Rec3PPT.pdf

8/18/2019 Rec3PPT.pdf

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TrussTruss – – Assumptions AssumptionsThere are four main assumptions made in the

analysis of trussTruss members are connected together at their

ends only.

Truss are connected together by frictionless

pins.

The truss structure is loaded only at the joints.

The weights of the members may be neglected.

1

2

3

4


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Simple Truss Simple Truss

The basic building block of atruss is a triangle. Large truss

are constructed by attaching

several triangles together Anew triangle can be added

truss by adding two members

and a joint. A trussconstructed in this fashion is

known as a simple truss.


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Method of Joints Method of Joints --TrussTrussThe truss is made up of single bars, which are

either in compression, tension or no-load. Themeans of solving force inside

of the truss use equilibrium

equations at a joint. Thismethod is known as the

method of joints.


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Method of Joints Method of Joints --TrussTrussThe method of joints uses the summation of

forces at a joint to solve the force in themembers. It does not use the

moment equilibrium equation

to solve the problem. In a twodimensional set of equations,

In three dimensions,

x y0 0 F F = =∑ ∑

z 0 F =

∑


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TrussTruss – – Example Problem Example Problem

Determine the loads ineach of the members by

using the method of

joints.


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Draw the free-bodydiagram. The summation

of forces and moment

about B result in

( ) ( ) ( )

x Ax B

y Ay Ay

A B

B

Ax

0

0 10 kips 10 kips 20 kips

0 5 ft 10 kips 10 ft 10 kips 20 ft

60 kips

60 kips

F R R

F R R

M R

R

R

= = +

= = − − ⇒ =

= = − −

⇒ =

⇒ = −

∑∑

∑


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Look at Joint B

x BC B BC BC

y BA BA

0 60 kips 60 kips

0 0 kips

F T R T T

F T T

= = + = + ⇒ = −

= = ⇒ =

∑∑


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Look at Joint D and findthe angle

1 o

x DC DA

y DA DA

DC

5 ft.tan 14.04

20 ft.0 cos

0 sin 10 kips 41.231 kips

40 kips

F T T

F T T

T

α

α

α

− = =

= = − −

= = − ⇒ =

= −

∑∑


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Look at Joint C and findthe angle

( ) ( ) ( ) ( )

1 o

y CA CA

x CD CA CB

o

5 ft.

tan 26.56510 ft.

0 sin 10 kips 22.361 kips

0 cos

40 kips 22.361 kips cos 26.565 60 kips

0

F T T

F T T T

β

β

β

−

= =

= = − ⇒ =

= = − −

= − − − −

=

∑

∑


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Example Problem Example Problem

Determine the forces in

members FH, DH,EG andBE in the truss using the

method of sections.


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Draw the free-body diagram. Thesummation of forces and moment

about H result in

( ) ( ) ( ) ( ) ( )

x Hx

Hx

y Hy I

H I

I

Hy

0 3 kips 3 kips 3 kips 3 kips

12 kips

0

0 15 ft 3 kips 10 ft 3 kips 20 ft 3 kips 30 ft 3 kips 40 ft

20 kips

20 kips

F R

R

F R R

M R

R

R

= = + + + +

⇒ = −

= = +

= = − − − −

⇒ =

⇒ = −

∑

∑∑


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Do a cut between BD and CE


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Take moment about A

( ) ( ) ( )

1 0

0

A CE

CE

10 fttan 53.137.5 ft

0 cos 53.13 20 ft 3 kips 10 ft

2.5 kips

M T

T

α −

= =

= = +

⇒ = −∑


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Do a cut between HD and GE


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Take the moment about I

Take the moment about D

( ) ( ) ( )I HD

HD

0 20 kips 15 ft 15 ft 3 kips 10 ft

18 kips

M T

T

= = − −

⇒ =

∑

( ) ( ) ( ) ( )D GE

GE

0 12 kips 20 ft 20 kips 15 ft 3 kips 10 ft 15 ft

6 kips

M T

T

= = − + + +

⇒ = −∑


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Do a cut between HD and HI


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Take the sum of forces in y

direction

( )

( )

1 0

0

y HF HD

HF 0

10 fttan 53.13

7.5 ft

0 sin 53.13 20 kips

20 kips 18 kips2.5 kips

sin 53.13

F T T

T

α −

= =

= = + −

−⇒ = =

∑

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