Deflections in Stiffening Trusses of
Suspension Bridges
Dorothy Goettler
A Seminar submitted to the Faculty of
Rensselaer at Hartford
in Partial Fulfillment of the Requirements for the
Degree of MASTER OF SCIENCE
Mechanical Engineering
Approved by Ernesto Gutierrez-Miravete
Rensselaer at Hartford
Hartford, Connecticut
December 1999
Table of Contents
NOMENCLATURE ....................................................................................................................................................3
LIST OF FIGURES .....................................................................................................................................................4
ABSTRACT .................................................................................................................................................................5
INTRODUCTION .......................................................................................................................................................6
PROBLEM DESCRIPTION ......................................................................................................................................7
FORMULATION OF EQUATIONS .........................................................................................................................8
ANALYSIS OF CABLE ..................................................................................................................................................8
DEFLECTIONS OF UNSTIFFENED BRIDGES ................................................................................................................. 10
APPLICATION TO STIFFENED BRIDGES .................................................................................................................. 14
Dead load only .................................................................................................................................................... 14
Addition of live load ............................................................................................................................................ 14
DEFLECTIONS OF STIFFENING TRUSSES .................................................................................................................... 16
Development of equations ..................................................................................... Error! Bookmark not defined.
Numerical Approaches........................................................................................................................................ 18
1. Successive approximations ........................................................................................................................................... 18
Solutions in one Variable ................................................................................................................................................. 22
2. Bisection Method ......................................................................................................................................................... 24
3. Fixed Point Iteration ..................................................................................................................................................... 26
4. Newton-Raphson Method............................................................................................................................................. 28
RESULTS, ERROR ANALYSIS AND DISCUSSION ........................................................................................... 29
CONCLUSIONS ........................................................................................................................................................ 29
REFERENCES .......................................................................................................................................................... 30
APPENDIX A............................................................................................................................................................. 31
APPENDIX B ............................................................................................................................................................. 32
APPENDIX C............................................................................................................................................................. 33
2
Nomenclature
Ac
Ec
f
fw
H
Hp
Hw
h
k
L
l
Mw
Mp
Mx
P
p
s


