Question 8 of 8: Shear flow, shear centre and warping torque in an open I-section

Question 8: Shear flow, shear centre and warping torque in 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 deep I-section, symmetric about the horizontal ($z$) axis but with unequal flange projections either side of the web:

Find. The complete shear-flow distribution when the load passes through the shear centre, and then the maximum shear stress when the same load is instead applied on the web.

Approach. Compute $I$, integrate the shear-flow equation outward from each free flange tip, close the walk at the web, and locate the shear centre from the couple formed by the unequal flange forces. For part (b), add the Saint-Venant torsional stress produced by moving the load onto the web.

1. Compute the second moment of area. The two flanges have identical total width $b_L + b_R = 350$ mm and sit at $y = \pm600$ mm, so$$I = \frac{t h^{3}}{12} + 2(b_L+b_R)t\left(\frac{h}{2}\right)^{2} = 360\times10^{6} + 630\times10^{6} = 990\times10^{6}\ \text{mm}^4 .$$Because the two flanges are identical, the section is symmetric about the horizontal axis, $I_{YZ} = 0$, and the ordinary shear-flow equation applies without any unsymmetrical-bending correction.
2. Walk the flange flows in from the free tips (part a). Starting from a tip, where $q = 0$, and integrating $q = -(S/I)\int y\,t\,\mathrm{d}s$ along a flange at constant $y = h/2$ gives a flow that grows linearly with distance from the tip. At the web junction,$$q_L = \frac{S}{I}\frac{h}{2}t\,b_L = \frac{23\,000}{990\times10^{6}}(600)(2.5)(200) = 6.97\ \text{N/mm},$$$$q_R = \frac{S}{I}\frac{h}{2}t\,b_R = 5.23\ \text{N/mm}.$$Both flow toward the web.
3. Continue down the web. The two flange flows merge at the top of the web, so it starts at$$q_{\text{top}} = q_L + q_R = 12.20\ \text{N/mm},$$and then grows parabolically as the web sweeps through the centroidal axis:$$q(y) = q_{\text{top}} + \frac{S t}{2I}\left[\left(\frac{h}{2}\right)^{2} - y^{2}\right].$$The maximum is at mid-height, $y = 0$:$$\boxed{q_{\max} = 12.20 + 10.45 = 22.65\ \text{N/mm},\qquad \tau_{\max} = \frac{22.65}{2.5} = 9.06\ \text{MPa}} .$$Integrating the web distribution confirms the statics — $q_{\text{top}}h + \tfrac23(q_{\max}-q_{\text{top}})h = 23\,000$ N — so the web alone carries the entire vertical shear, and the flanges carry none of it.
4. Locate the shear centre. The flange flows produce horizontal forces $H = Sht b^{2}/(4I)$ in each projection: $H_L = 697.0$ N in a long projection and $H_R = 392.0$ N in a short one. In the top flange both point away from the web, in the bottom flange both point toward it, so the projections form a couple of arm $h$. Equating that couple to $Se$,$$e = \frac{h^{2}t\left(b_L^{2} - b_R^{2}\right)}{4I} = \frac{(1200)^{2}(2.5)(200^{2}-150^{2})}{4(990\times10^{6})} = \boxed{15.91\ \text{mm toward the short flange}} .$$A cross-check from the couple directly, $(H_L - H_R)h/S = (697.0-392.0)(1200)/23\,000 = 15.91$ mm, agrees exactly. Note the direction: the shear centre moves toward the shorter projection, which is the same logic that puts a channel’s shear centre outside its web.
5. Move the load onto the web and find the torque (part b). Applying the same 23 kN on the web rather than 15.91 mm away from it superposes a torque$$T = S e = 23\,000(15.91) = 3.659\times10^{5}\ \text{N}\cdot\text{mm} = 366\ \text{N}\cdot\text{m}.$$An open section resists torque only by Saint-Venant shear, with torsion constant$$J = \frac{1}{3}\sum b_i t_i^{3} = \frac{(2\times350 + 1200)(2.5)^{3}}{3} = 9\,896\ \text{mm}^4 .$$
6. Add the torsional stress and report. The torsional shear stress on the wall surface is$$\tau_T = \frac{T\,t}{J} = \frac{3.659\times10^{5}(2.5)}{9\,896} = 92.4\ \text{MPa},$$which occurs on the surface of every 2.5 mm wall and therefore coincides at mid-web with the peak bending shear of 9.06 MPa. Adding them,$$\boxed{\tau_{\max} = 92.4 + 9.1 = 101.5\ \text{MPa, on the web surface at mid-height}} .$$The apparently trivial 15.91 mm offset multiplies the peak shear stress by more than eleven, because an open section is a catastrophically poor torsion member: $J$ is only 9 896 mm$^4$, some five orders of magnitude below the equivalent closed box.