Question 4 of 9: Castigliano's theorem — vertical deflection of an overhang tip

Question 4: Castigliano's theorem — vertical deflection of an overhang tip (18 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. Span $a = 3\ \text{m}$ from the pin A to the roller B; overhang $b = 2\ \text{m}$ from B to the free tip C; UDL $w = 12\ \text{kN/m}$ over the whole 5 m; $EI = 5000\ \text{kN}\cdot\text{m}^2$ constant.

Find. The vertical deflection $\delta_C$ of the free tip C, with its direction.

Approach. The beam is statically determinate, so Castigliano's first theorem can be applied directly: add a dummy vertical load $Q$ at C, express the bending moment throughout in terms of $Q$, differentiate, set $Q = 0$ and integrate. The trap is that $Q$ changes both reactions, so $\partial M/\partial Q$ is non-zero in the back span as well as on the overhang.

1. Reactions with the dummy load in place. With $Q$ acting downwards at C, moments about A give $$R_B = \frac{w(a+b)\frac{a+b}{2} + Q(a+b)}{a} = 50.0 + \frac{5}{3}Q ,\qquad R_A = w(a+b) + Q - R_B = 10.0 - \frac{2}{3}Q .$$ At $Q = 0$ these are $R_A = 10.0\ \text{kN}$ and $R_B = 50.0\ \text{kN}$, which already tells the story: the overhang is long enough that the pin is close to lifting.
2. Moment expressions and their derivatives. In the back span, with $x$ measured from A, $$M(x) = R_A x - \tfrac{1}{2}wx^{2},\qquad \frac{\partial M}{\partial Q} = \frac{\partial R_A}{\partial Q}\,x = -\frac{2}{3}x \quad (0 \le x \le 3).$$ On the overhang, with $s$ measured back from the free tip, $$M(s) = -\tfrac{1}{2}ws^{2} - Qs,\qquad \frac{\partial M}{\partial Q} = -s \quad (0 \le s \le 2).$$ The first of these is the step candidates most often miss: because the dummy load alters $R_A$, the back span contributes even though $Q$ is not applied there.
3. Apply Castigliano's first theorem. Setting $Q = 0$ after differentiating, $$\delta_C = \frac{1}{EI}\left[\int_0^{3}\!\left(10x - 6x^{2}\right) \left(-\tfrac{2}{3}x\right)\mathrm{d}x +\int_0^{2}\!\left(-6s^{2}\right)\left(-s\right)\mathrm{d}s\right].$$
4. Evaluate the two integrals. For the back span, $$-\tfrac{2}{3}\int_0^{3}\!\left(10x^{2} - 6x^{3}\right)\mathrm{d}x = -\tfrac{2}{3}\left[\frac{10(27)}{3} - \frac{6(81)}{4}\right] = -\tfrac{2}{3}\left(90 - 121.5\right) = +21.0\ \text{kN}^2\text{m}^3 .$$ For the overhang, $$6\int_0^{2} s^{3}\,\mathrm{d}s = 6\left(\frac{16}{4}\right) = +24.0\ \text{kN}^2\text{m}^3 .$$ Both terms are positive, so both act in the same sense as the dummy load: the tip goes down.
5. Deflection at C. Adding the two contributions, $$\delta_C = \frac{21.0 + 24.0}{EI} = \frac{45.0}{5000} = 9.00\times10^{-3}\ \text{m}$$ $$\boxed{\delta_C = 9.00\ \text{mm downward}}$$
6. Why the sign is worth checking. On a UDL-loaded overhanging beam the free tip does not always deflect down. The tip moves upward whenever the back span sags enough to rotate the beam over the roller faster than the overhang droops, and the changeover is at the universal ratio $b/a = 0.4343$, independent of $w$, of $a$ and of $EI$. Here $b/a = 2/3 = 0.667$ is well past that threshold, so the downward answer is the expected one — but on a shorter overhang the same arithmetic returns a negative number and it is not an error.
7. Supporting diagram values. The bending moment is hogging across the whole back span (there is no point of contraflexure between A and B other than the small sagging lobe near the pin): it rises to $M_{\max} = R_A^{2}/2w = +4.17\ \text{kN}\cdot\text{m}$ at $x = R_A/w = 0.833\ \text{m}$ from A, falls through zero and reaches $M_B = -wb^{2}/2 = -24.0\ \text{kN}\cdot\text{m}$ over the roller, then closes parabolically to zero at the free tip.