Question 2 of 7: Bar under combined axial, torsional and transverse load

Question 2: Bar under combined axial, torsional and transverse load

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. A cantilever bar; the transverse load $F$ produces bending at the critical section 400 mm from the free end, while $P$ and $T$ scale with $F$.

Find. The largest transverse load $F$ satisfying both the maximum-normal-stress and maximum-shear-stress limits.

Approach. Build the normal stress (axial + bending) and the torsional shear at the critical outer fibre, then impose the maximum-normal-stress theory ($\sigma_1\le85$) and the maximum-shear-stress theory ($\tau_{\max}\le45$); the smaller $F$ governs.

1. Section properties. For $d=60$ mm, $$A=\tfrac{\pi}{4}d^2=2827\ \text{mm}^2,\quad S=\tfrac{\pi}{32}d^3=2.121\times10^{4}\ \text{mm}^3,\quad Z_p=\tfrac{\pi}{16}d^3=4.241\times10^{4}\ \text{mm}^3$$
2. Normal stress at the outer fibre (axial tension plus bending, both maximal on the same fibre): $$\sigma=\frac{P}{A}+\frac{M}{S}=\frac{15F}{2827}+\frac{400F}{2.121\times10^{4}}=\left(0.005305+0.018863\right)F=0.024168\,F\ \text{MPa}$$
3. Torsional shear at the surface. Transverse (direct) shear is zero at that outer fibre, so $$\tau=\frac{T}{Z_p}=\frac{200F}{4.241\times10^{4}}=0.004716\,F\ \text{MPa}$$
4. Maximum-shear-stress theory. $\displaystyle\tau_{\max}=\sqrt{\left(\tfrac{\sigma}{2}\right)^2+\tau^2}=0.012972\,F\le45$ gives $$F=\frac{45}{0.012972}=3469\ \text{N}$$
5. Maximum-normal-stress theory. $\displaystyle\sigma_1=\frac{\sigma}{2}+\tau_{\max}=0.025056\,F\le85$ gives $$F=\frac{85}{0.025056}=3392\ \text{N}$$
6. Governing load. The smaller value controls: $$\boxed{F\approx3.39\ \text{kN}}$$ (the tensile limit governs). At this load the direct transverse shear at the neutral axis, $\tfrac{4}{3}F/A\approx1.6$ MPa, is negligible against the 45 MPa limit.