lecture truss [compatibility mode]

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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.


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


• 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.


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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


Problem 2

Using method of joints, determine the forces in the members of the truss shown below


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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.


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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


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


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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.


Problem 4

A Fink roof truss is loaded as shown. Use method of section to determine the force in members CD, CE and BD


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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

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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


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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


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


• 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


Inadequacy of external constraints

unstable

Simple Truss(internally rigid)




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


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