Strength of double cover butt joint, M20

Assumptions from the figure: main plates 10 mm and 18 mm, two cover plates 6 mm each (total 12 mm), plate width $=40+60+60+40=200$ mm; 6 bolts M20 (grade 4.6) on each side of the joint (3 across the width at 60 mm gauge, 2 along the load at 60 mm pitch), end distance 40 mm, $d_0=22$ mm. Steel E250: $f_y=250$, $f_u=410$ N/mm². The joint is in double shear; one shear plane is taken through the threads. An 8 mm packing is needed on the 10 mm plate side to make up the difference in thickness.

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

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{400}{\sqrt3\times1.25}(245+314)\times10^{-3}=103.28$ kN
• Packing of 8 mm (> 6 mm) reduces this: $\beta_{pkg}=1-0.0125\times8=0.90$, so $V_{dsb}=0.90\times103.28=92.95$ kN on the 10 mm plate side
• Bearing on the thinner main plate (10 mm < 12 mm of covers): $k_b=\min\left(\dfrac{40}{3\times22},\dfrac{60}{3\times22}-0.25,\dfrac{400}{410},1\right)=0.606$

$$V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.606\times20\times10\times410}{1.25}\times10^{-3}=99.39\ \text{kN}$$

• Bolt value $=\min(92.95,99.39)=92.95$ kN (shear with packing governs)

Strength of the joint

1. Bolts: $6\times92.95=557.7$ kN
2. Main plate 10 mm, gross section: $T_{dg}=\dfrac{200\times10\times250}{1.10}=454.5$ kN
3. Main plate net section (3 holes): $T_{dn}=\dfrac{0.9(200-3\times22)\times10\times410}{1.25}=395.6$ kN
4. Cover plates (2 × 6 = 12 mm): $T_{dg}=545.5$ kN, $T_{dn}=474.7$ kN
5. Block shear of the 10 mm plate (shear along the two outer bolt lines, $A_{vg}=2000$, $A_{vn}=1340$ mm²):
   • central block torn out ($A_{tg}=1200$, $A_{tn}=760$ mm²): $T_{db}=\min(486.8,\ 501.1)=486.8$ kN
   • two edge strips torn out ($A_{tg}=800$, $A_{tn}=580$ mm²): $T_{db}=\min(433.6,\ 410.2)=410.2$ kN

Answer: strength of the joint $=\min(\ldots)=395.6$ kN, governed by net-section rupture of the 10 mm plate (the 18 mm plate and the covers are stronger).