Question 8 of 8: Shear flow and corner bending stresses in a closed trapezoidal box

Question 8: Shear flow and corner bending stresses in a closed trapezoidal 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 box of trapezoidal outline and uniform wall thickness, loaded by a downward vertical force applied in the plane of the shallow right-hand web. Unlike questions 4 and 7, every wall carries direct stress as well as shear.

Find. (a) the shear flow distribution round the closed section, and (b) the direct stress at each of the four corners 300 mm inboard of the load.

Approach. Because the walls are effective in bending, the section properties must be integrated over the wall material rather than lumped into booms, and the basic shear flow varies continuously along each wall instead of stepping at booms. Compute the properties, solve part (b) from the bending moment, then differentiate the same stress field along the span for part (a).

1. Set out the wall geometry. The four median-line walls have lengths AB = 600, BC = 150, CD = 650 (the slope of 600 across and 250 up) and DA = 400 mm, so the total wall area is $A = 3\left(600+150+650+400\right) = 5\,400$ mm$^{2}$.
2. Locate the centroid. Taking first moments of the four wall areas about corner A,$$\bar{Z} = 258.3\ \text{mm}, \qquad \bar{Y} = 150.0\ \text{mm}$$The vertical centroid falls at exactly 150 mm, a coincidence of this geometry that makes corner C a zero-$Y$ point and simplifies the interpretation of the answers.
3. Evaluate the second moments. Each straight wall contributes its own second moment about its mid-point, $tL(\Delta Y)^{2}/12$ and $tL(\Delta Z)^{2}/12$ and $tL\Delta Z\Delta Y/12$, plus the parallel-axis transfer terms. Summing over the four walls,$$I_{YY} = 1.0350 \times 10^{8}, \qquad I_{ZZ} = 2.5162 \times 10^{8}, \qquad I_{YZ} = -5.250 \times 10^{7}\ \text{mm}^{4}$$The sloping top wall is what makes $I_{YZ}$ non-zero, and it is large enough that the neutral axis will not be horizontal even though the load is purely vertical.
4. Find the bending moment 300 mm inboard (part b). With a downward load 300 mm away,$$M_{Z} = \int\sigma Y\,\mathrm{d}A = (15\,000)(300) = 4.500 \times 10^{6}\ \text{N}\cdot\text{mm}, \qquad M_{Y} = 0$$positive because a downward load on a cantilever puts the upper fibres into tension.
5. Solve for the stress field and evaluate the corners. The pair of equations gives $a = 0.010145$ and $b = 0.048624$ N/mm$^{3}$. Note that $M_{Y}$ is zero yet $a$ is not: the product of inertia forces a horizontal stress gradient even under a purely vertical load. Substituting each corner's coordinates relative to the centroid,$$\sigma_{A} = -9.91, \quad \sigma_{B} = -3.83, \quad \sigma_{C} = 3.47, \quad \sigma_{D} = 9.54\ \text{MPa}$$$$\boxed{\sigma_{\max} = 9.54\ \text{MPa tension at D}, \quad \sigma_{\min} = -9.91\ \text{MPa compression at A}}$$The extremes fall on the tall left-hand web, the pair of corners furthest from the inclined neutral axis.
6. Differentiate the stress field along the span (part a). Repeating the same solve with a unit span offset gives the gradients $a' = 3.3817 \times 10^{-5}$ and $b' = 1.6208 \times 10^{-4}$ N/mm$^{4}$. Longitudinal equilibrium of a wall strip then gives the basic shear flow as a running integral rather than a set of steps,$$q_{b}(s) = \int_{0}^{s}\left(a'Z + b'Y\right)t\,\mathrm{d}s$$measured from a cut, taken here at corner A.
7. Integrate round the open section. Because $Z$ and $Y$ vary linearly along each wall, $q_{b}$ is quadratic in $s$ within each wall. Its values on arrival at each successive corner are$$q_{b}(B) = -41.23, \quad q_{b}(C) = -41.50, \quad q_{b}(D) = 0.76, \quad q_{b}(A) = 0.00\ \text{N/mm}$$Returning to zero at the cut confirms the integration.
8. Close the cell by moments about corner A. The applied force acts 600 mm from A, so$$M_{\text{ext}} = (600)(-15\,000) = -9.00 \times 10^{6}\ \text{N}\cdot\text{mm}, \qquad \oint q_{b}\,p\,\mathrm{d}s = -9.4821 \times 10^{6}\ \text{N}\cdot\text{mm}$$With an enclosed area $A_{\text{cell}} = 165\,000$ mm$^{2}$,$$q_{s,0} = \frac{M_{\text{ext}} - \oint q_{b}\,p\,\mathrm{d}s}{2A_{\text{cell}}} = 1.461\ \text{N/mm}$$
9. Report the closed-section flows. Adding the constant gives $q_{A} = 1.46$, $q_{B} = -39.76$, $q_{C} = -40.04$ and $q_{D} = 2.22$ N/mm, varying quadratically between those corner values. The largest magnitude anywhere on the circuit is$$\boxed{q_{\max} = 41.27\ \text{N/mm at } (Z, Y) = (600, 79)\ \text{mm, giving } \tau_{\max} = 13.76\ \text{MPa}}$$on the short right-hand web, close to its mid-height. That web is the critical one despite being the shallowest, because the torque and the vertical shear reinforce each other there and oppose each other on the tall left-hand web, where the flow drops almost to nothing.