Question 1 of 8: Schematic Shear Force and Bending Moment Diagrams

Question 1: Schematic Shear Force and Bending Moment Diagrams (8 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. Two structures of uniform, inextensible members. (a) Overhang \(L/2\), spans \(L,\;L,\;L\); rollers at supports 1, 2 and 3, built-in at support 4; \(w\) acting from the free tip to support 2 and \(2w\) acting over the last span. (b) A two-bay top beam of span \(L\) each side, carried by three columns of height \(h\) whose bases are hinged; the only action is a downward settlement \(\delta\) of the centre support B.

Find. The shape of the shear force and bending moment diagrams for each structure, with the sign of every region and the location of every point of contraflexure identified.

Approach. Structure (a) is a beam three degrees statically indeterminate, so obtain the four support moments by slope-deflection (or three-moment) and let statics give the shears; structure (b) has no load at all, so its diagram shapes follow from the imposed geometry alone — symmetry kills the centre-column moment and reduces the problem to one span with a rotational spring.

1. Structure (a): reduce the overhang to a known end action. The overhang carries \(w\) over \(L/2\) and is determinate, so it hands support 1 a known hogging moment and a known shear:$$M_1 = -\frac{w(L/2)^{2}}{2} = \boxed{-0.1250\,wL^{2}},\qquad V_{1^-} = -\tfrac{1}{2}wL$$Everything to the right of support 1 is then a three-span problem with a known applied end moment.
2. Solve the continuous beam. Writing slope-deflection for spans 1–2 (loaded with \(w\)), 2–3 (unloaded) and 3–4 (loaded with \(2w\), built in at 4), enforcing moment equilibrium at joints 1, 2 and 3 and solving the three simultaneous equations gives the support moments$$M_2 = -\tfrac{3}{208}wL^{2},\qquad M_3 = -\tfrac{7}{104}wL^{2},\qquad M_4 = \boxed{-\tfrac{45}{208}\,wL^{2} = -0.2163\,wL^{2}}$$all three hogging, as expected over interior supports and at a built-in end.
3. Recover the reactions from statics. Taking each span as a free body with its end moments known,$$R_1 = \tfrac{231}{208}wL = 1.1106\,wL,\qquad R_2 = \tfrac{35}{104}wL = 0.3365\,wL$$$$R_3 = \tfrac{47}{52}wL = 0.9038\,wL,\qquad R_4 = \tfrac{239}{208}wL = 1.1490\,wL$$and the four add to \(3.5\,wL\), which is exactly the total applied load \(w(1.5L)+2w(L)\) — the arithmetic check that must be made before any diagram is drawn.
4. Shear force diagram. The shear falls linearly under each loaded length and is constant over the unloaded middle span. Reading from the left tip: \(0\) to \(-0.500\,wL\) over the overhang, a jump to \(+0.611\,wL\) at support 1, down to \(-0.389\,wL\) at support 2, a constant \(-0.053\,wL\) across the unloaded span, then \(+0.851\,wL\) falling to \(-1.149\,wL\) at the wall.
Question 1(a) — shear force diagram; ordinates are multiples of wL.
5. Bending moment diagram. The moment is parabolic under each uniform load and straight over the unloaded span. It is hogging at every support, and sags only in the first and last spans, where$$M_{\max,\,1\text{-}2} = \boxed{+0.0614\,wL^{2}}\ \text{at }0.611L\ \text{right of support 1}$$$$M_{\max,\,3\text{-}4} = \boxed{+0.1137\,wL^{2}}\ \text{at }0.425L\ \text{right of support 3}$$The whole of the unloaded middle span hogs, so there is no point of contraflexure between supports 2 and 3 — a shape worth noticing, because a sketch drawn from habit usually puts a sagging bulge there.
Question 1(a) — bending moment diagram, sagging positive; ordinates are multiples of wL².
6. Structure (b): use symmetry before writing anything. The frame and the imposed displacement are both symmetric about column B, so there is no sway, joint B does not rotate, and column B translates vertically without any chord rotation. A column with \(\theta = 0\) at its head and no chord rotation carries no moment and no shear at all; the settlement is resisted entirely by bending of the beam and of the two outer columns.
7. Reduce to one span with a rotational spring. Each outer column is hinged at its base and rigid at its head, so it restrains the beam end with a spring \(3EI/h\). Writing slope-deflection for span A–B with \(\psi = \delta/L\), \(\theta_B = 0\) and joint A balanced against the spring gives, with \(r = h/L\),$$M_{\text{sag}}(A) = \boxed{-\frac{18\,EI\delta}{L^{2}(4r+3)}},\qquad M_{\text{sag}}(B) = \boxed{+\frac{6(2r+3)\,EI\delta}{L^{2}(4r+3)}}$$For the drawn proportions \(h \approx L\) these are \(-2.571\,EI\delta/L^{2}\) and \(+4.286\,EI\delta/L^{2}\). The two limits confirm the algebra: \(r \to 0\) (stiff short columns, beam ends effectively built in) returns the classical \(\mp 6EI\delta/L^{2}\), and \(r \to \infty\) (very flexible columns, beam ends effectively pinned) returns the propped-cantilever value \(3EI\delta/L^{2}\).
8. Shapes for structure (b). With no span load the beam moment is a straight line in each half: hogging at A, crossing zero, and sagging at B with the larger ordinate. The shear is constant in each half and reverses sign across the centre. Each outer column carries a moment growing linearly from zero at its hinged base to \(18EI\delta/[L^{2}(4r+3)]\) at its head, with constant shear equal to that head moment divided by \(h\); the centre column is completely unstressed in bending and simply pushes the settlement into the beam.