Question 1 of 8: Shear flow in a two-cell torsion box with a semicircular nose

Question 1: Shear flow in a two-cell torsion box with a semicircular nose (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. A closed box of overall depth 250 mm carrying a constant torque of 33 kN·m, divided by the vertical web 5 into a semicircular nose cell and a rectangular cell 500 mm long. The five wall gauges are:

Find. The shear flow carried by each of the five walls, then the largest shear stress in the section and the wall in which it acts.

[Figure not reproduced: Figure 1.1 — the two-cell torsion box. Wall 5 divides the semicircular nose cell (I) from the rectangular cell (II); the numbers 1–5 are the wall labels used on the exam paper. See the official exam paper.]

Approach. Treat the box as two closed cells sharing web 5: write the Bredt–Batho twist-rate expression for each cell, impose that both cells twist at the same rate, and close the system with torque equilibrium $T = 2\sum A_i q_i$.

1. Fix the cell geometry. The nose is a true semicircle of radius $R = 125$ mm, so its perimeter and enclosed area are$$s_1 = \pi R = \pi(125) = 392.70\ \text{mm},\qquad A_1 = \tfrac{1}{2}\pi R^{2} = 24\,544\ \text{mm}^2 .$$The rectangular cell encloses $A_2 = 500 \times 250 = 125\,000$ mm$^2$. The nose must be kept as an arc: replacing it by its 250 mm chord would collapse the nose cell onto web 5 and discard 16 % of the total enclosed area (even a two-chord triangular nose loses 6 %) and, more importantly, would shorten the compliance path in the softest wall.
2. Evaluate the line integrals $\delta = \oint \mathrm{d}s/t$ wall by wall. These are the compliances that decide how the torque splits between the cells:$$\delta_1 = \frac{392.70}{1.5} = 261.80,\qquad \delta_5 = \frac{250}{2.0} = 125.0,$$$$\delta_{\text{II,outer}} = \frac{500}{2.5} + \frac{250}{2.0} + \frac{500}{3.5} = 200.0 + 125.0 + 142.86 = 467.86 .$$All are dimensionless.
3. Write the equal-twist-rate condition for each cell. With $u = G\theta'$ the common rate of twist, and remembering that the internal web carries the difference of the two cell flows,$$\text{cell I:}\quad q_1(\delta_1 + \delta_5) - q_2\delta_5 = 2A_1 u,$$$$\text{cell II:}\quad q_2(\delta_{\text{II,outer}} + \delta_5) - q_1\delta_5 = 2A_2 u .$$Substituting the numbers gives $386.80\,q_1 - 125.0\,q_2 = 49\,087\,u$ and $-125.0\,q_1 + 592.86\,q_2 = 250\,000\,u$.
4. Add torque equilibrium and solve. The third equation is$$T = 2A_1 q_1 + 2A_2 q_2 = 49\,087\,q_1 + 250\,000\,q_2 = 33\times10^{6}\ \text{N}\cdot\text{mm}.$$Solving the three equations simultaneously for $q_1$, $q_2$ and $u$,$$\boxed{q_1 = 69.46\ \text{N/mm},\qquad q_2 = 118.36\ \text{N/mm}} .$$The rectangular cell takes the lion’s share because it encloses five times the area of the nose.
5. Distribute the flows onto the five walls (part a). Walls 2, 3 and 4 all bound cell II alone and therefore all carry $q_2$; the nose skin carries $q_1$; the internal web sees the difference$$q_5 = q_2 - q_1 = 118.36 - 69.46 = 48.90\ \text{N/mm}.$$A useful check: the two cell flows reproduce the applied torque to seven figures, $2(24\,544)(69.46) + 2(125\,000)(118.36) = 33.00\times10^{6}$ N·mm.
6. Convert each flow to a stress with its own gauge (part b). Dividing wall by wall, $\tau = q/t$:$$\tau_1 = \frac{69.46}{1.5} = 46.31,\quad \tau_2 = \frac{118.36}{2.5} = 47.34,\quad \tau_3 = \frac{118.36}{2.0} = 59.18,$$$$\tau_4 = \frac{118.36}{3.5} = 33.82,\quad \tau_5 = \frac{48.90}{2.0} = 24.45\ \text{MPa}.$$The peak is therefore$$\boxed{\tau_{\max} = 59.2\ \text{MPa in wall 3, the rear web}} ,$$because wall 3 combines the full cell-II flow with the thinnest gauge on that cell. Note that the wall carrying the largest flow is not the wall carrying the largest stress: walls 2, 3 and 4 share one flow but produce three different stresses.