CE 102 Statics

1
CE 102 Statics
Chapter 5 Forces in Beams and Cables
2
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
Sample Problem 7.4 Sample Problem 7.5 Cables With
Concentrated Loads Cables With Distributed
Loads Parabolic Cable Sample Problem 7.6 Catenary
3
Introduction

• Preceding chapters dealt with
• determining external forces acting on a
  structure and
• 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
• Beams - usually long, straight, prismatic members
  designed to support loads applied at various
  points along the member.
• Cables - flexible members capable of withstanding
  only tension, designed to support concentrated or
  distributed loads.

4
Internal Forces in Members

• Internal forces equivalent to F and -F are
  required for equilibrium of free-bodies AC and CB.

5
Sample Problem 5.1

• 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.

Determine the internal forces (a) in member ACF
at point J and (b) in member BCD at K.
6
Sample Problem 5.1
7
Sample Problem 5.1
8
Sample Problem 5.1
9
Sample Problem 5.1
10
Various Types of Beam Loading and Support

• 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
• determine shearing forces and bending moments
  produced by applied loads
• select cross-section best suited to resist
  shearing forces and bending moments

11
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.

12
Shear and Bending Moment in a Beam

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

13
Shear and Bending Moment Diagrams
14
Sample Problem 5.2

• 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.

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

• Plot results.

15
Sample Problem 5.2
16
Sample Problem 5.2

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

17
Sample Problem 5.3

• 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.

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.

• Plot results.

18
Sample Problem 5.3

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

19
Sample Problem 5.3
20
Sample Problem 5.3
21
Sample Problem 5.3

• Plot results.

From A to C
From C to D
From D to B
22
Relations Among Load, Shear, and Bending Moment
23
Relations Among Load, Shear, and Bending Moment
24
Sample Problem 5.4

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

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

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

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

25
Sample Problem 5.4

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

26
Sample Problem 5.4

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

27
Sample Problem 5.5

• 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.

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

• 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.

28
Sample Problem 5.5

• 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.

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

29
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.

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
• concentrated vertical loads on given vertical
  lines,
• weight of cable is negligible,
• cable is flexible, i.e., resistance to bending is
  small,
• 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.

31
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.

32
Cables With Distributed Loads

• For cable carrying a distributed load
• cable hangs in shape of a curve
• internal force is a tension force directed along
  tangent to curve.

• 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.

33
Parabolic Cable

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

34
Sample Problem 5.6

• 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.

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.

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

35
Sample Problem 5.6
36
Sample Problem 5.6
37
Sample Problem 5.6

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

38
Catenary

• Consider a cable uniformly loaded along the cable
  itself, e.g., cables hanging under their own
  weight.

39
Catenary

• To relate x and y cable coordinates,

which is the equation of a catenary.
40
Problem 5.7
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.
5 ft
5 ft
5 ft
D
A
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
41
Problem 7.158
5 ft
5 ft
5 ft
Solving Problems on Your Own
D
A
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)
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
the maximum tension in the cable, (c) the bending
moment at F and G.
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 S M 0 if you know the position, or S Fx
0 and S Fy 0 if you know the slope, to
generate needed equations of equilibrium.
42
Problem 7.158
5 ft
5 ft
5 ft
Solving Problems on Your Own
D
A
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)
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
the maximum tension in the cable, (c) the bending
moment at F and G.
3. The tension in each section can be determined
from the equations of equilibrium.
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.
43
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
Identify points of the cable where useful
information exists. Draw a free-body diagram of
one of the two portions of the cable. Use S M
0 to generate the equations of equilibrium.
A
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
Free Body Portion AB
Ay

S MB 0 T0(2 ft) - Ay(5 ft) 0
5 ft
A
Ay 0.4 T0
2 ft
T0
B
TBC
FBF
44
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
Identify points of the cable where useful
information exists. Draw a free-body diagram of
one of the two portions of the cable. Use S M
0 to generate the equations of equilibrium.
A
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
Free Body Portion ABC

S MC 0 T0(3 ft) - 0.4 T0 (10 ft)
FBF (5 ft) 0
0.4T0
5 ft
5 ft
A
TCD
T0
3 ft
FBF 0.2 T0
B
C
FBF
FCH
45
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
Identify points of the cable where useful
information exists. Draw a free-body diagram of
one of the two portions of the cable. Use S M
0 to generate the equations of equilibrium.
A
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
FBF 0.2 T0
Free Body Portion CD
Dy

S MC 0 Dy(5 ft) - T0 (3 ft) 0
5 ft
Dy 0.6 T0
T0
TBC
D
3 ft
C
FCH
46
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
Identify points of the cable where useful
information exists. Draw a free-body diagram of
one of the two portions of the cable. Use S M
0 to generate the equations of equilibrium.
A
2 ft
B
3 ft
C
F
H
G
E
7.5 ft
200 lb
FBF 0.2 T0
Free Body Portion BCD
0.6 T0

