--- title: "Unequal angle tension member with HSFG bolts" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-unequal-angle-hsfg-200kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-unequal-angle-hsfg-200kn site: BCT Engineering --- # Unequal angle tension member with HSFG bolts Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2079 Chaitra, 8 marks - Video: [Problem -1 | Class-18 | Bolts in angles (L sections) | Steel Structure | Civil Engineering | SEC](https://www.youtube.com/watch?v=xyJbpLE0JuM) — Saurov Nandi Majumdar - 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-unequal-angle-hsfg-200kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question Design a tensile member using single unequal angle section to carry design tensile load of 200kN. Use HSFG bolt of property class 6.8 to connect with gusset plate. ## Short answer ISA 100×75×8 connected by 6 M20 HSFG bolts (class 6.8) at 50 mm pitch. Slip resistance per bolt is 39.51 kN, giving 237.1 kN capacity >200 kN. ## Answer **Given:** design tension 200 kN; single unequal angle; HSFG bolts of property class 6.8 ($f_{ub}=600$ N/mm²); E250. The bolts are designed as a **friction-grip** connection (no slip at the design load): M20, $d_0=22$ mm, $\mu_f=0.48$ (sand-blasted surface), $K_h=1.0$, $n_e=1$ (single shear), $\gamma_{mf}=1.25$ (IS 800:2007 cl. 10.4.3). #### 1. Section Required $A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try **ISA 100×75×8**, $A_g=1336$ mm², connected through the 100 mm leg. #### 2. HSFG bolts - Proof load $F_o=0.7f_{ub}A_{nb}=0.7\times600\times245\times10^{-3}=102.9$ kN - Slip resistance $V_{dsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times1\times1.0\times102.9}{1.25}=39.51$ kN - Number of bolts $=\dfrac{200}{39.51}=5.1$ → **6 bolts** in one line, pitch 50 mm ($\ge2.5d$), end distance 40 mm ($\ge1.5d_0=33$ mm), gauge 55 mm from the heel. Capacity $237.1$ kN. #### 3. Member checks - Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}=303.6$ kN - Net rupture with shear lag: $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=w+w_1-t=75+55-8=122$ mm, $L_c=250$ mm, $\beta=1.188$: $T_{dn}=174.8+153.4=328.1$ kN - Block shear: $T_{db}=\min(384.7,312.2)=312.2$ kN All exceed 200 kN. The slenderness limit (400) is checked with the member length (not given): $r_{min}=15.9$ mm allows $L\le6.3$ m. **Answer: ISA 100×75×8 connected by 6 M20 HSFG bolts (class 6.8) in a line at 50 mm pitch.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What basis is used for HSFG bolt design here? A. Bearing only B. Friction-grip: no slip at design load C. Tension D. Welding Answer: B. Bolts are designed against slip using V_nsf. (Key point: Slip resistance) 2. How is the proof load found? A. F_o = f_ub A_nb B. F_o = 0.9 f_yb A_sb C. F_o = 0.7 f_ub A_nb = 102.9 kN D. F_o = f_u A_g Answer: C. F_o = 0.7 x 600 x 245 = 102.9 kN. (Key point: Proof load) 3. Slip resistance per bolt is 39.51 kN. Bolts needed for 200 kN? A. 6, as 200/39.51 = 5.06 B. 5 C. 4 D. 8 Answer: A. Round up: 6 bolts. (Key point: Number of bolts) 4. How is the angle selected? A. Any angle B. A_g >= 200 x 1.10/250 = 880 mm^2 C. A_g >= 200/410 D. A_g >= 2000 Answer: B. The yielding check gives the minimum gross area of 880 mm^2. 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