--- title: "Double angle tension member, M18, same side" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-double-angle-300kn-m18" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-double-angle-300kn-m18 site: BCT Engineering --- # Double angle tension member, M18, same side Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2076 Bhadra, 14 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/tension-double-angle-300kn-m18 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question Design a tensile member having length 2.5m, subjected to design axial load of 300 kN. Use double angle section provided at same side of gusset plate and M18 bolts to connect with gusset plate. ## Short answer Use 2 ISA 75×75×6 placed back-to-back on same side of gusset plate. Each angle connected with 5 M18 grade 4.6 bolts at 50 mm pitch (10 bolts total). Bolt capacity 354.7 kN governs and exceeds 300 kN. ## Answer **Assumptions:** the two angles are placed back to back (one leg of each touching) with their other legs bolted to the **same face** of the gusset plate, each angle with its own line of bolts. The bolts are therefore in **single shear**, and each angle carries half the load, $150$ kN. M18 bolts of grade 4.6, $d_0=20$ mm; gusset plate 10 mm; E250. #### 1. Bolts - Single shear, thread in the shear plane: $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}\times192\times10^{-3}=35.47$ kN - Bearing ($e=35$, $p=50$): $k_b=\min(35/60,\ 50/60-0.25,\ 0.976,\ 1)=0.583$; on the 6 mm angle leg $V_{dpb}=\dfrac{2.5\times0.583\times18\times6\times410}{1.25}\times10^{-3}=51.7$ kN (on the 10 mm gusset 86.1 kN) - Bolt value $=35.47$ kN → per angle $n=150/35.47=4.2$ → **5 bolts per angle** (10 in all), pitch 50 mm (min $2.5d=45$ mm), end distance 35 mm ($\ge1.5d_0=30$ mm); capacity $5\times35.47=177.4$ kN per angle. #### 2. Section $A_g\ge\dfrac{300\times1.10}{250}\times10^3=1320$ mm². Try **2 ISA 75×75×6** ($A=866$ mm² each, total 1732 mm²), bolt line 40 mm from the heel. - Yielding per angle: $T_{dg}=\dfrac{866\times250}{1.10}=196.8$ kN - Rupture per angle: $A_{nc}=(75-3-20)\times6=312$ mm², $A_{go}=(75-3)\times6=432$ mm², $b_s=75+40-6=109$ mm, $L_c=4\times50=200$ mm, $\beta=1.4-0.076\times\dfrac{75}{6}\times\dfrac{250}{410}\times\dfrac{109}{200}=1.084$ $$ T_{dn}=\frac{0.9\times312\times410}{1.25}+\frac{1.084\times432\times250}{1.10}=198.6\ \text{kN per angle} $$ - Block shear per angle ($A_{vg}=1410$, $A_{vn}=870$, $A_{tg}=210$, $A_{tn}=150$ mm²): $\min(229.3,\ 196.0)=196.0$ kN - Strength of the pair $=2\times\min(196.8,\ 198.6,\ 196.0,\ 177.4)=354.7$ kN $>300$ kN ✓ (bolts govern) #### 3. Slenderness and tacking The minimum radius of gyration of the pair (about the axis parallel to the gusset) equals that of one angle, $r\approx23$ mm, so $L/r=2500/23=109<400$ ✓. Tack the two angles together with bolts at intervals of not more than 1000 mm. **Answer: 2 ISA 75×75×6 placed back to back on the same side of the gusset, each connected with 5 M18 bolts (4.6) at 50 mm pitch (10 bolts in all).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the bolt in double shear here and which area is used for the shear capacity? A. Angles on both faces of the gusset; one plane through threads (Anb + Asb) B. Angles on one face only; use Asb twice C. Gusset is double thickness; use Anb only D. Bolt is in tension; use Asb Answer: A. With angles on opposite faces, the bolt has two shear planes; assuming one in threads gives Vdsb = fub/(sqrt3 gamma_mb)(Anb + Asb). (Key point: Double shear bolt value) 2. Bolt value = 82.40 kN (shear governs). How many bolts for 300 kN? A. 3 B. 4 C. 5 D. 6 Answer: B. 300/82.4 = 3.6, so 4 bolts are provided (capacity 329.6 kN). (Key point: Number of bolts) 3. How is the minimum gross area for the angles found? A. Ag = T x gamma_m0 / fy = 300 x 1.10 / 250 x 10^3 = 1320 mm2 B. Ag = T / fu C. Ag = T x 1.25 / fu D. Ag = T / (0.9 fy) Answer: A. Yielding of gross section: Tdg = Ag fy / gamma_m0, so Ag >= T gamma_m0 / fy. (Key point: Required gross area) 4. In the rupture and block shear checks, what load does each angle carry? A. 150 kN (half) B. Full 300 kN C. 100 kN D. 75 kN Answer: A. Two angles share the tension equally, so each carries 150 kN. 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