MACHINE DESIGN-I LECTURE NOTES (MEEN-422) (Handout to students by [email protected])

MACHINE DESIGN-I (MEEN-422) MECHANICAL ENGG. DEPT. ERITREA INSTITUTE OF TECHNOLOGY.

Meen 422

Machine Design - I UNIT-I Types of stresses in machine and failure theories

3 Cr.

A machine may be defined as a combination of stationary and moving parts constructed for the useful purpose of generating, transforming or utilizing mechanical energy. Machines can be classified in to: 1. Machines for generating mechanical energy: Converts some form of energy (electrical, heat, hydraulics etc.) into mechanical work. eg: steam engines, IC engines, water turbines etc. 2. Machines for transforming mechanical energy: known as converting machines. These types of machines transform mechanical energy into another form of energy. eg: Electric generators, hydraulic pumps etc. 3. Machines for utilizing mechanical energy: These machines receive mechanical energy and deliver and utilize it as such in the performance of useful work. eg: lathe, m/c tools etc. General Steps in Design Process : Market survey  Define specification of product  Feasibility study  Design synthesis  Preliminary Design and Development (CAD)  Analaysis (CAE)  Prototype  Testing  Final Product manufacture  Quality control In the design process first comes the stage to perform the market survey to identify the need for the product, which will be done by getting feedback from the customer and the requirements of customer. Then according to requirements of the product the specification of the product have to be specified. The next stage is to perform the economic analysis which is called feasibility study of the product in the market. If found optimal with the manufacturing of the product with profit the process will be proceeded to design synthesis stage, in which the dimensions of the product will be determined. The design calculations will be made for development of the product and Computer aided design (CAD) made for drafting (two dimensional) and modeling (three dimensional) of the product. The design made will be analyzed virtually by Computer Aided Engineering (CAE) tools to understand the failure of the component and predict the life of the component. In the analysis stage if the component gets failed, the design calculations have to 1 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


be made again iteratively till it is found safe in the analysis stage. After the success of the analysis stage the reduced (or) enlarged scale of the product will be fabricated and then tested by using sensors and hardware components. If the product is failing in the testing stage the iteration (or) recalculation of the design work should be made. If the product is found safe during testing stage then it can be preceded to Final product manufacture stage. After manufacturing the product it will be sent to quality control section to check the accuracy of the manufactured product, if the product doesn’t meet the required specifications it has to be forwarded to design calculation stage and iterative steps have to be performed till the product meets the required specifications. SIMPLE STRESSES IN MACHINE PARTS Stress: Force per unit area (or) the internal resistance by the body against the load applied. Stress () = P/A P = Force or load acting on the body A = Cross sectional area Units: In S.I. unit; stress is usually expressed as (Pa) 1 Pa = 1 N/m2 Strain: Deformation per unit length l Strain   l – change in length, l – original length l Young’s Modulus (based on Hooke’s law): Stress is directly proportional to strain (with in elastic limit)   = E l E =  / = Al E – Constant of proportionality: Young’s Modulus unit : GPa : GN/m2 or kN/mm2. SHEAR STRESS AND SHEAR STRAIN When the external force acting on a component tends to slide the adjacent planes w.r.t each other, the resulting stress on these planes are called direct shear stresses.  – shear stress A – Cross sectional area (mm2) r – Shear strain (radians)  = G.r. G – Modulus of rigidity=r (N/mm2) The relationship between the modulus of elasticity, the modulus of rigidity and Poisson’s ratio is given by E = 2G [1+ ]  - Poisson’s ratio  = strain in lateral dim. strain in axial dim. The permissible shear stress is given by Sy Sy0 = yield strength in shear (N/mm2)  0  fs 

2 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


STRESSES DUE TO BENDING MOMENT A beam is subjected to bending moment Mb then the governing equation from flexural formula

b 

Mb y I

b - bending stress at a distance of y from neutral axis (N/mm2)

STRESSES DUE TO TORSIONAL MOMENT The internal stresses, which are induced to resist the action of twist, are called torsional shear stress.



Mt .  J

- Torsional shear stress (N/mm2)

Mt – applied torque (N-mm) J – Polar moment of inertia (mm4) The angle of twist is given by



Mt l JG

 - angle of twist (radians) l – length of shaft (mm)

Material Properties Mechanical properties of a material generally determined through destructive testing samples under controlled loading conditions. Tensile test This is one of the simplest and basic test and determines values of number of parameters concerned with mechanical properties. A materials like strength, ductility and toughness. The information which can be obtained from the tests are: i) Proportional Limit ii) Elastic limit iii) Modulus of Elasticity iv) Yield strength A typical tensile specimen is shown in fig. below The tensile bar is mechanical from the material to be tested in one of the several standard diameters ‘do’ and gauge lengths ‘l0’. The gauge length is an arbitrary length defined along the small-diameter portion of the specimen by two indentations so that its increase can be measured during the test. The larger dia sections are threaded for insertion into a tensile test machine which is capable of applying either controlled loads or controlled deflections to the end of the bars; and gauge length portion in mirror polished to eliminate stress concentration from surface defects. The bar is stretched slowly in tension until it breaks, while the load and distance across the gauge length are monitored. Stress-strain Diagram In designing various parts of a machine, it is necessary to know how the material will function in service. For this, certain characteristics or properties of the material should be known. The mechanical properties mostly used in mechanical engineering practice are commonly determined from a standard tensile test. This test consists of gradually loading a standard specimen of a material and noting the corresponding values of load and elongation until the specimen fractures. The load is applied and measured by a testing machine. The stress is determined by dividing the load values by the original cross-sectional area of the specimen. 3 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


The elongation is measured by determining the amounts that two reference points on the specimen are moved apart by the action of the machine. The original distance between the two reference points is known as gauge length. The strain is determined by dividing the elongation values by the gauge length. The values of the stress and corresponding strain are used to draw the stress-strain diagram of the material tested. A stress-strain diagram for mild steel under tensile test is shown in Fig. (a). The various properties of the material are discussed below :

1. Proportional limit. We see from the diagram that from point O to A is a straight line, which represents that the stress is proportional to strain. Beyond point A, the curve slightly deviates from the straight line. It isthus obvious, that Hooke's law holds good up to point A and it is known as proportional limit. It is defined as that stress at which the stress-strain curve begins to deviate from the straight line. 2. Elastic limit. It may be noted that even if the load is increased beyond point A upto the point B, the material will regain its shape and size when the load is removed. This means that the material has elastic properties up to the point B. This point is known as elastic limit. It is defined as the stress developed in the material without any permanent set. 3. Yield point. If the material is stressed beyond point B, the plastic stage will reach i.e. on the removal of the load, the material will not be able to recover its original size and shape. A little consideration will show that beyond point B, the strain increases at a faster rate with any increase in the stress until the point C is reached. At this point, the material yields before the load and there is an appreciable strain without any increase in stress. In case of mild steel, it will be seen that a small load drops to D, immediately after yielding commences. Hence there are two yield points C and D. The points C and D are called the upper and lower yield points respectively. The stress corresponding to yield point is known as yield point stress. 4. Ultimate stress. At D, the specimen regains some strength and higher values of stresses are required for higher strains, than those between A and D. The stress (or load) goes on increasing till the point E is reached. The gradual increase in the strain (or length) of the specimen is followed with the uniform reduction of its cross-sectional area. The work done, during stretching the specimen, is transformed largely into heat and the specimen becomes hot. At E, the stress, which attains its maximum value is known as ultimate stress. It is defined as the largest stress obtained by dividing the largest value of the load reached in a test to the original cross-sectional area of the test piece. 5. Breaking stress. After the specimen has reached the ultimate stress, a neck is formed, which decreases the cross-sectional area of the specimen, as shown in Fig. (b). A little 4 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


consideration will show that the stress (or load) necessary to break away the specimen, is less than the maximum stress. The stress is, therefore, reduced until the specimen breaks away at point F. The stress corresponding to point F is known as breaking stress. Note: The breaking stress (i.e. stress at F which is less than at E) appears to be somewhat misleading. As the formation of a neck takes place at E which reduces the cross-sectional area, it causes the specimen suddenly to fail at F. If for each value of the strain between E and F, the tensile load is divided by the reduced cross sectional area at the narrowest part of the neck, then the true stress-strain curve will follow the dotted line EG. However, it is an established practice, to calculate strains on the basis of original cross-sectional area of the specimen. 6. Percentage reduction in area. It is the difference between the original cross-sectional area and cross-sectional area at the neck (i.e. where the fracture takes place). This difference is expressed as percentage of the original cross-sectional area. Let A = Original cross-sectional area, and a = Cross-sectional area at the neck.

7. Percentage elongation. It is the percentage increase in the standard gauge length (i.e. original length) obtained by measuring the fractured specimen after bringing the broken parts together. Let l = Gauge length or original length, and L = Length of specimen after fracture or final length.

Working Stress When designing machine parts, it is desirable to keep the stress lower than the maximum or ultimate stress at which failure of the material takes place. This stress is known as the working stress or design stress. It is also known as safe or allowable stress. FACTOR OF SAFETY - FOS While designing a component, it is necessary to ensure sufficient reserve strength in the case of an accident. It is ensured by taking a suitable factor of safety (fs) FOS can be defined as: [all = allowable stress]

Failure Stress Allowable Stress Failure load fs  Working load

fs 

(OR)

The allowable stress is the stress value which is used in design to determine the dimensions of the component. It is considered as a stress which the designer, expects will not exceed under normal operating conditions. For ductile material, the allowable stress (all) is given by,

all 

Syt  fs  5

Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


In case of ductile materials e.g. mild steel, where the yield point is clearly defined, the factor of safety is based upon the yield point stress. In such cases,

For brittle material:

all 

Sut Syt = yield tensile strength, Sut = ultimate tensile stress  fg 

In case of brittle materials e.g. cast iron, the yield point is not well defined as for ductile materials. Therefore, the factor of safety for brittle materials is based on ultimate stress.

Note : By failure it is not meant actual breaking of the material. Some machine parts are said to fail when they have plastic deformation set in them, and they no more perform their function satisfactory. Selection of Factor of Safety Before selecting a proper factor of safety, a design engineer should consider the following points: 1.) Effect of failure: Failure of the ball bearing in gear box. Failure of valve in pressure vessel 2.) Types of Load When external force acting on the m/c element is static - FOS is low. Impact load – FOS is high 3.) Degree of Accuracy in force analysis. When the force acting on the m/c element is precisely determined low FOS can be selected. Where as higher FOS is considered when the m/c component is subjected to a force whose magnitude or direction is uncertain and unpredictable. 4.) Material of Component When the component is made of homogenous ductile material, like steel, yield strength is the criterion of feature. FOS is small in such cases. Cast Iron component has non-homogenous structure and a higher FOS based on ultimate strength is chosen. Each of the above factors must be carefully considered and evaluated. The high factor of safety results in unnecessary risk of failure. The values of factor of safety based on ultimate strength for different materials and type of load are given in the following table:


MACHINE DESIGN-I (MEEN-422) MECHANICAL ENGINEERING DEPARTMENT, ERITREA INSTITUTE OF TECHNOLOGY.

Principal Stresses and Principal Planes It has been observed that at any point in a strained material, there are three planes, mutually perpendicular to each other which carry direct stresses only and no shear stress. It may be noted that out of these three direct stresses, one will be maximum and the other will be minimum. These perpendicular planes which have no shear stress are known as principal planes and the direct stresses along these planes are known as principal stresses. The planes on which the maximum shear stress act are known as planes of maximum shear. Determination of Principal Stresses for a Member Subjected to Bi-axial Stress When a member is subjected to bi-axial stress (i.e. direct stress in two mutually perpendicular planes accompanied by a simple shear stress), then the normal and shear stresses are obtained as discussed below: Consider a rectangular body ABCD of uniform cross-sectional area and unit thickness subjected to normal stresses 1 and 2 as shown in Fig. (a). In addition to these normal stresses, a shear stress also acts. It has been known from Strength of Materials concept that the normal stress across any oblique section such as EF inclined at an angle with the direction of 2, as shown in Fig. (a), is given by ……(i) and tangential stress (i.e. shear stress) across the section EF, ……………….(ii) Since the planes of maximum and minimum normal stress (i.e. principal planes) have no shear stress, therefore the inclination of principal planes is obtained by equating 1 = 0 in the above equation (ii), i.e.

…….(iii)



We know that there are two principal planes at right angles to each other. Let 1 and 2 be the inclinations of these planes with the normal cross-section. From Fig. shown below, we find that

So it implies for plane that

The maximum and minimum principal stresses may now be obtained by substituting the values of sin 2and cos 2in equation (i). Maximum principal (or normal) stress, ……(iv) and minimum principal (or normal) stress, ………(v) The planes of maximum shear stress are at right angles to each other and are inclined at 45° to the principal planes. The maximum shear stress is given by one-half the algebraic difference between the principal stresses, i.e. ……(vi) Note: 1. When a member is subjected to direct stress in one plane accompanied by a simple shear stress as shown in Fig.(b), then the principal stresses are obtained by substituting 2 = 0 in equation (iv), (v) and (vi).



Application of Principal Stresses in Designing Machine Members There are many cases in practice, in which machine members are subjected to combined stresses due to simultaneous action of either tensile or compressive stresses combined with shear stresses. In many shafts such as propeller shafts, Cframes etc., there are direct tensile or compressive stresses due to the external force and shear stress due to torsion, which acts normal to direct tensile or compressive stresses. The shafts like crank shafts, are subjected simultaneously to torsion and bending. In such cases, the maximum principal stresses, due to the combination of tensile or compressive stresses with shear stresses may be obtained. The results obtained in the previous article may be written as follows: 1. Maximum tensile stress, 2. Maximum compressive stress,

3. Maximum shear stress,

where

Note : 1. When = 0 as in the case of thin cylindrical shell subjected in internal fluid pressure, then 2. When the shaft is subjected to an axial load (P) in addition to bending and twisting moments as in the propeller shafts of ship and shafts for driving worm gears, then the stress due to axial load must be added to the bending stress (b). This will give the resultant tensile stress or compressive stress (t or c) depending upon the type of axial load (i.e. pull or push). Exercise Problems: 1.) A shaft, as shown in Fig, is subjected to a bending load of 3 kN, pure torque of 1000 N-m and an axial pulling force of 15 kN. Calculate the stresses at A and B.



Given : W = 3 kN = 3000 N ; T = 1000 N-m = 1 × 106 N-mm ; P = 15 kN = 15 × 103 N ; d = 50 mm; x = 250 mm We know that cross-sectional area of the shaft

Tensile stress due to axial pulling at points A and B,

Bending moment at points A and B,

Section modulus for the shaft,

Bending stress at points A and B,

This bending stress is tensile at point A and compressive at point B. Resultant tensile stress at point A,

and resultant compressive stress at point B, We know that the shear stress at points A and B due to the torque transmitted,

Stresses at point A We know that maximum principal (or normal) stress at point A,

Minimum principal (or normal) stress at point A,

and maximum shear stress at point A,



Stresses at point B We know that maximum principal (or normal) stress at point B,

Minimum principal (or normal) stress at point B,

=22 Mpa (compressive) and maximum shear stress at point B,

Theories of Failure Under Static Load It has already been discussed in the previous chapter that strength of machine members is based upon the mechanical properties of the materials used. Since these properties are usually determined from simple tension or compression tests, therefore, predicting failure in members subjected to uniaxial stress is both simple and straight-forward. But the problem of predicting the failure stresses for members subjected to bi-axial or tri-axial stresses is much more complicated. In fact, the problem is so complicated that a large number of different theories have been formulated. The principal theories of failure for a member subjected to bi-axial stress are as follows: 1. Maximum principal (or normal) stress theory (also known as Rankine’s theory). 2. Maximum shear stress theory (also known as Guest’s or Tresca’s theory). 3. Maximum principal (or normal) strain theory (also known as Saint Venant theory). 4. Maximum strain energy theory (also known as Haigh’s theory). 5. Maximum distortion energy theory (also known as Hencky and Von Mises theory). Since ductile materials usually fail by yielding i.e. when permanent deformations occur in the material and brittle materials fail by fracture, therefore the limiting strength for these two classes of materials is normally measured by different mechanical properties. For ductile materials, the limiting strength is the stress at yield point as determined from simple tension test and it is, assumed to be equal in tension or compression. For brittle materials, the limiting strength is the ultimate stress in tension or compression. 1. Maximum Principal or Normal Stress Theory (Rankine’s Theory) According to this theory, the failure or yielding occurs at a point in a member when the maximum principal or normal stress in a bi-axial stress system reaches the limiting strength of the material in a simple tension test. Since the limiting strength for ductile materials is yield point stress and for brittle materials (which do not have well defined yield point) the limiting strength is ultimate stress, therefore according 11 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


to the above theory, taking factor of safety (F.S.) into consideration, the maximum principal or normal stress (t1) in a bi-axial stress system is given by

Since the maximum principal or normal stress theory is based on failure in tension or compression and ignores the possibility of failure due to shearing stress, therefore it is not used for ductile materials. However, for brittle materials which are relatively strong in shear but weak in tension or compression, this theory is generally used. 2. Maximum Shear Stress Theory (Guest’s or Tresca’s Theory) According to this theory, the failure or yielding occurs at a point in a member when the maximum shear stress in a bi-axial stress system reaches a value equal to the shear stress at yield point in a simple tension test. Mathematically, …….(i)

Since the shear stress at yield point in a simple tension test is equal to one-half the yield stress in tension, therefore the equation (i) may be written as This theory is mostly used for designing members of ductile materials. 3. Maximum Principal Strain Theory (Saint Venant’s Theory) According to this theory, the failure or yielding occurs at a point in a member when the maximum principal (or normal) strain in a bi-axial stress system reaches the limiting value of strain (i.e. strain at yield point) as determined from a simple tensile test. The maximum principal (or normal) strain in a bi-axial stress system is given by

According to the above theory, …….(i)



From equation (i), we may write that

This theory is not used, in general, because it only gives reliable results in particular cases. 4. Maximum Strain Energy Theory (Haigh’s Theory) According to this theory, the failure or yielding occurs at a point in a member when the strain energy per unit volume in a bi-axial stress system reaches the limiting strain energy (i.e. strain energy at the yield point ) per unit volume as determined from simple tension test. We know that strain energy per unit volume in a bi-axial stress system,

and limiting strain energy per unit volume for yielding as determined from simple tension test,

This theory may be used for ductile materials. 5. Maximum Distortion Energy Theory (Hencky and Von Mises Theory) According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy (also called shear strain energy) per unit volume in a bi-axial stress system reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from a simple tension test. Mathematically, the maximum distortion energy theory for yielding is expressed as

This theory is mostly used for ductile materials in place of maximum strain energy theory. Note: The maximum distortion energy is the difference between the total strain energy and the strain energy due to uniform stress.



Exercise Problems: 2.) The load on a bolt consists of an axial pull of 10 kN together with a transverse shear force of 5 kN. Find the diameter of bolt required according to 1. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal strain theory; 4. Maximum strain energy theory; and 5. Maximum distortion energy theory. Take permissible tensile stress at elastic limit = 100 MPa and poisson’s ratio = 0.3. Let d = Diameter of the bolt in mm. Cross-sectional area of the bolt,

We know that axial tensile stress,

and transverse shear stress,

1. According to maximum principal stress theory We know that maximum principal stress,

According to maximum principal stress theory,

2. According to maximum shear stress theory We know that maximum shear stress,

According to maximum shear stress theory,



3. According to maximum principal strain theory We know that maximum principal stress, …….(As calculated before) and minimum principal stress,

We know that according to maximum principal strain theory,

4. According to maximum strain energy theory We know that according to maximum strain energy theory,

5. According to maximum distortion energy theory According to maximum distortion energy theory,



3.) A cylindrical shaft made of steel of yield strength 700 MPa is subjected to static loads consisting of bending moment 10 kN-m and a torsional moment 30 kN-m. Determine the diameterof the shaft using two different theories of failure, and assuming a factor of safety of 2. Take E = 210 GPa and poisson's ratio = 0.25. Let d = Diameter of the shaft in mm. First of all, let us find the maximum and minimum principal stresses. We know that section modulus of the shaft

Bending (tensile) stress due to the bending moment,

and shear stress due to torsional moment,

We know that maximum principal stress,

and minimum principal stress,

Let us now find out the diameter of shaft (d) by considering the maximum shear stress theory and maximum strain energy theory. 1. According to maximum shear stress theory We know that maximum shear stress,

We also know that according to maximum shear stress theory



2. According to maximum strain energy theory We know that according to maximum strain energy theory,

Variable stresses in Machine In real practise of machines only a few machine parts are subjected to static loading. Since many of the machine parts (such as axles, shafts, crankshafts, connecting rods, springs, pinion teeth etc.) are subjected to variable or alternating loads (also known as fluctuating or fatigue loads). Completely Reversed or Cyclic Stresses Consider a rotating beam of circular cross-section and carrying a load W, as shown in Fig. This load induces stresses in the beam which are cyclic in nature. A little consideration will show that the upper fibres of the beam (i.e. at point A) are under compressive stress and the lower fibres (i.e. at point B) are under tensile stress. After half a revolution, the point B occupies the position of point A and the point A occupies the position of point B. Thus the point B is now under compressive stress and the point A under tensile stress. The speed of variation of these stresses depends upon the speed of the beam.

From above we see that for each revolution of the beam, the stresses are reversed from compressive to tensile. The stresses which vary from one value of compressive to the same value of tensile or vice versa, are known as completely reversed or cyclic stresses. Note: 1. The stresses which vary from a minimum value to a maximum value of the same nature, (i.e. tensile or compressive) are called fluctuating stresses. 2. The stresses which vary from zero to a certain maximum value are called repeated stresses. 3. The stresses which vary from a minimum value to a maximum value of the opposite nature (i.e. from a certain minimum compressive to a certain maximum tensile or from a minimum tensile to a maximum compressive) are called alternating stresses. 17 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Fatigue and Endurance Limit It has been found experimentally that when a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue. The failure is caused by means of a progressive crack formation which are usually fine and of microscopic size. The failure may occur even without any prior indication. The fatigue of material is effected by the size of the component, relative magnitude of static and fluctuating loads and the number of load reversals.

In order to study the effect of fatigue of a material, a rotating mirror beam method is used. In this method, a standard mirror polished specimen, as shown in Fig. (a), is rotated in a fatigue testing machine while the specimen is loaded in bending. As the specimen rotates, the bending stress at the upper fibres varies from maximum compressive to maximum tensile while the bending stress at the lower fibres varies from maximum tensile to maximum compressive. In other words, the specimen is subjected to a completely reversed stress cycle. This is represented by a time-stress diagram as shown in Fig. (b). A record is kept of the number of cycles required to produce failure at a given stress, and the results are plotted in stress-cycle curve as shown in Fig. (c). A little consideration will show that if the stress is kept below a certain value as shown by dotted line in Fig. (c), the material will not fail whatever may be the number of cycles. This stress, as represented by dotted line, is known as endurance or fatigue limit (e). It is defined as maximum value of the completely reversed bending stress which a polished standard specimen can withstand without failure, for infinite number of cycles (usually 107 cycles). It may be noted that the term endurance limit is used for reversed bending only while for other types of loading, the term endurance strength may be used when referring the fatigue strength of the material. It may be defined as the safe maximum stress which can be applied to the machine part working under actual conditions.



We have seen that when a machine member is subjected to a completely reversed stress, the maximum stress in tension is equal to the maximum stress in compression as shown in Fig. (b). In actual practice, many machine members undergo different range of stress than the completely reversed stress. The stress verses time diagram for fluctuating stress having values min and max is shown in Fig.(e). The variable stress, in general, may be considered as combination of steady (or) mean (or) average stress and a completely reversed stress component v. The following relations are derived from Fig. (e): 1. Mean or average stress,

2. Reversed stress component or alternating or variable stress,

Note: For repeated loading, the stress varies from maximum to zero (i.e. σmin = 0) in each cycle as shown in Fig. (d).

3. Stress ratio, For completely reversed stresses, R = – 1 and for repeated stresses, R = 0. It may be noted that R cannot be greater than unity. Effect of Loading on Endurance Limit—Load Factor The endurance limit (e) of a material as determined by the rotating beam method is for reversed bending load. There are many machine members which are subjected to loads other than reversed bending loads. Thus the endurance limit will also be different for different types of loading. The endurance limit depending upon the type of loading may be modified as discussed below: Let Kb = Load correction factor for the reversed or rotating bending load. Its value is usually taken as unity. Ka = Load correction factor for the reversed axial load. Its value may be taken as 0.8. Ks = Load correction factor for the reversed torsional or shear load. Its value may be taken as 0.55 for ductile materials and 0.8 for brittle materials.

