User:Eml5526.s11.team3.sahin/Homework 3

=Problem3.9 Linear Trial Solution of Weak Form =

Problem Statement
Refer to lecture notes 17.2

This problem is extracted Problem 3.3 from the textbook of Fish J., and Belytschko T.,

Given
Problem 3.1 given in Fish and Belytschko

The strong form is

The weak form is

Find
Consider a trial (candidate) solution of the form $$u\left(x\right)=\alpha_0+ \alpha_1(x-3)$$ and a weight function of the same form.

Part 1 Obtain a solution to the weak form in Problem 3.1.

Part 2 Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Part 3 Check the natural boundary condition; is it satisfied?

Part 1
The trial solutions $$u\left(x\right)$$ must satisfy the essential boundary condition $$u\left(x=3\right)= 0.001$$ so Therefore, it is weight function $$w\left(x=3\right)= 0$$ as the weight function must vanish on the essential boundaries

To find a trial solution, we need to put Eq 9.3 and Eq 9.4 into the weak form Eq 9.2

Evaluating the integrals with the assumption of cross area $$ \begin{align} & \qquad A \end{align} $$ and Young’s modulus $$ \begin{align} & \qquad E \end{align} $$ are constants

and factoring out $$ \begin{align} & \qquad \beta_1 \end{align} $$ gives

As it must hold for all $$ \begin{align} & \qquad \beta_1 \end{align} $$, the term in the parentheses must vanish, so After arranging the equation above, we obtain $$ \begin{align} & \qquad \alpha_1 \end{align} $$ Substituting this result into Eq 9.3 gives linear trial solution of the weak solution,

Part 2
We know that To check the equilibrium equation in the strong form in Problem 3.1, we need to substitute Eq 9.11 into Eq 9.1a

The equilibrium equation in the strong form is not equal. Therefore, it is not satisified.

Part 3
We know that the natural boundary condition is

To check the natural boundary condition; we need to substitute Eq 9.11 into Eq 9.1b

The natural boundary condition is not equal. Therefore, it is not satisified.