IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Forces in Beams and Cables
22.3.2016
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
22.3.2016
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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 equilibrium under application of cable and
member contact forces.
• Internal forces equivalent to a forcecouple system are necessary for equilibrium of free-bodies JD and ABCJ.
• An internal force-couple system is
required for equilibrium of two-force
members which are not straight.
22.3.2016
Dr. Engin Aktaş
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.1
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.
Determine the internal forces (a) in
member ACF at point J and (b) in
member BCD at K.
22.3.2016
• Cut member BCD at K. Determine
force-couple system equivalent to
internal forces at K by applying
equilibrium conditions to either part.
Dr. Engin Aktaş
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample
Problem
7.1
SOLUTION:
• Compute reactions and connection forces.
Consider entire frame as a free-body:
ME  0:
 2400 N 3.6 m   F 4.8 m   0
F  1800 N
 Fy  0 :
 2400 N  1800 N  E y  0
 Fx  0 :
22.3.2016
Dr. Engin Aktaş
E y  600 N
Ex  0
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.1
Consider member BCD as free-body:
MB  0:
 2400 N 3.6 m  C y 2.4 m  0
C y  3600 N
 MC  0 :
 2400 N1.2 m  B y 2.4 m  0
B y  1200 N
 Fx  0 :
 Bx  C x  0
Consider member ABE as free-body:
MA  0:
 Fx  0 :
 Fy  0 :
Bx 2.4 m  0
Bx  0
Bx  Ax  0
Ax  0
 Ay  B y  600 N  0 Ay  1800 N
From member BCD,
 Fx  0 :
22.3.2016
Dr. Engin Aktaş
 Bx  C x  0
Cx  0
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.1
• Cut member ACF at J. The internal forces at J are
represented by equivalent force-couple system.
Consider free-body AJ:
MJ  0:
 1800 N 1.2 m   M  0
M  2160 N  m
 Fx  0 :
F  1800 N  cos 41.7  0
F  1344 N
 Fy  0 :
V  1800 N  sin 41.7  0
22.3.2016
Dr. Engin Aktaş
V  1197 N
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.1
• Cut member BCD at K. Determine a force-couple
system equivalent to internal forces at K .
Consider free-body BK:
MK  0:
1200 N 1.5 m   M  0
 Fx  0 :
M  1800 N  m
F 0
 Fy  0 :
 1200 N  V  0
22.3.2016
Dr. Engin Aktaş
V  1200 N
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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:
1) determine shearing forces and bending
moments produced by applied loads
2) select cross-section best suited to resist
shearing forces and bending moments
22.3.2016
Dr. Engin Aktaş
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
22.3.2016
Dr. Engin Aktaş
10
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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’.
22.3.2016
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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,
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.
22.3.2016
Dr. Engin Aktaş
12
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.2
SOLUTION:
• Taking entire beam as a free-body,
calculate reactions at B and D.
Draw the shear and bending moment
diagrams for the beam and loading
shown.
22.3.2016
• Find equivalent internal force-couple
systems for free-bodies formed by
cutting beam on either side of load
application points.
• Plot results.
Dr. Engin Aktaş
13
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.2
SOLUTION:
• 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.
 Fy  0 :  20 kN  V1  0
V1  20 kN
 M 2  0 : 20 kN 0 m   M 1  0
M1  0
Similarly,
22.3.2016
V2  20 kN
M 2  50 kN  m
V3  26 kN
V4  26 kN
M 3  50 kN  m
M 4  28 kN  m
V5  14 kN
V6  14kN
M 5  28 kN  m
M 6  0 kN  m
Dr. Engin Aktaş
14
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.2
• Plot results.
Note that shear is of constant value
between concentrated loads and
bending moment varies linearly.
22.3.2016
Dr. Engin Aktaş
15
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
0.75 m
2m
0.5 m
0.75 m
Sample Problem 7.3
• Taking entire beam as free-body,
calculate reaction B.
2 kN
3 kN/ m
A
E
C
D
• Determine equivalent internal forcecouple systems at sections cut within
segments AC, CD, and DB.
B
Draw the shear and bending moment
diagrams for the beam AB. The
distributed load of 3 kN/m. extends
over 2 m. of the beam, from A to C,
and the 2 kN load is applied at E.
22.3.2016
• Plot results.
Dr. Engin Aktaş
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
0.75 m
2m
0.5 m
0.75 m
Sample Problem 7.3
• Taking entire beam as a free-body, calculate
reaction at B .
2 kN
3 kN/ m
C
2m
3 kN/ m
A
D
1.25 m
 Fx  0 :
F
2 kN
B Bx
C D
1 kNm
22.3.2016
2kN0.75 m.  6kN3m.   M B  0
B
M B  19.5kNm
0.75 m
A
MB  0:
E
By
y
 0:
Bx  0
By  6kN  2kN  0
By  8kN
MB
• Note: The 2 kN load at E may be replaced by a 2 kN
force and 1 kNm couple at D.
Dr. Engin Aktaş
17
IZMIR INSTITUTE OF TECHNOLOGY
3 kN/ m
1.25 m
2 kN
19.5 kNm
B
A
C
1
2
D
M
3
8 kN
1 kNm
3x 12 x   M  0
 3x  V  0
 Fy  0 :
V
V
 Fy  0 :
2 kN
M
x
M
-8 kN
M
- 6 kNm
 6 V  0
- 9.5 kNm
- 10.5 kNm
- 19.5 kNm
3
V  6
6x 1  M  0 M  6  6x kN  m.
2m  x  2.75m
From D to B:
 Fy  0 :
V
- 6 kN
M2  0:
V
1 kNm
0  x  2m
M  1.5 x 2
V  3x
From C to D:
M
22.3.2016
• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From A to C:
 M1  0 :
x
x
AR231 Fall12/13
Sample Problem 7.3
0.75 m
2m
Department of Architecture
 6  2 V  0
V  8
 0 : 6x 1  2( x  2.75) 1  M  0
M  12.5  8x kN  m.
2.75m  x  4m
Dr. Engin Aktaş
18
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Relations Among Load, Shear, and Bending Moment
• Relations between load and shear:
V  V  V   wx  0
dV
V
 lim
 w
dx x 0 x
xD
VD  VC    w dx  area under load curve 
xC
• Relations between shear and bending moment:
M  M   M  Vx  wx x  0

