It is required to carry out one dimensional structural analysis of a circular bar of length 'L' fixed at one end and carries a point load 'P' at other end. Find the suitable differential equation with required boundary condition justify and solve it by using Rayleigh Ritz method for two linear element.

Cross Section Area = A; Modulus of Elasticity = E

Governing D.E is given by,

$\frac{d}{dx}[EA \frac{du}{dx}] = 0 $ for 0 $\leq$x$\leq$l

Boundary conditions are

u(0) = 0

$ \frac{du}{dx}|_{x=l} = \frac{P}{AE} $

1) Approximate solution is given by,

$u = C_0 + C_1x + C_2x^2 + C_3x^3 $ .....(1)

Differentiating above equation,

$ \frac{du}{dx} = C_1 + 2C_2x + 3C_3x^2 $ .....(2)

i) At x = 0, u = 0, then equation (1) becomes; 0 = C$_0$

ii) At x = l, $ \frac{du}{dx} = \frac{P}{AE} $

Eq. (2) becomes,

$ \frac{P}{AE} = C_1 + 2C_2l + 3C_3l^2 \\ \therefore C_1 = \frac{P}{AE} - 2C_2l - 3C_3l^2 $

Substituting th evalues of C$_0$ and C$_1$ in eq (1), we get,

$ u = [\frac{P}{AE} - 2C_2l - 3C_3l^2]x + C_2x^2 + C_3x^3 \\ \therefore u = (x^2-2lx)C_2 + (x^3-3l^2x)C_3 + \frac{P}{AE}x $

Differentiating the above eq w.r.t x

$ \frac{du}{dx} = (2x-2l)C_2 + (3x^2-3l^2)C_3 + \frac{P}{AE} $

Differentiating again,

$ \frac{d^2u}{dx^2} = 2C_2 + 6xC_3 $

2) To find residue

$ \frac{d}{dx}[EA \frac{du}{dx}] = R \\ EA \frac{d^2u}{dx^2} = R $

Substitute the value of $ \frac{d^2u}{dx^2} $ in above equation,

EA (2C$_2$ + 6xC$_3$) = R

3) Weighted integral form,

$ \int_0^1 w_iR \,\, dx = 0 \\ \int_0^l w_i[EA (2C_2 + 6xC_3)] \,\, dx = 0 $ .....(3)

Now, w = coefficient of constant c in u

$ w_1 = (x^2 - 2lx) \hspace{0.5cm} w_2 = (x^3 - 3lx) $

For i = 1, $ w_1 = (x^2 - 2lx) $, eq (3) becomes

$ \int_0^l (x^2 - 2lx)(2C_2 + 6xC_3) \,\, dx = 0 \\ \int_0^l [2x^2C_2 - 4lxC_2 + 6C_3x^3 - 12 lx^2C_3] \,\, dx = 0 \\ 2C_2\int_0^l x^2 \,\, dx - 4lC_2\int_0^l x \,\, dx + 6C_3\int_0^l x^3 \,\, dx - 12lC_3\int_0^l x^2 \,\, dx = 0 \\ 2C_2[\frac{x^3}{3}]_0^l - 4lC_2[\frac{x^2}{2}]_0^l + 6C_3[\frac{x^4}{4}]_0^l - 12lC_3[\frac{x^3}{3}]_0^l = 0 \\ 2C_2[\frac{l^3}{3}] - 4lC_2[\frac{l^2}{2}] + 6C_3[\frac{l^4}{4}] - 12lC_3[\frac{l^3}{3}] = 0 \\ C_2[\frac{2l^3}{3} - \frac{4l^3}{2}] + C_3[\frac{6l^4}{4} - \frac{12l^4}{3}] = 0 \\ -1.333l^3C_2 -2.5l^4C_3 = 0 \hspace{1cm} ...(A)$

For i = 2, $ w_2 = x^3-3xl^2 $

$ \int_0^l (x^3-3xl^2)(2C_2 + 6C_3x) \,\, dx = 0 \\ [\frac{2C_2 x^4}{4} - \frac{6C_2l^2x^2}{2} + \frac{6C_3x^4}{5} - \frac{6C_3x^3l^2}{1}]_0^l = 0 \\ C_2[\frac{l^4}{2} - \frac{3l^4}{1}] + C_3[\frac{6l^5}{5} - 6l^5] = 0 \\ -2.5l^4C_2 - 4.8l^5C_3 = 0 \hspace{1cm} ...(B) $

Bt solving eq(A) and eq(B), we get,

C$_2$ = C$_3$ = 0

Therefore approximate solution u becomes,

$ u = \frac{P}{AE}x $