Wood Industry Technology Program PDF

Wood Industry Technology Program Lectures notes: Resistant and tested wood By Dr. Hamouda Mohamed Dr/ Mohamed Ali Hassan 1  Chapter 1 Dimensional Analysis 2 Dimension: A dimension defines a physical quantity, which can be observed or...

Wood Industry Technology Program Lectures notes: Resistant and tested wood By Dr. Hamouda Mohamed Dr/ Mohamed Ali Hassan 1  Chapter 1 Dimensional Analysis 2 Dimension: A dimension defines a physical quantity, which can be observed or measured, qualitatively. For example; time, length, area, volume, mass, force, temperature and energy are all considered dimensions. Some dimensions are considered as primary and some are considered as secondary. Primary dimensions vary from one unit system to another. Time, length, mass and temperature are considered as primary dimensions. Primary dimensions are also called as base unit. Secondary dimensions are a combination of various dimensions. For example, volume is length cubed, velocity is distance divided by time and force includes the dimensions of mass, length and time. Unit: A unit expresses the quantitative magnitude of a dimension. For example, a unit of length may be measured as a meter, centimeter or a millimeter and a unit of mass may be measured as a kilogram, gram or a microgram. 3 Engineering units Physical quantities are measured using various systems. The most common unit systems are: ✓ The Imperial (English) system (English Engineering System, ees) ✓ Centimeter, gram, second (cgs) system ✓ Meter, kilogram, second (mks) system English system is used primarily by American and British chemical and food industries. Outside the United States and Britain, industry has adopted the mks system, while the sciences have adopted cgs system. The confusion that resulted from the use of various units led to the development of a common system of units. As a result of international agreement in 1960 by the General Conference on Weights and Measures, the “ Systeme International d’Unities” or the SI units, have emerged. SI is designed to meet the needs of both science and industry. Although the use of SI units is gaining momentum, it is still necessary to convert data from one system to another. The various systems in use are shown in Table 1. Table 1–Systems of measurements System Use Length Mass Time Temperature Force Energy English Industrial Foot Pound Second °F Pound BTU mass force Metric Cgs Scientific Centimeter Gram Second °C Dyne Calorie Mks Industrial Meter Kilogram Second °C Kilogram Kilo force calorie SI Universal Meter Kilogram Second K Newton Joule 4 The SI system is based on 7 units. These units, along with their symbols are summarized in Table 2. Table 2–Examples of SI-derived units expressed in terms of base units Quantity Name Symbol Length meter m Mass kilogram kg Time second s Electric current ampere A Temperature kelvin K Amount of substance mole mol Luminous intensity candela cd Derived units are algebraic combinations of base units expressed by means of multiplication and division. Definitions of some commonly used derived units are as follows: Newton (N) : The force that gives to a mass of 1 kg an acceleration of 1 m/s 2. (The force that gives m/s2 acceleration to a 1 kg mass is called 1 newton.) 1 (N) = 1 (kg) x 1 (m s–2) Joule (J) : The work done when a force of 1 N is displaced by a distance of 1 m in the direction of force. (The work done by a 1 N force traveling 1 m in its own direction is called 1 joule.) 1 (J) = 1 (N) x 1 (m) kg m = –––––– x m s2 = kg m2 s–2 (Joule) Heat, energy and work are all in the same dimension. Watt (W) : The power that gives rise to the production of energy at the rate of 1 J/s. 5  (The work done in unit time (s) is called power (J) or watt with its unit.) J N m kg m m kg m2 1 (W) = –––– = –––––– = –––––– x –––– = ––––––– s s s2 s s3 Electrical power unit 1 (W) = 1 (A) x 1 (V) equals to the heat and work units 1 (W) = 1 (J/s) Pascal (Pa) : The force exerted per unit area is called pressure. (The force acting on unit area is called pressure) N kg m s–2 kg 1 (Pa) = ––––– = –––––––– = –––––– (Pascal) m2 m2 m s2 Since the value of Pascal is very small, kpa (1000 times larger) or bar (105 times larger) units are used. 1 bar = 105 Pa Example of SI-derived units expressed in terms of base units, SI-derived units with special names, and SI-derived units expressed by means of special names are given in Table 3, 4 and 5, respectively. Table 3–Examples of SI-derived units expressed in terms of base units SI unit Quantity Name Symbol Area Square meter m2 Volume Cubic meter M3 Speed (velocity) Meter per second m/s Acceleration Meter per square second m/s2 Density Kilogram per cubic meter kg/m3 Concentration Mole per cubic meter mol/m3 Specific volume Cubic meter per kilogram m3/kg 6  Table 4–Examples of SI-derived units with special names Quantity Name Symbol Expression in terms Expression in of other units terms of SI base units Force newton N m kg s–2 Pressure Pascal Pa N m–2 kg m–1 s–2 Energy, work, heat joule J Nm kg m2 s–2 Power watt W J s–1 kg m2 s–3 Table 5–Examples of SI-derived units expressed by means of special names Quantity Name Symbol Expression in terms of SI base units Viscosity Pascal second Pa s kg m–1 s–1 Heat capacity Joule per kelvin J K–1 kg m2 s–2 K–1 Specific heat capacity Joule per kilogram kelvin J kg–1 K–1 m2 s–2 K–1 Thermal conductivity Watt per meter kelvin W m–1 K–1 m kg s–3 K–1 Prefixes Recommended for use in SI system Prefixes for general use are shown in Table 6. Table 6–Prefixes recommended for use in SI Prefix Multiple Symbol tera 1012 T giga 109 G mega 106 M kilo 1000 k hekto 102 h deka 101 da deci 10–1 d centi 10–2 c mili 10–3 m micro 10–6 μ nano 10–9 η pico 10–12 p femto 10–15 f Symbols for the prefixes are written in capital letters when the multiplying factor is 106 and larger. Prefixes designating multiplying factor less than 106 are written in lower case letters. 7  1 g = 103 mg = 106 μg = 109 ng = 1012 pg A dimension expressed as a numerical quantity and a unit must be such that the numerical quantity is between 0.1 and 1000. Examples: 1) 10,000 cm should be 100 m. 2) 0.0000001 m should be 1 μm. 3) 10,000 Pa should be 10 kPa. Dimensional equation The magnitude of a numerical quantity should be written both the number and its unit. An equation that contains both numerals and their units is called a dimensional equation. The units in a dimensional equation are treated just like algebraic terms. All mathematical operations applied on the numerals must also be applied on their corresponding units. Examples: (4 m)2 = 16 m2 J J. kg. K 5 [––––––] (10 kg) (5 K) = 5 (10) (5) [–––––––––] = 250 J kg. K kg. K Addition and subtraction of numerals and their units also follow the rules of algebra, i.e., only same terms can be added or subtracted. Thus, 5 m – 3 m = (5 – 3) m = 2 m, but 5 m– 3 cm cannot be simplified unless their units are expressed in the same units. Equations must be dimensionally consistent. Thus, if the dimension of the left-hand side of an equation is “length,” the dimension of the right-hand side must also be “length;” otherwise, the equation is incorrect. Conversion of units During unit conversion, first prepare an equation containing, the unit being converted, the unit of final answer and conversion factors. For “conversion factor,“ we need a unit conversion table given in Appendix Table A.1. 8 Steps in conversion of units a) Place the units of the final answer on the left-hand side of the equation and the number being converted and its unit are the first entry on the right hand side of equation. Example 1: Thermal conductivity of stainless steel in EES is 10 BTU/h ft °F. Convert this value in EES to SI unit system (J/m s K). First we need to prepare an equation, which is given below: (CF) [J/(m s K)] = BTU/(ft h °F) (CU) where; CF : Conversion factor (çevirme katsayısı) which will be calculated, CU : Conversion units from conversion table. b) Then, put an equal sign between two expressions. c) Conversion factors are found from conversion tables for the units to be converted in the right hand side of equation. In this example, conversion units were found from table and given below: 1 BTU = 1055 J, 1 ft = 0.3048 m, 1°C = 1.8°F (where, temperature difference occurs), 1°C = 1 K, 1 h = 3600 s. d) Set up the conversion factors as a ratio, using Appendix Table 1. In this example, the rations were given below: 1055 J/BTU, ft/0.3048 m, 1.8°F/°C and h/3600 s 9 e) Sequentially multiply the conversion factors such that the original units are systematically eliminated by cancellation replacement with the desired unit. J BTU 1055 J ft 1.8°F °C h (CF) ⎯⎯–⎯ = ⎯⎯⎯– ⎯⎯–⎯ –⎯⎯⎯⎯ ⎯⎯⎯ ––––– ⎯⎯⎯ ms K h ft °F BTU 0.3048 m °C K 3600 s J 1055 BTU J ft °F h J (CF) –⎯⎯⎯ = ⎯⎯–⎯⎯⎯⎯⎯– ⎯⎯⎯⎯⎯⎯⎯⎯– = 1.73 ⎯⎯⎯ msK 0.3048 x 5/9 x 360 h ft °F BTU m K s msK f) After cancellation of units in the right hand side of equation, appropriate conversion factor is calculated. After cancellation, the numerical value for conversion factor (CF) is 1.73. The conversion factor calculated is put in the equation: 1 BTU/(h ft °F) = (1.73) J/(m s K) g) The numerical value in front of the unit converted is now taken into consideration. This numerical value is put on both side of equation. 