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Spring

A spring is defined as an elastic body, whose function is to distort when loaded and to

recover its original shape when the load is removed.

The various important applications of springs are as follows:

To cushion, absorb or control energy due to either shock or vibration as in car

springs, railway buffers, air-craft landing gears, shock absorbers and vibration

dampers.

To apply forces, as in brakes, clutches and spring loaded valves.

To control motion by maintaining contact between two elements as in cams and

followers.

To measure forces, as in spring balances and engine indicators.

To store energy, as in watches, toys, etc.

Types of springs

Though there are many types of the springs, yet the following, according to their shape,

are important from the subject point of view.

Helical springs

The helical spring are made up of a wire coiled in the form of a helix and is primarily

intended for compressive or tensile loads.

The cross-section of the wire from which the spring is made may be circular, square or

rectangular. The two forms of helical springs are compression helical spring as shown

in Fig. (a) And tension helical spring as shown in Fig. (b).

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The helical springs are said to be closely coiled when the spring wire is coiled so close

that the plane containing each turn is nearly at right angles to the axis of the helix and the

wire is subjected to torsion.

In other words, in a closely coiled helical spring, the helix angle is very small; it is

usually less than 10°.

The major stresses produced in helical springs are shear stresses due to twisting. The load

applied is parallel to or along the axis of the spring.

In open coiled helical springs, the spring wire is coiled in such a way that there is a gap

between the two consecutive turns, as a result of which the helix angle is large.

Since the application of open coiled helical springs are limited, therefore our discussion

shall confine to closely coiled helical springs only.

The helical springs have the following advantages:

These are easy to manufacture.

These are available in wide range.

These are reliable.

These have constant spring rate.

Their performance can be predicted more accurately.

Their characteristics can be varied by changing dimensions.

Torsion springs

Torsion springs may be of helical or spiral type as shown in Fig.

The helical type may be used only in applications where the load tends to wind up the

spring and are used in various electrical mechanisms.

The spiral type is also used where the load tends to increase the number of coils and

when made of flat strip are used in watches and clocks.

The major stresses produced in torsion springs are tensile and compressive due to

bending.

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Laminated or leaf springs

The laminated or leaf spring (also known as flat spring or carriage spring) consists of a

number of flat plates (known as leaves) of varying lengths held together by means of

clamps and bolts, as shown in Fig.

These are mostly used in automobiles. The major stresses produced in leaf springs are

tensile and compressive stresses.

Disc or Belleville springs

These springs consist of a number of conical discs held together against slipping by a

central bolt or tube as shown in Fig.

These springs are used in applications where high spring rates and compact spring units

are required.

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The major stresses produced in disc or Belleville springs are tensile and compressive

stresses.

Special purpose springs

These springs are air or liquid springs, rubber springs, ring springs etc.

The fluids (air or liquid) can behave as a compression spring. These springs are used for

special types of application only.

Material for Helical Springs

The material of the spring should have high fatigue strength, high ductility, high

resilience and it should be creep resistant. It largely depends upon the service for which they are

used i.e. severe service, average service or light service.

Severe service means rapid continuous loading where the ratio of minimum to maximum

load (or stress) is one-half or less, as in automotive valve springs.

Average service includes the same stress range as in severe service but with only

intermittent operation, as in engine governor springs and automobile suspension springs.

Light service includes springs subjected to loads that are static or very infrequently

varied, as in safety valve springs.

The springs are mostly made from oil-tempered carbon steel wires containing 0.60 to 0.70

per cent carbon and 0.60 to 1.0 per cent manganese. Music wire is used for small springs.

Non-ferrous materials like phosphor bronze, beryllium copper, Monel metal, brass etc., may

be used in special cases to increase fatigue resistance, temperature resistance and corrosion

resistance.

Terms used in Compression Springs

The following terms used in connection with compression springs are important from the subject

point of view.

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Problems

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1. A helical spring is made from a wire of 6 mm diameter and has outside diameter of 75 mm.

If the permissible shear stress is 350 MPa and modulus of rigidity 84 kN/mm2, find the

axial load which the spring can carry and the deflection per active turn.

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2. Design a spring for a balance to measure 0 to 1000 N over a scale of length 80 mm. The

spring is to be enclosed in a casing of 25 mm diameter. The approximate number of turns is

30. The modulus of rigidity is 85 kN/mm2. Also calculate the maximum shear stress

induced.

