Bolted bracket on ISHB 300, 600 kN, M24

Given: factored $P=600$ kN at $e=300$ mm from the column face; the load is not in the plane of the bolts, so each bolt carries direct shear and tension from the moment. ISHB 300 @ 577 N/m: $b_f=250$ mm, $t_f=10.6$ mm. Bracket plate 12 mm; M24 bolts, grade 4.6, $d_0=26$ mm; E250.

Bolt capacities (IS 800:2007 cl. 10.3)

• M24: $A_{sb}=452$, $A_{nb}=353$ mm²
• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{353}{1.25}\times10^{-3}=65.22$ kN
• Tension: $T_{db}=\dfrac{\min(0.9f_{ub}A_{nb},\ f_{yb}A_{sb}\gamma_{mb}/\gamma_{m0})}{\gamma_{mb}}=98.62$ kN
• Bearing on the flange (10.6 mm, $e=40,\ p=80$): $k_b=0.513$, $V_{dpb}=107.0$ kN (not governing)

Layout

Two vertical lines of 8 bolts (16 bolts), gauge 100 mm (flange 250 mm), pitch 80 mm, end distance 40 mm ($\ge1.5d_0=39$ mm), bracket depth $=40+7\times80+40=640$ mm. The bottom bolt row is taken as the axis of rotation (conservative).

        |  o o  y=560
        |  o o  y=480
  600kN |  o o   ...
 ------>|  o o  y=80
  300   |  o o  y=0 (axis)

Forces

• Shear per bolt $V=\dfrac{600}{16}=37.5$ kN
• $M=600\times300=180\times10^3$ kN·mm; $\Sigma y^2=2\sum y_i^2=1792000$ mm²
• Maximum tension (top row): $T=\dfrac{My_{max}}{\Sigma y^2}=\dfrac{180000\times560}{1792000}=56.25$ kN

Interaction (cl. 10.3.6)

$$\left(\frac{V}{V_{db}}\right)^2+\left(\frac{T}{T_{db}}\right)^2=\left(\frac{37.5}{65.22}\right)^2+\left(\frac{56.25}{98.62}\right)^2=0.656\le1.0$$

Also $V<V_{dsb}$ and $T<T_{db}$. Safe.

Answer: 16 bolts M24 (4.6) in 2 lines of 8, pitch 80 mm, gauge 100 mm. The 10.6 mm flange should be checked for prying (provide a stiffener at the bracket level, or a thicker end plate) and the bracket should bear on a seat at the bottom.