Chapter 7: Forces in Beams and Cables
최해진
hjchoi@cau.ac.kr
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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 7.6
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 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 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.
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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.
• 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 7.1
SOLUTION:
• Compute reactions and connection forces.
Consider entire frame as a free-body:
åME = 0:
- (2400 N )(3.6 m ) + F (4.8 m ) = 0
F = 1800 N
å Fy = 0 :
- 2400 N + 1800 N + E y = 0
å Fx = 0 :
E y = 600 N
Ex = 0
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Sample Problem 7.1
Consider member BCD as free-body:
åMB = 0:
- (2400 N )(3.6 m ) + C y (2.4 m ) = 0
C y = 3600 N
å MC = 0 :
- (2400 N )(1.2 m ) + B y (2.4 m ) = 0
B y = 1200 N
å Fx = 0 :
- Bx + C x = 0
Consider member ABE as free-body:
åMA = 0:
å Fx = 0 :
å Fy = 0 :
B x (2.4 m ) = 0
Bx = 0
B x - Ax = 0
Ax = 0
- Ay + B y + 600 N = 0 Ay = 1800 N
From member BCD,
å Fx = 0 :
- Bx + C x = 0
Cx = 0
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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:
åMJ = 0:
- (1800 N )(1.2 m ) + M = 0
M = 2160 N × m
å Fx = 0 :
F - (1800 N ) cos 41.7° = 0
F = 1344 N
å Fy = 0 :
-V + (1800 N ) sin 41.7° = 0
V = 1197 N
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Sample Problem 7.1
• Cut member BCD at K. Determine a force-couple
system equivalent to internal forces at K .
Consider free-body BK:
åMK = 0:
(1200 N )(1.5 m ) + M = 0
M = -1800 N × m
å Fx = 0 :
F =0
å Fy = 0 :
- 1200 N - V = 0
V = -1200 N
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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 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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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,
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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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.
• 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 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.
å Fy = 0 : - 20 kN - V1 = 0
V1 = -20 kN
å M 2 = 0 : (20 kN )(0 m ) + M 1 = 0
M1 = 0
Similarly,
V3
V4
V5
V6
= 26 kN
= 26 kN
= -14 kN
= -14 kN
M3
M4
M5
M6
= -50 kN × m
= +28 kN × m
= +28 kN × m
=0
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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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Sample Problem 7.3
SOLUTION:
• Taking entire beam as free-body,
calculate reactions at A and B.
• Determine equivalent internal forcecouple 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 N/m extends over
0.3 m of the beam, from A to C, and the
400-N load is applied at E.
• Plot results.
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Sample Problem 7.3
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at A and B.
åMA = 0:
B y (0.8m ) - (12 N )(0.3m ) - (400 N )(0.55m ) = 0
B y = 365 N
åMB = 0:
(12N )(0.65m ) + (400 N )(0.25m ) - A(0.8m ) = 0
A = 135 N
å Fx = 0 :
Bx = 0
• Note: The 400 N load at E may be replaced by a
400 N force and 1600 Nm couple at D.
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Sample Problem 7.3
• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From A to C:
å Fy = 0 : 135 - 40 x - V = 0
V = 515 - 40 x
å M1 = 0 :
- 135 x - 40 x( 12 x ) + M = 0
M = 135 x - 20 x 2
From C to D:
å Fy = 0 :
135 - 12 - V = 0
V = 123 N
åM2 = 0:
- 135 x + 12( x - 0.15) + M = 0
M = (1.8 + 123x ) N × m
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Sample Problem 7.3
• Evaluate equivalent internal force-couple
systems at sections cut within segments AC,
CD, and DB.
From D to B:
å Fy = 0 :
135 - 12m - 400 - V = 0
V = -277 N
åM2 = 0:
- 135 x + 12( x - 0.15) - 1600 + 400( x - 0.45) + M
=0
M = (1781.8 - 277 x ) N × cm
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Sample Problem 7.3
• Plot results.
