Question 8 of 8: Cantilever with Partial UDL and Inclined Tip Load

Question 8: Cantilever with Partial UDL and Inclined Tip Load (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. Cantilever, total length $5\text{ m}$, fixed at the left end. UDL $w=3\text{ kN/m}$ over a $2\text{ m}$ length starting $1.5\text{ m}$ from the fixed end. Inclined point load $P=50\text{ kN}$ at the free tip, $60^{\circ}$ below horizontal (down and toward the wall). Unsymmetric I cross-section: bottom flange $50\times15\text{ mm}$, web $40\text{ mm}$ thick, top flange $100\times15\text{ mm}$, overall depth $300\text{ mm}$; point E is on the top surface of the top flange, at its left edge.

Find. Part (a): the normal-stress distribution at the base (combined axial + bending). Part (b): the maximum shear stress at the base. Part (c): the shear stress at point E.

Approach. Resolve the tip load into an axial component (compression, along the beam) and a transverse component (adds to the UDL reaction for shear/bending); find the section properties of the unsymmetric I-section (centroid not at mid-height), then superpose axial and bending stress for part (a), and use $\tau=VQ/(Ib)$ for parts (b) and (c).

1. Resolve the tip load. $F_y=P\sin60^{\circ}=43.30\text{ kN}$ (transverse, adds to the shear/bending); $F_x=P\cos60^{\circ}=25.0\text{ kN}$ (axial, directed toward the wall — compression).
2. Section properties (unsymmetric I, y measured from the bottom). Areas: bottom flange $750\text{ mm}^2$ at $\bar y=7.5$; web $10{,}800\text{ mm}^2$ at $\bar y=150$; top flange $1{,}500\text{ mm}^2$ at $\bar y=292.5$. Total $A=13{,}050\text{ mm}^2$; $\bar Y=\dfrac{\sum A_i\bar y_i}{\sum A_i}=\boxed{158.2\text{ mm}}$ (NOT mid-depth, since the flanges differ). $I=\sum\left[I_i+A_i(\bar y_i-\bar Y)^2\right]=\boxed{110.5\times10^{6}\text{ mm}^4}$.
3. Base reactions. $N=F_x=25.0\text{ kN}$ (compression); $V=w(2)+F_y=6+43.30=\boxed{49.30\text{ kN}}$; $M=w(2)(2.5)+F_y(5)=15+216.5=\boxed{231.5\text{ kN}\cdot\text{m}}$ (hogging: tension on top, compression on bottom, as for any cantilever under downward load).
4. Part (a) — combined stress. Bending alone: top fibre ($c=300-158.2=141.8\text{ mm}$) $\sigma_{bend,top}=Mc/I=+297.2\text{ MPa}$ (tension); bottom fibre ($c=158.2\text{ mm}$) $\sigma_{bend,bot}=-331.5\text{ MPa}$ (compression). Axial: $\sigma_{axial}=N/A=25{,}000/13{,}050=1.92\text{ MPa}$ (compression, uniform). Superposing, $\sigma_{top}=297.2-1.9=\boxed{+295.3\text{ MPa (tension)}}$; $\sigma_{bot}=-331.5-1.9=\boxed{-333.4\text{ MPa (compression, governs)}}$ — the bottom fibre governs because axial compression adds to (rather than relieves) the bending compression there.
5. Part (b) — maximum shear. Occurs at the neutral axis, in the web. $Q_{NA}=$ first moment of the area above the N.A. $=523{,}100\text{ mm}^3$ (upper web $+$ top flange, each times its own distance to the N.A.). $\tau_{\max}=\dfrac{VQ_{NA}}{Ib}=\dfrac{49{,}300\times523{,}100}{110.5\times10^6\times40}=\boxed{5.84\text{ MPa}}$ ($b=40\text{ mm}$, the web thickness at the N.A.).
6. Part (c) — shear at point E. Point E sits on the outer, TRACTION-FREE surface of the top flange (nothing lies above it), so $Q=0$ there by definition ($Q$ is the first moment of the area between the point and the free surface, which is zero at the surface itself). Hence $\tau_E=\boxed{0}$ — a conceptual result, not one requiring further calculation, and consistent with the boundary condition that shear stress must vanish at a free surface.