Machine Design
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MACHINE DESIGN
With some notion of arrangement of machine elements, we may begin the calculations. From data, such as the work done or power consumed, we compute forces on each part for a sequence of positions of the machines cycle using the principles of mechanics then we 1design each element so that it will wi ll perform its allotted function without failure. We must necessarily use the principles of strength of materials, but this course should not be thought of as a review of those principles. Rather, it is an application to engineering problems with the aim of finding suitable dimensions of the machine elements. In the process the designer makes a stress analysis, deciding upon the points in the various parts subjected to the maximum stress conditions (and the kind of stress). s tress). Since it is seldom possible to use any theoretical equation to determine a dimension and adopt the result unthinkingly, the important requisite now is judgement. Computed results only provide evidence for eventual decisions. FOR THE STUDENTS
Education without misrepresentation is impossible. There are so many things that need to be said at once to the beginner that many statements must be simplifications in order to be intelligible. For pedagogical reasons, safe design procedures are often given in this book usually into too much detail. Although it is a convenience to the teacher to be confronted with some uniformity of approach for grading purposes, there are likely to be other correct points of view. Your teacher or supervisor may ask for another. After few years of experience, your conclusions will rest more and more on your own background, but it is hoped that you will always be in search of better design approaches. Since there is no single correct answer to a design problems unless the procedure is specified correctly, including the design factor and the material, your instructor is more interested on how you attack a problem and in the decisions you make than in the results. Therefore, work for a good solution, not so much for an answer. In any subject, there is a certain amount of new language to be learned. Since difficulty with a new subject is often synonymous with an ignorance of the language of the subject, pay close attention to new words, making a real effort to master them. Comprehending study prior to working a problem is truly a time saver, student practice to the contrary notwithstanding. The attitudes that you might begin to acquire have been expressed in Report on Engineering Design from which the following is quoted: 1) Willingness to proceed in the face of incomplete and often contradictory data and incomplete knowledge of the problem. 2) Recognition of the necessity of developing and using engineering judgement. 3) Questioning attitude toward every piece of information, every specification, every method, every result. 4) Recognition of experiment as the ultimate arbiter. 1
5) Willingness to assume final responsibility for a useful result. Engineering is the art of applying the physical sciences to the solution of the problems of mankind. If, after completion of your study of this book, you feel somewhat knowledgeable yet uncertain, our aim will have been at least partially accomplished. The art is never perfected. Moreover, since uncertainty is the father of progress, only the ignorant can afford to certain. As you will see, machine design is engineering. The Essence of Designing
Thorough understanding of materials for design 1. Know the properties and strength of the materials (done by std. test) 2. Apply a suitable factor of ignorance (factor of safety) to produce a working or allowable or design strength. Design strength = strength of material / factor of safety 3. Equate the design strength with the induced i nduced stress due to the load. Sufficient knowledge on the strength theory and effects of load on a material: 1. Know the load on the material. 2. Compute the failure causing stress (induced stress).
the
3. For good design, equate the induced stress with the all owable or design strength of material.
DUCTILE MATERIAL
BRITTLE MATERIAL
Low and medium carbon steels
Cast iron, high carbon and alloy steel, glass, rubber
High resistance to deformation
Low resistance to deformation
Basically soft
Basically hard
High capacity for impact load
Low capacity for impact load
Fails by yielding or necking
Fails by structure
Has a defined yield point
No defined yield point
Low carbon 0.10% - 0.25% Medium carbon 0.25% - 0.50%
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5) Willingness to assume final responsibility for a useful result. Engineering is the art of applying the physical sciences to the solution of the problems of mankind. If, after completion of your study of this book, you feel somewhat knowledgeable yet uncertain, our aim will have been at least partially accomplished. The art is never perfected. Moreover, since uncertainty is the father of progress, only the ignorant can afford to certain. As you will see, machine design is engineering. The Essence of Designing
Thorough understanding of materials for design 1. Know the properties and strength of the materials (done by std. test) 2. Apply a suitable factor of ignorance (factor of safety) to produce a working or allowable or design strength. Design strength = strength of material / factor of safety 3. Equate the design strength with the induced i nduced stress due to the load. Sufficient knowledge on the strength theory and effects of load on a material: 1. Know the load on the material. 2. Compute the failure causing stress (induced stress).
