Question 1 of 7: A1 — Moments of resistance of a plated Z-section

Question 1: A1 — Moments of resistance of a plated Z-section (12 + 8 = 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.

Find. The factored moments of resistance $M_{r,aa}$ and $M_{r,bb}$ about the two centroidal axes drawn on the figure.

Approach. Locate the centroid (immediate, because the section is centro-symmetric), classify every plate element to S16 Table 2, and because the section is Class 2 compute the resistance from the plastic section modulus, $M_r=\phi Z F_y$, about each axis in turn.

1. Establish the centroid and confirm the drawn axes. The section is made of three rectangles: the 500 × 25 horizontal plate, a 25 × 200 leg hanging below its left end, and an identical leg rising above its right end. Rotating the section 180° about the mid-point of the plate maps it onto itself, so the section is centro-symmetric and its centroid is that mid-point. The a-a and b-b axes drawn on Figure A1 are therefore already the centroidal axes, as the question implies. The gross area is $$A_g = 500(25) + 2(25)(200) = 12\,500 + 10\,000 = \boxed{22\,500\ \text{mm}^2}$$
2. Second moments of area about a-a and b-b. Taking each rectangle about its own centroid and transferring, with the leg centroids 112.5 mm above and below a-a and 237.5 mm either side of b-b:$$I_{aa}=\frac{500(25)^3}{12}+2\left[\frac{25(200)^3}{12}+5000(112.5)^2\right]=160.5\times10^{6}\ \text{mm}^4$$$$I_{bb}=\frac{25(500)^3}{12}+2\left[\frac{200(25)^3}{12}+5000(237.5)^2\right]=825.0\times10^{6}\ \text{mm}^4$$ The extreme fibres lie 212.5 mm from a-a and 250 mm from b-b, so the elastic moduli are $S_{aa}=756\times10^{3}$ mm³ and $S_{bb}=3300\times10^{3}$ mm³.
3. Classify the elements (S16 Table 2). The governing element is a leg projecting 200 mm from the plate that supports it, so it is an outstand with $b/t = 200/25 = 8.00$. For $F_y=350$ MPa the outstand limits are $145/\sqrt{F_y}=7.75$ (Class 1) and $170/\sqrt{F_y}=9.09$ (Class 2). Since $7.75 < 8.00 \le 9.09$ the legs are Class 2. The horizontal plate has the neutral axis inside it for both cases and $h/w = 500/25 = 20$, far below the Class 1 web limit $1100/\sqrt{F_y}=58.8$. The section is therefore Class 2 overall, which still permits the plastic moment to be developed (only the rotation capacity for plastic analysis is lost).
4. Plastic section moduli by equal areas. The plastic neutral axis splits the area in half. About a-a, the material above the axis is the upper half of the plate plus the whole upper leg, $6250 + 5000 = 11\,250$ mm², exactly $A_g/2$ — so the plastic axis coincides with a-a, and by the same argument with b-b. (This is a direct consequence of the centro-symmetry; it does not hold for a general built-up section, where the equal-area axis and the elastic centroid differ.) Summing $A_i|y_i|$:$$Z_{aa}=2\left[6250(6.25)\right]+2\left[5000(112.5)\right]=\boxed{1203\times10^{3}\ \text{mm}^3}$$$$Z_{bb}=2\left[6250(125)\right]+2\left[5000(237.5)\right]=\boxed{3938\times10^{3}\ \text{mm}^3}$$ The shape factors are $Z_{aa}/S_{aa}=1.59$ and $Z_{bb}/S_{bb}=1.19$ — the first is very high because for bending about a-a almost the whole plate area sits on the neutral axis and contributes nothing elastically.
5. Moments of resistance. With full lateral restraint there is no lateral-torsional buckling to reduce the capacity, so S16 13.5 applies directly for a Class 2 section:$$M_{r,aa}=\phi Z_{aa}F_y = 0.90(1203\times10^{3})(350)=\boxed{379\ \text{kN}\cdot\text{m}}$$$$M_{r,bb}=\phi Z_{bb}F_y = 0.90(3938\times10^{3})(350)=\boxed{1240\ \text{kN}\cdot\text{m}}$$ Bending about the vertical axis b-b is over three times stronger, because both legs and the full width of the plate are then working at a long lever arm.