Question 2 of 9: Schematic shear force and bending moment diagrams (12 marks)

Question 2: 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. Structure (a) is a beam with an $L/4$ overhang at each end carrying a tip load $P=wL/4$, a pin at A, a roller at B one span $L$ away with a UDL $w$ over that span only, and a rigid joint C a further $L$ along, propped by a fixed-base column of height $L/2$. Structure (b) is a symmetric portal on fixed bases whose beam is lifted at mid-span by a jack; no dimensions are printed on it.

Find. The shape of the shear force and bending moment diagrams for both structures, with the maximum and minimum ordinates marked.

Approach. Show first that neither structure can sway, so each reduces to a small slope-deflection problem in joint rotations alone; the ordinates then follow from statics on each member.

1. (a) Show that nothing translates. The beam is horizontal and inextensible and A is a pin, so $u_{C}=u_{A}=0$; the column is vertical and inextensible on a fixed base, so $v_{C}=0$. With the roller holding B the only unknowns are $\theta_{A},\theta_{B},\theta_{C}$, and the frame behaves as a continuous beam carrying a rotational spring $4EI/(L/2)=8EI/L$ at C.
2. (a) Remove the overhangs as determinate cantilevers. Each delivers a known end moment to the beam line: $$M_{\text{overhang}}=-P\,\frac{L}{4}=-\frac{wL}{4}\cdot\frac{L}{4}=-\frac{wL^{2}}{16}$$ that is, $wL^{2}/16$ hogging at A and again at C.
3. (a) Solve the three joint equations. Writing $M_{ij}=(2EI/L_{ij})(2\theta_{i}+\theta_{j})+\mathrm{FEM}_{ij}$ for the four members and imposing $\sum M_{ij}=0$ at A, B and C gives end moments that are exact fractions of $wL^{2}$: $$M_{\text{sag}}(A)=-\frac{wL^{2}}{16},\qquad M_{\text{sag}}(B)=-\frac{3wL^{2}}{64},\qquad M_{\text{sag}}(C)^{-}=0$$
4. (a) Locate the peak sagging moment. The shear in span A–B starts at $33wL/64$ and falls at rate $w$, so it vanishes $33L/64$ from A and $$M_{\max}=-\frac{wL^{2}}{16}+\frac{33wL}{64}\cdot\frac{33L}{64}-\frac{w}{2}\left(\frac{33L}{64}\right)^{2}=\boxed{0.0704\,wL^{2}}$$

One feature of structure (a) deserves to be pointed out rather than smoothed over. The bending moment in span B–C falls to exactly zero at joint C, because the column absorbs the whole $wL^{2}/16$ delivered by the right-hand overhang. The diagram therefore steps at C from zero on the beam side to $wL^{2}/16$ hogging on the overhang side, and that step is the column moment — a diagram drawn without it will not satisfy joint equilibrium. The column also pushes horizontally, so the pin at A carries a $3wL/16$ horizontal reaction even though every applied load is vertical.

Structure (b) is easier, because a jack under mid-span is statically equivalent to an upward force $P$ there, and both structure and load are symmetric.

5. (b) Reduce by symmetry. Symmetry forbids sway and gives $\theta_{C}=-\theta_{B}$, so the beam term collapses to $2EI\theta_{B}/L$ and the single joint equation at B reads $$\left(\frac{4EI}{h}+\frac{2EI}{L}\right)\theta_{B}=\frac{PL}{8}$$
6. (b) Recover the ordinates. Substituting back, $$M_{BA}=\boxed{\frac{PL^{2}}{4(2L+h)}},\qquad M_{AB}=\frac{M_{BA}}{2},\qquad M_{\text{mid}}=\frac{PL}{4}-M_{BA}$$ with column shear $V=(M_{AB}+M_{BA})/h=3PL^{2}/[8h(2L+h)]$ and beam shear $P/2$ either side of the jack.

No dimensions are printed on structure (b), so the closed forms above are the real answer. The numerical illustration on the sketch adopts the drawn proportions $L=6$ m and $h=4$ m; the shapes, the zeros and the point of contraflexure at one third of the column height above the base do not depend on that choice at all.