User:Egm5526.s11.team2.sandhu

hw 5
User:Egm5526.s11.team2.sandhu/newpage =HW 4=

= Problem 4.3: Construct weak form of heat problem and find solution =

Given: Strong Form and Boundary Conditions
Statement:- Given the strong form for the heat conduction problem in a circular plate:
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$$  \displaystyle k\frac{d}\left( {r\frac} \right) + rs = 0,0 < r \leqslant R $$ (4.3.1) natural boundary condition :$$ \frac\left( {r = 0} \right) = 0$$
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essential boundary condition :$$T\left( {r = R} \right) = 0$$

where R is the total radius of the plate, s is the heat source per unit length along the plate radius, T is the temperature and k is the conductivity. Assume that k, s and R are given:

a. Weak Form
Construct the weak form for the above strong form.

b. Solution using quadratic polynomial trial function
Use quadratic trial (candidate) solutions of the form $$T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$ and weight functions of the same form to obtain a solution of the weak form.

c. Solve Differential equation
Solve the differential equation with the boundary conditions and showthat the temperature distribution along the radius is given by:


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$$  \displaystyle T = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.2)
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a. Weak Form Derivation
Strong Form Can be written as:


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$$  \displaystyle p\left( T \right) = 0 $$
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$$  \displaystyle p\left( T \right) = k\frac{d}\left( {r\frac} \right) + rs $$
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$$w\left( r \right)$$ is weight function Weighted Residual Form can be written as


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$$  \displaystyle \int\limits_0^R {w\left( r \right)p\left( T \right)dr = 0} $$     (4.3.3)
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$$  \displaystyle \int\limits_0^R {w\left( r \right)\left[ {k\frac{d}\left( {r\frac} \right) + rs} \right]dr} = 0 $$
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$$  \displaystyle \int\limits_0^R {w\left( r \right)k\frac{d}\left( {r\frac} \right)dr + } \int\limits_0^R {w\left( r \right)rsdr} = 0 $$
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Write $$w\left( r \right) = w$$ Integrating the first term by parts.


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$$  \displaystyle \left[ {wkr\frac} \right]_0^R - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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Assuming $$w\left( R \right) = 0$$ and using natural boundary condition $$\frac\left( {r = 0} \right) = 0$$


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$$  \displaystyle 0 - 0 - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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$$  \displaystyle \int\limits_0^R {\left( {\frackr\frac} \right)} dr = \int\limits_0^R {wrsdr} $$     (4.3.4)
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Writing in standard form:


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b. Solution using quadratic polynomial trial function
let $$T$$ and $$w$$ are of the following
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$$  \displaystyle T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2} $$ and $$ \displaystyle w = {\beta _0} + {\beta _1}r + {\beta _2}{r^2} $$     (4.3.6)
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Now using essential Boundary Condition i.e $$T\left( R \right) = 0$$


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$$  \displaystyle {\alpha _0} + {\alpha _1}R + {\alpha _2}{R^2} = 0 $$     (4.3.7)
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$$  \displaystyle {\alpha _0} = - \left( {{\alpha _1}R + {\alpha _2}{R^2}} \right) $$     (4.3.8)
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Now using Natural boundary condition i.e. $$\frac\left( {r = 0} \right) = 0$$


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$$  \displaystyle {\alpha _1} + 2{\alpha _2}\left( 0 \right) = 0 $$     (4.3.9)
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$$  \displaystyle {\alpha _1} = 0 $$     (4.3.10)
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$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} $$     (4.3.11)
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Now using homogeneous boundary condition $$w\left( R \right) = 0$$
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$$  \displaystyle {\beta _0} + {\beta _1}R + {\beta _2}{R^2} = 0 $$     (4.3.12)
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$$  \displaystyle {\beta _0} = - \left( {{\beta _1}R + {\beta _2}{R^2}} \right) $$     (4.3.13)
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Also


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$$  \displaystyle \frac = {\alpha _1} + 2{\alpha _2}r $$     (4.3.14)
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As $${\alpha _1} = 0$$
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$$  \displaystyle \frac = 2{\alpha _2}r $$     (4.3.15)
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Now differentiating $$w$$ with respect to $$r$$
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$$  \displaystyle \frac = {\beta _1} + 2{\beta _2}r $$     (4.3.16)
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Putting values of $$w,\frac,\frac$$ in the standard weak form.


