DESIGN OF MACHINE ELEMENTS -II - National Institute of ... · PDF fileSixth semester...

Sixth semester Mechanical Design of Machine Elements -II

Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

DESIGN OF MACHINE ELEMENTS -II WORK BOOK CUM LECTURE NOTES

(FOR SIXTH SEMESTER MECHANICAL STUDENTS)

(FOR PRIVATE CIRCULATION ONLY)

Part - A

JAGADEESHA T Associate Professor

Mechanical Engineering Department

ST. JOSEPH ENGINEERING COLLEGE VAMANJOOR, MANGALORE – 575 028,



CHAPTER 1: CURVED BEAMS LEARNING OBJECTIVES

] Introduction to curved beams

] Discuss the stress distribution pattern in curved beams when compared to

straight beam with sketches

] To derive the expression for the normal stress due to bending at the extreme

fibers of a curved beam

] To design curved beams for different cross sections

] To study and design stresses in closed rings.

Curved Beam A beam in which the neutral axis in the unloaded condition is curved instead of straight. Or if the beam is originally curved before applying the bending moment, are termed as “Curved Beams Curved beams find applications in many machine members such as c – clampers, crane hooks, frames of presses, chains, links, and rings

Straight Beam A beam is a straight structural member subjected to a system of external forces acting at right angles to its axis 1. If a member is fixed or built in one end while its other

end is free, the member is called cantilever beam.

2. If the ends of the beam are made to freely rest on

supports the beam is called a freely or simply supported beam.

3. If a beam is fixed at both its, it is called as built-in or fixed beam.

4. A beam which is provided with more than two

supports is called as continuous beam.



A beam is said to be statically determinate beam, if its reaction components can be determined by using equations of static equilibrium only. Commonly encountered statically determinate beams are cantilever beams, SS beam and over hanging beams. Beams are subjected to transverse loads such as concentrated load, UDL, UVL & applied moments. Beam transfer applied load to supports, the beam develop resistance to moments & transverse shear forces at all its cross-sections. Differences between Straight Beams & Curved Beams Straight Beams Curved Beams 1 Neutral axis of the cross-section

passes through the centroid of the section.

Neutral axis does not coincide with the cross-section, but is shifted towards the centre of curvature of the beam.

2 The variation of bending stress is linear, magnitude being proportional to the distance of a fiber from the neutral axis.

The distribution of the stress in the case of curved beam is non- linear (Hyper- bolic) because of the neutral axis is initially curved.

3 No stress concentration Stress concentration is higher at the inner fibers

4 Neutral axis remains undisturbed along the CG.

Neutral axis always shifts towards the centre of curvature.

5 We use Euler equation to calculate bending stress M/I = F/Y=E/R

ZM

CIM

==σ

We use o

ii AeR

Mc=σ or

o

oo AeR

Mc=σ to

calculate inner /outer fibre stress

Derive the expression for the normal stress due to bending at the extreme fibers of a curved beam. Assumptions:-

1. The beam is subjected to pure bending. 2. Material of the beam is isotropic & homogeneous & obeys hook’s law. 3. Plane sections perpendicular to the axis of the beam remain plane even after

bending.



4. The stress induced do not exceed elastic limit. 5. Each layer of the beam is free to expand OR contract, independent of the

layer above or below it. 6. Young’s modulus is same in tension &compression.

Let us use following standard symbols. F =Load M =Applied bending moment, N –mm e = distance from the centroidal axis to the neutral axis, measured towards centre of curvature, mm Ci = distance from neutral axis to inner fiber (radius) mm Co = distance from neutral axis to outer fiber (radius) mm. Ri = inner radius of curvature, mm Ro = outer radius of curvature, mm Rn = Radius of neutral axis, mm R (Rc) = Radius of centroidal axis, mm A = area of section, mm2

σi = Stress in the inner fiber, N/mm2

σo = Stress in the outer fiber, N/mm2 Consider a part of the curved beam between two radial planes ab & cd subtending an angle ‘θ’ at the centre of curvature when the beam is subjected to bending moment. ‘M’ as shown in fig, the plane ‘cd’ rotates with respect to ‘ab’ through an angle ‘θ’ & takes new position ‘fg’. The outer fiber are slanted due to compression and inner fibre are elongated due to tension. Now, consider a fiber of depth ‘dy’ & cross- sectional area ‘dA’ at a distance ‘y’ from the neutral axis. The original length of strip at a distance ‘y’ from the neutral axis is (Rn + y)θ. It is shortened by an amount ydθ.



