Question 1 of 9: Schematic shear-force and bending-moment diagrams

Question 1: Schematic shear-force and bending-moment diagrams (12 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. Three structures of uniform EI with inextensible members; loads and support types as drawn, no dimensions printed. The representative dimensions adopted are listed below.

Find. The shear-force and bending-moment diagrams of each structure, with the correct zeros, discontinuities and signs, and the ordinates that follow from the adopted dimensions.

Approach. Part (a) is a three-times indeterminate continuous beam: fix the determinate overhang first, then solve the remaining compatibility by the stiffness (or moment-distribution) method; part (b) is a two-pinned portal, one degree indeterminate, whose thrust follows from a single unit-load compatibility equation; part (c) is symmetric under a symmetric imposed displacement, which kills the centre column entirely and leaves one rotation.

1. Part (a): fix what statics already gives. The overhang beyond the first roller is a determinate cantilever, so its moment is known without any analysis: $$M_{A} = -\dfrac{w a^{2}}{2} = -\dfrac{12 \times 3^{2}}{2} = \boxed{-54.00\ \text{kN}\cdot\text{m}}$$ and the shear rises linearly from zero at the tip to −36.00 kN just left of the first roller. Nothing about the redundancy changes these two numbers — they are the first marks on the page.
2. Solve the redundant supports. With three roller reactions and a built-in end the beam is three degrees indeterminate. Assembling the slope-deflection equations for the four spans and enforcing moment equilibrium at the three interior joints gives the support reactions

whose sum, 168.0 kN, equals the total applied load \(w(a+L) + P = 12 \times 9 + 60\). That check costs one line and catches almost every arithmetic slip in a beam of this size.

3. Build the shear diagram from the reactions. Shear jumps by the reaction at every support and by −P under the point load, and it slopes at −w only where the distributed load acts. The distributed load stops at the second roller, so the shear is constant through the third span — a flat step that a hand sketch very often gets wrong.
4. Locate the sagging peak. In the loaded span the shear crosses zero at \(x = 43.44/12 = 3.62\) m from the first roller, and there $$M_{\max} = \boxed{+24.63\ \text{kN}\cdot\text{m}}$$
5. Read the remaining ordinates. The moment is −9.35 kN·m over the second roller, −16.62 kN·m over the third, +52.10 kN·m under the point load and $$M_{D} = \boxed{-59.19\ \text{kN}\cdot\text{m}}$$ at the built-in end. Between the second and third rollers the moment is a straight line (no load there), and it is straight again on each side of the point load, where it peaks with a kink.

6. Part (b): identify the redundancy. Two hinged bases give four reaction components against three equations of statics, so the frame is one degree indeterminate. Release the horizontal restraint at the right base and take the thrust X there as the redundant; the primary structure (a pin and a vertical roller) is then determinate.
7. Write the two unit-load integrals. With uniform EI the factor \(1/EI\) is common to both integrals and cancels out of the answer, which is why the question can be posed without giving EI: $$f_{11} = \int \dfrac{m_{1}^{2}}{EI}\,\mathrm{d}s = \dfrac{1}{EI}\left(2\dfrac{h^{3}}{3} + h^{2}L\right) = \dfrac{360}{EI}$$ $$\Delta_{10} = \int \dfrac{M_{0} m_{1}}{EI}\,\mathrm{d}s = \dfrac{3240 + 3888}{EI} = \dfrac{7128}{EI}$$ where 3240 is the loaded-column contribution and 3888 the beam contribution (the unloaded column carries no primary moment at all).
8. Enforce compatibility. The horizontal movement of the released base must vanish: $$\Delta_{10} + X f_{11} = 0 \;\Rightarrow\; X = -\dfrac{7128}{360} = \boxed{-19.80\ \text{kN}}$$ so the far base pushes back with 19.80 kN and the loaded base takes the balance, \(52.20\) kN, of the \(wh = 72\) kN total.
9. Get the ordinates. The column moments at the beam are $$M_{2} = \boxed{97.20\ \text{kN}\cdot\text{m}},\qquad M_{3} = \boxed{118.80\ \text{kN}\cdot\text{m}}$$ The loaded column carries a parabolic moment that peaks where its shear vanishes, at \(52.20/12 = 4.35\) m above the base, with \(M_{\max} = 113.54\) kN·m. The beam has no transverse load, so its moment runs straight from one joint value to the other and its shear is the constant \((97.20 + 118.80)/6 = 36.00\) kN — the same 36 kN couple that appears as equal and opposite vertical reactions at the two bases.

10. Part (c): use the symmetry before writing any equation. The structure is symmetric and so is the imposed displacement, so the rotation of the centre joint B is exactly zero. The centre column therefore has zero rotation at both ends and no chord rotation (the beams are inextensible, so nothing moves horizontally): it carries no moment and no shear at all. That single observation reduces the whole frame to one span with a rotational spring at its outer end.
11. One joint equation. With the outer column pinned at its base its modified stiffness is \(3EI/h\), and the beam sees a chord rotation \(\psi = -\delta/L\). Joint A gives $$\dfrac{3EI}{h}\theta_{A} + \dfrac{2EI}{L}\left(2\theta_{A} + \dfrac{3\delta}{L}\right) = 0 \;\Rightarrow\; \theta_{A} = -\dfrac{6\delta h}{L(3L + 4h)} = -2.857 \times 10^{-3}\ \text{rad}$$
12. Close the form. Substituting back, with \(r = h/L\), $$M_{A} = \dfrac{18EI\delta}{L^{2}(3 + 4r)} = \boxed{28.57\ \text{kN}\cdot\text{m}},\qquad M_{B} = \dfrac{6(3 + 2r)EI\delta}{L^{2}(3 + 4r)} = \boxed{47.62\ \text{kN}\cdot\text{m}}$$ The two limits are worth remembering: as \(r \to 0\) these tend to the fixed-end values \(6EI\delta/L^{2}\), and as \(r \to \infty\) to the propped value \(3EI\delta/L^{2}\). The beam shear is \((28.57 + 47.62)/6 = 12.70\) kN and each outer column carries a constant shear of \(28.57/6 = 4.76\) kN.