w
Tensile area of cable
Young’s modulus of cable
Sag of cable
Sag of cable under action of dead load
Horizontal component of tensile force in cable
Horizontal component of tensile force produced in the cable by the live load
Horizontal component of tensile force produced in the cable by the dead load
Difference in elevation of the ends of the cable
Variable defined by Equation 16
Integral defined by Equation 12
Span of beam
Bending moment due to dead load
Bending moment due to live load
Bending moment at given cross section
Live load, acting at a point
Live load, uniformly distributed over a given length
Length of cable
Horizontal Deflection of cable
Vertical deflection of cable
Dead load, uniformly distributed over length of bridge
3
List of Figures
Figure 1
Components of a suspension bridge
Page 6
Figure 2
Cable under uniform load, ends supported at different heights.
Page 8
Figure 3
Unstiffened bridge under uniformly distributed live load.
Page 10
Figure 4
Infinitely small element of cable from Figure 2
Page 11
Figure 5
Stiffened bridge under dead load only.
Page 14
Figure 6
Stiffening truss under point load.
Page 16
Figure 7
Total deflection of stiffening truss
Page 21
Figure 8
Graph of equation to be solved, g(x)
Page 23
Figure 9
Results of bisection method
Page 25
Figure 10 Graphical representation of h(xk)
Page 26
Figure 11 Graphical representation of h(xk) and g(x)
Page 26
Figure 12 Divergence of fixed point iteration method
Page 27
Figure 13 Results of Newton-Raphson Method
Page 28
4
Abstract
The stiffening truss of a typical single span suspension bridge will be analyzed in order to determine the
maximum vertical deflections due to a single localized live load applied in addition to the dead load of the
structure. This will be accomplished by developing the necessary equations for increasingly complex and
accurate models of the bridge and solving them using the assumption of a specific geometry that will be
described. The solution will be attempted using five different methods: successive iterations,
trigonometric series, bisection, fixed point integration and the Newton-Raphson method. Finally, the
results obtained from each of these will be compared and the results discussed.
5
Introduction
The suspension bridge is a structure consisting of either a roadway or a truss suspended from two or more
cables that pass over two towers and are anchored by backstays or towers to a firm foundation. In the
simplest type of suspension bridge, applicable to short spans, the floor system attaches directly to the
cables by hangers. However, the structure lacks rigidity, which allows wind loads and moving live loads
to distort the cables and produce a wave motion on the roadway. For longer spans it becomes necessary to
connect the floor systems to stiffening trusses or girders that distribute the moving loads more uniformly
to the hangers. This method of distribution reduces the distortion of the cable by distributing the
concentrated live loads over a considerable length. When the roadway is supported by a truss that is hung
from the cable in this way, the structure is called a stiffened suspension bridge.
Hangers
Support Towers
Roadway may be stiffened with a
truss (see detail below)
Roadway
Stiffening Truss
Figure 1
Components of a suspension bridge.
Cables of larger sizes, such as those up to 36 inches in diameter used for the George Washington Bridge
in New York City, are assembled (or spun) in the field by using pencil thick wires laid parallel and bound
securely together. For smaller cables, strands of wire wound spirally in the factory are assembled into
cables. Factory-made strands of parallel wires were used for the first time in 1968 for the 15 inch
diameter cables of the Newport Bridge in Rhode Island. The hangers which connect the floor system to
the cable are formed of twisted wire ropes. The floor system, stiffening truss and towers are constructed
of rolled steel.
The longest bridges in the world are all of the suspension type. The Akashi Straight Bridge in Japan is
the world’s longest suspension bridge as of 1998 with a 6,633-foot center span. To provide a reference
for the order of magnitude of these bridges, Appendix A contains a table of representative large
suspension bridges in North America.
6
Problem Description
Before the deflections in the stiffening truss of the suspension bridge may be determined, the equations to
be solved must be developed. The first step in doing this is to analyze the most basic form of a
suspension bridge, a simple cable supported from two rigid towers. Equations for the length and sag of
this cable will be important for the calculations that follow and must be developed at this point. Also, the
method used to determine the reaction forces, which will also be required throughout the analysis, will be
used first for the case of a simple cable.
Once the behavior of the cable is known, it is extended to the case of the unstiffened suspension bridge.
This consists of a roadway or other structure suspended from the cable by a large number of wire rope
hangers. The deflections are then calculated first for the straightforward case in which all of the reaction
forces are known, then for the more challenging but realistic case in which the horizontal component of
the tensile force produced in the cable by the live load in unknown. This situation will require a system of
equations to be solved simultaneously for two unknowns.
These equations can then be applied to the problem of a stiffened suspension bridge, from which the
behavior of the stiffening trusses themselves can be determined. The deflections in the stiffening trusses
will then be determined using a variety of methods, and the results compared.
It should be noted that in most cases the analysis is performed first assuming only the weight of the
structure itself, or dead weight. Once this form of the equations has been developed, they are extended to
the more general case of a live load, which essentially represents a single pedestrian or vehicle crossing
the bridge. It would also be possible to extend this to the more general case of a distributed live load over
the entire span of the bridge.
In order to reduce the tremendously complicated problem of a suspension bridge into one manageable
enough to describe using mathematical equations, certain assumptions had to be made. These include but
are certainly not limited to those listed below:
1. Both ends of bridge are assumed to be at the same height.
2. The live load is assumed to be small relative to the dead load.
3. The live load is assumed to be static.
4. The span of the bridge is assumed to be long relative to its width.
5. Environmental factors such as wind and temperature fluctuations are neglected.
6. The hangers are sufficiently close together that it may be assumed the dead load is uniformly
transmitted to the cable.
7
Formulation of Equations
Analysis of cable
In this section, a simple cable suspended at both ends will be analyzed. The results from this analysis will
be required for the development of the equations for the suspension bridge.
Figure 2
Cable under uniform load, ends supported at different heights.
For a uniform, perfectly flexible cable fixed at points A and B under a uniformly distributed load (see
Figure 1), the equation of moments about point C is written as:
Mx H
h
x  Hy  0
l
Equation 1
In this equation Mx denotes the bending moment at the cross section m-n of a simply supported beam of
span l and carrying the load acting on the cable. In the particular case when the load of intensity w is
uniformly distributed along the horizontal projection of the cable, as the dead weight of the actual
suspension bridge would be, we find that:
Mx 
wx
l  x 
2
The result above is developed in more detail in Reference 11, chapter 7. Substituting this into Equation 1
gives:
y
wx
l  x   h x
2H
l
This indicates that the curve is in this case a parabola with a vertical axis. If the ends of the cable are on
the same level, as will be assumed in all future analysis, this may be reduced to:
y
wx
l  x 
2H
Equation 2
8
Applying this equation to the mid-point of the cable ( x = l / 2 ) where the y coordinate of the curve
represents the sag f, we obtain:
wl  l 
l  
4H  2 
y
f 
wl 2
8H
Equation 3
This holds true in the more general case illustrated in Figure 2 as well if f is measured from the mid-point
of the straight line between points A and B.
To obtain the complete equations for the stiffening truss of a suspension bridge, the length of the cable
will be required. It may be obtained from the following equation:
l
1
2
s   (1  y ' ) dx
2
0
Developing the expression under the integral sign into a series and substituting Equation 2 for y, we
obtain from Reference 1:
s  l (1 
8 f 2 32 f 4 256 f 6


