# Deflections in Stiffening Trusses of Suspension Bridges

```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
Rensselaer at Hartford
Hartford, Connecticut
October 1999
Nomenclature
f
fw
H
Hw
h
l
Mw
Mx
P
p
s

w
Sag of cable
Horizontal component of tensile force in cable
Horizontal component of tensile force produced in the cable by the dead load
Difference in elevation of the ends of the cable
Span of beam
Bending moment at given cross section
Live load, acting at a point
Live load, uniformly distributed over a given length
Length of cable
Vertical deflection of cable
List of Figures
Abstract
The deflections in stiffening trusses of suspension bridges will be analyzed using a variety of numerical
methods, and the results will be compared.
Background
This will be a brief non-technical description of a suspension bridge. It will include sketches of the
assumed arrangement as well as a discussion of stiffened vs. unstiffened bridges.
Introduction
Before the deflections in the stiffening truss of the suspension bridge may be determined, the equations
must be developed. The first step in doing this is to analyze the most basic form of a suspension bridge,
the cable itself. Equations for the length and sag of this cable will be important for every calculation that
follows. 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.
The deflections are then calculated first for the straightforward case in which all of the reaction forces are
known, then for the more challenging case in which the horizontal component of the tensile force
2
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 something
manageable enough to describe using mathematical equations, certain assumptions has 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.
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.
Theory
Analysis of cable
Figure 1
For a uniform, perfectly flexible cable fixed at points A and B under a uniformly distributed load (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:
3
Mx 
wx
l  x 
2H
Substituting this into the 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, this may be reduced to:
y
wx
l  x 
2H
Equation 2
Applying this equation to the mid-point of the cable where the y coordinate of the curve represents the sag
f, we obtain:
f 
wl 2
8H
Equation 3
This holds true in the more general case illustrated in Figure 1 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:
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
4
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. In the case
of a suspension bridge of large span, the dead load uniformly distributed on a horizontal plane is usually
many times larger than that uniformly distributed along the cables. We can 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 2
Figure 2 illustrates a symmetrical case in which in which a load of intensity p is uniformly distributed
along the distance 2a of the span. The solid line 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:
y  
Mw  M p
Hw  H p
5
Subtracting Equation 5 from this equation, we obtain:

Mp Hpy
Hw  H p
Equation 6
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 3.
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:
ds 
ds  H p ds
Ac Ec dx
Equation 9
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:
6
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

Ac Ec
x
0
(
x
ds 3
1 x
) dx   y ' ' dx    '2 dx
0
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
l
Ac Ec
ds
 ( dx )
3
0
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
)
dx

(
1

y
'
)
dx

l
{
(

)(
1

)
0 dx
0
4 2
l2
l2
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
0
l
y ' ' dx  y ' 0   y ' 'dx 
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
Substituting Equations 12 and 13 into Equation n, 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
7
Equation 14, together with Equation 6, gives the system of equations necessary to calculate vertical
deflections of the cable.
Application to stiffened bridges
Figure 4 below illustrates the simplest type of stiffened suspension bridge: a single span cable stiffened
by a simply supported truss of constant cross section.
Figure 4
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 to the cable which takes a parabolic form. The solid
produced in both the cable and the truss are as seen by the dashed lines above. We assume that at any
given position along the length of the bridge the deflection of the cable is equal to the deflection of the
truss.
Also, the spacing of the hangers is assumed to be small compared to the length of the span so that the load
may be considered uniformly distributed along the span. This implies that the elongation of the hangers
and their small inclination to the vertical during deformation are neglected.
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
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
8
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
Deflections of stiffening trusses
Development of equations
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
9
Figure 6
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 &lt; 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 &gt; 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.
  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:
10
 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.
Solution of equations
A number of 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:
kl = 10
P = 2,000 lbf
I = 210,000 in4
Hw / AcEc = 0.002
Hw = 270,000,000 lbf
f / l = 0.1
E = 30,000,000 psi
c = 0.75l
A = 4500 in2
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:
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:
11
Hp 
9
l
P
64 f
Using the 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.
1 5
l. .
4 2
L
RHS
H p.
2
16. f .
1
2
l
8. f 1 c
P. . . . 1
l 2 l
Hw
Hp L
.
A c. E c l
1
1
2 2
2 2
16. f
2
l
c
l
3. l . 4. f
ln
32. f
l
1
1
. ( sinh( k. l)
16. f
2
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 = 2685.4814691737 lbf
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
12
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.
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
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
2  
Hp
Hw
For x &lt; l - c
For x &gt; l - c
 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 &lt; x &lt; l. Some of these results are
summarized in the table below so that they may be compared to the values that will be obtained using
various other methods.
