Lecture truss [compatibility mode]

11/29/2013
1
Truss
Engineering Mechanics- Truss
Important Features of a Truss
• A truss consists of long and slender straight
members connected at joints. No member is
continuous through a joint.
• Bolted or welded connections (joints) are assumed
to be pinned together. Forces acting at the member
ends reduce to a single force and no couple. Only
two-force members are considered.
Formation of a truss is nothing more than
assemblage of triangles!

11/29/2013
2
External Loads on a Truss
Members of a truss are slender and not capable of supporting large lateral loads.
Loads must be applied at the joints. Members are light in weight and economical.
Most structures are made of several trusses joined together to form a space
framework. Each truss carries those loads which act in its plane and may be
treated as a two-dimensional structure.
Various Types of Truss

11/29/2013
3
Simple Truss
• A simple truss is constructed by
successively adding two members and
one connection to the basic triangular
truss.
• A simple truss is not necessarily
made only of triangles.Although
triangles are always maintained in
practice
• In a simple truss, m = 2n - 3 where
m is the total number of members
and n is the number of joints.
basic truss
• A simple truss is always internally
rigid. Will not deform and collapse
under the action of load
Analysis of Trusses by the Method of Joints
• Dismember the truss and create a freebody
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, and
opposite sense.
• Forces exerted by a member on the pins or
joints at its ends are directed along the member
and equal and opposite.
• For simple Truss: m = 2n – 3
Total unknowns
member forces = m = 5
reactions = r = 3
Total no. of equilibrium equations
2n = 8 (2 force equilibrium equations per joint)
m+r=2n
The truss is said to be statically determinate
and completely constrained.

11/29/2013
4
Analysis of Trusses by the Method of Joints
Force Polygon for joint
equilibrium
Joints Under Special Loading Conditions
• Forces in opposite members intersectingin
two straight lines at a joint are equal.
• The forces in two opposite members are
equal when a load is aligned with a third
member. The thirdmember force is equal
to the load (includingzero load).
• The forces in two members connected at a
joint are equal if the members are aligned
and zero otherwise.
• Recognition of joints under special loading
conditions simplifies a truss analysis.
Engineers Mechanics- Truss

11/29/2013
5
Problem -1
For the given loading , determine the zero force member in the truss shown
Problem 2
Using method of joints, determine the forces in the members of the truss shown
below

11/29/2013
6
Problem 2
FAB=0=FBC
FAD= -5 (C)
FBD= FBF
2 FBD (1.6/3.4) + 20 + FBE = 0
FBD = FBF = -34 (C)
B
FEB=12 (T)
FED= FEF
D
FDE + FDB (3/3.4) = 0
FDE = 30 (T) = FEF
Check
+FAD + FBD (1.6/3.4) + 21 =0
21
2121
Problem 3
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 equilibrium
equations for the reactions at E and C.
• JointA 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
• All member forces and support reactions
are known at joint C. However, the joint
equilibriumrequirements may be applied
to check the results.

11/29/2013
7
Problem 3
SOLUTION:
• Based on a free-body diagram of the entire
truss, solve the 3 equilibrium equations for the
reactions at E and C.
       m6m12N1000m24N2000
0
E
MC


 N000,10E
  xx CF 0 0xC
  yy CF N10,000N1000-N20000
 N7000yC
Problem -3
• Joint A is subjected to only two unknown
member forces. Determine these from the
534
N2000 ADAB FF

CF
TF
AD
AB
N2500
N1500


• There are now only two unknown member
forces at joint D.
  DADE
DADB
FF
FF
5
32

CF
TF
DE
DB
N3000
N2500



11/29/2013
8
• There are now only two unknown member
forces at joint B. Assume both are in
tension.
 
N3750
250010000 5
4
5
4


BE
BEy
F
FF
CFBE N3750
   
N5250
3750250015000 5
3
5
3


BC
BCx
F
FF
TFBC N5250
• There is one unknown member force at
joint E. Assume the member is in
tension.
 
