Question 1 of 8: Shear flow, shear centre and torsion of an open I-section

Question 1: 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. An open I-section whose two flanges are identical but are not centred on the web: from the figure each flange runs 250 mm to one side of the web and 125 mm to the other, so the section is symmetric about the horizontal ($z$) axis only.

Find. (a) the complete shear-flow distribution in the flanges and web, (b) the horizontal distance from the web to the shear centre, and (c) the largest shear stress anywhere in the section once the same 25 kN is moved onto the web centre-line.

Approach. Because the section is symmetric about the horizontal axis, $I_{YZ}=0$ and the open-section shear flow reduces to $q_s = (S_y/I_{ZZ})\int_0^s t\,Y\,ds$ walked from each free edge; the four flange forces then form a couple whose arm is the web depth, and equating that couple to $S_y e$ locates the shear centre, after which moving the load onto the web adds a St Venant torque to the same bending shear flow.

1. Part (a) — second moment of area about the horizontal centroidal axis. The centroid sits at mid-depth by symmetry. Adding the web, the two flanges shifted to $Y=\pm h/2$, and the flanges’ own (negligible) inertia, $$I_{ZZ} = \frac{t h^{3}}{12} + 2\,(b_L+b_R)t\left(\frac{h}{2}\right)^{2} + 2\,\frac{(b_L+b_R)t^{3}}{12}$$Substituting $t=3$, $h=320$, $b_L+b_R=375$ mm gives $8.192\times10^{6} + 57.600\times10^{6} + 1.7\times10^{3}$, so $$\boxed{I_{ZZ} = 65.79\times10^{6}\ \text{mm}^{4}}$$
2. Flange shear flow, walked from each free tip. A flange lies at constant $Y = h/2 = 160$ mm, so the first moment grows linearly with the distance $s$ from the tip and $$q(s) = \frac{S_y}{I_{ZZ}}\,t\,\frac{h}{2}\,s = \frac{25\,000}{65.79\times10^{6}}(3)(160)\,s = 0.1824\,s \ \text{N/mm}$$The flow therefore rises from zero at each tip to $q_L = 0.1824(250) = 45.60$ N/mm where the long flange meets the web and to $q_R = 0.1824(125) = 22.80$ N/mm at the short-flange junction. Because both halves of the top flange drain toward the web, the flow entering the web at the top is their sum, $68.40$ N/mm; the bottom flange is the mirror image with the flow running outward.
3. Web shear flow and its peak. Continuing down the web from the top junction, each strip of web adds its own first moment, so $$q(Y) = q_{\text{top}} + \frac{S_y t}{2 I_{ZZ}}\left[\left(\frac{h}{2}\right)^{2} - Y^{2}\right]$$which is parabolic and peaks at mid-depth ($Y=0$): $q_{\max} = 68.40 + (3.800\times10^{-4})(3)(160^{2}/2)$, that is $$\boxed{q_{\max} = 82.99\ \text{N/mm}\quad\text{at web mid-depth}}$$Dividing by the 3 mm wall, the bending shear stress peaks at $\tau = 82.99/3 = 27.66$ MPa. As a check on the whole distribution, integrating $q(Y)$ over the 320 mm web returns 25.0 kN — the web carries the entire vertical shear, as it must for a section with horizontal flanges.
4. Part (b) — the flange forces that locate the shear centre. Integrating the linear flange flow gives a force $F = \tfrac{1}{2}q_{\text{junction}}\,b$ in each flange half. The long and short halves of one flange pull in opposite directions, so each flange delivers a net horizontal force $$\Delta F = \frac{S_y t h}{4 I_{ZZ}}\left(b_L^{2}-b_R^{2}\right) = 4\,275\ \text{N}$$The top flange pushes one way and the bottom flange the other, so the two form a couple of arm $h$.
5. Shear-centre offset. Equating that couple to the moment of the applied shear about the web, $S_y e = h\,\Delta F$, gives the closed form $e = h^{2}t\,(b_L^{2}-b_R^{2})/(4 I_{ZZ})$ and $$\boxed{e = 54.72\ \text{mm from the web, on the side of the 125 mm (short) flange}}$$The direction is the usual channel result seen from a different angle: the longer flange half carries the larger flange force, so the shear centre retreats away from it.
6. Part (c) — the torque introduced by moving the load to the web. Applying the same 25 kN on the web centre-line is statically the shear-centre case plus a torque $$T = S_y e = 25\,000 \times 54.72 = 1.368\times10^{6}\ \text{N}\cdot\text{mm}$$An open section resists this only in St Venant torsion, with $J = \sum b t^{3}/3 = (2\times375 + 320)(3)^{3}/3 = 9\,630$ mm⁴ — nearly four orders of magnitude below $I_{ZZ}$ (a ratio of about 6 800).
7. Maximum shear stress under the eccentric load. The torsional shear stress is uniform through the thickness in magnitude and peaks at the wall surface, $\tau_T = T t / J = (1.368\times10^{6})(3)/9\,630 = 426.1$ MPa. At mid-depth of the web the torsional and bending shear stresses act along the same line on one face, so they add: $$\boxed{\tau_{\max} = 426.1 + 27.7 = 453.8\ \text{MPa}}$$That is a factor of 16 above the shear stress the same load causes when it passes through the shear centre.