User:EML5526.S11.Team3.vnarayanan/Homework1

Solution 1.2
The Partial Differential Equation (PDE) of an elastic bar subjected to an axial load derived in the section (1.1) is


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|$$\displaystyle \frac{\partial}{\partial x}[E(x) A(x))\frac{\partial u}{\partial x}] + f(x,t) = \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)


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Now, a particular case is taken into consideration, where the bar has a rectangular cross-section. The breadth of the bar is a constant and is equal to 'b'. But the height of the bar varies along the length of the bar and hence it is function of x, which is 'h(x).' (For figure, refer page 6-1 of meeting 6).

Let us consider the inertial force at the center of the thin strip of the elastic bar. Therefore the mass at that location is given by,


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|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b h(x + \frac{dx}{2}) $$ $$
 * $$\displaystyle (Eq. 2)


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|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)] $$ $$
 * $$\displaystyle (Eq. 3)


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Performing the balance of forces similar to part 1.1 and rearranging the terms we get


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|$$\displaystyle \frac{\sigma (x+dx)bh(x+dx)-\sigma (x)bh(x)}{dx}+f(x,t)=\rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)


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Applying the limit as $${\frac{dx}{2}\to0}$$ in the above equation,by the definition of partial derivative

$$ \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

we get,
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|$$\displaystyle \frac{\partial (\sigma (x)bh(x+0))}{\partial x}+f(x,t)=\rho(x + 0) b \frac{1}{2} [h(x) + h(x+0)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 5)


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

Hence the PDE for 1 Dimensional elastic bar is
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|$$\displaystyle \frac{\partial (E(x)bh(x)\frac{\partial u}{\partial x})}{\partial x}+f(x,t)= \rho (x)bh(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 6)


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|$$\displaystyle \frac{\partial (E(x)h(x)\frac{\partial u}{\partial x})}{\partial x}+\frac{1}{b}f(x,t)= \rho (x)h(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 7)


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