User:Fem.tm7/HW

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.

Given


From lecture page 2-3 equation 3 ,the PDE for an elastic bar under dynamic loading conditions is:


 * $$\frac{\partial{}}{\partial{x}}[E(x)A(x)]\frac{\partial{u}}{\partial{x}}+f(x,t)=m\frac{\partial^2{u}}{\partial{t^2}}$$

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.

Solution
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 rod using the free body diagram above, the net force is equal to the internal forces plus the external, or applied, forces.


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$$ \displaystyle \Sigma \text{ Internal Force } + \text{ External Force } = \text{ Net Force}$$ (1.1)
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$$ \displaystyle \Sigma \text{ Internal Force }=-N(x) + N(x+\Delta x)$$ (1.2)
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$$ \displaystyle \text{ Applied Force }=f(x+\frac{\Delta x}{2},t)\Delta x$$ (1.3)
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$$ \displaystyle \text{ Net Force}= m*a =\rho (x)A(x)\frac{\partial^2{u}}{\partial{t^2}}\Delta x$$ (1.4) Note: Acceleration is a partial derivative since the displacement, u, is also dependent on the axial position, x.
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Substituting equations 1.2 through 1.4 back into the force balance equation, 1.1:


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$$ \displaystyle -N(x) + N(x+\Delta x) + f(x+\frac{\Delta x}{2},t)\Delta x =\rho (x)A(x)\frac{\partial^2{u}}{\partial{t^2}}\Delta x$$ (1.5)
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The balance of forces at position x can be expressed as:


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$$ \displaystyle N(x)=\sigma (x)n(x)A(x)=-\sigma (x)A(x)$$ (1.6)
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The corresponding force at x + Δx:


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$$ \displaystyle N(x+\Delta x)=\sigma (x+\Delta x)n(x+\Delta x)A(x+\Delta x)=\sigma (x+\Delta x)A(x\Delta x)$$ (1.7)
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Substituting equations 1.6 and 1.7 into 1.5 yields:


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$$ \displaystyle -\sigma (x)A(x) + \sigma (x+\Delta x)A(x+\Delta x) + f(x+\frac{\Delta x}{2},t)\Delta (x) = \rho (x)A(x)\Delta x\frac{\partial^2{u}}{\partial{t^2}}$$ (1.8)
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Dividing this equation and the internal forces (1.2) through by Δx and recognizing that as Δx approaches zero:


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$$ \displaystyle \lim_{\Delta x \to 0} \frac{N(x+\Delta x)-N(x)}{\Delta x}=\frac{dN}{dx}.$$ (1.9)
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Equation 1.8 can be rewritten as:
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$$ \displaystyle \frac{\partial{\sigma (x)A(x)}}{\partial{x}}+f(x,t)=\rho (x) A(x)\frac{\partial^2{u}}{\partial{t^2}}$$ (1.10)
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The strain-displacement equation is
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$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (1.11) By Hooke ‘ s Law
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (1.12)
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and substituting into equation 1.10 yields:


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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

$$     (1.13)
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Given
Drawing inserted here to show rectangular cross-section and define h(x).\





Solution
The cross section area
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$$ \displaystyle
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A\left( x \right) = b \cdot h\left( x \right)

$$     (1.14) The mass
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$$ \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

$$     (1.15) Take the limit $$\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

$$     (1.16) Substituting Equation 1.14 and 1.16 into Equation 1.13
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$$ \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

$$     (1.17)
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 * }

Authors
Johnathan Whittaker Bullard

Philip Flater

Brandon Hua

Jiang Jin

Braden Snook

Srilalithkumar Swaminathan

Zongyi Yang