--- title: "Check bolted bracket on SC 250 (25 cm eccentricity)" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:bracket-sc250-100kn-25cm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-sc250-100kn-25cm site: BCT Engineering --- # Check bolted bracket on SC 250 (25 cm eccentricity) Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2074 Bhadra, 10 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/bracket-sc250-100kn-25cm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question A bracket plate 12 mm thick transmits a load of 100 KN at an eccentricity of 25 cm to a column section SC 250 through 14-16 mm diameter. Product grade C and property class 4.6 bolts arranged in two vertical rows 10 cm apart. The pitch of the bolt is 8 cm and load lies in the plane of the bolts. Check the safety of the bolted joints. The grade of steel is Fe410. ## Short answer 100 kN at 250 mm eccentricity, 14 M16 grade 4.6 bolts, two rows, gauge 100 mm, pitch 80 mm. Maximum bolt force 18.42 kN (27.63 kN if working load), bolt shear capacity 29.01 kN. Joint is safe. ## Answer **Reading of the data:** "14-16 mm diameter" is taken as 14 bolts of 16 mm diameter (M16, product grade C, class 4.6) in two vertical rows 100 mm apart, 7 per row, pitch 80 mm. The 100 kN is first treated as the design (factored) load; the working-load case is checked at the end. Plate 12 mm, Fe410. #### Bolt value (IS 800:2007 cl. 10.3) - M16: $A_{nb}=157$ mm², $d_0=18$ mm (grade C hole: $d+2$). - Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN - Bearing ($e=40$, $p=80$): $k_b=\min(0.741,1.231,0.976,1)=0.741$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=116.6$ kN - Bolt value $=29.01$ kN #### Forces - $\Sigma(x^2+y^2)=14(50^2)+2\times2(80^2+160^2+240^2)=393400$ mm²; $T=100\times250=25000$ kN·mm - Direct shear per bolt $=100/14=7.14$ kN - Critical bolt (corner, $x=50$, $y=240$): $F_x=\dfrac{Ty}{\Sigma r^2}=15.25$ kN; torsional vertical part $=3.18$ kN - $R=\sqrt{15.25^2+(7.14+3.18)^2}=18.42$ kN $R=18.42\ \text{kN}<29.01$ kN (utilisation 0.63). If 100 kN is a working load, the factored force is $1.5\times18.42=27.63$ kN $<29.01$ kN (utilisation 0.95), still safe. **Answer: the joint is safe.** (End distance 40 mm ≥ $1.5d_0=27$ mm; pitch 80 mm lies between $2.5d=40$ mm and $\min(32t,300)$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does '14-16 mm diameter' mean in the question? A. 14 bolts of 16 mm diameter B. Bolts of diameter between 14 and 16 mm C. 16 bolts of 14 mm diameter D. One bolt of 14.16 mm Answer: A. The answer reads it as 14 bolts of 16 mm diameter, 7 per row in two rows. (Key point: Interpreting the data) 2. For M16 bolts of product grade C, what hole diameter d0 is used? A. 16 mm B. 17 mm C. 20 mm D. 18 mm (d + 2) Answer: D. Grade C bolts use clearance holes of d + 2 mm, so d0 = 18 mm. (Key point: Hole diameter) 3. Shear capacity of M16 is 29.01 kN and bearing is 116.6 kN. What governs and what check is made? A. Bearing governs; R is checked against 116.6 kN B. Shear governs; the corner bolt force 18.42 kN is checked against 29.01 kN C. Tension governs D. Both are added to give 145.6 kN Answer: B. Bolt value is the smaller one (shear, 29.01 kN) and the resultant bolt force of 18.42 kN is below it, so the joint is safe. (Key point: Governing bolt value) 4. If the 100 kN were a working load instead of a design load, what would change? A. Nothing B. Divide it by 1.5 C. Multiply it by 1.5 before checking D. Ignore the eccentricity Answer: C. Limit-state checks need factored load, so a working load is multiplied by 1.5. 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