Explanation for Shape Function (linear) 4

[amitdesh123]

A concept of shape function in FEA is basic and very important. It governs the displacement in the element. The accuracy of the solution depends upon the shape function. I am posting a simple explanation for the shape function concept.

Let us consider a simple element. Assuming linear displacement

u = ax + b -----------------eq 1

At the node i

u1 = ax1 + b ------------------ eq 2

At node j

u2 = ax2 + b ----------------- eq 3

From above two equations we can find the values of a and b.

Subtracting eq 2 from eq 3 we get,

u2 - u1 = ax2 - ax1

(u2 - u1) / (x2 - x1) = a ------------------ eq 4

Substituting eq 4 in eq 2 we get the value of b as,

u1- [(u2 - u1) / (x2 - x1)] * x1 = b -------------eq 5

Now eq 1 can be written as

u = [(u2 - u1) / (x2 - x1)] * x1 + u1- [(u2 - u1) / (x2 - x1)] * x1

After rearranging the terms you get,

u = [(u2 - u1) / (x2 - x1)] * x + [ (u1x2 - u2x1) / (x2 - x1)]

Further rearranging you get,

u = [ (x2 - x) / (x2 - x1) ] * u1 + [ ( x - x1) / (x2 -x1) ] * u2

i.e.

u = N1*u1 + N2*u2

where N1 = [ (x2 - x) / (x2 - x1) ]
N2 = [ ( x - x1) / (x2 -x1) ]
and are called as "Shape Functions."

It simply means that if you know the nodal displacements u1 and u2 over the element at x we can find the intermediate displacement values by interpolation using above Shape Functions.

Shape Functions in Natural Co-ord System

From above information N1 and N2 are:

N1 = [ (x2 - x) / (x2 - x1) ]

N2 = [ ( x - x1) / (x2 -x1) ]

Now if we introduce Natural co-ord system i.e. take i = -1 and j = +1
That means in natural co-ord system ? = -1 and ? = +1
Therefore terms in above two eq of N1 and N2 can be written as:
x1 = -1
x2 = +1
x = ?
Substituting above values in in the eq of N1 and N2 we get,

N1 = ( 1- ? ) / 2 & N2 = ( ? +1 ) / 2


- Thanks,

Amit Deshpande

 

[40818]

Very good......but why?

 

[EdR]

Agree with 40818.....

Also disagree with statement that:

"The accuracy of the solution depends upon the shape function"

Assuming that the shape function has no errors in it you can obtain just as accurate a solution using very different shape functions i.e. linear, parabolic, cubic .... etc.

Looks to me like an exercise from a first FEM class....

Ed.R.

 

[EngAddict]

Looks like my second or third FEA lecture from uni. Think that was around the time of the Rayleigh-Rite formulation and Lagrange Polynomial functions, also the Hermitian polynomial. Isn't is the Lagrange multiplier that is used to solve contact problems in direct solvers?

The problem I has at uni was there was no connection in the teachings between the computer lab classes and the theory. They really should be taught simulataneously with the theory explained then demonstrated on a computer. Means a lot of lab time though.

Back on subject though, what is the point in deriving a shape function if you are not going explain the importance that you stated first. You can look up most shape functions anyway.

 

[amitdesh123]

Thanks to All for your suggestions. Actually no question was raised in the forum regarding a shape function. But when I was searching on the net for the explanation of shape function I did not find any simple explanation.

So I referred some of the text books and made very simple derivation which I guess will be helpful for the beginners. That's why I have posted it.