User:Eml5526.s11.team3.ushnish12/Homework 3

The strong form of the governing differential equation is:-
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$$\displaystyle
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\frac{d}{dx}(AE\frac{du}{dx})+2x=0 ; 1
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The boundary conditions are given by the equations:-
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$$\displaystyle
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\sigma (1)={{(E\frac{du}{dx})}_{x=1}}=0.1

$$ (Eq )<2>
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$$\displaystyle
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u(3) = 0.001

$$ (Eq )<3>
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To form the weak form, we first consider the strong form and multiply the governing equation Eq(2) by an arbitrary function $$\displaystyle w(x)$$ and integrating over the domain on which they hold.

The function $$\displaystyle w(x)$$ is called the weight function.

For the governing equation, the pertinent domain is the interval [1,3].

The resulting equation after multiplying the governing equation with $$\displaystyle w(x)$$ takes the form -


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$$\displaystyle
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\int\limits_{1}^{3}{w[\frac{d}{dx}(AE\frac{du}{dx})+2x]}\text{dx} = 0 ; \forall w

$$ (Eq )<4>
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In the above, $$\forall w$$ denotes that $$\displaystyle w(x)$$ is an arbitrary function which holds for all functions $$\displaystyle w(x)$$. Furthermore, it is convenient to consider that the weight function to satisfy the condition


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$$\displaystyle
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w(3)=0

$$ (Eq )<5>
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For convenience, we write Eq(4) into the equivalent form :


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$$\displaystyle
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\int\limits_{1}^{3}{w[\frac{d}{dx}(AE\frac{du}{dx})]\text{dx}+\int\limits_{1}^{3}{2wx\text{dx}=0}}

$$ (Eq )<6>
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To the solve the first term of the above integral we use Integration by parts method:

The stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain$$(\frac{du}{dx})$$ as


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$$\displaystyle
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\sigma (x)=E(x)\frac{du}{dx}

$$ (Eq )<9>
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Using Eq(9), Eq(8) is written as


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$$\displaystyle
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{{(wA\sigma )}_{x=3}}-{{(wa\sigma )}_{x=1}}- \int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} +\int\limits_{1}^{3}{2wx\text{dx}=0}

$$ (Eq )<10>
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Now   $${{(w)}_{x=3}}=0\quad and\quad{{(\sigma )}_{x=1}}=0.1$$ as defined earlier.

Applying these conditions to Eq(10), we obtain


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$$\displaystyle
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-0.1{{(wA)}_{x=1}} - \int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} +\int\limits_{1}^{3}{2wx\text{dx}=0} $$ (Eq )<11>
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Rearranging the terms in Eq(11) we obtain the strong form which is given by:-


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$$\displaystyle
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\int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} = -0.1{{(wA)}_{x=1}}+\int\limits_{1}^{3}{2wx\text{dx}=0} \quad \forall w\quad\text{with}\quad w(3)=0

$$
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