Collocation solution for a cantilevered bar under a linearly varying axial load [20 marks]: 22-Mec-B10 Finite Element Analysis, December 2013, Q4

Question 4: Collocation solution for a cantilevered 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 bar fixed at $x=0$ and free at $x=L$, carrying a distributed axial load that grows linearly from the wall:

Given data — Question 4
Quantity	Value / expression
Governing equation	$EA\,u''(x)+cx=0$, $0\lt x\lt L$
Distributed axial load	$q(x)=cx$ (force per unit length), $c$ constant
Essential boundary condition	$u(0)=0$ (fixed end)
Natural boundary condition	$EA\,u'(L)=0$ (free end, zero axial force)
Trial function	Cubic polynomial in $x$
Collocation points	$x=\tfrac{1}{3}L$ and $x=\tfrac{2}{3}L$

Find. The approximate cubic displacement field $u_a(x)$ produced by the collocation form of the method of weighted residuals, and a statement of how it compares with the exact solution.

Figure 4.1 — Cantilevered bar of length L fixed at the left end, carrying the linearly varying axial distributed load q(x) = cx. The two green markers are the collocation (evaluation) points at x = L/3 and x = 2L/3.

Approach. Choose a cubic trial function that satisfies the essential boundary condition identically, form the residual of the governing differential equation, set that residual to zero at the two specified evaluation points, and close the system with the natural boundary condition — three equations for the three remaining coefficients.

Select an admissible cubic trial function. The essential (geometric) boundary condition $u(0)=0$ must be satisfied exactly by any admissible trial function, so drop the constant term and write $$u_a(x)=a_1x+a_2x^{2}+a_3x^{3}$$ which gives $u_a(0)=0$ for any coefficients. Three unknowns $a_1,a_2,a_3$ remain, so three independent conditions are required.
Form the residual. Differentiating twice, $u_a''(x)=2a_2+6a_3x$, and substituting into the governing equation gives the residual $$\mathcal{R}(x)=EA\,u_a''(x)+cx = EA\big(2a_2+6a_3x\big)+cx$$ The exact solution would make $\mathcal{R}$ vanish for all $x$; the collocation method instead forces it to vanish at selected points. Equivalently, the weight functions are Dirac delta functions, $w_i(x)=\delta(x-x_i)$, so that $\int_0^L \delta(x-x_i)\,\mathcal{R}(x)\,dx=\mathcal{R}(x_i)=0$.
Apply the first collocation condition at $x=L/3$. Substituting $x=L/3$ into the residual and setting it to zero, $$EA\left(2a_2+6a_3\frac{L}{3}\right)+\frac{cL}{3}=0 \quad\Longrightarrow\quad 2EA\,a_2+2EA\,L\,a_3=-\frac{cL}{3}$$
Apply the second collocation condition at $x=2L/3$. Likewise at $x=2L/3$, $$EA\left(2a_2+6a_3\frac{2L}{3}\right)+\frac{2cL}{3}=0 \quad\Longrightarrow\quad 2EA\,a_2+4EA\,L\,a_3=-\frac{2cL}{3}$$
Solve the two collocation equations. Subtracting the third-step equation from the fourth eliminates $a_2$ and leaves $2EA\,L\,a_3=-cL/3$, hence $$a_3=-\frac{c}{6EA}$$ Back-substituting into the first collocation equation, $2EA\,a_2 + 2EA\,L\left(-\dfrac{c}{6EA}\right) = -\dfrac{cL}{3}$, so $2EA\,a_2-\dfrac{cL}{3}=-\dfrac{cL}{3}$ and therefore $$a_2=0$$
Close the system with the natural boundary condition. The traction-free end requires $EA\,u_a'(L)=0$, i.e. $u_a'(L)=a_1+2a_2L+3a_3L^{2}=0$. With $a_2=0$ and $a_3=-c/(6EA)$, $$a_1+3L^{2}\left(-\frac{c}{6EA}\right)=0 \quad\Longrightarrow\quad a_1=\frac{cL^{2}}{2EA}$$
Assemble the approximate solution. Collecting the three coefficients, $$\boxed{\,u_a(x)=\frac{cL^{2}}{2EA}\,x-\frac{c}{6EA}\,x^{3}=\frac{c}{6EA}\Big(3L^{2}x-x^{3}\Big)\,}$$
Check the residual and compare with the exact solution. Substituting the coefficients back, the residual is $$\mathcal{R}(x)=EA\big(2(0)+6x\big)\left(-\frac{c}{6EA}\right)+cx=-cx+cx=0 \quad \text{for all } x$$ so the residual vanishes identically, not merely at the two collocation points. Integrating the governing equation directly, $u''=-cx/(EA)$ gives $u'=-cx^{2}/(2EA)+K_1$; the condition $u'(L)=0$ fixes $K_1=cL^{2}/(2EA)$, and $u(0)=0$ fixes the second constant at zero, so $$u_{\text{exact}}(x)=\frac{c}{6EA}\Big(3L^{2}x-x^{3}\Big)$$ The collocation cubic is exactly the analytical solution. The tip displacement follows as $u_a(L)=\dfrac{c}{6EA}(3L^{3}-L^{3})=\dfrac{cL^{3}}{3EA}$.

