--- title: "Double angle tension member 400 kN, M20" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-double-angle-400kn-class-c" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-double-angle-400kn-class-c site: BCT Engineering --- # Double angle tension member 400 kN, M20 Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2075 Bhadra, 10 marks - Video: [Design of Tension Member (Steel) Numerical](https://www.youtube.com/watch?v=iMxjQBHJop0) — Studyclass 247 - 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/tension-double-angle-400kn-class-c - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question Design a double angle section to carry a tension of 400 kN. The end connection to be made using M20 bolts at product class C and property class 5.6. Assuming the angle sections are provided on the both sides of gusset plate. The steel used are E 250. ## Short answer Use 2 ISA 75×75×8 placed on opposite sides of gusset plate. Connect with 4 M20 grade 5.6 bolts at 60 mm pitch (double shear). Bolt capacity 477 kN and section capacity 517 kN both exceed 400 kN. ## Answer **Given:** factored tension 400 kN; two angles on opposite faces of the gusset (bolts in double shear); M20 bolts, product grade C, class 5.6 ($f_{ub}=500$, $f_{yb}=300$); E250; gusset plate assumed 12 mm; $d_0=22$ mm. #### 1. Bolts - $V_{dsb}=\dfrac{500}{\sqrt3\times1.25}(245+314)\times10^{-3}=129.1$ kN (one plane in threads) - Bearing on gusset 12 mm ($e=40$, $p=60$): $k_b=\min(0.606,0.659,1.22,1)=0.606$; $V_{dpb}=119.3$ kN - Bolt value $119.3$ kN → $n=\dfrac{400}{119.3}=3.35$ → **4 bolts**, pitch 60 mm, end distance 40 mm; capacity $477.1$ kN. #### 2. Section $A_g\ge\dfrac{400\times1.10}{250}\times10^3=1760$ mm². Try **2 ISA 75×75×8** ($A=1138$ mm² each, $2276$ mm²), bolt gauge 40 mm from the heel. - Yielding: $T_{dg}=2\times\dfrac{1138\times250}{1.10}\times10^{-3}=517.3$ kN (258.6 kN per angle) - Rupture per angle (load 200 kN): $A_{nc}=(75-4-22)\times8=392$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=w+w_1-t=75+40-8=107$ mm, $L_c=180$ mm, $\beta=1.142$ → $T_{dn}=263.1$ kN - Block shear per angle: $\min(287.6,258.6)=258.6$ kN - Strength of one angle $=\min(258.6,\ 263.1,\ 258.6)=258.6$ kN; pair $=517.2$ kN $>400$ ✓ (member length not given; check $L/r\le400$: with $r_{min}\approx23$ mm this allows $L\le9$ m). Provide tack bolts at no more than 1000 mm centres, with packing. **Answer: 2 ISA 75×75×8 (one on each side of the gusset) with 4 M20 bolts (5.6) at 60 mm pitch.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which strength is used for the shear capacity of M20 class 5.6 bolts? A. fyb = 300 N/mm2 B. fub = 500 N/mm2 C. fu of plate = 410 D. fub = 800 Answer: B. Property class 5.6 means fub = 500 N/mm2, used in Vdsb = fub/(sqrt3 gamma_mb)(Anb + Asb). (Key point: Bolt shear capacity) 2. Bolt value = 119.3 kN. How many bolts are needed for 400 kN? A. 4 B. 3 C. 5 D. 6 Answer: A. 400/119.3 = 3.35, so 4 bolts (capacity 477.1 kN). (Key point: Number of bolts) 3. How is the strength of the double angle checked in rupture? A. Both angles are treated as one with the full area at fu B. Only one angle is effective C. Rupture is ignored for double angles D. Each angle carries half the load (200 kN) and its Tdn is doubled Answer: D. Tdn per angle = 252.2 kN, pair = 504.4 kN, which exceeds 400 kN. (Key point: Strength of pair) 4. Why are tack bolts with packing provided between the double angles? A. To make the two angles act together and limit slenderness of individual angles B. To increase bolt shear capacity C. To resist the tension directly D. To reduce shear lag factor beta Answer: A. Tack bolts at limited spacing connect the two angles so they behave as a single member. 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