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fiNFLUENCE OF LOWER CHQRD SE22 3%!
STRESSES EN TW’G-HERGED TRUSSEQ ARCH

Thesis hr the Qegmc a? M. S.
{MCMGAN STATE COLLEGE
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This is to certify that the .
thesis entitled v 4
Influence of Lower Chord Size on Stresses j

in Two-Hinged Trussed Arch ;

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presented by '
Hedayat Behbeheni
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has been accepted towards fulfillment -.‘_<
of the requirements for
Mdegree mi

fijor professor

  

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INFLUENCE OF LOWER CHORD‘SIZE ON
STRESSES IN TWO—HINGED
TRUSSED ARCH

BY
HEDAIAT EENBEHKNI

A THESIS
Submitted to the School of‘Graduate Studies of Michigan
StateacoIIege of AgricuIture and Applied Science~
in partial fulfillment of the requirements
fer the degree of

MASTER OF SCIENCE
Department of Civil Engineering

I95H

114-5"!

TABLE OF CONTENTS

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Page

12
11+

20

2.

STATEMENT OF PROBLEM
The purpose of this analysis is;

to show the influence of lower chord

size on stresses and

to determine the limiting cross-sectionalzw”'
of the lower chord in a two-hinged trussed
arch. In other words to investigate

whether or not this structure will approach
a three-hinged arch, by increasing the size

of the lower chord.

INTRODUCTION

The analysis of statically indeterminate structures
by classical methods has always been a very tedious Job,
since before the problem can be solved it is necessary to
make some assumptions which will have to be corrrected

later.

As an example, let it be required to find the stresses

in a statically indeterminate truss for a given loading.

Before the problem can be solved it will be necessary
to make an assumption regarding the cross-sectional area
of the members. Having done this, then the stresses can
be found by means of virtual work or any other known method.
Generally speaking, these stresses are either too great or
too small for the assumed areas and, it will be necessary
to revise the original assumption. It is clear now that,
when the cross-sectional area of the members are changed,

the stresses obtained previously will no longer be true.

From the brief discussion presented, it can be concluded
that in order to design an indeterminate truss for given
loading, one would have to use a trial and error procedure.
The number of trials required to get a close approximation
depends greatly on the amount of eXperience that a person

has with the particular kind of structure.

-4-

Time and energy will be saved if one could estimate
in advfiace the amount of variation in stress due to changes

in area, because it would permit making a much closer and

‘wiser assumption.

With this thought in mind, I have tried to show the
variation in stress by changing the cross-sectional area

of one group of members with respect to the others.

Although there are many different kinds of trusses for
which this analysis could be used, the author is merely

concerned with two—hinged trussed arch.

Since there are four reactions on this structure and
all members are required to carry direct stress, it is

statically indeterminate to the first degree.

In order to analyze this structure, it was necessary
to assume some values for the cross-sectional area of the
members. The assumption was made by setting a definite
ratio between the size of the lower chord and the other
members. By varying the size of lower chord and keeping
the rest constant, some stresses were obtained which are

plotted on the graph in dimensionless form.

The author sincerely hepes that the results of this

analysis will be of some value to the designers of such

structures.

ANALYSIS

The method of virtual work was used to obtain the
stresses. I do not intend to discuss this method since it
can be found in almost any textbook on Statically Indeter—
minate Structures. However, it might be well to point out

the equations that were used and define the terms in them.

The equation of virtual work for a statically indeter—

minate truss to the first degree is;
SuL uzL _
2A3: + RfiA—E - 0
But since E is constant it can be omitted.

2
€£§§%l+'RgE—E : 0

 

A
or
2e
R -‘— 2L
2%:-
Where;
S = Bar stress in the determinate truss.
R = Horizontal reaction which was assumed as the
redundant.
L : Length of member.
A : Cross—sectional area of member.
E : Modulus of elasticity.
u = Bar stress due to a unit load applied at the point

where the redundant was removed.

-6-

SuL
AE

u2

= Deflection due to the load.

33%:= Deflection due to a unit load.

2
Rig-EL- = Deflection due to R.

Having R, the actual bar stresses can be found very easily

since

Actual stress : S-+ Ru

For this analysis two trussed-arches were used, the

dimensions of which will be shown in the next few pages.