the
3. For good design, equate the induced stress with the all owable or design strength of material.
DUCTILE MATERIAL
BRITTLE MATERIAL
Low and medium carbon steels
Cast iron, high carbon and alloy steel, glass, rubber
High resistance to deformation
Low resistance to deformation
Basically soft
Basically hard
High capacity for impact load
Low capacity for impact load
Fails by yielding or necking
Fails by structure
Has a defined yield point
No defined yield point
Low carbon 0.10% - 0.25% Medium carbon 0.25% - 0.50%
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Last two digits in a four digit number give the approximate average carbon content in points or hundredths of percent. Example: AISI C1030 has 0.30% carbon or 30 points of carbon.
Standard Tensile Test
PEL proportional elastic limit Ou max. stress that would cause failure Oy max. stress without causing failure
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Super Imposed Stress-Strain Curve
Ductile Material: Tensile Stress Properties
Resilience (Modulus of Resilience)
ability to absorb impact loads within the elastic zone
Toughness (Modulus of Toughness) ability to absorb impact loads within the elastic zone. Hookes Law (ut tension sic vis) Within the PEL, the stress is directly proportional to the strain.
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Facts from the Stress-Strain Curve
Suggested Factors of Safety for Novice in Machine Design LOAD
DUCTILE
BRITTLE
Steady
1.5 2.0
3.0 4.0
Minor shock
3.0 4.0
6.0 8.0
Heavy shock
6.0 8.0
12.0 16.0
NOTE: sigma (fiber stresses) tau (shear stresses) Direct Loads: Axial or Normal (F A) Loads are applied at the neutral or symmetric axis
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Definitions Age Hardening
- (precipitation hardening) occurs in some metals, notably certain
stainless steel, aluminium, and copper alloys at ambient temperature after the solution heat treatment, the process being one of a constituent precipitating from solid solution. - is a substance w/ metallic properties, composed to be the metallic
Alloy
elements added which at least one is metal. Alloying elements - in steel are usually considered to be the metallic elements added for
the purpose of modifying the properties. Anisotropy
- is the characteristic of exhibiting different properties when tested in
different directions. Brittleness
- is a tendency to fra cture without appreciable deformation.
Charpy test
- is one in which a specimen supported at both ends as a simple beam is
broken by the impact of a falling pendulum. Cold shortness
- is a brittleness of metals at ordinary or low temperature.
Cold working
- is the process of deforming a metal plastically at a temperature below
the recrystallization temperature and at rate to pr oduce strain hardening. Damping capacity - is the ability of a material to absorb or dump vibrations, which is the
process of absorbing kinetic energy of vibration owing to hysteresis. De carburization - is a loss of carbon from the surface of steel, occurring during hot
rolling, forging and heat treating, when the surrounding medium reacts with the carbon. Ductility
- is that property that permits permanent deformation before fracture in
tension. Elasticity - is the ability of a material to be return and to return to the original shape. Embrittlement - involves the loss of ductility because of a physical of chemical charge of
a material. Force Carbon - is that part of the carbon content of steel or iron that is in the form of
graphite or temper carbon. Hard Drawn is a temper produce in a wire, rod or tu be by cold drawing. Homogeneous Materials - have the structure at all points.
