User:Fem.tm7/HW1

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 of area $$

\displaystyle

A\left( x \right) = b \cdot h\left( x \right)

$$.

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 (FBD) to the right. Details of the remaining forces will be discussed in the solution below. The dotted line in the figure represents the centerline, or axis, of the elastic bar.

Solution
Assumptions :
 * 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.


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

$$ \displaystyle \Sigma \text{ Internal Force } + \text{ Applied Force } = \text{ Net Force}$$ (1.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \Sigma \text{ Internal Force }=-N(x) + N(x+\Delta x)$$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \text{ Applied Force }=f(x+\frac{\Delta x}{2},t)\Delta x$$ (1.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \text{ Net Force}= m*a =\rho (x)A(x)\frac{\partial^2{u}}{\partial{t^2}}\Delta x$$ (1.4) Note: Acceleration is a partial derivative since the displacement, u, is also dependent on the axial position, x.
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Substituting equations 1.2 through 1.4 back into the force balance equation, 1.1:


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

$$ \displaystyle -N(x) + N(x+\Delta x) + f(x+\frac{\Delta x}{2},t)\Delta x =\rho (x)A(x)\frac{\partial^2{u}}{\partial{t^2}}\Delta x$$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The balance of forces at position x can be expressed as:


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

$$ \displaystyle -N(x)=\sigma (x)n(x)A(x)=-\sigma (x)A(x)$$ (1.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The corresponding force at x + Δx:


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

$$ \displaystyle N(x+\Delta x)=\sigma (x+\Delta x)n(x+\Delta x)A(x+\Delta x)=\sigma (x+\Delta x)A(x\Delta x)$$ (1.7)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Substituting equations 1.6 and 1.7 into 1.5 yields:


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

$$ \displaystyle -\sigma (x)A(x) + \sigma (x+\Delta x)A(x+\Delta x) + f(x+\frac{\Delta x}{2},t)\Delta (x) = \rho (x)A(x)\Delta x\frac{\partial^2{u}}{\partial{t^2}}$$ (1.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Dividing this equation and the internal forces (1.2) through by Δx and recognizing that as Δx approaches zero:


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

$$ \displaystyle \lim_{\Delta x \to 0} \frac{N(x+\Delta x)-N(x)}{\Delta x}=\frac{dN}{dx}.$$ (1.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Equation 1.8 can be rewritten as:


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

$$ \displaystyle \frac{\partial{\sigma (x)A(x)}}{\partial{x}}+f(x,t)=\rho (x) A(x)\frac{\partial^2{u}}{\partial{t^2}}$$ (1.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The strain-displacement equation is
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\varepsilon \left( x \right) = \frac

$$     (1.11) By Hooke ‘ s Law
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (1.12)
 * 
 * }

and substituting into equation 1.10 yields:


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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\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)
 * 
 * }

Given
The width of the bar (spatial dimension into the page of the free body diagram) is defined as b (see bar cross-section in below right figure). So rather than the area being an arbitrary value A(x) at position 'x' or A(x+Δx) at position 'x+Δx' as in the first part of the problem, the area can be defined more explicitly as
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

A\left( x \right) = b \cdot h\left( x \right)

$$     (1.14)
 * 
 * }





Solution
The mass
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

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$$
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\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)
 * <p style="text-align:right">
 * }

Authors
Johnathan Whittaker Bullard

Philip Flater

Brandon Hua

Jiang Jin

Braden Snook

Srilalithkumar Swaminathan

Zongyi Yang