Effect of Surface Finish on Endurance Limit—Surface Finish Factor When a machine member is subjected to variable loads, the endurance limit of the material for that member depends upon the surface conditions. Fig. shows the values of surface finish factor for the various surface conditions and ultimate tensile strength.



When the surface finish factor is known, then the endurance limit for the material of the machine member may be obtained by multiplying the endurance limit and the surface finish factor. We see that for a mirror polished material, the surface finish factor is unity. In other words, the endurance limit for mirror polished material is maximum and it goes on reducing due to surface condition. Effect of Size on Endurance Limit—Size Factor A little consideration will show that if the size of the standard specimen as shown in Fig. (a) is increased, then the endurance limit of the material will decrease. This is due to the fact that a longer specimen will have more defects than a smaller one. Note: 1. The value of size factor is taken as unity for the standard specimen having nominal diameter of 7.657 mm. 2. When the nominal diameter of the specimen is more than 7.657 mm but less than 50 mm, the value of size factor may be taken as 0.85. 3. When the nominal diameter of the specimen is more than 50 mm, then the value of size factor may be taken as 0.75. Relation Between Endurance Limit and Ultimate Tensile Strength It has been found experimentally that endurance limit (e) of a material subjected to fatigue loading is a function of ultimate tensile strength (u). Fig. shows the endurance limit of steel corresponding to ultimate tensile strength for different surface conditions. Following are some empirical relations commonly used in practice :



Factor of Safety for Fatigue Loading When a component is subjected to fatigue loading, the endurance limit is the criterion for faliure. Therefore, the factor of safety should be based on endurance limit. Mathematically,

Note:

Stress Concentration Whenever a machine component changes the shape of its cross-section, the simple stress distribution no longer holds good and the neighbourhood of the discontinuity is different. This irregularity in the stress distribution caused by abrupt changes of form is called stress concentration. It occurs for all kinds of stresses in the presence of fillets, notches, holes, keyways, splines, surface roughness or scratches etc. In order to understand fully the idea of stress concentration, consider a member with different cross-section under a tensile load as shown in Fig.

A little consideration will show that the nominal stress in the right and left hand sides will be uniform but in the region where the crosssection is changing, a re-distribution of the force within the member must take place. The material near the edges is stressed considerably higher than the average value. The maximum stress occurs at some point on the fillet and is directed parallel to the boundary at that point. 21 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Theoretical or Form Stress Concentration Factor The theoretical or form stress concentration factor is defined as the ratio of the maximum stress in a member (at a notch or a fillet) to the nominal stress at the same section based upon net area. Mathematically, theoretical or form stress concentration factor,

The value of Kt depends upon the material and geometry of the part. Note: 1. In static loading, stress concentration in ductile materials is not so serious as in brittle materials, because in ductile materials local deformation or yielding takes place which reduces the concentration. In brittle materials, cracks may appear at these local concentrations of stress which will increase the stress over the rest of the section. It is, therefore, necessary that in designing parts of brittle materials such as castings, care should be taken. In order to avoid failure due to stress concentration, fillets at the changes of section must be provided. 2. In cyclic loading, stress concentration in ductile materials is always serious because the ductility of the material is not effective in relieving the concentration of stress caused by cracks, flaws, surface roughness, or any sharp discontinuity in the geometrical form of the member. If the stress at any point in a member is above the endurance limit of the material, a crack may develop under the action of repeated load and the crack will lead to failure of the member. Stress Concentration due to Holes and Notches Consider a plate with transverse elliptical hole and subjected to a tensile load as shown in Fig.(a). We see from the stress-distribution that the stress at the point away from the hole is practically uniform and the maximum stress will be induced at the edge of the hole. The maximum stress is given by

When a/b is large, the ellipse approaches a crack transverse to the load and the value of Kt becomes very large. When a/b is small, the ellipse approaches a longitudinal slit as shown in Fig.(b) and the increase in stress is small. When the hole is circular as shown in Fig.(c), then a/b = 1 and the maximum stress is three times the nominal value. 22 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


The stress concentration in the notched tension member, as shown in Fig. below,

is influenced by the depth a of the notch and radius r at the bottom of the notch. The maximum stress, which applies to members having notches that are small in comparison with the width of the plate, may be obtained by the following equation,

Methods of Reducing Stress Concentration We have already discussed whenever there is a change in cross-section, such as shoulders, holes, notches or keyways and where there is an interference fit between a hub or bearing race and a shaft, then stress concentration results. The presence of stress concentration cannot be totally eliminated but it may be reduced to some extent. A device or concept that is useful in assisting a design engineer to visualize the presence of stress concentration and how it may be mitigated is that of stress flow lines, as shown in Fig. The mitigation of stress concentration means that the stress flow lines shall maintain their spacing as far as possible.

In Fig.(a) we see that stress lines tend to bunch up and cut very close to the sharp reentrant corner. In order to improve the situation, fillets may be provided, as shown in Fig. (b) and (c) to give more equally spaced flow lines. Figs. below show the several ways of reducing the stress concentration in shafts and other cylindrical members with shoulders, holes and threads respectively. It may be noted that it is not practicable to use large radius fillets as in case of ball and roller bearing mountings. In such cases, notches may be cut as shown in Fig.



Factors to be Considered while Designing Machine Parts to Avoid Fatigue Failure The following factors should be considered while designing machine parts to avoid fatigue failure: 1. The variation in the size of the component should be as gradual as possible. 2. The holes, notches and other stress raisers should be avoided. 3. The proper stress de-concentrators such as fillets and notches should be provided wherever necessary. 4. The parts should be protected from corrosive atmosphere. 5. A smooth finish of outer surface of the component increases the fatigue life. 6. The material with high fatigue strength should be selected. 7. The residual compressive stresses over the parts surface increases its fatigue strength. Fatigue Stress Concentration Factor When a machine member is subjected to cyclic or fatigue loading, the value of fatigue stress concentration factor shall be applied instead of theoretical stress concentration factor. Since the determination of fatigue stress concentration factor is not an easy task, therefore from experimental tests it is defined as Fatigue stress concentration factor,

Notch Sensitivity In cyclic loading, the effect of the notch or the fillet is usually less than predicted by the use of the theoretical factors as discussed before. The difference depends upon the stress gradient in the region of the stress concentration and on the hardness of the material. The term notch sensitivity is applied to this behaviour. It may be defined as the degree to which the theoretical effect of stress concentration is actually 24 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


reached. The stress gradient depends mainly on the radius of the notch, hole or fillet and on the grain size of the material. Since the extensive data for estimating the notch sensitivity factor (q) is not available, therefore the curves, as shown in Fig., may be used for determining the values of q for two steels.

When the notch sensitivity factor q is used in cyclic loading, then fatigue stress concentration factor may be obtained from the following relations:

Combined Steady and Variable Stress The failure points from fatigue tests made with different steels and combinations of mean and variable stresses are plotted in Fig. as functions of variable stress (v) and mean stress (m). The most significant observation is that, in general, the failure point is little related to the mean stress when it is compressive but is very much a function of the mean stress when it is tensile. In practice, this means that fatigue failures are rare when the mean stress is compressive (or negative). Therefore, the greater emphasis must be given to the combination of a variable stress and a steady (or mean) tensile stress.



There are several ways in which problems involving this combination of stresses may be solved, but the following are important from the subject point of view : 1. Gerber method, 2. Goodman method, and 3. Soderberg method. Gerber Method for Combination of Stresses The relationship between variable stress (v) and mean stress (m) for axial and bending loading for ductile materials are shown in Fig. The point e represents the fatigue strength corresponding to the case of complete reversal (m = 0) and the point u represents the static ultimate strength corresponding to v = 0. A parabolic curve drawn between the endurance limit (e) and ultimate tensile strength (u) was proposed by Gerber in 1874. Generally, the test data for ductile material fall closer to Gerber parabola as shown in Fig., but because of scatter in the test points, a straight line relationship (i.e. Goodman line and Soderberg line) is usually preferred in designing machine parts.

.......(i) where F.S. = Factor of safety, m = Mean stress (tensile or compressive), u = Ultimate stress (tensile or compressive), and e = Endurance limit for reversal loading. Considering the fatigue stress concentration factor (Kf), the equation (i) may be written as



Goodman Method for Combination of Stresses A straight line connecting the endurance limit (e) and the ultimate strength (u), as shown by line AB in Fig., follows the suggestion of Goodman. A Goodman line is used when the design is based on ultimate strength and may be used for ductile or brittle materials.

In Fig., line AB connecting e and u is called Goodman's failure stress line. If a suitable factor of safety (F.S.) is applied to endurance limit and ultimate strength, a safe stress line CD may be drawn parallel to the line AB. Let us consider a design point P on the line CD. Now from similar triangles COD and PQD,

........(i) This expression does not include the effect of stress concentration. It may be noted that for ductile materials, the stress concentration may be ignored under steady loads. Since many machine and structural parts that are subjected to fatigue loads contain regions of high stress concentration, therefore equation (i) must be altered to include this effect. In such cases, the fatigue stress concentration factor (Kf) is used to multiply the variable stress (v). The equation (i) may now be written as

........(ii) 27 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


. Considering the load factor, surface finish factor and size factor, the equation (ii) may be written as

.......(iii)

Note: 1. The equation (iii) is applicable to ductile materials subjected to reversed bending loads (tensile or compressive). For brittle materials, the theoretical stress concentration factor (Kt) should be applied to the mean stress and fatigue stress concentration factor (Kf) to the variable stress. Thus for brittle materials, the equation (iii) may be written as .........(iv) 2. When a machine component is subjected to a load other than reversed bending, then the endurance limit for that type of loading should be taken into consideration. Thus for reversed axial loading (tensile or compressive), the equations (iii) and (iv) may be written as



Soderberg Method for Combination of Stresses A straight line connecting the endurance limit (e) and the yield strength (y), as shown by the line AB in Fig. follows the suggestion of Soderberg line. This line is used when the design is based on yield strength.

Proceeding in the same way as discussed above the line AB connecting e and y, as shown in Fig., is called Soderberg's failure stress line. If a suitable factor of safety (F.S.) is applied to the endurance limit and yield strength, a safe stress line CD may be drawn parallel to the line AB. Let us consider a design point P on the line CD. Now from similar triangles COD and PQD,

........(i) For machine parts subjected to fatigue loading, the fatigue stress concentration factor (Kf) should be applied to only variable stress (v). Thus the equations (i) may be written as .................................(ii) Considering the load factor, surface finish factor and size factor, the equation (ii) may be written as

...........................(iii) 29 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Notes: 1. The Soderberg method is particularly used for ductile materials. The equation (iii) is applicable to ductile materials subjected to reversed bending load (tensile or compressive). 2. When a machine component is subjected to reversed axial loading, then the equation (iii) may be written as

3. When a machine component is subjected to reversed shear loading, then equation (iii) may be written as

where Kf s is the fatigue stress concentration factor for reversed shear loading. The yield strength in shear (y) may be taken as one-half the yield strength in reversed bending (y). Exercise Problems: 4.) A machine component is subjected to a flexural stress which fluctuates between + 300 MN/m2 and – 150 MN/m2. Determine the value of minimum ultimate strength according to 1. Gerber relation; 2. Modified Goodman relation; and 3. Soderberg relation. Take yield strength = 0.55 Ultimate strength; Endurance strength = 0.5 Ultimate strength; and factor of safety = 2.

Let u = Minimum ultimate strength in MN/m2. We know that the mean or average stress,

1. According to Gerber relation We know that according to Gerber relation



2. According to modified Goodman relation We know that according to modified Goodman relation,

3. According to Soderberg relation We know that according to Soderberg relation,

5.) A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Determine the diameter of bar using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Use Goodman straight line as basis for design.

Let d = Diameter of bar in mm.

We know that mean or average force,

We know that according to Goodman's formula,



6.) A circular bar of 500 mm length is supported freely at its two ends. It is acted upon by a central concentrated cyclic load having a minimum value of 20 kN and a maximum value of 50 kN. Determine the diameter of bar by taking a factor of safety of 1.5, size effect of 0.85, surface finish factor of 0.9. The material properties of bar are given by : ultimate strength of 650 MPa, yield strength of 500 MPa and endurance strength of 350 MPa.

Let d = Diameter of the bar in mm. We know that the maximum bending moment,

and minimum bending moment,

Mean or average bending moment,

and variable bending moment,

Section modulus of the bar,

Mean or average bending stress,

and variable bending stress,

We know that according to Goodman's formula,



and according to Soderberg's formula,

Taking larger of the two values, we have d = 62.1 mm 7.) A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630 MPa. It is subjected to a torque which fluctuates between 2000 N-m to – 800 N-m. Using Soderberg method, calculate the factor of safety. Assume suitable values for any other data needed. Given : d = 50 mm ; u = 630 MPa = 630 N/mm2 ; Tmax = 2000 N-m ; Tmin = – 800 N-m We know that the mean or average torque,

Mean or average shear stress,

Variable torque,



Let F.S. = Factor of safety. We know that according to Soderberg's formula,

Combined Variable Normal Stress and Variable Shear Stress When a machine part is subjected to both variable normal stress and a variable shear stress; then it is designed by using the following two theories of combined stresses : 1. Maximum shear stress theory, and 2. Maximum normal stress theory. according to Soderberg's formula,

Multiplying throughout by y, we get

The term on the right hand side of the above expression is known as equivalent normal stress due to reversed bending. Equivalent normal stress due to reversed bending,

Similarly, equivalent normal stress due to reversed axial loading,

and total equivalent normal stress,

that for reversed torsional or shear loading,

Multiplying throughout by y, we get

The term on the right hand side of the above expression is known as equivalent shear stress. Equivalent shear stress due to reversed torsional or shear loading,

The maximum shear stress theory is used in designing machine parts of ductile materials. According to this theory, maximum equivalent shear stress, 34 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


The maximum normal stress theory is used in designing machine parts of brittle materials. According to this theory, maximum equivalent normal stress,

Exercise Problems: 8.) A pulley is keyed to a shaft midway between two bearings. The shaft is made of cold drawn steel for which the ultimate strength is 550 MPa and the yield strength is 400 MPa. The bending moment at the pulley varies from – 150 N-m to + 400 N-m as the torque on the shaft varies from – 50 N-m to + 150 N-m. Obtain the diameter of the shaft for an indefinite life. The stress concentration factors for the keyway at the pulley in bending and in torsion are 1.6 and 1.3 respectively. Take the following values: Factor of safety = 1.5 Load correction factors = 1.0 in bending, and 0.6 in torsion Size effect factor = 0.85 Surface effect factor = 0.88

Let d = Diameter of the shaft in mm. First of all, let us find the equivalent normal stress due to bending. We know that the mean or average bending moment,

and variable bending moment,

Mean bending stress,

and variable bending stress,

Assuming the endurance limit in reversed bending as one-half the ultimate strength and since the load correction factor for reversed bending is 1 (i.e. Kb = 1), therefore endurance limit in reversed bending,



Since there is no reversed axial loading, therefore equivalent normal stress due to bending,

Now let us find the equivalent shear stress due to torsional moment. We know that the mean torque,

Mean shear stress,

and variable shear stress,

We know that equivalent shear stress,

and maximum equivalent shear stress,



UNIT-II DESIGN OF SHAFTS A shaft is a rotating machine element which is used to transmit power from one place to another. The power is delivered to the shaft by some tangential force and the resultant torque (or twisting moment) set up within the shaft permits the power to be transferred to various machines linked up to the shaft. In order to transfer the power from one shaft to another, the various members such as pulleys, gears etc., are mounted on it. These members along with the forces exerted upon them causes the shaft to bending. In other words, we may say that a shaft is used for the transmission of torque and bending moment. The various members are mounted on the shaft by means of keys or splines. Material Used for Shafts The material used for shafts should have the following properties : 1. It should have high strength. 2. It should have good machinability. 3. It should have low notch sensitivity factor. 4. It should have good heat treatment properties. 5. It should have high wear resistant properties. The material used for ordinary shafts is carbon steel of grades 40 C 8, 45 C 8, 50 C 4 and 50 C 12. The mechanical properties of these grades of carbon steel are given in the following table.

When a shaft of high strength is required, then an alloy steel such as nickel, nickelchromium or chrome-vanadium steel is used. Manufacturing of Shafts Shafts are generally manufactured by hot rolling and finished to size by cold drawing or turning and grinding. The cold rolled shafts are stronger than hot rolled shafts but with higher residual stresses. The residual stresses may cause distortion of the shaft when it is machined, especially when slots or keyways are cut. Shafts of larger diameter are usually forged and turned to size in a lathe. Types of Shafts The following two types of shafts are important from the subject point of view : 1. Transmission shafts. These shafts transmit power between the source and the machines absorbing power. The counter shafts, line shafts, over head shafts and all factory shafts are transmission shafts. Since these shafts carry machine parts such as pulleys, gears etc., therefore they are subjected to bending in addition to twisting. 2. Machine shafts. These shafts form an integral part of the machine itself. The crank shaft is an example of machine shaft. 37 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Standard Sizes of Transmission Shafts The standard sizes of transmission shafts are : 25 mm to 60 mm with 5 mm steps; 60 mm to 110 mm with 10 mm steps ; 110 mm to 140 mm with 15 mm steps ; and 140 mm to 500 mm with 20 mm steps. The standard length of the shafts are 5 m, 6 m and 7 m. Stresses in Shafts The following stresses are induced in the shafts : 1. Shear stresses due to the transmission of torque (i.e. due to torsional load). 2. Bending stresses (tensile or compressive) due to the forces acting upon machine elements like gears, pulleys etc. as well as due to the weight of the shaft itself. 3. Stresses due to combined torsional and bending loads Design of Shafts The shafts may be designed on the basis of 1. Strength, and 2. Rigidity and stiffness. In designing shafts on the basis of strength, the following cases may be considered : (a) Shafts subjected to twisting moment or torque only, (b) Shafts subjected to bending moment only, (c) Shafts subjected to combined twisting and bending moments, and (d) Shafts subjected to axial loads in addition to combined torsional and bending loads. Shafts Subjected to Twisting Moment Only When the shaft is subjected to a twisting moment (or torque) only, then the diameter of the shaft may be obtained by using the torsion equation. We know that ……..(i) where T = Twisting moment (or torque) acting upon the shaft, J = Polar moment of inertia of the shaft about the axis of rotation, = Torsional shear stress, and r = Distance from neutral axis to the outer most fibre = d / 2; where d is the diameter of the shaft. We know that for round solid shaft, polar moment of inertia,

The equation (i) may now be written as

……(ii) From this equation, we may determine the diameter of round solid shaft ( d ). We also know that for hollow shaft, polar moment of inertia,

where do and di = Outside and inside diameter of the shaft, and r = do / 2. Substituting these values in equation (i), we have ……..(iii) 38 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Let k = Ratio of inside diameter and outside diameter of the shaft = di / do Now the equation (iii) may be written as ………(iv) From the equations (iii) or (iv), the outside and inside diameter of a hollow shaft may be determined. It may be noted that 1. The hollow shafts are usually used in marine work. These shafts are stronger per kg of material and they may be forged on a mandrel, thus making the material more homogeneous than would be possible for a solid shaft. When a hollow shaft is to be made equal in strength to a solid shaft, the twisting moment of both the shafts must be same. In other words, for the same material of both the shafts,

2. The twisting moment (T) may be obtained by using the following relation : We know that the power transmitted (in watts) by the shaft,

where T = Twisting moment in N-m, and N = Speed of the shaft in r.p.m. 3. In case of belt drives, the twisting moment ( T ) is given by where T1 and T2 = Tensions in the tight side and slack side of the belt respectively, and R = Radius of the pulley. Shafts Subjected to Bending Moment Only When the shaft is subjected to a bending moment only, then the maximum stress (tensile or compressive) is given by the bending equation. We know that ...(i) where M = Bending moment, I = Moment of inertia of cross-sectional area of the shaft about the axis of rotation, b = Bending stress, and y = Distance from neutral axis to the outer-most fibre. We know that for a round solid shaft, moment of inertia, Substituting these values in equation (i), we have

From this equation, diameter of the solid shaft (d) may be obtained. We also know that for a hollow shaft, moment of inertia,



Again substituting these values in equation (i), we have

From this equation, the outside diameter of the shaft (do) may be obtained. Shafts Subjected to Combined Twisting Moment and Bending Moment When the shaft is subjected to combined twisting moment and bending moment, then the shaft must be designed on the basis of the two moments simultaneously. Various theories have been suggested to account for the elastic failure of the materials when they are subjected to various types of combined stresses. The following two theories are important from the subject point of view : 1. Maximum shear stress theory or Guest's theory. It is used for ductile materials such as mild steel. 2. Maximum normal stress theory or Rankine’s theory. It is used for brittle materials such as cast iron. Let = Shear stress induced due to twisting moment, and b = Bending stress (tensile or compressive) induced due to bending moment. According to maximum shear stress theory, the maximum shear stress in the shaft,

max 

1 2

 b 2  42

Substituting the values of b and  from above relations ……..(i) The expression is known as equivalent twisting moment and is denoted by Te. The equivalent twisting moment may be defined as that twisting moment, which when acting alone, produces the same shear stress () as the actual twisting moment. By limiting the maximum shear stress (max) equal to the allowable shear stress () for the material, the equation (i) may be written as …….(ii) From this expression, diameter of the shaft ( d ) may be evaluated. Now according to maximum normal stress theory, the maximum normal stress in the shaft, …..(iii)

…..(iv) The expression is known as equivalent bending moment and is denoted by Me. The equivalent bending moment may be defined as that moment 40 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


which when acting alone produces the same tensile or compressive stress (b) as the actual bending moment. By limiting the maximum normal stress [b(max)] equal to the allowable bending stress (b), then the equation (iv) may be written as ……(v) From this expression, diameter of the shaft ( d ) may be evaluated. Note: 1. In case of a hollow shaft, the equations (ii) and (v) may be written as

2. It is suggested that diameter of the shaft may be obtained by using both the theories and the larger of the two values is adopted. Exercise Problems: 1.) A solid circular shaft is subjected to a bending moment of 3000 N-m and a torque of 10 000 N-m. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and a ultimate shear stress of 500 MPa. Assuming a factor of safety as 6, determine the diameter of the shaft. We know that the allowable tensile stress,

and allowable shear stress,

Let d = Diameter of the shaft in mm. According to maximum shear stress theory, equivalent twisting mome We also know that equivalent twisting moment (Te),

According to maximum normal stress theory, equivalent bending moment,

We also know that the equivalent bending moment (Me),

Taking the larger of the two values, we have d = 86 say 90 mm 2.) A shaft supported at the ends in ball bearings carries a straight tooth spur gear at its mid span and is to transmit 7.5 kW at 300 r.p.m. The pitch circle diameter of the gear is 150 mm. The distances between the centre line of bearings and gear are 100 41 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


mm each. If the shaft is made of steel and the allowable shear stress is 45 MPa, determine the diameter of the shaft. Show in a sketch how the gear will be mounted on the shaft; also indicate the ends where the bearings will be mounted? The pressure angle of the gear may be taken as 20°.

Tangential force on the gear,

Tangential force on the gear,

and the normal load acting on the tooth of the gear, Since the gear is mounted at the middle of the shaft, therefore maximum bending moment at the centre of the gear, Let d = Diameter of the shaft. We know that equivalent twisting moment

We also know that equivalent twisting moment (Te)

3.) A line shaft is driven by means of a motor placed vertically below it. The pulley on the line shaft is 1.5 metre in diameter and has belt tensions 5.4 kN and 1.8 kN on the tight side and slack side of the belt respectively. Both these tensions may be assumed to be vertical. If the pulley be overhang from the shaft, the distance of the centre line of the pulley from the centre line of the bearing being 400 mm, find the diameter of the shaft. Assuming maximum allowable shear stress of 42 MPa. A line shaft with a pulley is shown in Fig. We know that torque transmitted by the shaft, 42 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


T = (T1 – T2) R = (5400 – 1800) 0.75 = 2700 N-m = 2700 × 103 N-mm

Neglecting the weight of shaft, total vertical load acting on the pulley, W = T1 + T2 = 5400 + 1800 = 7200 N Bending moment, M = W × L = 7200 × 400 = 2880 × 103 N-mm Let d = Diameter of the shaft in mm. We know that the equivalent twisting moment

We also know that equivalent twisting moment (Te),

4.) A shaft is supported by two bearings placed 1 m apart. A 600 mm diameter pulley is mounted at a distance of 300 mm to the right of left hand bearing and this drives a pulley directly below it with the help of belt having maximum tension of 2.25 kN. Another pulley 400 mm diameter is placed 200 mm to the left of right hand bearing and is driven with the help of electric motor and belt, which is placed horizontally to the right. The angle of contact for both the pulleys is 180° and = 0.24. Determine the suitable diameter for a solid shaft, allowing working stress of 63 MPa in tension and 42 MPa in shear for the material of shaft. Assume that the torque on one pulley is equal to that on the other pulley.