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Isotropic - materials have the same properties in all directions. Izod test -
is a test which a specimen supported at one end as a cantilever beam is
broken by the impact of a falling pendulum. Killed Steel - is a steel that has been deoxidized with a strong deoxylizing agent such as
silicon or aluminium in order to eliminate a reaction between the carbon and oxygen during solidification. Machinability - is a somewhat indefinite property that refers to the relative case with
which the material can be cut. Malleability
-
is a material susceptibility to extreme deformation in rolling or
hammering. Mechanical Properties - are those that have to with stress or strain. Ultimate strength
and percentage elongation for example; Percentage for Elongation
- is the extension in the vicinity of the fracture of a tensile
specimen , express a percentage of the origi nal gage length, as 20% in 2 inches. Percentage reduction or area is the smallest area at the point of rupture of a tensile
specimen divided by the original area. Physical Properties include not mechanical properties and other physical properties
such as density, conductivity, coefficient of thermal conduction. Plasticity is the ability of a metal to be deform considerably without rupture. Poisons Ratio is the ratio of lateral strain to the longitudinal strain with the element is
loaded with a longitudinal tensile force. Precipitation Heat treatment brings clout the precipitation of constituent from a super
saturated solid solution by holding a body at an elevated temperature , also called artificial aging. Proof stress is that stress which causes a specified permanent deformation of a
material. Red Shortness is the brittleness in the steel when it is red hot. Relaxation associated with creep, is the decreasing stress at the constant strain;
important for metals in high temperature service. Residual stresses are those not due to applied loads or temperature gradients; they
exist for various reasons as un equal reasons rates cold working etc.
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Rimmed steel is incompletely deoxidized steel. Solution heat treatment is the process of rolling an alloy at a suitably high temperature,
long enough to permit one or more constituents to pass into solid solution and then cooling fast enough to hold the constituents as a super saturated solution. Stiffness is the ability to resist deformation. Strain Hardening is increasing the hardness and strength by plastic deformation at
temperatures lower than the recrystallization range. Temper is a condition produce in a non ferrous metal by mechanical or thermal
treatment. Toughness is the capacity of a material to withstand a shock load without breaking. Transverse strength refers to the results of a transverse bend test the specimens being
mounted as a simple beam ; also called rupture modulus. Work Hardening is the same as strain hardening. Wrought steel is the steel that has been hammered. Rolled or drawn in the process of
manufacture: it may be plain carbon or alloy steel.
HEAT-TREATMENT TERMS Heat treatment - is an operation or combination of operations involving the heating and
cooling of metal or an alloy in the solid state for the purpose of altering the properties of the metal. Aging - is a change in a metal by which is structure recovers from an unstable or
metastable condition that has been produced by quenching or cold working. Annealing - a comprehensive term, is a heating and slow cooling of a solid metal, usually
done to soften it. Critical range - has the same meaning as transformation range. Drawing - is often use to mean tempering, but this usage conflicts with the meaning of the
drawing of a material through a die and is to be avoided. Graphitizing - causes the combined carbon to transform wholly or in parts into graphitic
or free carbon. Hardening - is the heating of certain steels above the transformation range and then
quenching, for the purpose of increasing the hardness. 8
Malleablizing - is an annealing process whereby combined carbon in white cast iron is
transform wholly or in part to temper carbon. Normalizing - is the heating of an iron-base alloy to some 100°F above the transformation
range with subsequent cooling to bellow that range in still air at room temperature. Spheroidizing - is any heating and cooling of steel that produces a rounded or globular
form of carbide. Stress relieving - is the heating of a metal body to a suitable temperature and holding it at
that temperature for a suitable time for the purpose of reducing internal residual stresses. Tempering is a reheating of hardened or normalized steel to a temperature below the
transformation range, followed by any desired rate of cooling. Transformation range for ferrous metals is the temperature interval during which
austenite is formed during heating.
TYPES OF MATERIALS ALLOY STEEL
Wrought alloy steel is a steel that contains significant quantities of recognized alloying metals, the most common being aluminium, chromium, cobalt, copper, manganese, molybdenum, nickel, phosphorous, silicon, titanium, tungsten and vanadium. WROUGHT IRON
Wrought iron is made by burning the carbon from molten iron and then putting the product through hammering and rolling operation. CAST IRON
Cast iron in a general sense includes white cast iron , malleable iron and nodular cast iron, but when cast iron is such in used without qualifying adjective, gray ir on is meant. MALLEABLE IRON
Malleable iron is a heat treated white cast iron.
