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

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. Three structures of uniform $EI$ with inextensible members.

Find. The shape of the shear force and bending moment diagrams for each structure, with the discontinuities, the points of zero shear and the points of contraflexure located correctly.

1. (a) Establish the shape rules before computing anything. The beam has three rollers plus a built-in end, so with five vertical/moment reaction components and two equations of equilibrium it is three times statically indeterminate. Whatever the redundants turn out to be, the diagrams must satisfy $$V=\frac{dM}{dx},\qquad \frac{dV}{dx}=-w .$$ Hence: the shear is linear under the distributed load on the overhang and constant everywhere else; it steps down by $P$ at each point load and steps up by the reaction at each support. The moment is parabolic on the overhang and piecewise linear elsewhere, is zero at the free tip, and is continuous at every support (there is no applied couple anywhere).
2. (a) Fix the ordinates. Solving the three-fold redundancy (illustrative case $w=P/L$, so $wL=P$) gives hogging peaks over the second and fourth supports, a sagging peak under the first point load and a small sagging moment at the built-in end because the last span carries no load and the fixed end must hold the beam down: $$M_{\text{sag,max}} \approx +0.205\,PL,\qquad M_{\text{hog}} \approx -0.113\,PL \text{ and } -0.158\,PL,$$ with $V_{\max}\approx+0.955P$ just right of the second support and $-1.045P$ just left of the third. The overhang moment is a pure cantilever parabola $-wx^{2}/2$ running from zero at the tip to $-wL^{2}/18$ over the first roller.

3. (b) Read the boundary conditions off the supports. The two outer bases are drawn with the “typical hinge” circle, so their bending moments are identically zero and the moment diagram in each outer column is a straight line from zero at the base to its value at the beam. The centre base is built in and takes a finite moment. The beam is inextensible, so all three column heads share one sway $\Delta$ and the storey-shear equation $$\sum_{\text{columns}} \frac{M_{\text{top}}+M_{\text{bottom}}}{h} = -P$$ closes the system.
4. (b) Interpret the resulting diagrams. Since $P$ may be moved along the inextensible beam, the loading is again anti-symmetric, so the two outer columns carry identical moments and the two beam spans carry equal and opposite end moments. Each column shear is constant, and the shear in each beam span is constant because no vertical load acts. Numerically (illustrative $P=1$, $L=1$, $h=1.5$) the centre column takes $0.548\,PL$ at its base and $0.490\,PL$ at its head, the outer columns take $0.231\,PL$ at their heads, and the beams reach $\pm0.245\,PL$ at the centre joint. The centre column carries no axial force — the signature of an anti-symmetric response in a member lying on the axis of symmetry — while the outer columns carry equal and opposite axial forces.

5. (c) Recognise that the imposed movement is a symmetric action. Only the relative displacement of the two bases matters, because superposing a rigid-body translation of the whole frame changes nothing. Moving the left base outward by $\Delta$ is therefore identical to spreading the two bases apart by $\Delta$, which is a symmetric action on a symmetric frame. Consequently the two column heads share the movement equally, $u_B=u_C=-\Delta/2$, the two columns bend in mirror image, and the two joint rotations are equal and opposite.
6. (c) Write one slope-deflection equation and close it. With $k_c=2EI/h$, $k_b=2EI/L$ and the chord rotation $\psi=\Delta/(2h)$ in each column, joint equilibrium at $B$ gives $k_c\!\left(2\theta-3\psi\right)+k_b\theta=0$, so $$\theta=\frac{3\Delta L}{2h\,(2L+h)} .$$ For the square portal drawn ($h=L$) this reduces to $\theta=\Delta/(2L)$ and $$\boxed{M_{\text{base}}=\frac{2EI\Delta}{h^{2}},\qquad M_{\text{column head}}=M_{\text{beam}}=\frac{EI\Delta}{h^{2}},\qquad V_{\text{column}}=\frac{3EI\Delta}{h^{3}} .}$$ The bending moment in the beam is the same at both ends and of the same sign, so the beam carries a constant moment and zero shear — the diagnostic feature that distinguishes a symmetric base spread from the anti-symmetric sway produced by a lateral load, where the beam moment reverses across the span and the beam shear is finite. Each column has one point of contraflexure, one third of the way up from the base.