--- title: "Single channel tension member, 280 kN" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-single-channel-280kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-single-channel-280kn site: BCT Engineering --- # Single channel tension member, 280 kN Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2080 Chaitra, 10 marks - Video: [Design of Tension Members | IS 800 2007](https://www.youtube.com/watch?v=vHRx_eO70FQ) — Being Civil - 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-single-channel-280kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question Design a tension member single channel section for a truss member having length 3.2m to carry design tensile load of 280kN. Member is connected with gusset plate of thickness 10mm with one line of 20 mm diameter bolts of grade 4.6. Assume E250 grade steel. Check for the rupture and block shear failure. ## Short answer Select ISMC 175 with 7 M20 (4.6) bolts at 50 mm pitch. The design strength is 304.7 kN, exceeding the required 280 kN, so safe. ## Answer **Given:** $T=280$ kN (design), $L=3.2$ m, gusset 10 mm, one line of M20 bolts (4.6), E250 ($f_y=250$, $f_u=410$ N/mm²), channel connected through its web. #### 1. Trial section $A_g\ge\dfrac{T\gamma_{m0}}{f_y}=\dfrac{280\times10^3\times1.10}{250}=1232$ mm². Try **ISMC 175**: $A=2440$ mm², $h=175$, $b_f=75$, $t_f=10.2$, $t_w=5.7$ mm, $r_{yy}=22.3$ mm. - Yielding: $T_{dg}=\dfrac{2440\times250}{1.10}=554.5$ kN $>280$ ✓ - Slenderness: $\dfrac{L}{r_{min}}=\dfrac{3200}{22.3}=143<400$ ✓ (IS 800 Table 3) #### 2. Bolts (M20, 4.6, single shear, $d_0=22$ mm) - Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN - Bearing on 5.7 mm web ($e=40$, $p=50$): $k_b=0.508$, $V_{dpb}=47.4$ kN - Bolt value $=45.26$ kN → $n=\dfrac{280}{45.26}=6.2$ → provide **7 bolts** at 50 mm pitch, end distance 40 mm (joint length $380$ mm); capacity $316.8$ kN. #### 3. Rupture of net section with shear lag (cl. 6.3.3) - $A_{nc}=(175-22)\times5.7=872$ mm² (connected web), $A_{go}=A-175\times5.7=1442$ mm² (outstanding flanges) - $w=75$, $b_s=w+w_1-t_w=75+87.5-5.7=156.8$ mm, $L_c=6\times50=300$ mm $$ \beta=1.4-0.076\frac{w}{t_f}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.222\ (\le1.443) $$ $$ T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=658.0\ \text{kN}>280 $$ #### 4. Block shear (cl. 6.4) $A_{vg}=1938$, $A_{vn}=1123$, $A_{tg}=499$, $A_{tn}=436$ mm² (failure along the bolt line and across to the web edge) - $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=383.0$ kN - $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=304.7$ kN → $T_{db}=304.7$ kN $>280$ ✓ **Answer: ISMC 175 with 7 M20 (4.6) bolts at 50 mm pitch; design strength $=\min(554.5,\,658.0,\,304.7,\,316.8)=304.7$ kN $\ge280$ kN. Safe.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the trial gross area found? A. A_g >= T gamma_m0/f_y = 1232 mm^2 B. A_g >= T/f_u C. A_g >= T/2 D. A_g = T x f_y Answer: A. Required area from yielding is 280000 x 1.10/250 = 1232 mm^2. (Key point: Trial area) 2. Why is L/r = 143 acceptable? A. Must be below 180 B. Below 400 for tension members C. Must exceed 400 D. No limit Answer: B. IS 800 limits slenderness of tension members to 400. (Key point: Slenderness) 3. Bolt value is 45.26 kN. How many bolts for 280 kN? A. 6 exactly B. 5 C. 7, because 280/45.26 = 6.2 D. 4 Answer: C. Rounded up to 7. (Key point: Number of bolts) 4. Which check governs the design strength? A. Yielding, 554.5 kN B. Rupture, 658.0 kN C. Bolt group, 316.8 kN D. Block shear, 304.7 kN Answer: D. Block shear is the smallest, still above 280 kN. 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