13
x
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
deflection1 (x &lt; l - c) deflection1 (x &gt; 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
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
Deflection
Total Deflection - Method of Successive
Iterations
0
-0.001 0
500
1000
1500
-0.002
-0.003
Distance along Bridge
Trigonometric series
Trigonometric series may also be used to determine the deflection curve of a stiffening truss. This
method simplifies the calculation of deflection while allowing Hp to be calculated more accurately, as the
second term on the right hand side of Equation 14 need not be omitted. We begin with the case of one
concentrated force acting on the stiffening truss, as in Figure 6. The differential Equation 15 becomes in
this case, for x &lt; l – c,
EI
d 2
4f
Pcx
 H w  H p   H p 2 xl  x  
2
l
dx
l
Equation (a)
14
For x &gt; l – c, this is:
EI
d 2
4f
Pl  c l  x 
 H w  H p   H p 2 xl  x  
2
l
dx
l
Equation (b)
The right sides of these equations may be represented for the entire length l of the truss by the
trigonometric series:
x
b1 sin
l
 b2 sin
2x
3x
 b3 sin

l
l
Equation (c)
The coefficients b1, b2, b3, … may be calculated from the formula
bm 
4f
2
Hp 2

l 
l
 xl  x  sin
l
0
Pl  c  l
mx
Pc l  c
mx
mx 

dx 
x sin
dx 
l  x  sin
dx 


l c
l
l 0
l
l
l

This gives:
bm 
16 H p f
m
3
3
1  cos m  
m l  c 
2 Pl
sin
2 2
l
m 
Equation (d)
As the series in Equation (c) applies to the entire length of the truss, Equations (a) and (b) can now be
replaced by one equation:
EI

d 2
mx



H

H


bm sin

w
p
2
l
dx
m 1
Equation (e)
The solution of this equation, satisfying the end conditions, will now be taken in the form of the following
series:

   a m sin
m 1
mx
l
Equation (f)
Substituting this series into Equation (e) and calculating for each value of m the coefficients of sin(mx/l),
we obtain:
bm l 2
am  
EI  m 2 2  H w  H p l 2
The series from Equation (f) now becomes:
bm l 2 sin mx / l 
  
2 2
2 2
m 1 EI m   k l



15
k2 
Hw  H p
Equation 16
EI
Substituting Equation (d) for bm, we are finally able to represent the deflection of the truss by the
following series:

32 H p fl 2
 EI
3
sin mx / l 
2 Pl 3

 3 2 2 2 2 EI 2
m 1, 3, 5 m m   k l





m 1, 2, 3

1
m m  k l
2
2
2
2 2

sin
m l  c  sin( mx)
l
l
This series will now be applied to the numerical example already evaluated using the method of
successive iteration.
…
Solutions in one Variable
For the following 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.
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. 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
 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
P
8 f 1 c  c 
1
sinh kl  sinh kc  sinh k l  c   0
1    2 2

l  2 l  l  k l sinh kl

To begin, 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.
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, it was necessary to separate
it into left and right hand sides of the equation. This does not, however, affect the final result.
16
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
From this graph, the viable methods for approximating k may be determined:
Bisection Method
Below is the Mathcad program used to implement the Bisection Method.
17
root( N )
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.
1
.
3
This is the beginning of a for loop. The short vertical line
a 0.0065 in
represents the length of the loop. The contents of the loop will
1
b 0.0066. in
be repeated until i has stepped through each integer between 1
and N. It is not necessary in Mathcad to manually increase the
for i  1 .. N
value of i with each step, the default action if no other is
b a
p a
specified is to increase it by one.
2
4 The variable p is defined to be the midpoint of the distance
FP g( p )
between a and b.
5
The variable FP is defined to be the value of the function g(k)
FA g( a )
at point p.
a p if FA . FP &gt; 0
6 The variable FA is defined to be the value of the function g(k)
( b p ) otherwise
at point a.
7
If FA*FP &gt; 0, indicating that point p was to the left of the zero
p
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.
18
root( i )
0.00656
1
12. ft
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
0.00655
root( i )
12
0.00654
0.00653
0.00652
0
10
20
i
To the left is a table of the values obtained for N = 20, or 20 iterations as output by
Mathcad. It appears that by the end of the column the values for k have converged.
The plot above illustrates this graphically.
Fixed Point Iteration
The next method of approximation used was Fixed Point Iteration. In order to ensure that this was indeed
a viable option, it was necessary to plot the function h(xk) = xk to ensure that such a point did in fact
exist.
xk
0.0065, 0.00651.. 0.0066
h( xk)
0.00665
xk
0.0066
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 of the plot of the function to
be analyzed using fixed point integration.
h( xk )
0.00655
0.0065
0.0065
0.00655
0.0066
0.00665
xk
19
k
1
1
0.0065. in , 0.00651. in .. 0.0066. in
g( k)
Equation20 RHS( k)
1
0.5
Equation20 LHS( k)
0
g( k )
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.
h( xk )
0.5
1
0.0065
0.00655
k
0.0066
0.00665
, xk
12
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.
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
0.0065
0.00655
0.0066
k
0.00665
0.0066
12
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.
20
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.
i
1.. 4
0.00654657 0.0065466
newton( i )
12. ft
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
As seen above, the Newton-Raphson method converges very rapidly to an approximation.
Results and Discussion
Conclusions
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.
21
```