S MB 0 0.6T0 (10 ft) - T0 (2 ft)
- FCH (5ft) 0
5 ft
5 ft
TAB
2 ft
T0
D
B
FCH 0.8 T0
C
FCH
0.2 T0
47
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
A
The tension in each section can be determined
from the equations of equilibrium.
2 ft
B
3 ft
C
F
H
FCH 0.8 T0
FBF 0.2T0
G
E
7.5 ft
200 lb
Free Body Beam EFH
S ME 0
0.2 T0
0.8 T0
Ey
0.2T0 (5 ft) 0.8 T0 (10 ft) - (200
lb)(7.5 ft) 0
5 ft
5 ft
H
F
Ex
G
E
T0 166.67 lb
7.5 ft
200 lb
FCH 0.8(166.67) FCH 133.33 lb T
FBF 0.2(166.67) FBF 33.33 lb T
48
5 ft
5 ft
5 ft
Problem 7.158 Solution
D
A
The maximum tension occurs in the steepest
portion of the cable.
2 ft
B
3 ft
C
F
H
T0 166.67 lb
G
E
7.5 ft
200 lb
Tm
Dy 0.6 T0
Tm T02 (0.6T0)2 1.1662T0
1.1662(166.67)
D
T0
Tm 194.4 lb
49
Beam EFH
Problem 7.158 Solution
FCH 133.33 lb
FBF 33.33 lb
The moments at points G and F are determined by
using free-body diagrams for two sections of the
beam.
Ey
5 ft
5 ft
H
F
Ex
G
E
7.5 ft
200 lb
133.33 lb
V
S MG 0 (133.33 lb)(2.5 ft) - MG 0
MG 333 lb-ft
MG
G
2.5 ft
S MF 0 (133.33 lb)(5 ft)
- (200 lb)(2.5 ft) - MF 0
MF 166. 7 lb-ft
133.33 lb
V
G
MF
2.5 ft
2.5 ft
200 lb
50
Problem 5.8
y
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.
A
x
B
L
51
Problem 7.159
Solving Problems on Your Own
y
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.
A
B
x
L
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).
52
y
Problem 7.159 Solution
The shear V can be obtained by integrating the
function -w(x) and the moment M can be obtained
by integrating V (x).
A
x
B
x
L
px 2L
-w -w0 cos
px 2L
V - wdx - w0 sin C1
ò
dM dx
px 2L
V - w0 sin C1
px 2L
M Vdx w0 cos C1x
C2
ò
53
y
Problem 7.159 Solution
The shear V can be obtained by integrating the
function -w(x) and the moment M can be obtained
by integrating V (x).
A
x
B
x
L
Boundary conditions
At x 0 V C1 0 C1 0 At
x 0 M w0 (2L/p)2 cos (0) C2 0

C2 -w0 (2L/p)2
54
y
Problem 7.159 Solution
The shear V can be obtained by integrating the
function -w(x) and the moment M can be obtained
by integrating V (x).
A
x
B
x
L
px 2L
M w0 cos - w0
px 2L
M w0 ( -1 cos )
Mmax w0 -1 0
Mmax at x L
4 p2
Mmax w0 L2
55
Problem 5.9
120 mm
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.
B
A
E
100 mm
J
F
100 mm
k
100 mm
D
C
280 mm
56
120 mm
Problem 7.161
B
A
Solving Problems on Your Own
E
100 mm
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.
J
F
100 mm
k
100 mm
D
C
280 mm
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.
57
120 mm
Problem 7.161 Solution
B
A
Cut the member at a point, and draw the free-body
diagram of each of the two portions.
Tx
100 mm
J
T
Ty
100 mm
k
A
MK 300 N-m
V
F
D
3002 1602 340
300
D
280 - 120 160
A
F
8
MK 300 N-m
T
15
17
Ty
V
k
D
100 mm
15 17
Tx
D
Ty T
C
8 17
Tx T
280 mm
58
120 mm
Problem 7.161 Solution
8 17
Tx T
B
A
Select one of the two free-body diagrams and use
it to write the equations of equilibrium.
J
200 mm
15 17
Ty T
k
Free Body ABK
MK 300 N-m
V
F
D
SMk 0 300 N-m - T(0.2 m) -
T(0.12 m) 0
T 1500 N
59
120 mm
Problem 7.161 Solution
B
A
8 17
Tx (1500) 705.88 N
100 mm
J
15 17
Ty (1500) 1323.53 N
MJ
V
F
Free Body ABJ
D
SMJ 0 MJ - (705.88 N)(0.1 m) - (1323.53
N)(0.12 m) 0
MJ 229 N-m

