Question 1 of 8: Shear flow and bending stresses in a closed trapezoidal box

Question 1: Shear flow and 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 single-cell closed trapezoidal box of constant wall thickness $t = 3$ mm, with median-line corner coordinates measured from the bottom of the left web, carrying a downward tip force applied on the line of the right web.

Find. (a) the shear-flow distribution around the closed section under the load as applied, and (b) the direct bending stress at each of the four corners of a section 300 mm inboard of the loaded one.

Approach. Locate the centroid and the three second moments of the median line, use them to write the bending field and its span-wise gradient, integrate that gradient round the wall to get the open shear flow, and close the cell with the single constant that makes the resultant pass through the actual line of the load.

1. Set the corner coordinates and wall lengths. Taking the origin at corner A, the median line runs $A(0,0) \to B(650,0) \to C(650,200) \to D(0,450) \to A$. The sloping top wall has length $$L_{CD} = \sqrt{650^{2} + 250^{2}} = 696.42\ \text{mm}$$ so the wall lengths are 650, 200, 696.42 and 450 mm, giving a total median-line area $A_{\text{wall}} = 3(650+200+696.42+450) = 5989.26$ mm². The area enclosed by the median line is that of a trapezium, $$A_{\text{cell}} = \tfrac{1}{2}(450+200)(650) = 211\,250\ \text{mm}^{2}$$ which will be needed to close the cell.
2. Locate the centroid. Each straight wall contributes its own area $tL$ acting at its mid-point, so $$\begin{aligned} \bar{Z} &= \frac{\sum tL\,z_{c}}{\sum tL} = \frac{1\,702\,758.9}{5989.26} = 284.30\ \text{mm}, \\ \bar{Y} &= \frac{1\,042\,758.9}{5989.26} = 174.10\ \text{mm} \end{aligned}$$ measured from A. The centroid sits well left of mid-width and well below mid-depth, because the deep left web and the sloping top wall both carry material toward the top left.
3. Compute the three second moments about the centroid. For a straight wall of length $L$ running from $(Z_{1},Y_{1})$ to $(Z_{2},Y_{2})$ the standard thin-wall results are $I = tL\left(c^{2} + d^{2}/12\right)$ about each axis and $tL\left(c_{Z}c_{Y} + \Delta Z\,\Delta Y/12\right)$ for the product term, with $c$ the mid-point coordinate and $d$ the projected length. Summing the four walls, $$\begin{aligned} I_{ZZ} &= 3.3826\times10^{8}\ \text{mm}^{4}, \\ I_{YY} &= 1.4914\times10^{8}\ \text{mm}^{4}, \\ I_{YZ} &= -6.5073\times10^{7}\ \text{mm}^{4} \end{aligned}$$ The product term is not zero, so the section is genuinely unsymmetrical and a vertical load alone will bend it about both axes. The determinant that appears in every later formula is $$D = I_{ZZ}I_{YY} - I_{YZ}^{2} = 4.6210\times10^{16}\ \text{mm}^{8}$$
4. Write the span-wise gradient of the bending field. With the load applied at the tip and the section a distance $x$ inboard, $M_{Z} = -S x$ and $M_{Y} = 0$ for a downward force, so the coefficients of $\sigma = aZ + bY$ grow linearly with $x$ at the rates $$b' = \frac{-S\,I_{ZZ}}{D} = \frac{-15\,000\times3.3826\times10^{8}}{4.6210\times10^{16}} = -1.09796\times10^{-4}\ \text{N/mm}^{4}$$ $$a' = \frac{S\,I_{YZ}}{D} = \frac{15\,000\times(-6.5073\times10^{7})}{4.6210\times10^{16}} = -2.11218\times10^{-5}\ \text{N/mm}^{4}$$ These two numbers are all that the shear-flow calculation needs.
5. Integrate the wall equilibrium equation from a cut. Longitudinal equilibrium of a wall element gives $\partial q/\partial s = -t\left(a'Z + b'Y\right)$, and because $Z$ and $Y$ vary linearly along a straight wall the increment over a whole wall is exact: $$\Delta q = -t\,L\left(a'\bar{Z}_{w} + b'\bar{Y}_{w}\right)$$ with $\bar{Z}_{w}$, $\bar{Y}_{w}$ the wall mid-point measured from the centroid. Cutting the section at A and walking $A \to B \to C \to D \to A$ gives the open (basic) flows $$\begin{aligned} \Delta q_{AB} &= -35.600, \\ \Delta q_{BC} &= -0.248, \\ \Delta q_{CD} &= +36.410, \\ \Delta q_{DA} &= -0.563\ \text{N/mm} \end{aligned}$$ whose sum is zero to five figures — the arithmetic check that the walk really closes.
6. Close the cell on the actual line of the load. A closed section is once redundant in shear, and the missing constant $q_{s,0}$ follows from moment equivalence: the whole flow field must have the same moment about any point as the applied force acting on its true line. Taking moments about corner A, with the force acting downward at $z = 650$ mm, $$q_{s,0} = \frac{z_{\text{load}}\,S_{Y} - M_{A}(q_{b})}{2A_{\text{cell}}} = \frac{-9.750\times10^{6} - (-1.06459\times10^{7})}{422\,500} = \boxed{+2.120\ \text{N/mm}}$$ Adding this constant to the open flows gives the corner values $q_{A} = +2.12$, $q_{B} = -33.48$, $q_{C} = -33.73$ and $q_{D} = +2.68$ N/mm.
7. Find where the flow peaks inside each wall. The flow is stationary where the integrand vanishes, that is where $a'Z + b'Y = 0$, or $Y = -\left(a'/b'\right)Z = -0.19237\,Z$. In the right web $Z = +365.70$ mm, which puts the turning point at $Y = -70.35$ mm, i.e. 103.8 mm above the bottom corner, where $$q_{\max} = \boxed{-35.25\ \text{N/mm}}\quad\text{giving}\quad \tau_{\max} = \frac{35.25}{3} = \boxed{11.75\ \text{MPa}}$$ In the left web the same rule puts the turning point 228.8 mm above the bottom, but there the flow only reaches $+10.74$ N/mm ($\tau = 3.58$ MPa). The shallow right web is therefore the critical wall, not the deep one — the whole 15 kN is being fed in at that web.
8. Report the shear centre, which explains the distribution. Repeating the closure with the no-twist condition $\oint q\,ds/t = 0$ instead of the moment condition gives $q_{s,0} = +15.88$ N/mm and a resultant that passes through $$z_{SC} = 262.39\ \text{mm}$$ from the left web. The load therefore acts $650 - 262.39 = 387.61$ mm outboard of the shear centre and applies a torque $T = 15\,000 \times 387.61 = 5.814\times10^{6}$ N·mm, worth a uniform $T/2A_{\text{cell}} = 13.76$ N/mm. That torsional part and the no-twist part sum to the 2.120 N/mm found in step 6, which is an independent check on the closure.

