Stiffness Coefficients 3-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Beam, Element, Generate, Stiffness, Coefficient, Deflection, Equations, Static, Equilibrium

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Assignment
Finite Element Methods
Q 5.21
For the beam element considered in problem 5.18 generate expressions for the
stiffness coefficients
K14, K24, K34 and K44.
Solution:
Direct approach is used to solve the problem i,e basic deflection relations and static equilibrium
equations. So first we do the solution comprising the deflection relations. For our convenience we
replace the symbol used in the figure by the most familiar symbols which are given below;
Va = q1
θa = q2
Vb = q3
θb = q4
So for the forth column of the stiffness matrix we have all the other constraints equal to
zero, while the θb =1. The θ is the rotation produced by the force V.
At any section along the length of the beam we have relation for bending moment which
is given as
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Assignment

Finite Element Methods

Q 5.

For the beam element considered in problem 5.18 generate expressions for the

stiffness coefficients

K 14 , K 24 , K 34 and K 44.

Solution:

Direct approach is used to solve the problem i,e basic deflection relations and static equilibrium equations. So first we do the solution comprising the deflection relations. For our convenience we replace the symbol used in the figure by the most familiar symbols which are given below; Va = q 1 θa = q 2 Vb = q 3 θb = q 4 So for the forth column of the stiffness matrix we have all the other constraints equal to zero, while the θb =1. The θ is the rotation produced by the force V. At any section along the length of the beam we have relation for bending moment which is given as

Now at x = 0 , θ = θa = 0 Thus at x = l, w = wb = ………………………………………. (3) ……………… (4) Now utilizing the second set of boundary conditions Now at x = 0 , θ = Wa = 0 So we have Hence we get ……………………………….(a) at x = l, w = θb =