22565 EMD : 1.9 : Fundamental of Design
1) A pull of 80 KN is given to rivet as shown in Figure If the
maximum permissible shear stress in the pin is 80 N/mm2, find
the diameter ‘d’ of the rivet.
Data : Load = 80KN , Permissible Shear stress = 80Mpa, d=?
Shear Stress = Force / Area
Area =  / 4 x d2
= F / A
80 = 80 x 1000 / ( / 4 x d2 )
d = 35.69  40 mm
22565 EMD : 1.9 : Fundamental of Design
2) A pull of 80 KN is given to rivet as shown in Figure If the
maximum permissible shear stress in the pin is 80 N/mm2, find
the diameter ‘d’ of the rivet.
Data : Load = 80KN , Permissible Shear stress = 80Mpa, d=?
Shear Stress = Force / Area
Area = 2 x ( / 4 x d2) (with double shear)
= F / A
80 = 80 x 1000 / (2 x ( / 4 x d2 ) )
d = 25.23  26 mm

22564 emd 1.11 numericals

  • 1.
    22565 EMD :1.9 : Fundamental of Design 1) A pull of 80 KN is given to rivet as shown in Figure If the maximum permissible shear stress in the pin is 80 N/mm2, find the diameter ‘d’ of the rivet. Data : Load = 80KN , Permissible Shear stress = 80Mpa, d=? Shear Stress = Force / Area Area =  / 4 x d2 = F / A 80 = 80 x 1000 / ( / 4 x d2 ) d = 35.69  40 mm
  • 2.
    22565 EMD :1.9 : Fundamental of Design 2) A pull of 80 KN is given to rivet as shown in Figure If the maximum permissible shear stress in the pin is 80 N/mm2, find the diameter ‘d’ of the rivet. Data : Load = 80KN , Permissible Shear stress = 80Mpa, d=? Shear Stress = Force / Area Area = 2 x ( / 4 x d2) (with double shear) = F / A 80 = 80 x 1000 / (2 x ( / 4 x d2 ) ) d = 25.23  26 mm