TRUSS ANALYSIS (TERM PAPER REPORT)

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TERM PAPER
ON
TTrruussss AAnnaallyyssiiss
SUBMITTED BY: SSUUBBMMIITTTTEEDD TTOO:
SUNIL KUMAR PINAKI NAYAK
Roll no: 140107133
System id: 2014016394
Branch: Civil (2B)
Department of Civil Engineering
School of Engineering and Technology
Sharda University, Gr. Noida, UP, INDIA
sunilsaurav222@gmail.com

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ABSTRACT:
The purpose of this term paper is to study about the basics of truss and its analysis and
calculation of reactions and internal forces of tension or compression in truss members of
statically determinate 2D trusses with arbitrary number of joints. All joints are supposed to be
hinged. The external forces and reactions (the amount of corresponding constraints must be equal
to 3) can be applied at arbitrary joint, but should be represented by their horizontal and vertical
components. The method of joints and method of section is used in the calculation. It is helpful
not only for structural engineers, but also for students because all main steps of the solution are
provided

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INTRODUCTION:
The truss as defined by Francis Brangan is a “framed structure consisting of a triangle or a group
of triangle arranged in such a manner that load applied at the points of intersection of the
member will cause only tension and compression in the members” . It is engineered to handle a
pre calculated load , and they are designed and constructed as the minimum structure which will
carry the designed load . The engineering principle behind the truss the shape of a triangle or a
series of a triangle . In geometry a triangle is aver strong ,and can withstand different loads better
than other shapes . In architecture and structural engineering , a truss is a structure comprising
one or more triangular units constructed with straights members whose ends are connected at
joints referred to as nodes . External forces and reaction to those force are considered to act only
at the nodes and result is forces in the members which are either tensile or compressive forces .
Moment (torques) are explicitly excluded because , and only because , all the joints in a truss are
treated as revolute .
DESCRIPTION:
In simple word we can say trusses formed from two forces member, i.e. straight member with
end point connection. A truss consists of straight members connected at joints .no member is
continuous through a joint. Different types of trusses are planer truss and space frame truss. We
will discuss this in detail. And various types of procedure or we can say that methods of analysis
of trusses are:
(a)Methods of joints.
(b)Methods of section.
(c)Graphical methods.
But we will discuss the two method only (a and b).

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There is no member AB; there are instead two distinct member AD and DB .
Truss structure are widely used in bridge, roofs and at many other places.
TYPES OF TRUSS:
1. PLANER TRUSS:
The simplest form of a truss is one single triangle. This type of truss is seen in a framed roof
consisting of rafters and a ceiling joist and in other mechanical structures such
as bicycles and aircraft. Because of the stability of this shape and the methods of analysis
used to calculate the forces within it, a truss composed entirely of triangles is known as a
simple truss e.g.
The traditional diamond-shape bicycle frame, which utilizes two conjoined triangles, is an
example of a simple truss.
2.SPACE FRAME TRUSS:
A space frame truss is a three-dimensional framework of members pinned at their ends.
A tetrahedron shape is the simplest space truss, consisting of six members which meet
at four joints. Large planar structures may be composed from tetrahedrons with
common edges and they are also employed in the base structures of large free-standing
power line pylons.
STATICAL DETERMINANCY:
A truss that is assumed to comprise members that are connected by means of pin joints, and
which is supported at both ends by means of hinged joints or rollers, is described as

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being statically determinate. Newton's Laws apply to the structure as a whole, as well as to each
node or joint.
Trusses that are supported at more than two positions are said to be statically indeterminate, and
the application of Newton's Laws alone is not sufficient to determine the member forces.
In order for a truss with pin-connected members to be stable, it must be entirely composed of
triangles. In mathematical terms, we have the following necessary condition for stability:
where m is the total number of truss members, j is the total number of joints and r is the number
of reactions (equal to 3 generally) in a 2-dimensional structure.
When , the truss is said to be statically determinate, because the (m+3) internal
member forces and support reactions can then be completely determined by 2j equilibrium
equations .
e.g.-Test the statically determinacy of the pin jointed trusses shown in below given figure:-
Solution:
The truss has five members and four joints. Thus m=5 and j=4 .
So, 2j-3=5=m .
So truss is statically determinate as it satisfy equation m=2j-3 .
Before going through analysis of trusses we must know some assumption which are made in the
theory that is going to be developed in this term paper.
ASSUMPTION:
1. The end of the member are pin connected (hinged),
2. The loads act only at the joints,
3. Self-weights of the members are negligible,
4. Cross-section of the member is uniform.
ANNALYSIS OF TRUSSES:
Analysis of trusses simply mean the analysis or calculation of forces and its nature in each
member of a trusses. The analysis of trusses often assume that loads are applied to joints only
and not at intermediate points along the member. The weight of the member is often insignificant
compared to the applied loads and so is often omitted.

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METHODS OF TRUSS ANALYSIS:
1.Methods of joints
2.Methods of section
3.Graphical The structure is explored and the forces on each member and joint are identified. At
each joints the forces in the members meeting and the loads acting , if any , constitute a system
of concurrent forces .
method
The first two are analytical methods and they are dealt in this term paper.
1.METHOD OF JOINTS:
RULES FOR THE METHOD OF JOINTS:
1. Find reactions at the supports
2. Initially we assume all members are in tension
3. If it turns out that they are in compression, the sign (direction) will tell us so
4. Forces are labelled e.g. TAC regardless of whether they go from A to C or C to A
5. Choose a joint with few unknown forces to start
SAMPLE EXAMPLE:-

