Cast-Iron Beam With an Unsymmetric Section: 04-BS-6, May 2017, Q8

Question 8: Cast-Iron Beam With an Unsymmetric 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. Simply supported cast-iron beam, span 4 m, UDL 25 kN/m. Unsymmetric I-section: bottom flange 300×20 mm (wide, tension side), web 30×200 mm, top flange 120×20 mm. Allowable tension 45 MPa, compression 100 MPa, shear 10 MPa.

Quantity	Value
Span, L	4 m
UDL, w	25 kN/m
Bottom flange	300×20 mm
Web	30×200 mm
Top flange	120×20 mm
Allow. tension / compression / shear	45 / 100 / 10 MPa

Find. Whether the beam fails in flexure or shear at 25 kN/m, and the maximum sustainable load.

Simply supported beam, 4 m span, 25 kN/m UDL.

Unsymmetric section: wide 300 mm bottom flange (tension side), narrow 120 mm top flange.

Approach. Locate the centroid of the unsymmetric section, compute the maximum moment and shear from statics, then check tensile, compressive, and shear stress against their respective allowables — scaling linearly to find the governing capacity.

Section properties. Bottom flange 300×20 mm, web 30×200 mm, top flange 120×20 mm, overall depth 240 mm. Locating the centroid from the bottom: $$\bar y=92.50\text{ mm},\qquad I=1.1103e+08\text{ mm}^4$$ The wide (300 mm) flange at the bottom pulls the centroid toward that side, giving $y_{bot}=92.50$ mm, less than $y_{top}=147.50$ mm — deliberately reducing the stress on the weaker tension side.
Design moment and shear. Simply supported, L=4 m, w=25 kN/m: $$M_{max}=\frac{wL^2}{8}=50.0\text{ kN}\!\cdot\!\text{m}\ (\text{sagging, bottom in tension}),\qquad V_{max}=\frac{wL}{2}=50\text{ kN}$$
(a) Flexural check. $$\sigma_{tension}(\text{bottom})=\frac{M_{max}\,y_{bot}}{I}=\boxed{41.66\text{ MPa}}\ (\lt\ 45\text{ MPa allow}),\qquad \sigma_{compression}(\text{top})=\frac{M_{max}\,y_{top}}{I}=\boxed{66.42\text{ MPa}}\ (\lt\ 100\text{ MPa allow})$$ Neither exceeds its allowable — the beam does NOT fail in flexure at 25 kN/m, though the tension side is close (ratio 0.93).
(b) Shear check. $Q_{NA}=573844$ mm³, giving $$\tau_{max}=\frac{V_{max}Q_{NA}}{It_w}=\boxed{8.614\text{ MPa}}\ (\lt\ 10\text{ MPa allow})$$ The beam does NOT fail in shear either.
(c) Maximum sustainable load. All three stresses scale linearly with w, so the governing (smallest) capacity is $$w_{tension}=27.01,\quad w_{compression}=37.64,\quad w_{shear}=29.02\ \text{kN/m}$$ $$\boxed{w_{max}=27.01\text{ kN/m}}\ \text{(governed by tensile flexural stress)}$$

Flexural stress distribution: +41.7 MPa tension at the bottom, −66.4 MPa compression at the top.

Quantity	Value
σ_tension (bottom, at 25 kN/m)	41.66 MPa (safe)
σ_compression (top, at 25 kN/m)	66.42 MPa (safe)
τ_max (at 25 kN/m)	8.61 MPa (safe)
Fails at 25 kN/m?	No (flexure and shear both safe)
Maximum load, w_max	27.01 kN/m (tension-governed)

Topic: Flexure of an unsymmetric section in a brittle (unequal tension/compression) material.

Key relations: centroid by first moments of area; $\sigma=My/I$ evaluated separately at $y_{top}$ and $y_{bot}$; $\tau_{max}=VQ_{NA}/(It_w)$.

Why this works: Placing the wider flange on the tension side moves the neutral axis toward it, shrinking $y_{bot}$ relative to $y_{top}$ and so reducing the tensile stress for a given moment — the standard design trick for brittle materials whose tensile strength is the weaker property.

Common pitfall: using the SAME distance from the centroid for both tension and compression checks (as one would for a symmetric section) when the section is deliberately unsymmetric; each side needs its own $y$.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6.6 (Bending of Unsymmetric/Composite Sections; Brittle Materials).

Question 8 of 8: Cast-Iron Beam With an Unsymmetric Section

Question 8: Cast-Iron Beam With an Unsymmetric Section (20 marks)