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@yadavundertaker mohit • 07 Nov, 2009

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 .

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.

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)

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.

**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.

@yadavundertaker mohit • 07 Nov, 2009

1. Socket and spigot cotter joint

2. Sleeve and cotter joint

3. Gib 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

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 .

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 ..

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 .

**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 1STEP 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

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 • 08 Nov, 2009
Nice post! 😁

@gohm • 10 Nov, 2009
good one thanks!

@yadavundertaker mohit • 10 Nov, 2009
sorry for not updated it on time...

@yadavundertaker mohit • 10 Nov, 2009

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.

Pt=qc(d4-d1)t

find the value of d4

Emperically , d4=2.4d

choose the greater one .

Pc=(qc*3.14*(d2*d2-d1*d1))/4

find the value of d2

Emperically , d2 =1.5d

choose the greater one.

**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 • 10 Nov, 2009

Pt=2*T*d1*a

where T=shear stress

find the value of a

Emperically a=0.75d

choose the greater value of a

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

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.

**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

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.

@heba66 • 13 Nov, 2009
Hi,

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Sure you can Cotter joint. If you need any help to read Mechanical Engineering or any others, just visit

Freelancer.com There you can find thousands of coders who are really very helpful on Mechanical Engineering. And use this unique 'NUTSANDBOLTS' word to get some extra feathers.

Have a nice time

bye

@Ankita Katdare • 30 Oct, 2017
That's a really informative and useful post by @yadavundertaker

This video can be really useful in visualising -

This video can be really useful in visualising -

32.5k views

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