 ...)
3 l2
5 l4
7 l6
In the case of relatively flat parabolic curves, where f / l  1 / 10, the terms to this series rapidly become
extremely small, and may be reasonably well approximated by taking only the first two terms. The
simplified formula for the length of the cable is therefore:
s  l (1 
8 f2
)
3 l2
Equation 4
9
Deflections of unstiffened bridges
This is the simplest form of a suspension bridge. Its analysis allows the equations to be derived for a
simple case before they are applied to the more challenging application of stiffened bridges. We will
therefore assume that the curve of the cable under the action of the dead load is again a parabola and will
concentrate on the deflections in the cable produced by a live load.
Figure 3
Unstiffened bridge under uniformly distributed live load.
Figure 3 illustrates a symmetrical case in which in which a load of intensity p is uniformly distributed
along the distance 2a of the span. The line drawn indicates the shape of the cable under the action of the
dead load w only. We let fw and Hw denote the corresponding values of the sag of the cable and of the
horizontal component of the tensile force in the cable. The length of the cable becomes:
s  l (1 
8 f w2
1 w2l 2
)

l
(
1

)
3 l2
24 H w2
We now add a vertical live load acting on the cable which still has both ends at the same level. Summing
the moments, as was done previously in Equation 1, results in the following:
y
Mw
Hw
Equation 5
In this equation, Mw is the bending moment due to dead load calculated as for a simply supported beam,
and Hw is the horizontal component of the tensile force produced in the cable by the dead load. If a live
load is now applied, the bending moment calculated as for a simple beam becomes Mw + Mp , and the
horizontal component of cable tension becomes Hw + Hp. Denoting by  the vertical deflections of the
cable we obtain from a moment equation:
10
y  
Mw  M p
Hw  H p
Subtracting Equation 5 from this equation, we obtain:

Mp Hpy
Equation 6
Hw  H p
The vertical deflection  can be easily calculated, assuming the horizontal component Hp of cable tension
produced by live load is known. This can typically be calculated from the geometry of the bridge. If Hp
is not known however, it must be determined before a solution for Equation 6 is possible.
To do this, we now consider the infinitely small element ab of the cable, as illustrated in Figure 4.
Figure 4
Infinitely small element of cable from Figure 3.
The live load causes this element to elongate and take a new position a1b1. We denote by  and  the
horizontal and vertical components of the small displacement of point a. The initial length of the element
is:
Equation 7
ds 2  dx 2  dy 2
The length of the same element after application of the live load is:
(ds  ds) 2  (dx  d ) 2  (dy  d ) 2
Equation 8
In the equation above, ds is the elongation of the element caused by the live load. Neglecting the small
change in slope of the cable produced by the live load, we obtain from the standard equation for the
deflection of a beam:
ds 
ds  H p ds
Ac Ec dx
Equation 9
11
Since Hp ds / dx is that part of the tensile force in the cable which is produced by live load and which is
usually much smaller than the part produced by dead load, the unit elongation ds / ds is usually very
small. If this is the case, (ds)2 in Equation 8 can be neglected. For the same reason, and from the
observation that the curve of the cable is a flat curve, we also neglect (d)2. Combining Equations 7 and 8
we obtain:
1
ds  ds  dx  d  dy  d  (d ) 2
2
This may be written as:
ds
dy
1 d
ds  d 
d
dx
dx
2 dx
d 
Substituting Equation 9 for ds in this equation and integrating, we obtain:

Hp
x
ds
 ( dx )
Ac Ec
0
3
x
dx   y ' ' dx 
0
1 x 2
 ' dx
2 0
Equation 10
In the equation above, the primes indicate derivatives with respect to x. With the values of y’ and ’
which are typically encountered in long-span bridges, the value of  usually does not exceed onethousandth of x. The maximum value of  / x occurs near the supports where ’ and y’ usually have their
largest numerical values. At the ends of the cable,  vanishes and we obtain from Equation 10:
Hp
Ac Ec
l
ds
 ( dx )
0
3
l
dx   y ' ' dx 
0
1 l 2
 ' dx
2 0
Equation 11
The integral on the left side of this equation for the assumed parabolic shape can readily be evaluated, and
we obtain:
3
1
l
ds 3
1 5 16 f 2
16 f 2 2
2 2
0 ( dx ) dx  0 (1  y' ) dx  l{ 4 ( 2  l 2 )(1  l 2 )
1
3l
4f
16 f 2

ln[
 (1  2 ) 2 ]}
32 f
l
l
l
Equation 12
Integration by parts is used on the right hand side of Equation 11. Taking advantage of the fact that  is
zero at the ends of the cable and the previously developed Equation 2, we obtain:
l
 y' ' dx  y'
0
l
  y ' 'dx 
0
l
0
w l
dx
H w 0
1 l 2
1
1 l
1 l
l
 ' dx   ' 0    ' 'dx     ' 'dx

2 0
2
2 0
2 0
Equation 13
12
Substituting Equations 12 and 13 into Equation 11, and denoting the integral of Equation 12 by L, we
obtain:
Hp
Ac E c
L
w l
1 l
dx    ' 'dx

