lecture truss [compatibility mode]
TRANSCRIPT
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TrussEngineering 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.
Engineering Mechanics- Truss
Formation of a truss is nothing more than assemblage of triangles!
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External Loads on a TrussMembers 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.
Engineering Mechanics- Truss
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 TrussEngineering Mechanics- Truss
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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
Engineering Mechanics- Truss
• 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 internallyrigid. 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 – 3Total unknownsmember forces = m = 5reactions = r = 3Total 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.
Engineering Mechanics- Truss
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Analysis of Trusses by the Method of JointsEngineering Mechanics- Truss
Force Polygon for joint equilibrium
Joints Under Special Loading Conditions• Forces in opposite members intersecting in
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 third member force is equal to the load (including zero 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
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Problem -1
For the given loading , determine the zero force member in the truss shown
Engineers Mechanics- Truss
Problem 2
Using method of joints, determine the forces in the members of the truss shown below
Engineers Mechanics- Truss
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Problem 2Engineers Mechanics- Truss
FAB=0=FBC
FAD= -5 (C)FBD= FBF2 FBD (1.6/3.4) + 20 + FBE = 0FBD = FBF = -34 (C)
B
FEB=12 (T)FED= FEF
D
FDE + FDB (3/3.4) = 0FDE = 30 (T) = FEF
Check+FAD + FBD (1.6/3.4) + 21 =0
212121
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.
• 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 results.
Engineers Mechanics- Truss
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Problem 3SOLUTION:
• Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.
m 6m 12N 1000m 24N 20000
EMC
N 000,10E
xx CF 0 0xC
yy CF N 10,000 N 1000 - N 20000
N 7000yC
Engineers Mechanics- Truss
Problem -3
• Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements.
534N 2000 ADAB FF
CFTF
AD
AB
N 2500 N 1500
• There are now only two unknown member forces at joint D.
DADE
DADB
FF
FF
532
CFTF
DE
DB
N 3000 N 2500
Engineers Mechanics- Truss
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• There are now only two unknown member forces at joint B. Assume both are in tension.
N 3750
250010000 54
54
BE
BEy
F
FFCFBE N 3750
N 5250
3750250015000 53
53
BC
BCx
F
FF
TFBC N 5250
• There is one unknown member force at joint E. Assume the member is in tension.
N 8750
375030000 53
53
EC
ECx
F
FFCFEC N 8750
Engineers Mechanics- Truss 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.
checks 087507000
checks 087505250
5453
y
x
F
F
Problem -3Engineers Mechanics- Truss
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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 diagram for 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.
Engineering Mechanics- Truss
Problem 4
A Fink roof truss is loaded as shown. Use method of section to determine the force in members CD, CE and BD
Engineers Mechanics- Truss
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Problem 4Engineers Mechanics- Truss
Problem 5
Use method of section to determine the force in members GJ and IK of the truss shown
Engineers Mechanics- Truss
100kN
100kN
100kN
Inverted K truss
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Problem-5Engineers Mechanics- Truss
100kN
100kN
100kN
G
Solution JX = -300KY = -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
Engineers Mechanics- Truss
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Engineers Mechanics- Truss Compound Truss
Determine the forces in bars 1. 2. and 3 of the plane truss supported and loaded as shown in the figure
Problem 6Engineers Mechanics- Truss
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
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Problem 6 Engineers Mechanics- Truss
By
Ax
Problem 6 Engineers Mechanics- Truss
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Engineers Mechanics- Truss Problem 6
Partially, Improperly and Properly Constrained Truss
Engineering Mechanics- 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.
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• Fewer unknowns than equations, partially constrained
• Equal number unknowns and equations but improperly constrained
Partially, Improperly and Properly Constrained Truss
Inadequacy of external constraints
unstable
Simple Truss(internally rigid)
Engineering Mechanics- Truss
Partially, Improperly and Properly Constrained Truss
Engineering Mechanics- Truss
Check if partially constrained, fully constrained or improperly constrainedP
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
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Engineers Mechanics- TrussProblem 7 : Partially, Improperly and Properly Constrained Truss
`
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 Constrained Truss
Engineering Mechanics- Truss
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.
BC