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Using the method of joints , determine the force in each member of the truss
SAMPLE EXAMPLE:
Using the method of joints, determine the force in each member of the truss
SOLUTION:
• Based on a free-body diagram of the entire truss, solve the 3
• Joint A is subjected to only two unknown member forces. Determine these from the joint
equilibrium requirements.
• In succession, determine unknown member forces at joints D, B, and E from joint
equilibrium requirements.
• All member forces and support reactions are known at joint C. However, the joint
equilibrium requirements may be applied to check the result

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• Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the
reactions at E and C.
       ft6ft12lb1000ft24lb2000
0
E
MC


,  lb000,10E
  xx CF 0
,
0xC
  yy CF lb10,000lb1000-lb20000
,
 lb7000yC
Joint A is subjected to only two unknown member forces. Determine these from the joint
equilibrium requirements

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534
lb2000 ADAB FF

CF
TF
AD
AB
lb2500
lb1500


• There are now only two unknown member forces at joint D.
  DADE
DADB
FF
FF
5
3
2

CF
TF
DE
DB
lb3000
lb2500


There are now only two unknown member forces at joint B. Assume both are in tension
 
lb3750
250010000 5
4
5
4


BE
BEy
F
FF
CFBE lb3750
   
lb5250
3750250015000 5
3
5
3


BC
BCx
F
FF
TFBC lb5250

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There is one unknown member force at joint E. Assume the member is in tension
 
lb8750
375030000 5
3
5
3


EC
ECx
F
FF
CFEC lb8750
All member forces and support reactions are known at joint C. However, the joint equilibrium
requirements may be applied to check the results
   
   checks087507000
checks087505250
5
4
5
3




y
x
F
F
2.METHOD OF SECTION:-
Used when we interested in calculating the load at certain members.In this case, the method of
joints becomes unnecessarily tedious when the truss exceeds a certain size.

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RULES FOR THE METHOD OF SECTION:-
1.If the joints are in equilibrium, so too is the truss as a whole hence, we can also split it into two
equilibrium part-trusses .
2.The method of sections is especially useful to find the force in a member in the middle of a large truss
SAMPLE EXAMPLE:-
Determine the force in members FH, GH, and GI.
SOLUTION:
• Take the entire truss as a free body. Apply the conditions for static equilibrium to solve
for the reactions at A and L.

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• Pass a section through members FH, GH, and GI and take the right-hand section as a free
body.
• Apply the conditions for static equilibrium to determine the desired member forces.
SOLUTION:
• Take the entire truss as a free body. Apply the conditions for static equilibrium to solve
for the reactions at A and L.
        
       







kN5.12
kN200
kN5.7
m25kN1m25kN1m20
kN6m15kN6m10kN6m50
A
ALF
L
L
M
y
A

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• Pass a section through members FH, GH, and GI and take the right-hand section as a free
body.
Apply the conditions for static equilibrium to determine the desired member forces
       
kN13.13
0m33.5m5kN1m10kN7.50
0



GI
GI
H
F
F
M
TFGI kN13.13

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        
  
kN82.13
0m8cos
m5kN1m10kN1m15kN7.5
0
07.285333.0
m15
m8
tan






FH
FH
G
F
F
M
GL
FG


CFFH kN82.13
 
        
kN371.1
0m10cosm5kN1m10kN1
0
15.439375.0
m8
m5
tan
3
2





GH
GH
L
F
F
M
HI
GI


CFGH kN371.1
ZERO-FORCEMEMBERS:
If a joint has only two non-collinear members and there is no external load or support reaction at
that joint, then those two members are zero-force members. In this example members DE, CD,
AF, and AB are zero force members.
You can easily prove these results by applying the equations of equilibrium to joints D and A.
Zero-force members can be removed (as shown in the figure) when analyzing the truss.

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If three members form a truss joint for which two of the members are collinear and there is no
external load or reaction at that joint, then the third non-collinear member is a zero force
member.
Again, this can easily be proven. One can also remove the zero-force member, as shown, on the
left, for analyzing the truss further.
Please note that zero-force members are used to increase stability and rigidity of the truss, and to
provide support for various different loading conditions.
MAJOR CAUSE OF TRUSS FAILURES:
1. Improperly installation or inadequate bracing.

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2. Overloading the trusses before permanent bracing or sheeting has been installed like stacks of
plywood placed on trusses , for example.
3 .Improper or inadequate connection to the supporting structure .
4. Improper or unauthorized field changes made to trusses.
5. Installing damaged, broken or improperly repaired trusses .
6. Also due to improper design, deterioration, overloading , inadequate repair , and weather
related influences.

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CONCLUSION:
The truss is a building invent that allows the weight of a roof to be distributed to the outer walls
for better support. Truss gives more support and allows the builders to use fewer materials to
achieve any construction. It allows distribution of load. It increases the span mean life span of
any construction like bridge or building Trusses have some disadvantages also. In truss bridge, it
takes up more space and can some-times become a distraction to drivers. It also have higher
maintenance demand of all joint and fitting, more calculation to determine that it will take the
maximum load. Determination of forces in member is carry out by two process, method of joint
& section. By this we conclude forces in each member and at each joint. Truss play a vital role in
our surrounding i.e. everywhere like in bridge, building, rooftop, etc.

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BIBLIOGRAPHY:-
1.F.P.Bear,E.RusselJontson. page-286,287,289,304(6.7)page-6.2,fig-6.16.
2.Engineering mechanics by D.P. Sharma.
3.Introduction to mechanical engineering b[1] Standard handbook of engineering
calculation/typerG.Hicks, editorS. David hicks coordinating editor,-3rd ed. Mc Graw-
Hill,inc,1994. p-1.12.
4.Vector mechanics for engineering (statics) ninth edition y onkarsingh(revised third edition).
5. Mechanics for mechanical engineering by Dr. Hanafyomar.

TRUSS ANALYSIS (TERM PAPER REPORT)

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TRUSS ANALYSIS (TERM PAPER REPORT)