Question 8 of 8: Cast-Iron Beam with an Unsymmetric I-Section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2017. Closed-book, 3 hours;
any FIVE of the eight questions constitute a complete paper (all eight are answered below as a
complete study resource). Aid sheet permitted; geometric properties of
W-shape sections were supplied as an appendix (not required for any of the eight questions solved
here).
top flange 120×20 mm; web 30 mm thick ×200 mm; bottom flange 300×20 mm (overall depth 240 mm)
Find. Whether the beam fails in flexure, whether it fails in shear (both at the
applied 25 kN/m), and the maximum UDL the section can actually carry.
Unsymmetric I-section: narrower flange in compression (top), wider flange in tension (bottom).
Approach. A simply supported UDL beam is sagging everywhere, so the bottom fibre
is always in tension and the top in compression — hence the wider (300 mm) flange is placed
at the bottom, where the lower allowable stress (45 MPa) needs the larger section modulus. Locate
the centroid and $I$ of the unsymmetric section, find the allowable moment from each
of the two (different) allowable normal stresses, compare with the actual applied moment and shear,
then back out the true load capacity.
Centroid (measuring y from the bottom of the 300×20 mm bottom flange):
$$\bar y=\frac{\sum A_i y_i}{\sum A_i}=92.5\text{ mm from the bottom}$$
so $c_{bot}=92.5$ mm, $c_{top}=240-92.5=147.5$ mm.
Moment of inertia about the centroid (parallel-axis theorem on each of the
three rectangles):
$$I=1.1103\times10^{8}\text{ mm}^4$$
Allowable moment from each stress limit.
$$M_{allow,tension}=\frac{\sigma_{t}I}{c_{bot}}=\frac{45(1.1103\times10^8)}{92.5}=54.01\text{ kN}\cdot\text{m}$$
$$M_{allow,comp}=\frac{\sigma_{c}I}{c_{top}}=\frac{100(1.1103\times10^8)}{147.5}=75.27\text{ kN}\cdot\text{m}$$
The smaller value governs flexure: $M_{allow}=54.01$ kN·m (bottom, tension side).
Applied moment and shear at 25 kN/m.
$$M_{applied}=\frac{wL^2}{8}=\frac{25(4)^2}{8}=50.0\text{ kN}\cdot\text{m}, \qquad V_{applied}=\frac{wL}{2}=50.0\text{ kN}$$
(a) Flexure check. Since $M_{applied}=50.0 kN·m,
$$\boxed{\text{the beam does NOT fail in flexure at 25 kN/m}}$$
Shear stress at the neutral axis. First moment of the area below the NA
(the full bottom flange plus the part of the web below the NA), divided by the web width:
$$Q=573{,}844\text{ mm}^3, \qquad \tau_{applied}=\frac{V_{applied}\,Q}{I\,t_{web}}=\frac{(50{,}000)(573{,}844)}{(1.1103\times10^8)(30)}=8.61\text{ MPa}$$
(b) Shear check. Since $\tau_{applied}=8.61<10$ MPa allowable,
$$\boxed{\text{the beam does NOT fail in shear at 25 kN/m either}}$$
(c) Maximum sustainable load. Scaling each allowable limit back to an
equivalent UDL:
$$w_{max,flexure}=\frac{8M_{allow}}{L^2}=\frac{8(54.01)}{16}=27.01\text{ kN/m}$$
$$w_{max,shear}=\frac{2V_{max}}{L}, \quad V_{max}=\frac{\tau_{allow}\,I\,t_{web}}{Q}=58.05\text{ kN}\ \Rightarrow\ w_{max,shear}=29.02\text{ kN/m}$$
Flexure (tension side) governs, being the smaller of the two:
$$\boxed{w_{max}=27.0\text{ kN/m}}$$
