Moments of resistance of a built-up channel-and-plate section: 07-Str-A2, May 2013, Q1

Question A1: Moments of resistance of a built-up channel-and-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. Two C310x45 channels (CISC properties below) placed back to back with their web faces in contact on the vertical axis of symmetry, welded to a 20 mm x 260 mm plate on the underside. Steel G40.21 350W, so $F_y = 350$ MPa and $\phi = 0.90$ (CSA S16 Cl 13.1).

Quantity	C310x45 (each)	Plate
Area $A$	5 690 mm²	260 x 20 = 5 200 mm²
Depth / thickness	$d = 305$ mm	$t_p = 20$ mm
Flange	$b = 80.5$ mm, $t = 12.7$ mm	—
Web	$w = 13.0$ mm	—
$I_x$ (own axis)	$67.3\times10^{6}$ mm⁴	$0.173\times10^{6}$ mm⁴
$I_y$ (own axis)	$2.12\times10^{6}$ mm⁴	$29.29\times10^{6}$ mm⁴
Centroid from web face	$\bar{x} = 17.1$ mm	—

Find. The factored moments of resistance $M_{rx}$ and $M_{ry}$ of the fabricated section about its two centroidal axes.

Figure A1 as built — two C310x45 channels back to back on a 20 x 260 plate. The elastic and plastic neutral axes for bending about x-x are 121.5 mm and 72.5 mm above the underside of the plate; the section is symmetric about y-y.

Approach. Locate the elastic centroid and compute $I_x$, $I_y$ by the parallel-axis theorem; classify every plate element to CSA S16 Table 2; then, because the section proves to be Class 1, take $M_r = \phi Z F_y$ about each axis, with the plastic moduli found from the equal-area axis.

Set up the geometry. Measuring $y$ upwards from the underside of the plate, the plate occupies $0 \le y \le 20$ and the channels $20 \le y \le 325$, so the overall depth is $h = 325$ mm. The two channel webs are in contact on the axis of symmetry, giving a combined web thickness of $2(13.0) = 26$ mm and a combined flange width of $2(80.5) = 161$ mm. Total area $A = 2(5\,690) + 5\,200 = 16\,580$ mm².
Locate the elastic centroid. Taking first moments about the underside of the plate, $$\bar{y}=\frac{\sum A_i y_i}{\sum A_i} =\frac{2(5\,690)(172.5)+5\,200(10)}{16\,580} =\frac{1\,963\,050+52\,000}{16\,580}=121.5\ \text{mm}$$ so the extreme fibres are $y_{bot} = 121.5$ mm and $y_{top} = 325 - 121.5 = 203.5$ mm.
Second moment of area about x-x. Applying the parallel-axis theorem to the channels and to the plate, $$I_x = 2\left[67.3\times10^{6}+5\,690(50.96)^2\right]+\frac{260(20)^3}{12}+5\,200(111.5)^2$$ $$I_x = 134.6\times10^{6}+29.6\times10^{6}+0.17\times10^{6}+64.7\times10^{6} =\boxed{229.0\times10^{6}\ \text{mm}^4}$$ The elastic section moduli are therefore $S_{top}=229.0\times10^{6}/203.5=1\,126\times10^{3}$ mm³ and $S_{bot}=229.0\times10^{6}/121.5=1\,884\times10^{3}$ mm³; the top fibre governs elastically because the plate pulls the centroid down.
Second moment of area about y-y. For a channel, $\bar{x}$ is measured from the outside face of the web, and back-to-back placement puts that face on the axis of symmetry, so each channel centroid sits 17.1 mm from y-y: $$\begin{aligned} I_y &= 2\left[2.12\times10^{6}+5\,690(17.1)^2\right]+\frac{20(260)^3}{12}\\ &= 7.57\times10^{6}+29.29\times10^{6}=\boxed{36.9\times10^{6}\ \text{mm}^4} \end{aligned}$$ with $S_y = 36.9\times10^{6}/130 = 284\times10^{3}$ mm³. The plate, not the channels, supplies four fifths of the weak-axis stiffness.
Classify the elements (CSA S16 Table 2). For $F_y = 350$ MPa the Class 1 limits are $145/\sqrt{F_y}=7.75$ for a projecting flange and $1100/\sqrt{F_y}=58.8$ for a web in flexural compression. The channel flange outstand gives $b/t = 80.5/12.7 = 6.34 < 7.75$; the combined web gives $h/w = (305-2(12.7))/26 = 10.8 < 58.8$; the plate outstand beyond the channel flanges gives $(260-161)/2/20 = 2.5$, far below any limit. Every element is Class 1, so the plastic moment may be developed about both axes provided the compression flange is laterally supported.
Plastic modulus about x-x. The plastic neutral axis divides the area equally, so $8\,290$ mm² must lie below it. The plate supplies $5\,200$ and the pair of bottom channel flanges $161(12.7) = 2\,045$, leaving $1\,045$ mm² to be taken from the 26 mm web over a height $1\,045/26 = 40.2$ mm. Hence the plastic axis lies at $y_p = 20+12.7+40.2 = 72.5$ mm. Summing $\sum A_i\,|y_i - y_p|$ over the plate, the two flange pairs and the two web portions gives $$Z_x = \boxed{1\,691\times10^{3}\ \text{mm}^3}$$ a shape factor $Z_x/S_{top} = 1.50$, which is large because the section is strongly monosymmetric.
Plastic modulus about y-y. The axis of symmetry is itself the equal-area axis, so $Z_y$ is twice the first moment of the half section about it: $$Z_y = 2\left[(130)(20)(65)+(13)(305)(6.5)+2(67.5)(12.7)(46.75)\right] = 550\times10^{3}\ \text{mm}^3$$
Factored moments of resistance. For a Class 1 section CSA S16 Cl 13.5(a) gives $M_r = \phi Z F_y$: $$M_{rx}=0.90(1\,691\times10^{3})(350)=\boxed{533\ \text{kN}\cdot\text{m}}$$ $$M_{ry}=0.90(550\times10^{3})(350)=\boxed{173\ \text{kN}\cdot\text{m}}$$