10 BTU/(h ft °F) = 10 (1.73) J/(m s K) Result : 10 BTU/(h ft °F) = 17.3 J/(m s K) Length units in English unit system 1 in = 2.54 cm 1 foot = 12 in 3 feet = 1 yard 3.28 feet = 1 m Volume units in English unit system 1 gal = 3.79 L 1 gal = 4 quarts 1 quart = 2 pints 1 pint = 16 fluid ounce (fl oz) 1 quart = 32 fluid ounce 10  Some conversion factor of mass: 1 Kilogram = 2.2046 pound 1 Pound = 453.6 gram 1 kilogram = 1000 gram 1 milligram = 1/1000 gram = 10-3 gram 1 centigram = 1/100 gram = 10-2 gram 1 decigram = 1/10 gram 1 quintal = 100 kg 1 metric ton = 1000 kilogram 11  ___________________________________________________________ Dimensions Dimensions: The powers, to which the fundamental units of mass, length and time written as M, L and T are raised, which include their nature and not their magnitude. For example Area = Length x Breadth = [ L1 ] × [L1 ] = [L2 ] = [M0L2T0 ] Power (0,2,0) of fundamental units are called dimensions of area in mass, length and time respectively. e.g. Density = mass/volume = [M] / [L3 ] = [ M1 L-3 T0 ] Dimensional formulae and si units of physical quantities Dimensional Formula:An expression along with power of mass, length & time which indicates how physical quantity depends upon fundamental physical quantity. e.g. Speed = Distance/Time = [L1] / [T1] =[M0 L1 T-1 ] It tells us that speed depends upon L & T. It does not depends upon M. Dimensional Equation: An equation obtained by equating the physical quantity with its dimensional formula is called dimensional equation. The dimensional equation of area, density & velocity are given as under:- Area = [M0 L2 T0] Density = [M1 L-3 T0] Velocity = [M0 L1 T-1] 12  Dimensional formula SI& CGS unit of Physical Quantities Sr. Physical Quantity Formula Dimensions Name of S.I unit No. 1 Force Mass × acceleration [M1L1T-2] Newton (N) 2 Work Force × distance [M1L2T-2] Joule (J) 3 Power Work / time [M1L2T-3] Watt (W) 4 Energy ( all form ) Stored work [M1L2T-2] Joule (J) 5 Pressure, Stress Force/area [M1L-1T-2] Nm-2 6 Momentum Mass × velocity [M1L1T-1] Kgms-1 7 Moment of force Force × distance [M1L2T-2] Nm 8 Impulse Force × time [M1L1T-1] Ns 9 Strain Change in dimension [M0L0T0] No unit / Original dimension 10 Modulus of Stress / Strain [M1L-1T-2] Nm-2 elasticity 11 Surface energy Energy / Area [M1L0T-2] Joule/m2 12 Surface Tension Force / Length [M1L0T-2] N/m 13 Co-efficient of Force × Distance/ [M1L-1T-1] N/m2 viscosity Area × Velocity 14 Moment of inertia Mass × (radius of [M1L2T0] Kg-m2 gyration)2 15 Angular Velocity Angle / time [M0L0T-1] Rad.per sec 16 Frequency 1/Time period [M0L0T-1] Hertz 17 Area Length × Breadth [M0L2T0] Metre2 18 Volume Length × breadth × [M0L3T0] Metre3 height 19 Density Mass/ volume [M1L-3T0] Kg/m3 20 Speed or velocity Distance/ time [M0L1T-1] m/s 21 Acceleration Velocity/time [M0L1T-2] m/s2 22 Pressure Force/area [M1L-1T-2] N/m2 Example.1 Derive the dimensional formula of following Quantity & write down their dimensions. (i) Density (ii) Power (iii) Co-efficient of viscosity (iv) Angle Sol. (i) Density = mass/volume 13  =[M]/[L3] = [M1L-3T0] (ii) Power = Work/Time =Force x Distance/Time =[M1L1T-2] x [L]/[T] =[M1L2T-3] Force x Distance (iii) Co-efficient of viscosity = Area x Velocity Mass x Acceleration x Distance x time length x length x Displacement =[M] x [LT-2] x [L] [T]/[L2] x [L] =[M1L-1T-1] (iv) Angle = arc (length)/radius (length) = [L]/[L] =[M0L0T0] = no dimension Example.2 Explain which of the following pair of physical quantities have the same dimension: (i) Work &Power (ii) Stress & Pressure (iii) Momentum &Impulse Sol. (i) Dimension of work = force x distance = [M1L2T-2] Dimension of power = work / time = [M1L2T-3] Work and Power have not the same dimensions. (ii) Dimension of stress = force / area = [M1L1T-2]/[L2] = [M1L-1T-2] Dimension of pressure = force / area = [M1L1T-2]/[L2] = [M1L-1T-2] Stress and pressure have the same dimension. (iii)Dimension of momentum = mass x velocity= [M1L1T-1] Dimension of impulse = force x time =[M1L1T-1] Momentum and impulse have the same dimension. 14 Principle of homogeneity of dimensions: It states that the dimensions of all the terms on both sides of an equation must be the same. According to the principle of homogeneity, the comparison, addition & subtraction of all physical quantities is possible only if they are of the same nature i.e., they have the same dimensions. If the power of M, L and T on two sides of the given equation are same, then the physical equation is correct otherwise not. Therefore, this principle is very helpful to check the correctness of a physical equation. Example: A physical relation must be dimensionally homogeneous, i.e., all the terms on both sides of the equation must have the same dimensions. In the equation, S = ut + ½ at2 The length (S) has been equated to velocity (u) & time (t), which at first seems to be meaningless, But if this equation is dimensionally homogeneous, i.e., the dimensions of all the terms on both sides are the same, then it has physical meaning. Now, dimensions of various quantities in the equation are: Distance, S = [L1] Velocity, u = [L1T-1] Time, t = [T1] Acceleration, a = [L1T-2] ½ is a constant and has no dimensions. Thus, the dimensions of the term on L.H.S. is S=[L1] and Dimensions of terms on R.H.S. ut + ½ at2 = [L1T-1] [T1] + [L1T-2] [T2] = [L1] + [L1] Here, the dimensions of all the terms on both sides of the equation are the same. Therefore, the equation is dimensionally homogeneous. 15 Dimensional equations, applications of dimensional equations; Dimensional Analysis: A careful examination of the dimensions of various quantities involved in a physical relation is called dimensional analysis. The analysis of the dimensions of a physical quantity is of great help to us in a number of ways as discussed under the uses of dimensional equations. Uses of dimensional equation: The principle of homogeneity & dimensional analysis has put to the following uses: (i) Checking the correctness of physical equation. (ii) To convert a physical quantity from one system of units into another. (iii) To derive relation among various physical quantities. 1. To check the correctness of Physical relations: According to principle of Homogeneity of dimensions a physical relation or equation is correct, if the dimensions of all the terms on both sides of the equation are the same.If the dimensions of even one term differs from those of others, the equation is not correct. Example 3. Check the correctness of the following formulae by dimensional analysis. (i) 𝐹 = 𝑚v2/r (ii) = 2𝜋√𝑙/𝑔 Where all the letters have their usual meanings. Sol. 𝑭 = 𝒎𝐯𝟐/𝐫 Dimensions of the term on L.H.S Force, F = [M1L1T-2] Dimensions of the term on R.H.S 𝒎𝐯𝟐/𝐫 = [M1][L1T-1]2 / [L] =[M1L2T-2]/ [L] =[M1L1T-2] The dimensions of the term on the L.H.S are equal to the dimensions of the term on R.H.S. Therefore, the relation is correct. (ii) 𝒕 = 𝟐𝝅√𝒍/𝒈 Here, Dimensions of L.H.S, t = [T1] = [M0L0T1] Dimensions of the terms on R.H.S Dimensions of (length) = [L1] Dimensions of g (acc due to gravity) = [L1T2] 2𝜋 being constant have no dimensions. 