07-Str-A2 · December 2013

Question 1 of 7: Moments of resistance of a built-up box-and-plate section

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

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Str-A2, Elementary Structural Design. Three hours; a "CLOSED BOOK" examination in which handbooks and textbooks are permitted. Seven questions in three parts — Part A steel (CAN/CSA-S16), Part B reinforced concrete (CAN/CSA-A23.3), Part C timber (CAN/CSA-O86). A candidate answers two questions from Part A, two from Part B and the single question in Part C, five in all, and page 1 records that all questions are of equal value, so each carries 20 marks with the split A1 (6 + 7 + 7), A2 (5 + 10 + 5), A3 (4 + 4 + 12), B1 (6 + 8 + 6), B2 (10 + 10), B3 (4 + 4 + 12), C1 (10 + 10) printed in the marking scheme. All seven questions are solved below.

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

Check — load factors used throughout. Page 1 note 6 states that all loads shown are unfactored but the paper nowhere splits them into dead and live. Every applied load in Figures A2, B1 and B3 is therefore treated as a specified live load and factored by 1.5, and self-weight, where a question asks for it, by 1.25, following NBCC load combination case 2, 1.25D + 1.5L. Question B3 names its 80 kN horizontal load as wind, so that question is additionally checked under case 4, 1.25D + 1.4W + 0.5L. If a grader intends a different dead/live split every factored action scales linearly and no design step below changes.

Check — steel grade in Figure A1. The A1 stem attaches "G40.21 300W" to the plates and is silent on the grade of the hollow section. All three components are taken as 300W, F_y = 300 MPa; if the HSS is in fact 350W the plates still yield first and the plastic moments quoted rise by less than 8 per cent. Corner radii of the hollow section are ignored (square corners), which is the usual hand-calculation idealisation and overstates the HSS area by about 4 per cent relative to the tabulated value.

Question A1: Moments of resistance of a built-up box-and-plate section (20 marks: 6 + 7 + 7)

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 hollow structural section with a wide plate welded to its top face and a narrower plate welded to its soffit, all of G40.21 300W steel.

Given data (Figure A1 and the question stem)
Quantity	Symbol	Value
Hollow section, outside dimension	D	304.8 mm square
Hollow section, wall thickness	t	12.7 mm
Top plate	b₁ × t₁	500 × 20 mm
Bottom plate	b₂ × t₂	400 × 16 mm
Yield strength	F_y	300 MPa
Resistance factor, steel	φ	0.90

Find. The factored moment of resistance of the section about the horizontal centroidal axis a–a and about the vertical centroidal axis b–b.

[Figure not reproduced: Figure A1 — the fabricated section, redrawn to scale. Axis b–b is the vertical axis of symmetry; axis a–a is the horizontal centroidal axis, which sits below mid-depth because the top plate is the heavier of the two. See the official exam paper.]

Approach. Locate the elastic centroid and compute the second moments of area about both axes, classify every plate element to CSA S16 Table 2 so that the correct resistance expression is known, and then — the section proving to be Class 1 about both axes — evaluate the plastic section moduli and take M_r = φZF_y.

(a) Section properties (6 marks)

Areas of the three components. With square corners the hollow section is the difference of two squares of side $D = 304.8$ mm and $D_i = D - 2t = 279.4$ mm: $$A_{\text{HSS}} = D^2 - D_i^2 = 304.8^2 - 279.4^2 = 14\,839\ \text{mm}^2$$ and the plates contribute $A_1 = 500(20) = 10\,000\ \text{mm}^2$ and $A_2 = 400(16) = 6\,400\ \text{mm}^2$, so the total area is $A = 31\,239\ \text{mm}^2$.
Elastic centroid, measured from the soffit. Taking first moments about the underside of the bottom plate, with component centroids at 8 mm, 168.4 mm and 330.8 mm, $$\bar y = \frac{\sum A_i y_i}{A} = \frac{6\,400(8) + 14\,839(168.4) + 10\,000(330.8)}{31\,239} = 187.5\ \text{mm}$$ The overall depth is $h = 16 + 304.8 + 20 = 340.8$ mm, so the extreme fibres lie 187.5 mm below and 153.3 mm above the axis a–a.
Second moments of area. The hollow section alone has $I_{\text{HSS}} = (D^4 - D_i^4)/12 = 211.4 \times 10^6\ \text{mm}^4$ about either of its own axes. Applying the parallel-axis theorem about a–a, $$I_{aa} = \sum \left( \tfrac{b_i t_i^3}{12} + A_i d_i^2 \right) = 628.9 \times 10^6\ \text{mm}^4$$ About b–b the section is symmetric and no transfer terms arise: $$I_{bb} = \frac{20(500)^3}{12} + \frac{16(400)^3}{12} + 211.4\times10^6 = 505.1 \times 10^6\ \text{mm}^4$$ The corresponding elastic section moduli are $S_{aa,\text{bot}} = 3.353\times10^6$, $S_{aa,\text{top}} = 4.103\times10^6$ and $S_{bb} = 2.020\times10^6\ \text{mm}^3$.
Element slenderness and section class (CSA S16 Table 2, with $F_y = 300$ MPa so that $\sqrt{F_y} = 17.32$). The top plate projects $(500-304.8)/2 = 97.6$ mm beyond each wall, a flange supported along one edge with $b/t = 97.6/20 = 4.88$ against a Class 1 limit of $145/\sqrt{F_y} = 8.37$. The part of that plate spanning between the two walls is supported along two edges, $279.4/20 = 14.0$ against $525/\sqrt{F_y} = 30.3$. The walls acting as webs in flexure give $279.4/12.7 = 22.0$ against $1100/\sqrt{F_y} = 63.5$. Every element is comfortably within its Class 1 limit for bending about either axis, so $$\boxed{\text{the section is Class 1 about a--a and about b--b}}$$ and the plastic moment may be developed. The box is torsionally closed, so lateral-torsional buckling does not reduce the resistance and the section capacity governs.