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NODULAR CAST IRON
It was a good resistance to thermal shock and its growth at high temperature is less than of gray iron. CAST STEEL
The combination of highest strength and highest ductility is a cast ferrous metal is obtained in cast steel. STAINLESS STEEL
Stainless steel is a relatively expensive but where the environment is significantly corrosive or at high of quite low temperatures. It provides an economical answer for many problems. COPPER ALLOYS
Copper and its alloy have characteristics that determine the advisability of their use; among these may be mentioned: electrical and thermal conductivity, resistance to corrosion, malleability and formability, ductility, strength and excellent machinability, non magnetic, pleasing finish, case of being plated and castability. ALUMINUM ALLOYS
The characteristics of aluminium alloys that suggest their use included: high electrical and thermal conductivity; resistance to some corrosive effects; ease of casting; working and high mechanical properties. PLASTICS
Plastics are divided into two main classes thermosetting, which undergo chemical change and harden on being heated, usually under pressure; and thermoplastic, which soften as the temperature rises and remain soft in the heated state.
SYSTEM OF SPECIFICATION STEEL
SAE
Plain Carbon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10XX Free Cutting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11XX Manganese . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13XX 10
Boron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14XX Nickel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2XXX Nickel chromium . . . . . . . . . . . . . . . . . . . . . . . . . . .3XXX Heat and corrosion resistant . . . . . . . . . . . . . . . . . 303XXX Molybdenum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4XXX Molybdenum-chromium. . . . . . . . . . . . . . . . . . . . .41XX Molybdenum-chromium-nickel. . . . . . . . . . . . . . . .43XX Molybdenum-nickel. . . . . . . . . . . . . . . . . . . . . . . . .46XX Molybdenum-chromium-nickel. . . . . . . . . . . . . . . .4 7XX Molybdenum-nickel. . . . . . . . . . . . . . . . . . . . . . . . .48XX Chromium. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5XXX Heat and corrosion resistant. . . . . . . . . . . . . . . . . 514XX Chromium vanadium. . . . . . . . . . . . . . . . . . . . . . . . 515XX Nickel-chromium-molybdenum. . . . . . . . . . . . . . . . .6XXX Nickel-chromium-molybdenum(except 92XX). . . . . . 9XXX Silicon-manganese. . . . . . . . . . . . . . . . . . . . . . . . . . .
92XX
1. A journal bearing with of 76.2 mm is subjected to a load of 4 900N while rotating at 100 rpm. If each coefficient of friction is 0.02 and the L/D=2.5, find its projected area in mm². Solution: L/D=2.5 L=D (2.5) L= 76.2(2.5) = 190.5mm
area= DxL =76.2x 190.5 area= 14516 mm²
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2. What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105m/m from a load producing a unit stress of 44000psi? Solution: E = Stress/Strain = 44000/0.00105 6
E = 41.905x10 psi
1
3. The shaft whose torque varies from 2000 to 6000 in-lbs. has 1 /2 inches in diameter and 60000psi yield strength. Compute for the shaft m ean average stress. Solution: Tave = (2000+6000)/2 = 4000 in-lbs Ss = (16T/ d3); d = 5 in. = [(16)(4000 in lb)]/ (1.5 in)3 Ss = 6036.099 psi 4.) How many 5/16 inch holes can be punch in one motion in a steel plate made of SAE 1010 steel, 7/16 inch thick using a force of 55tons. The ultimate strength for shear is 50ksi and use 2 factor of safety. Ss = 50 ksi = 50000 psi A = dt = (5/16 in)(7/16 in)
