Calculation of Stiffness Coefficients for a Beam Element, Exercises of Mathematical Methods for Numerical Analysis and Optimization

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Q: 5.20 For the beam element given in Pb 5.18, generate the stiffness

coefficient k13,k23,k33 and k43.

Solution

For third column we have below free body diagram

We know that for beam element

2

2

d w M

dx EI

Where

M  M (^) a V xa

Hence

2

2

( (^) a a )

d w M V x dx EI

And

2

1

a a

dw V x M x C dx EI

2 3

1 2

M x a V xa w C x C EI

Now putting boundary condition

at

a

x

dw

dx



Where 13 3

a

EI

V k l

2

a

EI

M k l

Now apply statics equation as

Fy^ ^0 ,^ Va^ ^ Vb^0 , this equation can also be written as

k 13  k 33  0 or k 13  k 33

So 33 3

12 EI

k l

Similarly

M^ b^0 ,^ M^ a ^ Mb^0 or k 23 ^ k 43 ^0

So 43 2

6 EI

k l

Hence

(^13 )

12 EI

k l

2

6 EI

k l

3

12 EI

k l

2

6 EI

k l