Question 7 of 8: Shear & Moment Diagrams (Overhang Beam)

Question 7: Shear & Moment Diagrams (Overhang Beam) (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.

Find. V(x) and M(x) along the whole beam, and the shear force and bending moment diagrams (using direct equilibrium, not superposition of existing solutions).

Approach. Find the reactions, then cut the beam in each of the two regions (A-B and B-C) and write V(x), M(x) directly from equilibrium of the cut segment, including the applied couple at A as a moment discontinuity.

1. Reactions. Summing moments about A (CCW+, including the applied couple): $$M_0 + R_B(8)-w(8)(4)-P_{end}(11)=0 \Rightarrow R_B=131.25\text{ kN}.$$ Then $$R_A=w(8)+P_{end}-R_B=68.75\text{ kN}.$$
2. Region 1 (0≤x≤8 m). $$V(x)=R_A-wx$$ $$M(x)=-M_0+R_A x-\dfrac{wx^2}{2}$$ (the $-M_0$ term is the step introduced by the applied CCW couple right at A: $M(0^+)=-30\text{ kN}\cdot\text{m}$).
3. Region 2 (8≤x≤11 m). $$V(x)=R_A-w(8)+R_B=40.0\text{ kN (constant)}$$ $$M(x)=-M_0+R_A x-w(8)(x-4)+R_B(x-8)$$
4. Key values. Zero-shear point in span A-B at $x_0=R_A/w=3437.500$ m, where the moment is maximum (sagging): $$\boxed{M_{max}=88.2\text{ kN}\cdot\text{m at }x=3.44\text{ m}}$$ At B, the overhang creates a large hogging moment: $$\boxed{M_B=-120\text{ kN}\cdot\text{m (hogging)}}$$ At the free end C, $M_C=0$ as required.
5. Diagrams. V(x) starts at R_A, decreases linearly to $x_0$ (crossing zero at the moment peak), continues linearly to just left of B, then jumps up by R_B and stays constant (=40 kN, matching the tip load) out to C. M(x) starts at $-M_0$, rises through the sagging peak, falls to the large hogging value at B, then returns linearly to zero at C.