Shear flow and peak shear stress in a two-cell torsion box: 22-Mec-B9 Advanced Engineering Structures, May 2014, Q5

Question 5: Shear flow and peak shear stress in a two-cell torsion box (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 two-cell closed thin-walled box in pure torsion, the cells sharing the vertical wall 5.

Quantity	Symbol	Value
Nose depth (= diameter of the semicircular wall 1)	$2R$	200 mm
Rear wall depth	$h_3$	100 mm
Spar-to-rear wall distance	$L$	300 mm
Wall thicknesses	$t_1 \ldots t_5$	1.0, 1.5, 3.0, 2.0, 4.0 mm
Applied torque	$T$	38 000 N·m = $38\times10^{6}$ N·mm

Find. (a) the shear flow in each of the five walls; (b) the maximum shear stress and the wall in which it occurs.

The two-cell section. Cell I is the semicircular nose closed by wall 5; cell II is the tapering trapezoid. Wall 5 is shared, and carries the difference of the two cell flows. Wall thicknesses are $t_1 = 1.0$, $t_2 = 1.5$, $t_3 = 3.0$, $t_4 = 2.0$ and $t_5 = 4.0$ mm.

Approach. A two-cell box in torsion has two unknown constant cell shear flows; torque equilibrium supplies one equation and the requirement that both cells twist at the same rate supplies the second, after which the shared wall carries the difference of the two flows.

Part (a) — wall lengths and enclosed areas. The nose is a semicircle of radius $R = 100$ mm and the skins taper from 200 mm to 100 mm over 300 mm, so each sloping skin rises 50 mm: $$\begin{aligned}s_1 &= \pi R = 314.16\ \text{mm}, \\ s_2 &= s_4 = \sqrt{300^{2}+50^{2}} = 304.14\ \text{mm}, \\ s_3 &= 100\ \text{mm}, \\ s_5 &= 200\ \text{mm}\end{aligned}$$$$\begin{aligned}A_{\mathrm{I}} &= \tfrac12\pi R^{2} = 15\,708\ \text{mm}^{2}, \\ A_{\mathrm{II}} &= \tfrac12(200+100)(300) = 45\,000\ \text{mm}^{2}\end{aligned}$$ Cell II encloses nearly three times the area of cell I, which is why it will attract the larger share of the torque.
Assemble the wall compliances. The twist integral needs $\delta = s/t$ for each wall: $$\begin{aligned}\delta_1 &= \frac{314.16}{1.0} = 314.16, \\ \delta_2 &= \frac{304.14}{1.5} = 202.76, \\ \delta_3 &= \frac{100}{3.0} = 33.33, \\ \delta_4 &= \frac{304.14}{2.0} = 152.07, \\ \delta_5 &= \frac{200}{4.0} = 50.00\end{aligned}$$ Wall 1 is by far the most compliant, being the longest and the thinnest.
Torque equilibrium: the Bredt-Batho sum over the cells. Each cell carries a constant circulating flow, and their torques add: $$T = 2A_{\mathrm{I}}q_{\mathrm{I}} + 2A_{\mathrm{II}}q_{\mathrm{II}} \;\Rightarrow\; 38\times10^{6} = 2(15\,708)q_{\mathrm{I}} + 2(45\,000)q_{\mathrm{II}}$$ This is one equation in two unknowns; the box is once redundant in torsion.
Compatibility: both cells must twist at the same rate. For cell $i$ the rate of twist is $\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{2A_iG}\oint\dfrac{q}{t}\mathrm{d}s$, with the shared wall carrying the difference of the two cell flows: $$\frac{1}{2A_{\mathrm{I}}}\Bigl[q_{\mathrm{I}}\delta_1 + (q_{\mathrm{I}}-q_{\mathrm{II}})\delta_5\Bigr] = \frac{1}{2A_{\mathrm{II}}}\Bigl[q_{\mathrm{II}}(\delta_2+\delta_3+\delta_4) + (q_{\mathrm{II}}-q_{\mathrm{I}})\delta_5\Bigr]$$ Substituting the numbers reduces this to $q_{\mathrm{I}} = 0.531815\,q_{\mathrm{II}}$.
Solve the pair. Substituting into the torque equation, $$\bigl[2(15\,708)(0.531815) + 2(45\,000)\bigr]q_{\mathrm{II}} = 38\times10^{6}$$$$\begin{aligned}q_{\mathrm{II}} &= \boxed{356.11\ \text{N/mm}}, \\ q_{\mathrm{I}} &= \boxed{189.39\ \text{N/mm}}\end{aligned}$$ and the shared wall carries the difference, $q_5 = q_{\mathrm{I}} - q_{\mathrm{II}} = -166.73$ N/mm, i.e. 166.73 N/mm circulating in the sense of cell II.
Assign the flows to the individual walls and check the torque. Walls 1 belongs to cell I alone, walls 2, 3 and 4 to cell II alone, and wall 5 is shared: $$\begin{aligned}q_1 &= 189.39, \\ q_2 &= q_3 = q_4 = 356.11, \\ q_5 &= 166.73\ \ \text{N/mm}\end{aligned}$$$$2A_{\mathrm{I}}q_{\mathrm{I}} + 2A_{\mathrm{II}}q_{\mathrm{II}} = 2(15\,708)(189.39) + 2(45\,000)(356.11) = 38.0\times10^{6}\ \text{N}\!\cdot\!\text{mm} = T\ \checkmark$$ Both cells return the same twist rate, $G\,\mathrm{d}\theta/\mathrm{d}z = 1.6285$ N/mm$^{3}$, confirming the compatibility solve.
Part (b) — convert each flow to a stress and find the largest. The shear stress in a wall is the flow divided by the wall thickness, so the thinnest wall of the more heavily loaded cell governs: $$\begin{aligned}\tau_1 &= \frac{189.39}{1.0} = 189.4, \\ \tau_2 &= \frac{356.11}{1.5} = 237.4, \\ \tau_3 &= \frac{356.11}{3.0} = 118.7, \\ \tau_4 &= \frac{356.11}{2.0} = 178.1, \\ \tau_5 &= \frac{166.73}{4.0} = 41.7\ \ \text{MPa}\end{aligned}$$$$\tau_{\max} = \boxed{237.4\ \text{MPa in wall 2}}$$ Wall 2 wins because it combines the larger cell flow with the second thinnest wall; note that wall 1 carries only half the flow of wall 2 yet is nearly as highly stressed, being only two-thirds as thick.

Quantity	Symbol	Result
Enclosed area, cell I (nose)	$A_{\mathrm{I}}$	15,708 mm²
Enclosed area, cell II	$A_{\mathrm{II}}$	45,000 mm²
Shear flow, cell I / wall 1	$q_1$	189.39 N/mm
Shear flow, cell II / walls 2, 3, 4	$q_2 = q_3 = q_4$	356.11 N/mm
Shear flow, shared wall 5	$q_5$	166.73 N/mm
Shear stress, wall 1	$\tau_1$	189.4 MPa
Shear stress, wall 2	$\tau_2$	237.4 MPa
Shear stress, wall 3	$\tau_3$	118.7 MPa
Shear stress, wall 4	$\tau_4$	178.1 MPa
Shear stress, wall 5	$\tau_5$	41.7 MPa
Maximum shear stress	$\tau_{\max}$	237.4 MPa, in wall 2
Torsional stiffness	$GJ / G$	23.33 × 10⁶ mm⁴

Check: the sense of the torque (clockwise) fixes only the sense of the circulating flows, not their magnitudes, so the stresses above are independent of it. Wall 5 carries the difference of the cell flows and is therefore lightly stressed at 41.7 MPa; if the two cells had been sized to twist equally on their own the shared wall would carry almost nothing, which is the usual design objective for a multi-cell box in pure torsion.

Topic. Torsion of a multi-cell closed thin-walled section.

Key relations. Bredt-Batho for each cell, $T = \sum_i 2A_i q_i$; equal twist rate $\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{2A_iG}\oint\dfrac{q}{t}\mathrm{d}s$; a wall shared by cells $i$ and $j$ carries $q_i - q_j$; $\tau = q/t$.

Why this works. In a closed cell the shear flow must be constant around the circuit, because any element of wall is in equilibrium along the span. With $n$ cells there are $n$ such constants, torque equilibrium gives one equation, and the remaining $n-1$ come from the fact that the cross-section rotates as a rigid body, so every cell must twist at the same rate.

Common pitfall. Chording the semicircular wall, or forgetting that wall 5 carries the difference rather than the sum of the cell flows. Using a straight 200 mm chord for wall 1 shortens $\delta_1$ by 36 per cent and shifts the load sharing noticeably; adding the flows in wall 5 instead of subtracting them overstates that wall by a factor of more than three.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., Ch. 23 (wing torsion, multi-cell sections) and Ch. 18; Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., Ch. 6.

Question 5 of 8: Shear flow and peak shear stress in a two-cell torsion box

Question 5: Shear flow and peak shear stress in a two-cell torsion box (20 marks)