Solution. Given : W = 1000 N ; δ = 80 mm ; n = 30 ; G = 85 kN/mm2 = 85 × 10

3 N/mm

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3. Design a helical compression spring for a maximum load of 1000 N for a deflection of 25

mm using the value of spring index as 5. The maximum permissible shear stress for spring

wire is 420 MPa and modulus of rigidity is 84 kN/mm2.

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Leaf Springs

Leaf springs (also known as flat springs) are made out of flat plates.

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The advantage of leaf spring over helical spring is that the ends of the spring may be

guided along a definite path as it deflects to act as a structural member in addition to

energy absorbing device.

Thus the leaf springs may carry lateral loads, brake torque, driving torque etc., in addition

to shocks.

Construction of Leaf Spring

A leaf spring commonly used in automobiles is of semi-elliptical form as shown in Fig.

It is built up of a number of plates (known as leaves).

The leaves are usually given an initial curvature or cambered so that they will tend to

straighten under the load.

The leaves are held together by means of a band shrunk around them at the centre or by a

bolt passing through the centre.

Since the band exerts stiffening and strengthening effect, therefore the effective length of

the spring for bending will be overall length of the spring minus width of band.

In case of a centre bolt, two-third distance between centers of U-bolt should be subtracted

from the overall length of the spring in order to find effective length.

The spring is clamped to the axle housing by means of U-bolts.

Materials for Leaf Springs

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The material used for leaf springs is usually a plain carbon steel having 0.90 to 1.0%

carbon.

The leaves are heat treated after the forming process.

The heat treatment of spring steel produces greater strength and therefore greater load

capacity, greater range of deflection and better fatigue properties.

According to Indian standards, the recommended materials are:

For automobiles : 50 Cr 1, 50 Cr 1 V 23, and 55 Si 2 Mn 90 all used in hardened

and tempered state.

For rail road springs: C 55 (water-hardened), C 75 (oil-hardened), 40 Si 2 Mn 90

(water hardened) and 55 Si 2 Mn 90 (oil-hardened).

The physical properties of some of these materials are given in the following

table. All values are for oil quenched condition and for single heat only

Problems

1. Design a leaf spring for the following specifications: Total load = 140 kN; Number of

springs supporting the load = 4; Maximum number of leaves = 10; Span of the spring =

1000 mm; Permissible deflection = 80 mm.

Take Young’s modulus, E = 200 kN/mm2 and allowable stress in spring material as 600 MPa.

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2. A truck spring has 12 numbers of leaves, two of which are full length leaves. The spring

supports are 1.05 m apart and the central band is 85 mm wide. The central load is to be

5.4 kN with a permissible stress of 280 MPa. Determine the thickness and width of the

steel spring leaves. The ratio of the total depth to the width of the spring is 3. Also

determine the deflection of the spring.

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3. A rail wagon of mass 20 tones is moving with a velocity of 2 m/s. It is brought to rest by two

buffers with springs of 300 mm diameter. The maximum deflection of springs is 250 mm.

The allowable shear stress in the spring material is 600 MPa. Design the spring for the

buffers.

Given: m = 20 t = 20 000 kg; v = 2 m/s; D = 300 mm; = 250 mm; = 600 MPa = 600 N/mm2

1. Diameter of the spring wire

Let d =Diameter of the spring wire.

We know that kinetic energy of the wagon

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Cotter Joint

A cotter is a flat wedge shaped piece of rectangular cross-section and its width is tapered

(either on one side or both sides) from one end to another for an easy adjustment.

The taper varies from 1 in 48 to 1 in 24 and it may be increased up to 1 in 8, if a locking

device is provided.

The cotter is usually made of mild steel or wrought iron.

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A cotter joint is a temporary fastening and is used to connect rigidly two co-axial rods or

bars which are subjected to axial tensile or compressive forces.

It is usually used in connecting a piston rod to the crosshead of a reciprocating steam

engine, a piston rod and its extension as a tail or pump rod, strap end of connecting rod

etc.

Types of Cotter Joints

Following are the three commonly used cotter joints to connect two rods by a cotter:

Socket and spigot cotter joint,

Sleeve and cotter joint, and

Gib and cotter joint.

Socket and Spigot Cotter Joint

In a socket and spigot cotter joint, one end of the rods (say A) is provided with a socket

type of end as shown in Fig. and the other end of the other rod (say B) is inserted into a

socket.

