Question 8 of 8: Shear-Force and Bending-Moment Diagrams

Question 8: Shear-Force and Bending-Moment Diagrams (20 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. The identical beam and loading as Question 1: span 8 m, UDL 20 kN/m over the full span, 500 kN·m clockwise couple applied at B; R_A=17.5 kN, R_B=142.5 kN (from Question 1's reaction solution).

Find. V(x) and M(x) over 0≤x≤8 m, and the resulting diagrams.

Approach. Cut the beam at a general x, sum forces/moments on the left segment; since the applied couple sits exactly at the right support, it enters only through the reactions, so a single pair of expressions covers the whole span.

1. Shear force. Only $R_A$ and the UDL act to the left of any interior cut: $$\boxed{V(x) = R_A - wx = 17.5 - 20x\ \text{kN}\quad(x\text{ in m},\ 0\le x\le 8)}$$ $V(0)=+17.5$ kN; $V(8^-)=17.5-160=-142.5$ kN, jumping back to 0 at $x=8^+$ as $R_B$ is added.
2. Bending moment. Integrating (or summing moments of the left segment about the cut): $$\boxed{M(x) = R_A x - \frac{wx^2}{2} = 17.5x - 10x^2\ \text{kN}\cdot\text{m}\quad(0\le x\le 8)}$$
3. Critical points. $V(x)=0$ at $x_0=R_A/w=17.5/20=0.875$ m, a LOCAL peak: $$M(0.875) = 17.5(0.875)-10(0.875)^2 = 7.66\ \text{kN}\cdot\text{m}$$ At the right end, $$M(8) = 17.5(8)-10(8)^2 = -500\ \text{kN}\cdot\text{m}$$ This large negative (hogging) value at B is exactly balanced by the applied 500 kN·m clockwise couple acting right at that same point, bringing the moment back to zero beyond the support — it is not a discontinuity error, it is the beam's response to carrying that end couple.