Jump to content
Mechanical Engineering

cadsanthanam

Members
  • Posts

    1
  • Joined

  • Last visited

  • Days Won

    1
  1. The link is subjected to 3 types of stresses 1. Bending of the link due to curvature and eccentricity, 2. Direct tensile load on the link, 3. Double shear at the contact point, 4. crushing at the link at the point of contct (but the links are same material, hence this may be neglected) of the above three, double shear corresponds to weakest portion, Now Allowable shear strength => 75/2 = P/2A where P = applied load = 50,000N, 2A - double shear area; solving A = 666 mm^2 => diameter = 29.14mm or dia = 30 mm
×
×
  • Create New...