07-Str-A2 · December 2014

Question 1 of 7: A1 — Moments of resistance of a built-up plate section

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Str-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, 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 in total, 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, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3 whose OCR text is unusable; all geometry below was read from the printed figure of the original PDF. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 1: A1 — Moments of resistance of a built-up plate section (10 + 10 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 doubly symmetric built-up section, 600 mm wide × 300 mm deep overall, assembled entirely from 25 mm plates of G40.21 350W steel.

Quantity	Value
Overall width, B	600 mm
Overall depth, D	300 mm
Plate thickness, t	25 mm
Flange plates	2 × (600 × 25)
Web plates	4 × (25 × 250), equally spaced
Yield strength, F_y	350 MPa
Resistance factor, φ	0.90

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

Figure A1 — the fabricated section: two 600 × 25 flange plates joined by four 25 × 250 web plates at 191.7 mm centres. All dimensions in millimetres.

Approach. Classify every plate element under CSA S16 Table 2; if all elements are Class 1 or 2 the section can develop its full plastic moment, so M_r = φZF_y about each axis, with the plastic modulus Z found from the equal-area neutral axis.

Establish the plate layout from the figure. The webs are equally spaced with the outer pair flush with the section edges, so the centre-to-centre pitch is $$p=\frac{B-t}{n_w-1}=\frac{600-25}{4-1}=191.67\ \text{mm}$$ placing the four web centrelines at 12.5, 204.17, 395.83 and 587.5 mm from the left face. Each web spans the clear depth between flanges, $h_w = 300-2(25)=250$ mm.
Compute the gross area. $$A_g = 2(600 \times 25) + 4(250 \times 25) = 30\,000 + 25\,000 = \boxed{55\,000\ \text{mm}^2}$$
Classify the elements (S16 Table 2). Each flange strip between two webs is supported along both longitudinal edges, so the relevant ratio is the clear span between webs over the thickness: $$\frac{b}{t}=\frac{191.67-25}{25}=6.67 \qquad\text{versus}\qquad \frac{525}{\sqrt{F_y}}=\frac{525}{\sqrt{350}}=28.1$$ The webs give $h_w/t = 250/25 = 10.0$, against a Class 1 limit of $1100/\sqrt{F_y}=58.8$ in flexure. Every element is comfortably Class 1, so the plastic moment is available about both axes and no local-buckling reduction applies.
Plastic modulus about x–x. The section is doubly symmetric, so the plastic neutral axis passes through O. Taking first moments of the two halves about that axis, with the flange centroid at $(300-25)/2 = 137.5$ mm and each half-web centroid at $h_w/4 = 62.5$ mm: $$Z_x = 2\left[(600)(25)(137.5) + 4(25)\left(\tfrac{250}{2}\right)(62.5)\right]$$ $$Z_x = 2\left[2\,062\,500 + 781\,250\right] = \boxed{5.6875 \times 10^{6}\ \text{mm}^3}$$ The equal-area requirement checks out: the material above the axis is $15\,000 + 12\,500 = 27\,500$ mm², exactly half of A_g.
Moment of resistance about x–x. With all elements Class 1 and the member laterally supported, $$M_{rx} = \phi Z_x F_y = 0.90\,(5.6875\times10^{6})(350) = 1.7916\times10^{9}\ \text{N}\cdot\text{mm}$$ $$\boxed{M_{rx} = 1792\ \text{kN}\cdot\text{m}}$$
Plastic modulus about y–y. Now the neutral axis is vertical through O. On the right-hand half sit half of each flange (300 × 25 at 150 mm from the axis) and the two webs whose centres lie at 395.83 and 587.5 mm, i.e. 95.83 mm and 287.5 mm from the axis: $$Z_y = 2\left[2(300)(25)(150) + (250)(25)(95.83) + (250)(25)(287.5)\right]$$ $$Z_y = 2\left[2\,250\,000 + 598\,958 + 1\,796\,875\right] = \boxed{9.2917 \times 10^{6}\ \text{mm}^3}$$
Moment of resistance about y–y. $$M_{ry} = \phi Z_y F_y = 0.90\,(9.2917\times10^{6})(350) = 2.9269\times10^{9}\ \text{N}\cdot\text{mm}$$ $$\boxed{M_{ry} = 2927\ \text{kN}\cdot\text{m}}$$

Although the section looks like a shallow beam, the four webs are spread across the full 600 mm width and act as very efficient flange material for bending about y–y. That is why the resistance about the vertical axis exceeds the resistance about the horizontal one by roughly 63 %, even though the section is twice as wide as it is deep.

Quantity	Value
Gross area, A_g	55 000 mm²
Element classification	Class 1 (both axes)
Plastic modulus, Z_x	5.6875 × 10⁶ mm³
Plastic modulus, Z_y	9.2917 × 10⁶ mm³
M_rx (about x–x)	1792 kN·m
M_ry (about y–y)	2927 kN·m

Topic: Plastic section properties and local-buckling classification of fabricated plate girders.

Key relations: equal-area plastic neutral axis; $Z=\sum A_i|y_i-y_{PNA}|$; $M_r=\phi Z F_y$ for Class 1 and 2 sections and $M_r=\phi S F_y$ for Class 3; S16 Table 2 limits $525/\sqrt{F_y}$ (element supported along two edges, uniform compression) and $1100/\sqrt{F_y}$ (web in flexure).

Why this works: Once every plate is stocky enough to reach yield before it buckles locally, the whole cross-section can be driven to the fully plastic stress distribution: a uniform $+F_y$ on one side of the neutral axis and $-F_y$ on the other. Force equilibrium then requires that the axis divide the section into two equal areas — not equal first moments, which is what the elastic neutral axis does — and the resisting couple is simply $F_y$ times the sum of each area multiplied by the distance of its own centroid from that axis. For a doubly symmetric assembly like this one the two axes coincide, which is why $Z$ falls out in a single line for each direction. Distributing the webs across the full width, rather than bunching them at mid-width, is what makes the y–y capacity so much larger.

Common pitfall: Locating the plastic neutral axis by equal first moments (that is, at the elastic centroid) instead of by equal areas — on unsymmetric built-up sections the two can be a hundred millimetres apart. A second frequent slip is classifying the flange strips as outstands using the $200/\sqrt{F_y}$ limit; here each strip is held by a web on both sides, so the far more generous two-edge limit applies.

Source: CSA S16, Design of Steel Structures, Clause 13.5 and Table 2; Kulak & Grondin, Limit States Design in Structural Steel, Ch. 5.

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