
0
515x 480
M
M
365
lb
11, 680 365 lb in.
x
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Vector Mechanics for Engineers: Statics
Sample Problem 7.3
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Plot results.
From A to C:
V 515 40x
M 515x 20x
From C to D:
V 35 lb
M
2
2880 35 lb in.
x
From D to B:
V 365
lb
M
11, 680 365 lb in.
x
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Vector Mechanics for Engineers: Statics
Relations Among Load, Shear, and Bending Moment
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Relations between load and shear:
V
V
dV
dx
D
V V
lim
x0 wx
0
V
x
x
D
V wdx
C
x
C
w
area under load curve
• Relations between shear and bending moment:
MM dM
dx
M
D
x
M Vx
wx
0
2
M
lim lim
x0
x
x0
2
x
D
M V dx
C
x
C
V1wxV area under shear curve
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Vector Mechanics for Engineers: Statics
Relations Among Load, Shear, and Bending Moment
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Reactions at supports,
• Shear curve,
V
V
VA
wdx
V
A
x
0
wx
• Moment curve,
M
M
M
M
x
0
max
A
x
0
L
w
xdx
2
wL
8
2
Vdx
wx
RA RB
wL
2
wL L
wx w
x
2 2
M
w
2
at
2 Lx
x
dM
dx
V
0
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Vector Mechanics for Engineers: Statics
Sample Problem 7.4
Draw the shear and bending-
moment diagrams for the beam
and loading shown.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
SOLUTION:
• Taking entire beam as a free-body, determine
reactions at supports.
• Between concentrated load application
points, dV dx w
0 and shear is
constant.
• With uniform loading between D and E, the
shear variation is linear.
• Between concentrated load application
points, dM dx V constant.
The change
in moment between load application points is
equal to area under shear curve between
points.
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
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Vector Mechanics for Engineers: Statics
Sample Problem 7.4
SOLUTION:
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Taking entire beam as a free-body,
determine reactions at supports.
M A 0 :
D 24 ft 20 kips 6 ft 12 kips
12 kips 28
ft
14 ft
0
D 26 kips
Fy
0
:
A 20 kips 12
kips 26 kips 12
kips 0
y
Ay
18 kips
• Between concentrated load application points,
dV dx w
0 and shear is constant.
• With uniform loading between D and E, the shear
variation is linear.
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Vector Mechanics for Engineers: Statics
Sample Problem 7.4
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• Between concentrated load application
points, dM dx V constant.
The change
in moment between load application points is
equal to area under the shear curve between
points.
M
M
M
M
B
C
D
E
M
M
M
M
A
B
C
D
108
16
140
48
M
M
M
M
108
kip ft
92
kip ft
48
kip ft
0
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
B
C
D
E
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Vector Mechanics for Engineers: Statics
Sample Problem 7.6
Sketch the shear and bending-
moment diagrams for the
cantilever beam and loading
shown.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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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Vector Mechanics for Engineers: Statics
Sample Problem 7.6
SOLUTION:
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
dV
at A, VA
0,
w
w
dx
V 1
B A 2 0
V 1
B 2 0
V w a
w a
dV
at B,
w
dx
0
• With zero load, change in shear between B and C is
zero.
0
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Vector Mechanics for Engineers: Statics
Sample Problem 7.6
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
dM
at A, M A
0,
V
dx
M
M
B
C
M
M
A
B
1
3
1
2
w
w
0
0
a
a
2
La M 1 w a3La
• 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.
0
M
B
C
1
3
6
w
0
0
a
2
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