--- title: "Tee bracket bolted to column flange, M24" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:bracket-tee-m24" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-tee-m24 site: BCT Engineering --- # Tee bracket bolted to column flange, M24 Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2080 Chaitra, 10 marks - Video: [Numerical on Bracket Connection Type II - Design of Connections](https://www.youtube.com/watch?v=8N3uFPHMW2A) — Ekeeda - 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-tee-m24 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question Design a bracket connection to transfer an end reaction of 225 kN due to factored load as shown in figure below. The end reaction from the girder acts at an eccentricity of 300 mm from the face of the column flange of thickness 12 mm. Design bolted joint connecting the Tee-flange of thickness 10 mm with the column flange. Use steel of grade E250 and M24 bolts of grade 4.6. [Figure: Tee bracket attached to a column flange, 225 kN load acting 300 mm from the column face] ## Short answer Eight M24 grade 4.6 bolts in 4 rows at 100 mm pitch, gauge 100 mm. Shear per bolt = 28.12 kN, max tension from moment = 72.32 kN. Interaction check (V/V_db)^2 + (T/T_db)^2 = 0.724 < 1.0. Design is safe; provide 8 M24 bolts. ## Answer **Assumptions:** 8 bolts M24 (grade 4.6) in two vertical lines with 4 rows at pitch $p=100$ mm; the bottom row is taken as the axis of rotation (the Tee stem presses on the column flange at the bottom). The load acts in a plane perpendicular to the bolt plane, so the bolts are in **shear + tension**. Factored $P=225$ kN, $e=300$ mm, $M=67.5$ kN·m. ```text 225 kN --------> |T| o o y=300 300 |T| o o y=200 |T| o o y=100 |T| o o y=0 (axis) ``` #### Bolt capacities (IS 800:2007 cl. 10.3) - M24: $A_{sb}=452$, $A_{nb}=353$ mm²; $f_{ub}=400$, $f_{yb}=240$ N/mm². - Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{353}{1.25}\times10^{-3}=65.22$ kN - Tension: $T_{nb}=\min(0.9\times400\times353,\ 240\times452\times1.25/1.10)=123.3$ kN; $T_{db}=T_{nb}/1.25=98.62$ kN - Bearing: hole $d_0=26$ mm, $e=40$, $p=100$ mm: $k_b=\min(40/78,\,100/78-0.25,\,0.976,\,1)=0.513$; $V_{dpb}=2.5k_b\,d\,t\,f_u/\gamma_{mb}=100.9$ kN with $t=10$ mm (Tee flange), larger than $V_{dsb}$. #### Forces - Shear per bolt $V_{sb}=225/8=28.12$ kN - $\Sigma y^2=2(100^2+200^2+300^2)=280000$ mm² - $T_{max}=\dfrac{M y_{max}}{\Sigma y^2}=\dfrac{67.5\times10^6\times300}{280000}=72.32$ kN #### Interaction check (IS 800:2007 cl. 10.3.6) $$ \left(\frac{28.12}{65.22}\right)^2+\left(\frac{72.32}{98.62}\right)^2=0.724<1.0 $$ Also $T_{max}