Shear stress in rivets B and C

Working stress method (IS 800:1984). Hole diameters are used for the rivet areas: $d_0=d+1.5$ mm, i.e. 17.5 mm for the 16 mm rivets (A, B) and 21.5 mm for the 20 mm rivet (C).

Reading of the figure: rivets A and B are 60 mm above and below the horizontal axis and 90 mm from the load line; rivet C is on the axis, 120 mm from the load line. $P=20$ kN, vertical.

Step 1 - rivet areas and centroid of the group

• $A_A=A_B=\dfrac{\pi}{4}(17.5)^2=240.5$ mm²; $A_C=\dfrac{\pi}{4}(21.5)^2=363.1$ mm²; $\Sigma A=844.1$ mm²
• Distance of centroid from load line: $\bar x=\dfrac{2A_A(90)+A_C(120)}{\Sigma A}=102.90$ mm
• So A, B are $12.90$ mm on the load side of the centroid and C is $17.10$ mm on the far side.

Step 2 - direct shear stress (same in all rivets)

$\tau_d=\dfrac{P}{\Sigma A}=\dfrac{20000}{844.1}=23.69$ N/mm² (vertical)

Step 3 - torsion

$M=Pe=20\times10^3\times102.90=2058\times10^3$ N·mm

$$J=\Sigma A r^2=2A_A(12.90^2+60^2)+A_C(17.10^2)=1918015\ \text{mm}^4$$

Torsional shear stress in a rivet: $\tau_t=\dfrac{Mr}{J}$ (perpendicular to the radius).

Step 4 - resultant stresses

Rivet B ($x=12.90$, $y=-60$): $\tau_x=\dfrac{My}{J}=64.38$ N/mm² (magnitude), $\tau_y=\tau_d+\dfrac{Mx}{J}=37.54$ N/mm²

$$\tau_B=\sqrt{64.38^2+37.54^2}=74.53\ \text{N/mm}^2$$

Rivet C ($y=0$): $\tau_x=0$, $\tau_y=\tau_d-\dfrac{M\times17.10}{J}=5.35$ N/mm²

$$\tau_C=5.35\ \text{N/mm}^2$$

Answer: shear stress in rivet B (and A) $\approx74.5$ N/mm²; in rivet C $\approx5.3$ N/mm², both below the permissible 100 N/mm² of power-driven shop rivets.