--- title: "Bracket plate on column flange, 5 M20 bolts" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:bracket-bolted-5-m20" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-bolted-5-m20 site: BCT Engineering --- # Bracket plate on column flange, 5 M20 bolts Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2081 Chaitra, 8 marks - Video: [Design of Bolted bracket connection 2-8|Example|IS 800| Design of Steel Structural elements |18cv61](https://www.youtube.com/watch?v=93-oQqn3QD0) — Dr Aravinthan. K - 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-5-m20 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question A bracket plate is bolted to the flange of a column using 5 Nos. of M 20 bolts of grade 4.6 and is loaded as shown in the figure. Determine the maximum value of working load 'P' which can be carried safely. [Figure: 10 mm thick bracket plate bolted to the flange of an ISMB 300 column; load P acts at 250 mm from the column face; plate width 300 mm; 5 bolts arranged with vertical pitch 80 mm and 80 mm and horizontal spacings 90 mm, 60 mm, 60 mm as dimensioned] ## Short answer For M20 grade 4.6 bolts, shear capacity 45.26 kN governs. Direct shear per bolt P/5 plus torsion from eccentricity 100 mm gives resultant 0.4031P_u per bolt. Equating to shear capacity gives P_u = 112.3 kN. Safe working load = P_u/1.5 ≈ 74.9 kN. ## Answer **Assumptions (figure values are ambiguous):** the bracket plate (10 mm) lies against the column flange (ISMB 300, $t_f=12.4$ mm) and the load $P$ acts in the plane of the bolts. The 5 bolts are M20, grade 4.6, in the pattern: four corner bolts at $(\pm60,\pm80)$ mm and one at the centre (gauges 60+60 and pitch 80+80; this makes the corner distance $r=\sqrt{60^2+80^2}=100$ mm). The bolt group centroid is 150 mm from the column face (90 mm edge + 60 mm), so the eccentricity of $P$ (250 mm from the face) is $e=250-150=100$ mm. Working load $P$ = factored load / 1.5. ```text o o y = +80 o y = 0 centre o o y = -80 x=-60 +60 P acts e = 100 mm from centroid ``` #### Bolt value (IS 800:2007, cl. 10.3) - M20: $A_{nb}=245$ mm², $f_{ub}=400$, $f_u=410$, $d_0=22$ mm. - Shear (single shear, thread in plane): $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN - Bearing (plate 10 mm, $e=40$, $p=80$): $k_b=\min(0.606,\,0.962,\,0.976,\,1)=0.606$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=99.4$ kN - Bolt value $V_{db}=45.26$ kN (shear governs). #### Force on the critical (corner) bolt, per unit factored load $P_u$ - $\Sigma r^2=4\times100^2=40000$ mm²; torsion $T=P_ue=100P_u$ - Direct shear (vertical) $F_D=P_u/5=0.2P_u$ - Torsional force $F_T=\dfrac{T r}{\Sigma r^2}=\dfrac{100P_u\times100}{40000}=0.25P_u$, with components $F_{Tx}=0.25P_u\times\tfrac{80}{100}=0.20P_u$ and $F_{Ty}=0.25P_u\times\tfrac{60}{100}=0.15P_u$ - Resultant $R=\sqrt{(0.20P_u)^2+(0.2P_u+0.15P_u)^2}=0.4031P_u$ #### Capacity $$ 0.4031P_u\le45.26\Rightarrow P_u=112.3\ \text{kN} $$ **Answer: factored load $P_u\approx112.3$ kN; safe working load $P=P_u/1.5\approx74.9$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the M20 grade 4.6 bolt, which governs the bolt value and what is it? A. Bearing, 99.4 kN B. Shear, 45.26 kN C. Tension, 99.4 kN D. Shear, 99.4 kN Answer: B. Shear strength is 45.26 kN versus bearing 99.4 kN, so shear governs. (Key point: Bolt value governing) 2. How is the eccentricity of the load obtained? A. 250 mm from face minus the 150 mm distance of bolt group centroid from the face = 100 mm B. Directly 250 mm C. Plate width 300 mm D. Half the bolt pitch Answer: A. The load acts at 250 mm; the centroid is at 150 mm from the column face, so e = 100 mm. (Key point: Eccentricity) 3. What is the formula for the torsional force on the critical bolt? A. P/n B. T r / sum r^2 C. M y / I D. T / r Answer: B. F_T = T r / Σr^2 with T = P e; here 0.25 Pu at the corner bolt. (Key point: Torsional force) 4. Which are the final values of factored and working loads? A. Pu = 74.9 kN, P = 112.3 kN B. Pu = 112.3 kN, P = 168.5 kN C. Pu = 45.26 kN, P = 30 kN D. Pu = 112.3 kN, P = 74.9 kN Answer: D. 0.4031 Pu <= 45.26 gives Pu = 112.3 kN, and working load P = Pu/1.5 = 74.9 kN. 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