Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 10M

Year: may 2016

$- \frac{d}{dx}\bigg[ (x - 1) \frac{du}{dx} \bigg] = x^2\\ - \bigg[ (x - 1) \frac{d^2y}{dx^2} + \frac{du}{dx} \bigg] = x^2\\ (x - 1) \frac{d^2y}{dx^2} + \frac{du}{dx} + x^2 = 0$

Appropriate solution

$u = C_0 + C_1x + C_2x^2 + C_3x^3\\ B.C.(1): u(5) = 10\\ 10 = C_0 + 5C_1 + 25C_2 + 125C_3\\ \therefore C_0 = 10 - 5C_1 - 25C_2 - 125 C_3\\ u = 10 - 5C_1 - 25C_2 - 125C_3+C_1x + C_2 x^2+ C_3x^3\\ u = C_1(x - 5) + C_2 (x^2 - 25) + C_3(x^3 - 125) + 10\\ \frac{dy}{dx} = C_1 + 2C_2x + 3 C_3 x^2$

B.C.(2) u'(3) = 5

$\therefore 5 = C_1 + 6C_2 + 27C_3$ $\therefore c_1 = 5-6 c_2-27c_3$

$u =10-5 c_1- 25c_2-125c_3 + c_1x+c_2x^2+c_3x^3$

$u = c_1(x-5)+c_2(x^2-25)+c_3(x^3-125)=10$

$\frac{du}{dx}=c_1+2c_2x+3c_3x^2$

B.c (2) u' (3) = 5

$\therefore 5= c_1+6c_2+27c_3$

$\therefore c_1 = 5 -6c_2-27c_3$

$\therefore u = (5-6c_2-27c_3)(x-5)+c_2(x^2-25)+c_3(x^3-125)+10$

$u = c_2(x^2-25-6x+30)+c_3(x^3-125-27x+135)+5(x-5)+10$

$\therefore u = c_2(x^2-6x+5)+c_3(x^3-27x+10)+5(x-5)+10$

$u'=c_2(2x-6)+c_3(3x^2-27)+ 5 $

$u''=2c_2+6xc_3$

$\therefore R = (x-1)(2c_2+6xc_3)+c_2(2x-6)+c_3(3x^2-27)+(5+x^2)$

$R=c_2(2x-2+2x-6)+c_3(6x^2-6x+3x^2-27)+(5-x^2)$

$=c_2(4x-8)+c_3(9x^2-6x-27)+(5+x^2)$

weighted integral form,

$\int\limits_3^5 w_i\hspace{0.2cm} R\hspace{0.2cm} dx= 0 $

$\int\limits_3^5 w_i[c_2(4x-8)=c_3(9x^2-6x-27)+(5+x^2)]dx$

$w_1 =x^2-6x+5\hspace{1cm}and\hspace{1cm}w_2=x^3-27x+10$

for i =1 :

$\int\limits_3^5(x^2-6x+5)[c_2(4x-8)+c_3(9x^2-6x-27)+(5+x^2)]dx = 0 $

$-37.33c_2-422.4c_3-102.933=0 $

$\hspace{0.9cm}37.33c_2+422.4c_3=-102.933 \hspace{2cm}...(1)$

for i = 2

$\int\limits_3^5(x^3-27x+10 )[c_2(4x-8)+c_3(9x^2-6x-27)+(5+x^2)]dx = 0 $

$-422.4c_2-4790.4c_3-1162.7=0$

$\hspace{1.3cm}422.4c_2+4790.4c_3=-1162.7\hspace{2cm}----(2)$

from (1) & (2) :$\therefore c_2=-4.87\hspace{0.3cm} and \hspace{0.3cm}c_3=0.186 $

$u = -4.87(x^2-6x+5)+ 0.186(x^3-27x+10)+5(x-5)+10$

$u(4) = 13.286 $

$u(5) =10$

Exact solution :

$-\frac{d}{dx}\bigg[(x-1)\frac{du}{dx}\bigg] = x^2$

$(x-1)\frac{du}{dx}= -\int\limits x^2 dx = \frac{x^3}{x}+6$

$\therefore\frac{du}{dx} = -\frac{x^3}{3(x-1)}+\frac{c}{x-1}$

At x = 3 : $\frac{du}{dx}= 5 .$

$ \hspace{1.5cm}5 = - \frac{3^3}{3(3-1)}+\frac{c}{3-1}$

$ \hspace{1.5cm}5= -\frac{9}{2}+\frac{c}{2}$

$ \hspace{1.5cm}c=19$

$\therefore\frac{du}{dx} = -\frac{x^3}{3(x-1)} +\frac{19}{x-1}$

$\int\limits \frac{du}{dx}dx= \int\limits-\frac{x^3}{3(x-1)}dx+ \int\limits\frac{19}{x-1}dx.$

$u = -\frac{1}{3}\int \limits \frac{x^3}{x-1}dx +19 \int\limits\frac{1}{x-1}dx.$

$u = - \frac{1}{3}\int\limits\bigg[(\frac{x^3-1}{x-1})+\frac{1}{x-1}\bigg]dr+ 19 log(x-1)+c_1$

$u=-\frac{1}{3}\int\limits(x^2+x+1) dx -\frac{1}{3} log (x-1)+19log(x-1)+c_1$

$u= -\frac{1}{3}\bigg(\frac{x^3}{3}+\frac{x^2}{2}+x\bigg)+\frac{56}{3}log(x-1)+c_1$

At x = 5 ; u = 10

$\therefore 10 = - \frac{1}{3}\bigg(\frac{125}{3}+\frac{25}{2}+5\bigg)+ \frac{56}{3}log 4 +c_1$

$\hspace{0.9cm}c_1 = 18.48$

$\therefore u = - \frac{1}{3}\bigg(\frac{x^3}{3}+\frac{x^2}{2}+x\bigg)+\frac{56}{3}log (x-1)+18.48$

$u(4) = 16.27$

$u(5) = 10 $