Cotter joint

Replies

• 

  yadavundertaker mohit

  DIFFERENT TYPES OF COTTER JOINTS
  
  1. Socket and spigot cotter joint
  2. Sleeve and cotter joint
  3. Gib and cotter joint
  
  
  DESIGN PROCEDURE FOR THE SPIGOT AND COTTER JOINT
  
  To understand the design steps let take a question and solve it step by step.
  
  Q. Design a cotter joint subjected to a tensile load of 35 kN and a compressive load of 40 KN . The allowable stresses are
  tensile stress qt = 70N/mm2
  compressive stress qc = 110 N/mm2
  shear stres t = 50 N/mm2
  
  
  SOLUTION
  
  STEP 1
  
  Failure of rod in tension or compression
  
  tensile stress = tensile load / area
  
  qt=(Pt*40/(3.14*d*d)
  
  where d is the diameter of the rod
  
  from above equation we will find out the value of d and round it off to the higher integer. Now we apply the Standard shaft rule..i.e the diameter of the rod should be of the given range
  
  diameter increment in steps
  
  1-10mm 1mm
  10-24mm 2mm
  24-45 mm 3mm
  45-100 mm 5mm
  above 100mm 10mm
  
  we will make the value of the d such that it will satisfy the above table.
  
  compressive stress = compressive load /area
  
  qc=(4*Pc)/(3.14*d*d)
  
  again find the value of d ,round it off ,make according to the above table and now compare the two values of d .Take the value which is greater .
  
  
  STEP 2
  
  Failure of spigot in tension across slot
  
  Pt=qt*[3.14/4*d1*d1 -d1*t]
  
  where t =d1/4
  find the value of d1 and then t
  
  
  
  Emperically d1=1.2*d
  
  again find the value of d1 and choose which one is larger ..
  
  STEP 3
  
  Failure of rod or cotter in crushing
  
  check for the condition
  
  Pt<= d1*t*qc
  
  if the above condition meet then the value of d1 and t is correct else increase the value of d1 and t to satisfy the condition .Because if the above condition is not satisfied then cotter will be get crushed due to load .
• 

  juma1987

  Nice post! 😁
• 

  gohm

  good one thanks!
• 

  yadavundertaker mohit

  sorry for not updated it on time...
• 

  yadavundertaker mohit

  STEP IV
  
  Tensile failure of socket across slot
  
  Pt=qt[3.14/4(d3*d3-d1*d1)2-(d3-d1)t]
  
  find the value of d3 from above equation
  
  Emperically , d3=1.75d
  
  choose the greater one.
  
  
  
  STEP V
  
  Crushing of cotter against collar of socket
  
  
  Pt=qc(d4-d1)t
  
  find the value of d4
  
  Emperically , d4=2.4d
  
  choose the greater one .
  
  
  STEP VI
  
  Crushing of collar of spigot to socket
  
  
  Pc=(qc*3.14*(d2*d2-d1*d1))/4
  
  find the value of d2
  
  Emperically , d2 =1.5d
  
  choose the greater one.
• 

  yadavundertaker mohit

  STEP IX
  
  Shear failure of spigot end against cotter
  
  Pt=2*T*d1*a
  
  where T=shear stress
  
  find the value of a
  
  Emperically a=0.75d
  
  choose the greater value of a
  
  
  STEP X
  
  Shear failure of spigot collar
  
  Pt=3.14*d*h*T
  
  find the value of h from the above equation
  
  Emperically ,h=0.75d
  
  Choose the greater value of h
  
  
  Step XI
  
  Double shear failure of collar of socket by cotter
  
  Pt=2*T*(d4-d1)*e
  
  
  find the value of e from the above equation
  
  Emperically e=0.75d
  
  Choose the greater value of e.
• 

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• 

  Ankita Katdare

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