Question 6 of 8: Beam with an Inclined Cable — Combined Axial + Bending

Question 6: Beam with an Inclined Cable — Combined Axial + Bending (20 marks, 4 bonus)

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. Normal-stress distribution (max/min) and maximum shear stress at x=1 m from A; the plastic moment capacity.

Approach. Because the cable is inclined, it pulls B both up and toward the wall; the pin at A must react that horizontal pull, which puts the whole beam into constant axial tension — so this is a combined axial-plus-bending problem, not pure bending. Solve reactions (including the cable's horizontal component), get the section forces at x=1 m, then superpose $\sigma = N/A \pm Mc/I$.

1. Cable tension and reactions. With $\cos\alpha=1.5/2.5=0.6,\ \sin\alpha=0.8$, moments about A ($w L\cdot L/2$ balanced by the cable's vertical component at B): $$T\sin\alpha\,L = wL\cdot\frac{L}{2} \;\Rightarrow\; T = \frac{60(4)/2}{0.8} = 150\ \text{kN}$$ $$T_x=90\ \text{kN},\quad T_y=120\ \text{kN},\quad A_y = wL-T_y = 240-120=120\ \text{kN}$$
2. Axial force in the beam. The pin reacts the cable's horizontal pull, $A_x=-T_x=-90$ kN; tracing this through the beam (cable pulls B toward the wall, pin holds A back) puts the ENTIRE beam in constant axial tension: $$\boxed{N = 90\ \text{kN (tension), constant along AB}}$$
3. Section forces at x = 1 m (only the UDL and $A_y$ act to the left of the cut): $$M(1\text{m}) = A_y(1000)-\frac{w(1000)^2}{2} = 120{,}000(1000)-\frac{60(1000)^2}{2}=90.0\times10^6\ \text{N}\cdot\text{mm}$$ $$V(1\text{m}) = A_y - w(1000) = 120-60=60\ \text{kN}$$
4. Section properties. $$A=2(100)(15)+(300-30)(20)=8400\ \text{mm}^2\qquad I=\frac{100(300)^3}{12}-\frac{80(270)^3}{12}=9.378\times10^7\ \text{mm}^4,\ c=150\ \text{mm}$$
5. Superpose axial and bending stress ($N/A$ tension is uniform; the sagging moment puts the top in compression, the bottom in tension): $$\sigma_{top} = \frac{N}{A}-\frac{Mc}{I} = 10.71-143.96 = \boxed{-133.2\ \text{MPa (compression, top)}}$$ $$\sigma_{bot} = \frac{N}{A}+\frac{Mc}{I} = 10.71+143.96 = \boxed{+154.7\ \text{MPa (tension, bottom)}}$$ Both stay under the 350 MPa yield.
6. Maximum shear stress at the neutral axis, $Q_{NA}=1500(142.5)+2700(67.5)=3.96\times10^5\ \text{mm}^3$: $$\boxed{\tau_{max} = \frac{VQ_{NA}}{I\,t_w} = \frac{60{,}000(3.96\times10^5)}{9.378\times10^7(20)} = 12.67\ \text{MPa}} \quad(\ll 60\ \text{MPa yield})$$
7. Bonus (c): plastic moment capacity of the section, $Z=b_ft_f(d-t_f)+t_w(d-2t_f)^2/4$: $$Z = 100(15)(285)+20\frac{(270)^2}{4}=792{,}000\ \text{mm}^3 \;\Rightarrow\; \boxed{M_p = \sigma_Y Z = 350(792{,}000) = 277.2\ \text{kN}\cdot\text{m}}$$