Question 7 of 8: T-Beam with Mid-Span Point Load

Question 7: T-Beam with Mid-Span Point 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. Simply supported, span $8\text{ m}$, concentrated load $P=30\text{ kN}$ at mid-span ($4\text{ m}$ from each support). T cross-section: top flange $150\text{ mm}$ wide $\times25\text{ mm}$ thick; web $30\text{ mm}$ wide; overall depth $150\text{ mm}$ (web depth $=125\text{ mm}$). Allowable $\sigma=210\text{ MPa}$, $\tau=70\text{ MPa}$.

Find. Part (a): maximum absolute normal stress and maximum shear stress in the beam. Part (b): shear stress at point E (flange tip) at the left support section, with justification.

Approach. Locate the T-section centroid and moment of inertia by parts, find the maximum moment (mid-span) and shear (constant magnitude on each half) for the point-load beam, then apply the flexure and shear-flow formulas at the governing fibre and section.

1. Section centroid (y from the bottom of the web). Web: $A_w=30\times125=3{,}750\text{ mm}^2$ at $\bar y_w=62.5\text{ mm}$. Flange: $A_f=150\times25=3{,}750\text{ mm}^2$ at $\bar y_f=137.5\text{ mm}$ — the two areas happen to be EQUAL, so $\bar Y=\tfrac{1}{2}(62.5+137.5)=\boxed{100\text{ mm}}$ from the bottom.
2. Moment of inertia. $I=\left[\tfrac{30(125)^3}{12}+3{,}750(62.5-100)^2\right]+\left[\tfrac{150(25)^3}{12}+3{,}750(137.5-100)^2\right]=\boxed{15.625\times10^{6}\text{ mm}^4}$.
3. Reactions and internal forces. By symmetry, each reaction $=P/2=15\text{ kN}$; $V=15\text{ kN}$ (constant magnitude over each half-span); $M_{max}=15(4)=\boxed{60\text{ kN}\cdot\text{m}}$ at mid-span.
4. Part (a) — maximum normal stress. The web tip ($c_{bot}=100\text{ mm}$) is farther from the centroid than the flange top ($c_{top}=50\text{ mm}$), so it governs: $\sigma_{max}=Mc_{bot}/I=60\times10^6(100)/15.625\times10^6=\boxed{384\text{ MPa (tension, at the web tip, mid-span)}}$. (The flange-top compressive stress is smaller, $60\times10^6(50)/15.625\times10^6=192\text{ MPa}$, and does not govern.) This EXCEEDS the stated allowable of $210\text{ MPa}$, i.e. the section as given is not adequate for this load — the question asks only for the computed maximum stress, so that value is reported as found.
5. Part (a) — maximum shear stress. Occurs at the neutral axis, in the web ($t=30\text{ mm}$). $Q_{NA}=$ (area of web above the N.A.)$\times$(its own centroid distance) $+$ (flange area)$\times$(its centroid distance) $=(25)(30)(12.5)+3{,}750(37.5)=150{,}000\text{ mm}^3$. $\tau_{max}=VQ_{NA}/(It)=15{,}000(150{,}000)/(15.625\times10^6\times30)=\boxed{4.80\text{ MPa}}$, well inside the $70\text{ MPa}$ allowable.
6. Part (b) — shear stress at point E. Point E sits at the very TIP (outer edge) of the flange width, a traction-free surface with no material beyond it. The shear-flow formula $\tau=VQ/(It)$ requires $Q$, the first moment of the area BEYOND the point of interest; at the flange tip that area is zero by definition, so $Q=0$ and hence $\tau_E=\boxed{0}$ — a boundary-condition result (shear stress must vanish at a free surface), not one requiring further calculation, and it holds at the left-support section exactly as it would anywhere else along the span.