16 Hence, the dimensions of terms 2𝜋√𝑙/𝑔 on R.H.S = (L1/ L1T-2] )1/2 = [T1] = [M0L0T1] Thus, the dimensions of the terms on both sides of the relation are the same i.e., [M0L0T1].Therefore, the relation is correct. Example 4. Check the correctness of the following equation on the basis of 𝐸 dimensional analysis, 𝑉 = √. Here V is the velocity of sound, E is the elasticity and d is the density 𝑑 of the medium. Sol. Here, Dimensions of the term on L.H.S V =[M0L1T-1] Dimensions of elasticity, E = [M1L-1T-2] & Dimensions of density, d = [M1L-3T0] Thus, dimensions on both sides are the same, therefore the equation is correct. Example 5. Using Principle of Homogeneity of dimensions, check the correctness of equation, h = 2Td /rgCos𝜃. Sol. The given formula is, h = 2Td /rgCos𝜃. Dimensions of term on L.H.S Height (h) = [M0L1T0] Dimensions of terms on R.H.S T= surface tension = [M1L0T-2] D= density = [M1L-3T0] r =radius = [M0L1T0] g=acc.due to gravity = [M0L1T-2] Cos𝜃 = [M0L0T0]= no dimensions So, Dimensions of 2Td/rgCos𝜃 = [M1L0T-2] x [M1L-3T0] / [M0L1T0] x [M0L1T-2] = [M2L-5T0] 17 Dimensions of terms on L.H.S are not equal to dimensions on R.H.S. Hence, formula is not correct. Example 6. Check the accuracy of the following relations: (i) E = mgh + ½ mv2; (ii) v3-u2 = 2as2. Sol. (i) E = mgh + ½ mv2 Here,dimensions of the term on L.H.S. Energy, E = [M1L2T-2] Dimensions of the terms on R.H.S, Dimensions of the term, mgh = [M] ×[LT-2] × [L] = [M1L2T-2] Dimensions of the term, ½ mv2= [M] × [LT-1]2= [M1L2T-2] Thus, dimensions of all the terms on both sides of the relation are the same, therefore, the relation is correct. (ii) The given relation is, (iii) v3-u2= 2as2 Dimensions of the terms on L.H.S v3 = [M0] × [LT-1]3= [M0L3T-3] u2 = [M0] × [LT-1]2= [M0L2T-2] Dimensions of the terms on R.H.S 2as2 = [M0] × [LT-2] ×[L]2 = [M0L3T-2] Substituting the dimensions in the relations, v3-u2 = 2as2 We get, [M0L3T-3] - [M0L2T-2] = [M0L3T-2] The dimensions of all the terms on both sides are not same; therefore, the relation is not correct. Example 7. The velocity of a particle is given in terms of time t by the equation v = At + b/t+c What are the dimensions of a, b and c? Sol. Dimensional formula for L.H.S V = [L1T-1] 18 In the R.H.S dimensional formula of At [T]= [L1T-1] A =[LT-1] / [T-1] = [L1T-2] t +c = time, c has dimensions of time and hence is added in t. Dimensions of t + c is [T] Now, b/t+c=v b = v(t + c) = [LT-1] [T] = [L] There dimensions of a= [L1T-2], Dimensions of b = [L] and that of c = [T] Example 8. In the gas equation (P + a/v2) (v – b) = RT, where T is the absolute temperature, P is pressure and v is volume of gas. What are dimensions of a and b? Sol. Like quantities are added or subtracted from each other i.e., (P + a/v2) has dimensions of pressure = [ML-1T-2] Hence, a/v2 will be dimensions of pressure = [ML-1T-2] a = [ML-1T-2] [volume]2= [ML-1T-2] [L3]2 a = [ML-1T-2] [L6]= [ML5T-2] Dimensions of a = [ML5T-2] (v – b) have dimensions of volume i.e., b will have dimensions of volume i.e., [L3] or [M0L3T0] 19 Example 9. Convert a force of 1 Newton to dyne. Sol. To convert the force from MKS system to CGS system, we need the equation Q=n1u1=n2u2 n u1 Thus n2 = u1 2 Here n1 =1, u1=1N, u2=dyne  M L T −2  n =n 1 1 1 2 1  M 2 L2T2−2   M  L  T −2 n2 = n1  1  1  1   M 2  L2  T2   kg  m  s −2 n2 = n1      gm  cm  s   1000gm  100cm  s −2 n2 = n1  gm cm  s      n2 = 1(1000)(100) n2 = 105 Thus 1N=105 dynes. Example 10.Convert work of 1 erg into Joule. Sol: Here we need to convert work from CGS system to MKS system Thus in the equation n1u1 n2= u2 n1=1 u1=erg (CGS unit of work)u2= joule (SI unit of work) nu n2= 1 1 u2 −2 M1 L12T n2 = n1 1 M 2L22T2−2 2 −2  M  L   T  n2 = n1  1  1   1   M 2  L2   T2   gm  cm 2  s −2 n =n      2 1  kg  m   s   gm  cm 2  s −2 n2 = n1       1000gm 100cm   s  n2 = 1(10−3 )(10−2 )2 n 2 = 10−7 Thus, 1 erg= 10−7 Joule. 20  EXERCISES Multiple Choice Questions 1. [ML-1T-2] is the dimensional formula of (A) Force (B) Coefficient of friction (C) Modulus of elasticity (D) Energy. 2. 105Fermi is equal to (A) 1 meter (B) 100 micron (C) 1 Angstrom (D) 1 mm 3. rad / sec is the unit of (A) Angular displacement (B) Angular velocity (C) Angular acceleration (D) Angular momentum 4. What is the unit for measuring the amplitude of a sound? 21  (A) Decibel (B) Coulomb (C) Hum (D) Cycles 5. The displacement of particle moving along x-axis with respect to time is x=at+bt2-ct3. The dimension of c is (A) LT-2 (B) T-3 (C) LT-3 (D) T-3 Short Answer Questions 1. Define Physics. 2. What do you mean by physical quantity? 3. Differentiate between fundamental and derived unit. 4. Write full form of the following system of unit (i) CGS (ii) FPS (iii) MKS 5. Write definition of Dimensions. 6. What is the suitable unit for measuring distance between sun and earth? 7. Write the dimensional formula of the following physical quantity - (i) Momentum (ii) Power (iii) Surface Tension (iv) Strain 8. What is the principle of Homogeneity of Dimensions? 9. Write the S.I & C.G.S units of the following physical quantities- (a) Force (b) Work 10. What are the uses of dimensions? Long Answer Questions 1. Check the correctness of the relation 𝜆 = h /mv; where𝜆is wavelength, h- Planck’sconstant, m is mass of the particle and v - velocity of the particle. 2. Explain different types of system of units. 3. Check the correctness of the following relation by using method of dimensions (i) v = u + at (ii) F = mv / r2 (iii) v2 – u2 = 2as 4. What are the limitations of Dimensional analysis? 5. Convert an acceleration of 100 m/s2 into km/hr2. Answers to multiple choice questions: 1 (C) 2 (C) 3 (B) 4 (A) 5 (C) 22 Chapter 2 Tolerance and Allowance 23 Introduction: Large number of parts for an assembly and results in a considerable saving in the cost of production. In order to control the size of finished part, with due allowance for error, for interchangeable parts is called limit system. It may be noted that when an assembly is made of two parts, the part which enters into the other, is known as enveloped surface (or shaft for cylindrical part) and the other in which one enters is called enveloping surface (or hole for cylindrical part). Notes: 1. The term shaft refers not only to the diameter of a circular shaft, but it is also used to designate any external dimension of a part. 2. The term hole refers not only to the diameter of a circular hole, but it is also used to designate any internal dimension of a part. Important Terms used in Limit System The following terms used in limit system (or interchangeable system) are important from the subject point of view: 1. Nominal size. It is the size of a part specified in the drawing as a matter of conve- nience. 2. Basic size. It is the size of a part to which all limits of variation (i.e. tolerances) are applied to arrive at final dimensioning of the mating parts. The nominal or basic size of a part is often the same. 24 3. Actual size. It is the actual measure dimension of the part. The difference between the basic size and the actual size should not exceed a certain limit, otherwise it will interfere with the interchangeability of the mating parts. 4. Limits of sizes. There are two extreme permissible sizes for a dimension of the part as shown in Fig. 3.1. The largest permissible size for a dimension of the part is called upper or high or maximum limit, whereas the smallest size of the part is known as lower or minimum limit. 5. Allowance. It is the difference between the basic dimensions of the mating parts. The allowance may be positive or negative. When the shaft size is less than the hole size, then the allowance is positive and when the shaft size is greater than the hole size, then the allowance is negative. 6. Tolerance. It is the difference between the upper limit and lower limit of a dimension. In other words, it is the maximum permissible variation in a dimension. The tolerance may be unilateral or bilateral. When all the tolerance is allowed on one side of the nominal size, e.g. 20 0.000 – 0.004 , then it is said to be unilateral system of tolerance. The unilateral system is mostly used in industries as it permits changing the tolerance value while still retaining the same allowance or type of fit. Fig. 3.2. Method of assigning tolerances. 25  When the tolerance is allowed on both sides of the nominal size, e.g. 20+0.002 – 0.002 , then it is said to be bilateral system of tolerance. In this case + 0.002 is the upper limit and – 0.002 is the lower limit. The method of assigning unilateral and bilateral tolerance is shown in Fig. 3.2 (a) and (b) respectively. 7. Tolerance zone. It is the zone between the maximum and minimum limit size, as shown in Fig. 3.3. Fig. 3.3. Tolerance zone. 8. Zero line. It is a straight line corresponding to the basic size. The deviations are measured from this line. The positive and negative deviations are shown above and below the zero line respectively. 