--- title: "Bolted bracket connection, 200 kN at 250 mm" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:bracket-bolted-200kn-250mm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-bolted-200kn-250mm site: BCT Engineering --- # Bolted bracket connection, 200 kN at 250 mm Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2075 Baisakh, 12 marks - Video: [How to determine the bolt size for connecting a bracket subjected to bending moment.](https://www.youtube.com/watch?v=TPWAcc9m6uo) — Structural Engineer Calcs - 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-bolted-200kn-250mm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question Design bolted connection for the bracket loaded as shown in figure. [Figure: 12 mm thick bracket plate bolted to a column flange; 200 kN load acts at 250 mm from the column face] ## Short answer 200 kN factored load at 250 mm eccentricity. Use 14 M20 grade 4.6 bolts in two vertical lines of seven, gauge 100 mm, pitch 70 mm, end 40 mm. Shear value 45.26 kN, critical bolt force 40.64 kN. Safe. ## Answer **Assumptions:** the 200 kN is the factored load, acting in the plane of the bolts at $e=250$ mm from the centroid of the bolt group (column face taken as the bolt line). The column flange is wide enough for a gauge of 100 mm. M20 bolts, grade 4.6, in two vertical lines, plate 12 mm, E250. #### Bolt value - Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN - Bearing ($e=40$, $p=70$, $t=12$ mm): $k_b=0.606$, $V_{dpb}=119.3$ kN - Bolt value $=45.26$ kN #### Trial and check The bolts are assumed at gauge 100 mm and pitch 70 mm. For $n$ bolts the critical corner bolt receives direct shear $P/n$ plus the torsion effect of $P e$ about the centroid: | Bolts (2 lines) | Max bolt force R (kN) | Check vs 45.26 kN | |---|---|---| | 2 × 5 = 10 | 69.7 | Not safe | | 2 × 6 = 12 | 52.3 | Not safe | | 2 × 7 = 14 | 40.6 | Safe | For 14 bolts: $\Sigma(x^2+y^2)=309400$ mm², $T=200\times250=50000$ kN·mm - Direct shear $=200/14=14.29$ kN - Corner bolt ($x=50$, $y=210$): $F_x=\dfrac{Ty}{\Sigma r^2}=33.94$ kN; vertical torsion part $=8.08$ kN - $R=\sqrt{33.94^2+(14.29+8.08)^2}=40.64$ kN $<45.26$ kN **Answer: 14 bolts M20 (4.6) in two vertical lines of 7, gauge 100 mm, pitch 70 mm, end distance 40 mm (bracket depth about 500 mm).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the number of bolts decided for this eccentric connection? A. Divide P by bolt value only B. Try a number of bolts and increase it until the critical bolt force is within bolt value C. Use the minimum number of bolts allowed by code D. Choose bolts so that pitch equals gauge Answer: B. The eccentric moment adds torsion, so trials are made and the corner bolt resultant is checked against the bolt value. (Key point: Trial method) 2. With 2 x 6 = 12 bolts the maximum bolt force is 52.3 kN. Why is this not acceptable? A. It exceeds the bolt value of 45.26 kN B. It is below the bearing value C. The number of bolts must be odd D. It is less than the end distance Answer: A. Maximum bolt force 52.3 kN is greater than the bolt value 45.26 kN, so the bolts are overstressed. (Key point: Why 12 bolts fail) 3. For the final 14-bolt arrangement carrying 200 kN, what is the direct shear per bolt? A. 200 x 250/14 = 3571 kN B. 200/14 = 14.29 kN C. 200/7 = 28.6 kN D. 200/12 = 16.7 kN Answer: B. Direct shear is shared equally by all bolts: P/n = 200/14 = 14.29 kN. (Key point: Direct shear per bolt) 4. Which arrangement is finally adopted? A. 10 bolts in two lines of 5 B. 12 bolts in two lines of 6 C. 16 bolts in two lines of 8 D. 14 bolts in two lines of 7 Answer: D. Only 14 bolts bring the corner bolt force (40.64 kN) below the bolt value of 45.26 kN. 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