Question 1 of 8: Shear flow, shear centre and torsional overload of an open thin-walled section

Question 1: Shear flow, shear centre and torsional overload of an open thin-walled section (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. An I-shaped open section of uniform wall thickness, unsymmetrical about its vertical axis but symmetric about the horizontal centroidal axis, carrying a vertical shear.

Find. (a) the shear flow distribution around the section; (b) the position of the shear centre measured from the web centreline; (c) the greatest shear stress when the same 15 kN acts through the web rather than through the shear centre.

Approach. Because the section is symmetric about the horizontal axis the product of inertia vanishes, so the vertical shear alone drives the ordinary open-section shear-flow integral walked from each free flange tip; the horizontal flange forces it produces supply the couple that locates the shear centre, and shifting the load to the web adds a pure torque that an open section can only resist by St Venant torsion.

1. Part (a) — second moment of area about the horizontal centroidal axis. The two flanges are identical, so the centroid sits at mid-depth and each flange contributes its full area at the lever arm $h/2$. With $b_f = b_L + b_R = 360$ mm, $$I_{zz} = \frac{t\,h^{3}}{12} + 2\,b_f t\left(\frac{h}{2}\right)^{2} + 2\,\frac{b_f t^{3}}{12} = \frac{3(360)^{3}}{12} + 2(360)(3)(180)^{2} + 2\frac{(360)(3)^{3}}{12}$$$$I_{zz} = 11.664\times10^{6} + 69.984\times10^{6} + 0.0016\times10^{6} = \boxed{81.65\times10^{6}\ \text{mm}^{4}}$$The flange self-inertia term is only 1 620 mm⁴, about two parts in a hundred thousand of the total, and is kept only to show that neglecting it is legitimate.
2. Shear flow along each flange grows linearly from its free tip. Walking a distance $s$ inward from a flange tip, the swept material sits at the constant height $y = h/2$, so the first moment grows in proportion to $s$: $$q(s) = \frac{S_y}{I_{zz}}\int_0^{s} y\,t\,\mathrm{d}s = \frac{S_y}{I_{zz}}\,t\,\frac{h}{2}\,s = \frac{15\,000}{81.65\times10^{6}}(3)(180)\,s = 0.09920\,s\ \ \text{N/mm}$$ At the web the two flange segments therefore deliver $q_{240} = 0.09920(240) = 23.81$ N/mm and $q_{120} = 0.09920(120) = 11.90$ N/mm. In the top flange the flow runs outward from the web toward both tips; in the bottom flange it runs inward toward the web.
3. The web carries the sum of the two flange flows plus its own parabolic build-up. Continuity at the top junction gives $q_{\text{web,top}} = 23.81 + 11.90 = \boxed{35.71\ \text{N/mm}}$, directed upward — the same sense as $S_y$. Continuing down the web from $y = h/2$, $$q(y) = q_{\text{web,top}} + \frac{S_y t}{2 I_{zz}}\left[\left(\frac{h}{2}\right)^{2} - y^{2}\right]$$ which is greatest at mid-depth, where $y = 0$: $$q_{\max} = 35.71 + \frac{15\,000(3)(180)^{2}}{2(81.65\times10^{6})} = \boxed{44.64\ \text{N/mm}}$$
4. Statical check on the distribution. Integrating the web flow over the full depth must return the applied shear, because the flange flows are horizontal and carry no vertical component: $$\int_{-h/2}^{h/2} q(y)\,\mathrm{d}y = 35.71(360) + \frac{S_y t}{2I_{zz}}\left[\frac{h^{3}}{4}-\frac{h^{3}}{12}\right] = 12\,857 + 2\,143 = 15\,000\ \text{N} = S_y \ \checkmark$$ The distribution is therefore consistent, which also validates $I_{zz}$.
5. Part (b) — the flange forces that create the twisting couple. Each flange force is the area under its triangular flow diagram: $$\begin{aligned}F_L &= \tfrac12(0.09920)(240)^{2} = 2857.1\ \text{N}, \\ F_R &= \tfrac12(0.09920)(120)^{2} = 714.3\ \text{N}\end{aligned}$$ The long flange carries four times the force of the short one, because the flange force varies with the square of the projection.
6. Moment equilibrium about the web locates the shear centre. The web flow passes through the web centreline and contributes no moment there. In the top flange the long-side force acts away from the web and the short-side force toward it; in the bottom flange both reverse, so the four forces form two couples that add: $$S_y\,e = 2\left(\frac{h}{2}\right)(F_L - F_R) = 360\bigl(2857.1 - 714.3\bigr) = 771413\ \text{N}\!\cdot\!\text{mm}$$$$e = \frac{771413}{15\,000} = \boxed{51.43\ \text{mm}}$$ measured from the web centreline, on the side of the short (120 mm) flange projection — the same sense as for a channel, where the shear centre lies away from the flange tips.
7. Part (c) — moving the load to the web introduces a pure torque. Statically the 15 kN through the web equals the same 15 kN through the shear centre plus a torque about the shear centre: $$T = S_y\,e = 15\,000\,(51.43) = 771.4\times10^{3}\ \text{N}\!\cdot\!\text{mm}$$ The bending shear flow computed in part (a) is unchanged; the torque is carried entirely by St Venant (uniform) torsion, since the section is open and warping is assumed unrestrained.
8. Torsion constant of the open section and the torsional shear stress. For thin rectangles joined at their ends the torsion constant is the sum of the individual $bt^{3}/3$ terms, with total developed length $\sum b = 360 + 2(360) = 1080$ mm: $$\begin{aligned}J &= \frac{1}{3}\sum b\,t^{3} = \frac{(1080)(3)^{3}}{3} = 9720\ \text{mm}^{4}, \\ \tau_{T} &= \frac{T\,t}{J} = \frac{771413(3)}{9720} = \boxed{238.1\ \text{MPa}}\end{aligned}$$ This stress is uniform along every wall and reverses through the thickness, being greatest at the wall surfaces.
9. Superposing bending and torsional shear gives the peak stress. The bending shear stress is largest at mid-depth of the web, $\tau_{b} = q_{\max}/t = 44.64/3 = 14.88$ MPa; the torsional stress there is $238.1$ MPa on the surface. Since every wall has the same thickness, the torsional part is the same everywhere, so the maximum lies where the bending part peaks: $$\tau_{\max} = \tau_{T} + \tau_{b} = 238.1 + 14.88 = \boxed{253.0\ \text{MPa}}$$ at the surface of the web at mid-depth — roughly seventeen times the stress that the same load produces when it is applied through the shear centre.