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The weak form for the bar

The weak form says the energy does not change in any admissible direction. You get it either from $\delta\Pi(u, v) = 0$ , or by multiplying the strong form by a test function $v$ and integrating by parts. The force condition at the end then appears by itself.

B(u, v) = F(v)\quad \text{for all } v,\ v(0) = 0

weak form

B(u, v) = \int_0^L EA\,u'v'\,dx

bilinear form

F(v) = \int_0^L F_x\,v\,dx + P\,v(L)

linear form

Symbols

$B$	symmetric, bilinear
$F$	linear (the loads)
$v$	test function

Example

Integration by parts: $\int_0^L -(EAu')'v\,dx = \int_0^L EAu'v'\,dx - [EAu'v]_0^L$ . With $v(0) = 0$ and $EAu'(L) = P$ the boundary term becomes $P\,v(L)$ .

The weak form only needs first derivatives – that is why piecewise linear functions are enough.

Practise weak form and shape functions for free →

← The Gateaux derivative · Master element and Jacobian →

Part of Finite Element Method: Weak form and shape functions.