F = Ss x A = 50000 lb/in2 (.4295 in2) F = 21475.73 lb
A = 0.4295 in2
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1. The simple beam shown, 30 inch long =(a+L+d), is mode of ASI C1022 steel, as rolled, left as forged. At a= 10, F1= 3000 lb is a dead load. At d= 100 inch, F 2= 2400 lb is a repeated, reversed load. For N= 15, indefinite life and h= 3b, determine b and h (Ignore stress concentration)
Given L = 30 in a = 10 in h = 3b
F1 = 30000 lb d = 10 in F2 = 2400 lb
F.S. = 15
AISI C1022 as rolled (Table AT 7; Faires p.576)
Oy = 52 ksi Ou = 72 ksi Required Dimension b and H Solution Since it is a ductile material use Soderberg equation 1/FS =O m/Oy + O a/O n Solving for O m
O m= (O max +O min)/2 Where
O max= Mmax C/1 O min= O min C/1
Solving for Mmax, Mmin 10in
10in
R1 A
10in R2
30
B 13
Solving for support reactions + A= 0 F1 (10) + F2 (20) - R2 (30)= 0 (3000)(10) + (2400)(20) - R2(30)=0 R2=2600lb FY =0
R1=3000 + 2400 R2 R1= 2800lb
M diagram Mmax = 28000 lb-in Mmin= 0
O m= O max +O min/2 O m= (2800lb-in)(3b/2)/(27b4/12)/2 O m=9333.335 lb-in/b3
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Solving for O n
O n = 0.8 O n (surface factor) Where O n = 0.5 O u (wrought iron) = 0.5 (72000 psi) O n= 36000 psi Surface factor = 0.5273 (forged) Fig. AF 5 (Faires, page 583)
O n= 0.8 (36psi)(0.5273) O n= 151186.24psi Substituting the value of FS, O m, O y, O a, and O n to Soderberg Equation 1/FS = O m/O y + O a/O n 3
2
3
2
1/1.5 = (9333.335 lb-in/b )/52000 lb/ in + (9333.335 lb- in/b )/15186.24 lb/in 3
3
1/1.5 = 0.79407877 in / b 3
3
b = 1.191118163 in b = 1.06 in h = 3.18 in
Therefore The dimension of the beam to satisfy the given conditions are b= 1.06 in and h= 3.18 in
2. The same as Problem 1, except that the cycle of F@ will not exceed 100,00 and all surface are machined. Given: Same value to problem 1 except of O n 5 FS= 1.5 nc= 10
O m= 933.335 lb-in/b3 O y= 52,000 lb/in2 O n= 0.8 O n1 (surface factor) O n1= O n (106/105) 0.09 O n1= 36000 psi (106/105) 0.09 O n1= 44,289.68 psi AF 5 (Faires , p. 583) For machined surface factor 0.8 9
15
O n= 44,289.68 psi (0.8)(0.89) O n= 31,524.2522 psi Substituting the value to Soderberg Equation 1/FS = O m/O y + O a/O n 3
3
1/1.5 = (9333.335 lb-in)/b (52000 psi) + ( 9333.335 lb- in)/ b 31,524.2522psi 3
1/1.5 = 0.4755/ b 3
3
b = 0.7132 in b = 0.8934 in h = 2.6803 in
Therefore The dimension of the beam to satisfy the given condition are b= 0.8 934 in and h= 2.6803 in
3. The beam shown has a circular section and supports a load F that varies from 1000 lb to 3000 lb, it is machined from AISI C1020 steel as soled. Determine the diameter D if r=0.2D and n=2; indefinite life.
Given: F= 3000lb r/d= 0.2 h= 1.5D F= 1000LB TO 3000 lb
Material C1020 as rolled Faires at 7 O u= 65 ksi O y= 49 ksi BHN= 143
Required: Diameter D
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Solution: Using Soderberg Equation for ductile material
1/FS = O m/O y + kf (O a/O n) O m = ((3000(16)/D3 )+ (1000(16)/ D3))/2 O m = 32 kips/ D3
O n = 0.8 O n (surface factor) O n = 0.5(65) O n = 32.5 ksi O n = 0.8 (32.5)(0. 9) = 23.4 ksi AF 5 (Faires, p. 583) for surface factor of 0. 9 For Fillet