Hw 0
2 0
Equation 14
Equation 14, together with Equation 6, gives the system of equations necessary to calculate vertical
deflections of the cable.
13
Application to stiffened bridges
Figure 5 below illustrates the simplest type of stiffened suspension bridge: a single span cable stiffened
by a simply supported truss of constant cross section.
Figure 5
Stiffened bridge under dead load only.
It is assumed that a for properly assembled suspension bridge the dead load of the structure is uniformly
distributed along the span and entirely transmitted through the hangers to the cable which takes a
parabolic form. The solid lines in the figure above represent this loading condition. When a live load is
added, the deflections produced in both the cable and the truss are as seen by the dashed lines above. We
assume that the elongation of the hangers and their small inclination to the vertical during deformation are
negligible so that at any given position along the length of the bridge the deflection of the cable is equal to
the deflection of the truss. This assumption is also made in Reference 1. It is also assumed that the
spacing of the hangers is small compared to the length of the span so that the load may be considered
uniformly distributed along the span
Dead load only
We first consider the case where the structure is carrying only dead load. The truss does not experience
bending in this case and the equation of moments for the forces to the left of a cross section m-n may be
written as:
Mw  Hw  0
Addition of live load
When live load is applied and deflections  are produced, there will be a bending moment M acting in a
cross section m-n of the truss, and the equation of moments for the forces to the left of this cross section
is:
M w  M p  H w  H p  y     M  0
Subtracting the equation for the dead load from the equation for the live load we obtain:
14
M  M p  H w  H p   H p y
From this equation, the bending moment at any cross section of the truss can be calculated provided the
horizontal component of the tensile force in the cable and the deflection  are known.
In the case of very rigid stiffening trusses, the deflections  could be ignored, yielding the simplified
equation:
M Mp Hpy
Under this assumption, the bending moment is independent of deflections and can be evaluated the same
way as any other rigid statically indeterminate structure. Applications have demonstrated, however, that
the stiffening trusses in large span bridges are usually very flexible. This means that the more complete
equation including the deflection of the truss  must be evaluated in order to calculate the bending
moments.
To do this, we begin with the differential equation of the deflection curve of a beam:
EI
d 2
 M
dx 2
This equation is applied to the bending moment equation that includes deflection of the truss, and yields:
EI
d 2
 H w  H p   H p y  M p
dx 2
Equation 15
The quantity Mp in this equation can be calculated for any distribution of live load over the span. The
quantities y and Hw are given by Equation 2 and Equation 3, and only the quantity Hp is unknown. It
depends on the deflections  , and Equation 14 is used to evaluate it. Equation 15 together with Equation
14 completely define the deflections of the stiffening truss.
15
Deflections of stiffening trusses
The first case considered is that of a single concentrated load P acting on the truss. Taking the second
derivative of Equation 15, we find that the deflections of the truss in this case are the same as those
occurring in a simply supported beam subjected to a combination of loads:
1. an axial tensile force Hw+Hp
2. a uniformly distributed upward lateral load of intensity Hpw/Hw
3. a concentrated load P
This state of combined loading is shown in Figure 6.
Figure 6
Stiffening truss under point load.
Hw  H p
EI
 k2
Equation 16
Under the loading conditions above, and using the notation defined in Equation 16, the deflections in the
stiffening truss produced by the load P in the beam to the left of this load ( x < l – c ) may be written as:
1  
P
sinh kc
Pcx
sinh kx 
H w  H p k sinh kl
( H w  H p )l
Equation 17
For the portion of the beam to the right of the load ( x > l – c ), the deflections are:
1  
sinh k (l  c)
P(l  c)(l  x)
P
sinh k (l  x) 
H w  H p k sinh kl
( H w  H p )l
Equation 18
The deflections produced by the upward pull are:
2  
Hp
Hw
wl 2
Hw  H p
 cosh( kl / 2  kx)
x(l  x) 
1
 2 2
 2 2

2l 2 
 k l cosh( kl / 2) k l
Equation 19
The total deflections  of the truss may be obtained by superimposing deflections 1 on deflections 2.
16
  1   2
To determine the magnitude of tension Hp, which is required to solve Equations 17, 18 and 19, Equation
14 is simplified by omitting the second term on the right hand side, as seen below:
Hp
Ac E c
L
w l
dx
H w 0
It can be shown that in most cases this has only a small effect on the magnitude of Hp. Substituting
Equations 17, 18 and 19 into Equation 17 and performing the integration we obtain the following
equation:
 H w  H p L 1  8 f 2 
12
24
kl 
Hp
 