SFx 0 705.88 N - V 0 V
706 N

SFy 0 -F - 1323.53 N 0 F
1324 N
60
Problem 5.10
6 m
9 m
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
A
2.25 m
a
B
C
60 kg/m
distance a, (b) the length of the cable, (c) the
components of the reaction at A.
61
Problem 7.162
6 m
9 m
Solving Problems on Your Own
A
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.
2.25 m
a
B
C
60 kg/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 S M 0 if you know the position, or S Fx
0 and S 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).
62
Free Body Portion AC
Problem 7.162 Solution
Identify points of the cable where useful
information exists. Cut the cable at these points
and draw a free-body diagram. Use S M 0 if you
know the position, or S Fx 0 and S Fy 0 if
you know the slope to generate needed equations
of equilibrium.
9 m
Ay
T0
A
a
T0
C
9w
4.5 m
S Fy 0 Ay - 9w 0,
Ay 9w

S MA 0 To a - (9w)(4.5 m) 0, To a
40.5w (1)
63
Free Body Portion CB
Problem 7.162 Solution
By
Identify points of the cable where useful
information exists. Cut the cable at these points
and draw a free-body diagram. Use S M 0 if you
know the position, or S Fx 0 and S Fy 0 if
you know the slope to generate needed equations
of equilibrium.
6 m
yB
B
T0
C
T0
3 m
6w
S Fy 0 By - 6w 0,
By 6w

S MB 0 (6w)(3 m) - To yB 0, To
yB 18w
64
Free Body Entire Cable
Problem 7.162 Solution
9 m
6 m
9w
6w
To a 40.5w (1)
A
2.25 m
T0
B
C
T0
yB a - 2.25

Identify points of the

cable where useful

information exists. Cut the cable at these
points and draw a free-body diagram. Use S M 0
if you know the position, or S Fx 0 and S Fy
0 if you know the slope, to generate needed
equations of equilibrium.
7.5 m
15w
S MA 0 6w (15 m) - 15w(7.5 m) To (2.25 m)
0
To 10w
(10w) a 40.5w a 4.05
m
Using (1)
65
9 m
6 m
Problem 7.162 Solution
9w
6w
10w
(b) Length of AC CB
A
2.25 m
The length of the cable can be determined from
(7.10).
B
4.05
C
10w
7.5 m
15w
Portion AC
xA 9 m, yA a 4.05 m
yA xA
yA xA
2 3
2
4
( )
( )

2 5
SAC xA 1 -
...
SAC 9 m 1 (2/3)(0.45)2 - (2/5)(0.45)2 ...
10.067 m
yB xB
Portion CB
xB 6 m, yB 4.05 - 2.25 1.8 m
0.3
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
(c) Components of Reaction at A
Problem 7.162 Solution
9 m
6 m
9w
6w
10w
A
2.25 m
B
4.05
C
10w
7.5 m
15w
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 N Ax 5890 N
67
Problem 5.13
1500 lb/ft
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.
B
A
C
6000 lb
4 ft
6 ft
68
Problem 7.163
Solving Problems on Your Own
1500 lb/ft
B
A
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.
C
6000 lb
4 ft
6 ft
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.
69
Problem 7.163 Solution
9 kips
3 ft
Draw a free-body diagram for the entire beam, and
use it to determine the reactions at the beam
supports.
C
B
A
4 ft
6 kips
6 ft
S MA 0 (6 kips)(10 ft) - (9 kips)(7 ft) B
(4 ft) 0
B 0.75 kips


S Fy 0 A 0.75 kips 6 kips - 9 kips 0

A 2.25 kips
70
Problem 7.163 Solution
Draw the shear diagram.
1.5 kips/ft
A
C
B
2.25 kips
6 kips
0.75 kips
6 ft
4 ft
V (kips)
3 kip
D
2.25 kip
x
A
B
C
VB w
3 kips 1.5 kips/ft
BD
- 6 kip
2 ft
4 ft
71
Draw the bending-moment diagram.
Problem 7.163 Solution
1.5 kips/ft
A
C
B
2.25 kips
6 kips
0.75 kips
6 ft
4 ft
V (kips)
3 kip
1 2
(3 kips)(2 ft) (3 kip-ft)
D
2.25 kip
(9 kip-ft)
x
A
B
C
(- 12 kip-ft)
VB w
3 kips 1.5 kips/ft
BD
- 6 kip
2 ft
4 ft
(12 kip-ft)
Mmax 12 kip-ft
(9 kip-ft)
M (kip-ft)
6 ft from A
x
A
B
D
C