Trusses - Method Of ts 2u3f

CIVL 3121

Trusses - Method of ts

Method of ts

Method of ts

 If a truss is in equilibrium, then each of its ts must be in equilibrium.  The method of ts consists of satisfying th equilibrium the ilib i equations ti for f forces f acting ti on each t.

 Fx

0

 Fy

 Recall, that the line of action of a force acting on a t is determined by the geometry of the truss member.  The line of f action acti n is formed f rmed by c connecting nnectin the two ends of each member with a straight line.  Since direction of the force is known, the remaining unknown is the magnitude of the force.

0

Method of ts

Method of ts t B

t A

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Upper chord

Verticals

Tension Force

t B

t A

Diagonals

Lower chord

Compression Force

Method of ts

Method of ts Upper chord in compression

gusset plate

weld

Idealized t – connected by a frictionless pin

Lower chord in tension

CIVL 3121

Trusses - Method of ts

Method of ts

Method of ts 

Upper chord in compression

Procedure for analysis - the following is a procedure for analyzing a truss using the method of ts: 1 1.

If possible possible, determine the reactions

2. Draw the free body diagram for each t. In general, assume all the force member reactions are tension (this is not a rule, however, it is helpful in keeping track of tension and compression ).

Lower chord in tension

Method of ts 

Method of ts

Procedure for analysis - the following is a procedure for analyzing a truss using the method of ts:



0

 Fy

Procedure for analysis - the following is a procedure for analyzing a truss using the method of ts: 4. If possible, begin solving the equilibrium equations at a t where only two unknown reactions exist. Work your way from t to t, selecting the new t using the criterion of two unknown reactions. 5. Solve the t equations of equilibrium simultaneously, typically using a computer or an advanced calculator.

3. Write the equations of equilibrium for each t,

 Fx

0

Method of ts 

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Method of ts

Example - Consider the following truss



Example - Consider the following truss

First, determine the reactions for the truss 500 lb

500 lb

Ax

Ay

500 lb

Cy = 500 lb

10 ft

10 ft

First, determine the reactions for the truss

(10ft  MA  0  500 lb500 lb )  C y (10ft )

Cy

10 ft

 Fy

 0  Ay  C y

 Fx

 0  Ax  500 lb

Ay = -500 lb 10 ft

Ax = -500 lb

Ax

Ay

Cy

CIVL 3121

Trusses - Method of ts

Method of ts

Method of ts

The equations of equilibrium for t A

The equations of equilibrium for t B 500 lb

FAB 500 lb

FAC

500 lb

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 Fx

 0  FAC  500 lb

FAC = 500 lb

 Fy

 0  FAB  500 lb

FAB = 500 lb

FAB

 Fx

 0  FBC cos 45  500 lb

FBC = -707.2 707.2 lb

FBC

The forces in the truss can be summarized as: FAB = 500 lb (T)

FBC = 707.2 lb (C)

FAC = 500 lb (T)

Method of ts 

Problem – Determine the force in each member of the truss shown below

Method of ts 

Problem – Determine the force in each member of the truss shown below B

A

4 ft

C

E

60

60

4 ft

D

4 ft 800 lb

In the notes on page 10

Method of ts 

Problem – Determine the force in each member of the truss shown below

Zero Force 

Truss analysis may be simplified by determining with no loading or zero–force.



These may provide stability or be useful if th loading the l di changes. h



Zero–force may be determined by inspection of the ts

CIVL 3121

Trusses - Method of ts

Zero Force

Zero Force

Case 1: If two are connected at a t and there is no external force applied to the t

 y

F1



 Fy

 0  F1 sin 

F1 = 0

 Fx

 0  F1 cos   F2

F2 = 0

Case 2: If three are connected at a t and there is no external force applied to the t and two of the are colinear



y



F3

F2 x

 Fy

Zero Force Determine the force in each member of the truss shown below:

Determine the force in each member of the truss shown below:



E

A

E

Method of ts The equations of equilibrium for t C

Determine the force in each member of the truss shown below:

800 lb

800 lb

4 3

C FBC

E

8 ft

F

8 ft

4 5

4 5

3 5

3 5

 Fx

 0   FBC  FCD

 Fy

 0   FBC  FCD  800 lb

4

FBC = FCD

3

FCD

FBC = -666.7 lb

D

G

8 ft

8 ft

Zero Force

A

F

G

8 ft

B

D

A

F

G

Using Case 1 FEF and FCF are zero zero-force force

C

B

D

8 ft

800 lb

Using g Case 1 FAGG and FCGG are zero-force

B

F1 = 0

Zero Force

C

The remaining non-zero forces can be found using th method the th d of f ts j i t

 0  F1 sin 

In the notes on page 11

800 lb

Using g Case 2 FBG and FDF are zero-force



F1

F2 x



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FBC = 666.7 lb (C)

CIVL 3121

Trusses - Method of ts

End of Trusses - Part 2

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