O a =(6Mmax-6Mmin)/2 = (6(3000lb)(6in)/D3)- (6(1000)(6) lb-in/D3)/2 O a = 36,000/D3 Solving for Kf Kf=q (Kt-1)+1 q=( 1/1+ a)/r =(1/1+0.01)/0.2D For Kt=1.72 Table AF 9 (Faires,P.583) Kf= (1/D+0.050)(1.72-1)+1 Kf= (0.72/D+0.05)+1 Substituting the values to Spderberg Equation 1/FS = O m/O y + kf (O a/O n) 3 3 1/1.5 = (32kips/D /49ksi + 36kips/ D /32.5ksi)(0.72/(D+ 0.05)+1) 3 3 1/1.5= (0.6531/ D )+(1.1077/D )(0.72/(D+0.05)(+1)) D= 1.4084 Therefore diameter D= 1.4084in to satisfy the g iven condition
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4. The same as Problem 3, except that the load Fos steady at F= 3000 lb and the beam rotates as a shaft. Given
O u= 65 ksi O y= 45 ksi
F= 3000 lb r= 0.2D h=1.5D
Kf= (0.72/D+0.05)+1 O n= 32.5 ksi
BHN=143
For reversed load O m = 0
O n= 0.6 O n (size factor)(surface factor) O n= 0.6 (32.5)(0.85)(0.5) O n= 14.9175 ksi For Fillet
O a= 6Mmax/tD2 O a = 6(3000lb)(6in)/D3 O a= 108 kips-in 1/FS = O m/O y + kf (O a/O n) 3
1/1.5= (108 kips in/D (14.9175ksi)) ((0.72/D+0.05)+1) D= 22.4122in 5. A connecting link is as shown, except that there is a 1/8 in radial hole drilled through it at the center section. It is machined from AISI 2330, WQT 1000F, and is subjected to a repeated, reversed axial load whose maximum value is 5 kips. For N=1.5, determine the 6 diameter of the link at the hole a) for indefinite life b) for a life of 10 repetitions (no column action) c) in the link found in (a) what is the maximum tensile stress.
F
D Given: Fmax= 5 kips Load= repeated and reversed axial load Material AISI 2330 WQT, 1000 F AT 7 (Faires, 576)
O u= 105 ksi O y= 85 ks BHN=207 18
Required Diameter D a. for indefinite life 6 b. for a life of 10 repitition
Solution: Using Soderberg Equation for ductile material
1/FS = O m/O y + kf (O a/O n) a. Kf=q(Kt-1)+1 q= 1/(1+(a/r)) ; a=0.0025 q= 1/(1+(0.0025/0.0625)) q= 0.9615 Use AF 8 B/h flat Kt=? d/D= o.125/1.25 = 0.1; assume D=1.25 Kt= 2.7 Kf=0.9615(2.7-1)+1 Kf=2.6346 For reversed load ; O m=0 Substituting the values;
Solving for O n
1/FS = O m/O y + kf (O a/O n) 1/FS= Kf(O a/O n) 4 1/1.5= ((5kips/(D /4)-(dD/8))/28.56)(2.6346) 4 28.56/(1.5(2.6346))= 5kips/(( D /4)-(D(1/8)) 4 7.2263(D /4)-(D(1/8))=5 2 5.676 D -0.09034 D-5= 0 D=1.25in
O n= 0.4O u=0.2(105) O n= 42 ksi O n=0.8 O n (surface factor) Surface factor = 0.85 at Table AF5 (Faire, p.583) O n=0.8(42)(0.85) O u=28.56ksi
Therefore; Diameter D= 1.25 in for indefinite life 6
b) For definite life 10 Solving for O n O n=0.8 O n1(surface factor) O n1= O n(106/nc) 0.09 O n1= 42(1)0.09 O n1= 42 19
Using the same value of Kf, FS,
O a substitute the data to the same equation
1/FS= Kf(O a/O n) 4 42/1.5= ((5kips/(D /4)-(dD/8))(2.6346) 4 (28ksi/2.6346)((D /4)-((D)(1/8)))-5=0 2 8.346D -1.3285 D-5= 0 D=0.8577in Therefore; 6 Daimeter D=0.8577 in for definite life of 10
6. Design the size of solid steel shaft to be used for a 500hp, 250 rpm application if the allowable tortional deflection is 1 degree and the allowable stress is 10000 psi and modulus 6 of rigidly is 13x10 psi. Given: P=500 ho N= 250 rpm Ss= 10,000 psi 6 G= 13x10 psi Solution: Solve for the shaft diameter P=2TN T=P/(2N) T= (500(33000/1)/(2(250)) T=10,504.2262 ft-lb = 126,050.7149 in-lb Thus, 3
Ss=16T/D
D= D=4.0041in Solve for the shaft diameter based on deflection (torsional) Let torsional deflection is per 20D length =TL/JG where: = 1x (180/)rad; 1 (/180) T= 126,050.7149 in-lb L= 20D 4 J= (/32)D 6 G=13x10 psi 20
4
6
1 (/180)= (126,050.7149(20D))/((/32)D (13x10 psi) D=
D= 4.8371 in
7.
