Double cover butt joint, M16 class 10.9

Data: working load 300 kN → factored $P_u=1.5\times300=450$ kN. Flats 100×20 mm, covers 12 mm each, M16 HSFG bolts of class 10.9 ($f_{ub}=1000$ N/mm²), $d_0=18$ mm, E250. "Slip permitted at design load" means the bolts are designed as bearing-type (for shear and bearing, IS 800:2007 cl. 10.3/10.4). Double shear, with one plane through the threads.

Bolt value

• Shear: $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{1000}{\sqrt3\times1.25}(157+201)\times10^{-3}=165.35$ kN
• Bearing on the 20 mm flat ($e=30$, $p=50$): $k_b=\min(0.556,\ 0.676,\ 2.44,\ 1)=0.556$

$$V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.556\times16\times20\times410}{1.25}\times10^{-3}=145.8\ \text{kN}$$

• Bolt value $=145.8$ kN (bearing governs)

Number of bolts

$n=\dfrac{450}{145.8}=3.08\Rightarrow$ provide 4 bolts on each side of the joint, in a single line along the centre of the flat at 50 mm pitch with 30 mm end distance, so the flat is weakened by only one hole.

Checks

• Flat gross yielding: $T_{dg}=\dfrac{100\times20\times250}{1.10}=454.5$ kN $\ge450$ ✓
• Flat net rupture (1 hole): $T_{dn}=\dfrac{0.9(100-18)\times20\times410}{1.25}=484.1$ kN $\ge450$ ✓
• Cover plates $2\times12=24>20$ mm: $T_{dg}=545.5$ kN ✓
• Bolts: $4\times145.8=583.1$ kN $\ge450$ ✓
• (For information, the slip resistance per bolt at service load is $V_{dsf}=\mu_f n_eK_hF_o/\gamma_{mf}=95.9$ kN with $\mu_f=0.48$, $n_e=2$, $K_h=1.0$, $F_o=0.7f_{ub}A_{nb}=109.9$ kN and $\gamma_{mf}=1.10$; the total of 384 kN exceeds 300 kN, so the joint does not slip under the working load either.)

Answer: 4 M16 (10.9) HSFG bolts on each side in a single line at 50 mm pitch, with 12 mm cover plates on both faces.