Question 1 of 7: Moments of resistance of a built-up Z-section

Question A1: Moments of resistance of a built-up Z-section (12 + 8 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 built-up Z-section from 25 mm G40.21 350W plates (F_y=350 MPa): a 500 mm long horizontal web plate (two 250 mm halves), with a 200 mm leg turned down at one end and a 200 mm leg turned up at the other – a doubly point-symmetric Z. Full lateral restraint is provided along the member.

Find. (a) M_r about the horizontal centroidal axis a–a; (b) M_r about the vertical centroidal axis b–b.

Approach. Because the section is point-symmetric about its own centroid, that centroid also splits the area equally on either side of both a–a and b–b – so the elastic and plastic neutral axes coincide for each direction. Compute A, I_xx, I_yy from first principles, classify the plate elements per CSA S16 Table 2, then take M_r=φZF_y for each axis (Class ≤2, and "fully lateral restraint" removes any LTB reduction).

1. Section properties. Web (500×25) plus two 200×25 legs: $$A = 500(25) + 2(200)(25) = \boxed{22\,500\text{ mm}^2}$$ By symmetry the centroid is at the geometric middle of the web, on both a–a and b–b.
2. Second moments of area. Summing each plate’s own inertia plus $$Ad^2$$ about the centroid (leg centroids sit 112.5 mm off a–a and 237.5 mm off b–b): $$I_{xx} = \frac{500(25)^3}{12} + 2\left[\frac{25(200)^3}{12} + 5000(112.5)^2\right] = \boxed{160.5\times10^6\text{ mm}^4}$$ $$I_{yy} = \frac{25(500)^3}{12} + 2\left[\frac{200(25)^3}{12} + 5000(237.5)^2\right] = \boxed{825\times10^6\text{ mm}^4}$$
3. Plastic section moduli. The equal-area axis is confirmed to lie on both a–a and b–b (each splits the 22 500 mm² exactly in half), so $$Z=\Sigma A_i|y_i|$$ using each half’s own centroidal distance: $$Z_{xx} = 2\left[5000(112.5)+3125(6.25)\right] = \boxed{1.203\times10^6\text{ mm}^3}$$ $$Z_{yy} = 2\left[5000(237.5)+3125(125)\right] = \boxed{3.9375\times10^6\text{ mm}^3}$$
4. Classification (CSA S16 Table 2). Each 200 mm leg is an unstiffened outstand of the 25 mm plate, $$b/t=200/25=8.0$$; with F_y=350 MPa the Class 1 limit is $$145/\sqrt{F_y}=7.75$$ and Class 2 is $$170/\sqrt{F_y}=9.09$$ – so the legs are Class 2. The 500 mm web, welded to a leg at each end, is a stiffened element with clear $$b/t=(500-50)/25=18.0$$, far below its Class 1 limit of 28.1 – Class 1. Class ≤2 throughout permits the full plastic M_r=φZF_y for both axes.
5. Moments of resistance. With φ=0.90 and full lateral restraint (no LTB reduction): $$M_{rx} = \phi Z_{xx}F_y = 0.9(1.203\times10^6)(350) = \boxed{379\text{ kN}\cdot\text{m}}\ \text{(about a-a)}$$ $$M_{ry} = \phi Z_{yy}F_y = 0.9(3.9375\times10^6)(350) = \boxed{1240\text{ kN}\cdot\text{m}}\ \text{(about b-b)}$$