Question 4 of 7: Collocation solution for an axially loaded cantilevered bar

Question 4: Collocation solution for an axially loaded cantilevered bar (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 prismatic bar of length $L$, cross-sectional area $A$ and Young’s modulus $E$, fixed at $x=0$ and free at $x=L$. It carries a distributed axial load $q(x)=cx$ growing linearly from zero at the wall, and a concentrated axial load $P$ at the tip. The governing equation is $EA\,u''+cx=0$ on $0\lt x\lt L$, with the essential boundary condition $u(0)=0$ and the natural boundary condition $EA\,u'(L)=P$. Collocation points are prescribed at $x=L/4$ and $x=3L/4$; the trial function is to be a cubic polynomial.

Find. The three coefficients of the cubic trial function, and hence the approximate displacement field $u(x)$.

Approach. Choose a cubic trial function that satisfies the essential boundary condition identically, form the residual of the differential equation, set that residual to zero at the two collocation points, close the system with the natural boundary condition, and finally test the result against the exact solution.

1. Choose an admissible trial function. A general cubic has four coefficients, but the essential (geometric) boundary condition $u(0)=0$ must be satisfied a priori by any trial function in a weighted-residual method, which removes the constant term: $$\tilde u(x)=a_1x+a_2x^{2}+a_3x^{3},\qquad \tilde u(0)=0\ \text{ identically.}$$ Three unknowns remain, so three independent equations are needed — the two collocation conditions plus the natural boundary condition.
2. Form the residual. Substituting the trial function into the governing equation leaves a residual that would vanish identically only for the true solution: $$R(x)=EA\,\frac{d^{2}\tilde u}{dx^{2}}+cx = EA\left(2a_2+6a_3x\right)+cx .$$ Note that $a_1$ does not appear: the first-order term is annihilated by the second derivative, so the residual alone can never determine $a_1$. That is precisely why the natural boundary condition is indispensable here.
3. Impose collocation at $x=L/4$. The collocation method is the weighted-residual method with Dirac delta weighting functions $W_k(x)=\delta(x-x_k)$, so $\int_0^L \delta(x-x_k)R(x)\,dx = R(x_k)=0$: the residual is forced to zero at the chosen points rather than in an averaged sense. At $x_1=L/4$, $$2EA\,a_2+6EA\,a_3\left(\tfrac{L}{4}\right)+c\left(\tfrac{L}{4}\right)=0 \quad\Longrightarrow\quad 2EA\,a_2+\tfrac32 EA\,a_3L+\tfrac14 cL = 0 .$$
4. Impose collocation at $x=3L/4$. Likewise at $x_2=3L/4$, $$2EA\,a_2+6EA\,a_3\left(\tfrac{3L}{4}\right)+c\left(\tfrac{3L}{4}\right)=0 \quad\Longrightarrow\quad 2EA\,a_2+\tfrac92 EA\,a_3L+\tfrac34 cL = 0 .$$
5. Solve for $a_3$. Subtracting the first collocation equation from the second eliminates $a_2$: $$3EA\,a_3L+\tfrac12 cL = 0 \quad\Longrightarrow\quad a_3 = -\,\frac{c}{6EA}.$$
6. Back-substitute for $a_2$. Returning to the equation at $x=L/4$, $$2EA\,a_2+\tfrac32 EA\left(-\frac{c}{6EA}\right)L+\tfrac14 cL = 2EA\,a_2-\tfrac14 cL+\tfrac14 cL = 0 \quad\Longrightarrow\quad a_2 = 0 .$$ The quadratic term vanishes because the exact solution contains no $x^{2}$ contribution — a first hint that the trial family is rich enough to contain the true answer.
7. Apply the natural boundary condition. The remaining coefficient follows from the traction condition at the free end, $$EA\,\tilde u'(L)=EA\left(a_1+2a_2L+3a_3L^{2}\right)=P .$$ With $a_2=0$ and $a_3=-c/(6EA)$, $$EA\,a_1+3EA\left(-\frac{c}{6EA}\right)L^{2}=P \quad\Longrightarrow\quad EA\,a_1 = P+\frac{cL^{2}}{2} \quad\Longrightarrow\quad a_1 = \frac{P}{EA}+\frac{cL^{2}}{2EA}.$$ Physically $P+cL^{2}/2$ is the total axial force carried at the wall — the tip load plus the resultant $\int_0^L cx\,dx = cL^{2}/2$ of the distributed load — so $a_1$ is simply the strain at $x=0$.
8. Assemble the approximate solution. Substituting the three coefficients, $$\boxed{\;\tilde u(x)=\left(\frac{P}{EA}+\frac{cL^{2}}{2EA}\right)x-\frac{c}{6EA}x^{3} =\frac{Px}{EA}+\frac{c}{6EA}\left(3L^{2}x-x^{3}\right)\;}$$ with tip displacement $$\tilde u(L)=\frac{PL}{EA}+\frac{cL^{3}}{3EA}.$$
9. Verify against the exact solution. Integrating the governing equation directly, $EA\,u''=-cx$ gives $EA\,u'=-\tfrac{c}{2}x^{2}+C_1$; the natural boundary condition fixes $C_1=P+cL^{2}/2$, and $u(0)=0$ removes the second constant, so $$u_{\text{exact}}(x)=\frac{Px}{EA}+\frac{c}{6EA}\left(3L^{2}x-x^{3}\right),$$ which is identical to the collocation result. Substituting back, $R(x)=EA(-cx/EA)+cx=0$ for all $x$, so the residual vanishes identically rather than only at the two chosen stations.
10. Interpret the coincidence. The exact solution of $EA\,u''=-cx$ is a cubic in $x$; the trial family chosen is the complete set of cubics satisfying $u(0)=0$, so the exact solution lies inside the trial space. Any consistent weighted-residual method — collocation, sub-domain, least squares or Galerkin — must then recover it exactly, and the particular choice of collocation points is immaterial: $x=L/4$ and $3L/4$, or any other two distinct points, give the same coefficients. This is the finite element analogue of a $p$-refinement that has already captured the solution, and it is the reason the marks here go to the method rather than to the arithmetic. Had the distributed load been $q=cx^{2}$, the exact solution would be quartic, the cubic trial function would be genuinely approximate, and the answer would then depend on where the collocation points were placed.