16-Civ-A2 Elementary Structural Design · May 2015

Question 3 of 7: A3 — Moments of resistance of the built-up plate section

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

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 3: A3 — Moments of resistance of the built-up plate 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.

Plate	Size (mm)	Area (mm²)	Position
Bottom plate	460 × 20	9200	soffit; centroid 10 mm above the base
Webs (2)	20 × 210	4200 each	centrelines at ±150 mm from b–b
Top plates (2)	100 × 20	2000 each	centred on the webs, 240 mm above the base
Overall	460 wide × 250 deep	21 600 total	symmetric about b–b only
Steel	G40.21 350W: F_y = 350 MPa

Find. The factored moments of resistance M_r about the horizontal axis a–a and the vertical axis b–b.

Figure A2 — built-up section from 20 mm G40.21 350W plates. The elastic centroid O lies 97.3 mm above the soffit; the plastic axis for a–a bending is 37 mm lower, at 60 mm.

Approach. Locate the elastic centroid, classify every plate element under both bending directions, then locate the equal-area (plastic) axis for each direction and sum $Z = \sum A_i |y_i - y_p|$.

Gross area and elastic centroid. Measuring y upward from the soffit, $$ A = 9200 + 2(4200) + 2(2000) = 21\,600~\text{mm}^2 $$ $$ \bar{y} = \frac{9200(10) + 8400(125) + 4000(240)}{21\,600} = \frac{2\,102\,000}{21\,600} = 97.3~\text{mm} $$ The section is symmetric about b–b but strongly unsymmetric about a–a, with the centroid sitting well below mid-depth because the 460 mm bottom plate is much larger than the two 100 mm top plates.
Classify every element (S16 Table 2, $F_y = 350$). The top-plate outstands are $(100-20)/2 = 40$ mm, giving $b/t = 2.0$ against a Class 1 limit of $145/\sqrt{350} = 7.75$. Each web has $h/w = 210/20 = 10.5$ against $1100/\sqrt{350} = 58.8$. The bottom plate spans 280 mm clear between the webs, $b/t = 14.0$ against the supported-both-edges limit $525/\sqrt{350} = 28.1$, and in its own plane $460/20 = 23.0$ against 58.8. Every element is Class 1 for both bending directions, so the plastic moment governs both answers and no effective-width reduction is needed.
Plastic neutral axis about a–a. The plastic axis divides the area equally, not the first moments. Half the area is 10 800 mm², of which the bottom plate supplies 9200 mm²; the remaining 1600 mm² must come from the two webs, whose combined width is 40 mm: $$ y_p = 20 + \frac{1600}{40} = \boxed{60~\text{mm above the soffit}} $$ Note that this sits 37 mm below the elastic centroid at 97.3 mm — a reminder that plastic analysis equalises areas while elastic analysis equalises first moments.
Plastic modulus about a–a. Summing $A_i|y_i - y_p|$ over the four regions:

Region A (mm²) Lever arm (mm) A × arm (mm³)

Bottom plate 9200 50.0 460 000

Webs below the axis 1600 20.0 32 000

Webs above the axis 6800 85.0 578 000

Top plates 4000 180.0 720 000

Z_a-a 1 790 000

$$ M_{r,a-a} = \phi Z_{a-a} F_y = 0.90(1.790\times10^6)(350) = \boxed{564~\text{kN}\cdot\text{m}} $$
Plastic modulus about b–b. Because the section is symmetric about b–b, the equal-area axis is the centreline and $Z$ is twice the first moment of the half-section about it. For the right-hand half: the bottom-plate half is 230 × 20 = 4600 mm² at 115 mm, one web is 4200 mm² at 150 mm and one top plate is 2000 mm² at 150 mm. $$ Z_{b-b} = 2\left[4600(115) + 4200(150) + 2000(150)\right] = 2(1\,459\,000) = 2.918\times10^6~\text{mm}^3 $$ $$ M_{r,b-b} = 0.90(2.918\times10^6)(350) = \boxed{919~\text{kN}\cdot\text{m}} $$
Interpret the comparison. The resistance about the vertical axis is 63 percent larger than about the horizontal axis. That is not an error: the section is 460 mm wide and only 250 mm deep, so the material is spread further from b–b than from a–a. For completeness the elastic properties are $I_{a-a} = 189.3\times10^6$ mm⁴ with $S_{a-a} = 1.240\times10^6$ mm³ (extreme fibre at the top, 152.7 mm from the centroid), and $I_{b-b} = 444.8\times10^6$ mm⁴ with $S_{b-b} = 1.934\times10^6$ mm³. The shape factors are 1.44 and 1.51 respectively — both far above the 1.12 typical of a rolled I-shape, because so much of this section's area sits near its own neutral axis.

Region	A (mm²)	Lever arm (mm)	A × arm (mm³)
Bottom plate	9200	50.0	460 000
Webs below the axis	1600	20.0	32 000
Webs above the axis	6800	85.0	578 000
Top plates	4000	180.0	720 000
Z_a-a	1 790 000

Check — lateral support. The question asks for the section moments of resistance, so the values above are the fully braced Class 1 capacities $\phi Z F_y$. If the member were used with an unbraced length in a-a bending, the lateral-torsional buckling provisions of S16 Clause 13.6 would reduce M_r,a-a; the b–b value is unaffected because bending about the weak-in-depth axis cannot buckle laterally.

Quantity	a–a (horizontal)	b–b (vertical)
Gross area	21 600 mm²
Elastic centroid	97.3 mm above the soffit	on the centreline
Plastic neutral axis	60.0 mm above the soffit	on the centreline
Second moment of area, I	189.3×10⁶ mm⁴	444.8×10⁶ mm⁴
Elastic modulus, S	1.240×10⁶ mm³	1.934×10⁶ mm³
Plastic modulus, Z	1.790×10⁶ mm³	2.918×10⁶ mm³
Class	Class 1 (all elements)
Moment of resistance, M_r	564 kN·m	919 kN·m

Topic. Plastic section properties and element classification of built-up plate sections.

Key relations. Equal-area condition $\sum A_{\text{below}} = \sum A_{\text{above}} = A/2$; $Z = \sum A_i|y_i - y_p|$; $M_r = \phi Z F_y$ for Class 1 and 2; S16 Table 2 width-to-thickness limits $145/\sqrt{F_y}$ (outstand), $525/\sqrt{F_y}$ (supported both edges) and $1100/\sqrt{F_y}$ (web in flexure).

Why this works. At the plastic limit every fibre has yielded, so the stress magnitude is F_y everywhere and only its sign changes across the neutral axis. Axial equilibrium then requires equal tension and compression areas, which is why the plastic axis is an equal-area axis rather than a centroidal one. Once its position is known, the moment is simply the yield stress times the first moment of the two halves about it. The whole procedure is valid only if no plate buckles locally before the section yields through — which is exactly what the Class 1 check establishes, and why classification must precede the calculation rather than follow it.

Common pitfall. Placing the plastic neutral axis at the elastic centroid. On a doubly symmetric rolled shape they coincide and the error is invisible; on an unsymmetric built-up section like this one they are 37 mm apart and the resulting Z is wrong. The second pitfall is assuming the deeper direction is automatically the stronger one — here the section is wider than it is deep, and b–b is decisively the strong axis.

Source. CSA S16, Design of Steel Structures, Clause 11 (Table 2) and Clause 13.5; CISC Handbook of Steel Construction, Part 6 (properties of built-up sections); Kulak & Grondin, Limit States Design in Structural Steel, Chapter 5.

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