Question 2 of 8: Shear Flow, Shear Centre and Maximum Shear Stress in a Channel

Question 2: Shear Flow, Shear Centre and Maximum Shear Stress in a Channel (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 channel with equal top and bottom flanges and a vertical web.

Find. (a) shear-flow distribution, (b) shear-centre location $e$, (c) location and magnitude of $\tau_{\max}$.

Approach. The section is symmetric about the horizontal ($x$) axis, so the centroid is at mid-height. Compute $I_x$, then the flexural shear flow $q=VQ/I_x$ (linear in each flange, parabolic in the web); the shear centre follows from the couple the flange forces make about the web.

1. Second moment of area (midline model). $$I_x=\frac{t_w h^3}{12}+2\,(b\,t_f)\!\left(\frac{h}{2}\right)^2=\frac{2(35)^3}{12}+2(25)(1.5)(17.5)^2=3.01\times10^{4}\ \text{mm}^4.$$
2. Flange shear flow (linear). With $s$ measured from the free tip, $Q(s)=t_f s\,(h/2)$, so $q$ rises from $0$ at the tip to $$q_{\text{junc}}=\frac{V\,t_f b\,(h/2)}{I_x}=\frac{1500(1.5)(25)(17.5)}{3.01\times10^{4}}=32.7\ \text{N/mm}.$$
3. Web shear flow (parabolic). Adding the web’s own first moment, the peak at the neutral axis is $$q_{NA}=\frac{V\!\left[b t_f\frac{h}{2}+t_w\frac{h}{2}\frac{h}{4}\right]}{I_x}=\frac{1500(962.5)}{3.01\times10^{4}}=47.9\ \text{N/mm}.$$
Q2(a). Flexural shear-flow: linear in each flange (0 at the tip, q = 32.7 N/mm at the junction) and parabolic in the web, peaking at the neutral axis.
4. Shear-centre location (part b). Each flange carries a horizontal force $H=\displaystyle\int_0^b q\,ds=\dfrac{V t_f h b^2}{4 I_x}=409\ \text{N}$. The two flange forces form a couple $H\,h$ that must equal $V e$: $$e=\frac{H h}{V}=\frac{V t_f h^2 b^2}{4 I_x V}=\boxed{9.53\ \text{mm}}$$ measured from the web, on the side away from the flanges (opposite the opening).
5. Maximum shear stress (part c). Convert flow to stress by dividing by the local thickness: at the neutral axis $\tau=q_{NA}/t_w=47.9/2=23.97\ \text{MPa}$; at the flange–web junction $\tau=q_{\text{junc}}/t_f=32.7/1.5=21.79\ \text{MPa}$. The larger governs: $$\boxed{\tau_{\max}=24.0\ \text{MPa at the neutral axis (in the web)}}$$