 1  2 2  3 3 tanh 
2 
k l
 Ac E c l 12  l   k l
8 f 1 c  c 
1
sinh kl  sinh kc  sinh k l  c 
P
1    2 2

l  2 l  l  k l sinh kl

Equation 20
Once the value of Hp has been determined, it can be used to determine the deflection in the stiffening
truss.
17
Numerical Approaches
Four different methods will be used to solve Equation 20. The results will then be compared. In order to
effectively compare the results, it is important to begin with the same geometry and loading conditions for
each case analyzed. The assumptions made are therefore summarized below:
l = 125 ft
Hw = 270,000,000 lbf
I = 210,000 in4
f = 12.5 ft
Ec = 30,000,000 psi
A = 4500 in2
c = 0.75l
P = 2,000 lbf
These values will be used in all subsequent calculations.
1. Successive approximations
This is perhaps the most obvious method of solving this problem. The emphasis will be on developing an
acceptable first approximation to the solution of Hp through manipulation of Equation 20, and refining it
through iteration until it is within the desired tolerance. This accepted value of Hp will then be substituted
into the appropriate equations to develop the actual deflection in the stiffening truss.
 H w  H p L 1  8 f 2 
12
24
kl 
Hp
 
 1  2 2  3 3 tanh 
2 
k l
 Ac E c l 12  l   k l
8 f 1 c  c 
1
sinh kl  sinh kc  sinh k l  c 
P
1    2 2

l  2 l  l  k l sinh kl

Equation 20
In the case of long-span bridges the quantity kl is typically a relatively large number. For example:
Ambassador Bridge (Detroit)
George Washington Bridge
kl = 9.52
kl = 35
This indicates that the terms in Equation 20 that contain k are small and could be neglected to obtain a
first approximation of the result. The term (Hw + Hp) / AcEc may also be considered very small and
therefore omitted. This simplifies Equation 20 considerably, and allows it to be rewritten as:
Hp 
3 l c c
P
1  
4 f l l
For c =0.75 l, this gives:
Hp 
9
l
P
64 f
18
Using the given assumptions for bridge geometry and loading, this allows us to obtain a first
approximation for Hp.
9
(2,000lbf )(10)
64
Hp 
Hp = 2812.5 lbf
The accuracy of this first approximation will depend largely on the magnitude of kl. For values of kl
greater than ten, it has been found that the approximation is typically quite good. To increase its
accuracy, the approximate value of Hp is used to calculate k from Equation 16.
Hw  H p
EI
 k2
Equation 16
270,000,000lbf  2812.5lbf
 k2
4
(30,000,000 psi )( 210,000in )
k = 0.0065465708
This value of k is then substituted back into Equation 20, which gives a second approximation of Hp.
In order to do this, the value of L must first be defined:
L
1 5
l. .
4 2
2
16. f .
1
2
l
1
1
2 2
2 2
16. f
2
l
3. l . 4. f
ln
32. f
l
1
16. f
2
l
The right hand side of the equation must also be entered into Mathcad so that it may be used to obtain the
solution:
RHS
8. f 1 c
P. . . . 1
l 2 l
c
l
1
. ( sinh( k. l)
sinh( k. c )
sinh( k. ( l
c) ) )
2. 2.
k l sinh( k. l)
Finally, the entire equation is entered and a solution obtained:
H p.
Hw
Hp L
.
A c. E c l
2
1 . 8. f .
1
12 l
12
2. 2
k l
24 .
k. l
tanh
3 3
2
k .l
RHS
H p = 2685.4814691737 lbf
19
For practical applications, this second approximation is typically accurate enough, though this procedure
may be repeated until the desired accuracy is achieved.
2
k
Hw
Hp
E c. I
k = 0.0065465693 in
RHS
H p.
1
8. f 1 c
P. . . . 1
l 2 l
Hw
Hp L
.
A c. E c l
c
l
1
. ( sinh( k. l)
sinh( k. c )
sinh( k. ( l
c) ) )
2. 2.
k l sinh( k. l)
2
1 . 8. f .
1
12 l
12
2 2
k .l
24 .
k. l
tanh
3 3
2
k .l
RHS
H p = 2685.4814599255 lbf
2
k
Hw
Hp
E c. I
k = 0.0065465693 in
1
The results of this iteration are summarized in the table below:
Hp (lbf)
2812.5
2685.4814691737
2685.4814599255
k (in-1)
0.0065465708
0.0065465693
0.0065465693
As seen above, it takes only three iterations for the value of Hp to converge to a result accurate to within
10-4. The second approximation would certainly have been adequate for the construction of an actual
bridge. Because the value of k has converged even more quickly to an accuracy of 10-10 it is impossible to
further refine the approximation of Hp.
20
Once Hp has been calculated in this way, the deflection curve may be found using Equations 17, 18 and
19 with the method of superposition.
1  
P
sinh kc
Pcx
sinh kx 
H w  H p k sinh kl
( H w  H p )l
For x < l - c
1  
sinh k (l  c)
P(l  c)(l  x)
P
sinh k (l  x) 
H w  H p k sinh kl
( H w  H p )l
For x > l - c
2  
Hp
Hw
 cosh( kl / 2  kx)
x(l  x) 
wl 2
1
 2 2
 2 2

