Single equal angle in compression, 50 kN

Approach. A single angle loaded through one leg and connected by at least two bolts at each end is designed with the equivalent slenderness ratio of IS 800:2007 cl. 7.5.1.2 (Table 12). $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$, $E = 2\times10^5$ N/mm².

Data

Factored load $P = 50$ kN; length between end connections $L = 2.0$ m; $\ge 2$ bolts at each end. End fixity is not stated, so a hinged connection is assumed (conservative): $k_1 = 0.70$, $k_2 = 0.60$, $k_3 = 5$ (Table 12).

Trial sections (equal angles from IS 808)

For each angle: $\lambda_{vv} = (L/r_{vv})/\sqrt{\pi^2E/250}$, $\lambda_\phi = (b/t)/\sqrt{\pi^2E/250}$ (since $b_1 = b_2 = b$), $\lambda_e = \sqrt{k_1 + k_2\lambda_{vv}^2 + k_3\lambda_\phi^2}$, then $KL/r = \lambda_e\sqrt{\pi^2E/250}$ and curve c. The section must also not be slender: for an angle in axial compression, $b/t \le 15.7\varepsilon$ and $(b+d)/t \le 25\varepsilon$ (Table 2).

ISA 75×75×5 has enough strength on paper, but $(b+d)/t = 150/5 = 30 > 25\varepsilon$, so it is a slender section and cannot use the full $f_{cd}$. The lightest safe, non-slender angle is ISA 70×70×6 (mass 6.3 kg/m).

Detailed check of ISA 70×70×6

$A = 8.06$ cm² $= 806$ mm², $r_{vv} = 1.35$ cm.

$$\begin{aligned} \sqrt{\pi^2E/250} &= 88.86 \\ \lambda_{vv} &= \frac{2000/13.5}{88.86} = 1.667,\quad \lambda_\phi = \frac{70/6}{88.86} = 0.131 \\ \lambda_e &= \sqrt{0.70 + 0.60(1.667)^2 + 5(0.131)^2} = 1.567 \\ KL/r &= 1.567\times88.86 = 139.2 < 180 \end{aligned}$$ $$\begin{aligned} f_{cc} &= 101.9\ \text{N/mm}^2,\ \lambda_n = 1.567,\ \phi = 2.062,\ \chi = 0.294 \\ f_{cd} &= 66.8\ \text{N/mm}^2 \\ P_d &= A f_{cd} = 806\times66.8/10^3 = 53.8\ \text{kN} > 50\ \text{kN} \end{aligned}$$

Section class: $b/t = 70/6 = 11.7 \le 15.7\varepsilon$ and $(b+d)/t = 140/6 = 23.3 \le 25\varepsilon$ (Table 2, semi-compact limits). OK, not slender.

Answer: provide ISA 70×70×6, connected with at least two M16 bolts at each end; $P_d = 53.8$ kN $> 50$ kN.