The strain across cross section dA would be Stress in the fibre is given by σ α Є within elastic limit or σ = E Є E- Young’s modulus



At the outer fiber, σo = - MCo A e Ro At inner fibre σi = + MbCi are the required equations. A e Ri Show that, for a thick curved beam of circular cross-section 2 √ rd = √ ri + √ ro Where rd mean reciprocal radius rd. Problem 1. The section of a crane hook is rectangular in shape whose width is 30mm & depth is 60mm. The centre of curvature of the section is at a distance of 125mm from the inside section & the load line is 100mm from the same point. Find the capacity of the hook if the allowable stress in tension is 75 N/mm2

Ans : 8480 N is the capacity of hook.

Problem 2. A crane hook of trapezoidal cross-section has an inner fiber width = 120mm, depth = 100mm & inner radius = 120mm. Calculate the width if stresses are numerically equal at inner & outer fibers. Also determine the capacity of the hook, if the permissible stress is 100MPa.

Ans : width = 28mm Capacity of hook =



Problem 3. A crane hook shown in figure below is made of 30mm diameter steel rod. The distance between the centroidal axis of the rod & the centre of curvature of the hook is 50mm. Determine the load ‘F’ so that the maximum stress in the rod is not to exceed 40 N/mm2.

Ans : Capacity of F = 1557 N Problem 4. . A crane hook has trapezoidal cross-section. The maximum tensile stress occurs at point P as shown in figure. Determine i) the distance of centre of curvature to centroidal axis. ii) B M for section AA iii) Distance from centre of curvature to the neutral axis. iv) Area v) Maximum tensile stress ( point P) vi) maximum stress at point Q.



Problem 5. A section of frame for a punch press is shown in figure below. Determine the capacity of the press if the maximum tensile stresses in the frame is not to exceed 60MPa.

Problem 6. A section of a C clamp is shown in figure. What force F can be exerted by the screw if the max tensile stress in clamp is limited to 140 MPa.

Problem 7 A portable hydraulic inverter has a maximum riveting force of 70 kN. The U frame is made cast steel with an ultimate stress of 480 MPa and a yield point in tension of 240 MPa. Consider only the secion AA and determine the following 1. Bending moment 2, Distance from centroidal axis to neutral axis 3. Direct tensile force 4, Maximum tensile force and location 5. Maximum shear stress and location



Problem 8 The C frame of a 100 KN capacity press is shown in figure. The material of frame is grey cast iron whose ultimate tensile stress is 200 MPA and FOS =3. Determine the maximum stress at both inner fibre and outer fibre.

Problem 9.Ring is made from a 75mm dia bar. The inside dia of the ring is 100mm. if the load acting on the ring is 20kN as shown in figure. calculate the maximum shear stress and its location.

Problem 10 : An off set bar is loaded as shown in figure. The weight of the bar can be neglected what is the maximum offset ( dimension x) if the allowable stress in tension is limited to 70 MPa. Where will the maximum tensile and shear stress occur.


Jagadeesha T, Associate Professor, St Joseph Engineerin

Problem 11: A supporting structure for a movable crane has dimensions shown . Find the load P of the crane , if the maximum stress in critical section AA is not to exceed 35 MPa. Problem 12: same as problem 11. Take P=10000N determine the maximum stress. Problem 13: same as problem 11. Take circular cross section of dia 100,mm instead of rectangular cross section. Problem 14: An S link of unequal radii is made20mm diameter and it is loaded as shown. Detethe location and magnitude of maximum tensileshear stress. Mean radius of small radii is 80mmthat of bigger loop is 100mm.

Problem 15: Figure indicates the section of a framagnitude of the force F that can be exerted if tMPa. In section XX.

Problem 16 The G-clamp illustrated above is made from a steel whose design stress is 50 MPa. It is supposed to sustain a force, P, of 750 N. Can it ?

Drill Press

g College, Vamanjoor, Mangalore

from rmine and and

me of a drill press, indicating the he maximum tensile stress is 80

80 mm

100 mm

20mm

2000 N

2000 N



CHAPTER 2 LEARNING OBJECTIVES

] Introduction to different types of spring used

] Discuss the applications of springs

] To derive the expression for the shear stress and direct shear stress

] To derive the expression for deflection of helical springs

] To design compression and tension helical springs

] To design springs for fluctuating loads

] To design concentric springs

] To design leaf spring and concept of equalized stress in leaf springs

Introduction

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. A spring is a mechanical device which is used for efficient storage and release of energy. Application of springs

] To absorb or control energy due to either shock or vibration as in automotives, railways, aircrafts, landing gears and vibration dampers etc.

