Talk:Nonlinear finite elements/Axially loaded bar
The last paragraph states that "this is tedious since we have to repeat the process for every point in the bar".
I don't think this is true like that, i.e. we can just pick one point 'x' that stands for all points. Then
![{\displaystyle \mathbf {f} _{x}=\mathbf {R} +\int _{x}^{L}\mathbf {q} (\mathbf {x'} )~dx'=\mathbf {R} +\int _{x}^{L}a\mathbf {x'} ~dx'=\mathbf {R} +\left[{\frac {a}{2}}\mathbf {x'} ^{2}\right]=\mathbf {R} +{\frac {aL^{2}}{2}}-{\frac {a\mathbf {x} ^{2}}{2}}=\mathbf {R} +{\frac {a(L^{2}-\mathbf {x} ^{2})}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db8a22caedffba210610145cdf519eff9e68949a)
And then go from there?
If the steps after this one become "tedious" then maybe that should be stated like that. (The preceding unsigned comment was added by Thuldai (talk • contribs) 09:06, 27 February 2008)