Check — lateral support. These are section moments of resistance, which is what the question asks for. They are valid only where the compression flange is continuously braced. If the top flange of the 325 mm deep section were unbraced over a long span, lateral-torsional buckling to CSA S16 Cl 13.6 would reduce $M_{rx}$; the weak-axis value $M_{ry}$ is never reduced by LTB. Should a grader prefer an elastic (Class 3) treatment, the corresponding values are $\phi S_{top}F_y = 355$ kN·m and $\phi S_y F_y = 89.3$ kN·m.

Quantity	Value
Total area $A$	16 580 mm²
Elastic centroid above plate soffit	121.5 mm
$I_x$	$229.0\times10^{6}$ mm⁴
$S_x$ (governing, top fibre)	$1\,126\times10^{3}$ mm³
$Z_x$	$1\,691\times10^{3}$ mm³
$I_y$	$36.9\times10^{6}$ mm⁴
$S_y$	$284\times10^{3}$ mm³
$Z_y$	$550\times10^{3}$ mm³
Classification	Class 1 about both axes
$M_{rx}$	533 kN·m
$M_{ry}$	173 kN·m

Topic. Section properties and classification of a fabricated monosymmetric section.

Key relations. $\bar{y}=\sum A_i y_i/\sum A_i$; $I = \sum (I_i + A_i e_i^2)$; $S = I/c$; $Z = \sum A_i |y_i - y_p|$ with $y_p$ the equal-area axis; $M_r = \phi Z F_y$ (Class 1 and 2) or $\phi S F_y$ (Class 3).

Why this works. A cross-section reaches its plastic moment only if every plate element can reach yield without buckling locally first, which is exactly what the width-to-thickness limits of Table 2 police. Here the channels are heavy relative to their outstands and the added plate is short and thick, so all elements are stocky and the full plastic modulus is available. The plastic and elastic neutral axes differ (72.5 mm against 121.5 mm) precisely because the section is monosymmetric: the elastic axis balances first moments of area, the plastic axis balances areas. That difference is what makes the shape factor 1.50 rather than the 1.12 typical of a rolled I-shape, and it is why quoting an elastic answer here understates the strength by a third.

Common pitfall. Placing the channel centroids at their full $\bar{x}$ plus half the web thickness. Back-to-back means the outside web faces touch, and $\bar{x}$ is already measured from that face, so the lever arm is 17.1 mm, not 23.6 mm. The other common slip is to take $S_x$ at the bottom fibre, where the plate makes the modulus 67 per cent larger; the smaller modulus always governs elastic design.

Source. CSA S16 Cl 13.5 and Table 2; CISC Handbook of Steel Construction, channel property tables; Salmon, Johnson and Malhas, Steel Structures, Ch. 7.

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

Question A1: Moments of resistance of a built-up channel-and-plate section (10 + 10 marks)