The space diagram of the shaft is shown in Fig.(a). Let T1 = Tension in the tight side of the belt on pulley C = 2250 N ...(Given) T2 = Tension in the slack side of the belt on pulley C. We know that

Vertical load acting on the shaft at C, WC = T1 + T2 = 2250 + 1058 = 3308 N and vertical load on the shaft at D = 0 The vertical load diagram is shown in Fig. (c). 43 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


We know that torque acting on the pulley C, T = (T1 – T2) RC = (2250 – 1058) 0.3 = 357.6 N-m The torque diagram is shown in Fig. (b). Let T3 = Tension in the tight side of the belt on pulley D, and T4 = Tension in the slack side of the belt on pulley D. Since the torque on both the pulleys (i.e. C and D) is same, therefore

By solving the above equations, we find that T3 = 3376 N, and T4 = 1588 N Horizontal load acting on the shaft at D, WD = T3 + T4 = 3376 + 1588 = 4964 N and horizontal load on the shaft at C = 0 The horizontal load diagram is shown in Fig. (d). Now let us find the maximum bending moment for vertical and horizontal loading. First of all, considering the vertical loading at C. Let RAV and RBV be the reactions at the bearings A and B respectively. We know that RAV + RBV = 3308 N Taking moments about A, We know that B.M. at A and B,

The bending moment diagram for vertical loading in shown in Fig. (e). Now considering horizontal loading at D. Let RAH and RBH be the reactions at the bearings A and B respectively. We know that Taking moments about A,

We know that B.M. at A and B,

The bending moment diagram for horizontal loading is shown in Fig.( f ). Resultant B.M. at C, and resultant B.M. at D,

The resultant bending moment diagram is shown in Fig. (g). We see that bending moment is maximum at D. Maximum bending moment, M = MD = 819.2 N-m 44 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


Let d = Diameter of the shaft. We know that equivalent twisting moment,


Again we know that equivalent bending moment,

We also know that equivalent bending moment (Me),

Taking larger of the two values, we have d = 51.7 say 55 mm

5.) A steel solid shaft transmitting 15 kW at 200 r.p.m. is supported on two bearings 750 mm apart and has two gears keyed to it. The pinion having 30 teeth of 5 mm module is located 100 mm to the left of the right hand bearing and delivers power horizontally to the right. The gear having 100 teeth of 5 mm module is located 150 45 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


mm to the right of the left hand bearing and receives power in a vertical direction from below. Using an allowable stress of 54 MPa in shear, determine the diameter of the shaft.

The space diagram of the shaft is shown in Fig. (a). in the next page We know that the torque transmitted by the shaft,

The torque diagram is shown in Fig. (b). in the next page Radius of gear C,

and radius of pinion D, Assuming that the torque at C and D is same (i.e. 716 × 103 N-mm), therefore tangential force on the gear C, acting downward, and tangential force on the pinion D, acting horizontally, The vertical and horizontal load diagram is shown in Fig. (c) and (d) respectively. Now let us find the maximum bending moment for vertical and horizontal loading. First of all, considering the vertical loading at C. Let RAV and RBV be the reactions at the bearings A and B respectively. We know that Taking moments about A, we get

and We know that B.M. at A and B,

The B.M. diagram for vertical loading is shown in Fig. (e). Now considering horizontal loading at D. Let RAH and RBH be the reactions at the bearings A and B respectively. We know that Taking moments about A, we get

We know that B.M. at A and B,



The B.M. diagram for horizontal loading is shown in Fig.( f ). We know that resultant B.M. at C,

and resultant B.M. at D,

The resultant B.M. diagram is shown in Fig. (g). We see that the BM is max. at D. Maximum bending moment, M = MD = 829 690 N-mm Let d = Diameter of the shaft. We know that the equivalent twisting moment, We also know that equivalent twisting moment (Te),

d = 47 say 50 mm

Shafts Subjected to Fluctuating Loads In the previous articles we have assumed that the shaft is subjected to constant torque and bending moment. But in actual practice, the shafts are subjected to fluctuating torque and bending moments. In order to design such shafts like line shafts and 47 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


counter shafts, the combined shock and fatigue factors must be taken into account for the computed twisting moment (T ) and bending moment (M ). Thus for a shaft subjected to combined bending and torsion, the equivalent twisting moment, and equivalent bending moment,

The following table shows recommended values of Km and Kt .

Exercise Problems: 6.) A mild steel shaft transmits 20 kW at 200 r.p.m. It carries a central load of 900 N and is simply supported between the bearings 2.5 metres apart. Determine the size of the shaft, if the allowable shear stress is 42 MPa and the maximum tensile or



compressive stress is not to exceed 56 MPa. What size of the shaft will be required, if it is subjected to gradually applied loads?

Size of the shaft Let d = Diameter of the shaft, in mm. We know that torque transmitted by the shaft, and maximum bending moment of a simply supported shaft carrying a central load, We know that the equivalent twisting moment, We also know that equivalent twisting moment (Te),

We know that the equivalent bending moment,


Taking the larger of the two values, we have d = 53.4 say 55 mm Size of the shaft when subjected to gradually applied load Let d = Diameter of the shaft. From Table, for rotating shafts with gradually applied loads, Km = 1.5 and Kt = 1 We know that equivalent twisting moment,


We know that the equivalent bending moment,




Taking the larger of the two values, we have d = 57.7 say 60 mm 7.) Design a shaft to transmit power from an electric motor to a lathe head stock through a pulley by means of a belt drive. The pulley weighs 200 N and is located at 300 mm from the centre of the bearing. The diameter of the pulley is 200 mm and the maximum power transmitted is 1 kW at 120 r.p.m. The angle of lap of the belt is 180° and coefficient of friction between the belt and the pulley is 0.3. The shock and fatigue factors for bending and twisting are 1.5 and 2.0 respectively. The allowable shear stress in the shaft may be taken as 35 MPa.

The shaft with pulley is shown in Fig.

We know that torque transmitted by the shaft,

Let T1 and T2 = Tensions in the tight side and slack side of the belt respectively in newtons.

We know that

By solving above two equations, we get, T1 = 1303 N, and T2 = 507 N We know that the total vertical load acting on the pulley, Bending moment acting on the shaft, Let d = Diameter of the shaft. We know that equivalent twisting moment,




Shafts Subjected to Axial Load in addition to Combined Torsion and Bending Loads When the shaft is subjected to an axial load (F) in addition to torsion and bending loads as in propeller shafts of ships and shafts for driving worm gears, then the stress due to axial load must be added to the bending stress (b). We know that bending equation is

and stress due to axial load

Resultant stress (tensile or compressive) for solid shaft, …………….(i)

In case of a hollow shaft, the resultant stress,

In case of long shafts (slender shafts) subjected to compressive loads, a factor known as column factor (α) must be introduced to take the column effect into account. Stress due to the compressive load,

The value of column factor (α) for compressive loads may be obtained from the following relation :



Note: The value of column factor (α) for tensile load is unity. This expression is used when the slenderness ratio (L / K) is less than 115. When the slenderness ratio (L / K) is more than 115, then the value of column factor may be obtained from the following relation: ( It is an Euler’s formula for long columns.) where L = Length of shaft between the bearings, K = Least radius of gyration, y = Compressive yield point stress of shaft material, and C = Coefficient in Euler's formula depending upon the end conditions. The following are the different values of C depending upon the end conditions. C =1, for hinged ends, = 2.25, for fixed ends, = 1.6, for ends that are partly restrained as in bearings. Note: In general, for a hollow shaft subjected to fluctuating torsional and bending load, along with an axial load, the equations for equivalent twisting moment (Te) and equivalent bending moment (Me) may be written as

It may be noted that for a solid shaft, k = 0 and d0 = d. When the shaft carries no axial load, then F = 0 and when the shaft carries axial tensile load, then α= 1. 8.) A hollow shaft of 0.5 m outside diameter and 0.3 m inside diameter is used to drive a propeller of a marine vessel. The shaft is mounted on bearings 6 metre apart and it transmits 5600 kW at 150 r.p.m. The maximum axial propeller thrust is 500 kN and the shaft weighs 70 kN. Determine : 1. The maximum shear stress developed in the shaft, and 2. The angular twist between the bearings. Given: do = 0.5 m ; di = 0.3 m ; P = 5600 kW = 5600 × 103 W ; L = 6 m ; N = 150 r.p.m. ; F = 500 kN = 500 × 103 N ; W = 70 kN = 70 × 103 N 1. Maximum shear stress developed in the shaft Let = Maximum shear stress developed in the shaft. We know that the torque transmitted by the shaft,



and the maximum bending moment,

Now let us find out the column factor α. We know that least radius of gyration,

Slenderness ratio,

Assuming that the load is applied gradually, therefore from Table, we find that

We know that the equivalent twisting moment for a hollow shaft,

We also know that the equivalent twisting moment for a hollow shaft (Te),

2. Angular twist between the bearings Let = Angular twist between the bearings in radians. We know that the polar moment of inertia for a hollow shaft,

From the torsion equation,



Design of Shafts on the basis of Rigidity Sometimes the shafts are to be designed on the basis of rigidity. We shall consider the following two types of rigidity. 1. Torsional rigidity. The torsional rigidity is important in the case of camshaft of an I.C. engine where the timing of the valves would be effected. The permissible amount of twist should not exceed 0.25° per metre length of such shafts. For line shafts or transmission shafts, deflections 2.5 to 3 degree per metre length may be used as limiting value. The widely used deflection for the shafts is limited to 1 degree in a length equal to twenty times the diameter of the shaft. The torsional deflection may be obtained by using the torsion equation,

2. Lateral rigidity. It is important in case of transmission shafting and shafts running at high speed, where small lateral deflection would cause huge out-of-balance forces. The lateral rigidity is also important for maintaining proper bearing clearances and for correct gear teeth alignment. Exercise Problem: 9.) Compare the weight, strength and stiffness of a hollow shaft of the same external diameter as that of solid shaft. The inside diameter of the hollow shaft being half the external diameter. Both the shafts have the same material and length. Given : do = d ; di = do / 2 or k = di / do = 1 / 2 = 0.5 Comparison of weight We know that weight of a hollow shaft, WH = Cross-sectional area × Length × Density …….(i)

and weight of the solid shaft, 54 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


……..(ii) Since both the shafts have the same material and length, therefore by dividing equation (i) by equation (ii), we get

Comparison of strength We know that strength of the hollow shaft, …….(iii) and strength of the solid shaft, ………………..(iv) Dividing equation (iii) by equation (iv), we get

Comparison of stiffness We know that stiffness

Stiffness of a hollow shaft, …….(v) and stiffness of a solid shaft, ………………….(vi) Dividing equation (v) by equation (vi), we get

Shafts in Series and Parallel When two shafts of different diameters are connected together to form one shaft, it is then known as composite shaft. If the driving torque is applied at one end and the resisting torque at the other end, then the shafts are said to be connected in series as shown in Fig. (a). In such cases, each shaft transmits the same torque and the total angle of twist is equal to the sum of the angle of twists of the two shafts. Mathematically, total angle of twist,

If the shafts are made of the same material, then C1 = C2 = C. 55 Prepared by Kiran Kumar.K, Lecturer. (E-mail:- [email protected])


When the driving torque (T) is applied at the junction of the two shafts, and the resisting torques T1 and T2 at the other ends of the shafts, then the shafts are said to be connected in parallel, as shown in Fig. (b). In such cases, the angle of twist is same for both the shafts, i.e.

If the shafts are made of the same material, then C1 = C2

Assignment Question: 10.) Fig. below shows a shaft carrying a pulley A and a gear B and supported in two bearings C and D. The shaft transmits 20 kW at 150 r.p.m. The tangential force Ft on the gear B acts vertically upwards as shown. The pulley delivers the power through a belt to another pulley of equal diameter vertically below the pulley A. The ratio of tensions T1/T2 is equal to 2.5. The gear and the pulley weigh 900 N and 2700 N respectively. The permissible shear stress for the material of the shaft may be taken as 63 MPa. Assuming the weight of the shaft to be negligible in comparison with the other loads, determine its diameter. Take shock and fatigue factors for bending and torsion as 2 and 1.5 respectively.



UNIT-III DESIGN OF 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. The various important applications of springs are as follows: 1. 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. 2. To apply forces, as in brakes, clutches and spring loaded valves. 3. To control motion by maintaining contact between two elements as in cams and followers. 4. To measure forces, as in spring balances and engine indicators.

Types of Springs 1. Helical springs. The helical springs 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).

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: (a) These are easy to manufacture. (b) These are available in wide range. (c) These are reliable. (d) These have constant spring rate. (e) Their performance can be predicted more accurately. 2. Conical and volute springs. The conical and volute springs, as shown in Fig. are used in special applications where a telescoping spring or a spring with a spring rate that increases with the load is desired. The conical spring, as shown in Fig. (a), is wound with a uniform pitch whereas the volute springs, as shown in Fig. (b), are wound in the form of paraboloid with constant pitch and lead angles.



The springs may be made either partially or completely telescoping. In either case, the number of active coils gradually decreases. The decreasing number of coils results in an increasing spring rate. This characteristic is sometimes utilised in vibration problems where springs are used to support a body that has a varying mass. The major stresses produced in conical and volute springs are also shear stresses due to twisting. 3. Torsion springs. These 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.

4. 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.

5. Disc or bellevile 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. The major stresses produced in disc or bellevile springs are tensile and compressive stresses.

6. 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. 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. The helical springs are either cold formed or hot formed depending upon the size of the wire. Wires of small sizes (less than 10 mm diameter) are usually wound cold whereas larger size wires are wound hot. The strength of the wires varies with size, smaller size wires have greater strength and less ductility, due to the greater degree of cold working.

Standard Size of Spring Wire The standard size of spring wire may be selected from the following table:

Terms used in Compression Springs 1. Solid length. When the compression spring is 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 and the diameter of the wire. Mathematically, Solid length of the spring, LS = n'.d where n' = Total number of coils, and d = Diameter of the wire. 2. Free length. The free length of a compression spring, as shown in Fig, is the length of the spring in the free or unloaded condition. It is equal to the solid length plus the maximum deflection or compression of the spring and the clearance between the adjacent coils (when fully compressed). Mathematically,



Free length of the spring, LF = Solid length + Maximum compression + Clearance between adjacent coils (or clash allowance) In actual practice, the compression springs are seldom designed to close up under the maximum working load and for this purpose a clearance (or clash allowance) is provided between the adjacent coils to prevent closing of the coils during service. It may be taken as 15 per cent of the maximum deflection.

The following relation may also be used to find the free length of the spring, i.e. In this expression, the clearance between the two adjacent coils is taken as 1 mm. 3. Spring index. The spring index is defined as the ratio of the mean diameter of the coil to the diameter of the wire. Mathematically, where D = Mean diameter of the coil, and d = Diameter of the wire. 4. Spring rate. The spring rate (or stiffness or spring constant) is defined as the load required per unit deflection of the spring. Mathematically, where W = Load, and δ= Deflection of the spring. 5. Pitch. The pitch of the coil is defined as the axial distance between adjacent coils in uncompressed state. Mathematically,

The pitch of the coil may also be obtained by using the following relation, i.e.

where LF = Free length of the spring, LS = Solid length of the spring, n' = Total number of coils, and d = Diameter of the wire In choosing the pitch of the coils, the following points should be noted : (a) The pitch of the coils should be such that if the spring is accidently or carelessly compressed, the stress does not increase the yield point stress in torsion. (b) The spring should not close up before the maximum service load is reached. Note : In designing a tension spring, the minimum gap between two coils when the spring is in the free state is taken as 1 mm. Thus the free length of the spring, LF = n.d + (n – 1) and pitch of the coil,

Stresses in Helical Springs of Circular Wire Consider a helical compression spring made of circular wire and subjected to an axial load W, as shown in Fig. (a). Let D = Mean diameter of the spring coil, d = Diameter of the spring wire, n = Number of active coils, G = Modulus of rigidity for the spring material, W = Axial load on the spring, = Maximum shear stress induced in the wire, C = Spring index = D/d, p = Pitch of the coils, and δ= Deflection of the spring, as a result of an axial load W.



Now consider a part of the compression spring as shown in Fig. (b). The load W tends to rotate the wire due to the twisting moment ( T ) set up in the wire. Thus torsional shear stress is induced in the wire. A little consideration will show that part of the spring, as shown in Fig. 23.10 (b), is in equilibrium under the action of two forces W and the twisting moment T. We know that the twisting moment,

…….(i) The torsional shear stress diagram is shown in Fig. (a). In addition to the torsional shear stress (1) induced in the wire, the following stresses also act on the wire : 1. Direct shear stress due to the load W, and 2. Stress due to curvature of wire. We know that direct shear stress due to the load W,

……….(ii) The direct shear stress diagram is shown in Fig. (b) and the resultant diagram of torsional shear stress and direct shear stress is shown in Fig. (c).



We know that the resultant shear stress induced in the wire,

The positive sign is used for the inner edge of the wire and negative sign is used for the outer edge of the wire. Since the stress is maximum at the inner edge of the wire, therefore Maximum shear stress induced in the wire, = Torsional shear stress + Direct shear stress

... (Substituting D/d = C)

in the above equation gives …..(iii)

From the above equation, it can be observed that the effect of direct shear is appreciable for springs of small spring index C. Also we have neglected the effect of wire curvature in equation (iii). It may be noted that when the springs are subjected to static loads, the effect of wire curvature may be neglected, because yielding of the material will relieve the stresses. In order to consider the effects of both direct shear as well as curvature of the wire, a Wahl’s stress factor (K) introduced by A.M. Wahl may be used. The resultant diagram of torsional shear, direct shear and curvature shear stress is shown in Fig. (d). Maximum shear stress induced in the wire,

……(iv) Where wahl’s factor (K)

Note: The Wahl’s stress factor (K) may be considered as composed of two sub-factors, KS and KC, such that K = KS × KC where KS = Stress factor due to shear, and KC = Stress concentration factor due to curvature.

The values of K for a given spring index (C) may be obtained from the graph as shown in Fig. We see from Fig. that Wahl’s stress factor increases very rapidly as the spring index decreases. The spring mostly used in machinery have spring index above 3.



Deflection of Helical Springs of Circular Wire We know that Total active length of the wire, l = Length of one coil × No. of active coils = πD × n Let = Angular deflection of the wire when acted upon by the torque T. Axial deflection of the spring, ....(i) We also know that

Now substituting the values of l and J in the above equation, we have

......(ii) Substituting this value of in equation (i), we have

and the stiffness of the spring or spring rate,

Eccentric Loading of Springs Sometimes, the load on the springs does not coincide with the axis of the spring, i.e. the spring is subjected to an eccentric load. In such cases, not only the safe load for the spring reduces, the stiffness of the spring is also affected. The eccentric load on the spring increases the stress on one side of the spring and decreases on the other side. When the load is offset by a distance e from the spring axis, then the safe load on the spring may be obtained by multiplying the axial load by the factor where D is the mean diameter of the spring.



Buckling of Compression Springs It has been found experimentally that when the free length of the spring (LF) is more than four times the mean or pitch diameter (D), then the spring behaves like a column and may fail by buckling at a comparatively low load. The critical axial load (Wcr) that causes buckling may be calculated by using the following relation, i.e.

It may be noted that a hinged end spring is one which is supported on pivots at both ends as incase of springs having plain ends where as a built-in end spring is one in which a squared and ground end spring is compressed between two rigid and parallel flat plates. In order to avoid the buckling of spring, it is either mounted on a central rod or located on a tube. When the spring is located on a tube, the clearance between the tube walls and the spring should be kept as small as possible, but it must be sufficient to allow for increase in spring diameter during compression.

Surge in Springs When one end of a helical spring is resting on a rigid support and the other end is loaded suddenly, then all the coils of the spring will not suddenly deflect equally, because some time is required for the propagation of stress along the spring wire. A little consideration will show that in the beginning, the end coils of the spring in contact with the applied load takes up whole of the deflection and then it transmits a large part of its deflection to the adjacent coils. In this way, a wave of compression propagates through the coils to the supported end from where it is reflected back to the deflected end. This wave of compression travels along the spring indefinitely. If the applied load is of fluctuating type as in the case of valve spring in internal combustion engines and if the time interval between the load applications is equal to the time required for the wave to travel from one end to the other end, then resonance will occur. This results in very large deflections of the coils and correspondingly very high stresses. Under these conditions, it is just possible that the spring may fail. This phenomenon is called surge. It has been found that the natural frequency of spring should be atleast twenty times the frequency of application of a periodic load in order to avoid resonance. The natural frequency for springs clamped between two plates is given by

where d = Diameter of the wire, D = Mean diameter of the spring, n = Number of active turns, G = Modulus of rigidity, g = Acceleration due to gravity, and = Density of the material of the spring.



The surge in springs may be eliminated by using the following methods : 1. By using friction dampers on the centre coils so that the wave propagation dies out. 2. By using springs of high natural frequency. 3. By using springs having pitch of the coils near the ends different than at the centre to have different natural frequencies. Exercise Problems: 1.) A mechanism used in printing machinery consists of a tension spring assembled with a preload of 30 N. The wire diameter of spring is 2 mm with a spring index of 6. The spring has 18 active coils. The spring wire is hard drawn and oil tempered having following material properties: 2 Design shear stress = 680 MPa, Modulus of rigidity = 80 kN/mm Determine : 1. the initial torsional shear stress in the wire; 2. spring rate; and 3. the force to cause the body of the spring to its yield strength. 2 Given : Wi = 30 N ; d = 2 mm ; C = D/d = 6 ; n = 18 ; = 680 MPa = 680 N/mm ; G = 80× 103 1. Initial torsional shear stress in the wire We know that Wahl’s stress factor,

Initial torsional shear stress in the wire, = 143.5 MPa 2. Spring rate We know that spring rate (or stiffness of the spring),

3. Force to cause the body of the spring to its yield strength Let W = Force to cause the body of the spring to its yield strength. We know that design or maximum shear stress (),

W = 680 / 4.78 = 142.25 N 2.) 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. Given: W = 1000 N; δ= 25 mm ; C = D/d = 5 ; = 420 MPa = 420 N/mm2 ; G = 84 × 103 N/mm2 1. Mean diameter of the spring coil Let D = Mean diameter of the spring coil, and d = Diameter of the spring wire. We know that Wahl’s stress factor,

and maximum shear stress (),

From Table, we shall take a standard wire of size SWG 3 having diameter (d ) = 6.401 mm. Mean diameter of the spring coil, D = C.d = 5 d = 5 × 6.401 = 32.005 mm (∵ C = D/d = 5) and outer diameter of the spring coil, Do = D + d = 32.005 + 6.401 = 38.406 mm 2. Number of turns of the coils Let n = Number of active turns of the coils. We know that compression of the spring (δ),



For squared and ground ends, the total number of turns, n' = n + 2 = 14 + 2 = 16 3. Free length of the spring We know that free length of the spring = n'.d + δ+ 0.15*δ= 16 × 6.401 + 25 + 0.15 × 25 =131.2 mm 4. Pitch of the coil We know that pitch of the coil

3.) Design a valve spring of a petrol engine for the following operating conditions : Spring load when the valve is open = 400 N Spring load when the valve is closed = 250 N Maximum inside diameter of spring = 25 mm Length of the spring when the valve is open = 40 mm Length of the spring when the valve is closed = 50 mm Maximum permissible shear stress = 400 MPa Given : W1 = 400 N ; W2 = 250 N ; Di = 25 mm ; l1 = 40 mm ; l2 = 50 mm ; = 400 MPa = 400 N/mm2 1. Mean diameter of the spring coil Let d = Diameter of the spring wire in mm, And D = Mean diameter of the spring coil = Inside dia. of spring + Dia. of spring wire = (25 + d) mm Since the diameter of the spring wire is obtained for the maximum spring load (W1), therefore maximum twisting moment on the spring,

We know that maximum twisting moment (T ),

Solving this equation by hit and trial method, we find that d = 4.2 mm. From Table, we find that standard size of wire is SWG 7 having d = 4.47 mm. Now let us find the diameter of the spring wire by taking Wahl’s stress factor (K) into consideration. We know that spring index,

Wahl’s stress factor,

We know that the maximum shear stress (),

Taking larger of the two values, we have d = 4.54 mm From Table, we shall take a standard wire of size SWG 6 having diameter (d ) = 4.877 mm. Mean diameter of the spring coil, D = 25 + d = 25 + 4.877 = 29.877 mm and outer diameter of the spring coil, Do = D + d = 29.877 + 4.877 = 34.754 mm 2. Number of turns of the coil Let n = Number of active turns of the coil. We are given that the compression of the spring caused by a load of (W1 – W2), i.e. 400 – 250 = 150 N is l2 – l1, i.e. 50 – 40 = 10 mm. In other words, the deflection () of the spring is 10 mm for a load(W) of 150 N.