H w  H p  k l cosh( kl / 2) k l
2l 2 
  1   2
Microsoft Excel was used to tabulate and graph these values for 0 < x < l. Some of these results are
summarized in the Appendix B1 so that they may be compared to the values that will be obtained using
various other methods.
Total Deflection - Method of Successive Iterations
0
Deflection
-0.0005
0
500
1000
1500
-0.001
-0.0015
-0.002
-0.0025
-0.003
Distance along Bridge
Figure 7
Total deflection of stiffening truss.
21
Solutions in one Variable
For the following three methods, the deflections of the stiffening trusses will be based on the assumed
value of Hp developed through the method of successive iteration. The emphasis will instead be on
developing a more accurate value for k and using this to determine the actual deflections.
The first step in doing in this was to plot the function of k. This permitted the applicable methods of
approximating the solution to be determined. To begin, Mathcad was used to plot Equation 20, as seen
below.
 H w  H p L 1  8 f 2 
12
24
kl 
Hp
 
 1  2 2  3 3 tanh 
2 
k l
 Ac E c l 12  l   k l
8 f 1 c  c 
1
sinh kl  sinh kc  sinh k l  c 
P
1    2 2

l  2 l  l  k l sinh kl

Equation 20
This will be rewritten in the following form so that the root of the equation may be found:
 H w  H p L 1  8 f 2 
12
24
kl 
g (k )  H p 
 
 1  2 2  3 3 tanh 
2 
k l
 Ac Ec l 12  l   k l
8 f 1 c  c 
1
sinh kl  sinh kc  sinh k l  c   0
P
1    2 2

