25-Nav-A5 Ship Design · May 2016

Question 8 of 8: Open Channel Cross-Section — Shear Flow, Shear Centre, and Maximum Shear Stress

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 8: Open Channel Cross-Section — Shear Flow, Shear Centre, and Maximum Shear Stress (equal value)

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. Thin open channel, symmetric about the horizontal centroidal axis: web height 40 mm (thickness 3 mm), flanges 30 mm long top and bottom (thickness 2 mm each, equal by the stated symmetry), thin-wall (centreline) dimensions.

Find. (a) shear-flow distribution; (b) shear-centre location; (c) maximum shear stress.

Approach. Compute $I_{xx}$ about the horizontal (symmetry) axis, integrate $q=VQ/I_{xx}$ from each flange's free tip inward to get the flange shear flow and its value entering the web, continue through the web to the neutral axis for the maximum, then locate the shear centre by equating the moment of the flange shear-flow forces to $Ve$ about the web line.

1. Section moment of inertia. Thin-wall approximation, each flange treated as area $bt_f$ at $\pm H/2$ from the axis, web as a thin rectangle centred on the axis: $$I_{xx}=\frac{t_wH^3}{12}+2\left[b t_f\left(\frac{H}{2}\right)^2\right]=1.60\times10^{-8}+2(2.40\times10^{-8})=6.404\times10^{-8}\text{ m}^4=64{,}040\text{ mm}^4$$
2. Part (a) — flange shear flow. Starting from the free tip ($q=0$) and integrating toward the web, at a distance $\xi$ from the tip ($0\le\xi\le b$): $Q(\xi)=t_f\xi(H/2)$, so $q(\xi)=VQ(\xi)/I_{xx}$ rises linearly to its value at the web junction: $$q_{\text{junction}}=\frac{V\,t_fb(H/2)}{I_{xx}}=\frac{2500(0.002)(0.03)(0.02)}{6.404\times10^{-8}}=46{,}846\text{ N/m}$$ $$\boxed{q_{\text{junction}}\approx46.85\text{ N/mm}}\ \text{(same in both flanges, by symmetry)}$$
3. Web shear flow. Continuing from the top of the web ($Q=Q_{\text{junction}}$) down to the neutral axis ($z=0$), $Q(z)=Q_{\text{junction}}+\tfrac{t_w}{2}\!\left[(H/2)^2-z^2\right]$ is parabolic, maximised at $z=0$: $$Q_{\max}=t_fb(H/2)+\frac{t_w(H/2)^2}{2}=1.2\times10^{-6}+0.6\times10^{-6}=1.8\times10^{-6}\text{ m}^3$$ $$\boxed{q_{\max}=\frac{VQ_{\max}}{I_{xx}}=\frac{2500(1.8\times10^{-6})}{6.404\times10^{-8}}\approx70.27\text{ N/mm}}\ \text{(at the neutral axis, in the web)}$$
4. Part (b) — shear centre. The resultant force carried by one flange is $F_f=\int_0^bq(\xi)\,d\xi=\dfrac{Vt_fb^2(H/2)}{2I_{xx}}=702.7\text{ N}$. Equating the moment of the top+bottom flange force couple about the web line to $Ve$: $$e=\frac{F_f\,H}{V}=\frac{702.7(0.04)}{2500}$$ $$\boxed{e\approx11.24\text{ mm, measured from the web centreline, on the side AWAY from the flanges}}$$ (the classical result that a channel's shear centre lies outside the open section, behind the web).
5. Part (c) — maximum shear stress. Converting shear flow to stress ($\tau=q/t$) at the two candidate locations: $$\tau_{\text{web, N.A.}}=\frac{q_{\max}}{t_w}=\frac{70{,}269}{0.003}=23.42\text{ MPa}\qquad \tau_{\text{flange-web junction}}=\frac{q_{\text{junction}}}{t_f}=\frac{46{,}846}{0.002}=23.42\text{ MPa}$$ $$\boxed{\tau_{\max}\approx23.42\text{ MPa, at the neutral axis in the web (and equal, for these proportions, at the flange–web junction)}}$$