The governing differential equation for steady state one dimensional conduction heat transfer with internal heat generation is given by $\frac{d}{dx}[K \frac{dT}{dx}]=q \hspace{1cm} \text{for} \hspace{1cm} 0 \leq x \leq 1$ were develop the finite element termulation of linear element. use R-R method, mapped over general element.

Given D.E. $ \frac{d}{dx}[K \frac{dT}{dx}] = q $ for 0 $\leq$ x $\leq$ i.e. $ \frac{d}{dx}[K \frac{dT}{dx}] - q = 0 $

In local co-ordinates, replace $dx$ by $d \bar{x}$

$ \frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q = 0 $

With local boundary conditions as,

$ K \frac{dT}{d \bar{x}}|_{\bar{x}=0} = - \frac{Q_1}{A} $

$ K \frac{dT}{d \bar{x}}|_{\bar{x}=he} = \frac{Q_2}{A} $

Since the solution is approximate, residue is given by,

$ \frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q = R $

Weighted integral form,

$ \int_0^1 w_iR \,\, dx = 0 $

$ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q ) \,\, d \bar{x} = 0 \\ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}]) \,\, d \bar{x} - \int_0^{he}w_iq \,\,d \bar{x} = 0 \hspace{0.5cm} ....(1) $

Weak formulation

Consider equation, $ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}]) \,\, d \bar{x} $

Integrating by parts, we get,

$ [w_i(K \frac{dT}{d \bar{x}})]_0^{he} - \int_0^{he} \frac{dw_i}{d \bar{x}}[w_i(K \frac{dT}{d \bar{x}})] \,\, d \bar{x} $

Substituting above value in eq(1), we get,

$ [w_i(K \frac{dT}{d \bar{x}})]_0^{he} - \int_0^{he} \frac{dw_i}{d \bar{x}}[w_i(K \frac{dT}{d \bar{x}})] \,\, d \bar{x} - \int_0^{he} w_iq \,\, d \bar{x} = 0 \hspace{0.5cm} ....(2) $

Let the approximate solution be,

$ T = T_1 \phi_1 + T_2 \phi_2 $

where $ \phi_1 = 1 - \frac{\bar{x}}{he}; \hspace{0.5cm} \phi_2 = \frac{\bar{x}}{he} $

Rayleigh-Ritz method, $ w_i = \phi_1 $

For i = 1, $ w_1 = 1 - \frac{\bar{x}}{he}; \hspace{0.5cm} \frac{dw_1}{d \bar{x}} = - \frac{1}{he} $

Substituting in eq(2), we get,

$ \frac{Q_1}{A} - K \int_0^{he} ( \frac{-1}{he}) (- \frac{T_1}{he} + \frac{T_2}{he}) \,\, d \bar{x} - \int_0^{he} (1 - \frac{\bar{x}}{he}).q \,\, d \bar{x} = 0 \\ \frac{Q_1}{A} = \frac{-K}{he}[\frac{-T_1 \bar{x}}{he} + \frac{T_2 \bar{x}}{he}]_0^{he} - q[\bar{x} - \frac{(\bar{x})^2}{2he}]_0^{he} \\ \frac{Q_1}{A} = \frac{-K}{he}[-T_1 + T_2] - q[\frac{he}{2}] $

or

$ \frac{Q_1}{A} = \frac{K}{he}[T_1 - T_2] - q \frac{he}{2} \\ \frac{K}{he} [T_1 - T_2] = \frac{Q_1}{A} + q \frac{he}{2} \hspace{0.5cm} ...(3) $

For i = 2, $ w_2 = \frac{\bar{x}}{he}, \hspace{0.5cm} \frac{dw_2}{d \bar{x}} = \frac{1}{he} $

Substituting in eq(2), we get,

$ \frac{Q_2}{A} - K \int_0^{he} \frac{1}{he}( \frac{-T_1}{he} + \frac{T_2}{he}) d \bar{x} - \int_0^{he} \frac{ \bar{x}}{he}q \,\, d \bar{x} \\ \frac{Q_2}{A} = \frac{K}{he} [ \frac{-T_1 \bar{x}}{he} + \frac{T_2 \bar{x}}{he}]_0^{he} - \frac{q}{he}[ \frac{ ( \bar{x})^2}{2}]_0^{he} \\ \frac{Q_2}{A} = \frac{K}{he}[-T_1 + T_2] - \frac{q(he)}{2} $

or

$ \frac{K}{he}[-T_1 + T_2] = \frac{Q_2}{A} +\frac{q(he)}{2} \hspace{0.5cm} ....(4) $

Putting eq (3) and (4) in matrix form, we get,

$ \frac{K}{he} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} T_1 \\ T_2 \end{Bmatrix} = \begin{Bmatrix} \frac{Q_1}{A} \\ \frac{Q_2}{A} \end{Bmatrix} + \frac{q(he)}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $