Question 1 of 8: Shear flow and bending stress in a trapezoidal monocoque box

Question 1: Shear flow and bending stress in a trapezoidal monocoque 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 closed single-cell trapezoidal box of constant wall thickness, built in at the root and loaded by a vertical tip force. Because the walls are “fully effective in bending as well as shear”, the whole wall material carries direct stress, so the section is not idealised into booms and the shear flow varies continuously (and quadratically) along every wall.

Find. (a) the shear flow $q$ everywhere around the closed cell at the root, and (b) the direct bending stress at each of the four corners A, B, C and D of the root section.

Approach. Locate the centroid and evaluate all three second moments, solve the two unsymmetrical-bending equations for the stress field at the root, then integrate $\mathrm{d}q/\mathrm{d}s$ around the cell from an arbitrary cut and close it by requiring the shear flows to reproduce the moment of the applied load about the centroid.

1. Set out the wall geometry and locate the centroid. Taking the bottom-left corner A as a temporary origin with $z$ to the right and $y$ upward, the four median wall lengths are $L_{AB}=650$, $L_{BC}=200$, $L_{CD}=\sqrt{650^{2}+100^{2}}=657.65$ and $L_{DA}=300$ mm, so the total wall area is $A_w = t\sum L = 2(1807.65) = 3615.29$ mm². First moments of the four straight strips about A give $$\bar z = \frac{\sum t L \bar z_i}{A_w} = 307.02\ \text{mm}, \qquad \bar y = \frac{\sum t L \bar y_i}{A_w} = 126.91\ \text{mm}.$$
2. Evaluate the three centroidal second moments. For a straight thin strip running from $(Z_1,Y_1)$ to $(Z_2,Y_2)$ of length $\ell$ the exact thin-wall integrals are $\int Y^{2}\,\mathrm{d}s = \tfrac{\ell}{3}(Y_1^{2}+Y_1Y_2+Y_2^{2})$ and $\int ZY\,\mathrm{d}s = \tfrac{\ell}{6}(2Z_1Y_1+Z_1Y_2+Z_2Y_1+2Z_2Y_2)$, with the analogous form for $\int Z^{2}\mathrm{d}s$. Summing over the four walls, $$\begin{aligned} I_{ZZ} &= 1.9654\times10^{8}\ \text{mm}^{4}, \\ I_{YY} &= 4.8405\times10^{7}\ \text{mm}^{4}, \\ I_{YZ} &= -1.5125\times10^{7}\ \text{mm}^{4}. \end{aligned}$$ The product term is small but not zero — ignoring it would misplace the neutral axis.
3. Write the root bending moment. The tip force is purely vertical, so at the root it produces a bending moment about the horizontal centroidal axis only; its offset line of action produces a torque about the span axis, not a second bending moment. Hence $M_Z = PL = 18\,000 \times 5000 = 9.00\times10^{7}\ \text{N}\cdot\text{mm}$ and $M_Y = 0$.
4. Solve for the direct-stress field. Substituting into the pair of equations quoted in the preamble, $$a = \frac{-M_Z I_{YZ}}{I_{ZZ}I_{YY}-I_{YZ}^{2}} = 0.14662, \qquad b = \frac{M_Z I_{ZZ}}{I_{ZZ}I_{YY}-I_{YZ}^{2}} = 1.90512 \ \ \text{MPa/mm},$$ so that $\sigma_x = 0.14662\,Z + 1.90512\,Y$ with $Z$ and $Y$ in mm.
5. Part (b) — evaluate the stress at the four corners. Substituting each corner's centroidal coordinates, $$\boxed{\sigma_A = -286.8,\quad \sigma_B = -191.5,\quad \sigma_C = +189.5,\quad \sigma_D = +284.7 \ \ \text{MPa}}$$ Both bottom corners are in compression and both top corners in tension, which is the expected sense for a downward tip load on a cantilever, and the deep-web side carries the larger magnitude at each face because it lies further from the neutral axis.
6. Locate the neutral axis. Setting $\sigma_x = 0$ gives $Y = -(a/b)Z = -0.07696\,Z$, a line through the centroid inclined at $4.40^{\circ}$ below the horizontal centroidal axis. The same line is where $\mathrm{d}q/\mathrm{d}s$ changes sign, so it also locates every turning point of the shear-flow distribution — one number does both jobs.
7. Part (a) — integrate the open shear flow. Longitudinal equilibrium of a wall element gives $\mathrm{d}q/\mathrm{d}s = (t/L)(aZ+bY)$ for this linearly tapering moment distribution. Cutting the cell just past corner A and walking A → B → C → D → A, numerical integration of that expression returns $q_b$ values of $0$, $-62.18$, $-62.26$ and $+0.12$ N/mm at A, B, C and D respectively, and the walk closes on itself to within $2\times10^{-14}$ N/mm, which confirms the integration.
8. Close the cell by moment equivalence. The load acts at $Z = 650-\bar z = 342.98$ mm, so its moment about the centroid is $-P(342.98) = -6.1736\times10^{6}\ \text{N}\cdot\text{mm}$, while the open flows contribute $\oint q_b\,(Z\,\mathrm{d}Y - Y\,\mathrm{d}Z) = -9.3571\times10^{6}$. With the enclosed area giving $2A_{\text{cell}} = 3.25\times10^{5}$ mm², $$q_{s,0} = \frac{M_{\text{ext}} - M(q_b)}{2A_{\text{cell}}} = \frac{-6.1736\times10^{6} + 9.3571\times10^{6}}{3.25\times10^{5}} = +9.795\ \text{N/mm}.$$ This closure needs neither the thickness nor the shear modulus, because the line of action of the load is given.
9. Assemble the total shear flow and find its peak. Adding $q_{s,0}$ to the open flows gives the corner values collected below. Because $q$ is a running integral of a linear function it varies quadratically along each wall, so the largest value need not occur at a corner — and here it does not. The turning point in the shallow web sits where the neutral axis crosses it, $100.5$ mm above corner B, giving $$\boxed{q_{\max} = -56.23\ \text{N/mm} \ \Rightarrow\ \tau_{\max} = q_{\max}/t = 28.12\ \text{MPa}}$$ A second, much smaller interior peak of $18.43$ N/mm occurs in the deep web $150.5$ mm above A.
10. Check the result statically. Integrating the final flows around the cell returns a horizontal resultant of $0.0$ N, a vertical resultant of $-18\,000$ N and a moment about the centroid of $-6.174\times10^{6}\ \text{N}\cdot\text{mm}$ — that is, exactly the applied tip load acting on its true line of action. These two checks catch sign errors that no amount of formula-matching will.