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

] It is used to return the mechanical part to its orginal position , when it has temporarily displaced like springs used in valves, clutches and linkages.

] To control motion by maintaining control between two elements as in CAMS & followers.

] To measure forces as in spring balances and engine indicators. ] To store energy as in watches, toys movie cameras.

TYPES OF SPRINGS 1.HELICAL SPRINGS Helical springs are made of wire coiled in the form of helix and are primarily intended for compressive or tensile loads. The cross-section of wire from which the spring is made may be circular, square or rectangular. The two forms of helical springs are compression helical spring and tension helical spring as shown in figure.



Helical springs are said to be closely coiled, when the helix angle is very small (< 10o), where as in open coil helical spring the helix angle is large Advantages.

] These springs are easy to manufacture. ] They are available in wide range. ] They are highly reliable. ] They have constant spring rates. ] Their performance can be predicted more accurately. ] There characteristics can be varied by changing dimensions.

2. Conical and Volute springs The conical and volute spring shown in the figure are used in special applications where the spring rate increases in increase in load. Another feature of these types of springs is the decreasing number of coils results in an increasing spring rate. This characteristic is some times utilized in vibrations problems where springs are used to support to body that have varying mass. 3. Torsion springs These springs may be of helical or spiral type as shown in figure. Helical types of springs are used where the load tends to wind up the springs and are used in electrical mechanisms. Spiral type is used where the loads tends to increase the number of coils and are used in watches and clocks.

4. Laminated or Leaf springs.

The laminated or leaf spring (also known as flat spring) consists of a number of flat plates (known as leaves) of varying lengths held together by means of clamps and bolts. These types of springs are most used in automobiles.

5. Disc springs These springs consists of a number of conical discs held together by a central bolt or tube as shown in figure. These springs are used in applications where high spring rates and compact spring units are required



TERMS USED IN COMPRESSION SPRINGS Solid length: When the springs are compressed until the coils come in contact with each other, then the spring is said to be solid. The solid length of a spring is the product of total number of coils & the diameter of the wire. Mathematically, Solid length = Free length Free length of a compression spring is the length of the spring in the free or unloaded condition & is equal to the solid length plus the maximum deflection or compression of the spring & the clearance between the adjacent coils. Free length - Lf = solid length + max. Deflection + clearance between adjacent coils. Spring index It is defined as the ratio of the man diameter of the coil to the diameter of the wire. Spring index = C = D/d Spring Rate spring rate (stiffness/spring constant) is the defined as the load required per unit deflection of the spring. Spring Rate, K= F/ δ F- load, N δ - Defection, mm Pitch: Pitch of the coils is defined as the axial distance between adjacent coils in un compressed state.

STRESS IN HELICAL SPRING AND CIRCULAR WIRE.



Consider a helical compression spring made of circular wire & subjected to an axial load F, as shown in figure. Let, D = Mean diameter of the coil d = Diameter of the spring wire, n = number of active coils, G = Modulus of Rigidity for the spring material, F = Axial load on the spring, τ = Max. Shear stress induced in the wire, C = spring index = D/d p = pitch of the coils δ = deflection of the spring. Consider a point of the spring shown in fig (b). The load ‘F’ tends to rotate the wire & as a result twisting moment (T) is developed in the wire, & thus torsional shear stress is induce d in the wire. Let us consider that part of spring is in equilibrium under the action of two forces ‘F’ & twisting ‘T’. In addition to the torsional shear stress (τ1) induced in the wire, the following stresses also act on the wire. Direct stress due to the load. F, & Stress due to curvature of wire

Shear stress = 8FD .K

πd3

Where K = 4C - 1 + 0.615 ; Wahl stress Conc. Factor

4C - 4 C



Deflection of helical springs of circular wire.

Let l = total active length of wire = πD × n

θ = Angular deflection of the wire due to Torque t.

Therefore, Axial deflection of the spring δ = θ × D/2 ----------- (1)

Also from the Torsion equation, we have,

Therefore δ = 8FD3n = 8FC

3n where n is the number of active coils.

Gd4 G.d

Design Procedure for Helical Springs.



