Max tension in 220x12 plate with rivets

Basis: working stress method (IS 800:1984), Fe410 ($f_y=250$, permissible tension $0.6f_y=150$ N/mm²). 20 mm power-driven shop rivets: hole $d_0=21.5$ mm, $\tau_{vf}=100$, $\sigma_{pf}=300$ N/mm². The six rivets lie in the lap and act in single shear. The gusset (10 mm) is thinner than the flat (12 mm), so the gusset governs bearing.

Reading of the figure: the dimensions 50 + 60 + 60 + 50 = 220 mm are across the width, so the rivets lie on three gauge lines 60 mm apart (at 50, 110 and 170 mm from one edge), staggered at 30 mm along the load.

Rivet value

• Shear: $R_s=\tau_{vf}\dfrac{\pi}{4}d_0^2=100\times363.1\times10^{-3}=36.31$ kN
• Bearing: $R_b=\sigma_{pf}d_0t=300\times21.5\times10\times10^{-3}=64.50$ kN
• Rivet value $=36.31$ kN → six rivets: $217.8$ kN

Strength of the flat (on the net area)

• Solid plate (for reference): $150\times220\times12\times10^{-3}=396.0$ kN
• Straight section through two holes (outer gauge lines): $A_n=(220-2\times21.5)\times12=2124$ mm² → $P=318.6$ kN
• Zig-zag section through three holes (one on each gauge line, $g=60$, $s=30$): $A_n=\left(220-3\times21.5+2\times\dfrac{30^2}{4\times60}\right)\times12=1956$ mm² → $P=293.4$ kN

Answer: maximum tension the flat can carry $=\min(217.8,\ 318.6,\ 293.4)=217.8$ kN, governed by shearing of the rivets.