The end of the rod which goes into a socket is also called spigot. A rectangular hole is

made in the socket and spigot.

A cotter is then driven tightly through a hole in order to make the temporary connection

between the two rods.

The load is usually acting axially, but it changes its direction and hence the cotter joint

must be designed to carry both the tensile and compressive loads. The compressive load

is taken up by the collar on the spigot.

Design of Socket and Spigot Cotter Joint

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1. Failure of the rods in tension

The rods may fail in tension due to the tensile load P. We know that

From this equation, diameter of the rods (d) may be determined.

2. Failure of spigot in tension across the weakest section (or slot):

Since the weakest section of the spigot is that section which has a slot in it for the cotter,

as shown in Fig. therefore

Area resisting tearing of the spigot across the slot and tearing strength of the spigot across

the slot

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Equating this to load (P), we have

From this equation, the diameter of spigot or inside diameter of socket (d2) may be

determined.

From this equation, the diameter of spigot or inside diameter of socket (d2) may be

determined.

3. Failure of the rod or cotter in crushing

We know that the area that resists crushing of a rod or cotter

=d2 × t

∴ Crushing strength = d2 × t × σc

Equating this to load (P), we have

P =d2 × t × σc

From this equation, the induced crushing stress may be checked.

4. Failure of the socket in tension across the slot

We know that the resisting area of the socket across the slot, as shown in Fig.

∴ Tearing strength of the socket across the slot

Equating this to load (P), we have

From this equation, outside diameter of socket (d1) may be determined

5. Failure of cotter in shear

Considering the failure of cotter in shear since the cotter is in double shear, therefore

shearing area of the cotter

= 2 b × t

and shearing strength of the cotter

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=2 b × t × τ

Equating this to load (P), we have

P =2 b × t × τ

From this equation, width of cotter (b) is determined

6. Failure of the socket collar in crushing

Considering the failure of socket collar in crushing as shown in Fig. We know that area

that resists crushing of socket collar

=(d4 – d2) t

and crushing strength =(d4 – d2) t × σc

Equating this to load (P), we have

P =(d4 – d2) t × σc

From this equation, the diameter of socket collar (d4) may be obtained.

7. Failure of socket end in shearing

Since the socket end is in double shear, therefore area that resists shearing of socket

collar

=2 (d4 – d2) c

and shearing strength of socket collar

=2 (d4 – d2) c × τ

Equating this to load (P), we have

P =2 (d4 – d2) c × τ

From this equation, the thickness of socket collar (c) may be obtained.

8. Failure of rod end in shear

Since the rod end is in double shear, therefore the area resisting shear of the rod end

= 2 a × d2

And shear strength of the rod end

= 2 a × d2 × τ

Equating this to load (P), we have

P = 2 a × d2 × τ

From this equation, the distance from the end of the slot to the end of the rod (a) may be

obtained.

9. Failure of spigot collar in crushing

Considering the failure of the spigot collar in crushing We know that area that resists

crushing of the collar

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And crushing strength of the collar

Equating this to load (P), we have

From this equation, the diameter of the spigot collar (d3) may be obtained.

10. Failure of the spigot collar in shearing

Considering the failure of the spigot collar in shearing We know that area that resists

shearing of the collar

= π d2 × t1

And shearing strength of the collar,

= π d2 × t1 × τ

Equating this to load (P) we have

P = π d2 × t1 × τ

From this equation, the thickness of spigot collar (t1) may be obtained.

11. Failure of cotter in bending

The maximum bending moment occurs at the Centre of the cotter and is given by

We know that section modulus of the cotter,

Z = t × b2 / 6

∴ Bending stress induced in the cotter,

This bending stress induced in the cotter should be less than the allowable bending stress of the

cotter.

12. The length of cotter (l) is taken as 4 d.

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13. The taper in cotter should not exceed 1 in 24. In case the greater taper is required, then a

locking device must be provided.

14. The draw of cotter is generally taken as 2 to 3 mm.

Problems

1. Design and draw a cotter joint to support a load varying from 30 kN in compression to 30 kN

in tension. The material used is carbon steel for which the following allowable stresses may

be used. The load is applied statically. Tensile stress = compressive stress = 50 MPa ; shear

stress = 35 MPa and crushing stress = 90 MPa.