Part (b) needs only the bending field itself, evaluated at the single station asked for.

1. Evaluate the bending moment at the section 300 mm inboard. The 15 kN force acts at the free end, so $$M_{Z} = S\,x = 15\,000 \times 300 = 4.50\times10^{6}\ \text{N}\cdot\text{mm}$$ with $M_{Y} = 0$; the moment vector is horizontal because the force is vertical.
2. Solve the two-equation pair for the stress coefficients. Substituting the section constants into $a I_{ZZ} + b I_{YZ} = 0$ and $a I_{YZ} + b I_{YY} = M_{Z}$, $$\begin{aligned} a &= -\frac{M_{Z}I_{YZ}}{D} = 6.3365\times10^{-3}, \\ b &= \frac{M_{Z}I_{ZZ}}{D} = 3.2939\times10^{-2}\ \text{N/mm}^{3} \end{aligned}$$ The non-zero $a$ is entirely due to the product term: a vertical load on this section also bends it sideways.
3. Evaluate $\sigma = aZ + bY$ at the four corners. Using centroidal coordinates, $$\sigma_{A} = 6.3365\times10^{-3}(-284.30) + 3.2939\times10^{-2}(-174.10) = \boxed{-7.54\ \text{MPa}}$$ $$\begin{aligned} \sigma_{B} &= \boxed{-3.42\ \text{MPa}}, \\ \sigma_{C} &= \boxed{+3.17\ \text{MPa}}, \\ \sigma_{D} &= \boxed{+7.29\ \text{MPa}} \end{aligned}$$ Tension is positive. The top corners are in tension and the bottom corners in compression, which is the correct sense for a downward tip load on a member built in further inboard.
4. Sanity-check the neutral axis. Setting $\sigma = 0$ gives $Y = -\left(a/b\right)Z = -0.19237\,Z$, the same line that located the shear-flow turning points — as it must be, since both come from the same field. That line is inclined about 10.9° to the horizontal, so the extreme fibre is corner D rather than the geometric top of the section; on a symmetric section it would have been horizontal.