9. Upper deviation. It is the algebraic difference between the maximum size and the basic size. The upper deviation of a hole is represented by a symbol ES (Ecart Superior) and of a shaft, it is represented by es. 10. Lower deviation. It is the algebraic difference between the minimum size and the basic size. The lower deviation of a hole is represented by a symbol EI (Ecart Inferior) and of a shaft, it is represented by ei. 11. Actual deviation. It is the algebraic difference between an actual size and the corresponding basic size. 12. Mean deviation. It is the arithmetical mean between the upper and lower deviations. 13. Fundamental deviation. It is one of the two deviations which is conventionally chosen to define the position of the tolerance zone in relation to zero line, 26 3.14 Fits The degree of tightness or looseness between the two mating parts is known as a fit of the parts. The nature of fit is characterised by the presence and size of clearance and interference. The clearance is the amount by which the actual size of the shaft is less than the actual size of the mating hole in an assembly as shown in Fig. 3.5 (a). In other words, the clearance is the difference between the sizes of the hole and the shaft before assembly. The difference must be positive. Fig. 3.5. Types of fits. The interference is the amount by which the actual size of a shaft is larger than the actual finished size of the mating hole in an assembly as shown in Fig. 3.5 (b). In other words, the interference is the arithmetical difference between the sizes of the hole and the shaft, before assembly. The difference must be negative. 3.15 Types of Fits According to Indian standards, the fits are classified into the following three groups : 1. Clearance fit. In this type of fit, the size limits for mating parts are so selected that clearance between them always occur, as shown in Fig. 3.5 (a). It may be noted that in a clearance fit, the tolerance zone of the hole is entirely above the tolerance zone of the shaft. In a clearance fit, the difference between the minimum size of the hole and the maximum size of the shaft is known as minimum clearance whereas the difference between the maximum size of the hole and minimum size of the shaft is called maximum clearance as shown in Fig. 3.5 (a). 27  The clearance fits may be slide fit, easy sliding fit, running fit, slack running fit and loose running fit. 2. Interference fit. In this type of fit, the size limits for the mating parts are so selected that interference between them always occur, as shown in Fig. 3.5 (b). It may be noted that in an interference fit, the tolerance zone of the hole is entirely below the tolerance zone of the shaft. In an interference fit, the difference between the maximum size of the hole and the minimum size of the shaft is known as minimum interference, whereas the difference between the minimum size of the hole and the maximum size of the shaft is called maximum interference, as shown in Fig. 3.5 (b). The interference fits may be shrink fit, heavy drive fit and light drive fit. 3. Transition fit. In this type of fit, the size limits for the mating parts are so selected that either a clearance or interference may occur depending upon the actual size of the mating parts, as shown in Fig. 3.5 (c). It may be noted that in a transition fit, the tolerance zones of hole and shaft overlap. The transition fits may be force fit, tight fit and push fit. 3.16 Basis of Limit System The following are two bases of limit system: 1. Hole basis system. When the hole is kept as a constant member (i.e. when the lower deviation of the hole is zero) and different fits are obtained by varying the shaft size, as shown in Fig. 3.6 (a), then the limit system is said to be on a hole basis. 2. Shaft basis system. When the shaft is kept as a constant member (i.e. when the upper deviation of the shaft is zero) and different fits are obtained by varying the hole size, as shown in Fig. 3.6 (b), then the limit system is said to be on a shaft basis. Fig. 3.6. Bases of limit system. The hole basis and shaft basis system may also be shown as in Fig. 3.7, with respect to the zero line. Fig. 3.7. Bases of limit system. 28 29  Table 3.4. Manufacturing processes and IT grades produced. S.No. Manufacturing IT grade produced S.No. Manufacturing IT grade produced process process 1. Lapping 4 and 5 9. Extrusion 8 to 10 2. Honing 4 and 5 10 Boring 8 to 13 3. Cylindrical 5 to 7 11. Milling 10 to 13 grinding 4. Surface grinding 5 to 8 12. Planing and 10 to 13 shaping 5. Broaching 5 to 8 13. Drilling 10 to 13 6. Reaming 6 to 10 14. Die casting 12 to 14 7. Turning 7 to 13 15. Sand casting 14 to 16 8. Hot rolling 8 to 10 16. Forging 14 to 16 For hole, H stands for a dimension whose lower deviation refers to the basic size. The hole H for which the lower deviation is zero is called a basic hole. Similarly, for shafts, h stands for a dimension whose upper deviation refers to the basic size. The shaft h for which the upper deviation is zero is called a basic shaft. This view along the deck of a liquefied natural gas (LNG) carrier shows the tops of its large, insulated steel tanks. The tanks contain liquefied gas at-162°C. A fit is designated by its basic size followed by symbols representing the limits of each of its two components, the hole being quoted first. For example, 100 H6/g5 means basic size is 100 mm and the tolerance grade for the hole is 6 and for the shaft is 5. Some of the fits commonly used in engineering practice, for holes and shafts are shown in Tables 3.5 and 3.6 respectively according to IS : 2709 – 1982 (Reaffirmed 1993).  Table 3.5. Commonly used fits for holes according to IS : 2709 – 1982 (Reaffirmed 1993). Type Class With holes Remarks and uses of fit of shaft H6 H7 H8 H11 a — — — a11 Large clearance fit and widely used. b — — — b11 c — c8 *c 9 c 11 Slack running fit. d — d8 *d 8 d 11 Loose running fit—used for plummer d 9, d10 block bearings and loose pulleys. e e7 e8 *e 8-e 9 — Easy running fit—used for properly lubricated bearings requiring appreciable clearance. In the finer grades, it may be Clearance used on large electric motor and fit turbogenerator bearings according to the working condition. f *f 6 f7 *f 8 — Normal running fit—widely used for grease lubricated or oil lubricated bearings where no substantial temperature differences are encountered—Typical applications are gear box shaft bearings and the bearings of small electric motors, pumps, etc. g *g 5 *g 6 g7 — Close running fit or sliding fit—Also fine spigot and location fit—used for bearings for accurate link work and for piston and slide valves. h *h 5 *h 6 *h 7–h 8 *h11 Precision sliding fit. Also fine spigot and location fit—widely used for non- running parts. j *j5 *j6 *j7 — Push fit for very accurate location with easy assembly and dismantling—Typical applications are coupling, spigots and recesses, gear rings clamped to steel hubs, etc. Transition k *k 5 *k 6 k7 — True transition fit (light keying fit)—used fit for keyed shaft, non-running locked pins, etc. m *m 5 *m 6 m7 — Medium keying fit. n n5 *n 6 n7 — Heavy keying fit—used for tight assembly of mating parts. * Second preference fits.  Type Class With holes Remarks and uses of fit of shaft H6 H7 H8 H11 p p5 *p6 — — Light press fit with easy dismantling for non-ferrous parts. Standard press fit with easy dismantling for ferrous and non- ferrous parts assembly. Interference r r5 *r6 — — Medium drive fit with easy dismantling for fit ferrous parts assembly. Light drive fit with easy dismantling for non-ferrous parts assembly. s s5 *s6 s7 — Heavy drive fit on ferrous parts for permanent or semi-permanent assembly. Standard press fit for non-ferrous parts. t t5 t6 *t7 — Force fit on ferrous parts for permanent assembly. u u5 u6 *u7 — Heavy force fit or shrink fit. v, x — — — — Very large interference fits — not recommended for use y, z — — — — Table 3.6. Commonly used fits for shafts according to IS : 2709 – 1982 (Reaffirmed 1993). Type Class With shafts Remarks and uses of fit of hole *h5 h6 h7 *h8 h9 h11 A — — — — — A11 Large clearance fit and widely used. B — — — — — B11 C — — — — — C11 Slack running fit. Clearance D — *D9 — D10 D10 *D11 Loose running fit. fit E — *E8 — E8* E9 — Easy running fit. F — *F7 — F8 *F8 — Normal running fit. G *G6 G7 — — — — Close running fit or sliding fit, also spigot and location fit. H *H6 H7 H8 H8 H8, H9 H11 Precision sliding fit. Also fine spigot and location fit. Js *Js6 Js7 *Js8 — — — Push fit for very accurate location with easy assembly and disassembly. * Second preference fits.  