From A to C:
V = 135 - 40 x
M = 135 x - 20 x 2
From C to D:
V = 123 N
M = (1.8 + 123 x ) N × m
From D to B:
V = -277 N
M = (1781.8 - 277 x ) N × m
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Relations Among Load, Shear, and Bending
Moment
• Relations between load and shear:
V - (V + DV ) - wDx = 0
dV
DV
= lim
= -w
dx Dx ®0 Dx
xD
VD - VC = - ò w dx = -(area under load curve)
xC
• Relations between shear and bending moment:
(M + DM ) - M - VDx + wDx Dx = 0
2
dM
DM
= lim
= lim V - 12 wDx = V
dx Dx ®0 Dx Dx ®0
(
)
xD
M D - M C = ò V dx = (area under shear curve)
xC
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Relations Among Load, Shear, and Bending
Moment
• Reactions at supports,
R A = RB =
wL
2
• Shear curve,
x
V - V A = - ò w dx = - wx
0
V = V A - wx =
wL
æL
ö
- wx = wç - x ÷
2
è2
ø
• Moment curve,
x
M - M A = ò Vdx
0
x
(
w
æL
ö
M = ò wç - x ÷dx = L x - x 2
2
ø
0 è2
wL2
M max =
8
)
dM
æ
ö
M
at
=
V
=
0
ç
÷
dx
è
ø
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Sample Problem 7.4
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.
Draw the shear and bendingmoment diagrams for the beam
and loading shown.
• 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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Sample Problem 7.4
SOLUTION:
• Taking entire beam as a free-body,
determine reactions at supports.
åMA = 0:
D(7.2 m ) - (20 kN )(1.8 m ) - (12 kN )(4.2 m )
- (12 kN )(8.4 m ) = 0
D = 26 kN
å F y =0 :
Ay - 20 kN - 12 kN + 26 kN - 12 kN = 0
Ay = 18 kN
• 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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Sample Problem 7.4
• 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 B - M A = +108 M B = +108 kN × m
M C - M B = -16
M C = +92 kN × m
M D - M C = -140 M D = -48 kN × m
M E - M D = +48
ME = 0
• With a linear shear variation between D
and E, the bending moment diagram is a
parabola.
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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.
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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.
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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.
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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.
• 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 å M D = 0.
• For other points on cable,
å M C2 = 0 yields y 2
å Fx = 0, å Fy = 0 yield Tx , T y
• Tx = T cosq = Ax = constant
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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 C to given point D. Forces are
horizontal force T0 at C and tangential force T at D.
• From force triangle:
T cosq = T0
T sin q = 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.
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T = T02 + W 2
tan q =
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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 W = wx , internal tension force
magnitude and direction are
wx
T = T02 + w 2 x 2
tan q =
T0
• Summing moments about D,
x
M
=
0
:
wx
- T0 y = 0
å D
2
or
wx 2
y=
2T0
The cable forms a parabolic curve.
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Sample Problem 7.8
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 freebody and summing moments about C.
The cable AE supports three vertical
loads from the points indicated. If
point C is 1.5 m below the left
support, determine (a) the elevation
of points B and D, and (b) the
maximum slope and maximum
tension in the cable.
• Calculate elevation of B by considering
AB as a free-body and summing
moments B. Similarly, calculate
elevation of D using ABCD as a freebody.
• Evaluate maximum slope and
maximum tension which occur in DE.
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Sample Problem 7.8
SOLUTION:
• 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,
åM
E
= 0:
and from taking cable portion ABC as a
free-body and summing moments about C.
åM
C
= 0:
- 1.5 Ax - 9 Ay + 18 = 0
Solving simultaneously,
Ax = -18 kN
Ay = 5 kN
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Sample Problem 7.8
• Calculate elevation of B by considering AB as
a free-body and summing moments B.
åM
B
= 0:
y B (18) - (5)(6 )
y B = -1.67 m
Similarly, calculate elevation of D using
ABCD as a free-body.
åM = 0:
- (18) y D - (5) (13.5) + (6)(7.5) + (12)(4.5) = 0
y D = 1.75 m
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Sample Problem 7.8
• Evaluate maximum slope and
maximum tension which occur in DE.
tan q =
14.7
15
Tmax =
18 kN
cosq
q = 43.4°
Tmax = 24.8 kN
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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 W = ws , the internal tension force
magnitude is
T = T02 + w 2 s 2 = w c 2 + s 2
c=
T0
w
• To relate horizontal distance x to cable length s,
T
ds
dx = ds cosq = 0 cosq =
T
q + s2 c2
s
x=ò
0
ds
q + s2 c2
= c sinh -1
s
c
and s = c sinh
x
c
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Catenary
• To relate x and y cable coordinates,
W
s
x
dy = dx tan q = dx = dx = sinh dx
T0
c
c
x
y - c = ò sinh
0
x
x
dx = c cosh - c
c
c
x
y = c cosh
c
which is the equation of a catenary.
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