Solution. Given: P = 30 kN = 30 × 103 N; σt = 50 MPa = 50 N / mm

2; τ = 35 MPa = 35 N /

mm2; σc = 90 MPa = 90 N/mm

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1. Diameter of the rods

Let d = Diameter of the rods.

Considering the failure of the rod in tension. We know that load (P),

2. Diameter of spigot and thickness of cotter

Let d2 = Diameter of spigot or inside diameter of socket, and

t = Thickness of cotter. It may be taken as d2 / 4.

Considering the failure of spigot in tension across the weakest section. We know that load (P),

Since this value of σc is more than the given value of σc = 90 N/mm2, therefore the dimensions

d2 = 34 mm and t = 8.5 mm are not safe. Now let us find the values of d2 and t by substituting

the value of σc = 90 N/mm2 in the above expression, i.e.

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3. outside diameter of socket

Let d1 = Outside diameter of socket.

Considering the failure of the socket in tension across the slot. We know that load (P),

4. Width of cotter

Let b = Width of cotter.

Considering the failure of the cotter in shear. Since the cotter is in double shear, therefore load

(P),

5. Diameter of socket collar

Let d4 = Diameter of socket collar.

Considering the failure of the socket collar and cotter in crushing. We know that load (P),

30 × 103 = (d4 – d2) t × σc = (d4 – 40)10 × 90 = (d4 – 40) 900

∴ d4 – 40 = 30 × 103 / 900 = 33.3 or d4 = 33.3 + 40 = 73.3 say 75 mm Ans.

6. Thickness of socket collar

Let c = Thickness of socket collar.

Considering the failure of the socket end in shearing. Since the socket end is in double shear,

therefore load (P),

30 × 103 = 2(d4 – d2) c × τ = 2 (75 – 40) c × 35 = 2450 c

∴ c = 30 × 103 / 2450 = 12 mm Ans.

7. Distance from the end of the slot to the end of the rod

Let a = Distance from the end of slot to the end of the rod.

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Considering the failure of the rod end in shear. Since the rod end is in double shear, therefore

load (P),

30 × 103 = 2 a × d2 × τ = 2a × 40 × 35 = 2800 a

∴ a = 30 × 103 / 2800 = 10.7 say 11 mm Ans.

8. Diameter of spigot collar

Let d3 = Diameter of spigot collar.

Considering the failure of spigot collar in crushing. We know that load (P),

9. Thickness of spigot collar

Let t1 = Thickness of spigot collar.

Considering the failure of spigot collar in shearing. We know that load (P),

30 × 103 = π d2 × t1 × τ = π × 40 × t1 × 35 = 4400 t1

∴ t1 = 30 × 103 / 4400 = 6.8 say 8 mm Ans.

10. The length of cotter ( l ) is taken as 4 d.

∴ l = 4 d = 4 × 28 = 112 mm Ans.

11. The dimension e is taken as 1.2 d.

∴ e = 1.2 × 28 = 33.6 say 34 mm Ans.

Gib and Cotter Joint

A Gib and cotter joint is usually used in strap end (or big end) of a connecting rod as

shown in Fig. below.

when the cotter alone (i.e. without Gib) is driven, the friction between its ends and the

inside of the slots in the strap tends to cause the sides of the strap to spring open (or

spread) outwards as shown dotted in Fig. 12.11 (a).

In order to prevent this, gibs as shown in Fig. 12.11 (b) and (c), are used which hold

together the ends of the strap. Gibs provide a larger bearing surface for the cotter to slide

on, due to the increased holding power.

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Thus, the tendency of cotter to slacken back owing to friction is considerably decreased.

The jib, also, enables parallel holes to be used.

Design of Gib and Cotter Joint for Square Rods

Consider a gib and cotter joint for square rods as shown in Fig. above. The rods may be

subjected to a tensile or compressive load.

All components of the joint are assumed to be of the same material

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Problem

Design a Gib and cotter joint as shown in Fig. 12.13, to carry a maximum load of 35 kN.

Assuming that the Gib, cotter and rod are of same material and have the following

allowable stresses:

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Knuckle Joint

A knuckle joint is used to connect two rods which are under the action of tensile loads.

However, if the joint is guided, the rods may support a compressive load.

A knuckle joint may be readily disconnected for adjustments or repairs.

Its use may be found in the link of a cycle chain, tie rod joint for roof truss, valve rod

joint with eccentric rod, pump rod joint, tension link in bridge structure and lever and rod

connections of various types.

