16-Civ-A2 Elementary Structural Design · December 2019

Question 1 of 7: A1 — Truss compression member in back-to-back angles

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

Question 1: A1 — Truss compression member in back-to-back angles (20 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. The design starts from the truss geometry and support conditions of Figure A1, the specified panel load P1 (75 kN live, 90 kN dead), a 3.0 m long member FG, 300W steel and a 10 mm gap between the angles.

Find. The factored compression in FG, a back-to-back double-angle section that carries it under CSA S16 Clause 13.3, and the spacing of the interconnectors needed to make the two angles act as one member.

Approach. Factor the panel load, solve the determinate truss by joint equilibrium to get the compression in FG, then select a double-angle strut and check it with the S16 Clause 13.3.1 column curve about both principal axes before setting the interconnector spacing from Clause 19.1.4.

1. Factor the panel load. Note 6 says the figure loads are unfactored, and the governing combination for gravity-only framing is NBCC case 2, $1.25D + 1.5L$: $$P_{1f} = 1.25(90) + 1.5(75) = 112.5 + 112.5 = \boxed{225\ \text{kN}}$$
2. Confirm the truss is determinate. With eight joints, thirteen members and three reaction components, $m + r = 13 + 3 = 16 = 2n$. Dropping the vertical CF would leave fifteen unknowns for sixteen equations, i.e. a mechanism — which is the check that settles whether the line from F to C is a member or only the load arrow. It is a member.
3. Reactions. Taking moments about A for the three 225 kN loads at x = 3, 6 and 6 m: $$R_D = \frac{225(3) + 225(6) + 225(6)}{9} = 375\ \text{kN},\qquad R_A = 3(225) - 375 = 300\ \text{kN}$$ and $H_A = 0$ because every applied load is vertical.
4. Pick off the zero-force members first. Joint H carries only AH and HG at right angles with no load, so both are zero; the same argument at E makes DE zero. That removes the whole left-hand and right-hand end panels from the load path and is worth stating, because it tells the designer FG is fed only through joint G.
5. Method of sections for FG. Cut vertically between B and C, through FG, BF and BC, and take moments about B (where BF and BC intersect) on the left free body. That free body carries $R_A = 300$ kN up at A and the 225 kN load at G, so $$\sum M_B = 0:\quad F_{FG}(4) + 300(3) - 225(0) = 0 \;\Rightarrow\; F_{FG} = -225\ \text{kN}$$ The negative sign is compression, and a full joint-equilibrium solve of all thirteen members confirms it: $$\boxed{C_f = 225\ \text{kN (compression) in FG}}$$
6. Trial section. Try two L89×89×9.5 angles back to back with the 10 mm gap. Per angle $A = 1600$ mm², $I = 1.16\times10^{6}$ mm⁴, $r_x = 27.0$ mm, $r_z = 17.4$ mm and $\bar{x} = 25.9$ mm, so for the pair $A_g = 3200$ mm². About the axis parallel to the connected legs $r_x$ is unchanged at 27.0 mm; about the axis through the gap the parallel-axis theorem gives $$I_y = 2\left[1.16\times10^{6} + 1600\,(25.9 + 5)^{2}\right] = 5.375\times10^{6}\ \text{mm}^{4},\qquad r_y = \sqrt{\tfrac{5.375\times10^{6}}{3200}} = 41.0\ \text{mm}$$
7. Slenderness. The panel is 3.0 m long and pinned at both ends within the truss, so $K = 1.0$ and the weaker axis is x-x: $$\frac{KL}{r_x} = \frac{3000}{27.0} = 111.1 \;\;>\;\; \frac{3000}{41.0} = 73.2$$ The back-to-back arrangement has deliberately made the gap axis the stronger one; that is the whole point of pairing the angles.
8. Compressive resistance, S16 Clause 13.3.1. $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^{2}E}} = \frac{111.1}{\pi}\sqrt{\frac{300}{200\,000}} = 1.370$$ $$C_r = \phi A F_y\left(1 + \lambda^{2n}\right)^{-1/n},\quad \phi = 0.90,\ n = 1.34$$ $$C_r = 0.90(3200)(300)\left(1 + 1.370^{2.68}\right)^{-1/1.34} = \boxed{353\ \text{kN}}$$ Since $C_f/C_r = 225/353 = 0.64$, the pair is adequate with sensible reserve.
9. Interconnector spacing, S16 Clause 19.1.4. The connectors must keep the slenderness of one angle about its own weak (z-z) axis below three quarters of the governing slenderness of the built-up member: $$\frac{s}{r_z} \le 0.75\frac{KL}{r} \;\Rightarrow\; s \le 0.75(111.1)(17.4) = 1450\ \text{mm}$$ Use three interconnectors at 750 mm centres (quarter points of the 3.0 m panel), giving $s/r_z = 750/17.4 = 43.1$ against the limit of 83.3. Each is a 10 mm filler plate with two M20 A325 bolts, matching the 10 mm gap called for in the question.