Example: Axially loaded Elastic Bar

Strong Form: $ \frac{d}{dx}\big(AE\frac{du}{dx}\big)+b=0 $

with boundary conditions:

at $ x=0 $, traction/force per unit area $ \bar{t} $, is prescribed: $ \sigma(x=0) = \big(E\frac{du}{dx}\big)_{x=0}=\frac{p(0)}{A(0)}=-\bar{t} $

at $ x=l $, displacement $ \bar{u} $ is prescribed: $ u(x=l) = \bar{u} $

Weak Form: $ w(x) $ is an arbitrary weight function, multiplied by the governing equation and boundary conditions and integrated over the domain.

Governing equation: $ \int^l_0 w\big[\frac{d}{dx}\big(AE\frac{du}{dx}\big)+b\big]=0 $

Boundary condition (traction force): $ \big(wA\big(E\frac{du}{dx}+\bar{t}\big)\big)_{x=0}=0 $

Boundary condition (end-point displacement): $ w(l) = 0 $

Reminder:

$ \frac{d}{dx}(wf) = w\frac{df}{dx}+f\frac{dw}{dx} \Rightarrow w\frac{df}{dx}=\frac{d}{dx}(wf)-f\frac{dw}{dx} $

Therefore, when we integrate the equation over $ [0,l] $:

$ \int^l_0 w\frac{df}{dx} dx = \int^l_0 \frac{d}{dx}(wf) dx – \int^l_0 f\frac{dw}{dx} dx $

$ \Rightarrow \int^l_0 w\frac{df}{dx} dx = wf\Big|^l_0 – \int^l_0 f\frac{dw}{dx} dx $

Substitute in $ f = AE\frac{du}{dx} $:

$ \int^l_0 w\frac{d}{dx}(AE\frac{du}{dx}) dx = (wAE\frac{du}{dx})\Big|^l_0 – \int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx $

1. From the governing equation, the first term becomes: $ -\int^l_0 wb dx $
2. From the relation $ \sigma = E\frac{du}{dx} $, the middle term (first term on the right hand side) becomes: $ (wA\sigma)_{x=l}-(wA\sigma)_{x=0} $

And the equation becomes:

$ (wA\sigma)_{x=l}-(wA\sigma)_{x=0} – \int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx + \int^l_0 wb dx = 0 $

Consider the boundary conditions:

1. Boundary condition (traction force): $ \big(wA\big(E\frac{du}{dx}=-\bar{t}\big)\big)_{x=0} $
2. Boundary condition (end-point displacement): $ w(l) = 0 $

Now, let’s say hello to the slimmed and toned equation:

$ \int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx = \bar{t}\big)\big)_{x=0} + \int^l_0 wb dx $

where we only need to recover $ u(x) $ such that it satisfies $ u(l) = \bar{u} $.

This form has a weaker continuity requirements than the strong form, hence the name “weak form”.

Sources:

• J. Fish and T. Belytschko – First Course in Finite Elements
• Krishnan Suresh – Why Weak Form?