User:Eml5526.s11.team01/Hwk1

=Problem 1:= Obtain the dynamic partial differential equation of an elastic bar by doing balance of forces From Lecture Slide.

Given
An elastic bar of arbitrary cross section $$A\left( x \right)$$ with Young's modulus $$E\left( x \right)$$ under dynamic conditions, restricted to move in longitudinal $$x$$ direction only.

Find
The governing partial differential equation of the elastic bar under dynamic conditions.

Solution
In this document is deduced the differential equation of a bar made of linear elastic material, subject to small strain under dynamic conditions. Only strain in the longitudinal direction of the bar is considered.

Fig. 1.1 shows the bar which, in general, has variable cross-section $$A\left( x \right)$$. $$u\left( x,t \right)$$ is the displacement in the longitudinal direction. Fig. 1.2 shows the free body diagram of a small piece of the bar of length $$dx$$. On that fig. the stresses $$\sigma $$ acting over each face of the element are shown.






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$$ -A\sigma\ + \left(\sigma\ + \frac{\partial \sigma}{\partial x}dx\right)\left(A + \frac{\partial A}{\partial x}dx\right) +fdx = dxA\rho\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 1.1) $$
 * }
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In the latter equation, $$\rho $$ is the density (mass per unit of volume). Doing the indicated product we have:


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$$ -A\sigma\ + A\sigma\ + A\frac{\partial \sigma}{\partial x}dx +\sigma\frac{\partial A}{\partial x}dx + \cancelto{0}\frac{\partial \sigma}{\partial x}\frac{\partial A}{\partial x}dxdx + fdx = dxA\rho\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 1.2) $$
 * }
 * }

Neglecting the second order factor, and dividing by $$dx$$ we have:


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$$ A\frac{\partial \sigma}{\partial x} +\sigma\frac{\partial A}{\partial x} + f = A\rho\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 1.3) $$
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Where, the product $$\rho A$$ has been replaced by $$m$$. The first two terms in the last equation are the derivative of a product, therefore:


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$$ \frac{\partial A\sigma}{\partial x} + f = m\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 1.4) $$
 * }
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Using the Hooke's law to replace the stress as function of the infinitesimal strain $$\varepsilon $$ we have:


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$$ \frac{\partial AE\epsilon}{\partial x} + f = m\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 1.5) $$
 * }
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The strain-displacement equation is obtained by applying the engineering definition of strain for an infinitesimal segment of the bar. The elongation of the segment is given by $$u\left( x + \Delta x \right) - u\left( x \right)$$ and the original length is $$\Delta x$$, therefore, the strain is given by:


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$$\epsilon\left( x \right) = \frac{elongation}{original\ length} = \frac{u\left( x + \Delta x \right) - u\left( x \right)}{\Delta x}$$
 * $$\displaystyle (Eq. 1.6) $$
 * }
 * }

Taking the limit of the above as $$\Delta x \to 0$$, we recognize that the right-hand side is the deriative of u(x). Therefore, the strain-displacement equation is:


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$$\epsilon\left( x \right) = \frac{\partial u}{\partial x}$$
 * $$\displaystyle (Eq. 1.7) $$
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Replacing the infinitesimal deformation in terms of the displacement, we have:


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$$ \frac{\partial AE\frac{\partial u}{\partial x}}{\partial x} + f = m\frac{\partial^2 u}{\partial t^2}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.8) $$
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=Problem 2:= Discuss the case in which the bar from problem 1 has rectangular cross section. From Lecture Slide.

Given
The crossection of the bar is given by:
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$$ A = bh\left(x\right)$$
 * $$\displaystyle (Eq. 2.1) $$
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Find
Discuss the case when the bar has constant width and variable height

Solution
Going back to Eq. 1.3 we can replace the area by using Eq. 2.1 to obtain:


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$$ \frac{\partial \left(bh\left(x\right)\sigma\right)}{\partial x} + f = \frac{1}{2}\left(h\left(x\right)+h\left(x + dx\right)\right)\rho\left(x + dx\right)\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 2.2) $$
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 * }

Where the factor $$\frac{1}{2}\left(h\left(x\right)+h\left(x + dx\right)\right)\rho\left(x + dx\right)$$ has been used to approximate the mass per unit length in the middle of the segment $$\left(x + \frac{dx}{2}\right)$$; the factor $$\frac{1}{2}\left(h\left(x\right)+h\left(x + dx\right)\right)$$ is the average height between the beginning and the end of the segment of length dx (Fig. 2) which is being used as an approximation of the height at the middle of the segment: $$h\left(x + \frac{dx}{2}\right)$$. But as the segment is small this height can be further approximated to the height at the beginning of the segment $$h\left(x\right)$$. Doing that approximation and considering that $$b$$ is constant and can be taken out of the derivative, Eq. 2.2 becomes:


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$$b \frac{\partial \left(h\left(x\right)\sigma\right)}{\partial x} + f = h\rho\frac{\partial^2 u}{\partial t^2}$$
 * $$\displaystyle (Eq. 2.3) $$
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Finally, replacing the stress $$\sigma$$ as function of strain by using Hooke's Law, and the strain as a function of the displacement by using Eq. 1.6 and Eq. 1.7 we have:


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$$b \frac{\partial hE\frac{\partial u}{\partial x}}{\partial x} + f = h\rho\frac{\partial^2 u}{\partial t^2}$$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\,(Eq. 2.4) $$
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=Contributing Team Members=
 * James Roark
 * Harrison Sheffield
 * Abhijeet Bhalerao
 * Cedric Adam
 * Fernando Casanova
 * Yu Hao