User:Eml5526.s11.team7.jin/HW1

=Problem 1.1=

Problem Description

 * Part 1: Derive the one-dimensional partial differential equation (PDE) for an elastic bar under dynamic loading conditions. (Refer to equation 3 of lecture page 2-3.)


 * Part 2: Discuss the case in which the bar has a rectangular cross-section.

Part 1


From lecture page 2-3 equation 3 ,the PDE for an elastic bar under dynamic loading conditions is:
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$$ \displaystyle
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\frac{\partial }\left[ {E\left( x \right)A\left( x \right)\frac} \right] + f\left( {x,t} \right) = m\left( x \right)\frac

$$     (Eq 1.1) Since the elastic bar is loaded dynamically, the applied force, f, is a function of distance, x , as well as time, t. This is shown in the free body diagram to the right. Details of the remaining forces will be discussed in the solution below.
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Part 2
The cross section is shown in the following figure

Part 1
Assumptions :
 * The rod is in stress equilibrium.
 * The rod satisfies Hooke's Law, $$\sigma (x) = E(x)\epsilon (x) $$.
 * Compatible displacement field.
 * The rod obeys the strain displacement equation.

Following Newton's Second Law:


 * $$\displaystyle F=m*a $$

Where:


 * F=Force


 * m=Mass


 * a=Acceleration

Additionally,


 * $$\displaystyle m=\rho * V$$

Where $$\displaystyle\rho = $$ density and V=Volume.

and,


 * $$a=\frac{d^2u}{dt^2}$$

Where u is displacement and t is time.

Performing a force balance on the body in the x-axis direction
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$$ \displaystyle
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N\left( {x + \Delta x} \right) + N\left( x \right) + f\left( {x + \frac{2},t} \right)\Delta x = \rho \left( x \right)A\left( x \right)\Delta x\frac

$$     (Eq 1.2) Where
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$$ \displaystyle
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N\left( x \right) = \sigma \left( x \right)A\left( x \right)n\left( x \right) = - \sigma \left( x \right)A\left( x \right)

$$     (Eq 1.3) And
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$$ \displaystyle
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N\left( {x + \Delta x} \right) = \sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right)n\left( {x + \Delta x} \right) = \sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right)

$$     (Eq 1.4) Plugging in Equation 1.2
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$$ \displaystyle
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\sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right) - \sigma \left( x \right)A\left( x \right) + f\left( {x + \frac{2},t} \right)\Delta x = \rho \left( x \right)A\left( x \right)\Delta x\frac

$$     (Eq 1.5) Dividing by $$\Delta x$$ on each side
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$$ \displaystyle
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\frac + f\left( {x + \frac{2},t} \right) = \rho \left( x \right)A\left( x \right)\frac

$$     (Eq 1.6) If we take the limit as $$\Delta x \to 0$$
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$$ \displaystyle
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\frac{\partial }\left[ {\sigma \left( x \right)A\left( x \right)} \right] + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac

$$     (Eq 1.7) The strain-displacement equation is
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$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (Eq 1.8) By Hooke ‘ s Law
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (Eq 1.9) Substituting Equation 1.9 into Equation 1.7
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$$ \displaystyle
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\frac{\partial }\left[ {E\left( x \right)A\left( x \right)\frac} \right] + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac

$$     (Eq 1.10) That is
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$$ \displaystyle
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\frac{\partial }\left[ {E\left( x \right)A\left( x \right)\frac} \right] + f\left( {x,t} \right) = m\left( x \right)\frac

$$     (Eq 1.11) Where
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$$ \displaystyle
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m\left( x \right) = \rho \left( x \right)A\left( x \right)

$$     (Eq 1.12)
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Part 2
The cross section area
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$$ \displaystyle
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A\left( x \right) = b \cdot h\left( x \right)

$$     (Eq 1.13) The mass
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$$ \displaystyle
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m\left( {x + \frac{2}} \right) = \rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + \Delta x} \right)} \right]b

$$     (Eq 1.14) Take the limit $$\Delta x \to 0$$
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$$ \displaystyle
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m\left( x \right) = \rho \left( x \right)h\left( x \right)b

$$     (Eq 1.15) Substituting Equation 1.13 and 1.14 into Equation 1.10
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$$ \displaystyle
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\frac{\partial }\left[ {E\left( x \right)h\left( x \right)b\frac} \right] + f\left( {x,t} \right) = \rho \left( x \right)h\left( x \right)b\frac

$$     (Eq 1.16)
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Author
Braden Snook, Jiang Jin, Bullard Johnathan, Philip Flater, Brandon Hua, Zongyi Yang, Srilalithkumar Swaminathan