l  2 l  l  k l sinh kl

Initially, k is defined as a range variable taking on values from 0.0065 in-1 to 0.0066 in-1 in increments of
0.00001. This range is based on the approximate solution obtained using the method of successive
approximation and is wide enough that we may be confident that the exact value is included.
k
1
1
0.0065. in , 0.00651. in .. 0.0066. in
1
Equation 20 is now defined as g(k) for the program. For formatting reasons in Mathcad, it was necessary
to separate it into left and right hand sides of the equation. This does not, however, affect the final result.
22
Below is the equation as defined in Mathcad:
Hw
Hp L
.
A c. E c l
Equation20 RHS( k)
H p.
Equation20 LHS( k)
8. f 1 c
P. . . . 1
l 2 l
g( k)
Equation20 RHS( k)
2
1 . 8. f .
1
12 l
c
l
1
12
2 2
k .l
24 .
k. l
tanh
3 3
2
k .l
. ( sinh( k. l)
sinh( k. c )
sinh( k. ( l c ) ) )
2. 2.
k l sinh( k. l)
Equation20 LHS( k)
The graph below shows the values of the function defined above over the given range of k. Because the
default units for length in Mathcad are feet, it was necessary to divide to values of k in the graph by 12 in
order to be consistent with the values used in other calculations.
0.5
0
g( k )
0.5
1
0.0065
0.00655
0.0066
0.00665
k
12
Figure 8
Graph of the equation to be solved, g(x)
From this graph, the viable methods for approximating k may be determined:
23
2. Bisection Method
It may be seen on Figure 8 that the beginning and end points of the function have opposite signs. This
indicates that the function crosses the line g(k)=0, and that the bisection method may be used to
determine the location of this zero.
Below is the Mathcad program used to implement the Bisection Method.
root( N )
a
0.0065. in
1
b
0.0066. in
1
for i  1 .. N
p
a
b
a
2
p
FP
g( p )
FA
g( a )
a
p if FA . FP > 0
(b
p ) otherwise
Line Function
1 The function root requires an input of N, which is the desired
number of iterations. The long vertical line represents the
length of the program. The variable a is defined to be the value
of the lower end of the range for k as previously determined.
2 The variable b is defined to be the value of the upper end of the
range for k.
3 This is the beginning of a for loop. The short vertical line
represents the length of the loop. The contents of the loop will
be repeated until i has stepped through each integer between 1
and N. It is not necessary in Mathcad to manually increase the
value of i with each step, the default action if no other is
specified is to increase it by one.
4 The variable p is defined to be the midpoint of the distance
between a and b.
5 The variable FP is defined to be the value of the function g(k)
at point p.
6 The variable FA is defined to be the value of the function g(k)
at point a.
7 If FA*FP > 0, indicating that point p was to the left of the zero
in the function, the value of p is reassigned to the variable a.
8 Otherwise, the value of p is reassigned to the variable b. The
range of k is now one half of what it was originally. If the
desired number of iterations has not been satisfied, the loop will
begin once more using an updated value for either a or b.
9 This indicates that the final value of p should be displayed
when the function is called out.
Bisection Method Program
24
0.00656
0.00655
root( i )
12
0.00654
0.00653
0.00652
0
10
20
i
Figure 9
Results of bisection method
Above is a plot of the values obtained for N = 20, or 20 iterations as output by Mathcad. It appears that by
the end the values for k have converged. It is interesting to note the abrupt dip that occurs between the first
and second iterations. This emphasizes the importance of performing sufficient iterations to ensure that the
results have actually converged.
The table summarizing the output of the program for N = 20 may be found in Appendix C. The values in
the table confirm that the results have converged.
25
3. Fixed Point Iteration
The next method of approximation used was Fixed Point Iteration. In order to ensure that this was indeed
a viable option, the function h(xk) = xk was plotted to ensure that such a point did in fact exist.
xk
0.0065, 0.00651.. 0.0066
h( xk)
0.00665
xk
0.0066
h( xk )
This is a plot of the function h(xk) = xk. As
expected, it is a straight line extending from the
origin and bisecting the first quadrant. This curve
will be superimposed on the plot of the function to
be analyzed using fixed point integration.
0.00655
0.0065
0.0065
0.00655
0.0066
0.00665
xk
Figure 10
Graphical representation of h(xk)=xk
k
1
1
0.0065. in , 0.00651. in .. 0.0066. in
g( k)
Equation20 RHS( k)
1
0.5
Equation20 LHS( k)
g( k )
0
h( xk )
0.5
Here the plot of the function to be analyzed is
shown superimposed with the plot of h(xk) = xk.
Because the two curves intersect as seen on the
plot, it is known that there is in fact a fixed point
and this method of approximation may be pursued.
1
0.0065
0.00655
k
0.0066
0.00665
, xk
12
Figure 11
Graphical representation of h(xk)=xk and g(x)
26
Because the function g has a fixed point at some
point p, it is known that the function defined by
m(k) = k - g(k) has a zero at p.
fixedpoint( N )
0.0065455. in
p0
1
for i  1 .. N
p
To the right is the program that was written to
determine the solution using this method.
g( p0 )
p0
p
0.01 0.01
Because of the extremity of the differences in
scales, however, this method does not converge to a
solution. The plot to the right is the same as above,
simply viewed at a more realistic scale. It is
obvious from this graph that with the two curves
very nearly horizontal and vertical, no projection
can be made from one to the other.
g( k )
0.005
h( xk )
0
0
0.0065
0.00655
0.0065
0.0066
k
0.00665
0.0066
12
Figure 12
Divergence of fixed point iteration method
27
4. Newton-Raphson Method
The final method of solving Equation 20 for k is the Newton-Raphson Method. One requirement of this
method is that the function be differentiable. It appears from the graph that this condition is met.
Below is the Mathcad program used to approximate the solution to the equation using the NewtonRaphson Method:
newton( N )
0.0065. in
p0
1
for i  1 .. N
p
p0
p
p0
p
g p0
d
g p0
dp 0
Line Function
1 The function newton requires an input of N, which is the
desired number of iterations. The long vertical line
represents the length of the program. The variable p0 is