Type Class With shafts Remarks and uses of fit of hole *h5 h6 h7 *h8 h9 h11 Transi- K *K6 K7 *K8 — — — Light keying fit (true transition) tion fit for keyed shafts, non-running locked pins, etc. M *M6 *M7 *M8 — — — Medium keying fit. N *N6 N7 *N8 — — — Heavy keying fit (for tight assembly of mating surfaces). Interfer- P *P6 P7 — — — — Light press fit with easy ence fit dismantling for non-ferrous parts. Standard press fit with easy dismantling for ferrous and non- ferrous parts assembly. R *R6 R7 — — — — Medium drive fit with easy dismantling for ferrous parts assembly. Light drive fit with easy dismantling for non-ferrous parts assembly. S *S6 S7 — — — — Heavy drive fit for ferrous parts permanent or semi- permanent assembly, standard press fit for non-ferrous parts. T *T6 T7 — — — — Force fit on ferrous parts for permanent assembly. 3.18 Calculation of Fundamental Deviation for Shafts We have already discussed that for holes, the upper deviation is denoted by ES and the lower deviation by EI. Similarly for shafts, the upper deviation is represented by es and the lower deviation by ei. According to Indian standards, for each letter symbol, the magnitude and sign for one of the two deviations (i.e. either upper or lower deviation), which is known as fundamental deviation, have been determined by means of formulae given in Table 3.7. The other deviation may be calculated by using the absolute value of the standard tolerance (IT) from the following relation: ei = es – IT or es = ei + IT It may be noted for shafts a to h, the upper deviations (es) are considered whereas for shafts j to Zc, the lower Computer simulation of stresses on a jet deviation (ei) is to be considered. engine blades. * Second preference fits.  The fundamental deviation for Indian standard shafts for diameter steps from 1 to 200 mm may be taken directly from Table 3.10 (page 76). Table 3.7. Formulae for fundamental shaft deviations. Upper deviation (es) Lower deviation (ei) Shaft designation In microns (for D in mm) Shaft designation In microns (for D in mm) a = – (265 + 1.3 D) J 5 to j 8 No formula for D  120 k 4 to k 7 = + 0.6 3 D = – 3.5 D for D > 120 k for grades =0  3 and  8 b = – (140 + 0.85 D) m = + (IT 7 – IT 6) for D  160 = – 1.8 D n = + 5 (D)0.34 for D > 160 p = + IT 7 + 0 to 5 c = – 52 (D)0.2 r = Geometric mean of values of ei for D  40 for shaft p and s = – (95 + 0.8 D) s = + (IT 8 + 1 to 4) for D  50 for D > 40 = + (IT 7 + 0.4 D) for D > 50 d = – 16 (D)0.44 t = + (IT 7 + 0.63 D) e = – 11 (D)0.41 u = + (IT 7 + D) f = – 5.5 (D)0.41 v = + (IT 7 + 1.25 D) x = + (IT 7 + 1.6 D) g = – 2.5 (D)0.34 y = + (IT 7 + 2 D) z = + (IT 7 + 2.5 D) h =0 za = + (IT 8 + 3.15 D) zb = + (IT 9 + 4 D) zc = + (IT 10 + 5 D) For js, the two deviations are equal to ± IT/2. 3.19 Calculation of Fundamental Deviation for Holes The fundamental deviation for holes for those of the corresponding shafts, are derived by using the rule as given in Table 3.8.  Manufacturing Considerations in Machine Design ◼ 75 Table 3.8. Rules for fundamental deviation for holes. General rule Hole limits are identical with the shaft limits of the same symbol (letter and grade) but disposed on All deviation except those below the other side of the zero line. EI = Upper deviation es of the shaft of the same letter symbol but of opposite sign. N 9 and coarser ES = 0 grades Special rule For sizes J, K, M Upto grade 8 ES = Lower deviation ei of the shaft of the same above and N inclusive letter symbol but one grade finer and of opposite 3 mm sign increased by the difference between the P to ZC upto grade 7 tolerances of the two grades in question. inclusive The fundamental deviation for Indian standard holes for diameter steps from 1 to 200 mm may be taken directly from the following table. Table 3.9. Indian standard ‘H’ Hole Limits for H5 to H13 over the range 1 to 200 mm as per IS : 919 (Part II) -1993. Diameter steps Deviations in micron (1 micron = 0.001 mm) in mm H5 H6 H7 H8 H9 H10 H11 H12 H13 H5 – H13 Over To High High High High High High High High High Low + + + + + + + + + 1 3 5 7 9 14 25 40 60 90 140 0 3 6 5 8 12 18 30 48 75 120 180 0 6 10 6 9 15 22 36 58 90 150 220 0 10 14 8 11 18 27 43 70 110 180 270 0 14 18 18 24 9 13 21 33 52 84 130 210 330 0 24 30 30 40 11 16 25 39 62 100 160 250 460 0 40 50 50 65 13 19 30 46 74 120 190 300 390 0 65 80 80 100 15 22 35 54 87 140 220 350 540 0 100 120 120 140 18 25 40 63 100 160 250 400 630 0 140 160 160 180 20 29 46 72 115 185 290 460 720 0 180 200 78 ◼ A Textbook of Machine Design Example 3.1. The dimensions of the mating parts, according to basic hole system, are given as follows : Hole : 25.00 mm Shaft : 24.97 mm 25.02 mm 24.95 mm Find the hole tolerance, shaft tolerance and allowance. Solution. Given : Lower limit of hole = 25 mm ; Upper limit of hole = 25.02 mm ; Upper limit of shaft = 24.97 mm ; Lower limit of shaft = 24.95 mm Hole tolerance We know that hole tolerance = Upper limit of hole – Lower limit of hole = 25.02 – 25 = 0.02 mm Ans. Shaft tolerance We know that shaft tolerance = Upper limit of shaft – Lower limit of shaft = 24.97 – 24.95 = 0.02 mm Ans. Allowance We know that allowance = Lower limit of hole – Upper limit of shaft = 25.00 – 24.97 = 0.03 mm Ans. Example 3.2. Calculate the tolerances, fundamental deviations and limits of sizes for the shaft designated as 40 H8 / f7. Solution. Given: Shaft designation = 40 H8 / f 7 The shaft designation 40 H8/ f 7 means that the basic size is 40 mm and the tolerance grade for the hole is 8 (i.e. I T 8) and for the shaft is 7 (i.e. I T 7). Tolerances Since 40 mm lies in the diameter steps of 30 to 50 mm, therefore the geometric mean diameter, D = 30  50 = 38.73 mm We know that standard tolerance unit, i = 0.45 3 D + 0.001 D = 0.45 + 0.001 × 38.73 = 0.45 × 3.38 + 0.03873 = 1.559 73 or 1.56 microns = 1.56 × 0.001 = 0.001 56 mm...(Q 1 micron = 0.001 mm) From Table 3.2, we find that standard tolerance for the hole of grade 8 (I T 8) = 25 i = 25 × 0.001 56 = 0.039 mm Ans. and standard tolerance for the shaft of grade 7 (I T 7) = 16 i = 16 × 0.001 56 = 0.025 mm Ans. Note : The value of I T 8 and I T 7 may be directly seen from Table 3.3.  Manufacturing Considerations in Machine Design ◼ 79 Fundamental deviation We know that fundamental deviation (lower deviation) for hole H, EI = 0 From Table 3.7, we find that fundamental deviation (upper deviation) for shaft f, es = – 5.5 (D)0.41 = – 5.5 (38.73)0.41 = – 24.63 or – 25 microns = – 25 × 0.001 = – 0.025 mm Ans.  Fundamental deviation (lower deviation) for shaft f, ei = es – I T = – 0.025 – 0.025 = – 0.050 mm Ans. The –ve sign indicates that fundamental deviation lies below the zero line. Limits of sizes We know that lower limit for hole = Basic size = 40 mm Ans. Upper limit for hole = Lower limit for hole + Tolerance for hole = 40 + 0.039 = 40.039 mm Ans. Upper limit for shaft = Lower limit for hole or Basic size – Fundamental deviation (upper deviation)...(Q Shaft f lies below the zero line) = 40 – 0.025 = 39.975 mm Ans. and lower limit for shaft = Upper limit for shaft – Tolerance for shaft = 39.975 – 0.025 = 39.95 mm Ans. Example 3.3. Give the dimensions for the hole and shaft for the following: (a) A 12 mm electric motor sleeve bearing; (b) A medium force fit on a 200 mm shaft; and (c) A 50 mm sleeve bearing on the elevating mechanism of a road grader. Solution. (a) Dimensions for the hole and shaft for a 12 mm electric motor sleeve bearing From Table 3.5, we find that for an electric motor sleeve bearing, a shaft e 8 should be used with H 8 hole. Since 12 mm size lies in the diameter steps of 10 to 18 mm, therefore the geometric mean diameter, D = 10  18 = 13.4 mm We know that standard tolerance unit, i = 0.45 3 D + 0.001 D = 0.45 + 0.001 × 13.4 = 1.07 + 0.0134 = 1.0834 microns  *Standard tolerance for shaft and hole of grade 8 (IT 8) = 25 i...(From Table 3.2) = 25 × 1.0834 = 27 microns = 27 × 0.001 = 0.027 mm...(Q 1 micron = 0.001 mm) From Table 3.7, we find that upper deviation for shaft ‘e’, es = – 11(D)0.41 = – 11 (13.4)0.41 = – 32 microns = – 32 × 0.001 = – 0.032 mm * The tolerance values may be taken directly from Table 3.3. 