1. Diameter of the wire:

2. Mean diameter of the coil:

(a) Outer diameter of the coil:

(b) Inner diameter of the coil:

3. Number of coils:

4. Free length

5. Stiffness or Spring Rate

6. Pitch:

PROBLEMS

(1) Design a helical compression spring to support an axial load of N. The deflection under

load is limited to 60 mm. The spring index is 6. The spring is made of chrome-vanadium

steel & FOS = 2.

Spring specifications (i) Wire diameter (ii) Mean diameter (iii) Free length (iv) Total no of coils – (v) Style of ends = squared & ground (vi) Pitch - p – (vii)Spring rate – (viii)Material – Chrome- vanadium (2) Design a helical compression spring for a max. load of 1000N for a deflection of 25mm using the spring index as 5. the max permissible shear stress for spring wire is 420 N/mm2 & G = 84×103 N/mm2



(3) A railway carriage weighing 40KN & moving at 8km/hr is to be brought to rest by 2 buffer springs. The compression between the coils must be twice the wire diameter. Assume spring index as 8. And allowable shear stress for the spring material = 450N/mm. Take G = 0.8 * 10 N/mm. Design the spring?

(4) A railway carriage weighing 20KN & moving at 2.5 km/hr is to be brought to rest by buffer springs. Find how many springs each of 15 coils will be required to store the energy of motion during a compression of 150mm. The available size of wire is 20mm. The mean radius of coil is 100mm. (5) A bumper spring consists of 2 helical steel springs of square cross section brings to rest a railway carriage weighing 50KN & moving at 1.5 km/hr. In doing so the springs are compressed by 200mm. The mean diameter of coil is 6 times the side of square section. Design the spring if the permissible shear stress is 345 MPa and Modulus of rigidity is 78 MPa. 6) A loaded narrow cart weighing 7.5 kN and moving with a velocity of 1m/second is brought to rest by a bumper consisting of 2 helical steel springs of square section. The mean coil diameter of spring is 6 times the side of square . the spring are compressed by 150mm. While bringing the car to rest. The permissible shear stress in spring is not to exceed 400 MPa. Find the following Mean load on each spring Side of square section of wire Mean dia of coil No of active coils, take G = 82 Gpa. 7) A load of 2KN is dropped axially on a helical spring from a height of 250mm. the spring has 20 turns, & it is made of 25mm diameter wire. The spring index is 8. Find the max. Shear stress induced in the spring 7 the amount of compression produced. Take G = 82.7GN/mm 8) Design a helical spring for a spring loaded safety valve for the following conditions: (i) Diameter of the valve = 65mm (ii)Operating pressure = 0.7N/mm2 (iii) Max. Pressure on the valve = 0.75N/mm2 (iv) Max. lift of the valve when pressure = 3.5mm rises from 0.7 to 0.75 N/mm2 (v) Max. Allowable stress = 550MPa (vi) Spring index = 6 9) The valve spring of an I.C Engine is 40mm long, when the valve is open & 48mm long when the valve is closed . The spring loads are 250N when the valve is closed & 400n when the valve is open. The inside diameter of the spring is not to be less than 25mm & take FOS


Jagadeesha T, Associate Professor, St Jos

= 2. Assume spring index to be 6 &G = 79.34×103MPa & yield shear stress = 690N/mm2. Design the spring Assignment Questions

(1) A helical spring made from 6.3mm diameter steel wire has an outside diameter of 57.3mm with squared & ground ends and has 12 coils. The allowable shear stress is 827MPa. Determine the following (i) Spring rate (ii) Free length (iii) Pitch (2) The following data refers to the valve of a petrol engine Length of the spring when the valve is open – 40mm Length of the spring when the valve is closed – 48mm Spring load when the valve is closed – 350N Spring load when the valve is open – 220N Spring index – 6.8 (3) The maximum shear stress allowed is 150MPa & the modulus of rigidity is 84GPa. The ends are squared & ground and the gap between the adjacent coils is 0.1 times the wire diameter. Determine the following (i) Wire dia (ii)Mean dia (iii) Number of coils (iv) Free length (v) Pitch Design of springs for fluctuating loads Force acting on the spring in many applications varies in magnitude in time. Let us consider a spring subjected to fluctuating force as shown in the figure. The load varies from Fmax to Fmin The mean force =

Fmin

Fmax

Force

TIME

eph Engineering College, Vamanjoor, Mangalore



A helical compression spring is subjected to purely compressive forces. In general, the spring wires are subjected to pulsating shear stresses. Let Sen be the endurance limit in

shear. For cold drawn steel wires Sen = 0.21x enσ and uy 42.0 σ=τ For oil hardened

tempered steel wires Sen = 0.22x enσ and uy 45.0 σ=τ where enσ is the ultimate tensile

strength is shown below:

The mean stress mτ is plotted on the abscissa while the stress amplitude aτ on the ordinate.

