Compression capacity of strut of two L75x75x8

Approach. Two angles ISA 75×75×8 connected back to back on the same side of a gusset by more than one bolt in each angle. For such a discontinuous double-angle strut, IS 800:2007 cl. 7.5.2.2 says it is to be designed as for angles loaded through one leg (cl. 7.5.1.2), using the equivalent slenderness ratio from Table 12. The angles are stitched together (cl. 7.8, 10.2.5), so the capacity is twice that of one angle. $f_y = 250$ N/mm² (E250), $\gamma_{m0} = 1.10$.

Data (IS 808)

ISA 75×75×8: $A = 11.4$ cm² = 1140 mm², minimum radius of gyration $r_{vv} = 1.45$ cm = 14.5 mm, $b_1 = b_2 = 75$ mm, $t = 8$ mm. Length between intersections $L = 3000$ mm. $\varepsilon = 1$.

Equivalent slenderness ratio (cl. 7.5.1.2)

Constants for two or more bolts at each end (Table 12). The end fixity is not stated, so take the hinged gusset connection (conservative): $k_1 = 0.7$, $k_2 = 0.6$, $k_3 = 5$.

$$\begin{aligned} \sqrt{\pi^2E/250} &= 88.86 \\ \lambda_{vv} &= \frac{L/r_{vv}}{\varepsilon\sqrt{\pi^2E/250}} = \frac{3000/14.5}{88.86} = 2.328 \\ \lambda_{\phi} &= \frac{(b_1+b_2)/2t}{\varepsilon\sqrt{\pi^2E/250}} = \frac{150/16}{88.86} = 0.106 \\ \lambda_e &= \sqrt{k_1 + k_2\lambda_{vv}^2 + k_3\lambda_\phi^2} = \sqrt{0.7 + 0.6(2.328)^2 + 5(0.106)^2} = 2.002 \end{aligned}$$

Equivalent slenderness ratio $KL/r = \lambda_e\times88.86 = 177.9$, which is less than 180 (Table 3), so it is permitted.

Compressive strength of one angle (cl. 7.1.2.1, curve c, $\alpha = 0.49$)

$$\begin{aligned} f_{cc} &= \pi^2E/\lambda^2 = 62.4\ \text{N/mm}^2 \\ \lambda_n &= \sqrt{f_y/f_{cc}} = 2.002 \\ \phi &= 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 2.946 \\ \chi &= 1/(\phi + \sqrt{\phi^2 - \lambda_n^2}) = 0.196 \\ f_{cd} &= \chi f_y/\gamma_{m0} = 44.5\ \text{N/mm}^2 \\ P_{d1} &= A f_{cd} = 1140\times44.5/10^3 = 50.7\ \text{kN} \end{aligned}$$

Capacity of the strut

$$P_d = 2\times50.7 = 101.5\ \text{kN}$$

Answer: design compressive strength $\approx 101$ kN (about 68 kN service load).

If the gusset connection is taken as fixed against in-plane rotation ($k_1 = 0.20$, $k_2 = 0.35$, $k_3 = 20$), the same working gives $P_d = 2\times79.6 = 159$ kN. The true value lies between these; the hinged value is the safe one.