Question 1 of 7: Moments of Resistance of a Built-Up Section

Question A1: Moments of Resistance of a Built-Up Section (6 + 7 + 7 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 square HSS with a wide plate welded to the top face and a narrower plate welded to the bottom face, read from Figure A1:

Steel is G40.21 300W, so $F_y = 300$ MPa and $E = 200\,000$ MPa. Overall depth $H = 20 + 304.8 + 16 = 340.8$ mm; overall width (top plate) 500 mm. Axis a–a is the horizontal centroidal axis, axis b–b the vertical centroidal axis.

Find. The factored moments of resistance $M_{r,aa}$ and $M_{r,bb}$ of the fabricated section under CSA S16.

Approach. Locate the elastic centroid and compute $I$ about each axis, check every plate and wall against the Class 1 width-to-thickness limits, and — because all elements are Class 1 — take the moment of resistance as $M_r = \phi Z F_y$, with the plastic modulus found from the equal-area neutral axis.

1. Locate the elastic centroid about a–a. Taking first moments about the bottom face, with element centroids at 8 mm, 168.4 mm and 330.8 mm, $$\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{6400(8) + 14\,839(168.4) + 10\,000(330.8)}{31\,239} = 187.5 \text{ mm}$$ measured from the bottom, i.e. 153.3 mm below the top face. The centroid sits above mid-depth because the 500 mm plate is the heavier flange.
2. Second moment of area about a–a. Applying the parallel-axis theorem element by element, with $I_{HSS} = (304.8^4 - 279.4^4)/12 = 211.4 \times 10^6$ mm⁴, $$I_{aa} = \sum \left( \frac{b_i t_i^3}{12} + A_i d_i^2 \right) = 628.9 \times 10^6 \text{ mm}^4$$ which gives elastic section moduli $S_{top} = 628.9 \times 10^6 / 153.3 = 4103 \times 10^3$ mm³ and $S_{bot} = 628.9 \times 10^6 / 187.5 = 3353 \times 10^3$ mm³. The bottom fibre governs elastically.
3. Classify the elements. With $F_y = 300$ MPa the Class 1 limits are $145/\sqrt{F_y} = 8.37$ for a plate supported along one edge, $420/\sqrt{F_y} = 24.2$ for a plate supported along two edges, and $1100/\sqrt{F_y} = 63.5$ for a web in flexure. The actual ratios are $$\frac{(500-304.8)/2}{20} = 4.88, \qquad \frac{279.4}{20} = 13.97, \qquad \frac{279.4}{12.7} = 22.0$$ Every element is comfortably Class 1, so the section can develop its full plastic moment and $M_r = \phi Z F_y$ applies about both axes.
4. Plastic neutral axis about a–a. Because $F_y$ is uniform, the plastic axis divides the area equally: $A/2 = 15\,620$ mm². The top plate (10 000 mm²) plus the HSS top wall (304.8 × 12.7 = 3871 mm²) accounts for 13 871 mm², so the balance of 1749 mm² is taken from the two side walls of combined width 25.4 mm: $$y_p = 20 + 12.7 + \frac{1749}{25.4} = \boxed{101.5 \text{ mm from the top face}}$$ Note how far this sits above the elastic axis at 153.3 mm — the plastic axis balances areas, the elastic axis balances first moments.
5. Plastic modulus and $M_{r,aa}$. Summing $\sum A_i |y_i - y_p|$ over the six sub-areas above and below the plastic axis gives $Z_{aa} = 4149 \times 10^3$ mm³, a shape factor of $4149/3353 = 1.24$. Hence $$M_{r,aa} = \phi Z_{aa} F_y = 0.90 \times 4149 \times 10^3 \times 300 = \boxed{1120 \text{ kN}\cdot\text{m}}$$
6. Properties about b–b. Both plates are centred on the vertical axis, so the section is doubly symmetric about b–b and the centroid lies on the web centreline. Bending the plates about their own strong direction, $$I_{bb} = \frac{20(500)^3}{12} + \frac{16(400)^3}{12} + 211.4\times10^6 = 505.1 \times 10^6 \text{ mm}^4$$ so $S_{bb} = 505.1 \times 10^6 / 250 = 2020 \times 10^3$ mm³.
7. Plastic modulus and $M_{r,bb}$. With the plastic axis on the line of symmetry, each half-plate contributes $(b/2)(t)(b/4)$ and the HSS contributes $(D^3 - d^3)/4$: $$Z_{bb} = 2\left[\tfrac{500}{2}(20)\tfrac{500}{4}\right] + 2\left[\tfrac{400}{2}(16)\tfrac{400}{4}\right] + \frac{304.8^3 - 279.4^3}{4} = 3516 \times 10^3 \text{ mm}^3$$ $$M_{r,bb} = 0.90 \times 3516 \times 10^3 \times 300 = \boxed{949 \text{ kN}\cdot\text{m}}$$ The section is roughly 18 % stronger about a–a than about b–b, which is what the deeper 340.8 mm profile against the 500 mm width would suggest once the thin plates are accounted for.