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 . 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.

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 .

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.

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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