Structural Analysis - Beam-Flex, Study notes of Structural Analysis

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Flexibility Method

Beam with Concentrated

Loads

DISPLACEMENTS DUE TO VARIOUS

LOADINGS

a b A

B

B A

A

B

A B

ƟA = PAB x (2L-a)/(6LEI)

ƟB = PAB x (2L-b)/(6LEI)

ƟA = - ƟB = PAB x L^3 /

(24EI)

ƟA = 7PAB x L^3 / 360EI

ƟB = 8PAB x L^3 / 360EI

ƟA = ƟB = -ML /

24EI

Similar to stiffness method the unknown reactions in flexibility method is given by F*Q = DQ - DQL

Where ,

• F = Flexibility Matrix
• Q = Unknown Reactions
• DQ = Reactions at supports or points of consideration
• DQL = Fixed end forces and moments

In the above problem, DQ = 0

(since there is no sinking of supports or rotation)

DQL1 = ƟB = PLAB^2 / 16EIAB = 24*102 / 16EI = 150 / EI

DQL2 = ƟC = PLAB^2 / 16EIAB + PLBC^2 / 16EIBC = 225 / EI

A B^ C

F 11 F 21

F 11 =L /(3EI) = 10 / 3EI F 21 =L /(6EI) = 10 / 6EI

A B^ C

F 22

F 12

F 12 =L /(6EI) = 10 / 6EI F 22 =L /(3EI) + L/(3EI)= 20 / 3EI

Applying Unit Moment at A :

Applying Unit Moment at B :

F*Q = DQ - DQL

Now,

And, DQ = { 0 }

Hence we have,

F*Q = - DQL

10 / 3EI 10 / 6EI

10 / 6EI 20 / 3EI

MA

MB

150 / EI

225 / EI

MA

MB

Solving simultaneously, we get :

After substituting the respective values for the F and DQL in the above

equations, the values of moments at A and B are obtained as follows:

MA = -32.14 (Hogging)

MB = -25.71 (Hogging)

Reactions are calculated using Free Body Diagrams.

25.71 B

C

RB RC

12 KN

Taking Sum of moments at B = 0, -25.71 = RC * 10 – 12 * 5 Thus, RC = 3.429 KN (Upwards)

Shear Force Diagram:

12.643 KN

-3.429 KN

-11.375 KN

7.736 KN

A B C

Bending Moment Diagram:

31.075 KNm

-32.14 KNm -25.71 KNm

17.645 KNm

A B C