(b) Moment of resistance about a–a (7 marks)

Locate the plastic neutral axis. The plastic axis divides the area equally, so it must carry $A/2 = 15\,619\ \text{mm}^2$ below it. Working up from the soffit, the bottom plate supplies 6 400 mm² and the lower wall of the hollow section a further $304.8(12.7) = 3\,871\ \text{mm}^2$; the balance of $5\,348\ \text{mm}^2$ is taken by the two vertical walls over a height of $5\,348/(2 \times 12.7) = 210.6$ mm. Hence $$y_p = 16 + 12.7 + 210.6 = 239.3\ \text{mm above the soffit}$$ which lies within the walls, confirming the assumed distribution.
Plastic section modulus. Summing the first moments of every component area about that axis, $Z_{aa} = \sum A_i \lvert y_i - y_p \rvert$, gives $$Z_{aa} = 4.149 \times 10^{6}\ \text{mm}^3$$ The shape factor $Z_{aa}/S_{aa,\text{bot}} = 1.24$ is the expected value for a section whose material is concentrated near the extreme fibres.
Factored moment of resistance (CSA S16 Clause 13.5(a), Class 1 and 2 sections): $$M_{r,aa} = \phi Z_{aa} F_y = 0.90 \left(4.149\times10^{6}\right)(300) = 1\,120 \times 10^{6}\ \text{N}\cdot\text{mm}$$ $$\boxed{M_{r,aa} = 1\,120\ \text{kN}\cdot\text{m}}$$ For comparison, first yield at the soffit would occur at $\phi S_{aa,\text{bot}} F_y = 905\ \text{kN}\cdot\text{m}$, so roughly a quarter of the resistance comes from plastification of the cross-section.

(c) Moment of resistance about b–b (7 marks)

Plastic section modulus about the axis of symmetry. Because b–b is an axis of symmetry the plastic and elastic neutral axes coincide, and each component contributes its own plastic modulus directly: $$Z_{bb} = \frac{t_1 b_1^2}{4} + \frac{t_2 b_2^2}{4} + \frac{D^3 - D_i^3}{4} = 1.250\times10^{6} + 0.640\times10^{6} + 1.626\times10^{6}$$ $$Z_{bb} = 3.516 \times 10^{6}\ \text{mm}^3$$ The shape factor is now 1.74, much higher than about a–a, because bending about b–b stresses the two plates as deep rectangles whose material is mostly close to the axis.
Factored moment of resistance. Again from Clause 13.5(a), $$M_{r,bb} = \phi Z_{bb} F_y = 0.90\left(3.516\times10^{6}\right)(300) = 949 \times 10^{6}\ \text{N}\cdot\text{mm}$$ $$\boxed{M_{r,bb} = 949\ \text{kN}\cdot\text{m}}$$ against a first-yield value of $\phi S_{bb} F_y = 546\ \text{kN}\cdot\text{m}$.
Comment on the result. The two resistances differ by only about 18 per cent even though $I_{aa}$ exceeds $I_{bb}$ by 25 per cent, precisely because the plate material that is inefficient in the a–a direction becomes fully effective once the whole section yields about b–b. A designer relying on elastic moduli alone would conclude, wrongly, that the section is twice as strong one way as the other.

Final results — Question A1
Quantity	Value
Total area, A	31 239 mm²
Centroid above soffit, ȳ	187.5 mm (overall depth 340.8 mm)
I_aa / I_bb	628.9 × 10⁶ / 505.1 × 10⁶ mm⁴
Section class (both axes)	Class 1
Z_aa (plastic axis 239.3 mm above soffit)	4.149 × 10⁶ mm³
Z_bb	3.516 × 10⁶ mm³
M_r about a–a	1 120 kN·m
M_r about b–b	949 kN·m

Topic. Classification of built-up steel sections and the plastic moment of resistance.

Key relations. $\bar y = \sum A_iy_i/\sum A_i$; parallel-axis theorem $I = \sum(I_i + A_id_i^2)$; equal-area rule for the plastic neutral axis; Class 1 and 2 limits $145/\sqrt{F_y}$ (flange outstand), $525/\sqrt{F_y}$ (flange supported two edges), $1100/\sqrt{F_y}$ (web in flexure); and $M_r = \phi Z F_y$ for Class 1 and 2 against $M_r = \phi S F_y$ for Class 3.

Why this works. The moment of resistance a section can deliver is set by how much of it can reach yield before a plate element buckles locally. Classification is therefore the first step, not an afterthought: it decides whether the answer is a plastic modulus, an elastic modulus, or an effective-width calculation. Once every element is Class 1, the full rectangular stress block may be assumed, tension yield below the equal-area axis and compression yield above it, and the resisting moment is simply the first moment of the whole area about that axis multiplied by the yield stress. Because a closed box has enormous torsional stiffness, no lateral-torsional check is needed and the section resistance is the member resistance.

Common pitfall. Two errors dominate. The first is to place the plastic neutral axis at the elastic centroid; the two coincide only when the section is symmetric about the bending axis, and here they are 52 mm apart. The second is to use the same neutral axis for both directions of bending, which silently converts an unsymmetric section into a symmetric one and can overstate $Z_{aa}$ by ten per cent or more.

Source. CSA S16, Clauses 11 (classification, Table 2) and 13.5 (bending resistance); Kulak and Grondin, Limit States Design in Structural Steel, Chapter 5; CISC Handbook of Steel Construction, Part 6 for HSS properties.

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