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Structural Analysis -TRUSS BY FLXIBILITY METHOD, Study notes of Structural Analysis

TRUSS BY FLEXIBILITY METHOD.Determine forces in members and support reactions.

Typology: Study notes

2010/2011

Uploaded on 09/01/2011

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TRUSS BY FLEXIBILITY
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TRUSS BY FLEXIBILITY

METHOD

2P

P

A

B

C

D

Determine forces in members and support

reactions.

Now we find the forces in members due to application of

external forces.

2P

P

A

B

C

D

p

p

2p √2p

p

3p p

A

C

D

B

Now we find the forces in members by application of unit horizontal

force at C.

Deflection in a truss is given by

∆=∑P.p.L/(A.E)

So various forces in members and their deflection is

shown in the table

Member Length Member

forces due

to external

forces

S

Member

forces due

to unit

load at C

S 1

Member

forces due

to unit

force

along DB

S 2

L.S. S

1

A.E

L.S. S

2

A.E

AB L/A.E 0 O -1/√2 0 0

BC L/A.E P -1 -1/√2 -PL/A.E -PL/√2AE

CD L/A.E P 0 -1/√2 0 -PL/√2AE

AD L/A.E -2P 0 -1/√2 0 √2PL/AE

AC L/A.E -√2P √ 2 1 -2√2PL/AE^ -2PL/AE

BD √2L/A.E 0 0 1 0 0

FOR FLEXIBILITY METHOD

D

Q−

D

QL=

F.Q

D

QL

= L/AE×{-2√2P-P}= -3.828PL/AE

D

QL

= L/AE×{-P/√2- P/√2+√2P-2P}=-2PL/AE

F = Q =

D

QL

Q

1

= Reaction at C.

Q

2

= Force in member DB.

D

QL

= Deflection at C due to applied load at C.

D

QL

= Deflection in member DB due to applied load in

member BD

F = L/A.E ×

D

QL

=PL/AE ×

F.Q = D

Q

- D

QL

× =

L/AE -^ L/AE

SOLVING THE ABOVE MATRIX BY CONVERTING IT INTO SIMULTANEOUS

EQUATION FORM WE GET

Q

1

= 1.172P

Q

2

= -0.243 P

Now we find forces in all the members by applying

unknown forces.

2P

P

A

B

D C

1.17p

0.55p

1.82p 0.77p

0.17p

0.71p

1.27p 0.72p

0.24p

0.24p

1.72p