Point A will co-ordinate ( 2

S,

2S enen ) indicate the failure point of spring wire in a pulsating

stress cycle fatigue test. Point B indicates failure under static conditions i.e. when the stress reaches the torsional yield strength yτ . The line AB is called the line of failure. To consider

the effect of FOS a line of CD is drawn from point B in such a way that N

yτ=OD. The line

CD is parallel to AB. The line CD is called design line. Any point on CD such as X represents a situation with the same FOS. Considering the similar triangles XFD and AEB. This is called MODIFIED SODERBERG’S EQUATION Problem : Design a helical coil spring for engine valve to exert force of 700 N in the open and 450 N in the closed position. The lift of the valve is 16mm. Keep the Outside spring coil



diameter between 4 cm and 5cm. The spring ends are square and ground. The allowable shear stress of spring material is 450 MPa. For the material of the spring yτ =700 MPa.

Sen =365 MPa. Take G= 80 GPa. Helical tension springs Helical tension springs are usually wound with the coils closed and under initial tension. They are also called closed wound springs. The different styles of ends for the helical springs are shown in figure.

When loops or end hooks are provided the small radius where the hook joins the 1st coil is a region of highly local stress. The end should be designed in such a way that , the effect of stress concentration at the bend is minimum. Sometimes the effect of stress concentration in springs is so severe that the spring body becomes stronger that the end then the failure occurs in the end coils. For helical extension springs, all coils are active, The number of active coils is same as the total number of coils. Figure 2 shows the relation between the external force and spring elongation.

Here the force F must exceed the initial tension Fi before a

Elongation

External force

Fi



deflection y is experienced. The free length lo of an extension spring is equal to the body length + two times the hood distance lo= d( i+1) = body length Figure shows safety device operated by a lever and a tension spring. During the normal closed position of the valve the pressure is 0.4 MPa. The maximum lift valve is 4mm at a pressure of 0.55 Mpa. Diameter of the valve is 50mm. take allowable shear stress as 560 MPa for the material of the spring. Spring Index C=7 and G=80 GPa. Design the spring. Spring surge ( critical frequency of helical springs) If one end of a compression spring is held against in a flat surface and the other end is disturbed by repeated load, a helical compression wave is created that travels back and forth from one end to other end. Surging may occur in spring which is subjected to loads whose frequency is close to the natural frequency of the spring. When the frequency of repeated load is equal to natural frequency of spring resonance occur. To avoid this possible it is advisable that the natural frequency of spring is atleast 20 times the frequency of the external repeated loads. If the frequency is not high enough, the spring should be redesigned to increase the k or decrease w.

Natural frequency of spring = fn = wgF

21 o

π Hz

Problem A disc cam 250mm diameter rotates off centre with eccentricity of 20 mm and operates the roller follower that is carried by the arm as shown in the figure. The follower is held against the cam by means of an extension spring. Assuming that the force between the follower and the cam is 300 N at the lowest position and 500 N at the highest position of follower. The spring index is 6. determine the wire diameter, outside diameter of spring and no. of active coils. The maximum shear stress may be taken as 250 MPa and Modulus of rigidity – 82.7 Gpa.



Problem A helical coil spring made from 6.3 diameter steel wire has an outside diameter of 57.3mm with square and ground ends has 12 coils. The length of the spring is such what when it is compressed the torsional stress is 827 MPa. Determine the spring rate , Free length , critical frequency . The density of material is 7800 kg/m3 and G = 0.8x105 MPa.

Concentric springs Concentric springs are closed coil helical springs forced one inside the other. They are used to obtain greater load carrying capacity or to obtain certain load deflection characteristics. Figure shows two springs placed one inside the other. When springs are nested , the mechanism contnues even if one of the springs break. The conditions to be satisfied by the concentric spring as follows.

1. when the springs are of equal free length and made of same material, then maximum shear stress in the springs are equal.