We know that the deflection of the spring (δ),

... (Taking G = 80 × 103 N/mm2)

n = 10 / 0.707 = 14.2 say 15 Taking the ends of the springs as squared and ground, the total number of turns of the spring, n' = 15 + 2 = 17 3. Free length of the spring Since the deflection for 150 N of load is 10 mm, therefore the maximum deflection for the maximum load of 400 N is

Free length of the spring,

4. Pitch of the coil We know that pitch of the coil

4.) A rail wagon of mass 20 tonnes 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. 2 Given: m = 20; t = 20 000 kg ; v = 2 m/s ; D = 300 mm ; δ= 250 mm ; = 600 MPa = 600 N/mm 1. Diameter of the spring wire Let d =Diameter of the spring wire. We know that kinetic energy of the wagon

……(i) Let W be the equivalent load which when applied gradually on each spring causes a deflection of 250 mm. Since there are two springs, therefore Energy stored in the springs

.............(ii) From equations (i) and (ii), we have 6 3 W = 40 × 10 / 250 = 160 × 10 N We know that torque transmitted by the spring,

We also know that torque transmitted by the spring (T ),

2. Number of turns of the spring coil Let n = Number of active turns of the spring coil. We know that the deflection of the spring (δ), 3

2

... (Taking G = 84 MPa = 84 × 10 N/mm )

n = 250 / 31.7 = 7.88 say 8 Assuming square and ground ends, total number of turns, n' = n + 2 = 8 + 2 = 10



3. Free length of the spring We know that free length of the spring, 4. Pitch of the coil We know that pitch of the coil

Springs in Series Consider two springs connected in series as shown in Fig.

Let W = Load carried by the springs, δ1 = Deflection of spring 1, δ2 = Deflection of spring 2, k1 = Stiffness of spring 1 = W / δ1, and k2 = Stiffness of spring 2 = W / δ2 A little consideration will show that when the springs are connected in series, then the total deflection produced by the springs is equal to the sum of the deflections of the individual springs. Total deflection of the springs,

where k = Combined stiffness of the springs.



Springs in Parallel Consider two springs connected in parallel as shown in Fig.

Let W = Load carried by the springs, W1 = Load shared by spring 1, W2 = Load shared by spring 2, k1 = Stiffness of spring 1, and k2 = Stiffness of spring 2. A little consideration will show that when the springs are connected in parallel, then the total deflection produced by the springs is same as the deflection of the individual springs. We know that W = W1 + W2 δ.k =δ.k1+δ.k2 k = k1 + k2 where k = Combined stiffness of the springs, and δ= Deflection produced.

Concentric or Composite Springs A concentric or composite spring is used for one of the following purposes : 1. To obtain greater spring force within a given space. 2. To insure the operation of a mechanism in the event of failure of one of the springs.

The concentric springs for the above two purposes may have two or more springs and have the same free lengths as shown in Fig. (a) and are compressed equally. Such springs are used in automobile clutches, valve springs in aircraft, heavy duty diesel engines and rail-road car suspension systems. Sometimes concentric springs are used to obtain a spring force which does not increase in a direct relation to the deflection but increases faster. Such springs are made of different lengths as shown in Fig.(b). The shorter spring begins to act only after the longer spring is compressed to a certain amount. These springs are used in governors of variable speed engines to take care of the variable centrifugal force.



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

Consider a single plate fixed at one end and loaded at the other end as shown in Fig. This plate may be used as a flat spring.

Let t = Thickness of plate, b = Width of plate, and L = Length of plate or distance of the load W from the cantilever end. We know that the maximum bending moment at the cantilever end A, M = W.L

Bending stress in such a spring,

…(i) The maximum deflection for a cantilever with concentrated load at the free end is given by

……..(ii)

It may be noted that due to bending moment, top fibres will be in tension and the bottom fibres are in compression, but the shear stress is zero at the extreme fibres and maximum at the centre, as shown in Fig. Hence for analysis, both stresses need not to be taken into account simultaneously. We shall consider the bending stress only.



If the spring is like a simply supported beam, with length 2L and load 2W in the centre, as shown in Fig, (Flat spring (simply supported beam type)

Maximum bending moment in the centre, M = W.L

We know that maximum deflection of a simply supported beam loaded in the centre is given by

From above we see that a spring such as automobile spring (semi-elliptical spring) with length 2L and loaded in the centre by a load 2W, may be treated as a double cantilever. If the plate of cantilever is cut into a series of n strips of width b and these are placed as shown in Fig, then equations (i) and (ii) may be written as

…..(iii) and …….(iv)

The above relations give the stress and deflection of a leaf spring of uniform cross-section. The stress at such a spring is maximum at the support. If a triangular plate is used as shown in Fig. (a), the stress will be uniform throughout. If this triangular plate is cut into strips of uniform width and placed one below the other, as shown in Fig. (b) to form a graduated or laminated leaf spring, then



….(v) …..(vi) where n = Number of graduated leaves. A little consideration will show that by the above arrangement, the spring becomes compact so that the space occupied by the spring is considerably reduced. When bending stress alone is considered, the graduated leaves may have zero width at the loaded end. But sufficient metal must be provided to support the shear. Therefore, it becomes necessary to have one or more leaves of uniform cross-section extending clear to the end. We see from equations (iv) and (vi) that for the same deflection, the stress in the uniform cross-section leaves (i.e. full length leaves) is 50% greater than in the graduated leaves, assuming that each spring element deflects according to its own elastic curve. If the suffixes F and G are used to indicate the full length (or uniform cross section) and graduated leaves, then

…….(vii) Adding 1 to both sides, we have

….(viii) where W = Total load on the spring = WG + WF WG = Load taken up by graduated leaves, and WF = Load taken up by full length leaves. From equation (vii), we may write

…….(ix) Bending stress for full length leaves,



The deflection in full length and graduated leaves is given by equation (iv), i.e.

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 a 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 centres 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. The longest leaf known as main leaf or master leafhas its ends formed in the shape of an eye through which the bolts are passed to secure the spring to its supports. Usually the eyes, through which the spring is attached to the hanger or shackle, are provided with bushings of some antifriction material such as bronze or rubber. The other leaves of the spring are known as graduated leaves. In order to prevent digging in the adjacent leaves, the ends of the graduated leaves are trimmed in various forms as shown in Fig. Since the master leaf has to with stand vertical bending loads as well as loads due to sideways of the vehicle and twisting, therefore due to the presence of stresses caused by these loads, it is usual to provide two full length leaves and the rest graduated leaves as shown in Fig. Rebound clips are located at intermediate positions in the length of the spring, so that the graduated leavesalso share the stresses induced in the full length leaves when the spring rebounds.

Equalised Stress in Spring Leaves (Nipping) We have already discussed that the stress in the full length leaves is 50% greater than the stress in the graduated leaves. In order to utilise the material to the best advantage, all the leaves should be equally stressed. This condition may be obtained in the following two ways : 1. By making the full length leaves of smaller thickness than the graduated leaves. In this way, the full length leaves will induce smaller bending stress due to small distance from the neutral axis to the edge of the leaf. 2. By giving a greater radius of curvature to the full length leaves than graduated leaves, as shown in Fig, before the leaves are assembled to form a spring. By doing so, a gap or clearance will be left



between the leaves. This initial gap, as shown by C in Fig, is called nip. When the central bolt, holding the various leaves together, is tightened, the full length leaf will bend back as shown dotted in Fig. and have an initial stress in a direction opposite to that of the normal load. The graduated leaves will have an initial stress in the same direction as that of the normal load. When the load is gradually applied to the spring, the full length leaf is first relieved of this initial stress and then stressed in opposite direction. Consequently, the full length leaf will be stressed less than the graduated leaf. The initial gap between the leaves may be adjusted so that under maximum load condition the stress in all the leaves is equal, or if desired, the full length leaves may have the lower stress. This is desirable in automobile springs in which full length leaves are designed for lower stress because the full length leaves carry additional loads caused by the swaying of the car, twisting and in some cases due to driving the car through the rear springs. Let us now find the value of initial gap or nip C. Consider that under maximum load conditions, the stress in all the leaves is equal. Then at maximum load, the total deflection of the graduated leaves will exceed the deflection of the full length leaves by an amount equal to the initial gap C. In other words,

……(i) Since the stresses are equal, therefore

and Substituting the values of WG and WF in equation (i), we have

……….(ii) The load on the clip bolts (Wb) required to close the gap is determined by the fact that the gap is equal to the initial deflections of full length and graduated leaves.

........(iii) The final stress in spring leaves will be the stress in the full length leaves due to the applied load minus the initial stress. Final stress,



……(iv)

... (Substituting n = nF + nG)

Note : 1. The final stress in the leaves is also equal to the stress in graduated leaves due to the applied load plus the initial stress. 2. The deflection in the spring due to the applied load is same as without initial stress. Length of Leaf Spring Leaves The length of the leaf spring leaves may be obtained as discussed below : Let 2L1 = Length of span or overall length of the spring, l = Width of band or distance between centres of U-bolts. It is the ineffective length of the spring, nF = Number of full length leaves, nG = Number of graduated leaves, and n = Total number of leaves = nF + nG. We have already discussed that the effective length of the spring,

Materials for Leaf Springs 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. For automobiles : 50 Cr 1, 50 Cr 1 V 23, and 55 Si 2 Mn 90 all used in hardened and tempered state.

Exercise Problems: 5.) A truck spring has 12 number 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. Given : n = 12 ; nF = 2 ; 2L1 = 1.05 m = 1050 mm ; l = 85 mm ; 2W = 5.4 kN = 5400 N or W = 2700 N ; F =280 MPa = 280 N/mm2 Thickness and width of the spring leaves Let t = Thickness of the leaves, and b = Width of the leaves. Since it is given that the ratio of the total depth of the spring (n × t) and width of the spring (b) is 3, therefore

We know that the effective length of the spring,

and number of graduated leaves,

nG = n – nF = 12 – 2 = 10



Assuming that the leaves are not initially stressed, therefore maximum stress or bending stress for full length leaves (F),

and b = 4 t = 4 × 10 = 40 mm Deflection of the spring We know that deflection of the spring,

3

2

... (Taking E = 210 × 10 N/mm )

= 16.7 mm 6.) A locomotive semi-elliptical laminated spring has an overall length of 1 m and sustains a load of 70 kN at its centre. The spring has 3 full length leaves and 15 graduated leaves with a central band of 100 mm width. All the leaves are to be stressed to 400 MPa, when fully loaded. The ratio of the total spring depth to that of width is 2. E = 210 kN/mm2. Determine : 1. The thickness and width of the leaves. 2. The initial gap that should be provided between the full length and graduated leaves before the band load is applied. 3. The load exerted on the band after the spring is assembled. Given : 2L1 =1 m =1000 mm ; 2W = 70 kN or W = 35 kN = 35 × 103 N ; nF = 3 ; nG = 15 ; l = 100 mm; 2 2 3 2 = 400 MPa = 400 N/mm ; E = 210 kN/mm = 210 × 10 N/mm 1. Thickness and width of leaves Let t = Thickness of leaves, and b = Width of leaves. We know that the total number of leaves, n = nF + nG = 3 + 15 = 18 Since it is given that ratio of the total spring depth (n × t) and width of leaves is 2, therefore

We know that the effective length of the leaves, 2L = 2L1 – l = 1000 – 100 = 900 mm or L = 900 / 2 = 450 mm Since all the leaves are equally stressed, therefore final stress (),

and b = 9 t = 9 × 12 = 108 mm

2. Initial gap We know that the initial gap (C) that should be provided between the full length and graduated leaves before the band load is applied, is given by

3. Load exerted on the band after the spring is assembled We know that the load exerted on the band after the spring is assembled,



UNIT IV DESIGN OF FLYWHEELS A flywheel used in machines serves as a reservoir which stores energy during the period when the supply of energy is more than the requirement and releases it during the period when the requirement of energy is more than supply. In case of steam engines, internal combustion engines, reciprocating compressors and pumps, the energy is developed during one stroke and the engine is to run for the whole cycle on the energy produced during this one stroke. For example, in I.C. engines, the energy is developed only during power stroke which is much more than the engine load, and no energy is being developed during suction, compression and exhaust strokes in case of four stroke engines and during compression in case of two stroke engines. The excess energy developed during power stroke is absorbed by the flywheel and releases it to the crankshaft during other strokes in which no energy is developed, thus rotating the crankshaft at a uniform speed. A little consideration will show that when the flywheel absorbs energy, its speed increases and when it releases, the speed decreases. Hence a flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In machines where the operation is intermittent like punching machines, shearing machines, riveting machines, crushers etc., the flywheel stores energy from the power source during the greater portion of the operating cycle and gives it up during a small period of the cycle. Thus the energy from the power source to the machines is supplied practically at a constant rate throughout the operation. Note: The function of a governor in engine is entirely different from that of a flywheel. It regulates the mean speed of an engine when there are variations in the load, e.g. when the load on the engine increases, it becomes necessary to increase the supply of working fluid. On the other hand, when the load decreases, less working fluid is required. The governor automatically controls the supply of working fluid to the engine with the varying load condition and keeps the mean speed within certain limits. As discussed above, the flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In other words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation. It does not control the speed variations caused by the varying load. Coefficient of Fluctuation of Speed The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed. Let N1 = Maximum speed in r.p.m. during the cycle, N2 = Minimum speed in r.p.m. during the cycle, and

Coefficient of fluctuation of speed,

The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. It variesdepending upon the nature of service to which the flywheel is employed. Note: The reciprocal of coefficient of fluctuation of speed is known as coefficient of steadiness and it is denoted by m.



Fluctuation of Energy The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider a turning moment diagram for a single cylinder double acting steam engine as shown in Fig. The vertical ordinate represents the turning moment and the horizontal ordinate (abscissa) represents the crank angle. A little consideration will show that the turning moment is zero when the crank angle is zero. It rises to a maximum value when crank angle reaches 90º and it is again zero when crank angle is 180º. This is shown by the curve abc in Fig. and it represents the turning moment diagram for outstroke. The curve cde is the turning moment diagram for instroke and is somewhat similar to the curve abc. Since the work done is the product of the turning moment and the angle turned, therefore the area of the turning moment diagram represents the work done per revolution. In actual practice, the engine is assumed to work against the mean resisting torque, as shown by a horizontal line AF. The height of the ordinate aA represents the mean height of the turning moment diagram. Since it is assumed that the work done by the turning moment per revolution is equal to the work done against the mean resisting torque, therefore the area of the rectangle aA Fe is proportional to the work done against the mean resisting torque.

We see in Fig, that the mean resisting torque line AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from ‘a’ to ‘p’ the work done by the engine is equal to the area aBp, whereas the energy required is represented by the area aABp. In other words, the engine has done less work (equal to the area aAB) than the requirement. This amount of energy is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of energy is represented by the area pBCq. Therefore the engine has done more work than the requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the flywheel increases while the crank moves from p to q. Similarly when the crank moves from q to r, more work is taken from the engine than is developed. This loss of work is represented by the area CcD. To supply this loss, the flywheel gives up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank moves from r to s, excess energy is again developed given by the area DdE and the speed again increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The variations of energy above and below the mean resisting torque line are called fluctuation of energy. The areas BbC, CcD, DdE etc. represent fluctuations of energy.



A little consideration will show that the engine has a maximum speed either at q or at s. This is due to the fact that the flywheel absorbs energy while the crank moves from p to q and from r to s. On the other hand, the engine has a minimum speed either at p or at r. The reason is that the flywheel gives out some of its energy when the crank moves from a to p and from q to r. The difference between the maximum and the minimum energies is known as maximum fluctuation of energy. A turning moment diagram for a four stroke internal combustion engine is shown in Fig. We know that in a four stroke internal combustion engine, there is one working stroke after the crank has turned through 720º (or 4πradians). Since the pressure inside the engine cylinder is less than the atmospheric pressure during suction stroke, therefore a negative loop is formed as shown in Fig. During the compression stroke, the work is done on the gases, therefore a higher negative loop is obtained. In the working stroke, the fuel burns and the gases expand, therefore a large positive loop is formed. During exhaust stroke, the work is done on the gases, therefore a negative loop is obtained. A turning moment diagram for a compound steam engine having three cylinders and the resultant turning moment diagram is shown in Fig. The resultant turning moment diagram is the sum of the turning moment diagrams for the three cylinders. It may be noted that the first cylinder is the high pressure cylinder, second cylinder is the intermediate cylinder and the third cylinder is the low pressure cylinder. The cranks, in case of three cylinders are usually placed at 120º to each other.



Maximum Fluctuation of Energy A turning moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig. The horizontal line AG represents the mean torque line. Let a1, a3, a5 be the areas above the mean torque line and a2, a4 and a6 be the areas below the mean torque line. These areas represent some quantity of energy which is either added or subtracted from the energy of the moving parts of the engine.

Let the energy in the flywheel at A = E, then from Fig., we have Energy at B = E + a1 Energy at C = E + a1 – a2 Energy at D = E + a1 – a2 + a3 Energy at E = E + a1 – a2 + a3 – a4 Energy at F = E + a1 – a2 + a3 – a4 + a5 Energy at G = E + a1 – a2 + a3 – a4 + a5 – a6 = Energy at A Let us now suppose that the maximum of these energies is at B and minimum at E. Maximum energy in the flywheel = E + a1 and minimum energy in the flywheel = E + a1 – a2 + a3 – a4 Maximum fluctuation of energy, E = Maximum energy – Minimum energy = (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4 Coefficient of Fluctuation of Energy It is defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is usually denoted by CE. Mathematically, coefficient of fluctuation of energy,

The workdone per cycle may be obtained by using the following relations: 1. Workdone / cycle = Tmean ×  where Tmean = Mean torque, and = Angle turned in radians per revolution = 2π, in case of steam engines and two stroke internal combustion engines. = 4π, in case of four stroke internal combustion engines. The mean torque (Tmean) in N-m may be obtained by using the following relation i.e.

where P = Power transmitted in watts, N = Speed in r.p.m., and ω= Angular speed in rad/s = 2πN / 60 2. The workdone per cycle may also be obtained by using the following relation: Workdone / cycle =P × 60/n where n = Number of working strokes per minute. = N, in case of steam engines and two stroke internal combustion engines. = N / 2, in case of four stroke internal combustion engines.



Energy Stored in a Flywheel A flywheel is shown in Fig. We have already discussed that when a flywheel absorbs energy its speed increases and when it gives up energy its speed decreases.

Let m = Mass of the flywheel in kg, k = Radius of gyration of the flywheel in metres, I = Mass moment of inertia of the flywheel about the axis of rotation in kg-m2 = m.k2, N1 and N2 = Maximum and minimum speeds during the cycle in r.p.m., ω1 and ω 2 = Maximum and minimum angular speeds during the cycle in rad/s, N = Mean speed during the cycle in r.p.m. = (N1+N2)/2 ω= Mean angular speed during the cycle in rad / s = (ω1+ ω2)/2 CS = Coefficient of fluctuation of speed = (N1-N2)/N (or) (ω1-ω2)/ω We know that mean kinetic energy of the flywheel,

….(i)

.......(ii) …….(iii) The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because the thickness of rim is very small as compared to the diameter of rim. Therefore substituting k = R in equation (ii), we have From this expression, the mass of the flywheel rim may be determined. Note: 1. In the above expression, only the mass moment of inertia of the rim is considered and the mass moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion of weight of the flywheel is in the rim and a small portion is in the hub and arms. Also the hub and arms are nearer to the axis of rotation, therefore the moment of inertia of the hub and arms is very small. 2. The density of cast iron may be taken as 7260 kg/m3 and for caststeel, it may taken as 7800 kg /m3.



3. The mass of the flywheel rim is given by m = Volume × Density = 2πR×A×. From this expression, we may find the value of the cross-sectional area of the rim. Assuming the cross-section of the rim to be rectangular, then A=b×t where b = Width of the rim, and t = Thickness of the rim. Knowing the ratio of b / t which is usually taken as 2, we may find the width and thickness of rim. 4. When the flywheel is to be used as a pulley, then the width of rim should be taken 20 to 40 mm greater than the width of belt.

Stresses in a Flywheel Rim A flywheel, as shown in Fig, consists of a rim at which the major portion of the mass or weight of flywheel is concentrated, a boss or hub for fixing the flywheel on to the shaft and a number of arms for supporting the rim on the hub. The following types of stresses are induced in the rim of a flywheel: 1. Tensile stress due to centrifugal force, 2. Tensile bending stress caused by the restraint of the arms, and 3. The shrinkage stresses due to unequal rate of cooling of casting. These stresses may be very high but there is no easy method of determining. This stress is taken care of by a factor of safety. We shall now discuss the first two types of stresses as follows: 1. Tensile stress due to the centrifugal force The tensile stress in the rim due to the centrifugal force, assuming that the rim is unstrained by the arms, is determined in a similar way as a thin cylinder subjected to internal pressure. Let b = Width of rim, t = Thickness of rim, A = Cross-sectional area of rim = b × t, D = Mean diameter of flywheel R = Mean radius of flywheel, = Density of flywheel material, ω= Angular speed of flywheel, v = Linear velocity of flywheel, and t = Tensile or hoop stress.



2. Tensile bending stress caused by restraint of the arms The tensile bending stress in the rim due to the restraint of the arms is based on the assumption that each portion of the rim between a pair of arms behaves like a beam fixed at both ends and uniformly loaded, as shown in Fig., such that length between fixed ends,

The uniformly distributed load (w) per metre length will be equal to the centrifugal force between a pair of arms. We know that maximum bending moment,

Bending stress,

...........(Substituting ω= v/R) Now total stress in the rim,

If the arms of a flywheel do not stretch at all and are placed very close together, then centrifugal force will not set up stress in the rim. In other words, t will be zero. On the other hand, if the arms are stretched enough to allow free expansion of the rim due to centrifugal action, there will be no restraint due to the arms, i.e. b will be zero. It has been shown by G. Lanza that the arms of a flywheel stretch about ¾ th of the amount necessary for free expansion. Therefore the total stress in the rim,



Stresses in Flywheel Arms The following stresses are induced in the arms of a flywheel. 1. Tensile stress due to centrifugal force acting on the rim. 2. Bending stress due to the torque transmitted from the rim to the shaft or from the shaft to the rim. 3.Shrinkage stresses due to unequal rate of cooling of casting. These stresses are difficult to determine. We shall now discuss the first two types of stresses as follows: 1. Tensile stress due to the centrifugal force Due to the centrifugal force acting on the rim, the arms will be subjected to direct tensile stress whose magnitude is same as discussed in the previous article. Tensile stress in the arms,

2. Bending stress due to the torque transmitted Due to the torque transmitted from the rim to the shaft or from the shaft to the rim, the arms will be subjected to bending, because they are required to carry the full torque load. In order to find out the maximum bending moment on the arms, it may be assumed as a centilever beam fixed at the hub and carrying a concentrated load at the free end of the rim as shown in Fig.

Let T = Maximum torque transmitted by the shaft, R = Mean radius of the rim, r = Radius of the hub, n = Number of arms, and Z = Section modulus for the cross-section of arms. We know that the load at the mean radius of the rim

and maximum bending moment which lies on the arm at the hub,

Bending stress in arms,

Total tensile stress in the arms at the hub end,



Design of Flywheel Arms The cross-section of the arms is usually elliptical with major axis as twice the minor axis, as shown in Fig., and it is designed for the maximum bending stress.

