---
title: "Single equal angle in compression, 50 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:compression-single-equal-angle-50kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/compression-single-equal-angle-50kn
site: BCT Engineering
---

# Single equal angle in compression, 50 kN

Design of Steel and Timber Structure · Chapter 6: Compression Members

- Asked in: 2070 Magh, 10 marks
- Video: [Design of steel structures# Compression member design#Single Equal angle section#Revised Section#](https://www.youtube.com/watch?v=8Kz9BoVEHbg) — CIVIL ENGINEERING COURSE Theory & Practical
- 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/compression-single-equal-angle-50kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a single equal angle to carry a compression of 50 KN. The centre to centre distance between the end connections is 2.0 M. Assume that at least two bolts are used for the end connections.

## Short answer

Use a single equal angle ISA 70×70×6 in Fe 250 steel, with at least two bolts at each end. Its design compressive strength is 53.8 kN, greater than the required 50 kN.

## Answer

**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).

| Angle | A (cm²) | $r_{vv}$ (cm) | KL/r | $P_d$ (kN) | $(b+d)/t$ | Result |
|---|---|---|---|---|---|---|
| ISA 60×60×6 | 6.84 | 1.15 | 155 | 38.2 | 20.0 | Not safe |
| ISA 65×65×6 | 7.44 | 1.26 | 146 | 46.2 | 21.7 | Not safe |
| ISA 70×70×5 | 6.77 | 1.36 | 140 | 45.0 | 28.0 | Not safe (and slender) |
| ISA 75×75×5 | 7.27 | 1.46 | 134 | 51.6 | 30.0 | Slender ($> 25$), reject |
| ISA 70×70×6 | 8.06 | 1.35 | 139 | 53.8 | 23.3 | Safe |

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is a single angle connected by at least two bolts at each end designed in compression?
   A. With the equivalent slenderness ratio of cl. 7.5.1.2 (Table 12)
   B. As a concentrically loaded column with plain KL/r = L/r_min
   C. Using the net area as in tension
   D. By ignoring slenderness for angles
   Answer: A. Loading through one leg is eccentric, so IS 800 cl. 7.5.1.2 uses an equivalent slenderness ratio with constants from Table 12. (Key point: Single angle with two bolts)

2. From the trial table, why is ISA 75x75x5 selected over ISA 70x70x6?
   A. Both are safe, but 75x75x5 is lighter (5.7 kg/m)
   B. ISA 70x70x6 is not safe
   C. ISA 75x75x5 has the larger capacity
   D. ISA 70x70x6 fails the slenderness limit of 180
   Answer: A. Both 70x70x6 (53.8 kN) and 75x75x5 (51.6 kN) exceed 50 kN, and the lightest safe section is chosen: 75x75x5. (Key point: Trial selection)

3. What gives the design capacity of ISA 75x75x5 and does it satisfy the 50 kN load?
   A. P_d = A x f_cd = 727 x 71.0 = 51.6 kN > 50 kN
   B. P_d = A x f_y = 727 x 250 = 181.8 kN
   C. P_d = 71.0 kN taken directly from f_cd
   D. P_d = 45 kN, which is less than 50 kN
   Answer: A. f_cd comes from curve c at the equivalent slenderness (133.8), and capacity is area times f_cd = 51.6 kN, which exceeds 50 kN. (Key point: Capacity check)

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