User:Eml5526.s11.team3.akj/Homework 1

 Homework 1

= Homework 1 = Formulation of Partial Differential Equation (PDE) for a 1- Dimensional Problem.

Refer to lecture notes 6.1

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Problem Statement
1.1 To find the 1-Dimensional Partial Differential Equation(PDE) of an elastic bar of length L and varying cross-section, subjected to a body force or distributed force f(x,t) along the axis of the bar.

1.2 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.

Solution 1.1
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 $$n\left(x+dx\right)=1$$

For the element of the elastic bar shown, we have $$m\left(x\right)=\rho(x) A(x)dx$$

By using Newton's 2nd law of motion, we can have


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle N(x+dx)-N(x)+f(x)dx= \rho(x) A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)


 * }
 * }

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


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle N\left(x+dx\right)=\sigma(x+dx)A(x+dx)n(x+dx)$$ $$
 * $$\displaystyle (Eq. 2)


 * }
 * }

$$N\left(x\right)$$ 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


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \sigma (x+dx)A(x+dx)-\sigma (x)A(x)+f(x,t)dx=\rho (x)A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 3)


 * }
 * }

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


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \frac{\sigma (x+dx)A(x+dx)-\sigma (x)A(x)}{dx}+f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)


 * }
 * }

Applying the limit as $${dx}{\to0}$$ in the above equation,by the definition of partial derivative, as we know that

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

the first term on the LHS reduces to $$ \frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} $$, the second term on the LHS remains the same as there is no $$d\left(x\right)$$ term. the RHS also remains the same.

Equation (4) reduces to


 * {| style="width:100%" border="0" align="left"

\frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} +f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)


 * }
 * }

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

$$ \frac{\partial\left[E(x)A(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$ $$
 *  $$ \displaystyle
 *  $$ \displaystyle (Eq. 6)

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

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 [[media:fe1.s11.mtg6.djvu|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,

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

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

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

we get,

By Hooke's Law, the stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain $$\frac{\partial u}{\partial x}$$ as

$$\displaystyle \sigma (x)=E(x)\frac{\partial u}{\partial x} $$

Hence the PDE for 1 Dimensional elastic bar is

$$ \frac{\partial \left[E(x)bh(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)bh(x)\frac{\partial^2u }{\partial t^2}$$ $$
 *  $$ \displaystyle (Eq. 6)

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

Reference
1. Newton's 2nd law of motion 2. Partial derivative 3. Hooke's Law

Contributing Authors
EML5526.S11.Team3.risher 21:35, 24 January 2011 (UTC)

Eml5526.s11.team3.akj 16:50, 24 January 2011 (EST)

EML5526.S11.Team3.vnarayanan 16:52, 24 January 2011 (EST)

Eml5526.s11.team3.ushnish12 16:54, 24 January 2011 (EST)

Eml5526.s11.team3.hylon 16:56, 24 January 2011 (EST)

Eml5526.s11.team3.perry 11:57, 26 January 2011 (UTC)

Eml5526.s11.team3.sahin 09:27, 26 January 2011 (EST)