07-Str-A2 · May 2017

Question 2 of 7: A2 — Moments of resistance of a built-up plate section

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: A2 — Moments of resistance of a built-up plate section (8 + 12 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. From the figure, all plates 25 mm thick, G40.21-350W (F_y = 350 MPa):

Find. M_rx and M_ry, the factored moments of resistance of the cross-section about its two centroidal axes.

Approach. Locate the elastic centroid, compute I_x and I_y, classify every plate element under S16 Table 2, then — the section being Class 1 — find each plastic modulus from the equal-area axis and take M_r = φZF_y.

1. Locate the centroid. Measuring y upward from the underside of the bottom flanges, $$\bar{y} = \frac{10\,000(12.5) + 12\,500(150) + 15\,000(287.5)}{37\,500} = \boxed{168.3\ \text{mm}}$$ so the x–x axis sits 168.3 mm above the bottom and 131.7 mm below the top. The section is symmetric about y–y, so that axis is the vertical centreline.
2. Second moments of area. Summing I_own + Ad² over the five plates, $$I_x = 526.5\times10^{6}\ \text{mm}^{4}, \qquad I_y = 1384.0\times10^{6}\ \text{mm}^{4}$$ This is the first thing to notice about the section: it is wide and shallow, so I_y is 2.6 times I_x and y–y is the strong axis, not x–x. The corresponding radii of gyration are r_x = 118.5 mm and r_y = 192.1 mm, both needed in Question A3.
3. Classify the elements. With √F_y = 18.71, the Class 1 limits of S16 Table 2 are 145/√F_y = 7.75 for a flange outstand, 525/√F_y = 28.1 for a plate supported on both edges in uniform compression, and 1100/√F_y = 58.8 for a web in flexure. The actual ratios are

   Element	b/t or h/w	Class 1 limit
   Top-flange outstand past a web, 87.5/25	3.5	7.75
   Top flange between the webs, 375/25	15.0	28.1
   Web, 250/25	10.0	58.8
   Bottom-flange outstand, 87.5/25	3.5	7.75

   Every element is comfortably Class 1, so the full plastic moment may be used about both axes.
4. Locate the plastic neutral axis for x–x. The plastic axis divides the section into equal areas, not equal first moments. Half the area is 18 750 mm²; the two bottom flanges supply 10 000 mm², and the two webs then supply 50 mm² for every millimetre of height: $$y_p = 25 + \frac{18\,750 - 10\,000}{2 \times 25} = \boxed{200\ \text{mm from the bottom}}$$ This sits 31.7 mm above the elastic centroid — the two axes coincide only in a doubly symmetric section.
5. Plastic modulus about x–x. Taking each part's area times the distance from its own centroid to y_p, $$Z_x = 10\,000(187.5) + 8750(87.5) + 3750(37.5) + 15\,000(87.5) = \boxed{4.094\times10^{6}\ \text{mm}^{3}}$$
6. Plastic modulus about y–y. Symmetry puts the plastic axis on the centreline, so $$Z_y = 2(7500)(150) + 2(6250)(200) + 2(5000)(200) = \boxed{6.750\times10^{6}\ \text{mm}^{3}}$$
7. Moments of resistance. The question asks for the resistance of the cross-section, so no lateral-torsional reduction applies: $$M_{rx} = \phi Z_x F_y = 0.90 \times 4.094\times10^{6} \times 350 = \boxed{1290\ \text{kN}\cdot\text{m}}$$ $$M_{ry} = \phi Z_y F_y = 0.90 \times 6.750\times10^{6} \times 350 = \boxed{2126\ \text{kN}\cdot\text{m}}$$
8. Sanity-check with the elastic moduli. S_x = 526.5×10⁶/168.3 = 3.128×10⁶ mm³ to the bottom fibre and S_y = 1384.0×10⁶/300 = 4.613×10⁶ mm³, giving shape factors of 1.31 and 1.46. Both are far above the 1.12 typical of a rolled I-shape, which is exactly what one expects when so much of the area sits close to the neutral axis.