Question 4 of 9: Least work — propped cantilever under a partial distributed load

Question 4: Least work — propped cantilever under a partial distributed load (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. A prismatic propped cantilever: built in at ①, roller at ③ 6 m away, with point ② marking mid-length. The uniformly distributed load of 6.4 kN/m covers only the first 3 m, ① to ②; the remaining 3 m is unloaded.

Find. The redundant reaction, then the shear-force and bending-moment diagrams from ① to ③ with their maximum and minimum ordinates labelled.

Approach. One degree indeterminate. Take the roller reaction R as the redundant, express the moment field in terms of R measured from the free (roller) end, and apply least work, \(\partial U/\partial R = \frac{1}{EI}\int M\,(\partial M/\partial R)\,\mathrm{d}s = 0\), which is simply the statement that the roller does not move.

1. Moment field measured from the roller. With s running back from ③, the unloaded half sees only R, and the loaded half picks up the distributed load once \(s\) passes \(L - d = 3\) m: $$M = Rs \quad (0 \le s \le L-d), \qquad M = Rs - \dfrac{w(s-(L-d))^{2}}{2} \quad (L-d \le s \le L)$$ In both branches \(\partial M/\partial R = s\), which is what makes this integral so cheap.
2. Apply least work. EI is constant and divides out: $$R\int_{0}^{L}s^{2}\,\mathrm{d}s = \dfrac{w}{2}\int_{L-d}^{L}\left(s-(L-d)\right)^{2}s\,\mathrm{d}s$$ Substituting \(u = s - (L-d)\) turns the right-hand integral into \(\int_{0}^{d}u^{2}(u + L - d)\,\mathrm{d}u = d^{4}/4 + (L-d)d^{3}/3\).
3. Close the form. Both sides collapse to a tidy result worth carrying away: $$R = \dfrac{w d^{3}}{2L^{3}}\left(L - \dfrac{d}{4}\right) = \dfrac{6.4 \times 27}{2 \times 216}\left(6 - 0.75\right) = \boxed{R_{3} = 2.10\ \text{kN}}$$ Note how little of the 19.2 kN total the prop attracts — the load is bunched at the stiff built-in end, so the far support does almost nothing.
4. Back-substitute for the built-in end. Vertical equilibrium gives \(R_{1} = wd - R_{3} = 17.10\) kN, and moments about ① give $$M_{1} = R_{3}L - wd\dfrac{d}{2} = 12.6 - 28.8 = \boxed{-16.20\ \text{kN}\cdot\text{m}}$$ (a hog, as a built-in end under a load close to it must be).
5. Shear diagram. The shear starts at +17.10 kN, falls linearly at 6.4 kN/m across the loaded 3 m to −2.10 kN, and then stays constant at −2.10 kN all the way to the roller, which returns it to zero. The constant tail is the tell that the load really does stop at ②.
6. Bending-moment diagram. The moment is parabolic over the loaded length and straight beyond it. Shear vanishes at \(x = 17.10/6.4 = 2.672\) m, giving $$M_{\max} = -16.20 + 17.10(2.672) - 3.2(2.672)^{2} = \boxed{+6.645\ \text{kN}\cdot\text{m}}$$ At point ② the moment has fallen back slightly to +6.30 kN·m, and from there it runs straight to zero at the roller with slope −2.10 kN. Contraflexure occurs at 1.231 m from the built-in end.