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Cotter joint

Discussion in 'Mechanical | Automobile | Aeronautics' started by yadavundertaker, Nov 8, 2009.

  1. yadavundertaker

    yadavundertaker Apprentice

    Engineering Discipline:
    Mechanical
    COTTER JOINT

    A cotter joint is used to connect rigidly two co-axial rods or bars which are subjected to axial tensile or compressive forces . It is a temporary fastening .
    [​IMG][​IMG]
    [​IMG]
    COTTER


    A cotter is a flat wedge shaped piece of rectangular cross section and its width is tapered (either on one side or on both sides) from one end to another for an easy adjustment.

    APPLICATIONS OF COTTER

    1. Connection of the piston rod with the cross heads
    2. Joining of tail rod with piston rod of a wet air pump
    3. Foundation bolt
    4. Connecting two halves of fly wheel (cotter and dowel arrangement)

    COMPARISON BETWEEN KEY AND COTTER

    1. Key is usually driven parallel to the axis of the shaft which is subjected to torsional or twisting stress. Whereas cotter is normally driven at right angles to the axis of the connected part which is subjected to tensile or compressive stress along its axis.

    2. A key resists shear over a longitudinal section whereas a cotter resist shear over two transverse section.
     
  2. yadavundertaker

    yadavundertaker Apprentice

    Engineering Discipline:
    Mechanical
    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 .
     
  3. juma1987

    juma1987 Certified CEan

    Nice post! :D
     
  4. gohm

    gohm Addict

    Engineering Discipline:
    Mechanical
    good one thanks!
     
  5. yadavundertaker

    yadavundertaker Apprentice

    Engineering Discipline:
    Mechanical
    sorry for not updated it on time...
     
  6. yadavundertaker

    yadavundertaker Apprentice

    Engineering Discipline:
    Mechanical
    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.
     
  7. yadavundertaker

    yadavundertaker Apprentice

    Engineering Discipline:
    Mechanical
    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.
     
  8. heba66

    heba66 Certified CEan

    Hi,
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