Question 7 of 8: Square bar under axial load plus torque

Question 7: Square bar under axial load plus torque (equal value)

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. P = 75 kN axial (compressive, at centroid), T = 11 kN·m, square side b; σ_Y = 330 MPa; safety factor N = 2 ⇒ allowable equivalent stress 165 MPa. For a solid square in torsion, τ_max = T/(0.208 b³) at the mid-side.

Find. Minimum b by (a) Tresca and (b) von Mises.

Approach. At the mid-side of the square the state is a uniform axial stress σ = P/b² plus the peak torsional shear τ; combine them into each yield criterion and solve the resulting equation for b.

1. Critical-point stresses. Axial and torsional shear, both functions of b: $$\sigma=\frac{P}{b^2}=\frac{75\,000}{b^2},\qquad \tau=\frac{T}{0.208\,b^3}=\frac{11\times10^{6}}{0.208\,b^3}$$
2. Maximum-shear (Tresca) design — part (a). For combined normal and shear stress the Tresca equivalent is √(σ² + 4τ²): $$\sqrt{\sigma^2+4\tau^2}=\frac{\sigma_Y}{N}=165\ \text{MPa}$$ Solving this single equation numerically (torsion dominates, so b is set mainly by τ): $$\boxed{b_{\text{Tresca}}=86.3\ \text{mm}}$$
3. Von-Mises design — part (b). The distortion-energy equivalent uses 3τ²: $$\sqrt{\sigma^2+3\tau^2}=165\ \text{MPa}\;\Rightarrow\;\boxed{b_{\text{VM}}=82.2\ \text{mm}}$$
4. Interpret. Von Mises is less conservative for shear-dominated states, so it permits a smaller bar (82.2 mm vs. 86.3 mm). Because no member length is given, column buckling and bending from the offset are outside the question’s scope; the design is a local-strength check at the fixed end.