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Figure below shows a mass consisting of concentric attached cylinders Derive the natural frequency of undamped free vibrations.

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

Marks: 10M

Year: may 2016

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ddx[(x1)dudx]=x2[(x1)d2ydx2+dudx]=x2(x1)d2ydx2+dudx+x2=0

Appropriate solution

u=C0+C1x+C2x2+C3x3B.C.(1):u(5)=1010=C0+5C1+25C2+125C3C0=105C125C2125C3u=105C125C2125C3+C1x+C2x2+C3x3u=C1(x5)+C2(x225)+C3(x3125)+10dydx=C1+2C2x+3C3x2

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

5=C1+6C2+27C3 c1=56c227c3

u=105c125c2125c3+c1x+c2x2+c3x3

u=c1(x5)+c2(x225)+c3(x3125)=10

dudx=c1+2c2x+3c3x2

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

5=c1+6c2+27c3

c1=56c227c3

u=(56c227c3)(x5)+c2(x225)+c3(x3125)+10

u=c2(x2256x+30)+c3(x312527x+135)+5(x5)+10

u=c2(x26x+5)+c3(x327x+10)+5(x5)+10

u=c2(2x6)+c3(3x227)+5

u

\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

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