User:Eml5526.s11.team2/hwk1

=Problem 1.1.1=

Problem Statement
To derive the second order differential equation for elastic bar problem the static model was introduced in the Meeting 5-6 and for the dynamic problem the dynamic model was introduced in Meeting 6-1 in the lectures. Using a free-body diagram apply the appropriate force balance and verify that the following differential equation holds for the dynamic case of a 1-D axially loaded bar as shown below:

Where boundary conditions are

and the initial conditions



Solution (Method I)
Let $$\displaystyle \vec{n}(x) $$ be defined as the unit vector having positive orientation in the direction of $$ \sigma $$ on the element we are considering. This means that the sign convention of our unit vectors will be

Which corresponds to the convention depicted in the free-body diagram shown above. Let us also define $$ \displaystyle \sigma(x) $$ as the force for a given area as,

Upon the rearrangement of terms we arrive at

Classical mechanics allows us to write the sum of the forces as the mass times the acceleration

We will define the mass and acceleration for this system in equation (1.7) as

Using the free-body diagram, let us now consider all of the forces acting on the elastic bar $$\displaystyle f(x), N(x), $$ and $$\displaystyle  N(x+dx) $$. Balancing the forces acting on our finite element results in the following relation,

Next we will substitute (1.5) and (1.6) into (1.10) to get

Recall that a Taylor's Series expansion (See:Taylor Series) about $$ a $$ near $$ x $$ has the following form:

Next we will need to determine a relationship for the displaced area and stress, namely $$\displaystyle \sigma(x+dx)$$ and $$ \displaystyle   A(x+dx)$$. To accomplish this we will performing a Taylor Series expansion choosing $$\displaystyle a=x+dx $$ and neglect all higher order terms.

Taking the product of (1.13) and (1.14) gives us

Substitute (1.15) into (1.11) results in

This can be simplified by canceling the like terms and dividing through by $$ \displaystyle dx $$ as shown below,

to get the following simplified relation,

Next we must apply the product rule in the reverse directions. Meaning we will consider (1.19) the expansion obtained by applying the product rule and recombine terms to get the unexpanded form shown here,

Young's modulus, E, is generally defined as the ratio of the tensile stress to tensile strain:



where

Rewriting the definition of Young's Modulus in (1.21) we obtain a differential relationship in terms of the tensile stress according to Hooke's Law

Substituting (1.22) into our simplified and unexpanded form of our force balance shown in (1.20) we arrive at the solution,

Solution (Method II)
Divide Equation 1.11 in the above method 1 by $$ \displaystyle dx $$.

Take the $$ \lim_{x \to 0} $$ so as to obtain the differential form in space.

Notice that equation 1.27 matches equation 1.20 from Method 1 above. Recall the following definition for stress.

Now insert equation 1.28 into equation 1.27 to obtain the final desired form.

=Problem 1.1.2=

Problem Statement
Discuss the case where the elastic bar has a rectangular cross-section and derive the appropriate PDE for the 1-D axially loaded bar.

Solution

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$$ \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b $$  (2.1)
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multiply both sides by $$dx$$


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b} \right]dx $$     (2.2)
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expanding $$ h(x+dx)$$ with taylor series


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( x \right) + \fracdx \ldots } \right]b} \right]dx $$     (2.3)
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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (2.4)
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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( {x + \frac{2}} \right)} \right]b} \right]dx $$     (2.5)
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now expanding $$ \rho \left( {x + \frac{2}} \right)$$ and $$h\left( {x + \frac{2}} \right)$$ by taylor series


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\left[ {\rho \left( x \right) + \frac\frac{2} \ldots } \right]\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (2.6)
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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \rho \left( x \right)\frac\frac{2} + h\left( x \right)\frac\frac{2} + \frac\frac{2}\frac\frac{2} \ldots } \right]bdx $$     (2.7)
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Neglecting higer order terms and considering first three terms of expansion


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \frac\frac{2}} \right]bdx $$     (2.8)
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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx + \frac\frac{2}} \right]b $$     (2.9)
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Neglecting second term due to presence of $$d{x^2}$$


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx} \right]b $$     (2.10)
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as $$bh\left( x \right) = A\left( x \right)$$


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$$ \displaystyle m\left( {x + \frac{2}} \right)dx = \rho \left( x \right)A\left( x \right)dx $$     (2.11)
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$$ \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right) $$     (2.12)
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It shows that regardless of the cross-section in the differential element RHS of the equation will be same for all distances. The high order terms will vanish all the time and this will lead that only order one terms can stay in the equation.


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$$ \displaystyle m\left( {x + \frac{2}} \right) = m\left( x \right) $$     (2.13)
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Now PDE for the general cross section as proved in 1.1.1


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$$ \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\frac $$     (2.14)
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For given rectangular cross section put $$m = m\left( {x + \frac{2}} \right)$$
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$$ \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\left( {x + \frac{2}} \right)\frac $$     (2.15)
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From equation (2.13) $$m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right)$$


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$$ \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac $$     (2.16)
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Now put $$A\left( x \right) = bh\left( x \right)$$


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$$ \displaystyle \frac{\partial }\left( {E\left( x \right)bh\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)bh\left( x \right)\frac $$     (2.17)
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As $$b$$ is a constant, taking it out of partial differential


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$$ \displaystyle b\frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + f\left( {x,t} \right) = b\rho \left( x \right)h\left( x \right)\frac $$     (2.18)
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Dividing both sides of equation with $$b$$


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$$ \displaystyle \frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + \frac{b} = \rho \left( x \right)h\left( x \right)\frac $$     (2.19)
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This partial differential equation represent a bar with varying rectangular cross section.

=Contributing Members=

= References =