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$$  \displaystyle \int\limits_0^R {kr\left( {{\beta _1} + 2{\beta _2}r} \right)} \left( {2{\alpha _2}r} \right)dr = \int\limits_0^R {sr\left( {{\beta _0} + {\beta _1}r + {\beta _2}{r^2}} \right)} dr $$
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$$  \displaystyle k\int\limits_0^R {\left( {2{\beta _1}{\alpha _2}{r^2} + 4{\beta _2}{\alpha _2}{r^3}} \right)dr} = s\int\limits_0^R {\left( {{\beta _0}r + {\beta _1}{r^2} + {\beta _2}{r^3}} \right)dr} $$
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$$  \displaystyle \left[ {\frac{3} + k{\beta _2}{\alpha _2}{r^4}} \right]_0^R = \left[ {\frac{2} + \frac{3} + \frac{4}} \right]_0^R $$
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Putting value of $$$$ from equation


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$$  \displaystyle \frac{3} + k{\beta _2}{\alpha _2}{R^4} = \frac{2} + \frac{3} + \frac{4} $$
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Rearranging terms


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$$  \displaystyle {\beta _1}\left[ {\frac{3} + \frac{2} - \frac{3}} \right] + {\beta _2}\left[ {k{\alpha _2}{R^4} + \frac{2} - \frac{4}} \right] = 0 $$
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As $${\beta _1},{\beta _2}$$ are arbitrary they can't be zero, putting bracketed terms equal to zero both equation give same value of $${\alpha _2}$$.


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$$  \displaystyle {\alpha _2} = \frac $$     (4.3.17)
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Using equation (4.3.10) $${\alpha _0}$$
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$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} = \frac $$     (4.3.18)
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Putting these values in $$T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$


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$$  \displaystyle T = \frac - \frac $$
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Now checking this solution by putting in given strong form given in equation.....


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$$  \displaystyle k\frac{d}\left( {r\frac{d}\left( {\frac{s}\left( {{R^2} - {r^2}} \right)} \right)} \right) + rs = 0 $$
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$$  \displaystyle k\frac{d}\left( { - \frac} \right) + rs = 0 $$
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$$  \displaystyle k\left( { - \frac{k}} \right) + rs = 0 $$
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$$  \displaystyle - \cancel{sr} + \cancel{sr} = 0 $$     (4.3.20)
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c. Exact solution for given differential equation
Starting from 4.3.1 and rearranging we get


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$$  \displaystyle d\left( {r\frac} \right) = - \frac{s}{k}rdr $$     (4.3.21) Integrating this we get
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$$  \displaystyle r\frac = - \frac{s}{r^2} + {c_1} $$     (4.3.22) Using the boundary condition and solving
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$$  \displaystyle \frac(r = 0) = 0 $$     (4.3.23)
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$$  \displaystyle \Rightarrow {c_1} = 0 $$     (4.3.24) Next we rearrange again and integrate
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$$  \displaystyle dT = - \frac{s}rdr $$     (4.3.25)
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$$  \displaystyle T = - \frac{s}{r^2} + {c_2} $$     (4.3.26) Again using the boundary condition to solve for the constant of integration
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$$  \displaystyle T(R) = 0 $$     (4.3.27)
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$$  \displaystyle \Rightarrow {c_2} = \frac{s}{R^2} $$     (4.3.28) We arrive at the exact solution to the strong form which was found from the weak form
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$$  \displaystyle T(r) = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.29)
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=Problem 4.6: Elastic bar with a variable distributed spring force p(x)=

Given: bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x)
Consider an elastic bar with a variable distributed spring p(x) along its length as shown in Figure 1. The distributed spring imposes an axial force on the bar in proportion to the displacement. Consider a bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x) and boundary conditions as shown in Figure 1

a. Derivation of Strong Form
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 in static system is zero

Using the free-body diagram, let us now consider all of the forces acting on the elastic bar $$\displaystyle b(x),p(x), N(x), $$ 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,

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,
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$$  \displaystyle \frac{d}[E(x)\cdot A(x)\frac] + b(x)-p(x) = 0 $$     (4.6.19) Standard form of the strong form can be written as
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b. Derivation of weak form
Now strong form can be written as
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$$  \displaystyle P\left( u \right) = 0 $$     (4.6.21)
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Where
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$$  \displaystyle p\left( u \right) = \frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right) $$     (4.6.22)
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Let $${w\left( x \right)}$$ is a weight function, then writing Weighted Residual Form
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$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {P\left( u \right)} \right]dx = 0} $$     (4.6.23)
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$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {\frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right)} \right]dx = 0} $$     (N)
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Writing $$w\left( x \right) = w,E\left( x \right) = E,A\left( x \right) = A,b\left( x \right) = b,p\left( x \right) = p$$
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$$  \displaystyle \int\limits_0^l {w\left[ {\frac{d}\left[ {EA\frac} \right] + \left( {b - p} \right)} \right]dx = 0} $$     (N)
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Expanding integral
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$$  \displaystyle \int\limits_0^l {w\frac{d}\left[ {EA\frac} \right]dx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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$$  \displaystyle \left[ {wEA\frac} \right]_0^l - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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Now assuming $$w\left( 0 \right) = 0$$ and also from natural boundary condition $$E(l)\frac = \bar t$$ above integral will take form as
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$$  \displaystyle w\left( l \right)A\bar t - 0 - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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Rearranging terms


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$$  \displaystyle \int\limits_0^l {EA\frac\fracdx} = \int\limits_0^l {w\left( {b - p} \right)dx}  + w\left( l \right)A\bar t $$ (N)
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Now writing in standard format


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