Question 4 of 7: Idealised box section — shear centre and panel shear flows

Question 4: Idealised box section — shear centre and panel shear flows (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 rectangular box built from four skins and stiffened by an angle at each corner, with the two left corners carrying the heavier angle and the right web rolled thinner than the other three walls.

Find. (A) the horizontal position of the shear centre, and (B) the shear flow carried by each of the four panels when a 10 kN vertical shear acts through it.

Approach. Replace each skin by an equivalent boom area at the corners it spans, compute the second moment from the booms alone, cut the cell to obtain the open shear flows, close it with a constant flow that produces zero twist, and finally take moments of the resulting flow to locate the shear centre.

1. Fix the median-line geometry. Working to the wall centre-lines, the boom spacings are $$b=600-\tfrac{10+6}{2}=592\ \text{mm},\qquad h=250-\tfrac{10+10}{2}=240\ \text{mm}$$ so the enclosed cell area is $A_{cell}=592\times240=142\,080\ \text{mm}^2$. Number the booms 1 (top left), 2 (top right), 3 (bottom right) and 4 (bottom left).
2. Size the angles. An equal-angle of leg $\ell$ and thickness $t$ has area $(2\ell-t)t$, so $$A_{50}=(50+50-15)(15)=1275\ \text{mm}^2,\qquad A_{40}=(40+40-10)(10)=700\ \text{mm}^2$$
3. Convert the skins into boom area. Idealisation moves the direct-stress capacity of each panel into the booms at its ends using $$B_r=A_{angle}+\sum \frac{t\,\ell}{6}\left(2+\frac{\sigma_2}{\sigma_1}\right)$$ For a horizontal skin the two end booms are at the same height, so $\sigma_2/\sigma_1=+1$ and the contribution is $t\ell/2$. For a web the ends are at $\pm h/2$, so $\sigma_2/\sigma_1=-1$ and the contribution collapses to $t\ell/6$. Hence $$B_1=B_4=1275+\frac{10(592)}{2}+\frac{10(240)}{6}=4635\ \text{mm}^2$$ $$B_2=B_3=700+\frac{10(592)}{2}+\frac{6(240)}{6}=3900\ \text{mm}^2$$ The skins contribute far more than the angles do, which is why the idealisation cannot be done with the angle areas alone.
4. Compute the second moment. All four booms lie at $y=\pm120$ mm, so $$I_{xx}=\sum B_ry_r^2=2(4635+3900)(120)^2$$ $$\boxed{I_{xx} = 2.4581{\times}10^{8}\ \text{mm}^4}$$
5. Find the open shear flows. Cut the top skin and walk 1→2→3→4→1, accumulating $$q_{b,n}=q_{b,n-1}-\frac{S_y}{I_{xx}}B_ny_n$$ With $S_y=10\,000$ N this gives $q_b=0$ in the top skin, $-19.04$ N/mm in the right web, $0$ in the bottom skin and $+22.63$ N/mm in the left web, and the walk closes on zero back at the cut, which is the first check.
6. Part (B) — close the cell and obtain the panel flows. The applied load acts through the shear centre, so the section does not twist and the constant closing flow follows from $$\oint\frac{(q_b+q_{s,0})}{t}\,ds=0\quad\Longrightarrow\quad q_{s,0}=-\frac{\sum q_b\,\ell/t}{\sum \ell/t}=\frac{218.5}{182.4}$$ $$\boxed{q_{s,0} = +1.198\ \text{N/mm}}$$ Adding it to the open flows gives the final panel values tabulated below.
7. Check equilibrium before going further. Resolving the four flows along their walls, the vertical components sum to exactly 10 000 N and the horizontal components cancel to zero. The two skins, which look inactive at 1.198 N/mm, are precisely what makes the horizontal balance close.
8. Part (A) — locate the shear centre. Taking moments of the closed flow about boom 4 — where two of the four walls contribute nothing, since their lines pass through the point — the flow system is statically equivalent to the applied 10 kN acting at $$e=\frac{\sum q_i\,(2A_i)}{S_y}=\frac{2.365{\times}10^{6}}{10\,000}$$ $$\boxed{e = 236.5\ \text{mm from the left web (355.5 mm from the right web)}}$$ The shear centre lies well to the left of mid-width, 59.5 mm off centre, because the left side combines the larger booms with the thicker web.
9. Convert the flows to stresses. Dividing each flow by its own wall thickness, $$\tau=\frac{q}{t}\quad\Longrightarrow\quad \tau_{\text{left web}}=2.38\ \text{MPa},\quad \tau_{\text{right web}}=2.97\ \text{MPa},\quad \tau_{\text{skins}}=0.12\ \text{MPa}$$ The largest shear flow is in the left web at 23.83 N/mm, but the largest shear stress is in the right web, because 6 mm of material must carry 17.84 N/mm. Reporting $q_{\max}/t_{\text{web}}$ without dividing wall by wall would understate the critical stress by twenty-five per cent.