80 ◼ A Textbook of Machine Design We know that lower deviation for shaft ‘e’, ei = es – IT = – 0.032 – 0.027 = – 0.059 mm  Dimensions for the hole (H 8) + 0.027 = 12 + 0.000 Ans. and dimension for the shaft (e 8) = 12––0.032 0.059 Ans. (b) Dimensions for the hole and shaft for a medium force fit on a 200 mm shaft From Table 3.5, we find that shaft r 6 with hole H 7 gives the desired fit. Since 200 mm lies in the diameter steps of 180 mm to 250 mm, therefore the geometric mean diameter, D = 180  250 = 212 mm We know that standard tolerance unit, i = 0.45 3 D + 0.001 D = 0.45 3 212 + 0.001 × 212 = 2.68 + 0.212 = 2.892 microns  Standard tolerance for the shaft of grade 6 (IT6) from Table 3.2 = 10 i = 10 × 2.892 = 28.92 microns = 28.92 × 0.001 = 0.02892 or 0.029 mm and standard tolerance for the hole of grade 7 (IT 7) = 16 i = 16 × 2.892 = 46 microns = 46 × 0.001 = 0.046 mm We know that lower deviation for shaft ‘r’ from Table 3.7 1 ei = (IT 7 + 0.4 D) + (IT 7 + 0 to 5) 2 1 = (46 + 0.4  212) + (46 + 3) = 90 microns 2 = 90 × 0.001 = 0.09 mm and upper deviation for the shaft r, es = ei + IT = 0.09 + 0.029 = 0.119 mm  Dimension for the hole H 7 = 200++0.046 0.00 Ans. and dimension for the shaft r 6 = 200++0.119 0.09 Ans. (c) Dimensions for the hole and shaft for a 50 mm sleeve bearing on the elevating mechanism of a road grader From Table 3.5, we find that for a sleeve bearing, a loose running fit will be suitable and a shaft d 9 should be used with hole H 8. Since 50 mm size lies in the diameter steps of 30 to 50 mm or 50 to 80 mm, therefore the geometric mean diameter, D = 30  50 = 38.73 mm  Manufacturing Considerations in Machine Design ◼ 81 We know that standard tolerance unit, i = 0.45 3 D + 0.001 D = 0.45 + 0.001 × 38.73 = 1.522 + 0.03873 = 1.56073 or 1.56 microns Standard tolerance for the shaft of grade 9 (IT 9) from Table 3.2 = 40 i = 40 × 1.56 = 62.4 microns = 62.4 × 0.001 = 0.0624 or 0.062 mm and standard tolerance for the hole of grade 8 (IT 8) = 25 i = 25 × 1.56 = 39 microns = 39 × 0.001 = 0.039 mm We know that upper deviation for the shaft d, from Table 3.7 es = – 16 (D)0.44 = – 16 (38.73)0.44 = – 80 microns = – 80 × 0.001 = – 0.08 mm and lower deviation for the shaft d, ei = es – IT = – 0.08 – 0.062 = – 0.142 mm  Dimension for the hole H 8 = 50++0.039 0.000 Ans. and dimension for the shaft d 9 – 0.08 = 50 – 0.142 Ans. Example 3.4. A journal of nominal or basic size of 75 mm runs in a bearing with close running fit. Find the limits of shaft and bearing. What is the maximum and minimum clearance? Solution. Given: Nominal or basic size = 75 mm From Table 3.5, we find that the close running fit is represented by H 8/g 7, i.e. a shaft g 7 should be used with H 8 hole. Since 75 mm lies in the diameter steps of 50 to 80 mm, therefore the geometric mean diameter, D = 50  80 = 63 mm We know that standard tolerance unit, i = 0.45 3 D + 0.001 D = 0.45 + 0.001 × 63 = 1.79 + 0.063 = 1.853 micron = 1.853 × 0.001 = 0.001 853 mm  Standard tolerance for hole ‘H’ of grade 8 (IT 8) = 25 i = 25 × 0.001 853 = 0.046 mm and standard tolerance for shaft ‘g’ of grade 7 (IT 7) = 16 i = 16 × 0.001 853 = 0.03 mm From Table 3.7, we find that upper deviation for shaft g, es = – 2.5 (D)0.34 = – 2.5 (63)0.34 = – 10 micron = – 10 × 0.001 = – 0.01 mm 82 ◼ A Textbook of Machine Design  Lower deviation for shaft g, ei = es – I T = – 0.01 – 0.03 = – 0.04 mm We know that lower limit for hole = Basic size = 75 mm Upper limit for hole = Lower limit for hole + Tolerance for hole = 75 + 0.046 = 75.046 mm Upper limit for shaft = Lower limit for hole – Upper deviation for shaft...(Q Shaft g lies below zero line) = 75 – 0.01 = 74.99 mm and lower limit for shaft = Upper limit for shaft – Tolerance for shaft = 74.99 – 0.03 = 74.96 mm We know that maximum clearance = Upper limit for hole – Lower limit for shaft = 75.046 – 74.96 = 0.086 mm Ans. and minimum clearance = Lower limit for hole – Upper limit for shaft = 75 – 74.99 = 0.01 mm Ans. 3.20 Surface Roughness and its Measurement A little consideration will show that surfaces Internet damper absorbs shock produced by different machining operations (e.g. turning, milling, shaping, planing, grinding and Hydraulic cylinder superfinishing) are of different characteristics. They folds wheels for storage show marked variations when compared with each other. The variation is judged by the degree of smoothness. A surface produced by superfinishing is the smoothest, while that by planing is the roughest. In the assembly of two mating parts, it becomes Liquid spring absolutely necessary to describe the surface finish in quantitative terms which is measure of micro- irregularities of the surface and expressed in microns. In order to prevent stress concentrations and proper functioning, it may be necessary to avoid or to have certain surface roughness. There are many ways of expressing the sur- Tyres absorb face roughness numerically, but the following two some energy methods are commonly used : 1. Centre line average method (briefly known as CLA method), and Landing Gear : When an aircraft comes in to 2. Root mean square method (briefly known land, it has to lose a lot of energy in a very short time. the landing gear deals with this as RMS method). and prevents disaster. First, mechanical or The centre line average method is defined as liquid springs absorb energy rapidly by being the average value of the ordinates between the compressed. As the springs relax, this energy will be released again, but in a slow surface and the mean line, measured on both sides controlled manner in a damper-the second of it. According to Indian standards, the surface finish energy absorber. Finally, the tyres absorb is measured in terms of ‘CLA’ value and it is denoted energy, getting hot in the process. by Ra.  Manufacturing Considerations in Machine Design ◼ 83 y1 + y2 + y3 +... yn CLA value or Ra (in microns) = n where, y1, y2,...yn are the ordinates measured on both sides of the mean line and n are the number of ordinates. The root mean square method is defined as the square root of the arithmetic mean of the squares of the ordinates. Mathematically, y12 + y22 + y32 +... yn 2 R.M.S. value (in microns) = n According to Indian standards, following symbols are used to denote the various degrees of surface roughness : Symbol Surface roughness (Ra) in microns  8 to 25  1.6 to 8  0.025 to 1.6  Less than 0.025 The following table shows the range of surface roughness that can be produced by various manufacturing processes. Table 3.11. Range of surface roughness. S.No. Manufacturing Surface roughness S.No. Manufacturing Surface roughness process in microns process in microns 1. Lapping 0.012 to 0.016 9 Extrusion 0.16 to 5 2. Honing 0.025 to 0.40 10. Boring 0.40 to 6.3 3. Cylindrical grinding 0.063 to 5 11. Milling 0.32 to 25 4. Surface grinding 0.063 to 5 12. Planing and shaping 1.6 to 25 5. Broaching 0.40 to 3.2 13. Drilling 1.6 to 20 6. Reaming 0.40 to 3.2 14. Sand casting 5 to 50 7. Turning 0.32 to 25 15. Die casting 0.80 to 3.20 8. Hot rolling 2.5 to 50 16. Forging 1.60 to 2.5 3.21 Preferred Numbers When a machine is to be made in several sizes with different powers or capacities, it is necessary to decide what capacities will cover a certain range efficiently with minimum number of sizes. It has been shown by experience that a certain range can be covered efficiently when it follows a geometrical progression with a constant ratio. The preferred numbers are the conventionally rounded off values derived from geometric series including the integral powers of 10 and having as common ratio of the following factors: 5 10, 10 10, 20 10 and 40 10 These ratios are approximately equal to 1.58, 1.26, 1.12 and 1.06. The series of preferred numbers are designated as *R5, R10, R20 and R40 respectively. These four series are called basic series. The other series called derived series may be obtained by simply multiplying or dividing the basic sizes by 10, 100, etc. The preferred numbers in the series R5 are 1, 1.6, 2.5, 4.0 and 6.3. Table 3.12 shows basic series of preferred numbers according to IS : 1076 (Part I) – 1985 (Reaffirmed 1990). * The symbol R is used as a tribute to Captain Charles Renard, the first man to use preferred numbers. 84 ◼ A Textbook of Machine Design Notes : 1. The standard sizes (in mm) for wrought metal products are shown in Table 3.13 according to IS : 1136 – 1990. The standard G.P. series used correspond to R10, R20 and R40. 