The following assumptions were made regarding the two

 

structures;

1. The trusses are symmetrical.

2. The load on each panel point is equal to P.

3. The members are pin—connected.

4. The bottom chord panel points lie on a parabola.
‘EQUATION OF PARABOLA

The equation of the parabola is;

Y a h — 3%? x2

Where;

y = Distance from the x—axis to the parabola.

x : Distance from the y—axis.

1.: Length of the truss.

h.: Maximum height of the parabola measured from

the x-axis.

The following diagrams represent the trusses and the
loading used in this analysis.

We U1 U2: r03 Wu J“; "6 I“? We

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

? Li L5 -
2‘ - L 5’,
L1 - ‘2‘- 2.
OR C’ t~ :r L
04 F‘ (m 7
Lo f H V l r .__‘L_

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. .48.
8 at 20' = 160' _J
'1
TRUSBED-ARCH 'a'
P P P P P P P
10 W1 He In 1’4 as We
L3
L -
2 1+ 58
ET 3 :19 L5
‘3
Lo L
1%
L‘ 6 at 30' = 180'
ll

 

 

TRUSSEDqARCH"b'

As it can readily be seen both structures are

statically indeterminate to the first degree. Following
the method of virtual work, it is necessary to remove the
redundant, which in this case was taken as the horizontal

reaction at the right support. This redundant must be

replaced with a force, namely R, as shown below.

P P P IP P P P P .P

00, U1 1:12 lag m, is; 1115 :17. as

 

 

 

 

 

 

 

 

 

LL; L5
2 3 L6
n1 ‘ L7-
x50 L8 ‘——R
TRUSSED-ARCH 'a'
P P P P P P P
No r31 132 1% 1% U 5 r06
L2 L3 LL»
L1 2L5

 

 

TRUSSED—ARCH 'b'

Now from the previous discussion, reaction.R and the
stress in the members can be calculated by assuming a cross—

sectional area for each member.

Let,

Abe = cross—sectional area of each bottom chord

member and

A cross—sectional area of each other member.

In order to show the influence of bottom chord on
stresses, it is necessary to keep Abe constant and vary A.
This is to say that;

Abe = KA
where K is a factor which can take on values from O-+-¢>
or O< KCCD . K can not reach a value of zero, because if
K = 0 then

Abe = (O)A = O
and if Abe = 0, the structure will become unstable.

K, however, can be equal to infinity without endangering
the stability of the structure, but it is neither practical
nor economical to make Abe so very much greater than.A.
Actually K never approaches zero or infinity in practice,
but for the purpose of this analysis let us assume that it

does.

If K is allowed to vary between the interval O<K<09

the stresses can be computed for each given K. The results

-10...

a sirs§§-@;Ki___
2; shown on graphs by plotting stress @ K : l
c

--A-— = K1. Note that in either case the numerator and

the denominator have the same units, which is to say that

against

the quantity is in dimensionless form.

..]_1...

 

 

gscussrop

 

Referring to the graphs, it can readily be seen that
all but the vertical members reach a value of zero stress
at some value of K. For example;
1. The horizontal and the diagonal members have zero
stress when K is equal tolO and 8 for trusses
"a“ and “bl respectively.
2. The stress in the lower chord members becomes equal

to zero when O<K<%.

It is evident that since the structure is statically
indeterminate to the first degree, the knowledge of stress
in any two of the members will make it statically determinate.

This knowledge can be of great help when making assump—
tions regarding the area of the members, because one can
Just as well assume Abc and A in such a manner that the
stress in one of the members would be equal to zero. The

problem can then be solved by the equations of equiblirum.

In order to be able to discuss the variation in stress
due to changes in K more specifically, it might be well
to consider the horizontal, vertical, diagonal and the lower

chord members separately.

.11}...

HORIZONTAL and DIAGONAL MEMBERS. These two sets of members

K
are discussed together since the §§§9§§-§L-l--- versus

stress @ K = 1
K1 curve is the same for both of them. This is caused by

the fact that the horizontal compoefit of a stress in a
diagonal is equal to the resulting horizontal stress at
the Joint with an Opposite sign.

The stress variation curve seems very abrupt, but it
is not unreasonable, since the greater the area of a member

the more stress it will take.

The stress in the horizontal and diagonal members is
very small when K is greater than four. When K gets smaller
than two, the stresses start to increase very rapidly since
the lower chord members tend to take a smaller amount of
the load. This stress variation is not unusual as the lower
chord members seem to transfer their stress to the other

members as K decreases.