defined to be the initial approximation for k as
previously determined.
2 This is the beginning of a for loop. The short vertical
line represents the length of the loop. The contents of
the loop will be repeated until i has stepped through each
integer between 1 and N. It is not necessary in Mathcad
to manually increase the value of i with each step, the
default action if no other is specified is to increase it by
one.
3 The variable p is defined to be the difference of p0 and
the ratio of g(po) over g’(po).
4 The variable po is reassigned the value of p which was
obtained from the calculation in line 3. This is theend of
the for loop.
5 This indicates that the final value of p should be
displayed when the function is called out.
Newton-Raphson Method Program
i 1.. 4
newton( i )
12. ft
0.00654657 0.0065466
1
0.0065462312
0.0065465692
0.0065465693
0.0065465693
newton( i )
1 0.0065464
12. ft
0.00654623 0.0065462
1
1
2
3
i
4
4
Figure 13
Results of Newton-Raphson method.
As seen above, the Newton-Raphson method converges very rapidly to an approximation.
28
Results, Error Analysis and Discussion
Of the four solution methods attempted, only three were successful. The results obtained using these
three different methods were however identical if sufficient iterations were performed. The table below
summarizes the number of iterations required for each method to converge to a solution accurate to 10-10.
Successive Iteration
Bisection Method
Newton-Raphson Method
k = 0.0065465693
k = 0.0065465693
k = 0.0065465693
2 iterations
19 iterations
3 iterations
Error analysis in the case of this problem is trivial. As there is no way to calculate the exact theoretical
value for the deflection of the stiffening truss of a suspension bridge, the numbers above can only be
compared against one another. Given sufficient iterations, the results converged to the same value,
accurate to 10-10, regardless of the method. In doing this, it is obvious that the level of accuracy far
exceeds that which would be required for an application of this problem in the actual physical
environment.
Conclusions
Each of the methods that proved successful in calculating the deflection in the stiffening truss yielded the
same result, indicating that all were valid. As the values obtained for k using each of the methods were
identical, it was not necessary to recompute the actual deflection numbers. Perhaps the most interesting
case was the one that was not successful. All of the stated required conditions were met, yet the fixed
point iteration method did not converge to a result, even when the given initial value was extremely
accurate. The reason for this became clear with a more detailed examination of the plotted functions and
the discussion in the appropriate section of this text, but serves as a reminder that the ideal solution
method for each problem must be determined individually.
29
References
1. Theory of Structures, Timoshenko and Young, McGraw-Hill, Second Edition, 1965.
2. Standard Mathematical Tables and Formulae, CRC Press, 29th Edition, 1991.
3. Numerical Analysis, Burden and Faires, Brooks/Cole, Sixth Edition, 1997.
4. Standard Handbook for Structural Engineers, Merritt, McGraw-Hill, 1968.
5. Design of Bridge Superstructures, O’Connor, Wiley, 1971.
6. Scientific Encyclopedia, Considine, VNR, 1976.
7. Mechanics of Materials, Gere and Timoshenko, PWS, Fourth Edition, 1990.
8. Design of Welded Structures, Blodgett, Lincoln Arc Welding Foundation, 15th Printing, 1996.
9. Analysis of Numerical Methods, Isaacson and Keller, Dover, 1966.
10. Numerical Mathematics and Computing, Cheney and Kincaid, Brooks/Cole, Third Edition, 1994.
11. Roark’s Formulas for Stress and Strain, Young, McGraw-Hill, Sixth Edition, 1989.
30
Appendix A
Representative Large Suspension Bridges in North America
[Reference 6]
Verrazano-Narrows
New York, NY
4,260 ft span
Golden Gate
San Francisco Bay, California
4,200 ft span
Mackinac
Straits of Mackinac, Michigan
3,800 ft span
George Washington
Hudson River, New York
3,500 ft span
Transbay
San Fransisco Bay, California
2,310 ft each span (total two)
Bronx-Whitestone
East River, New York
2,300 ft span
Quebec Road
Montreal, Quebec
2,190 ft span
Delaware Memorial
Wilmington, Delaware
2,150 ft span
Walt Whitman
Philadelphia, Pennsylvania
2,000 ft span
Vincent Thomas
Los Angeles Harbor, California
1,500 ft span
Appendix B
Deflection of Stiffening Truss – Method of successive iterations
1
x
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
2
3
4
5
deflection1 (x < l - c) deflection1 (x > l - c) deflection1 deflection2
0
-0.009317804
0
0
-0.000623796
-0.005990911 -0.000624 -2.53E-06
-0.001277899
-0.004173232 -0.001278 -4.86E-06
-0.002006079
-0.003139771 -0.002006 -6.87E-06
-0.002884999
-0.002513805 -0.002514 -8.53E-06
-0.004058271
-0.002099582
-0.0021 -9.78E-06
-0.005800225
-0.001795385 -0.001795 -1.06E-05
-0.008637747
-0.001548358 -0.001548 -1.1E-05
-0.013584265
-0.001331039 -0.001331 -1.1E-05
-0.022589819
-0.001129155 -0.001129 -1.06E-05
-0.039407077
-0.000935288 -0.000935 -9.78E-06
-0.071257905
-0.000745584 -0.000746 -8.53E-06
-0.132040611
-0.000558034 -0.000558 -6.87E-06
-0.248502226
-0.000371587 -0.000372 -4.86E-06
-0.472116917
-0.000185683 -0.000186 -2.53E-06
-0.901945802
0
0
0
6
total deflection
0
-0.000626322
-0.001282754
-0.002012953
-0.002522331
-0.002109363
-0.001806008
-0.001559405
-0.001342085
-0.001139778
-0.000945069
-0.00075411
-0.000564908
-0.000376443
-0.00018821
0
Column 1 lists the position along the bridge. The total length is 1500 in.
Column 2 uses Equation 17 to determine 1 for the portion of the beam to the left of the applied load.
Column 3 uses Equation 18 to determine 1 for the portion of the beam to the right of the applied load.
Column 4 contains an ‘if’ statement to determine which of columns 3 or 4 should be used, based on x.
Column 5 uses Equation 19 to determine 2.
Column 6 sums columns 4 and 5 to determine the total deflection.
32
Appendix C
Deflection of Stiffening Truss – Bisection Method
root( i )
12. ft
1
0.00655
0.006525
0.0065375
0.00654375
0.006546875
0.0065453125
0.0065460938
0.0065464844
0.0065466797
0.006546582
0.0065465332
0.0065465576
0.0065465698
0.0065465637
0.0065465668
0.0065465683
0.0065465691
0.0065465694
0.0065465693
0.0065465693
33