Question 2 of 9: Schematic shear force and bending moment diagrams

Question 2: Schematic shear force and bending moment diagrams (11 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) A beam built in at A, carried on rollers at B and C and overhanging past C to a free end D, with a uniformly distributed load $w$ acting only between the two rollers and a point load $P$ at the free end. (b) A single-bay portal with both column bases built in, unloaded, in which every member elongates because of a uniform rise in temperature. No dimensions or load magnitudes are printed — the question asks for the shapes of the diagrams, not for ordinates.

Find. The shear force and bending moment diagrams of both structures, drawn schematically with the correct curvature, discontinuities and points of contraflexure.

Approach. For (a) read the diagram shapes straight off the loading — constant shear and linear moment where there is no load, linear shear and parabolic moment under the UDL, a jump in shear at every support — and fix the two ends from statics. For (b) recognise that a uniform temperature rise is a prescribed deformation, not a load: the free thermal elongation of the beam is imposed on the frame and the columns resist it, so the whole problem reduces to a symmetric sway.

1. Part (a) — classify the three segments. Segment AB carries no transverse load, so its shear is constant and its moment varies linearly. Segment BC carries the UDL, so its shear falls linearly at $-w$ per metre and its moment is a parabola opening downwards. Segment CD carries only the tip load, so its shear is constant at $P$ and its moment is linear.
2. Part (a) — fix the two ends exactly. The overhang is determinate, so cutting just inside C gives, for an overhang of length $a$, $$\boxed{M_C = -P\,a}$$ which is a hogging moment and is the one ordinate that does not depend on the (unstated) spans. At the built-in end A there is no load between A and B, so the moment there is the carry-over from B on a propped cantilever, $M_A = \tfrac{1}{2}M_B$ in the slope-deflection sense; because the carry-over reverses the physical curvature, A is in sagging while B is in hogging. That sign reversal is the feature the examiner is looking for.
3. Part (a) — assemble the diagrams. The shear steps up at each roller by that roller's reaction and is continuous elsewhere; it crosses zero once inside span BC, and that crossing locates the single sagging peak of the parabola. The moment diagram is therefore: a short sagging region from A, crossing to hogging before B; a hogging peak at B; a sagging hump inside BC; a hogging peak at C of magnitude $P\,a$; and a straight run back to zero at the free end D. Two points of contraflexure appear in span BC and one in span AB.
4. Part (b) — recognise what a temperature rise does. The columns elongate freely, because nothing holds the beam down; they simply lift it, and no force is generated. The beam is the member that matters: its free elongation $\delta_T = \alpha\,\Delta T\,L$ must be accommodated while the two built-in bases stay a fixed distance apart, so each column top is pushed outwards by $$\Delta = \tfrac{1}{2}\delta_T$$ and the entire response is the frame's resistance to that prescribed sway.
5. Part (b) — deduce the shapes from symmetry. The structure and the imposed deformation are mirror-symmetric, so the two beam-end rotations are equal and opposite. Writing the beam slope-deflection equations with $\theta_3 = -\theta_2$ gives $$M_{23} = \frac{2EI}{L}(2\theta_2 + \theta_3) = \frac{2EI}{L}\theta_2, \qquad M_{32} = -\frac{2EI}{L}\theta_2$$ so the two end moments are equal and opposite in the slope-deflection sense, which means the beam carries a constant bending moment and therefore zero shear. The joint condition $M_{21} + M_{23} = 0$ then gives $$\theta_2 = \frac{3\,\Delta\,L}{h\,(2L + h)}$$
6. Part (b) — draw them. Each column has no transverse load, so its moment is linear; substituting $\theta_2$ back shows both column end moments carry the same sign, so the column bends in single curvature with no point of contraflexure, the base moment being the larger. The column shear is constant and equals the axial thrust delivered along the beam, so the beam is in axial compression while carrying constant moment and no shear.
7. Part (b) — sanity check the whole picture. Global horizontal equilibrium is satisfied because the two column shears are equal and opposite; global vertical equilibrium is trivial because nothing is loaded vertically; and if the beam were released axially every internal force would vanish, which is the correct limiting behaviour of a statically determinate frame under a uniform temperature change.