Double cover butt joint for boiler plates

Joint type: single-bolted (one row) double cover butt joint; bolts in double shear. Plates 12 mm, Fe410; M16 bolts grade 4.6, $d_0=18$ mm. Covers: two plates of 8 mm ($2\times8=16\ge12$ mm).

Strength of one bolt (IS 800:2007 cl. 10.3)

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}(157+201)\times10^{-3}=66.14$ kN
• Bearing on the 12 mm main plate: with a single row there is no pitch in the direction of load, so only the end distance $e=30$ mm ($\ge1.5d_0=27$ mm) controls: $k_b=\min(30/54,\ 0.976,\ 1)=0.556$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{\gamma_{mb}}=87.5$ kN
• Bolt value $=66.14$ kN

Pitch for maximum efficiency

Efficiency is highest when the net strength of the plate between bolt holes equals the bolt value, so that no material is wasted:

$$\frac{0.9(p-d_0)tf_u}{\gamma_{m1}}=V_b\Rightarrow p=18+\frac{66.14\times10^3}{0.9\times12\times410/1.25}=36.7\ \text{mm}$$

This is below the minimum pitch $2.5d=40$ mm, so take $p=40$ mm. Check:

• Net strength of plate in a pitch strip: $T_{dn}=\dfrac{0.9(40-18)\times12\times410}{1.25}=77.9$ kN $>V_b$
• Gross strength of plate in a pitch strip: $T_{dg}=\dfrac{40\times12\times250}{1.10}=109.1$ kN

Efficiency

$$\eta=\frac{\text{lesser of bolt value and }T_{dn}}{T_{dg}}=\frac{66.14}{109.1}\times100=60.6\%$$

Answer: M16 bolts at 40 mm pitch in one row, end distance 30 mm, with 8 mm cover plates on both sides; efficiency $\approx61\%$. (A single row cannot do better than this; efficiency can be raised by using more rows of bolts in a zig-zag pattern.)