User:Egm5526.s11.team4.zou/HW3

= Problem 3.5 - Equavalent stiffness of a spring=

The following problem is problem 2.2 as taken from Fish and Belytschko's 'A first course in Finite Elements'

Given:
The geometry of bar of thickness $$\displaystyle t$$ with a centered square hole shown in figure 1

Find:
Prove that the equivalent stiffness of a spring aligned in x direction is

Where $$\displaystyle E$$ is the Young's modulus and $$\displaystyle t$$ is the width of the bar.



Solution:
First divided the system into four parts (1,2,3,4)shown as below



Every part can be equivalently viewed as a spring:



Where the equivalent spring stiffness of part1 is:

And equivalent spring stiffness of part4:$$\displaystyle {{K}_{4}}=\frac=\frac{Eat}{(l/10)}=\frac{10Eat}{l}={{K}_{1}}$$

Similarily:

The equivalent spring stiffness of part3 is equals to the equivalent spring stiffness of part3

Since $$\displaystyle {{K}_{2}}$$and $$\displaystyle {{K}_{3}}$$are parallel–connected and$$\displaystyle {{K}_{2}}={{K}_{3}}$$

So the equivalent spring stiffness of this parallelly–connected is$$\displaystyle {{K}_{2-3}}={{K}_{2}}+{{K}_{3}}=2{{K}_{2}}=\frac{5Eat}{l}$$

Then the system can be drawn like the following:



Since $$\displaystyle {{K}_{1}}$$, $$\displaystyle {{K}_{2-3}}$$and $$\displaystyle {{K}_{4}}$$are series–connected and$$\displaystyle {{K}_{1}}={{K}_{4}}$$

So the equivalent spring stiffness of this series–connected is

Substituting $$\displaystyle {{K}_{1}}$$, $$\displaystyle {{K}_{2-3}}$$and$$\displaystyle {{K}_{4}}$$yield:

Problem Statement
Consider a trial (candidate) solution of the form $$\displaystyle u(x)={{\alpha }_{0}}+{{\alpha }_{1}}(x-3)+{{\alpha }_{2}}{{(x-3)}^{2}}$$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1. Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Check the natural boundary condition; is it satisfied?

Problem 3.1 Statement
From problem 3.1 in Fish and Belytschko :

Show that the weak form of

is given by

Acquiring the solution to the weak form
Consider general quadratic trial solution to the weak form,

Since the unknown solution and the weight function should have the same basis functions, we assume the weight function to be,

Plug Eq 10.3 into essential boundary condition, we have

The weak form requires that \omega (3)=0, which means,

Substitute Eq 10.3 ~ Eq 10.6 into the weak form Eq 10.2,

Since $$\displaystyle {{\beta }_{1}}$$ and $$\displaystyle {{\beta }_{2}}$$ are arbitrary, the coefficient of them must be zero, which yields,

Solve Eqs 10.8 we have,

Conbine Eqs 10.9 with Eq 10.5 we have the solution to weak form $$\displaystyle u(x)$$,

where $$\displaystyle I=\int_{1}^{3}{EAdx}$$, $$\displaystyle J=\int_{1}^{3}{(x-3)EAdx}$$, $$\displaystyle K=\int_{1}^{3}{{{(x-3)}^{2}}EAdx}$$ respectively.

Check the validity of the quadratic solution
Looking back to the strong form in Eq 10.1, we know that the quadradic solution will reduce to a constant after twice derivation. On the other hand the properties of $$\displaystyle E$$ and $$\displaystyle A$$ are unknown to us. Thus it is unlikely that the equality in Eq 10.1 should be satisfied. Hence we can almost conclude that any quadratic solution to this problem will fail to satisfy the strong form.

Moreover, since the weak form has absolve the natural boundary condition into one equation, the "error" produced by the solution can affect the equality of the natural boundary condition. Thus we have this possibility that the natural boundary condition is NOT satisfied.

We can illustrate this point by assuming $$\displaystyle E$$ and $$\displaystyle A$$ to be constants. In this case, $$\displaystyle I=2AE$$, $$\displaystyle J=-2AE$$, $$\displaystyle K=\frac{8}{3}AE$$, substitute them into Eq 10.10 we have,

Thus, natural B.C. becomes:

It is obvious that the trial solution don't satisfy the natural boundary condition. It is observed that the trial solution is only valid when the area A is infinite.

Secondly, adopting the trial solution, strong form becomes:

It is obvious that the trial solution don't exactly satisfy strong form.

And the strong form is valid only at $$\displaystyle x=2$$.

Discussion
We note that the Example3.3 in the textbook also used quadratic solution to solve a similar strong form(thus similar weak form). Though the quadratic solution fails to satisfy the strong form, it does satisfy the natural boundary condition(see figure 3.7(b) in the book). However, a mistake in the book is found: The third equation in page 57 is not valid, the term $$\displaystyle \int_{0}^{2}{({{\beta }_{1}}x+{{\beta }_{2}}{{x}^{2}})}10dx$$ should be $$\displaystyle \int_{0}^{2}{({{\beta }_{1}}x+{{\beta }_{2}}{{x}^{2}})}10xdx$$. Since the following conclusion in the book is based on that equation, it's reasonable to doubt the fact that their quadratic solution doesn't violate the natural b.c.