User talk:Eml5526.s11.team3

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HW 1
 Formulation of Partial Differential Equation (PDE) for a 1- Dimensional Problem 

--- Problem Statement

A bar is having a trapezoidal cross-section where the height and width is varying with distance along the x-axis.

a. Do the balancing of the forces acting on the cross-section and formulate the 1-D PDE for the problem.

b. Now if the bar has a rectangular cross-section where the breadth is constant, perform the balance of forces to formulate the 1-D PDE   for this case.

Use the same mass per unit length in the inertia force. Refer to meeting 6.1 for the figure.

Solution

a.

Consider an element of length $$d\left(x\right)$$ of the elastic bar, the free body diagram of the bar is shown as below.

In the above figure $$A\left(x\right)$$, $$ \sigma\left (x\right)$$ and  $$n\left(x\right)$$ represent the area, stress and normal to the plane on the left cross-section of the bar and

$$A\left(x+dx\right)$$, $$\sigma\left(x+dx\right)$$ and   $$n\left(x+dx\right)$$ represent the area, stress and normal to the plane on the right cross-section of the bar.

for the left cross-section $$n\left(x\right)=-1$$ and for the right cross-section $$n\left(x+dx\right)=1$$

by using newtons law of motion we can write    $$\sum(Forces)=mass*acceleration$$

by using newtons law for the element of the elastic bar shown we can write

$$ N(x+dx)-N(x)+f(x)dx= \rho A(x)dx\frac{\partial^2u }{\partial t^2} $$

where; $$N\left(x+dx\right)$$ is the net force due to the stress $$\sigma\left(x+dx\right)$$ which is given by

$$N\left(x+dx\right)=\sigma(x+dx)A(x+dx)n(x+dx)$$

and

$$N\left(x\right)$$ is the net force due to the stress $$\sigma\left(x\right)$$ which is given by $$N\left(x\right)=\sigma(x)A(x)n(x)$$

substituting for $$N\left(x+dx\right)$$,$$N\left(x\right)$$,$$n\left(x+dx\right)$$,$$n\left(x\right)$$, we get

$$ \sigma (x+dx)A(x+dx)-\sigma (x)A(x)+f(x,t)dx=\rho (x)A(x)dx\frac{\partial^2u }{\partial t^2} $$

dividing the above equation by $$d\left(x\right)$$, we get

$$ \frac{\sigma (x+dx)A(x+dx)-\sigma (x)A(x)}{dx}+f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2}.......(1) $$

Now if we take the limit of the above expression as $${dx\to0}$$, the first term on the left hand side reduces to $$ \frac{\partial (\sigma (x)A(x))}{\partial x} $$as we know that

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

the second term on the left hand side remains the same as there is no $$d\left(x\right)$$ term. the right hand side also remains the same.

equation (1) reduces to $$ \frac{\partial (\sigma (x)A(x))}{\partial x} +f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$

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 $$\frac{\partial (E(x)A(x)\frac{\partial u}{\partial x})}{\partial x}+f(x,t)= \rho (x)A(x)\frac{\partial^2u }{\partial t^2}$$