Let a1 = Major axis, and b1 = Minor axis. Section modulus, .....................(i) We know that maximum bending moment,

Maximum bending stress, ................(ii) Assuming a1 = 2 b1, the dimensions of the arms may be obtained from equations (i) and (ii). Note: 1. The arms of the flywheel have a taper from the hub to the rim. The taper is about 20 mm per metre length of the arm for the major axis and 10 mm per metre length for the minor axis. 2. The number of arms are usually 6. Sometimes the arms may be 8, 10 or 12 for very large size flywheels. 3. The arms may be curved or straight. But straight arms are easy to cast and are lighter. 4. Since arms are subjected to reversal of stresses, therefore a minimum factor of safety 8 should be used. In some cases like punching machines amd machines subjected to severe shock, a factor of safety 15 may be used. 5. The smaller flywheels (less than 600 mm diameter) are not provided with arms. They are made web type with holes in the web to facilitate handling.

Design of Shaft, Hub and Key The diameter of shaft for flywheel is obtained from the maximum torque transmitted. We know that the maximum torque transmitted,

where d1 = Diameter of the shaft, and = Allowable shear stress for the material of the shaft. The hub is designed as a hollow shaft, for the maximum torque transmitted. We know that the maximum torque transmitted,

where d = Outer diameter of hub, and d1 = Inner diameter of hub or diameter of shaft. The diameter of hub is usually taken as twice the diameter of shaft and length from 2 to 2.5 times the shaft diameter. It is generally taken equal to width of the rim. A standard sunk key is used for the shaft and hub. The length of key is obtained by considering the failure of key in shearing. We know that torque transmitted by shaft,

where L = Length of the key, = Shear stress for the key material, and d1 = Diameter of shaft.



Construction of Flywheels The flywheels of smaller size (upto 600 mm diameter) are casted in one piece. The rim and hub are joined together by means of web as shown in Fig.(a). The holes in the web may be made for handling purposes. In case the flywheel is of larger size (upto 2.5 metre diameter), the arms are made instead of web, as shown in Fig.(b). The number of arms depends upon the size of flywheel and its speed of rotation. But the flywheels above 2.5 metre diameter are usually casted in two piece. Such a flywheel is known as split flywheel. A split flywheel has the advantage of relieving the shrinkage stresses in the arms due to unequal rate of cooling of casting. A flywheel made in two halves should be spilt at the arms rather than between the arms, in order to obtain better strength of the joint. The two halves of the flywheel are connected by means of bolts through the hub, as shown in Fig. The two halves are also joined at the rim by means of cotter joint (as shown in Fig.). The width or depth of the shrink link is taken as 1.25 to 1.35 times the thickness of link. The slot in the rim into which the link is inserted is made slightly larger than the size of link.

Split Type Flywheel



Exercise Problems: 1.) A multi-cylinder engine is to run at a constant load at a speed of 600 r.p.m. On drawing the crank effort diagram to a scaleof 1 m = 250 N-m and 1 mm = 3º, the areas in sq mm above and below the mean torque line are as follows: + 160, – 172, + 168, – 191, + 197, – 162 sq mm The speed is to be kept within ± 1% of the mean speed of the engine. Calculate the necessary moment of inertia of the flywheel. Determine suitable dimensions for cast iron flywheel with a rim whose breadth is twice its radial thickness. The density of cast iron is 7250 kg / m 3, and its working stress in tension is 6 MPa. Assume that the rim contributes 92% of the flywheel effect. Given : N = 600 r.p.m (or) ω= 2π× 600 / 60 = 62.84 rad/s ; = 7250 kg/m3 ; t = 6 MPa=6×106 N/m2 Moment of inertia of the flywheel Let I = Moment of inertia of the flywheel. First of all, let us find the maximum fluctuation of energy. The turning moment diagram is shown in Fig.

2

1 mm on the turning moment diagram

Let the total energy at A = E. Therefore from Fig, we find that Energy at B = E + 160 Energy at C = E + 160 – 172 = E – 12 Energy at D = E – 12 + 168 = E + 156 Energy at E = E + 156 – 191 = E – 35 Energy at F = E – 35 + 197 = E + 162 Energy at G = E + 162 – 162 = E = Energy at A From above, we find that the energy is maximum at F and minimum at E. Maximum energy = E + 162 and minimum energy = E – 35 We know that the maximum fluctuation of energy, E = Maximum energy – Minimum energy = (E + 162) – (E – 35) = 197mm2 = 197×13.1 = 2581 N-m Since the fluctuation of speed is ± 1% of the mean speed (ω), therefore total fluctuation of speed,  and coefficient of fluctuation of speed,

We know that the maximum fluctuation of energy (E),



Dimensions of a flywheel rim Let t = Thickness of the flywheel rim in metres, and b = Breadth of the flywheel rim in metres = 2 t ...(Given) First of all let us find the peripheral velocity (v) and mean diameter (D) of the flywheel. We know that tensile stress (t ),

We also know that peripheral velocity (v),

Now let us find the mass of the flywheel rim. Since the rim contributes 92% of the flywheel effect, therefore the energy of the flywheel rim (Erim) will be 0.92 times the total energy of the flywheel (E). We know that maximum fluctuation of energy (E),

and energy of the flywheel rim, Let m = Mass of the flywheel rim. We know that energy of the flywheel rim (Erim),

Note: The mass of the flywheel rim may also be obtained by using the following relations. Since the rim contributes 92% of the flywheel effect, therefore using the relation below

We also know that mass of the flywheel rim (m),

t2 = 143.5 / 41 686 = 0.003 44 t = 0.0587 say 0.06 m = 60 mm and b = 2 t = 2 × 60 = 120 mm

2.) Design a cast iron flywheel used for a four stroke I.C engine developing 180 kW at 240 r.p.m. The hoop or centrifugal stress developed in the flywheel is 5.2 MPa, the total fluctuation of speed is to be limited to 3% of the mean speed. The work done during the power stroke is 1/3 more than the average work done during the whole cycle. The maximum torque on the shaft is twice the mean torque. The 3 density of cast iron is 7220 kg/m . 3 6 2 Given: P = 180 kW = 180 × 10 W; N = 240 r.p.m. ; t = 5.2 MPa = 5.2 × 10 N/m ; N1 – N2 = 3% N ; 3 = 7220 kg/m First of all, let us find the maximum fluctuation of energy (E). The turning moment diagram of a four stroke engine is shown in Fig. We know that mean torque transmitted by the flywheel,

workdone per cycle = Tmean × = 7161 × 4π= 90 000 N-m



Note: The workdone per cycle may also be obtained as discussed below :

Since the workdone during the power stroke is 1/3 more than the average workdone during the whole cycle, therefore, Workdone during the power (or working) stroke

……(i) The workdone during the power stroke is shown by a triangle ABC in Fig. in which the base AC = πradians and height BF = Tmax.

Workdone during power stroke = 1/2 × π× Tmax From equations (i) and (ii), we have

………(ii)

Height above the mean torque line, Since the area BDE shown shaded in Fig. 22.18 above the mean torque line represents the maximum fluctuation of energy (E), therefore from geometrical relation,

Maximum fluctuation of energy (i.e. area of BDE)

Note: The approximate value of maximum fluctuation of energy may be obtained as discussed below : Workdone per cycle = 90 000 N-mm ...(as calculated above) Workdone per stroke = 90 000 / 4 = 22 500 N-m ...(Q of four stroke engine) and workdone during power stroke = 120 000 N-m Maximum fluctuation of energy, E = 120 000 – 22 500 = 97 500 N-m



1. Diameter of the flywheel rim Let D = Diameter of the flywheel rim in metres, and v = Peripheral velocity of the flywheel rim in m/s. We know that the hoop stress developed in the flywheel rim ( t),

We also know that peripheral velocity (v),

2. Mass of the flywheel rim Let m = Mass of the flywheel rim in kg. We know that angular speed of the flywheel rim,

and coefficient of fluctuation of speed,

We know that maximum fluctuation of energy (E),

3. Cross-sectional dimensions of the rim Let t = Depth or thickness of the rim in metres, and b = Width of the rim in metres = 2 t ...(Assume) Cross-sectional area of the rim, We know that mass of the flywheel rim (m),

4. Diameter and length of hub Let d = Diameter of the hub, d1 = Diameter of the shaft, and l = Length of the hub. Since the maximum torque on the shaft is twice the mean torque, therefore maximum torque acting on the shaft, We know that the maximum torque acting on the shaft (Tmax),

d1 = 122 say 125 m The diameter of the hub is made equal to twice the diameter of shaft and length of hub is equal to width of the rim. d = 2 d1 = 2 × 125 = 250 mm = 0.25 m and l = b = 470 mm = 0.47 mm



5. Cross-sectional dimensions of the elliptical arms Let a1 = Major axis, b1 = Minor axis = 0.5 a1 ...(Assume) n = Number of arms = 6 ...(Assume) 2 b = Bending stress for the material of arms = 15 MPa = 15 N/mm ...(Assume) We know that the maximum bending moment in the arm at the hub end, which is assumed as cantilever is given by

and section modulus for the cross-section of the arm,

We know that the bending stress (b),

6. Dimensions of key The standard dimensions of rectangular sunk key for a shaft of diameter 125 mm are as follows: Width of key, w = 36 mm and thickness of key = 20 mm The length of key (L) is obtained by considering the failure of key in shearing. We know that the maximum torque transmitted by the shaft (Tmax),

Let us now check the total stress in the rim which should not be greater than 15 MPa. We know that total stress in the rim,

Since it is less than 15 MPa, therefore the design is safe. 3.) A single cylinder double acting steam engine delivers 185 kW at 100 r.p.m. The maximum fluctuation of energy per revolution is 15 per cent of the energy developed per revolution. The speed variation is limited to 1 per cent either way from the mean. The mean diameter of the rim is 2.4 m. Design the flywheel used in this application. Given : P = 185 kW = 185 × 103 W ; N = 100 r.p.m ; E = 15% E = 0.15 E ; D = 2.4 m (or) R = 1.2 m 1. Mass of the flywheel rim Let m = Mass of the flywheel rim in kg. We know that the work done (or) energy developed per revolution,



Maximum fluctuation of energy Since the speed variation is 1% either way from the mean, therefore the total fluctuation of speed, N1 – N2 = 2% of mean speed = 0.02 N and coefficient of fluctuation of speed,

Velocity of the flywheel,

We know that the maximum fluctuation of energy (E),

2. Cross-sectional dimensions of the flywheel rim Let t = Thickness of the flywheel rim in metres, and b = Width of the flywheel rim in metres = 2 t ...(Assume) Cross-sectional area of the rim, We know that mass of the flywheel rim (m),

3. Diameter and length of hub Let d = Diameter of the hub, d1 = Diameter of the shaft, and l = Length of the hub, We know that mean torque transmitted by the shaft,

Assuming that the maximum torque transmitted (Tmax) by the shaft is twice the mean torque, therefore 6 Tmax = 2 × Tmean = 2 × 17 664 = 35328 N-m = 35.328 × 10 N-mm We also know that maximum torque transmitted by the shaft (Tmax),

The diameter of the hub (d) is made equal to twice the diameter of the shaft (d1) and length of the hub (l ) is equal to the width of the rim (b). d = 2 d1 = 2 × 165 = 330 mm ; and l = b = 440 mm 4. Cross-sectional dimensions of the elliptical arms Let a1 = Major axis, b1 = Minor axis = 0.5 a1 ...(Assume) n = Number of arms = 6 ...(Assume) 2 b = Bending stress for the material of the arms = 14 MPa = 14 N/mm ...(Assume)



We know that the maximum bending moment in the arm at the hub end which is assumed as cantilever is given by

section modulus for the cross-section of the arm,

We know that the bending stress (b),

a1 = 193.6 say 200 mm and b1 = 0.5a1 = 0.5 × 200 = 100 mm 5. Dimensions of key The standard dimensions of rectangular sunk key for a shaft of 165 mm diameter are as follows: Width of key, w = 45 mm and thickness of key, t = 25 mm The length of key (L) is obtained by considering the failure of key in shearing. We know that the maximum torque transmitted by the shaft (Tmax),

Let us now check the total stress in the rim which should not be greater than 14 MPa. We know that the total stress in the rim,

Since it is less than 14 MPa, therefore the design is safe.



UNIT-V DESIGN OF I.C. ENGINE PARTS As the name implies, the internal combustion engines (briefly written as I. C. engines) are those engines in which the combustion of fuel takes place inside the engine cylinder. The I.C. engines use either petrol or diesel as their fuel. In petrol engines (also called spark ignition engines or S.I engines), the correct proportion of air and petrol is mixed in the carburettor and fed to engine cylinder where it is ignited by means of a spark produced at the spark plug. In diesel engines (also called compression ignition engines or C.I engines), only air is supplied to the engine cylinder during suction stroke and it is compressed to a very high pressure, thereby raising its temperature from 600°C to 1000°C. The desired quantity of fuel (diesel) is now injected into the engine cylinder in the form of a very fine spray and gets ignited when comes in contact with the hot air. The operating cycle of an I.C. engine may be completed either by the two strokes or four strokes of the piston. Thus, an engine which requires two strokes of the piston or one complete revolution of the crankshaft to complete the cycle, is known as two stroke engine. An engine which requires four strokes of the piston or two complete revolutions of the crankshaft to complete the cycle, is known as four stroke engine. The two stroke petrol engines are generally employed in very light vehicles such as scooters, motor cycles and three wheelers. The two stroke diesel engines are generally employed in marine propulsion. The four stroke petrol engines are generally employed in cars, jeeps and also in aeroplanes. The four stroke diesel engines are generally employed in heavy duty vehicles such as buses, trucks, tractors, diesel locomotive and in the earth moving machinery.

Principal Parts of an Engine The principal parts of an I.C engine, as shown in Fig. are as follows : 1. Cylinder and cylinder liner, 2. Piston, piston rings and piston pin or gudgeon pin, 3. Connecting rod with small and big end bearing, 4. Crank, crankshaft and crank pin, and 5. Valve gear mechanism.



Cylinder and Cylinder Liner The function of a cylinder is to retain the working fluid and to guide the piston. The cylinders are usually made of cast iron or cast steel. Since the cylinder has to withstand high temperature due to the combustion of fuel, therefore, some arrangement must be provided to cool the cylinder. The single cylinder engines (such as scooters and motorcycles) are generally air cooled. They are provided with fins around the cylinder. The multi-cylinder engines (such as of cars) are provided with water jackets around the cylinders to cool it. In smaller engines the cylinder, water jacket and the frame are made as one piece, but for all the larger engines, these parts are manufactured separately. The cylinders are provided with cylinder liners so that in case of wear, they can be easily replaced. The cylinder liners are of the following two types: 1. Dry liner, and 2. Wet liner.

A cylinder liner which does not have any direct contact with the engine cooling water, is known as dry liner, as shown in above Fig.(a). A cylinder liner which have its outer surface in direct contact with the engine cooling water, is known as wet liner, as shown in above Fig. (b). The cylinder liners are made from good quality close grained cast iron (i.e. pearlitic cast iron), nickel cast iron, nickel chromium cast iron. In some cases, nickel chromium cast steel with molybdenum may be used. The inner surface of the liner should be properly heat-treated in order to obtain a hard surface to reduce wear.

Design of a Cylinder In designing a cylinder for an I. C. engine, it is required to determine the following values : 1. Thickness of the cylinder wall. The cylinder wall is subjected to gas pressure and the piston side thrust. The gas pressure produces the following two types of stresses : (a) Longitudinal stress, and (b) Circumferential stress.



Since these two stressess act at right angles to each other, therefore, the net stress in each direction is reduced. The piston side thrust tends to bend the cylinder wall, but the stress in the wall due to side thrust is very small and hence it may be neglected. Let D0 = Outside diameter of the cylinder in mm, D = Inside diameter of the cylinder in mm, p = Maximum pressure inside the engine cylinder in N/mm2, t = Thickness of the cylinder wall in mm, and 1/m = Poisson’s ratio. It is usually taken as 0.25. The apparent longitudinal stress is given by

and the apparent circumferential stresss is given by

... (where l is the length of the cylinder and area is the projected area)

The thickness of a cylinder wall (t) is usually obtained by using a thin cylindrical formula,i.e.,

where p = Maximum pressure inside the cylinder in N/mm2, D = Inside diameter of the cylinder or cylinder bore in mm, c = Permissible circumferential or hoop stress for the cylinder material in MPa or N/mm2. Its value may be taken from 35 MPa to 100 MPa depending upon the size and material of the cylinder. C = Allowance for reboring. The allowance for reboring (C ) depending upon the cylinder bore (D) for I. C. engines is given in the following table :

The thickness of the cylinder wall usually varies from 4.5 mm to 25 mm or more depending upon the size of the cylinder. The thickness of the cylinder wall (t) may also be obtained from the following empirical relation, i.e. t = 0.045 D + 1.6 mm Thickness of the dry liner = 0.03 D to 0.035 D Thickness of the water jacket wall = 0.032 D + 1.6 mm or t / 3 m for bigger cylinders and 3t /4 for smaller cylinders Water space between the outer cylinder wall and inner jacket wall = 10 mm for a 75 mm cylinder to 75 mm for a 750 mm cylinder (or) 0.08 D + 6.5 mm 2. Bore and length of the cylinder. The bore (i.e. inner diameter) and length of the cylinder may be determined as discussed below : Let pm = Indicated mean effective pressure in N/mm2, D = Cylinder bore in mm, 2 A = Cross-sectional area of the cylinder in mm2 = πD /4 l = Length of stroke in metres, N = Speed of the engine in r.p.m., and n = Number of working strokes per min = N, for two stroke engine = N/2, for four stroke engine.



We know that the power produced inside the engine cylinder, i.e. indicated power,

From this expression, the bore (D) and length of stroke (l) is determined. The length of stroke is generally taken as 1.25 D to 2D. Since there is a clearance on both sides of the cylinder, therefore length of the cylinder is taken as 15 percent greater than the length of stroke. In other words, Length of the cylinder, L = 1.15 × Length of stroke = 1.15 l Note : (a) If the power developed at the crankshaft, i.e. brake power (B. P.) and the mechanical efficiency (m) of the engine is known, then

(b) The maximum gas pressure (p) may be taken as 9 to 10 times the mean effective pressure (pm). 3. Cylinder flange and studs. The cylinders are cast integral with the upper half of the crankcase or they are attached to the crankcase by means of a flange with studs or bolts and nuts. The cylinder flange is integral with the cylinder and should be made thicker than the cylinder wall. The flange thickness should be taken as 1.2 t to 1.4 t, where t is the thickness of cylinder wall. The diameter of the studs or bolts may be obtained by equating the gas load due to the maximum pressure in the cylinder to the resisting force offered by all the studs or bolts. Mathematically,

where D = Cylinder bore in mm, 2 p = Maximum pressure in N/mm , ns = Number of studs. It may be taken as 0.01 D + 4 to 0.02 D + 4 dc = Core or minor diameter, i.e. diameter at the root of the thread in mm, 2 t = Allowable tensile stress for the material of studs or bolts in MPa or N/mm . It may be taken as 35 to 70 MPa. The nominal or major diameter of the stud or bolt (d ) usually lies between 0.75 tf to tf Where tf is the thickness of flange. In no case, a stud or bolt less than 16 mm diameter should be used. The distance of the flange from the centre of the hole for the stud or bolt should not be less than d + 6 mm and not more than 1.5 d, where d is the nominal diameter of the stud or bolt. In order to make a leak proof joint, the pitch of the studs or bolts should lie between 19 d to 28.5 d, where d is in mm. 4. Cylinder head. Usually, a separate cylinder head or cover is provided with most of the engines. It is, usually, made of box type section of considerable depth to accommodate ports for air and gas passages, inlet valve, exhaust valve and spark plug (in case of petrol engines) or atomiser at the centre of the cover (in case of diesel engines). The cylinder head may be approximately taken as a flat circular plate whose thickness (th) may be determined from the following relation :

where D = Cylinder bore in mm, p = Maximum pressure inside the cylinder in N/mm2, c = Allowable circumferential stress in MPa or N/mm2. It may be taken as 30 to 50 MPa, and C = Constant whose value is taken as 0.1. The studs or bolts are screwed up tightly along with a metal gasket or asbestos packing to provide a leak proof joint between the cylinder and cylinder head. The tightness of the joint also depends upon the pitch of the bolts or studs, which should lie between 19 d to 28.5 d. The pitch circle diameter (Dp) is usually taken as D + 3d. The studs or bolts are designed in the same way as discussed above.



Design Problem-1: A four stroke diesel engine has the following specifications : 2 Brake power = 5 kW ; Speed = 1200 r.p.m. ; Indicated mean effective pressure = 0.35 N / mm ; Mechanical efficiency = 80 %. Determine : 1. bore and length of the cylinder ; 2. thickness of the cylinder head ; and 3. size of studs for the cylinder head. 2 Given: B.P. =5kW =5000 W; N = 1200 r.p.m. (or) n = N / 2 = 600; pm = 0.35 N/mm ; m = 80% = 0.8 1. Bore and length of cylinder Let D = Bore of the cylinder in mm, A = Cross-sectional area of the cylinder = (π*D^2)/4 l =Length of the stroke in m. = 1.5 D mm = 1.5 D / 1000 m ....(Assumption) We know that the indicated power, IP = BP / m= 5000/0.8 = 6250 W We also know that the indicated power (I.P.),

...For four stroke engine, n = N/2)

l = 1.5 D = 1.5 × 115 = 172.5 mm Taking a clearance on both sides of the cylinder equal to 15% of the stroke, therefore length of the cylinder, L = 1.15 l = 1.15 × 172.5 = 198 say 200 mm 2. Thickness of the cylinder head Since the maximum pressure ( p) in the engine cylinder is taken as 9 to 10 times the mean effective pressure ( pm), therefore let us take 2 p = 9*pm = 9 × 0.35 = 3.15 N/mm We know that thickness of the cylinder head,

3. Size of studs for the cylinder head Let d = Nominal diameter of the stud in mm, dc = Core diameter of the stud in mm. It is usually taken as 0.84 d. t = Tensile stress for the material of the stud which is usually nickel steel. ns = Number of studs. We know that the force acting on the cylinder head (or on the studs)

………..(i) The number of studs (ns ) are usually taken between 0.01 D + 4 (i.e. 0.01 × 115 + 4 = 5.15) and 0.02 D + 4 (i.e. 0.02 × 115 + 4 = 6.3). Let us take ns = 6. We know that resisting force offered by all the studs ...(ii) 2

...(Taking t = 65 MPa = 65 N/mm )

From equations (i) and (ii), 2 d = 32 702 / 216 = 151 (or) d = 12.3 say 14 mm The pitch circle diameter of the studs (Dp) is taken D + 3d. Dp = 115 + 3 × 14 = 157 mm We know that pitch of the studs

where d is the nominal diameter of the stud.



Minimum pitch of the studs and maximum pitch of the studs Since the pitch of the studs obtained above (i.e. 82.2 mm) lies within 71.1 mm and 106.6 mm, therefore, size of the stud (d ) calculated above is satisfactory. d =14 mm

Piston The piston is a disc which reciprocates within a cylinder. It is either moved by the fluid or it moves the fluid which enters the cylinder. The main function of the piston of an internal combustion engine is to receive the impulse from the expanding gas and to transmit the energy to the crankshaft through the connecting rod. The piston must also disperse a large amount of heat from the combustion chamber to the cylinder walls.

The piston of internal combustion engines are usually of trunk type as shown in Fig. Such pistons are open at one end and consists of the following parts : 1. Head or crown. The piston head or crown may be flat, convex or concave depending upon the design of combustion chamber. It withstands the pressure of gas in the cylinder. 2. Piston rings. The piston rings are used to seal the cyliner in order to prevent leakage of the gas past the piston. 3. Skirt. The skirt acts as a bearing for the side thrust of the connecting rod on the walls of cylinder. 4. Piston pin. It is also called gudgeon pin or wrist pin. It is used to connect the piston to the connecting rod.

Design Considerations for a Piston In designing a piston for I.C. engine, the following points should be taken into consideration : 1. It should have enormous strength to withstand the high gas pressure and inertia forces. 2. It should have minimum mass to minimise the inertia forces. 3. It should form an effective gas and oil sealing of the cylinder. 4. It should provide sufficient bearing area to prevent undue wear. 5. It should disprese the heat of combustion quickly to the cylinder walls. 6. It should have high speed reciprocation without noise.



7. It should be of sufficient rigid construction to withstand thermal and mechanical distortion. 8. It should have sufficient support for the piston pin.

