Question 1 of 7: Collocation solution of a bar under a linearly varying axial load

Question 1: Collocation solution of a bar under a linearly varying axial load (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A uniform bar of axial rigidity $EA$ fixed at $x=0$ and free at $x=L$, carrying the distributed axial load $q(x)=cx$, together with the two collocation stations $x = L/4$ and $x = 3L/4$.

Find. The cubic trial function that satisfies the essential boundary condition, the three equations that fix its coefficients, and the resulting approximate displacement field $u(x)$.

Approach. Choose a cubic trial function that satisfies the essential boundary condition a priori, force its residual to vanish at the two nominated collocation points, and close the system with the natural (traction-free) boundary condition at the free end.

1. Select a trial function that already satisfies the essential BC. The collocation method is a weighted-residual method whose weight functions are Dirac deltas, so the trial function must satisfy every essential boundary condition exactly. Writing a general cubic and imposing $u(0)=0$ deletes the constant term:$$u(x) \approx \tilde u(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3}$$ Three unknown coefficients remain, so three independent conditions are required.
2. Form the residual. Substituting the trial function into the governing differential equation leaves a residual $R(x)$ that would be identically zero only for the exact solution:$$R(x) = EA\,\frac{d^{2}\tilde u}{dx^{2}} + cx = EA\bigl(2a_{2} + 6a_{3}x\bigr) + cx$$
3. Collocate at the two nominated points. Collocation enforces $R(x_{i}) = 0$ at each chosen station $x_{i}$, which is the same as using $w_{i}(x) = \delta(x - x_{i})$ in $\int_{0}^{L} w_{i} R \, dx = 0$. At $x = L/4$:$$2EA\,a_{2} + \frac{3}{2}EA\,L\,a_{3} + \frac{cL}{4} = 0$$ and at $x = 3L/4$:$$2EA\,a_{2} + \frac{9}{2}EA\,L\,a_{3} + \frac{3cL}{4} = 0$$
4. Add the natural boundary condition. The traction-free end is a natural condition, which a collocation (strong-form) statement does not satisfy automatically, so it must be imposed as the third equation:$$\left.EA\frac{d\tilde u}{dx}\right|_{x=L} = EA\bigl(a_{1} + 2a_{2}L + 3a_{3}L^{2}\bigr) = 0$$
5. Solve the three equations. Subtracting the first collocation equation from the second eliminates $a_{2}$ and gives $3EA\,L\,a_{3} + cL/2 = 0$, hence$$a_{3} = -\frac{c}{6EA}$$ Back-substituting into the $x=L/4$ equation gives $2EA\,a_{2} - cL/4 + cL/4 = 0$, so$$a_{2} = 0$$ and the natural condition then yields $a_{1} = -3a_{3}L^{2} = cL^{2}/(2EA)$.
6. Assemble the approximate solution. Collecting the three coefficients:$$\boxed{\;u(x) \approx \frac{c}{6EA}\bigl(3L^{2}x - x^{3}\bigr)\;}$$ The free-end displacement follows as $u(L) = cL^{3}/(3EA)$.

Two features of this answer are worth stating explicitly, because they carry most of the marks. First, substituting the coefficients back into the residual gives $R(x) = EA(0 + 6a_{3}x) + cx = -cx + cx \equiv 0$: the residual vanishes identically, not merely at $x = L/4$ and $x = 3L/4$. Second, integrating the governing equation directly gives$$u_{\text{exact}}(x) = \frac{c}{6EA}\bigl(3L^{2}x - x^{3}\bigr)$$ which is the same cubic. The collocation solution is therefore the exact solution of this problem.

The reason is structural rather than lucky: the exact solution of $EA\,u^{\prime\prime} = -cx$ with these boundary conditions is itself a cubic polynomial, and the assumed basis contains that cubic. Whenever the trial space contains the exact solution, every weighted-residual method — collocation, least squares, Galerkin, subdomain — recovers it, and the placement of the collocation points becomes immaterial. Any two distinct stations in $0 \lt x \lt L$ produce the same coefficients.