2. The hoisting capacities (in tonnes) of cranes are in R10 series, while the hydraulic cylinder diameters are in R40 series and hydraulic cylinder capacities are in R5 series. 3. The basic thickness of sheet metals and diameter of wires are based on R10, R20 and R40 series. Wire diameter of helical springs are in R20 series. Table 3.12. Preferred numbers of the basic series, according to IS : 1076 (Part I)–1985 (Reaffirmed 1990). Basic series Preferred numbers R5 1.00, 1.60, 2.50, 4.00, 6.30, 10.00 R10 1.00, 1.25, 1.60, 2.00, 2.50, 3.15, 4.00, 5.00, 6.30, 8.00, 10.00 R20 1.00, 1.12, 1.25, 1.40, 1.60, 1.80, 2.00, 2.24, 2.50, 2.80, 3.15, 3.55, 4.00, 4.50, 5.00, 5.60, 6.30, 7.10, 8.00, 9.00, 10.00 R40 1.00, 1.06, 1.12, 1.18, 1.25, 1.32, 1.40, 1.50, 1.60, 1.70, 1.80, 1.90, 2.00, 2.12, 2.24, 2.36, 2.50, 2.65, 2.80, 3.00, 3.15, 3.35, 3.55, 3.75, 4.00, 4.25, 4.50, 4.75, 5.00, 5.30, 5.60, 6.00, 6.30, 6.70, 7.10, 7.50, 8.00, 8.50, 9.00, 9.50, 10.00 Table 3.13. Preferred sizes for wrought metal products according to IS : 1136 – 1990. Size range Preferred sizes (mm) 0.01 – 0.10 mm 0.02, 0.025, 0.030, 0.04, 0.05, 0.06, 0.08 and 0.10 0.10 – 1 mm 0.10, 0.11, 0.12, 0.14, 0.16, 0.18, 0.20, 0.22, 0.25, 0.28, 0.30, 0.32, 0.35, 0.36, 0.40, 0.45, 0.50, 0.55, 0.60, 0.63, 0.70, 0.80, 0.90 and 1 1 – 10 mm 1, 1.1, 1.2, 1.4, 1.5, 1.6, 1.8, 2.22, 2.5, 2.8, 3, 3.2, 3.5, 3.6, 4, 4.5, 5, 5.5, 5.6, 6, 6.3, 7, 8, 9 and 10 10 – 100 mm 10 to 25 (in steps of 1 mm), 28, 30, 32, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48,50, 52, 53, 55, 56, 58, 60, 62, 63, 65, 67, 68, 70, 71, 72, 75, 78, 80, 82, 85, 88, 90, 92, 95, 98 and 100 100 – 1000 mm 100 to 200 (in steps of 5 mm), 200 to 310 (in steps of 10 mm), 315, 320, 330, 340, 350, 355, 360, 370, 375, 380 to 500 (in steps of 10 mm), 520, 530,550, 560, 580, 600, 630, 650, 670, 700, 710 and 750 – 1000 (in steps of 50 mm) 1000 – 10 000 mm 1000, 1100, 1200, 1250, 1400, 1500, 1600, 1800, 2000, 2200, 2500, 2800, 3000, 3200, 3500, 3600, 4000, 4500, 5000, 5500, 5600, 6000, 6300, 7000, 7100, 8000, 9000 and 10 000 EXERCISES 1. A journal of basic size of 75 mm rotates in a bearing. The tolerance for both the shaft and bearing is 0.075 mm and the required allowance is 0.10 mm. Find the dimensions of the shaft and the bearing bore. [Ans. For shaft : 74.90 mm, 74.825 mm ; For hole : 75.075 mm, 75 mm] 2. A medium force fit on a 75 mm shaft requires a hole tolerance and shaft tolerance each equal to 0.225 mm and average interference of 0.0375 mm. Find the hole and shaft dimensions. [Ans. 75 mm, 75.225 mm ; 75.2625 mm, 75.4875 mm]  Manufacturing Considerations in Machine Design ◼ 85 3. Calculate the tolerances, fundamental deviations and limits of size for hole and shaft in the following cases of fits : (a) 25 H 8 / d 9; and (b) 60 H 7 / m 6 [Ans. (a) 0.033 mm, 0.052 mm; 0, – 0.064 mm, – 0.116 mm; 25 mm, 25.033 mm, 24.936 mm, 24.884 mm (b) 0.03 mm, 0.019 mm; 0.011 mm, – 0.008 mm; 60 mm, 60.03 mm, 59.989 mm, 59.97 mm] 4. Find the extreme diameters of shaft and hole for a transition fit H7/n6, if the nominal or basic diameter is 12 mm. What is the value of clearance and interference? [Ans. 12.023 mm, 12.018 mm; 0.006 mm, – 0.023 mm] 5. A gear has to be shrunk on a shaft of basic size 120 mm. An interference fit H7/u6 is being selected. Determine the minimum and maximum diameter of the shaft and interference. [Ans. 120.144 mm, 120.166 mm; 0.109 mm, 0.166 mm] QUESTIONS 1. Enumerate the various manufacturing methods of machine parts which a designer should know. 2. Explain briefly the different casting processes. 3. Write a brief note on the design of castings? 4. State and illustrate two principal design rules for casting design. 5. List the main advantages of forged components. 6. What are the salient features used in the design of forgings? Explain. 7. What do you understand by ‘hot working’ and ‘cold working’ processes? Explain with examples. 8. State the advantages and disadvantages of hot working of metals. Discuss any two hot working processes. 9. What do you understand by cold working of metals? Describe briefly the various cold working processes. 10. What are fits and tolerances? How are they designated? 11. What do you understand by the nominal size and basic size? 12. Write short notes on the following : (a) Interchangeability; (b) Tolerance; (c) Allowance; and (d) Fits. 13. What is the difference in the type of assembly generally used in running fits and interference fits? 14. State briefly unilateral system of tolerances covering the points of definition, application and advantages over the bilateral system. 15. What is meant by ‘hole basis system’ and ‘shaft basis system’? Which one is preferred and why? 16. Discuss the Indian standard system of limits and fits. 17. What are the commonly used fits according to Indian standards? 18. What do you understand by preferred numbers? Explain fully. OBJECTIVE TYPE QUESTIONS 1. The castings produced by forcing molten metal under pressure into a permanent metal mould is known as (a) permanent mould casting (b) slush casting (c) die casting (d) centrifugal casting 2. The metal is subjected to mechanical working for (a) refining grain size (b) reducing original block into desired shape (c) controlling the direction of flow lines (d) all of these 86 ◼ A Textbook of Machine Design 3. The temperature at which the new grains are formed in the metal is called (a) lower critical temperature (b) upper critical temperature (c) eutectic temperature (d) recrystallisation temperature 4. The hot working of metals is carried out (a) at the recrystallisation temperature (b) below the recrystallisation temperature (c) above the recrystallisation temperature (d) at any temperature 5. During hot working of metals (a) porosity of the metal is largely eliminated (b) grain structure of the metal is refined (c) mechanical properties are improved due to refinement of grains (d) all of the above 6. The parts of circular cross-section which are symmetrical about the axis of rotation are made by (a) hot forging (b) hot spinning (c) hot extrusion (d) hot drawing 7. The cold working of metals is carried out................ the recrystallisation temperature. (a) above (b) below 8. The process extensively used for making bolts and nuts is (a) hot piercing (b) extrusion (c) cold peening (d) cold heading 9. In a unilateral system of tolerance, the tolerance is allowed on (a) one side of the actual size (b) one side of the nominal size (c) both sides of the actual size (d) both sides of the nominal size 10. The algebraic difference between the maximum limit and the basic size is called (a) actual deviation (b) upper deviation (c) lower deviation (d) fundamental deviation 11. A basic shaft is one whose (a) lower deviation is zero (b) upper deviation is zero (c) lower and upper deviations are zero (d) none of these 12. A basic hole is one whose (a) lower deviation is zero (b) upper deviation is zero (c) lower and upper deviations are zero (d) none of these 13. According to Indian standard specifications, 100 H 6 / g 5 means that the (a) actual size is 100 mm (b) basic size is 100 mm (c) difference between the actual size and basic size is 100 mm (d) none of the above 14. According to Indian standards, total number of tolerance grades are (a) 8 (b) 12 (c) 18 (d) 20 15. According to Indian standard specification, 100 H6/g 5 means that (a) tolerance grade for the hole is 6 and for the shaft is 5 (b) tolerance grade for the shaft is 6 and for the hole is 5 (c) tolerance grade for the shaft is 4 to 8 and for the hole is 3 to 7 (d) tolerance grade for the hole is 4 to 8 and for the shaft is 3 to 7 ANSWERS 1. (c) 2. (d) 3. (d) 4. (c) 5. (d) 6. (b) 7. (b) 8. (d) 9. (b) 10. (b) 11. (b) 12. (a) 13. (b) 14. (c) 15. (a)  Chapter 3 The Basic Mechanical Properties of Wood Introduction The mechanical or strength properties of wood measure its fitness and ability to resist applied or external forces. By external forces meant any force outside of a given piece of material that tends to alter its size or shape or deform it in any manner. 