Question 8 of 8: Shear flow, shear centre and torsion of an open I-section

Question 8: Shear flow, shear centre and torsion of an open I-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. A thin-walled open I-section, symmetric about the horizontal z axis, with unequal flange overhangs. All dimensions are to the mid-planes of the walls.

Find. (a) the shear flow distribution in the walls, (b) the position of the shear centre relative to the vertical web, and (c) the maximum shear stress if the same force is applied in the plane of the web instead.

Approach. Symmetry about the z axis makes \(I_{yz} = 0\), so ordinary shear-flow theory applies. Build the flow from the free flange tips inward, take moments about the web to find the shear centre, and treat part (c) as the same bending shear plus a St Venant torque on an open section.

1. Second moment of area. The web contributes \(t h^3/12\) and each flange its area at 500 mm from the axis: $$I_{xx} = \frac{2.5(1000)^3}{12} + 2\left[(420)(2.5)(500)^2\right] = 2.083\times10^{8} + 5.250\times10^{8} = 7.333\times10^{8}\ \text{mm}^4$$ The flanges provide 72% of the stiffness even though they hold only 46% of the material.
2. Shear flow in the flanges (part a). Starting from a free tip, where the flow must be zero, $$q(s) = \frac{S}{I_{xx}}\int_0^s t\,y\,\mathrm{d}s = \frac{S\,t\,(h/2)}{I_{xx}}\,s$$ so the flow grows linearly to the web junction: $$q_{240} = \frac{20\,000(2.5)(500)(240)}{7.333\times10^{8}} = 8.18\ \text{N/mm}, \qquad q_{180} = 6.14\ \text{N/mm}$$
3. Shear flow in the web. At the top of the web the two flange flows merge, and the web adds its own parabolic term: $$q_{\text{web,top}} = 8.18+6.14 = 14.32\ \text{N/mm}, \qquad q_{\text{NA}} = 14.32 + \frac{S\,t}{I_{xx}}\frac{(h/2)^2}{2} = \boxed{22.84\ \text{N/mm}}$$ so the largest shear stress under this loading is $$\tau_{\max} = \frac{22.84}{2.5} = 9.14\ \text{MPa}$$ Integrating the web flow over the depth returns 20,000 N — the web carries the entire vertical shear, the flanges carrying only horizontal self-balancing forces.
4. Flange forces. Each triangular flow block delivers $$F = \tfrac{1}{2}q_{\text{max}}b: \qquad F_{240} = \tfrac{1}{2}(8.18)(240) = 981.8\ \text{N}, \qquad F_{180} = \tfrac{1}{2}(6.14)(180) = 552.3\ \text{N}$$ They act in opposite directions along the flange, so each flange carries a net horizontal force of 429.5 N, and the top and bottom flanges carry it in opposite senses — a couple.
5. Shear centre (part b). Equating the moment of that couple about the web centre-line to \(S\,e\), $$e = \frac{(F_{240}-F_{180})\,h}{S} = \frac{(429.5)(1000)}{20\,000} = \boxed{21.5\ \text{mm}}$$ and the closed form \(e = t h^2\left(b_1^2-b_2^2\right)/4I_{xx}\) gives the same value. The shear centre lies 21.5 mm from the web on the side of the shorter (180 mm) overhang, because the longer flange carries the larger force and the resultant must be moved away from it to balance the couple.
6. Torque when the load moves to the web (part c). Applying the same 20 kN in the plane of the web is the load through the shear centre plus a torque about it: $$T = S\,e = 20\,000(21.5) = 4.295\times10^{5}\ \text{N}\cdot\text{mm}$$
7. St Venant torsion of an open section. An open section has no closed cell to carry the torque as a circulating flow, so it resists only by a through-thickness couple, with $$J = \tfrac{1}{3}\sum s\,t^3 = \tfrac{1}{3}(2.5)^3\left[1000+2(420)\right] = 9583\ \text{mm}^4, \qquad \tau_{\text{tor}} = \frac{T\,t}{J}$$ $$\tau_{\text{tor}} = \frac{(4.295\times10^{5})(2.5)}{9583} = 112.1\ \text{MPa}$$
8. Combine (part c). The torsional stress peaks at the wall surface everywhere; adding it to the bending shear at the neutral axis, which is where the bending shear peaks, $$\tau_{\max} = 112.1 + 9.1 = \boxed{121.2\ \text{MPa}}$$ at the surface of the web at mid-depth.

The comparison between parts (a) and (c) is the lesson of the question. Moving the line of action of the load by 21.5 mm — about two per cent of the section depth — multiplies the maximum shear stress by more than thirteen. An open thin-walled section is enormously less efficient in torsion than in bending, because its torsion constant scales with \(t^3\) rather than with the enclosed area; closing the same section with a light skin would drop that 112 MPa by two orders of magnitude. This is why aircraft and vehicle structures use closed boxes wherever a torque has to be carried, and why an open channel or I-beam must be loaded through its shear centre or restrained against twist.