2

dM
M
 lim
 lim V  12 wx  V
dx x 0 x x 0
xD
M D  M C   V dx  area under shear curve 
xC
22.3.2016
Dr. Engin Aktaş
19
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Relations Among Load, Shear, and Bending Moment
• Reactions at supports,
R A  RB 
wL
2
• Shear curve,
x
V  V A    w dx   wx
0
V  V A  wx 
• Moment curve,
wL
L

 wx  w  x 
2
2

x
M  M A   Vdx
0
x

w
L

M   w  x dx  L x  x 2
2

0 2
wL2
M max 
8
22.3.2016
Dr. Engin Aktaş

dM


M
at
 V  0

dx


20
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 7.6
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.
Sketch the shear and bendingmoment diagrams for the
cantilever beam and loading
shown.
• 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.
22.3.2016
Dr. Engin Aktaş
21
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample
Problem
7.6
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.
at A, V A  0,
dV
  w   w0
dx
VB  V A   12 w0 a
at B,
VB   12 w0 a
dV
 w  0
dx
• With zero load, change in shear between B and C is
zero.
22.3.2016
Dr. Engin Aktaş
22
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample
Problem
7.6
• 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.
at A, M A  0,
dM
V  0
dx
M B  M A   13 w0 a 2
M B   13 w0 a 2
M C  M B   12 w0 a L  a  M C   16 w0 a3L  a 
• 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.
22.3.2016
Dr. Engin Aktaş
23
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
22.3.2016
Dr. Engin Aktaş
24
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
• 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  M D  0.
• For other points on cable,
 M C2  0 yields y2
 Fx  0,  Fy  0 yield Tx , Ty
• Tx  T cos  Ax  constant
22.3.2016
Dr. Engin Aktaş
25
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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 C to given point D. Forces are
horizontal force T0 at C and tangential force T at D.
• From force triangle:
T cos  T0
T sin   W
W
T0
• 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.
T  T02  W 2
22.3.2016
Dr. Engin Aktaş
tan 
26
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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 W  wx , internal tension force
magnitude and direction are
wx
T  T02  w 2 x 2
tan 
T0
• Summing moments about D,
x
M

