User:Eml5526.s11.team7.snook/HW4

=Problem 4.6: Elastic Bar With a Variable Distributed Spring: F&B 3.9=

Problem Description
An elastic bar is subjected along its length to a variable distributed spring p(x) as seen in the figure below. An axial force proportional to the displacement is imposed by the spring on the bar. Adapted from "A First Course in Finite Elements" by Fish and Belytschko p. 74 pb.3.9

a. Construct the strong form.

b. Construct the weak form.



Given
This bar is of length l, cross-sectional area A(x), Young's modulus E(x) with body force b(x) and with boundary conditions:


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$$\displaystyle u(0)=0$$
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$$\displaystyle E(l)A(l)\frac{du(l)}{dx}=t$$
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As seen in the figure above.

Solution
Assumptions:
 * The bar must be in equilibrium
 * It satisfies Hooke's law
 * Compatible Displacement Field
 * The Strain-Displacement Equation must be Satisfied

a. Construct the Strong Form
Following a similar solution method as Team 2 used here with Eq. 1.1-1.10. Using the following diagram from Team 7 homework 1 to start creating the strong form:



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. Note, in this specific problem we will designate our body force as b(x+dx/2) (instead of f(x+dx/2 as seen in the diagram from Team 7) and have the additional body force in the opposite direction caused by the springs (as seen by the figure in the problem description) designated as p(x+dx/2)


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$$ \displaystyle \Sigma \text{ Internal Force } + \text{ Applied Force } = \text{ Net Force}$$ (4.6.1)
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$$ \displaystyle \Sigma \text{ Internal Force }=N(x) + N(x+\Delta x)$$ (4.6.2)
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$$ \displaystyle \text{ Applied Force }=b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx$$ (4.6.3)
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Since we are looking at the static case:


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$$ \displaystyle \text{ Net Force}= m*a =0$$ (4.6.4)
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Substituting equations 4.6.2-4.6.4 into the force balance equation, 4.6.1:


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$$ \displaystyle N(x) + N(x+dx) + b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx =0$$ (4.6.5)
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Plugging in for the direction of the surface normal, the balance of forces at position x can be expressed as:


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$$ \displaystyle N(x)=n(x)\sigma (x)A(x)=-\sigma (x)A(x)$$ (4.6.6)
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Also plugging in for the surface normal, the corresponding force at x + dx:


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$$ \displaystyle N(x+dx)=n(x+dx)\sigma (x+dx)A(x+dx)=\sigma (x+dx)A(x+dx)$$ (4.6.7)
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Using a Taylor Series:


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$$ \displaystyle \sigma (x+dx)=\sigma (x)+\frac{d\sigma (x)}{dx}dx+...$$ (4.6.8)
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$$ \displaystyle A(x+dx)=A(x)+\frac{dA(x)}{dx}dx+...$$ (4.6.9)
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Additionally, in Eq. 4.6.3 the body force $$b(x+\frac{dx}{2})$$ has been averaged for the differential volume we are considering to formulate the governing differential equation. The same has been done for the force caused by the variable distributed spring, given as $$p(x+\frac{dx}{2})$$. Taylor expanding for these as well:


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$$\displaystyle b(x+\frac{dx}{2})=b(x)+\frac{db(x)}{dx}\frac{dx}{2}....$$. (4.6.10)
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$$\displaystyle p(x+\frac{dx}{2})=p(x)+\frac{dp(x)}{dx}\frac{dx}{2}....$$. (4.6.11)
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Plugging in Eq.'s 4.6.8 and 4.6.9 into Eq. 4.6.7, and then plugging this and Eq.'s 4.6.6, 4.6.10 and 4.6.11 into Eq. 4.6.5 gives:
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$$\displaystyle \sigma (x)A(x) + \sigma (x)\frac{dA(x)}{dx}dx+A(x)\frac{d\sigma (x)}{dx}-\sigma (x)A(x)+b(x)dx-p(x)dx$$. (4.6.12)
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Using the product rule, canceling out terms, and dividing through by dx gives:
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$$\displaystyle \frac{d}{dx}\sigma (x)A(x)+b(x)-p(x)=0$$. (4.6.13)
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Noting that the strain-displacement equation is
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$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (4.6.14) And by Hooke ‘ s Law
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (4.6.15)
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Gives the following strong form:
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$$\displaystyle \frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)=0$$. (4.6.16)
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b. Construct the Weak Form
Following the same solutions as Team 7's previous work here.

In order to reduce the requirements or constraints on what trial solutions could be substituted in for u, we will generate the weak form which requires the basis solutions only be once differentiable. This occurs from the natural boundary condition being absorbed in the weak form, as will be seen shortly. Also, this creates a symmetric stiffness matrix.

As with Team 7's solution, the strong form and natural boundary condition are used in an inner product with an arbitrary weighting function giving the following equations:


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$$\displaystyle \int_{0}^{l}w(x)\big [\frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)\big ]dx=0$$. (4.6.17)
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$$\displaystyle w(l)E(l)A(l)\frac{du(l)}{dx}-w(l)t=0$$. (4.6.18)
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Distributing the weighting function w(x) and the integral of Eq. 4.6.17 and looking at the first term:


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$$\displaystyle \int_{0}^{l}w\frac{d}{dx}E(x)A(x)\frac{du}{dx}dx=w(x)E(x)A(x)|_{0}^{l}-int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.19)
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In Eq. 4.6.19 integration by parts (derived and example of use here) was used. Now to remove the unknown flux at the essential boundary location (x=0), we will make the arbitrary weighting function be zero at x equal to zero, or


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$$\displaystyle w(0)=0$$. (4.6.20) Using Eq. 4.6.7 and 4.6.9, Eq. 4.6.8 becomes:
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$$\displaystyle w(x)E(x)A(x)|_{0}^{l}-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx=w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.21)
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Combining Equations 4.6.18 and 4.6.21 gives the following weak form:


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$$\displaystyle w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx+\int_{0}^{l}w(x)p(x)dx+\int_{0}^{l}w(x)b(x)dx$$. (4.6.22)
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Note in equation 4.6.22 the weighting function must be zero at the essential boundary condition (w(0)=0). Also, the weighting function and the displacement are both first order derivatives and now a symmetric stiffness matrix can be formed.

Author
Braden Snook