Using R-R method mapped over general element solve $\frac{d}{dx}(a \frac{du}{dx})+bu+c = 0; \hspace{1cm} 0 \leq x \leq 1$

Solve BCs are u(0)=u0 and $a(\frac{du}{dx})|_{x = L} =0$

Use Lagrange's linear shape function

Subject: Finite Element Analysis

Topic: FEA Procedure

Difficulty: Medium

Given D.E is $ \frac{d}{dx}(a \frac{du}{dx}) + bu + c = 0; \hspace{0.5cm} 0 \leq x \leq 1 $

Boundary conditions are

u(0) = u$_0$ $ a \frac{du}{dx}|_{x=L} = 0 $

1) Discretization: Dividing the domain into 3 elements

2) Take a general element of length he and set its local co-ordinate

$ x = x_A + \bar{x} $

Differentiating above eq, we get,

$ dx = d \bar{x} $

3) Converting governing differential equation into local co-ordinate

$ dx = d \bar{x} \\ \therefore \frac{d}{d \bar{x}} (a \frac{du}{d \bar{x}}) + cu + q = 0 $

Local boundaries are

$ a \frac{du}{d \bar{x}}|_{\bar{x} = 0} = - Q_1 \\ a \frac{du}{d \bar{x}}|_{\bar{x} = he} = Q_2 $

4) To find residue

$ \frac{d}{d \bar{x}} (a \frac{du}{d \bar{x}}) + cu + q = R $

5) Weighted Integral form

$ \int_0^1 w_i R \,\, dx = 0 \\ \int_0^{he} w_i[\frac{d}{d \bar{x}} (a \frac{du}{d \bar{x}}) + cu + q] \,\, d \bar{x} = 0 \\ \int_0^{he} w_i \frac{d}{d \bar{x}}(a \frac{du}{d \bar{x}}) \,\, d \bar{x} + \int_0^{he} w_i cu \,\, d \bar{x} + \int_0^{he} w_i q \,\, d \bar{x} = 0 \hspace{0.5cm} ....(A) $

Consider $ \int_0^{he} w_i \frac{d}{d \bar{x}}(a \frac{du}{d \bar{x}}) \,\, d \bar{x} $

Integrating by parts, above equation becomes,

$ []_0^{he} + []_0^{he} $