User:Eml5526.s11.team3.akj/Homework 4

=Problem4.3 Quadratic Trial Solution of Weak Form for the heat conduction problem in a circular plate =

Given:Strong form for the heat conduction problem in a circular plate
The Strong form for the heat conduction problem in a circular plate is

Where R is the 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, and k,s and R are given

a) Construct the weak form for the given Strong form
Construct the weak form for the above strong form.

b) Solution to the weak form
Use the quadratic trial (candidate) solution of the form $$\alpha _{0}+\alpha _{1}r+\alpha _{2}r^{2}$$  and weight function of the same form to obtain the solution of the weak form.

c) Solution to the strong form
Solve the differential equation with the boundary conditions ans show that the temperature distribution along the radius is given by

a) Weak form of the Strong form for the heat conduction in a circular plate
Multiplying the given strong from with the weight function w and integrating it over the interval $$0< r\leq R$$  we get,

Integrating first term of Equation 4.3.3 by parts:

We know that the weight function should vanish at $$r=R$$  i.e it should satisfy the homogeneous essential boundary condition  $$(w(R)=0)$$  and from the natural boundary condition  $$\frac{\mathrm{d} T}{\mathrm{d} r}(r=0)=0$$

hence substituting the above boundary conditions and rearranging the terms we get,

hence the weak form for the strong form in Eq.4.3.1 is

b) Solution of the weak form using quadratic trial solution
Given the trial solution is of the form $$T=\alpha _{0}+\alpha _{1}r+\alpha _{2}r^{2}$$  and the weight function is also of the same form and can be written as  $$w=\beta _{0}+\beta _{1}r+\beta _{2}r^{2}$$

From the given essential boundary condition we can write,

differentiating the trial solution of T we get

given natural boundary condition as $$\frac{\mathrm{d} T}{\mathrm{d} r}(r=0)=0$$ hence, $$\alpha_{1}=0$$ hence the trial solution is such that

from the homogeneous essential boundary condition we know $$w(R)=0$$ hence $$\beta _{0}+\beta _{1}R+\beta _{2}R^{2}=0$$ hence the weighting function is such that

substituting Eq 4..9 and 4.3.10 into the weak form derived in Eq 4.3.6 we get,

Integrating the left hand side of Eq 4.3.11 we get

Integrating the right hand side of Eq 4.3.11 we get

Hence combining Eq 4.3.13 and Eq 4.3.12 we can write,

Rearranging the terms in the above equation we get,

The above equation should be true for any value of $$\beta_{1}$$  and  $$\beta_{2}$$  hence equating each of  $$\beta_{1}$$  and  $$\beta_{2}$$  to zero we get, both the equations yield the same result

Substituting $$\alpha_{2}$$  into Eq 4.3.7 we get,  $$\alpha_{0}=\frac{sR^{2}}{4k}$$ Hence the trial solution to the weak form in Eq 4.3.6 is

c) Solution of the strong form
The Strong form for the heat conduction problem in a circular plate from Eq 4.3.1 is

Rearranging the terms in the strong form we get,

integrating the above Eq we get,

using the natural boundary condition we get, $$c_{1}=0$$ Agian integrating Eq 4.3.19 we get,

Applying essential boundary condition $$(T(r=R)=0)$$  we get,  $$c_{2}=\frac{R^{2}s}{4k}$$

Hence the temperature distribution of the circular plate along the radius is given by

Note:- From Eq 4.3.17 and Eq 4.3.21 it is observed that the solution to the weak form and the strong form are the same in this case

=Problem 4.8 Find mass matrix and force matrix=

Given : The strong form and a cubic trial solution
The strong form is

Given $$A=1$$ ,  $$E=2$$  ,$$\tilde{m}=3$$. The cubic trial solution is given as $$u^{h}(x)=\alpha _{0}+\alpha _{1}(x-3)+\alpha _{2}(x-3)^{2}+\alpha _{3}(x-3)^{3}$$

Where the basis function is given by  $$\left\{b_{i}\right\}=\left\{1,(x-3),(x-3)^{2},(x-3)^{3};i=0,1,2,3\right\}$$

a) Mass Matrix
Mass matrix $$\mathbf{\tilde{M}}$$

b) Force Matrix
Force matrix $$\mathbf{F}$$  for dynamics with  $$u^{h}(\beta ,t)=g(t)=sin(2t)$$

a) Mass Matrix
The mas matrix $$\mathbf{\tilde{M}}$$  is given by

where $$\mathbf{M_{EE}}=[M_{00}]$$ ,  $$M_{00}=\tilde{M}_{00}$$  and  $$\tilde{M}_{00}=\tilde{m}(b_{0},b_{0})$$

$$\tilde{m}(b_{0},b_{0})=\int_{1}^{3}b_{0}\bar{m}b_{0}dx$$ as given in Lecture 19-1

Substituting for $$b_{0}\quad and\quad \bar{m} $$  we get $$\tilde{m}(b_{0},b_{0})=\int_{1}^{3}3dx$$ hence

Similarly $$\mathbf{M_{EF}}=[M_{0j};j=1,2,3]$$ ,  $$M_{0j}=\tilde{M}_{0j}$$  and  $$\tilde{M}_{0j}=\tilde{m}(b_{0},b_{j})$$

Clearly we can see that $$\mathbf{M_{EF}}$$  is row matrix

hence we can write $$\tilde{m}(b_{0},b_{1})=\int_{1}^{3}3(x-3)dx$$ which on integration gives $$\tilde{m}(b_{0},b_{1})=-6$$

similarly $$\tilde{m}(b_{0},b_{2})=\int_{1}^{3}3(x-3)^{2}dx=8$$

and $$\tilde{m}(b_{0},b_{3})=\int_{1}^{3}3(x-3)^{3}dx=-12$$

hence

it is observed that $$\mathbf{M_{EF}}=\mathbf{M_{FE}^{T}}$$  hence  $$\mathbf{M_{FE}}$$  is column matrix given by

and $$\mathbf{M_{FF}}=[M_{ij};i,j=1,2,3]$$,  $$M_{ij}=\tilde{M}_{ij}$$  and  $$\tilde{M}_{ij}=\tilde{m}(b_{i},b_{j})$$ where $$\tilde{m}(b_{1},b_{1})=\int_{1}^{3}(x-3)3(x-3)dx=8$$ $$\tilde{m}(b_{1},b_{2})=\int_{1}^{3}(x-3)3(x-3)^{2}dx=-12$$ $$\tilde{m}(b_{1},b_{3})=\int_{1}^{3}(x-3)3(x-3)^{3}dx=\frac{96}{5}$$ $$\tilde{m}(b_{2},b_{1})=\int_{1}^{3}(x-3)^{2}3(x-3)dx=-12$$ $$\tilde{m}(b_{2},b_{2})=\int_{1}^{3}(x-3)^{2}3(x-3)^{2}dx=\frac{96}{5}$$ $$\tilde{m}(b_{2},b_{3})=\int_{1}^{3}(x-3)^{2}3(x-3)^{3}dx=-32$$ $$\tilde{m}(b_{3},b_{1})=\int_{1}^{3}(x-3)^{3}3(x-3)dx=\frac{96}{5}$$ $$\tilde{m}(b_{3},b_{2})=\int_{1}^{3}(x-3)^{3}3(x-3)^{2}dx=-32$$ $$\tilde{m}(b_{3},b_{3})=\int_{1}^{3}(x-3)^{3}3(x-3)^{3}dx=\frac{384}{7}$$

hence

Substituting Eq 4.8.3,Eq 4.8.4,Eq 4.8.5 and Eq 4.8.6 into Eq 4.8.2 we get

b) Force Matrix
The force matrix is given by

where $$\mathbf{\tilde{F}}=\left \{ \frac{\mathbf{F_{E}}}{\mathbf{F_{F}}} \right \}$$ and $$\mathbf{F_{F}}=\left \{ F_{i},i=1,2,3 \right \}$$ ,  $$ F_{i}=\tilde{F_{i}}$$  and  $$\tilde{F_{i}}=\tilde{f}(b_{i})$$  as given in Lecture 19-1

$$\tilde{f}(b_{i})=b_{i}(\alpha)h+\int_{1}^{3}b_{i}fdx$$

at $$\alpha$$  natural boundary condition is specified, in  this case is  $$\alpha=1$$  and  $$h$$  is the value of the of the natural boundary condition given by  $$h=0.1$$  and  $$f$$  in this problem is  $$2x$$

hence $$\tilde{f}(b_{1})=(1-3)0.1+\int_{1}^{3}(x-3)2xdx=\frac{-103}{15}$$ $$\tilde{f}(b_{2})=(1-3)^{2}0.1+\int_{1}^{3}(x-3)^{2}2xdx=\frac{42}{5}$$ $$\tilde{f}(b_{3})=(1-3)^{3}0.1+\int_{1}^{3}(x-3)^{3}2xdx=-12$$

stiffness matrix matrix is given by $$\mathbf{K_{EF}}=[K_{0j};j=1,2,3]$$ ,  $$K_{0j}=\tilde{K}_{0j}$$  and  $$\tilde{K}_{0j}=\tilde{k}(b_{0},b_{j})$$ and $$\tilde{k}(b_{0},b_{j})=\int_{1}^{3}{b_{0}}'AE{b_{j}}'dx$$

but $$b_{0}=1$$  hence  $${b_{0}}'=0$$ therefore  $$\tilde{k}(b_{0},b_{j})=0$$ $$\implies\quad\mathbf{K_{EF}}=\begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$$

since $$\mathbf{K_{EF}}=\mathbf{K_{FE}^{T}}$$

from Eq 4.8.1 $$g=sin(2t)$$ $$\implies \quad {g}''=-4sin(2t)$$

hence substituting Eq.4.8.5, Eq4.8.9 and Eq 4.8.10 into Eq 4.8.8 we get