N8750
375030000 5
3
5
3


EC
ECx
F
FF
CFEC N8750
Problem -3
• 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
Problem -3

11/29/2013
9
Analysis of Trusses by the Method of Sections
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagramfor the left side.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.
Problem 4
A Fink roof truss is loaded as shown. Use method of section to determine the
force in members CD, CE and BD

11/29/2013
10
Problem 4
Problem 5
Use method of section to determine the force in members GJ and IK of the
truss shown
100kN
100kN
100kN
Inverted K truss

11/29/2013
11
Problem-5
100kN
100kN
100kN
G
Solution JX = -300
KY = -JY = 100*(2.7 + 5.4 +8.1)/7.5 = 1620/7.5 = 216
Section XX, lower FBD,
∑MG = 0
-FKI [ 5.9( ] + JY(0.8) –JX(2.7) – KY(6.7) = 0
FKI = -143.19 kN (C )
∑MI = 0
FJG[ 5.9 ( ] + JY(6.7) –JX(2.7) – KY(0.8) = 0
FJG = 143.19 kN (T )
Compound Truss
• Compound trusses are composed of two or more
simple trusses . The trusses shown are statically
determinant, rigid, and completely constrained,
nrm 2

11/29/2013
12
Compound Truss
Determine the forces in bars 1. 2. and 3 of the plane truss supported and
loaded as shown in the figure
Problem 6
SOLUTION:
• Two simple trusses
ADE & BCF
• Detach to expose 6
unknowns (three
member forces at 1,
2,3 and three support
reactions)
• Draw free body
diagrams to find
forces in 1, 2 &3
E F
3

11/29/2013
13
Problem 6
By
Ax
Problem 6

11/29/2013
14
Problem 6
Partially, Improperly and Properly Constrained Truss
• For a truss to be properly constrained:
– It should be able to stay in equilibrium for any combination
of loading.
– Equilibrium implies both global equilibrium and internal
equilibrium.
• Note that if m + r < 2n, the truss is most definitely
partially constrained (and is unstable to certain
loadings). But m + r ≥ 2n, is no guarantee that the
truss is stable, and may be termed improperly
constrained.
• If m + r < 2n, the truss can never be statically
determinate.

11/29/2013
15
• Fewer unknowns than
equations, partially
constrained
• Equal number unknowns
and equations but
improperlyconstrained
Inadequacy
of external
constraints
unstable
Simple Truss
(internally
rigid)
Check if partially constrained, fully constrained or improperly constrained
P
P
P
Statically determinate, properly constrained
A Simple Truss is always a rigid truss
m + r = 2n m=9, r = 3, n=6
m=9, r = 3, n=6m + r = 2n
improperly constrained (non rigid,
may collapse)
m + r < 2n
Partially constrained (non rigid, may
collapse)
m=8, r = 3, n=6

11/29/2013
16
Problem 7 : Partially, Improperly and Properly ConstrainedTruss
`
2n = 20, m = 16, r = 4.
The circled part is a simple truss that
is adequately supported with 3
reactions. So it is completely
constrained
We only have to worry about the
remaining portion. We can easily
show that no matter whatever loading
is applied at the joints of the
remaining part can be supported.
Completely constrained & Statically
determinate
Problem 7 : Partially, Improperly and Properly ConstrainedTruss
a) j = 7, m = 10, r = 3, 2n > m + r.
Partially Constrained
b) j = 8, r = 3, m = 13, 2n = m + r.
Clearly the truss cannot be equilibrium
with this loading.
Take equilibrium of joint C. BC is
tensile
Take a section as shown and moment
about left hinge is not balanced
Improperly Constrained.
B
C

Lecture truss [compatibility mode]

More Related Content

What's hot

Similar to Lecture truss [compatibility mode]

More from sharancm2009

Recently uploaded

Lecture truss [compatibility mode]