--- title: "Compression capacity of strut of two L75x75x8" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:strut-two-l75x75x8" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/strut-two-l75x75x8 site: BCT Engineering --- # Compression capacity of strut of two L75x75x8 Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2069 Bhadra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/strut-two-l75x75x8 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question The center to center distance between the end connections of a discontinuous strut consisting of two L75 75×8 is 3.0m. Calculate the design load carrying capacity in compression if angles are connected to the same side of a gusset by more than one bolt in each angle. The grade of the steel is E250. ## Short answer Design compressive strength is about 101 kN (factored), assuming hinged gusset connection; equivalent slenderness ratio 177.9. Based on two L75×75×8 angles on the same side of gusset. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A discontinuous strut of two angles on the same side of a gusset with more than one bolt in each angle is designed how, per the answer? A. As angles loaded through one leg, using the equivalent slenderness ratio of Table 12 B. As a doubly symmetric column using plain KL/r C. As a tension member using net area D. As a laced column with 1.05 times slenderness Answer: A. Cl. 7.5.2.2 refers such struts to cl. 7.5.1.2, where the eccentric one-leg connection is handled by an equivalent slenderness ratio. (Key point: Design method for double-angle strut) 2. Which expression gives the equivalent slenderness lambda_e used for the angle? A. sqrt(k1 + k2*lambda_vv^2 + k3*lambda_phi^2) B. k1 + k2*lambda_vv + k3*lambda_phi C. lambda_vv + lambda_phi D. sqrt(lambda_vv^2 - lambda_phi^2) Answer: A. Cl. 7.5.1.2 combines the minor-axis slenderness lambda_vv and the leg-slenderness lambda_phi with constants k1, k2, k3 from Table 12 under a square root. (Key point: Equivalent slenderness formula) 3. Since the end fixity was not stated, which constants were chosen and why? A. Hinged: k1 = 0.7, k2 = 0.6, k3 = 5, as it is conservative B. Fixed: k1 = 0.2, k2 = 0.35, k3 = 20, as it is conservative C. Fixed, because two bolts always give fixity D. Hinged, because it gives the larger capacity Answer: A. The hinged constants give a larger equivalent slenderness and so a lower capacity (101 kN against 159 kN for fixed), making them the safe choice. (Key point: Conservative end condition) 4. How is the final strut capacity (about 101 kN) obtained from the single-angle strength? A. Twice the capacity of one angle, since the angles are stitched together (2 x 50.7) B. Equal to one angle's capacity of 50.7 kN C. Half of one angle's capacity D. One angle's capacity multiplied by 1.05 Answer: A. The two angles are connected together by stitching, so they act together and the capacity is twice that of one angle: 2 x 50.7 = 101.5 kN. 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