VERTICALLMEMBERS. The stress variation in these members is
more or less the same as in the diagonal and the horizontal
members, with the exception that they do not decrease quite

as much when K is increased.

Nets that the center member keeps the same stress
regardless of K. The stress in this member is equal to the

load applied directly above it.

-15-

#P

The rest of the vertical members reach a stress equal

to the load, applied directly above them, if K is made large

enough, namely 10 and 8 in trusses 'a' and I'b" respectively.

LOWER CHORD MEMBERS. These members act very much the same

as expected. The stresses increase directly as K increases.

If trussed arch I'a" is considered for K = 10 we can

immediately see that the following stress condition exists;

1.
2.

The stress in each top chord is zero.

The stress in each diagonal is zero.

 

 

 

 

3. The stress in the vertical is equal to the tap
chord panel load.

a. The horizontal component of stress in each bottom
chord is the same and equal to the horizontal
reaction.

P P P P P P P P

47 .42 {0 0
Q. t
% ‘ 14+ L
Q ’ 5 L
. 6
1 L7
-__,. ’to DB hP
c.5019 Thea?

TRUBSEDqARCH "a'I

-16-

It can therefore be said that the trussed—arch 'a' is

equivalent to the three-hinged trussed arch shown below.

Bar UuU connected at its ends
in such a manner that it carries
no direct stress.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P P P P P
UO U1 U2 1U3 U7 U8
1
1
“A g; :i 5%
RxL = E;—
hP—O-" ‘ L8 «1— RxLa = 4P
{‘ ' 8 at 20' : 160' _}
rRyLo '4 l~lv.,50P fl HYLB = ”.50P

The vertical reactions of this structure can be determined
by taking moments about points Lo and L3. To obtain the
horizontal reaction at Lo, the summation of moments of the
left half of the structure about the hinge at La can be set
equal to zero. The horizontal and the vertical reactions
are n.00P and #.50P respectively. Calculating the bar stresses
the following is found to exist;
1. Zero stress in top chord members,
2. Zero stress in diagonal members,
3. Stress equal to tap chord panel load in verticals, and
h. (-5.32P), (—n.7OP), (—h.28P), (—h.10) in members

LOLl, L1L2, L293 and L3Lu respectively.

-17-

 

Therefore, the two trusses mentioned are equivalent and
it can be concluded that when -fE9- : 10 in the trussed arch
'a', it is possible to analyze it as a three-hinged arch and
calculate the stresses in that manner. The reason for this
is the fact that, when the stress in UuUS becomes equal to

zero, the hinge at La becomes effective.

The above discussion also holds true for the trussed

arch I'b", except in this case K = 8 instead of 10.

Referring back to trussed arch “a", when K = 10, the
stress in diagonal and horizontal members is zero. But these
bars cannot be removed if the structure is to remain stable,
because the lower chord members are pin-connected. The
structure can and perhaps should be replaced by a three-hinged

parabolic arch, which would be more economical.

The diagram below represents a three-hinged arch that

could be used as a substitute for the two-hinged trussed arch
P P

 

 

 

THREE-HINGED PARABOLIC ARCH

..18..

The equation of parabola is;
2
= no - X
y 180

The reactions can be found by the equations of equilibrium.

-19-

CONCLUSION

 

As a result of this analysis it can be concluded that
in any two-hinged trussed arch there exists a definite
relationship between the stress in members and the area of

the lower chord.

A two-hinged trussed arch is effective as such only
when the difference in the cross-sectional area of the bottom
chord and the other members is relatively small. When this
difference gets very large, the lower chord tends to take
a much greater amount of the load, which in turn reduces the
load on the remaining members. In other words, the lower
chord carries the most load while the others help to keep
the stability of the structure. Just how large this difference
should be allowed to get for an economical design was not
considered in this analysis, as that would involve many

factors.

Two-hinged trussed arch can be treated as three-hinged
trussed arch when K reaches a certain value, this value is
different for each truss. When K does reach this value,
however, the two structures become equivalent as the stress

condition that was mentioned previously exists in the two.

At this point the structure may be replaced by a three-
hinged arch with the same horizontal and vertical reactions
resulting.

-20..

Perhaps the most important thing to be said, is that

when K gets relatively large, the structure can be

L

as a determinateIfwithout an appreciable amount of

analyzed

error.

This is caused by the fact that as K increases the stress

in the horizontal members get closer and closer to zero,

which in turn tends to make the center pin approach a hinge.

-21-

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