0
:
wx
 T0 y  0
 D
2
or
wx 2
y
2T0
The cable forms a parabolic curve.
22.3.2016
Dr. Engin Aktaş
27
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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 W  ws , the internal tension force
magnitude is
T  T02  w2 s 2  w c 2  s 2
c
T0
w
• To relate horizontal distance x to cable length s,
T
ds
dx  ds cos  0 cos 
T
q  s2 c2
s
x
0
22.3.2016
ds
q  s2 c2
Dr. Engin Aktaş
 c sinh 1
s
c
and s  c sinh
x
c
28
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Catenary
• To relate x and y cable coordinates,
W
s
x
dy  dx tan   dx  dx  sinh dx
T0
c
c
x
y  c   sinh
0
y  c cosh
x
x
dx  c cosh  c
c
c
x
c
which is the equation of a catenary.
22.3.2016
Dr. Engin Aktaş
29
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Examples
22.3.2016
Dr. Engin Aktaş
30
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer & Johnston)
SOLUTION:
1.4 m
1.6 m
• Taking entire beam as free-body,
calculate reactions A&B.
1.8 kN/ m
A
D
C
B
• Determine equivalent internal forcecouple systems at sections cut within
segments AC, CD, and DB.
4 kN
Draw the shear and bending moment
diagrams for the beam AB. The
distributed load of 1.8 kN/m. extends
over 2.6 m. of the beam, from A to D,
and the 4 kN load is applied at C.
22.3.2016
• Plot results.
Dr. Engin Aktaş
31
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer & Johnston)
SOLUTION:
• Taking entire beam as a free-body, calculate
reaction at B .
1.4 m
1.6 m
1.8 kN/ m
D
A
B
C
MB  0:
1.8kN / m2.6 m2.7m  4kN 2.4m  Ay 4m  0
4 kN
Ay  5.559kN
1.4 m
1.6 m
 Fx  0 :
1.8 kN/ m
D
Ax A
C
Ay
22.3.2016
4 kN
B
F
y
 0:
Ax  0
By  5.559kN  4kN  1.8kN / m(2.6m)  0
B y  3.121kN
By
Dr. Engin Aktaş
32
IZMIR INSTITUTE OF TECHNOLOGY
• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From A to C: 0  x  1.6m
1.8 kN/ m
D
A
C
2
3
1
4 kN
By
M
x
F
4 kNV
V
4 kN
2.68 kN
-1.32 kN
-3.12 kN
6.59 kNm
M
22.3.2016
M  5.56 x  0.9 x 2
V  5.56 1.8x
5.56 1.8x   4  V  0 V  1.56 1.8x
 x
M 2  0 :  5.56 x  1.8 x   4 x  1.6   M  0

M
2
M  0.9 x 2  1.56 x  6.4
From D to B: 2.6m  x  4m
-3.12 kN
 Fy  0 : 5.56  4.68  4 V  0 V  3.12 kN
y
x
V
 5.56 x  1.8 x 12 x   M  0
From C to D: 1.6m  x  2.6m
M
5.56 kN
 M1  0 :
 Fy  0 : 5.56 1.8x V  0
V
x
AR231 Fall12/13
Example (Beer & Johnston)
1.4 m
1.6 m
Department of Architecture
4.37 kNm
0 kNm
M
 0:
3
 0 :  5.56x  1.8(2.6)x 1.3  4x 1.6  M 
M  3.12 x  12.484 kN  m.
Dr. Engin Aktaş
33
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer & Johnston)
SOLUTION:
2 kN/ m
2 kN/ m
G
C
A
• Taking entire beam as free-body,
calculate reactions A&G.
D
E
F
B
Draw the shear and bending moment
diagrams for the beam AB.
22.3.2016
• Determine equivalent internal forcecouple systems at sections cut within
segments AC, CD, and DE. Use
symmetry for the rest of the beam.
• Plot results.
Dr. Engin Aktaş
34
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer & Johnston)
SOLUTION:
2 kN/ m
• Taking entire beam as a free-body, calculate
reaction at G .
2 kN/ m
G
C
D
E
B
F
M
G
 0:
2kN / m0.25 m0.725m  0.5kN 0.45m  1.50.30  0.15
 2kN / m 0.25m0.125m   Ay 0.6m  0
Ay  1.375kN
2 kN/ m
 Fx  0 :
2 kN/ m
G
C
D
Ax
F
y
Ay
 0:
E
Ax  0
B
F
Gy
Gy  1.375kN  22kN / m(0.25m)  0.5kN  0.5kN  0.75kN  0
G y  1.375kN
22.3.2016
Dr. Engin Aktaş
35
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
2 kN/ m
2 kN/ m
G
C
D
E
B
F
1.375 kN
1.375 kN
0.875 kN
V
AR231 Fall12/13
0.5 kN
0.375 kN
0
0
-0.375 kN
-0.5 kN
-0.875 kN
0.125 kNm
0.06875 kNm
M
0
0
-0.0625 kNm
22.3.2016
0.06875 kNm
-0.0625 kNm
Dr. Engin Aktaş
36