Material for Pistons The most commonly used materials for pistons of I.C. engines are cast iron, cast aluminium, forged aluminium, cast steel and forged steel. The cast iron pistons are used for moderately rated engines with piston speeds below 6 m / s and aluminium alloy pistons are used for highly rated engines running at higher piston speeds. It may be noted that 1. Since the coefficient of thermal expansion for aluminium is about 2.5 times that of cast iron, therefore, a greater clearance must be provided between the piston and the cylinder wall (than with cast iron piston) in order to prevent seizing of the piston when engine runs continuously under heavy loads. But if excessive clearance is allowed, then the piston will develop ‘piston slap’ while it is cold and this tendency increases with wear. The less clearance between the piston and the cylinder wall will lead to siezing of piston. 2. Since the aluminium alloys used for pistons have high heat conductivity (nearly four times that of cast iron), therefore, these pistons ensure high rate of heat transfer and thus keeps down the maximum temperature difference between the centre and edges of the piston head or crown. Note: (a) For a cast iron piston, the temperature at the centre of the piston head (TC) is about 425°C to 450°C under full load conditions and the temperature at the edges of the piston head (TE) is about 200°C to 225°C. (b) For aluminium alloy pistons, TC is about 260°C to 290°C and TE is about 185°C to 215°C. 3. Since the aluminium alloys are about three times lighter than cast iron, therfore, its mechanical strength is good at low temperatures, but they lose their strength (about 50%) at temperatures above 325°C. Sometimes, the pistons of aluminium alloys are coated with aluminium oxide by an electrical method. Note: 1.) The coefficient of thermal expansion for aluminium is 0.24 × 10–6 m / °C and for cast iron it is 0.1 × 10–6 m / °C. 2.) The heat conductivity for aluminium is 174.75 W/m/°C and for cast iron it is 46.6 W/m /°C. 3.) The density of aluminium is 2700 kg/m3 and for cast iron it is 7200 kg/m3.

Piston Head or Crown The piston head or crown is designed keeping in view the following two main considerations, i.e. 1. It should have adequate strength to withstand the straining action due to pressure of explosion inside the engine cylinder, and 2. It should dissipate the heat of combustion to the cylinder walls as quickly as possible. On the basis of first consideration of straining action, the thickness of the piston head is determined by treating it as a flat circular plate of uniform thickness, fixed at the outer edges and subjected to a uniformly distributed load due to the gas pressure over the entire cross-section. The thickness of the piston head (tH ), according to Grashoff’s formula is given by

.....................(i) where p = Maximum gas pressure or explosion pressure in N/mm2, D = Cylinder bore or outside diameter of the piston in mm, and t = Permissible bending (tensile) stress for the material of the piston in MPa or N/mm 2. It may be taken as 35 to 40 MPa for grey cast iron, 50 to 90 MPa for nickel cast iron and aluminium alloy and 60 to 100 MPa for forged steel. On the basis of second consideration of heat transfer, the thickness of the piston head should be such that the heat absorbed by the piston due combustion of fuel is quickly transferred to the cylinder walls. Treating the piston head as a flat circular plate, its thickness is given by ....................(ii)



where H = Heat flowing through the piston head in kJ/s or watts, k =Heat conductivity factor in W/m/°C. Its value is 46.6 W/m/°C for grey cast iron, 51.25 W/m/°C for steel and 174.75 W/m/°C for aluminium alloys. TC = Temperture at the centre of the piston head in °C, and TE = Temperature at the edges of the piston head in °C. The temperature difference (TC – TE) may be taken as 220°C for cast iron and 75°C for aluminium. The heat flowing through the piston head (H) may be determined by the following expression, i.e., H = C × HCV × m × B.P. (in kW) where C = Constant representing that portion of the heat supplied to the engine which is absorbed by the piston. Its value is usually taken as 0.05. HCV = Higher calorific value of the fuel in kJ/kg. HCV may be taken as 45 × 103 kJ/kg for diesel and 47 × 103 kJ/ kg for petrol, m = Mass of the fuel used in kg per brake power per second, and B.P. = Brake power of the engine per cylinder Note: 1. The thickness of the piston head (tH) is calculated by using equations (i) and (ii) and larger of the two values obtained should be adopted. 2. When tH is 6 mm or less, then no ribs are required to strengthen the piston head against gas loads. But when tH is greater then 6 mm, then a suitable number of ribs at the centre line of the boss extending around the skirt should be provided to distribute the side thrust from the connecting rod and thus to prevent distortion of the skirt. The thickness of the ribs may be takes as tH/3 to tH/2. 3. For engines having length of stroke to cylinder bore (L/D) ratio up to 1.5, a cup is provided in the top of the piston head with a radius equal to 0.7 D. This is done to provide a space for combustion chamber.

Piston Rings The piston rings are used to impart the necessary radial pressure to maintain the seal between the piston and the cylinder bore. These are usually made of grey cast iron or alloy cast iron because of their good wearing properties and also they retain spring characteristics even at high temperatures. The piston rings are of the following two types : 1. Compression rings or pressure rings, and 2. Oil control rings or oil scraper. The compression rings or pressure rings are inserted in the grooves at the top portion of the piston and may be three to seven in number. These rings also transfer heat from the piston to the cylinder liner and absorb some part of the piston fluctuation due to the side thrust. The oil control rings or oil scrapers are provided below the compression rings. These rings provide proper lubrication to the liner by allowing sufficient oil to move up during upward stroke and at the same time scraps the lubricating oil from the surface of the liner in order to minimise the flow of the oil to the combustion chamber. The compression rings are usually made of rectangular cross-section and the diameter of the ring is slightly larger than the cylinder bore. A part of the ring is cut- off in order to permit it to go into the cylinder against the liner wall. The diagonal cut or step cut ends, as shown in Fig. (a) and (b) respectively, may be used. The gap between the ends should be sufficiently large when the ring is put cold so that even at the highest temperature, the ends do not touch each other when the ring expands, otherwise there might be buckling of the ring.



The radial thickness (t1) of the ring may be obtained by considering the radial pressure between the cylinder wall and the ring. From bending stress consideration in the ring, the radial thickness is given by

where D = Cylinder bore in mm, pw= Pressure of gas on the cylinder wall in N/mm2. Its value is limited from 0.025 N/mm2 to 2 0.042 N/mm , and t = Allowable bending (tensile) stress in MPa. Its value may be taken from 85 MPa to 110 MPa for cast iron rings. The axial thickness (t2) of the rings may be taken as 0.7 t1 to t1. The minimum axial thickness (t2) may also be obtained from the following empirical relation:

Where nR = Number of rings. The width of the top land (i.e. the distance from the top of the piston to the first ring groove) is made larger than other ring lands to protect the top ring from high temperature conditions existing at the top of the piston, Width of top land, b1 = tH to 1.2 tH The width of other ring lands (i.e. the distance between the ring grooves) in the piston may be made equal to or slightly less than the axial thickness of the ring (t2). Width of other ring lands, b2 = 0.75 t2 to t2 The depth of the ring grooves should be more than the depth of the ring so that the ring does not take any piston side thrust. The gap between the free ends of the ring is given by 3.5t1 to 4t1. The gap, when the ring is in the cylinder, should be 0.002D to 0.004D.

Piston Barrel It is a cylindrical portion of the piston. The maximum thickness (t3) of the piston barrel may be obtained from the following empirical relation : t3 = 0.03D+b+4.5 mm where b = t1 + 0.4 mm i.e., Radial depth of piston ring groove which is taken as 0.4 mm larger than the radial thickness of the piston ring(t1) Thus, the above relation may be written as t3 = 0.03 D + t1 + 4.9 mm The piston wall thickness(t4) towards the open end is decreased and shall be taken as 0.25t3 to 0.35 t3.

Piston Skirt The portion of the piston below the ring section is known as piston skirt. In acts as a bearing for the side thrust of the connecting rod. The length of the piston skirt should be such that the bearing pressure on the piston barrel due to the side thrust does not exceed 0.25 N/mm2 of the projected area for low speed engines and 0.5 N/mm2 for high speed engines. It may be noted that the maximum thrust will be during the expansion stroke. The side thrust (R) on the cylinder liner is usually taken as 1/10 of the maximum gas load on the piston. We know that maximum gas load on the piston,

Maximum side thrust on the cylinder, …..(i) where p = Maximum gas pressure in N/mm2, and D =Cylinder bore in mm.



The side thrust (R) is also given as R =Bearing pressure × Projected bearing area of the piston skirt = pb × D × l where l = Length of the piston skirt in mm. ...(ii) From equations (i) and (ii), the length of the piston skirt (l) is determined. In actual practice, the length of the piston skirt is taken as 0.65 to 0.8 times the cylinder bore. Now the total length of the piston (L) is given by L = Length of skirt + Length of ring section + Top land The length of the piston usually varies between D and 1.5 D. It may be noted that a longer piston provides better bearing surface for quiet running of the engine, but it should not be made unnecessarily long as it will increase its own mass and thus the inertia forces.

Piston Pin The piston pin (also called gudgeon pin or wrist pin) is used to connect the piston and the connecting rod. It is usually made hollow and tapered on the inside, the smallest inside diameter being at the centre of the pin, as shown in Fig.

The piston pin passes through the bosses provided on the inside of the piston skirt and the bush of the small end of the connecting rod. The centre of piston pin should be 0.02 D to 0.04 D above the centre of the skirt, in order to off-set the turning effect of the friction and to obtain uniform distribution of pressure between the piston and the cylinder liner. The material used for the piston pin is usually case hardened steel alloy containing nickel, chromium, molybdenum or vanadium having tensile strength from 710 MPa to 910 MPa.

The connection between the piston pin and the small end of the connecting rod may be made either full floating type or semi-floating type. In the full floating type, the piston pin is free to turn both in the piston bosses and the bush of the small end of the connecting rod. The end movements of the piston pin should be secured by means of spring circlips, as shown in Fig. above in order to prevent the pin from touching and scoring the cylinder liner. Note: The mean diameter of the piston bosses is made 1.4d0 for cast iron pistons and 1.5d0 for aluminium pistons, where d0 is the outside diameter of the piston pin. The piston bosses are usually tapered, increasing the diameter towards the piston wall. In the semi-floating type, the piston pin is either free to turn in the piston bosses and rigidly secured to the small end of the connecting rod, or it is free to turn in the bush of the small end of the connecting rod and is rigidly secured in the piston bosses by means of a screw, as shown in Fig. The piston pin should be designed for the maximum gas load or the inertia force of the piston, whichever is larger. The bearing area of the piston pin should be about equally divided between the piston pin bosses and the connecting rod bushing. Thus, the length of the pin in the connecting rod bushing will be about 0.45 of the cylinder bore or piston diameter (D), allowing for the end clearance of the pin etc. The outside diameter of the piston pin (d0 ) is determined by equating the load on the



piston due to gas pressure (p) and the load on the piston pin due to bearing pressure ( pb1 ) at the small end of the connecting rod bushing.

Let d0 = Outside diameter of the piston pin in mm l1 = Length of the piston pin in the bush of the small end of the connecting rod in mm. Its value is usually taken as 0.45 D. pb1 = Bearing pressure at the small end of the connecting rod bushing in N/mm 2. Its value for the bronze bushing may be taken as 25 N/mm2. We know that load on the piston due to gas pressure or gas load .......(i) and load on the piston pin due to bearing pressure or bearing load = Bearing pressure × Bearing area = pb1×d0× l1 ...(ii) From equations (i) and (ii), the outside diameter of the piston pin (d0) may be obtained. The piston pin may be checked in bending by assuming the gas load to be uniformly distributed over the length l1 with supports at the centre of the bosses at the two ends. From Fig. given below, we find that the length between the supports,

Now maximum bending moment at the centre of the pin,



We have already discussed that the piston pin is made hollow. Let d0 and di be the outside and inside diameters of the piston pin. We know that the section modulus,

We know that maximum bending moment,

where b = Allowable bending stress for the material of the piston pin. It is usually taken as 84 MPa for case hardened carbon steel and 140 MPa for heat treated alloy steel. Assuming di = 0.6 d0, the induced bending stress in the piston pin may be checked. Design Problem-2: Design a cast iron piston for a single acting four stroke engine for the following data: 2; Cylinder bore = 100 mm; Stroke = 125 mm; Maximum gas pressure = 5 N/mm Indicated mean 2 effective pressure = 0.75 N/mm ; Mechanical efficiency = 80 %; Fuel consumption = 0.15 kg per 3 brake power per hour ; Higher calorific value of fuel = 42 × 10 kJ/kg ; Speed = 2000 r.p.m. Any other data required for the design may be assumed. 2 2 Given: D = 100 mm ; L = 125 mm = 0.125 m ; p = 5 N/mm ; pm = 0.75 N/mm ; m = 80% = 0.8 ; –6 3 m = 0.15 kg / BP / h = 41.7 × 10 kg/BP/s; HCV = 42 × 10 kJ / kg ; N = 2000 r.p.m. The dimensions for various components of the piston are determined as follows: 1. Piston head or crown The thickness of the piston head or crown is determined on the basis of strength as well as on the basis of heat dissipation and the larger of the two values is adopted. We know that the thickness of piston head on the basis of strength,

2

...(Taking t for C.I.= 38 MPa = 38 N/mm )

Since the engine is a four stroke engine, therefore, the number of working strokes per minute, n = N / 2 = 2000 / 2 = 1000 and cross-sectional area of the cylinder

We know that indicated power,

Brake power, BP = IP × m = 12.27 × 0.8 = 9.8 kW .........(m = BP / IP) We know that the heat flowing through the piston head, H = C × HCV × m × BP = 0.05 × 42 × 103 × 41.7 × 10–6 × 9.8 = 0.86 kW = 860 W ....(Taking C = 0.05)

Thickness of the piston head on the basis of heat dissipation,

Taking the larger of the two values, we shall adopt tH = 16 mm



Since the ratio of L / D is 1.25, therefore a cup in the top of the piston head with a radius equal to 0.7 D (i.e. 70 mm) is provided. 2. Radial ribs The radial ribs may be four in number. The thickness of the ribs varies from tH/3 to tH/2. Thickness of the ribs, tR = 16 / 3 to 16 / 2 = 5.33 to 8 mm Let us adopt tR = 7 mm 3. Piston rings Let us assume that there are total four rings (i.e. nr = 4) out of which three are compression rings and one is an oil ring. We know that the radial thickness of the piston rings,

2

...(Taking pw = 0.035 N/mm , and t = 90 MPa) and axial thickness of the piston rings t2 = 0.7 t1 to t1 = 0.7 × 3.4 to 3.4 mm = 2.38 to 3.4 mm Let us adopt t2 = 3 mm We also know that the minimum axial thickness of the pistion ring,

Thus the axial thickness of the piston ring as already calculated (i.e. t2 = 3 mm)is satisfactory. The distance from the top of the piston to the first ring groove, i.e. the width of the top land, b1 = tH to 1.2 tH = 16 to 1.2 × 16 mm = 16 to 19.2 mm and width of other ring lands, b2 = 0.75 t2 to t2 = 0.75 × 3 to 3 mm = 2.25 to 3 mm Let us adopt b1 = 18 mm ; and b2 = 2.5 mm We know that the gap between the free ends of the ring, G1 = 3.5 t1 to 4 t1 = 3.5 × 3.4 to 4 × 3.4 mm = 11.9 to 13.6 mm and the gap when the ring is in the cylinder, G2 = 0.002 D to 0.004 D = 0.002 × 100 to 0.004 × 100 mm = 0.2 to 0.4 mm Let us adopt G1 = 12.8 mm ; and G2 = 0.3 mm 4. Piston barrel Since the radial depth of the piston ring grooves (b) is about 0.4 mm more than the radial thickness of the piston rings (t1), therefore, b = t1 + 0.4 = 3.4 + 0.4 = 3.8 mm We know that the maximum thickness of barrel, t3 = 0.03 D + b + 4.5 mm = 0.03 × 100 + 3.8 + 4.5 = 11.3 mm and piston wall thickness towards the open end, t4 = 0.25 t3 to 0.35 t3 = 0.25 × 11.3 to 0.35 × 11.3 = 2.8 to 3.9 mm Let us adopt t4 = 3.4 mm 5. Piston skirt Let l = Length of the skirt in mm. We know that the maximum side thrust on the cylinder due to gas pressure (p),

...(Taking = 0.1) We also know that the side thrust due to bearing pressure on the piston barrel ( pb ), R = pb × D × l = 0.45 × 100 × l = 45 l N ......(Taking pb = 0.45 N/mm2) From above, we find that 45 l = 3928 or l = 3928 / 45 = 87.3 say 90 mm Total length of the piston , L = Length of the skirt + Length of the ring section + Top land = l + (4 t2 + 3b2) + b1 = 90 + (4 × 3 + 3 × 3) + 18 = 129 say 130 mm



6. Piston pin Let d0 = Outside diameter of the pin in mm, l1 = Length of pin in the bush of the small end of the connecting rod in mm, and pb1 = Bearing pressure at the small end of the connecting rod bushing in N/mm2. Its value for bronze bushing is taken as 25 N/mm2. We know that load on the pin due to bearing pressure = Bearing pressure × Bearing area = pb1 × d0 × l1 = 25 × d0 × 0.45 × 100 = 1125 d0 N ...(Taking l1 = 0.45 D) We also know that maximum load on the piston due to gas pressure or maximum gas load

From above, we find that 1125 d0 = 39 275 or d0 = 39 275/1125 = 34.9 say 35 mm The inside diameter of the pin (di) is usually taken as 0.6 d0. di = 0.6 × 35 = 21 mm Let the piston pin be made of heat treated alloy steel for which the bending stress (b)may be taken as 140 MPa. Now let us check the induced bending stress in the pin. We know that maximum bending moment at the centre of the pin,

We also know that maximum bending moment (M),

Since the induced bending stress in the pin is less than the permissible value of 140 MPa (i.e. 140 N/mm2), therefore, the dimensions for the pin as calculated above (i.e. d0 = 35 mm and di = 21 mm) are satisfactory.

Connecting Rod The connecting rod is the intermediate member between the piston and the crankshaft. Its primary function is to transmit the push and pull from the piston pin to the crankpin and thus convert the reciprocating motion of the piston into the rotary motion of the crank. The usual form of the connecting rod in internal combustion engines is shown in Fig. It consists of a long shank, a small end and a big end. The cross-section of the shank may be rectangular, circular, tubular, I-section or Hsection. Generally circular section is used for low speed engines while I-section is preferred for high speed engines.



The length of the connecting rod ( l ) depends upon the ratio of l / r, where r is the radius of crank. It may be noted that the smaller length will decrease the ratio l / r. This increases the angularity of the connecting rod which increases the side thrust of the piston against the cylinder liner which in turn increases the wear of the liner. The larger length of the connecting rod will increase the ratio l / r. This decreases the angularity of the connecting rod and thus decreases the side thrust and the resulting wear of the cylinder. But the larger length of the connecting rod increases the overall heightof the engine. Hence, a compromise is made and the ratio l / r is generally kept as 4 to 5. The small end of the connecting rod is usually made in the form of an eye and is provided with a bush of phosphor bronze. It is connected to the piston by means of a piston pin. The big end of the connecting rod is usually made split (in two halves) so that it can be mounted easily on the crankpin bearing shells. The split cap is fastened to the big end with two cap bolts. The bearing shells of the big end are made of steel, brass or bronze with a thin lining (about 0.75 mm) of white metal or babbit metal. The wear of the big end bearing is allowed for by inserting thin metallic strips (known as shims) about 0.04 mm thick between the cap and the fixed half of the connecting rod. As the wear takes place, one or more strips are removed and the bearing is trued up. Note: 1.) Length of the connecting rod is the distance between centres of small end and big end of the connecting rod. 2.) Connecting rod big end is plit into two halves, one half is fixed with the connecting rod and the other half (known as cap) is fastened with two cap bolts.

The connecting rods are usually manufactured by drop forging process and it should have adequate strength, stiffness and minimum weight. The material mostly used for connecting rods varies from mild carbon steels (having 0.35 to 0.45 percent carbon) to alloy steels (chrome-nickel or chromemolybdenum steels). The carbon steel having 0.35 percent carbon has an ultimate tensile strength of about 650 MPa when properly heat treated and a carbon steel with 0.45 percent carbon has a ultimate tensile strength of 750 MPa. These steels are used for connecting rods of industrial engines. The alloy steels have an ultimate tensile strength of about 1050 MPa and are used for connecting rods of aeroengines and automobile engines. The bearings at the two ends of the connecting rod are either splash lubricated or pressure lubricated. The big end bearing is usually splash lubricated while the small end bearing is pressure lubricated. In the splash lubrication system, the cap at the big end is provided with a dipper or spout and set at an angle in such a way that when the connecting rod moves downward, the spout will dip into the lubricating oil contained in the sump. The oil is forced up the spout and then to the big end bearing. Now when the connecting rod moves upward, a splash of oil is produced by the spout. This splashed up lubricant find its way into the small end bearing through the widely chamfered holes provided on the upper surface of the small end. In the pressure lubricating system, the lubricating oil is fed under pressure to the big end bearing through the holes drilled in crankshaft, crankwebs and crank pin. From the big end bearing, the oil is fed to small end bearing through a fine hole drilled in the shank of the connecting rod. In some cases, the small end bearing is lubricated by the oil scrapped from the walls of the cyinder liner by the oil scraper rings.

Forces Acting on the Connecting Rod The various forces acting on the connecting rod are as follows : 1. Force on the piston due to gas pressure and inertia of the reciprocating parts, 2. Force due to inertia of the connecting rod or inertia bending forces, 3. Force due to friction of the piston rings and of the piston, and 4. Force due to friction of the piston pin bearing and the crankpin bearing. 1. Force on the piston due to gas pressure and inertia of reciprocating parts Consider a connecting rod PC as shown in Fig. Let p = Maximum pressure of gas, D = Diameter of piston, 2 A = Cross-section area of piston = π*D /4 mR = Mass of reciprocating parts = Mass of piston, gudgeon pin etc. + 1/3 rd mass of connecting rod, ω= Angular speed of crank, ϕ= Angle of inclination of the connecting rod with the line of stroke, = Angle of inclination of the crank from top dead centre, r = Radius of crank, l = Length of connecting rod, and n = Ratio of length of connecting rod to radius of crank = l/r



We know that the force on the piston due to pressure of gas, and

inertia

force

of

reciprocating

parts,

It may be noted that the inertia force of reciprocating parts opposes the force on the piston when it moves during its downward stroke (i. e. when the piston moves from the top dead centre to bottom dead centre). On the other hand, the inertia force of the reciprocating parts helps the force on the piston when it moves from the bottom dead centre to top dead centre. Net force acting on the piston or piston pin (or gudgeon pin or wrist pin),

The –ve sign is used when piston moves from TDC to BDC and +ve sign is used when piston moves from BDC to TDC. When weight of the reciprocating parts (WR = mR*g) is to be taken into consideration, then The force FP gives rise to a force FC in the connecting rod and a thrust FN on the sides of the cylinder walls. From Fig, we see that force in the connecting rod at any instant, The force in the connecting rod will be maximum when the crank and the connecting rod are perpendicular to each other (i.e. when = 90°). But at this position, the gas pressure would be decreased considerably. Thus, for all practical purposes, the force in the connecting rod (FC) is taken equal to the maximum force on the piston due to pressure of gas (FL), neglecting piston inertia effects. 2. Force due to inertia of the connecting rod or inertia bending forces Consider a connecting rod PC and a crank OC rotating with uniform angular velocity ωrad/s. In order to find the accleration of various points on the connecting rod, draw the Klien’s acceleration diagram CQNO as shown in Fig. 32.11 (a). CO represents the acceleration of C towards O and NO represents the accleration of P towards O. The acceleration of other points such as D, E, F and G etc., on the connecting rod PC may be found by drawing horizontal lines from these points to intresect CN at d, e, f, and g respectively. Now dO, eO, fO and gO respresents the acceleration of D, E, F and G all towards O. The inertia force acting on each point will be as follows: Inertia force at C = m ×ω2×CO Inertia force at D = m ×ω2×dO Inertia force at E = m × ω2×eO, and so on.



