User:Eml5526.s11.team6.tupsakhare/hwk1

=Elastic bar PDE, Dynamic case=

Find
Derive the following equation by performing a balance of forces:

$$ \frac{\partial}{\partial x}\left [ A(x)E(x)\frac{\partial u}{\partial x} \right ]+f(x,t)=A(x) \rho(x)\cdot \frac{\partial ^{2}u}{\partial t^{2}}\ $$

Given
An infinitesimally small section of length dx of the 1-D elastic bar with varying material properties at a distance x from the origin

A(x) - Varying cross section of the elastic bar

E(x) - Modulus of elasticity

$$\rho(x) $$ - Mass density

u - Displacement constrained only to x- axis (1-D bar)

t - Time




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Where, $$\; {A}'=A(x) (dx)\cdot \rho (x)\cdot\frac{\partial^{2}u }{\partial t^{2}} =mass\cdot acceleration=Inertia\, Force$$
 * $$\displaystyle (Eq. 1.1) $$
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Solution
As per Newton's law of motion, the involved forces should contribute to inertia force.

Force equilibrium in the x-direction yields:

$$\sum F_{x}=0 $$


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$$\sum F_{x}=-N(x,t)+N(x+dx,t)+f(x,t) dx-m(x)\cdot \frac{\partial ^{2}u}{\partial t^{2}}=0$$
 * $$\displaystyle (Eq. 1.2) $$
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Where, $$\;\; m(x)=A(x)(dx)\rho (x)$$
 * $$\displaystyle (Eq. 1.3) $$
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Including Taylor series expansion for $$N(x+dx,t)$$ and neglecting the higher order terms (h.o.t) yields:

$$ \cancel {-N(x,t)}+[\cancel {N(x,t)}+\frac {\partial }{\partial x}(N(x,t))+(h.o.t)]+f(x,t)=\frac{m(x)}{dx}\cdot\frac{\partial ^{2} u}{\partial t^{2}}$$


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$$\frac{\partial }{\partial x}(N(x,t))+f(x,t)=\frac{m(x)}{dx}\cdot\frac{\partial ^{2} u}{\partial t^{2}}$$
 * $$\displaystyle (Eq. 1.4) $$
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$$Also,\; N(x,t)=A(x) \cdot \sigma (x)=A(x) \cdot [E(x)\cdot \epsilon (x,t)] $$
 * $$\displaystyle (Eq. 1.5) $$
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$$\;\;\;\;\;\;\;\;\;\;\;\Rightarrow N(x,t)= A(x)\cdot E(x) \cdot\frac{\partial u}{\partial x}(x,t)$$
 * $$\displaystyle (Eq. 1.6) $$
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Thus proving:


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 * $$\displaystyle (Eq. 1.7) $$
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= Modified Case=

Find
Discuss the case in which the bar has a rectangular cross section shown below.

Solution
We have from reference of class notes for mtg 2, eq(3) pg.2-3,


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$$m=\rho (x) \cdot A(x) $$
 * $$\displaystyle (Eq. 2.1) $$
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Also with reference to class notes for mtg 6, pg 6-1, point 2), for A(x)=bˑh(h),


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$$m=\rho \left ( x+\frac{dx}{2} \right )\cdot \frac{1}{2}\left [ h(x)+h(x+dx) \right ]b $$
 * $$\displaystyle (Eq. 2.2) $$
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Substituting Eq 2.1 and Eq 2.2 into Eq 1.7 for case of A(x)=bˑh(x),
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$$\frac{\partial}{\partial x}\left [ \left ( b\cdot h(x)\cdot E(x)\cdot \frac{\partial u}{\partial x}\right ) \right ]+f(x,t)=\frac{1}{2}\left [ h(x)+h(x+dx) \right ]\rho (x+dx)\cdot \frac{\partial^{2} u}{\partial t^{2}}\cdot b$$
 * $$\displaystyle (Eq. 2.3) $$
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Note:

1) $$\left [ \frac{1}{2}\left [ h(x)+h(x+dx) \right ]\rho (x+dx) \right ]$$ denotes mass per unit length at a distance of (x+dx/2), which is the middle of the elementary section considered in Figure 1.2.

2) $$\frac {1}{2}\left [h(x)+h(x+dx) \right ]$$ denotes average height of elementary section of figure 1.2.

3) Since we are considering an infinitesimally small section (dx) of the entire bar, we can write,

$$h(x)\approx h(x+dx)$$

$$\Rightarrow \frac{1}{2}\left [ h(x)+h(x+dx) \right ]\approx h(x)$$

4) Also, b is the width of the cross section and is constant and ρ(x+dx) is approximately equal to ρ(x) for a small section as in 3) above.

Eq 2.3 converts to the following:


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 * $$\displaystyle (Eq. 2.4) $$
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Eq 2.4 is the modified form of Eq 1.7 for the specific case of A(x)=bˑh(x).