Maximum factored bracket load on a round HSS column: 07-Str-A2, May 2013, Q3

Question A3: Maximum factored bracket load on a round HSS column (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. Circular hollow section, outside diameter $D = 406.4$ mm, wall $t = 9.53$ mm, G40.21 350W Class H, so $F_y = 350$ MPa, $E = 200\,000$ MPa, $\phi = 0.90$ and the column-curve exponent $n = 2.24$ (CSA S16 Cl 13.3.1, Class H). Height $L = 6\,000$ mm, pinned at the top and fixed at the base; bracket eccentricity $e = 0.8$ m.

Find. The largest factored vertical bracket load $P_f$ the column can carry.

The column as a propped cantilever. The bracket applies P at 0.8 m eccentricity at the top, so M = Pe there and, by carry-over to the fixed base, Pe/2 of opposite sign; the member bends in double curvature.

Approach. Compute the section properties and classify the tube; get $C_r$ from the CSA S16 column equation with $K = 0.8$ and $M_r = \phi Z F_y$; establish the moment diagram produced by the eccentric bracket; then solve the beam-column interaction equations of Cl 13.8.2 for the load that makes the utilisation exactly unity.

Section properties. With $d_i = D - 2t = 406.4 - 19.06 = 387.34$ mm, $$A=\frac{\pi}{4}\left(D^2-d_i^2\right)=11\,882\ \text{mm}^2,\qquad I=\frac{\pi}{64}\left(D^4-d_i^4\right)=234.1\times10^{6}\ \text{mm}^4$$ $$r=\sqrt{I/A}=140.4\ \text{mm},\qquad Z=\frac{D^3-d_i^3}{6}=1\,501\times10^{3}\ \text{mm}^3$$
Classify the tube. $D/t = 406.4/9.53 = 42.6$. CSA S16 Table 2 sets the class limits for a circular hollow section in flexure at $13\,000/F_y = 37.1$ (Class 1), $18\,000/F_y = 51.4$ (Class 2) and $66\,000/F_y = 188.6$ (Class 3), so the section is Class 2 and the plastic moment may still be used. For axial compression the non-slender limit is $23\,000/F_y = 65.7 > 42.6$, so the whole area is effective.
Compressive resistance. A column pinned at one end and fixed at the other takes the recommended design value $K = 0.8$, so $$\frac{KL}{r}=\frac{0.8(6\,000)}{140.4}=34.2,\qquad \lambda=\frac{KL}{r}\sqrt{\frac{F_y}{\pi^2E}}=34.2\sqrt{\frac{350}{1.974\times10^{6}}}=0.455$$ $$C_r=\phi A F_y\left(1+\lambda^{2n}\right)^{-1/n} =0.90(11\,882)(350)(1+0.455^{4.48})^{-1/2.24}=\boxed{3\,695\ \text{kN}}$$
Moment resistance. Being Class 2, and with no lateral-torsional buckling possible in a circular tube, $$M_r=\phi Z F_y = 0.90(1\,501\times10^{3})(350)=473\ \text{kN}\cdot\text{m}$$
The moment the bracket produces. The bracket delivers $M = P_f e$ to the top of the column. The pin there restrains translation but not rotation, so the member is a propped cantilever loaded by an end moment: the base carries half that moment with the opposite sign, and the member bends in double curvature with $M_f = P_f(0.8)$ at the top. Since the frame is braced, $\kappa = +0.5$ and $\omega_1 = 0.6-0.4(0.5) = 0.4$; because Cl 13.8.4 does not permit $U_1$ below unity for the member check, take $U_{1x} = 1.0$. For information, $C_e = \pi^2EI/L^2 = 12\,834$ kN, so the second-order magnification is small in any case.
Cross-sectional strength (Cl 13.8.2 a). With $C_r$ replaced by $\phi A F_y = 3\,743$ kN, $$\begin{aligned} \frac{P_f}{3\,743}+\frac{0.85(1.0)(0.8P_f)}{473} &\le 1.0\\ P_f\left(2.672\times10^{-4}+1.438\times10^{-3}\right) &\le 1.0 \quad\Rightarrow\quad P_f \le 587\ \text{kN} \end{aligned}$$
Overall member strength (Cl 13.8.2 b). Using the buckling resistance, $$\begin{aligned} \frac{P_f}{3\,695}+\frac{0.85(1.0)(0.8P_f)}{473} &\le 1.0\\ P_f\left(2.706\times10^{-4}+1.438\times10^{-3}\right) &\le 1.0 \quad\Rightarrow\quad \boxed{P_f \le 585\ \text{kN}} \end{aligned}$$ The member check governs, though only just, because the column is stocky.
Confirm the bending check separately. At $P_f = 585$ kN the applied moment is $M_f = 585(0.8) = 468$ kN·m against $M_r = 473$ kN·m, so the tube is at 99 per cent of its bending capacity and at only 16 per cent of its compressive capacity. The answer is therefore governed almost entirely by flexure from the 0.8 m bracket arm, which is worth saying explicitly: halving the eccentricity would roughly double the permissible load.

Quantity	Value
$A$ / $I$ / $r$ / $Z$	11 882 mm² / $234.1\times10^{6}$ mm⁴ / 140.4 mm / $1\,501\times10^{3}$ mm³
$D/t$ and class	42.6, Class 2 in flexure, non-slender in compression
$KL/r$ with $K = 0.8$	34.2 ($\lambda = 0.455$)
$C_r$	3 695 kN
$M_r$	473 kN·m
Cross-sectional strength limit	587 kN
Member strength limit (governs)	585 kN
$M_f$ at that load	468 kN·m (0.99 $M_r$)

Topic. Beam-columns: axial force and bending interaction to CSA S16.

Key relations. $\lambda = (KL/r)\sqrt{F_y/\pi^2E}$; $C_r=\phi AF_y(1+\lambda^{2n})^{-1/n}$ with $n = 2.24$ for Class H and 1.34 for Class C; $M_r=\phi ZF_y$ (Class 1 and 2); $C_f/C_r + 0.85\,U_{1x}M_{fx}/M_{rx}\le 1.0$; $U_{1x}=\omega_1/(1-C_f/C_e)$ with $\omega_1 = 0.6-0.4\kappa \ge 0.4$.

Why this works. The interaction equation is an empirical straight-line fit to the true failure surface of a member carrying both actions, and CSA S16 requires it to be satisfied three ways: for the cross-section at its most heavily loaded point, for the member as a whole including buckling, and for lateral-torsional buckling. A circular tube escapes the third check entirely because every axis is a principal axis with the same stiffness, which is exactly why tubes are attractive for bracket columns. Here the fixed base and pinned head give $K = 0.8$, and at $KL/r = 34$ the column is stocky, so $C_r$ falls barely below the squash load and the two governing checks land within 0.3 per cent of each other. What actually limits the load is the 0.8 m lever arm.

Common pitfall. Using $K = 0.7$ (the theoretical value) rather than the recommended design value 0.8, and taking the moment as constant over the height. The pin at the head is a support, not a rigid connection, so it sheds no moment but does prevent sway; the moment therefore falls linearly from $Pe$ at the top to $Pe/2$ of opposite sign at the base. A third slip is to use the Class C exponent $n = 1.34$, which for this section would understate $C_r$ by about 3 per cent.

Source. CSA S16 Cl 13.3.1, Cl 13.5, Cl 13.8.2, Cl 13.8.4 and Table 2; CISC Handbook of Steel Construction, HSS property tables and effective length recommendations.

Question 3 of 7: Maximum factored bracket load on a round HSS column

Question A3: Maximum factored bracket load on a round HSS column (12 + 8 marks)