The inertia forces will be opposite to the direction of acceleration or centrifugal forces. The inertia forces can be resolved into two components, one parallel to the connecting rod and the other perpendicular to rod. The parallel (or longitudinal) components adds up algebraically to the force acting on the connecting rod (FC) and produces thrust on the pins. The perpendicular (or transverse) components produces bending action (also called whipping action) and the stress induced in the connecting rod is called whipping stress. It may be noted that the perpendicular components will be maximum, when the crank and connecting rod are at right angles to each other. The variation of the inertia force on the connecting rod is linear and is like a simply supported beam of variable loading as shown in Fig. (b) and (c). acting on the connecting rod (FC) and produces thrust on the pins. The perpendicular (or transverse) components produces bending action (also called whipping action) and the stress induced in the connecting rod is called whipping stress. It may be noted that the perpendicular components will be maximum, when the crank and connecting rod are at right angles to each other. The variation of the inertia force on the connecting rod is linear and is like a simply supported beam of variable loading as shown in Fig. 32.11 (b) and (c). Assuming that the connecting rod is of uniform cross-section and has mass m1 kg per unit length, therefore,



Now the bending moment acting on the rod at section X – X at a distance x from P,

.......(i) Note:

For maximum bending moment, differentiate MX with respect to x and equate to zero, i.e.

Maximum bending moment,

..............[From equation (i)]

and the maximum bending stress, due to inertia of the connecting rod,

where Z = Section modulus. From above we see that the maximum bending moment varies as the square of speed, therefore, the bending stress due to high speed will be dangerous. It may be noted that the maximum axial force and the maximum bending stress do not occur simultaneously. In an I.C. engine, the maximum gas load occurs close to top dead centre whereas the maximum bending stress occurs when the crank angle =65° to 70° from top dead centre. The pressure of gas falls suddenly as the piston moves from dead centre. Thus the general practice is to design a connecting rod by assuming the force in the connecting rod (FC) equal to the maximum force due to pressure (FL), neglecting piston inertia effects and then checked for bending stress due to inertia force (i.e. whipping stress). 3. Force due to friction of piston rings and of the piston The frictional force ( F ) of the piston rings may be determined by using the following expression



where D = Cylinder bore, tR = Axial width of rings, nR = Number of rings, pR = Pressure of rings (0.025 to 0.04 N/mm2), and = Coefficient of friction (about 0.1). Since the frictional force of the piston rings is usually very small, therefore, it may be neglected. The friction of the piston is produced by the normal component of the piston pressure which varies from 3 to 10 percent of the piston pressure. If the coefficient of friction is about 0.05 to 0.06, then the frictional force due to piston will be about 0.5 to 0.6 of the piston pressure, which is very low. Thus, the frictional force due to piston is also neglected. 4. Force due to friction of the piston pin bearing and crankpin bearing The force due to friction of the piston pin bearing and crankpin bearing is to bend the connecting rod and to increase the compressive stress on the connecting rod due to the direct load. Thus the maximum compressive stress in the connecting rod will be c (max) = Direct compressive stress + Maximum bending or whipping stress due to inertia bending stress

Design of Connecting Rod In designing a connecting rod, the following dimensions are required to be determined: 1. Dimensions of cross-section of the connecting rod, 2. Dimensions of the crankpin at the big end and the piston pin at the small end, 3. Size of bolts for securing the big end cap, and 4. Thickness of the big end cap. The procedure adopted in determining the above mentioned dimensions is discussed as below: 1. Dimensions of cross-section of the connecting rod A connecting rod is a machine member which is subjected to alternating direct compressive and tensile forces. Since the compressive forces are much higher than the tensile forces, therefore, the cross-section of the connecting rod is designed as a strut and the Rankine’s formula is used. A connecting rod, as shown in Fig., subjected to an axial load W may buckle with X-axis as neutral axis (i.e. in the plane of motion of the connecting rod) or Y-axis as neutral axis (i.e. in the plane perpendicular to the plane of motion). The connecting rod is considered like both ends hinged for buckling about X-axis and both ends fixed for buckling about Y-axis.

A connecting rod should be equally strong in buckling about both the axes.



In order to thave a connecting rod equally strong in buckling about both the axes, the buckling loads must be equal, i.e.

This shows that the connecting rod is four times strong in buckling about Y-axis than about X-axis. If Ixx > 4 Iyy, then buckling will occur about Y- axis and if Ixx < 4 Iyy, buckling will occur about X-axis. In actual practice, Ixx is kept slightly less than 4 Iyy. It is usually taken between 3 and 3.5 and the connecting rod is designed for bucking about X-axis. The design will always be satisfactory for buckling about Y-axis. The most suitable section for the connecting rod is I-section with the proportions as shown in Fig.(a).

Let thickness of the flange and web of the section = t Width of the section, B = 4 t and depth or height of the section, H = 5t From Fig.(a), we find that area of the section, A = 2 (4 t × t) + 3 t × t = 11*t2 Moment of inertia of the section about X-axis,

and moment of inertia of the section about Y-axis,

After deciding the proportions for I-section of the connecting rod, its dimensions are determined by considering the buckling of the rod about X-axis (assuming both ends hinged) and applying the Rankine’s formula. We know that buckling load,

The buckling load (WB) may be calculated by using the following relation, i.e. WB = Max. gas force×Factor of safety The factor of safety may be taken as 5 to 6.



Note: (1) The I-section of the connecting rod is used due to its lightness and to keep the inertia forces as low as possible specially in case of high speed engines. It can also withstand high gas pressure. (2) Sometimes a connecting rod may have rectangular section. For slow speed engines, circular section may be used. (3) Since connecting rod is manufactured by forging, therefore the sharp corner of I-section are rounded off as shown in Fig. (b) for easy removal of the section from dies.

The dimensions B = 4 t and H = 5 t, as obtained above by applying the Rankine’s formula, are at the middle of the connecting rod. The width of the section (B) is kept constant throughout the length of the connecting rod, but the depth or height varies. The depth near the small end (or piston end) is taken as H1 = 0.75 H to 0.9H and the depth near the big end (or crank end) is taken H2 = 1.1H to 1.25H. 2. Dimensions of the crankpin at the big end and the piston pin at the small end Since the dimensions of the crankpin at the big end and the piston pin (also known as gudgeon pin or wrist pin) at the small end are limited, therefore, fairly high bearing pressures have to be allowed at the bearings of these two pins. The crankpin at the big end has removable precision bearing shells of brass or bronze or steel with a thin lining (1 mm or less) of bearing metal (such as tin, lead, babbit, copper, lead) on the inner surface of the shell. The allowable bearing pressure on the crankpin depends upon many factors such as material of the bearing, viscosity of the lubricating oil, method of lubrication and the space 2 2 limitations. The value of bearing pressure may be taken as 7 N/mm to 12.5 N/mm depending upon the material and method of lubrication used. The piston pin bearing is usually a phosphor bronze bush of about 3 mm thickness and the allowable 2 2 bearing pressure may be taken as 10.5 N/mm to 15 N/mm . Since the maximum load to be carried by the crankpin and piston pin bearings is the maximum force in the connecting rod (FC), therefore the dimensions for these two pins are determined for the maximum force in the connecting rod (FC) which is taken equal to the maximum force on the piston due to gas pressure (FL) neglecting the inertia forces. We know that maximum gas force,

.......(i) where D = Cylinder bore or piston diameter in mm, and p = Maximum gas pressure in N/mm2 Now the dimensions of the crankpin and piston pin are determined as discussed below : Let dc = Diameter of the crank pin in mm, lc = Length of the crank pin in mm, pbc = Allowable bearing pressure in N/mm2, and dp, lp and pbp = Corresponding values for the piston pin,

......(ii) ......(iii) Equating equations (i) and (ii), we have FL= dc*lc*pbc Taking lc = 1.25 dc to 1.5 dc, the value of dc and lc are determined from the above expression. Again, equating equations (i) and (iii), we have FL = dp*lp*pbp Taking lp = 1.5*dp to 2*dp, the value of dp and lp are determined from the above expression. 3. Size of bolts for securing the big end cap The bolts and the big end cap are subjected to tensile force which corresponds to the inertia force of the reciprocating parts at the top dead centre on the exhaust stroke. We know that inertia force of the reciprocating parts,



We also know that at the top dead centre, the angle of inclination of crank with line of stroke (= 0)

where mR = Mass of the reciprocating parts in kg, ω= Angular speed of the engine in rad/s, r = Radius of the crank in metres, and l = Length of the connecting rod in metres. The bolts may be made of high carbon steel or nickel alloy steel. Since the bolts are under repeated stresses but not alternating stresses, therefore, a factor of safety may be taken as 6. Let dcb = Core diameter of the bolt in mm, t = Allowable tensile stress for the material of the bolts in MPa, and nb = Number of bolts. Generally two bolts are used. Force on the bolts

Equating the inertia force to the force on the bolts, we have

From this expression, dcb is obtained. The nominal or major diameter (db) of the bolt is given by

4. Thickness of the big end cap The thickness of the big end cap (tc) may be determined by treating the cap as a beam freely supported at the cap bolt centres and loaded by the inertia force at the top dead centre on the exhaust stroke (i.e. FI when ). This load is assumed to act in between the uniformly distributed load and the centrally concentrated load. Therefore, the maximum bending moment acting on the cap will be taken as

where x = Distance between the bolt centres. = Dia. of crankpin or big end bearing(dc) + 2 × Thickness of bearing liner (3 mm) + Clearance (3 mm) Let bc = Width of the cap in mm. It is equal to the length of the crankpin or big end bearing (lc), and b = Allowable bending stress for the material of the cap in MPa. We know that section modulus for the cap,

From this expression, the value of tc is obtained. Note: The design of connecting rod should be checked for whipping stress (i.e. bending stress due to inertia force on the connecting rod).

Design Problem-3: Design a connecting rod for an I.C. engine running at 1800 r.p.m. and developing 2 a maximum pressure of 3.15 N/mm . The diameter of the piston is 100 mm ; mass of the reciprocating parts per cylinder 2.25 kg; length of connecting rod 380 mm; stroke of piston 190 mm and compression ratio 6 : 1. Take a factor of safety of 6 for the design. Take length to diameter ratio for big end bearing as 1.3 and small end bearing as 2 and the corresponding bearing pressures as 10 N/mm2 and 15 N/mm2. The density of material of the rod may be taken as 8000 kg/m3 and the allowable stress in the bolts as 60 N/mm2 and in cap as 80 N/mm2. The rod is to be of I-section for which you can choose your own proportions.



Use Rankine formula for which the numerator constant may be taken as 320 N/mm2 and the denominator constant 1 / 7500. Given : N = 1800 r.p.m. ; p = 3.15 N/mm2 ; D = 100 mm ; mR = 2.25 kg ; l = 380 mm = 0.38 m ; Stroke = 190 mm ; Compression ratio = 6 : 1 ; F. S. = 6. The connecting rod is designed as discussed below: 1. Dimension of I- section of the connecting rod Let us consider an I-section of the connecting rod, as shown in Fig.(a), with the following proportions:

Flange and web thickness of the section = t Width of the section, B = 4t and depth or height of the section, H = 5t First of all, let us find whether the section chosen is satisfactory or not. We have already discussed that the connecting rod is considered like both ends hinged for buckling about X-axis and both ends fixed for buckling about Y-axis. The connecting rod should be equally strong in buckling about both the axes. We know that in order to have a connecting rod equally strong about both the axes,

Now, for the section as shown in Fig. (a), area of the section,

Now let us find the dimensions of this I-section. Since the connecting rod is designed by taking the force on the connecting rod (FC) equal to the maximum force on the piston (FL) due to gas pressure, therefore,



We know that the connecting rod is designed for buckling about X-axis (i.e. in the plane of motion of the connecting rod) assuming both ends hinged. Since a factor of safety is given as 6, therefore the buckling load, We know that radius of gyration of the section about X-axis,

Length of crank,

Length of the connecting rod, l = 380 mm ...(Given) Equivalent length of the connecting rod for both ends hinged, L = l = 380 mm Now according to Rankine’s formula, we know that buckling load (WB),

t = 6.9 say 7 mm Thus, the dimensions of I-section of the connecting rod are : Thickness of flange and web of the section = t = 7 mm Width of the section, B = 4 t = 4 × 7 = 28 mm and depth or height of the section, H = 5 t = 5 × 7 = 35 mm These dimensions are at the middle of the connecting rod. The width (B) is kept constant throughout the length of the rod, but the depth (H) varies. The depth near the big end or crank end is kept as 1.1H to 1.25H and the depth near the small end or piston end is kept as 0.75H to 0.9H. Let us take Depth near the big end, H1 = 1.2H = 1.2 × 35 = 42 mm and depth near the small end, H2 = 0.85×H = 0.85 × 35 = 29.75 say 30 mm Dimensions of the section near the big end = 42 mm × 28 mm and dimensions of the section near the small end = 30 mm × 28 mm Since the connecting rod is manufactured by forging, therefore the sharp corners of I-section are rounded off, as shown in Fig. (b), for easy removal of the section from the dies. 2. Dimensions of the crankpin or the big end bearing and piston pin or small end bearing Let dc = Diameter of the crankpin or big end bearing, lc = length of the crankpin or big end bearing = 1.3×dc ...(Given) pbc = Bearing pressure = 10 N/mm2 ...(Given) We know that load on the crankpin or big end bearing = Projected area × Bearing pressure = dc×lc×pbc = dc × 1.3 dc × 10 = 13× (dc)2 Since the crankpin or the big end bearing is designed for the maximum gas force (FL), therefore, equating the load on the crankpin or big end bearing to the maximum gas force, i.e. 2 13× (dc) = FL = 24 740 N 2 (dc ) = 24 740 / 13 = 1903 or dc = 43.6 say 44 mm



and lc = 1.3×dc = 1.3 × 44 = 57.2 say 58 mm The big end has removable precision bearing shells of brass or bronze or steel with a thin lining (1mm or less) of bearing metal such as babbit. Again, let dp = Diameter of the piston pin or small end bearing, lp = Length of the piston pin or small end bearing = 2dp ...(Given) 2 pbp = Bearing pressure = 15 N/mm ..(Given) We know that the load on the piston pin or small end bearing = Project area × Bearing pressure = dp . lp . pbp = dp × 2 dp × 15 = 30 (dp)2 Since the piston pin or the small end bearing is designed for the maximum gas force (FL), therefore, equating the load on the piston pin or the small end bearing to the maximum gas force, i.e. 30*(dp)2 = 24 740 N (dp)2 = 24 740 / 30 = 825 or dp = 28.7 say 29 mm and lp = 2×dp = 2×29 = 58 mm The small end bearing is usually a phosphor bronze bush of about 3 mm thickness. 3. Size of bolts for securing the big end cap Let dcb = Core diameter of the bolts, 2 t = Allowable tensile stress for the material of the bolts = 60 N/mm ...(Given) and nb = Number of bolts. Generally two bolts are used. We know that force on the bolts

The bolts and the big end cap are subjected to tensile force which corresponds to the inertia force of the reciprocating parts at the top dead centre on the exhaust stroke. We know that inertia force of the reciprocating parts,

We also know that at top dead centre on the exhaust stroke, = 0

Equating the inertia force to the force on the bolts, we have

and nominal diameter of the bolt,

4. Thickness of the big end cap Let tc = Thickness of the big end cap, bc = Width of the big end cap. It is taken equal to the length of the crankpin or big end bearing (lc) = 58 mm (calculated above) σb = Allowable bending stress for the material of the cap = 80 N/mm2 ...(Given) The big end cap is designed as a beam freely supported at the cap bolt centres and loaded by the inertia force at the top dead centre on the exhaust stroke (i.e. FI when θ = 0). Since the load is assumed to act in between the uniformly distributed load and the centrally concentrated load, therefore, maximum bending moment is taken as



where x = Distance between the bolt centres

Maximum bending moment acting on the cap,

Section modulus for the cap

We know that bending stress ( b ),

Let us now check the design for the induced bending stress due to inertia bending forces on the connecting rod (i.e. whipping stress). We know that mass of the connecting rod per metre length,

Maximum bending moment,

Crankshaft A crankshaft (i.e. a shaft with a crank) is used to convert reciprocating motion of the piston into rotatory motion or vice versa. The crankshaft consists of the shaft parts which revolve in the main bearings, the crankpins to which the big ends of the connecting rod are connected, the crank arms or webs (also called cheeks) which connect the crankpins and the shaft parts. The crankshaft, depending upon the position of crank, may be divided into the following two types : 1. Side crankshaft or overhung crankshaft, as shown in Fig.(a), and 2. Centre crankshaft, as shown in Fig.(b).



The crankshaft, depending upon the number of cranks in the shaft, may also be classfied as single throw or multi-throw crankshafts. A crankhaft with only one side crank or centre crank is called a single throw crankshaft whereas the crankshaft with two side cranks, one on each end or with two or more centre cranks is known as multi-throw crankshaft. The side crankshafts are used for medium and large size horizontal engines.

Material and manufacture of Crankshafts The crankshafts are subjected to shock and fatigue loads. Thus material of the crankshaft should be tough and fatigue resistant. The crankshafts are generally made of carbon steel, special steel or special cast iron. In industrial engines, the crankshafts are commonly made from carbon steel such as 40 C 8, 55 C 8 and 60 C 4. In transport engines, manganese steel such as 20 Mn 2, 27 Mn 2 and 37 Mn 2 are generally used for the making of crankshaft. In aero engines, nickel chromium steel such as 35 Ni 1 Cr 60 and 40 Ni 2 Cr 1 Mo 28 are extensively used for the crankshaft. The crankshafts are made by drop forging or casting process but the former method is more common. The surface of the crankpin is hardened by case carburizing, nitriding or induction hardening.

Bearing Pressures and Stresses in Crankshaft The bearing pressures are very important in the design of crankshafts. The maximum permissible bearing pressure depends upon the maximum gas pressure, journal velocity, amount and method of lubrication and change of direction of bearing pressure. The following two types of stresses are induced in the crankshaft. 1. Bending stress; and 2. Shear stress due to torsional moment on the shaft. Most crankshaft failures are caused by a progressive fracture due to repeated bending or reversed torsional stresses. Thus the crankshaft is under fatigue loading and, therefore, its design should be based upon endurance limit. Since the failure of a crankshaft is likely to cause a serious engine destruction and neither all the forces nor all the stresses acting on the crankshaft can be determined accurately, therefore a high factor of safety from 3 to 4, based on the endurance limit, is used. The following table shows the allowable bending and shear stresses for some commonly used materials for crankshafts :

Design Procedure for Crankshaft The following procedure may be adopted for designing a crankshaft. 1. First of all, find the magnitude of the various loads on the crankshaft. 2. Determine the distances between the supports and their position with respect to the loads. 3. For the sake of simplicity and also for safety, the shaft is considered to be supported at the centres of the bearings and all the forces and reactions to be acting at these points. The distances between the supports depend on the length of the bearings, which in turn depend on the diameter of the shaft because of the allowable bearing pressures.



4. The thickness of the cheeks or webs is assumed to be from 0.4 ds to 0.6 ds, where ds is the diameter of the shaft. It may also be taken as 0.22D to 0.32 D, where D is the bore of cylinder in mm. 5. Now calculate the distances between the supports. 6. Assuming the allowable bending and shear stresses, determine the main dimensions of the crankshaft. Note:

1. The crankshaft must be designed or checked for at least two crank positions. Firstly, when the crankshaft is subjected to maximum bending moment and secondly when the crankshaft is subjected to maximum twisting moment or torque. 2. The additional moment due to weight of flywheel, belt tension and other forces must be considered. 3. It is assumed that the effect of bending moment does not exceed two bearings between which a force is considered.

Design of Centre Crankshaft We shall design the centre crankshaft by considering the two crank possitions, i.e. when the crank is at dead centre (or when the crankshaft is subjected to maximum bending moment) and when the crank is at angle at which the twisting moment is maximum. These two cases are discussed in detail as below: 1. When the crank is at dead centre. At this position of the crank, the maximum gas pressure on the piston will transmit maximum force on the crankpin in the plane of the crank causing only bending of the shaft. The crankpin as well as ends of the crankshaft will be only subjected to bending moment. Thus, when the crank is at the dead centre, the bending moment on the shaft is maximum and the twisting moment is zero.

Consider a single throw three bearing crankshaft as shown in Fig. Let D = Piston diameter or cylinder bore in mm, 2 p = Maximum intensity of pressure on the piston in N/mm , W = Weight of the flywheel acting downwards in N, and T1 + T2 = Resultant belt tension or pull acting horizontally in N. Note: T1 is the belt tension in the tight side and T2 is the belt tension in the slack side. The thrust in the connecting rod will be equal to the gas load on the piston (FP ). We know that gas load on the piston,



Now the various parts of the centre crankshaft are designed for bending only, as discussed below: (a) Design of crankpin Let dc = Diameter of the crankpin in mm, lc = Length of the crankpin in mm, b = Allowable bending stress for the crankpin in N/mm2. We know that bending moment at the centre of the crankpin, MC = H1*b2 ...(i) We also know that

…..(ii) From equations (i) and (ii), diameter of the crankpin is determined. The length of the crankpin is given by

2

where pb = Permissible bearing pressure in N/mm . (b) Design of left hand crank web The crank web is designed for eccentric loading. There will be two stresses acting on the crank web, one is direct compressive stress and the other is bending stress due to piston gas load (FP).



This total stress should be less than the permissible bending stress. (c) Design of right hand crank web The dimensions of the right hand crank web (i.e. thickness and width) are made equal to left hand crank web from the balancing point of view. (d) Design of shaft under the flywheel Let ds = Diameter of shaft in mm. We know that bending moment due to the weight of flywheel, MW = V3*c1 and bending moment due to belt tension, ’ MT = H3 *c1 These two bending moments act at right angles to each other. Therefore, the resultant bending moment at the flywheel location,

………..(i) We also know that the bending moment at the shaft,

…………..(ii) where b = Allowable bending stress in N/mm2. From equations (i) and (ii), we may determine the shaft diameter (ds). 2. When the crank is at an angle of maximum twisting moment The twisting moment on the crankshaft will be maximum when the tangential force on the crank (FT) is maximum. The maximum value of tangential force lies when the crank is at angle of 25° to 30º from the dead centre for a constant volume combustion engines (i.e., petrol engines) and 30º to 40º for constant pressure combustion engines (i.e., diesel engines). Consider a position of the crank at an angle of maximum twisting moment as shown in Fig. (a). If p’is the intensity of pressure on the piston at this instant, then the piston gas load at this position of crank,

and thrust on the connecting rod,

where ϕ= Angle of inclination of the connecting rod with the line of stroke PO. The thrust in the connecting rod (FQ) may be divided into two components, one perpendicular to the crank and the other along the crank. The component of FQ perpendicular to the crank is the tangential force (FT) and the component of FQ along the crank is the radial force (FR) which produces thrust on the crankshaft bearings. From Fig.(b), we find the forces acting on the crank



It may be noted that the tangential force will cause twisting of the crankpin and shaft while the radial force will cause bending of the shaft.

The reactions at the bearings 2 and 3, due to the flywheel weight (W) and resultant belt pull (T1 + T2) will be same as discussed earlier. Now the various parts of the crankshaft are designed as discussed below : (a) Design of crankpin Let dc = Diameter of the crankpin in mm. We know that bending moment at the centre of the crankpin, MC = HR1×b2 and twisting moment on the crankpin, TC = HT1 × r

Equivalent twisting moment on the crankpin,

......(i) We also know that twisting moment on the crankpin,

..........(ii) where From equations (i) and (ii), the diameter of the crankpin is determined.



(b) Design of shaft under the flywheel Let ds = Diameter of the shaft in mm. We know that bending moment on the shaft, MS = R3 × c1 and twisting moment on the shaft, TS = FT × r Equivalent twisting moment on the shaft,

.....(i) We also know that equivalent twisting moment on the shaft,

............(ii) From equations (i) and (ii), the diameter of the shaft is determined. (c) Design of shaft at the juncture of right hand crank arm Let ds1 = Diameter of the shaft at the juncture of right hand crank arm. We know that bending moment at the juncture of the right hand crank arm,

and the twisting moment at the juncture of the right hand crank arm, Equivalent twisting moment at the juncture of the right hand crank arm,

......(i) We also know that equivalent twisting moment,

......(ii) From equations (i) and (ii), the diameter of the shaft at the juncture of the right hand crank arm is determined. (d) Design of right hand crank web The right hand crank web is subjected to the following stresses: (i)Bending stresses in 2 planes normal to each other due to the radial and tangential components of FQ (ii) Direct compressive stress due to FR, and (iii) Torsional stress. The bending moment due to the radial component of FQ is given by,

.................(i) .....(ii)

From equations (i) and (ii), the value of bending stress bR is determined. The bending moment due to the tangential component of FQ is maximum at the juncture of crank and shaft. It is given by

.......(iii) where



.........(iv) Where From equations (iii) and (iv), the value of bending stress bT is determined. The direct compressive stress is given by,

The maximum compressive stress (c) will occur at the upper left corner of the cross-section of the crank. Now, the twisting moment on the arm,

We know that shear stress on the arm,

Where

Maximum or total combined stress,

The value of (c)max should be within safe limits. If it exceeds the safe value, then the dimension w may be increased because it does not affect other dimensions. (e) Design of left hand crank web Since the left hand crank web is not stressed to the extent as the right hand crank web, therefore, the dimensions for the left hand crank web may be made same as for right hand crank web. ( f ) Design of crankshaft bearings The bearing 2 is the most heavily loaded and should be checked for the safe bearing pressure. We know that the total reaction at the bearing 2,

where l2 = Length of bearing 2.

