Question 8 of 8: Shear Flow and Shear Centre of an Unequal-Flange Z-Section

Question 8: Shear Flow and Shear Centre of an Unequal-Flange Z-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 Z-section (median-line dimensions) carries a 3000 N vertical shear through its shear centre. Because the flanges are unequal the section is fully unsymmetric, so the product of inertia $I_{xy}$ enters the shear-flow law.

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

Approach. Locate the centroid and compute $I_x,I_y,I_{xy}$; apply the unsymmetric shear-flow formula (with $V_x=0$) walking from a free flange tip; the peak flow divided by $t$ gives $\tau_{\max}$, and the moment of the flow about the centroid locates the shear centre.

1. Centroid and inertias. Areas: web 180, top flange 90, bottom flange 150 mm². The centroid is $\bar{x}=5.71\ \text{mm}$ (right of the web), $\bar{y}=25.71\ \text{mm}$ (above the bottom flange). About centroidal axes, $I_x=2.625\times10^{5}$, $I_y=1.384\times10^{5}$, $I_{xy}=-1.427\times10^{5}\ \text{mm}^4$.
2. Unsymmetric shear-flow law. For $V_y$ only, $$q(s)=-\frac{V_y}{I_xI_y-I_{xy}^2}\big(I_y\,Q_x-I_{xy}\,Q_y\big),$$ where $Q_x=\int y\,t\,ds$, $Q_y=\int x\,t\,ds$ are accumulated (centroidal coordinates) from a free edge.
3. Flange flows (linear). Walking in from each tip, the flow builds linearly to the web junctions: $q=30.3\ \text{N/mm}$ at the top-flange/web corner and $q=22.8\ \text{N/mm}$ at the bottom-flange/web corner.
4. Web flow and its peak. Through the web the flow is roughly parabolic, peaking where $I_y\,y=I_{xy}\,\bar{x}_{\text{web}}$, i.e. $5.9\ \text{mm}$ above the centroid ($31.6\ \text{mm}$ from the bottom): $$\boxed{q_{\max}=61.7\ \text{N/mm}}$$ The computed flow closes to zero at both free tips and its resultant checks as $(F_x,F_y)=(0,\,3000)\ \text{N}$.
5. Maximum shear stress. Since $t$ is uniform, $\tau_{\max}=q_{\max}/t$: $$\boxed{\tau_{\max}=\frac{61.7}{3}=20.6\ \text{MPa}\ \text{in the web, }31.6\ \text{mm above the bottom}}$$
6. Shear centre. Taking moments of the shear flow about the centroid gives the horizontal offset $e_x=-0.32\ \text{mm}$ (just left of the centroid); repeating with a horizontal shear gives $e_y=-16.25\ \text{mm}$. $$\boxed{\text{Shear centre: }(e_x,e_y)=(-0.32,\,-16.2)\ \text{mm from the centroid}}$$ i.e. about 5.4 mm right of the web and 9.5 mm above the bottom flange.