2. The deflection is same for both the spring



2

2

1

2

1

dd

FF

= also F = F1 +F2 , where F is the applied load. F1 is the load taken by

spring 1 and F2 is the load taken by spring 2. Problem : The larger of the two concentric springs made of 38 mm diameter round bar has a mean coil dia of 228mm and 6 active coils. The inner spring has a wire of 25mm, spring index 5 and no of active coils = 9. Free height of outer spring is 19mm more than inner; Find the deflection of each spring for a load of 90 kN. Take G = 77000 MPa. Calculate the load carried by each spring. Problem One helical spring is nested inside the other . The dimensions of the spring are tabulated. Both springs have the same free length and carry a total maximum load of 2500 N Description

Outer spring

Inner spring

Number of active coils

6

10

Wire diameter

12.5mm

9mm

Mean coil diameter 100mm 70mm

Determine the maximum load carried by each spring. Total deflection of each spring. Maximum stress in 2 springs , take G=83 GPa.



Problem A concentric spring for an aircraft engine valve is to exert a maximum for of 5000N under an axial deflection of 40mm. Both springs have same free length and same solid length and are subjected to equal maximum shear stress of 850 MPa. If the spring index for both the spring is 6 find the load shared by each spring ii) maximum dimension of springs iii) No of active coils in each. Assume G= 80 and dimensional clearance is equal to difference between the wire diameter. Leaf springs The term flat spring is commonly applied to a wide variety of shapes made out of flat strip. The advantages of flat spring varies over the helical spring is that the end of the spring may be guided along a definite path as it deflects . Thus the spring may act as structural member as well as an emery absorbing device. Semi Elliptical leaf springs. A multi leaf spring used in automobile is shown in the figure. It is of semi elliptical form and hence the name semi elliptical leaf spring . It is built up of a number of plates known as leaves, The leaves are usually given an initial curvature or cambered.

The top leaf is known as the master leaf. The eye is provided for attaching the spring with another machine member. The amount of bend that is given to the spring from the central line, passing through the eyes, is known as camber. The camber is provided so that even at the maximum load the deflected spring should not touch the machine member to which it is attached. The camber shown in the figure is known as positive camber. The central clamp is required to hold the leaves of the spring. However, the bolt holes required to engage the bolts to clamp the leaves weaken the spring to some extent. Rebound clips help to share the load from the master leaf to the graduated lea If the leaf spring has a shape of uniformly varying width (say Lozenge shape) then the bending stress at all section remains uniform. The situation is also identical as before in case of varying thickness, the thickness should vary non-uniformly with length to make a beam of uniform strength (L/h2 = constant). These leaves require lesser material, have more resilience compared to a constant width leaf. These types of springs are called leaf springs of uniform strength. In general the differential curvature between the master leaf and the next leaves is provided in a laminated spring, where, radius of curvature being more for the master leaf. This construction reduces the stress in the master leaf as compared to the other leaves of the spring in a laminated spring. This type of constructional feature is termed as nipping



To prove that bending stresses in full length leaves are 50% more than those in graduated leaves. Full length leaves are nothing but leaves of uniform cross section. Whereas graduated leaves are of uniform strength. Graduated length leaves can be treated as a triangular plate as shown in figure (a)

it is assumed that individual leaves are separated and placed as shown in figure(2) . the leaves are cut longitudinally into two halves each of width d/2 and placed on each side. The bending stress in graduated leaves. ( derivation we will do it in class)

( )fg

3

3

n3n2EbtFl12+

=δ is the final required answer.

Equalised stress in the leaf spring ( nipping of the leaf spring) longitudinally into two halves each of width d/2 and placed on each side. The bending stress in graduated leaves.



The stress in the full length leaves are 50 % greater that the stresses in graduated length leaf. One of the methods of equalizing the stress in leaf spring is to pre- stress the spring. The pre stressing is achieved by bending the leaves to different radii of curvature before they are assembled to centre. The full length leaves is given a greater radius of curvature than a graduated leaf . The initial gap C between the full length and graduated length leaf before the assembling is called nip. Such pre stressing achieved by a difference in radii of curvature is known as nipping. Other ways to reduce the stresses in the master leaves are