Side or Overhung Crankshaft The side or overhung crankshafts are used for medium size and large horizontal engines. Its main advantage is that it requires only two bearings in either the single or two crank construction. The design procedure for the side or overhung crankshaft is same as that for centre crankshaft. Let us now design the side crankshaft by considering the two crank positions, i.e. when the crank is at dead centre (or when the crankshaft is subjected to maximum bending moment) and when the crank is at an angle at which the twisting moment is maximum. These two cases are discussed in detail as below: 1. When the crank is at dead centre. Consider a side crankshaft at dead centre with its loads and distances of their application, as shown in Fig.



The various parts of the side crankshaft, when the crank is at dead centre, are now designed as discussed below: (a) Design of crankpin. The dimensions of the crankpin are obtained by considering the crankpin in bearing and then checked for bending stress. Let dc = Diameter of the crankpin in mm, lc = Length of the crankpin in mm, and pb = Safe bearing pressure on the pin in N/mm2. It may be between 9.8 to 12.6 N/mm 2. We know that FP = dc*lc*pb



From this expression, the values of dc and lc may be obtained. The length of crankpin is usually from 0.6 to 1.5 times the diameter of pin. The crankpin is now checked for bending stress. If it is assumed that the crankpin acts as a cantilever and the load on the crankpin is uniformly distributed, then maximum bending moment will be (FPlc)/2. But in actual practice, the bearing pressure on the crankpin is not uniformly distributed and may, therefore, give a greater value of bending moment ranging between (FPlc)/2 and (FPlc). So, a mean value of bending moment, i.e. 3/4(FPlc) may be assumed. Maximum bending moment at the crankpin,

Section modulus for the crankpin,

Bending stress induced, b = M/Z This induced bending stress should be within the permissible limits. (b) Design of bearings. The bending moment at the centre of the bearing 1 is given by M = FP (0.75lc + t + 0.5l1) ...(i) where lc = Length of the crankpin, t = Thickness of the crank web = 0.45 dc to 0.75 dc, and l1 = Length of the bearing = 1.5 dc to 2 dc. We also know that ..............(ii) From equations (i) and (ii), the diameter of the bearing 1 may be determined. Note : The bearing 2 is also made of the same diameter. The length of the bearings are found on the basis of allowable bearing pressures and the maximum reactions at the bearings. (c) Design of crank web. When the crank is at dead centre, the crank web is subjected to a bending moment and to a direct compressive stress. We know that bending moment on the crank web, M = FP(0.75lc+0.5t)

We also know that direct compressive stress,

Total stress on the crank web, This total stress should be less than the permissible limits. (d) Design of shaft under the flywheel. The total bending moment at the flywheel location will be the resultant of horizontal bending moment due to the gas load and belt pull and the vertical bending moment due to the flywheel weight. Let ds = Diameter of shaft under the flywheel. We know that horizontal bending moment at the flywheel location due to piston gas load, M1 = FP(a + b2) – H1b2 = H2b1 and horizontal bending moment at the flywheel location due to belt pull,

Total horizontal bending moment, MH=M1+M2



We know that vertical bending moment due to flywheel weight,

Resultant bending moment, ........(i) We also know that .................(ii) From equations (i) and (ii), the diameter of shaft (ds) may determined. 2. When the crank is at an angle of maximum twisting moment. Consider a position of the crank at an angle of maximum twisting moment as shown in Fig. We have already discussed in the design of a centre crankshaft that the thrust in the connecting rod (FQ) gives rise to the tangential force (FT) and the radial force (FR).

The reactions at the bearings 1 and 2 due to the flywheel weight (W) and resultant belt pull (T1 + T2) will be same as discussed earlier. Now the various parts of the crankshaft are designed as discussed below: (a) Design of crank web. The most critical section is where the web joins the shaft. This section is subjected to the following stresses : (i) Bending stress due to the tangential force FT ; (ii) Bending stress due to the radial force FR ; (iii) Direct compressive stress due to the radial force FR ; and (iv) Shear stress due to the twisting moment of FT. We know that bending moment due to the tangential force,

where d1 = Diameter of the bearing 1.

.....(i)



.......(ii)

.............(iii)

.............(iv)

Where

Now the total or maximum stress is given by

............(v) This total maximum stress should be less than the maximum allowable stress. (b) Design of shaft at the junction of crank Let ds1 = Diameter of the shaft at the junction of the crank. We know that bending moment at the junction of the crank, M = FQ*(0.75*lc + t) and twisting moment on the shaft T = FT × r Equivalent twisting moment, ................(i) We also know that equivalent twisting moment, .................(ii) From equations (i) and (ii), the diameter of the shaft at the junction of the crank (ds1) may be determined. (c) Design of shaft under the flywheel Let ds = Diameter of shaft under the flywheel. The resultant bending moment (MR) acting on the shaft is obtained in the similar way as discussed for dead centre position. We know that horizontal bending moment acting on the shaft due to piston gas load,



..............(i)

...................(ii) From equations (i) and (ii), the diameter of shaft under the flywheel (ds) may be determined. Design Problem-4: Design a plain carbon steel centre crankshaft for a single acting four stroke single cylinder engine for the following data: 2 Bore = 400 mm; Stroke=600 mm; Engine speed = 200 r.p.m; Mean effective pressure = 0.5 N/mm ; 2 Maximum combustion pressure = 2.5 N/mm ; Weight of flywheel used as a pulley = 50 kN; Total belt pull = 6.5 kN. When the crank has turned through 35° from the top dead centre, the pressure on the piston is 1 N/mm2 and the torque on the crank is maximum. The ratio of the connecting rod length to the crank radius is 5. Assume any other data required for the design. 2 Given : D = 400 mm ; L = 600 mm (or) r = 300 mm ; pm = 0.5 N/mm2 ; p = 2.5 N/mm ; W = 50 kN ; 2 T1 + T2 = 6.5 kN ; = 35° ; p= 1 N/mm ; l/r = 5 We shall design the crankshaft for the two positions of the crank, i.e. firstly when the crank is at the dead centre; and secondly when the crank is at an angle of maximum twisting moment. 1. Design of the crankshaft when the crank is at the dead centre We know that the piston gas load,

Assume that the distance (b) between the bearings 1 and 2 is equal to twice the piston diameter (D). b = 2D = 2 × 400 = 800 mm

We know that due to the piston gas load, there will be two horizontal reactions H1 and H2 at bearings 1 and 2 respectively, such that

Assume that the length of the main bearings to be equal, i.e., c1 = c2 = c / 2. We know that due to the weight of the flywheel acting downwards, there will be two vertical reactions V2 and V3 at bearings 2 and 3 respectively, such that



Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin Let dc = Diameter of the crankpin in mm ; lc = Length of the crankpin in mm ; and 2 b = Allowable bending stress for the crankpin. It may be assumed as 75 MPa or N/mm . We know that the bending moment at the centre of the crankpin, MC = H1× b2 = 157.1 × 400 = 62840 kN-mm ...(i) We also know that

....(ii) Equating equations (i) and (ii), we have

We know that length of the crankpin,

(b) Design of left hand crank web We know that thickness of the crank web, t = 0.65×dc + 6.35 mm = 0.65 × 205 + 6.35 = 139.6 say 140 mm and width of the crank web, w = 1.125×dc+12.7 mm = 1.125 × 205 + 12.7 = 243.3 say 245 mm We know that maximum bending moment on the crank web,

Since the total stress on the crank web is less than the allowable bending stress of 75 MPa, therefore, the design of the left hand crank web is safe. (c) Design of right hand crank web From the balancing point of view, the dimensions of the right hand crank web (i.e. thickness and width) are made equal to the dimensions of the left hand crank web.



(d) Design of shaft under the flywheel Let ds = Diameter of the shaft in mm. Since the lengths of the main bearings are equal, therefore

Assuming width of the flywheel as 300 mm, we have c = 365 + 300 = 665 mm Allowing space for gearing and clearance, let us take c = 800 mm.

We know that bending moment due to the weight of flywheel, and bending moment due to the belt pull, Resultant bending moment on the shaft,

We also know that bending moment on the shaft (MS),

2. Design of the crankshaft when the crank is at an angle of maximum twisting moment We know that piston gas load,

In order to find the thrust in the connecting rod (FQ), we should first find out the angle of inclination of the connecting rod with the line of stroke (i.e. angle ϕ). We know that

We know that thrust in the connecting rod,

Tangential force acting on the crankshaft,

Due to the tangential force (FT), there will be two reactions at bearings 1 and 2, such that

Due to the radial force (FR), there will be two reactions at bearings 1 and 2, such that



Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin Let dc = Diameter of crankpin in mm. We know that the bending moment at the centre of the crankpin, MC = HR1× b2 = 47.3 × 400 = 18920 kN-mm and twisting moment on the crankpin, TC = HT1×r = 42 × 300 = 12600 kN-mm Equivalent twisting moment on the crankpin,

We know that equivalent twisting moment (Te),

Since this value of crankpin diameter (i.e. dc = 149 mm) is less than the already calculated value of dc = 205 mm, therefore, we shall take dc = 205 mm. (b) Design of shaft under the flywheel Let ds = Diameter of the shaft in mm. The resulting bending moment on the shaft will be same as calculated eariler, i.e. 6 MS = 10.08 × 10 N-mm and twisting moment on the shaft, 6 TS = FT × r = 84 × 300 = 25200 kN-mm = 25.2 × 10 N-mm Equivalent twisting moment on shaft,


From above, we see that by taking the already calculated value of ds = 135 mm, the induced shear stress is more than the allowable shear stress of 31 to 42 MPa. Hence, the value of ds is calculated by taking = 35 MPa or N/mm2 in the above equation, i.e.

(c) Design of shaft at the juncture of right hand crank arm Let ds1 = Diameter of the shaft at the juncture of the right hand crank arm. We know that the resultant force at the bearing 1,

Bending moment at the juncture of the right hand crank arm,

and twisting moment at the juncture of the right hand crank arm, TS1 = FT × r = 84 × 300 = 25 200 kN-mm = 25.2 × 106 N-mm Equivalent twisting moment at the juncture of the right hand crank arm,




(d) Design of right hand crank web Let bR = Bending stress in the radial direction ; and bT = Bending stress in the tangential direction. We also know that bending moment due to the radial component of FQ,

......(i) We also know that bending moment,

We know that bending moment due to the tangential component of FQ,

We also know that bending moment,

Direct compressive stress,

and total compressive stress,  We know that twisting moment on the arm,



and shear stress on the arm,

We know that total or maximum combined stress,

Since the maximum combined stress is within the safe limits, therefore, the dimension w = 245 mm is accepted. (e) Design of left hand crank web The dimensions for the left hand crank web may be made same as for right hand crank web. ( f ) Design of crankshaft bearings Since the bearing 2 is the most heavily loaded, therefore, only this bearing should be checked for bearing pressure. We know that the total reaction at bearing 2,

Total bearing pressure

2

Since this bearing pressure is less than the safe limit of 5 to 8 N/mm , therefore, the design is safe.

Valve Gear Mechanism The valve gear mechanism of an I.C. engine consists of those parts which actuate the inlet and exhaust valves at the required time with respect to the position of piston and crankshaft Fig.(a) shows the valve gear arrangement for vertical engines. The main components of the mechanism are valves, rocker arm, valve springs, push rod, cam and camshaft.

The fuel is admitted to the engine by the inlet valve and the burnt gases are escaped through the exhaust valve. In vertical engines, the cam moving on the rotating camshaft pushes the cam follower and push rod upwards, thereby transmitting the cam action to rocker arm. The camshaft is rotated by the toothed belt from the crankshaft. The rocker arm is pivoted at its centre by a fulcrum pin. When one end of the rocker arm is pushed up by the push rod, the other end moves downward. This pushes down the valve stem causing the valve to move down, thereby opening the port. When the cam follower moves over the circular portion of cam, the pushing action of the rocker arm on the valve is released and the valve returns to its seat and closes it by the action of the valve spring.



In some of the modern engines, the camshaft is located at cylinder head level. In such cases, the push rod is eliminated and the roller type cam follower is made part of the rocker arm. Such an arrangement for the horizontal engines is shown in Fig. (b).

Valves The valves used in internal combustion engines are of the following three types; 1. Poppet or mushroom valve; 2. Sleeve valve; 3. Rotary valve. Out of these three valves, poppet valve, as shown in Fig., is very frequently used. It consists of head, face and stem. The head and face of the valve is separated by a small margin, to aviod sharp edge of the valve and also to provide provision for the regrinding of the face. The face angle generally varies from 30° to 45°. The lower part of the stem is provided with a groove in which spring retainer lock is installed. Since both the inlet and exhaust valves are subjected to high temperatures of 1930°C to 2200°C during the power stroke, therefore, it is necessary that the material of the valves should withstand these temperatures. Thus the material of the valves must have good heat conduction, heat resistance, corrosion resistance, wear resistance and shock resistance. It may be noted that the temperature at the inlet valve is less as compared to exhaust valve. Thus, the inlet valve is generally made of nickel chromium alloy steel and the exhaust valve (which is subjected to very high temperature of exhaust gases) is made from silchrome steel which is a special alloy of silicon and chromium. In designing a valve, it is required to determine the following dimensions: (a) Size of the valve port Let ap = Area of the port, vp = Mean velocity of gas flowing through the port, a = Area of the piston, and v = Mean velocity of the piston. We know that ap.vp = a.v

Poppet (or) mushroom valve Conical poppet valve in the port The mean velocity of the gas ( vp) may be taken from the following table.

Sometimes, inlet port is made 20 to 40 percent larger than exhaust port for better cylinder charging.



(b) Thickness of the valve disc The thickness of the valve disc (t), as shown in Fig., may be determined empirically from the following relation, i.e.

where k = Constant = 0.42 for steel and 0.54 for cast iron, dp = Diameter of the port in mm, 2 p = Maximum gas pressure in N/mm , and b = Permissible bending stress in MPa or N/mm2 = 50 to 60 MPa for carbon steel and 100 to 120 MPa for alloy steel. (c) Maximum lift of the valve h = Lift of the valve. The lift of the valve may be obtained by equating the area across the valve seat to the area of the port. For a conical valve, as shown in Fig., we have

where α= Angle at which the valve seat is tapered = 30° to 45° In case of flat headed valve, the lift of valve is given by

The valve seats usually have the same angle as the valve seating surface. But it is preferable to make the angle of valve seat 1/2° to 1° larger than the valve angle as shown in Fig. below. This results in more effective seat.

(d) Valve stem diameter The valve stem diameter (ds) is given by

Note: The valve is subjected to spring force which is taken as concentrated load at the centre. Due to this spring force (Fs), the stress in the valve (t) is given by

Design Problem-5: The conical valve of an I.C. engine is 60 mm in diameter and is subjected to a maximum gas pressure of 4 N/mm2. The safe stress in bending for the valve material is 46 MPa. The valve is made of steel for which k = 0.42. The angle at which the valve disc seat is tapered is 30°. Determine : 1. thickness of the valve head ; 2. stem diameter ; and 3. maximum lift of the valve. 2 Given : dp = 60 mm ; p = 4 N/mm2 ; b = 46 MPa = 46 N/mm ; k = 0.42 ; α= 30° 1. Thickness of the valve head We know that thickness of the valve head,



2. Stem diameter We know that stem diameter

3. Maximum lift of the valve We know that maximum lift of the valve,

Rocker Arm The rocker arm is used to actuate the inlet and exhaust valves motion as directed by the cam and follower. It may be made of cast iron, cast steel, or malleable iron. In order to reduce inertia of the rocker arm, an I-section is used for the high speed engines and it may be rectangular section for low speed engines. In four stroke engines, the rocker arms for the exhaust valve is the most heavily loaded. Though the force required to operate the inlet valve is relatively small, yet it is usual practice to make the rocker arm for the inlet valve of the same dimensions as that for exhaust valve. A typical rocker arm for operating the exhaust valve is shown in Fig. The lever ratio a / b is generally decided by considering the space available for rocker arm. For moderate and low speed engines, a / b is equal to one. For high speed engines, the ratio a / b is taken as 1/ 1.3. The various forces acting on the rocker arm of exhaust valve are the gas load, spring force and force due to valve acceleration.



Since the maximum load on the rocker arm for exhaust valve is more than that of inlet valve, therefore, the rocker arm must be designed on the basis of maximum load on the rocker arm for exhaust valve, as discussed below : 1. Design for fulcrum pin. The load acting on the fulcrum pin is the total reaction (RF) at the Fulcrumpoint. Let d1 = Diameter of the fulcrum pin, and l1 = Length of the fulcrum pin. Considering the bearing of the fulcrum pin. We know that load on the fulcrum pin, RF = d1*l1*pb The ratio of l1 / d1 is taken as 1.25 and the bearing pressure ( pb ) for ordinary lubrication is taken from 2 2 3.5 to 6 N/mm and it may go upto 10.5 N/mm for forced lubrication. The pin should be checked for the induced shear stress. The thickness of the phosphor bronze bush may be taken from 2 to 4 mm. The outside diameter of the boss at the fulcrum is usually taken twice the diameter of the fulcrum pin. 2. Design for forked end. The forked end of the rocker arm carries a roller by means of a pin. For uniform wear, the roller should revolve in the eyes. The load acting on the roller pin is Fc. Let d2 = Diameter of the roller pin, and l2 = Length of the roller pin. Considering the bearing of the roller pin. We know that load on the roller pin, Fc = d2*l2*pb The ratio of l2 / d2 may be taken as 1.25. The roller pin should be checked for induced shear stesss. The roller pin is fixed in eye and the thickness of each eye is taken as half the length of the roller pin. Thickness of each eye = l2 / 2 The radial thickness of eye (t3) is taken as d1/2 . Therefore overall diameter of the eye, D1 = 2 d1 The outer diameter of the roller is taken slightly larger (atleast 3 mm more) than the outer diameter of the eye. A clearance of 1.5 mm between the roller and the fork on either side of the roller is provided. 3. Design for rocker arm cross-section. The rocker arm may be treated as a simply supported beam and loaded at the fulcrum point. We have already discussed that the rocker arm is generally of I-section but for low speed engines, it can be of rectangular section. Due to the load on the valve, the rocker arm is subjected to bending moment. Let l = Effective length of each rocker arm, and b = Permissible bending stress. We know that bending moment on the rocker arm, M = Fe × l ...(i) We also know that bending moment, M = b × Z ...(ii) where Z = Section modulus. From equations (i) and (ii), the value of Z is obtained and thus the dimensions of the section are determined.



4. Design for tappet. The tappet end of the rocker arm is made circular to receive the tappet which is a stud with a lock nut. The compressive load acting on the tappet is the maximum load on the rocker arm for the exhaust valve (Fe).

From this expression, the core diameter of the tappet is determined. The outer or nominal diameter of the tappet (dn) is given as

5. Design for valve spring: valve spring is used to provide sufficient force during the valve lifting process in order to overcome the inertia of valve gear and to keep it with the cam without bouncing. The spring is generally made from plain carbon spring steel. The total load for which the spring is designed is equal to the sum of initial load and load at full lift. Let W1 = Initial load on the spring = Force on the valve tending to draw it into the cylinder on suction stroke, W2 = Load at full lift = Full lift × Stiffness of spring Total load on the spring,W = W1 + W2

Design Problem-6: Design a rocker arm, and its bearings, tappet, roller and valve spring for the exhaust valve of a four stroke I.C. engine from the following data: Diameter of the valve head = 80 mm; Lift of the valve = 25 mm; Mass of associated parts with the valve = 0.4 kg ; Angle of action of camshaft = 110° ; R. P. M. of the crankshaft = 1500. From the probable indicator diagram, it has been observed that the greatest back pressure when the exhaust 2 valve opens is 0.4 N/mm2 and the greatest suction pressure is 0.02 N/mm below atmosphere. The rocker arm is to be of I-section and the effective length of each arm may be taken as 180 mm ; the angle between the two arms being 135°. The motion of the valve may be assumed S.H.M., without dwell in fully open position. Choose your own materials and suitable values for the stresses. Draw fully dimensioned sketches of the valve gear. A rocker arm for operating the exhaust valve is shown in Fig. First of all, let us find the various forces acting on the rocker arm of the exhaust valve. We know that gas load on the valve,





Let us now design the various parts of the rocker arm. 1. Design of fulcrum pin

Now external diameter of the boss, D1 = 2d1 = 2 × 30 = 60 mm Assuming a phosphor bronze bush of 3 mm thick, the internal diameter of the hole in the lever, dh = d1 + 2 × 3 = 30 + 6 = 36 mm Let us now check the induced bending stress for the section of the boss at the fulcrum which is shown in Fig.

2. Design for forked end



Let us now check the roller pin for induced shearing stress. Since the pin is in double shear, therefore, load on the roller pin (Fc),

This induced shear stress is quite safe. The roller pin is fixed in the eye and thickenss of each eye is taken as one-half the length of the roller pin. Thickness of each eye,

Let us now theck the induced bending stress in the roller pin. The pin is neither simply supported in fork nor rigidly fixed at the end. Therefore, the common practice is to assume the load distrubution as shown in Fig.

The maximum bending moment will occur at Y–Y. Neglecting the effect of clearance, we have



Since the radial thickness of eye (t3) is taken as d2 / 2, therefore, overall diameter of the eye, D2 = 2 d2 = 2 × 18 = 36 mm The outer diameter of the roller is taken slightly larger (at least 3 mm more) than the outer diameter of the eye. In the present case, 42 mm outer diameter of the roller will be sufficient. Providing a clearance of 1.5 mm between the roller and the fork on either side of the roller, we have

3. Design for rocker arm cross-section The cross-section of the roker arm is obtained by considering the bending of the sections just near the boss of fulcrum on both sides, such as section A – A and B – B. We know that maximum bending moment at A – A and B – B.

The rocker arm is of I-section. Let us assume the proportions as shown in Fig. below, we know that section modulus,

Bending stress (b),

Width of flange = 2.5 t = 2.5 × 8 = 20 mm Depth of web = 4 t = 4 × 8 = 32 mm and depth of the section = 6 t = 6 × 8 = 48 mm Normally thickness of the flange and web is constant throughout, whereas the width and depth is tapered. 4. Design for tappet screw The adjustable tappet screw carries a compressive load of Fe = 2460 N. Assuming the screw is made of mild steel for which the compressive stress (c) may be taken as 50 MPa.



D3 = 2 × 10 = 20 mm and t4 = 2 × 10 = 20 mm 5. Design for valve spring First of all, let us find the total load on the valve spring. We know that initial load on the spring, W1 = Initial spring force (Fs) = 96.6 N ...(Already calculated) and load at full lift, W2 = Full valve lift × Stiffness of spring (s) = 25 × 10 = 250 N ...(Assuming stiffness of the spring (s) = 10 N/mm)

Total load on the spring, W = W1 + W2 = 96.6 + 250 = 346.6 N Now let us find the various dimensions for the valve spring, as discussed below: (a) Mean diameter of spring coil Let D = Mean diameter of the spring coil, and d = Diameter of the spring wire. We know that Wahl’s stress factor,

The standard size of the wire is SWG 7 having diameter ( d ) = 4.47 mm. Mean diameter of the spring coil, D = C · d = 8 × 4.47 = 35.76 mm and outer diameter of the spring coil, Do = D + d = 35.76 + 4.47 = 40.23 mm (b) Number of turns of the coil Let n = Number of active turns of the coil. We know that maximum compression of the spring,

For squared and ground ends, the total number of the turns, n’= n + 2 = 10 + 2 = 12 (c) Free length of the spring

(d) Pitch of the coil



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MACHINE DESIGN-I LECTURE NOTES (MEEN-422) (Handout to students by [email protected])

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