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Dimensions of Various Parts of the Knuckle Joint

The dimensions of various parts of the knuckle joint are fixed by empirical relations as

given below.

It may be noted that all the parts should be made of the same material i.e. mild steel or

wrought iron.

If d is the diameter of rod, then diameter of pin,

d1 = d

Outer diameter of eye,

d2 = 2 d

Diameter of knuckle pin head and collar,

d3 = 1.5 d

Thickness of single eye or rod end,

t = 1.25 d

Thickness of fork, t1 = 0.75 d

Thickness of pin head, t2 = 0.5 d

Other dimensions of the joint are shown above.

Methods of Failure of Knuckle Joint

Consider a knuckle joint as shown in Fig.

Let P = Tensile load acting on the rod,

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d = Diameter of the rod,

d1 = Diameter of the pin,

d2 = Outer diameter of eye,

t = Thickness of single eye,

t1 = Thickness of fork.

σt , τ and σc = Permissible stresses for the joint material in tension, shear and

rushing respectively.

In determining the strength of the joint for the various methods of failure, it is assumed that

1. There is no stress concentration, and

2. The load is uniformly distributed over each part of the joint.

Following are the various methods of failure of the joint:

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Problem

1. Design a knuckle joint to transmit 150 kN. The design stresses may be taken as 75 MPa in

tension, 60 MPa in shear and 150 MPa in compression.

Given: P = 150 kN = 150 × 103 N; σt = 75 MPa = 75 N/mm

2; τ = 60 MPa = 60 N/mm2; σc = 150

MPa = 150 N/mm2

The joint is designed by considering the various methods of failure as discussed below:

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2. Design a knuckle joint for a tie rod of a circular section to sustain a maximum pull of 70 kN. The

ultimate strength of the material of the rod against tearing is 420 MPa. The ultimate tensile and

shearing strength of the pin material are 510 MPa and 396 MPa respectively. Determine the tie

rod section and pin section. Take factor of safety = 6.

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Sleeve and cotter joint

A sleeve and cotter joint as shown in Fig. is used to connect two round rods or bars.

In this type of joint, a sleeve or muff is used over the two rods and then two cotters (one

on each rod end) are inserted in the holes provided for them in the sleeve and rods.

The taper of cotter is usually 1 in 24.

It may be noted that the taper sides of the two cotters should face each other as shown in

Fig. below.

The clearance is so adjusted that when the cotters are driven in, the two rods come closer

to each other thus making the joint tight.

The various proportions for the sleeve and cotter joint in terms of the diameter of rod (d ) are as

follows :

Outside diameter of sleeve,

d1 = 2.5 d

Diameter of enlarged end of rod,

d2 = Inside diameter of sleeve = 1.25 d

Length of sleeve, L = 8 d

Thickness of cotter, t = d2/4 or 0.31 d

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Width of cotter, b = 1.25 d

Length of cotter, l = 4 d

Distance of the rod end (a) from the beginning to the cotter hole (inside the sleeve end)=

Distance of the rod end (c) from its end to the cotter hole

= 1.25 d

Design of Sleeve and Cotter Joint

1. Failure of the rods in tension

2. Failure of the rod in tension across the weakest section (i.e. slot)

3. Failure of the rod or cotter in crushing

4. Failure of sleeve in tension across the slot.

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5. Failure of cotter in shear

6. Failure of rod end in shear

Since the rod end is in double shear, therefore area resisting shear of the rod end

= 2 a × d2

and shear strength of the rod end

= 2 a × d2 × τ

Equating this to load (P), we have

P = 2 a × d2 × τ

From this equation, distance (a) may be determined.

7. Failure of sleeve end in shear

Problem:

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Design a sleeve and cotter joint to resist a tensile load of 60 kN. All parts of the joint are

made of the same material with the following allowable stresses: σt = 60 MPa; τ = 70 MPa;

and σc = 125 MPa.

Solution: Given: P = 60 kN = 60 × 103 N; σt = 60 MPa = 60 N/mm

2; τ = 70 MPa = 70

N/mm2; σc = 125 MPa = 125 N/mm2

1. Diameter of the rods

2. Diameter of enlarged end of rod and thickness of cotter

3. Outside diameter of sleeve

Let d1 = Outside diameter of sleeve.

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4. Width of cotter

5. Distance of the rod from the beginning to the cotter hole (inside the sleeve end)

6. Distance of the rod end from its end to the cotter hole