Tee bracket bolted to column flange, M24

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.

  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 $T_{max}<T_{db}$ and $V_{sb}<V_{dsb}$. Safe.

Detailing

• 8 bolts M24, grade 4.6, in 2 lines at gauge $\ge 2.5d=60$ mm (use 100 mm), pitch 100 mm (min $2.5d=60$ mm), edge distance 40 mm ($\ge1.5d_0=39$ mm).
• The Tee flange (10 mm) should be checked for prying; if required, increase its thickness to 12-16 mm or provide stiffeners, and provide a bearing seat at the bottom.

Answer: 8 nos. M24 bolts of grade 4.6 in 4 rows at 100 mm pitch; utilisation 0.72.