The reason the approximation is exact deserves a sentence, because it is the pedagogical point of the question rather than a coincidence. The exact solution of $EA\,u''+cx=0$ is a cubic polynomial, so the chosen trial space contains the exact solution. Any consistent weighted-residual method — collocation, subdomain, least squares or Galerkin — must then recover it, because the only member of the trial space with an identically zero residual and the correct boundary conditions is the exact solution itself. The choice of the collocation points at $L/3$ and $2L/3$ is therefore immaterial to the answer here: any two distinct interior points give the same coefficients. On a problem whose exact solution is not a cubic, the choice of collocation points would matter a great deal, and collocation would be noticeably less accurate than Galerkin for the same trial space because it enforces the differential equation at isolated points rather than in an averaged sense.

Question 4 — final results
Quantity	Result
Trial function	$u_a=a_1x+a_2x^{2}+a_3x^{3}$ (satisfies $u(0)=0$)
Residual	$\mathcal{R}(x)=EA(2a_2+6a_3x)+cx$
$a_1$	$\dfrac{cL^{2}}{2EA}$
$a_2$	$0$
$a_3$	$-\dfrac{c}{6EA}$
Approximate solution	$u_a(x)=\dfrac{c}{6EA}\big(3L^{2}x-x^{3}\big)$
Comparison with exact	Identical — residual vanishes for all $x$
Tip displacement	$u_a(L)=\dfrac{cL^{3}}{3EA}$

Topic. The collocation method as a member of the weighted-residual family, applied to a second-order axial bar equation with one essential and one natural boundary condition.

Key relations. Residual $\mathcal{R}=\mathcal{L}(u_a)-f$; weighted-residual statement $\int_\Omega w_i\mathcal{R}\,d\Omega=0$; collocation weights $w_i=\delta(x-x_i)$, giving $\mathcal{R}(x_i)=0$; number of conditions = number of undetermined coefficients.

Why this works. Every weighted-residual method makes the same bargain: an approximation cannot satisfy the differential equation everywhere, so it is required instead to satisfy a finite number of weighted averages of the error. Collocation makes the most literal choice of weight — a spike at a point — which makes it the cheapest of the family to apply by hand, since no integration is needed at all. The price is that the residual between collocation points is uncontrolled, and the resulting equations are not symmetric, so collocation is rarely used to build element stiffness matrices. It remains valuable as a teaching device and in spectral or meshless methods where the trial functions are globally smooth.

Common pitfall. Using only the two collocation equations and stopping there, leaving a one-parameter family of solutions. A cubic trial function that already satisfies the essential condition still has three free coefficients, so a third equation — here the natural boundary condition $EA\,u'(L)=0$ — is mandatory. The mirror-image error is to force the natural condition into the trial function as well and then have too few free coefficients for the two collocation points.

Source. Reddy, An Introduction to the Finite Element Method, 4th ed., §2.5.2 (the collocation method) and Example 2.5.1; Logan, 6th ed., §3.12 (residual methods and comparison of collocation, subdomain, least squares and Galerkin).

Question 4 of 7: Collocation solution for a cantilevered bar under a linearly varying axial load [20 marks]

Question 4: Collocation solution for a cantilevered bar under a linearly varying axial load [20 marks]