'Deformation may also be brought about by forces acting entirely Within a piece , such as are set up “ in wood by changes in its moisture content, but these forces are concerned chiefly with physical properties of wood other than those strictly pertaining to strength. It is largely the mechanical properties that determine the fitness Of wood for structural or building purposes and in numerable other uses , of which furniture, vehicles , implements, and tool handles are but a few common examples. In fact, there is hardly a single use for wood that does not depend, at least to some degree, on one or more of its strength properties. Knowledge of the mechanical properties of wood is obtained through experimentation, either by means of service tests in solving the use oi the material under actual conditions met with in practice, or by laboratory experiments requiring the use of special testing apparatus. From the standpoint of general efficiency, “however , laboratory tests , if properly conducted , are decidedly to be preferred. For one thing, they are more economical of time and material than service determinations and hence far better adapted to the extensive testing that must be carried on in studying the natural variation in wood. Further, _ it is possible in laboratory experiments to establish standard methods of testing , as well as to express the results obtained in_ defined units , thus affording a practical means of comparing different kinds of wood and also making it - possible to specify definite ’ sizes of material for given uses. With laboratory methods it is also possible to control most of the disturbing factors that may seriously affect the results obtained in service tests. One of the principal objectives of laboratory investigations is to determine values per unit area for the various strength proper ties of wood. These, however , because of the complex structure of wood , cannot have a constant value that will be exactly repeated in each test, even though no error be made. The most that can be accomplished is to find average values, the amount of variation above and below this average, and the laws that govern the variation.  1. Dead or steady load. A load is said to be a dead or steady load, when it does not change in magnitude or direction. 2. Live or variable load. A load is said to be a live or variable load, when it changes continually. 3. Suddenly applied or shock loads. A load is said to be a suddenly applied or shock load, when it is suddenly applied or removed. 4. Impact load. A load is said to be an impact load, when it is applied with some initial velocity. Note: A machine part resists a dead load more easily than a live load and a live load more easily than a shock load. 4.1 Stress When some external system of forces or loads act on a body, the internal forces (equal and opposite) are set up at various sections of the body, which resist the external forces. This internal force per unit area at any section of the body is known as unit stress or simply a stress. It is denoted by a Greek letter sigma (). Mathematically, Stress,  = P/A where P = Force or load acting on a body, and A = Cross-sectional area of the body. In S.I. units, the stress is usually expressed in Pascal (Pa) such that 1 Pa = 1 N/m 2. In actual practice, we use bigger units of stress i.e. megapascal (MPa) and gigapascal (GPa), such that 1 MPa = 1 × 106 N/m2 = 1 N/mm2 and 1 GPa = 1 × 109 N/m2 = 1 kN/mm2 4.2 Strain When a system of forces or loads act on a body, it undergoes some deformation. This deformation per unit length is known as unit strain or simply a strain. It is denoted by a Greek letter epsilon (). Mathematically, Strain,  = l / l or l = .l where l = Change in length of the body, and l = Original length of the body. 4.3 Tensile Stress and Strain Fig. 4.1. Tensile stress and strain. When a body is subjected to two equal and opposite axial pulls P (also called tensile load) as shown in Fig. 4.1 (a), then the stress induced at any section of the body is known as tensile stress as shown in Fig. 4.1 (b). A little consideration will show that due to the tensile load, there will be a decrease in cross-sectional area and an increase in length of the body. The ratio of the increase in length to the original length is known as tensile strain.  Simple Stresses in Machine Parts ◼ 89 Let P = Axial tensile force acting on the body, A = Cross-sectional area of the body, l = Original length, and l = Increase in length.  Tensile stress, t = P/A and tensile strain,  t = l / l 4.4 Compressive Stress and Strain When a body is subjected to two equal and opposite axial pushes P (also called compressive load) as shown in Fig. 4.2 (a), then the stress induced at any section of the body is known as compressive stress as shown in Fig. 4.2 (b). A little consideration will show that due to the compressive load, there will be an increase in cross-sectional area and a Shock absorber of a motorcycle absorbs stresses. decrease in length of the body. The ratio Note : This picture is given as additional information and is of the decrease in length to the original not a direct example of the current chapter. length is known as compressive strain. Fig. 4.2. Compressive stress and strain. Let P = Axial compressive force acting on the body, A = Cross-sectional area of the body, l = Original length, and l = Decrease in length.  Compressive stress, c = P/A and compressive strain, c = l /l Note : In case of tension or compression, the area involved is at right angles to the external force applied. 4.5 Young's Modulus or Modulus of Elasticity Hooke's law* states that when a material is loaded within elastic limit, the stress is directly proportional to strain, i.e.   or  = E.  E =  = Pl  A  l * It is named after Robert Hooke, who first established it by experiments in 1678. 90 ◼ A Textbook of Machine Design where E is a constant of proportionality known as Young's modulus or modulus of elasticity. In S.I. units, it is usually expressed in GPa i.e. GN/m2 or kN/mm2. It may be noted that Hooke's law holds good for tension as well as compression. The following table shows the values of modulus of elasticity or Young's modulus (E) for the materials commonly used in engineering practice. Table 4.1. Values of E for the commonly used engineering materials. Material Modulus of elasticity (E) in GPa i.e. GN/m2 or kN/mm2 Steel and Nickel 200 to 220 Wrought iron 190 to 200 Cast iron 100 to 160 Copper 90 to 110 Brass 80 to 90 Aluminium 60 to 80 Timber 10 Example 4.1. A coil chain of a crane required to carry a maximum load of 50 kN, is shown in Fig. 4.3. Fig. 4.3 Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa. Solution. Given : P = 50 kN = 50 × 103 N ; t = 75 MPa = 75 N/mm2 Let d = Diameter of the link stock in mm. π  Area, A = × d 2 = 0.7854 d 2 4 We know that the maximum load (P), 50 × 103 = t. A = 75 × 0.7854 d 2 = 58.9 d 2  d 2 = 50 × 103 / 58.9 = 850 or d = 29.13 say 30 mm Ans. Example 4.2. A piece of wood connect link, as shown in Fig. 4.4, is required to transmit a steady tensile load of 45 kN. Find the tensile stress induced in the link material at sections A-A and B-B. Fig. 4.4. All dimensions in mm.  Solution. Given : P = 45 kN = 45 × 103 N Tensile stress induced at section A-A We know that the cross-sectional area of link at section A-A, A1 = 45 × 20 = 900 mm2  Tensile stress induced at section A-A, = 45  10 P 3 = t1 = 50 N/mm2 = 50 MPa Ans. A1 900 Tensile stress induced at section B-B We know that the cross-sectional area of link at section B-B, A2 = 20 (75 – 40) = 700 mm2  Tensile stress induced at section B-B, = 45  10 P 3 t2 = = 64.3 N/mm2 = 64.3 MPa Ans. A2 700 Example 4.3. A hydraulic press exerts a total load of 3.5 MN. This load is carried by two steel rods, supporting the upper head of the press. If the safe stress is 85 MPa and E = 210 kN/mm 2, find : 1. diameter of the rods, and 2. extension in each rod in a length of 2.5 m. Solution. Given : P = 3.5 MN = 3.5 × 106 N ; t = 85 MPa = 85 N/mm2 ; E = 210 kN/mm2 = 210 × 103 N/mm2 ; l = 2.5 m = 2.5 × 103 mm  Example 4.4. A rectangular wood plate is fixed at each of its four corners by a 20 mm diameter bolt and nut as shown in Fig. 4.5. The plate rests on washers of 22 mm internal diameter and 50 mm external diameter. Copper washers which are placed between the nut and the plate are of 22 mm internal diameter and 44 mm external diameter. If the base plate carries a load of 120 kN (including self-weight, which is equally distributed on the four corners), calculate the stress on the lower washers before the nuts are tightened. What could be the stress in the upper and lower washers, when the nuts are tightened so as to produce a tension of 5 kN on each bolt?  Example 4.5. The piston rod of a steam eng