] Master leaf is made of stronger material than the other leaves. ] Master leaf is made thinner than the other leaves. This will reduce the bending stress

as evident from stress equation. But common practice is to increase the radius of curvature of the master leaf than the next Leaf because it is much easier and economical compared to other two methods. Problem A locomotive spring has an overall length of 1100mm, sustains of 1kn at its centre. The springs has 3 full length leaves and 15 graduated leaves with a central band of 100mm in width. All the leaves are stressed to 420 MPa. When fully loaded. The ratio of the total spring depth to width is 2. Find the breadth and depth b) Initial space that should be provided between full length and graduated leaf before it is assembled Problem A semi elliptical spring has an effective length of 1.2 m . The maximum stress is not to exceed 400 MPa. The deflection under maximum load of 9kN is 100mm. Find the thickness and width of leaves if

i) the leaves are pre-stressed ii) the leaves are not pre-stressed

take no of extra full length leaves as 2 and no of graduated leaves as 10. E = 200 GPa. Problem The free end of a horizontal , uniform strength cantilever beam is directly over and in contact with a vertical coil spring as shown in figure. The width of the beam at its fixed end is 600mm . It length is 800mm and its thickness is 12mm. The coil spring has 10 active coils of 12.5mm diameter wire and has an outside diameter of 100mm .Take G = 83 GPa and E=200 GPa.

a) what force if gradually applied to the end of cantilever beam is required to pass a defection of 40mm.

b) what is the bending stress in beam at a section of 400mm from the fixed end c) how much energy is absorbed by the coil spring.



Problem A semi elliptical leaf spring consists of extra length leaves and 15 graduated lenth leaves including the master leaf. The centre to centre distance between the 2 eyes of the spring is 1m. The maximum force that can act on the spring is 75 kN. For each leaf the ratio of width to thickness is 9:1. The modulus of elasticity of leaf material is 2x105 N/mm2 . The leaves are prestressed in such a way that when the force is maximum the stress induced in both leaves is 450 MPa. Determine the

• width and thickness of leaves • The initial nip • The initial pre load required to close the gap between the extra full length and

graduated leaves.

DISC SPRINGS.

These are made up from tapered washers as shown in figure. These springs are used where space is limited and where one requires high spring rate. These are characterised by non –linear load deflection curves. These springs may be stacked in series, parallel or a combination of parallel – series.

In series arrangement deflection is proportional to the no of discs while springs placed in parallel arrangement load carrying capacity is proportional to the no of discs. These springs are cheap and can be replaced after failure. Disc type are used for impact and suddenly



applied loads as in buffers , shearing and drawing machine , supports of carriages , They are also used for frame structures as vibration isolators. The relation between the load F and the axial deflection of each disc is given by equation 11.44. maximum stress at inner and outer edge is given by equation 11.45a and 11.45b Problem A disc spring is made of 3mm sheet steel with an outside dia 125 mm and inside dia of 50mm. The spring is dished out by 5mm. the maximum stress is 500 MPa. Determine the deflection of at the load carried by the spring. Take E= 210 GPa and 3.0=µ

HELICAL TORSION SPRINGS

Helical torsion spring is used to transmit a torque to a component of machine or a mechanism . It is used in doors hinges. Brush holders , automobile starters and door locks. The construction of helical torsion spring is similar to that of compression or extension spring except that the ends are formed in such a way that spring is loaded by a torque about the axis of the coils as shown in figure. The helical torsion spring resists bending moment (F.r) which winds up the spring. Therfore the primary stresses in these springs are bending stresses. Each individual section of torsion springs is treated as a portion of the curved beam. Derivation will be done in class room

i

4

o D64Ed

F =



Rubber springs These springs has 2 functions the first being it acts as spring and second it acts as a damper. Or shock absorber. Rubber springs are difficult to analyse because the behaviour of rubber under stress is not linear. The rubber springs are influenced by the kind of material and hardness. Rubber has increasing stiffness with increasing deformation. Rubber springs can be either in compression or in shear. Rubber under compression A simple rectangular block of rubber can be used as spring as shown in figure Rubber under shear Figure shows a rubber spring loaded under shear. It comprises of two blocks of rubber moulded together with a steel plates. Hooks law for rubber in shear holds fairly well. Hence the shearing deformation

Gulusmodrigiditysshearstres τ

==γ

AG2F

G=

τ=γ for small deflection γ=δ h

AG2Fh

=δ

δ

γ

DESIGN OF MACHINE ELEMENTS -II - National Institute of ... · PDF fileSixth semester...

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Transcript of DESIGN OF MACHINE ELEMENTS -II - National Institute of ... · PDF fileSixth semester...