A short stub shaft, made of SAE 1035 as rolled, receives 30 hp at 300 rpm via 12 in spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed midway between the bearings. The pressure angle of the gear teeth is 20 degrees, N= 1.5 based on the octahedral shear stress theory with varying stresses. A) neglecting the radial component R of the tooth load W, determine the shaft diameter B) considering both the tangential and the radial components, compute the shaft diameter.
Material SAE 1035 as rolled AT 7 Faires, p.576 O u=85 ksi O y= 55 ksi Repeated: O m= O a Given: P= 30 hp n= 300 rpm d super gear= 12 in = 20 FS= 1.5 Required: shaft diameter Solution: Using equivalent stress theory 2 2 1/FS= ¥ (O e/O n) + e/n) Where O e=(O m/O y)(O m)+Kf O a e= (n/y)(m)+Kf a 21
O n= 0.80n (0.85)(surface factor) O n= 0.5O u =0.5 (85000 psi) O n= 42,500 psi Surface factor: .8827 (machined)
O n=0.8(42500psi)(0.85)(0.8827) O n=25510.03 psi O m= O max/2 =O a O max= 32Mmax/D3 Mmax=(Rmax)(L)/4 Rmax=Fmax/cos from cos= Fmax/Rmax Fmax= Tmax/rgear Tmax= 63025P/n Tmax=63025x30hp/300rpm Tmax= 6302.5lb-in
Fmax= Tmax/ rgear Fmax=6302.5lb-in/6in Fmax=1050.416667 lb Rmax= Fmax/cos Rmax=1050.416667lb/cos20 Rmax= 1117.830069 lb Mmax=( (Rmax)(L))/4 = 1117.830069 lb (16in)/4 = 4471.320275 lb-in
O max=32Mmax/D3 O max= (32(4471.320275lb-in))/D3
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O max=(143082.2488lb-in/ D3)/2 O m=O max/2 =( 71541.12441 lb-in/D3)/2 = 71541.12441 lb-in/D3=O a Kf= 1.6 from Faires table AT 13 Substituting O e=(O n)( O n)+ Kf O a O e= (25510.03psi/55000psi)(71541.12441lb-in/D3)+1.6(71541.124441lb in/ D3) O e= (33182.11327+114465.7991)/D3 e=(n/y)m+ kf sa y= 0.6O y =0.6(55000psi) y=33000psi n= 0.6 O n (0.85)(surface factor) = 0.6 (42500psi)(0.85)(0.8827) n= 19132.5225 psi 3
e=16Tmax/ D m=max/2 3 = 16(6302.2)/ D 3 3 m=max/2 = (100840 lb-in/ D )/2 = 50420 lb-in/ D =a Kf s=1.3 from table AT13
Substituting; e=(n/y)m+ kf sa 3
3
=(19132.5225psi/33000psi)(50420 lb-in/ D ) +1.3(50420lb-in/ D ) 3 =(29232.1759+65546)/ D e=94778.17529 lb-in/ D
3
Substituting to the equivalent stress theory equation: 1/FS= ¥ (O e/O n) + e/n) 2
2
1/1.5= (1/1.5=¥ )
1/2.25=
)
23
2 6
2 6
1/ 2.25=(33.49906476/ D )+(24.53986793/ D ) 1/2.25=
2 6
6
D = 130.5875985 in 6 6 D = 13.23129005 in D= 1.537919835 in Therefore; Use standard size, D= 1
8. A beam 2 ft long is made of AISI C1045 as rolled. The dimensions of the beam are 1x3. At the midpoint is repeated, reversed load of 4000 lb. What is the factor of safety? 4000 1ft R
1ft
3 R 1
Given: Material AISI C1045 as rolled O u= 96 ksi L= 2ft b= 1 in h= 3 in Solution
O u/FS= Mc/I ; I= bh 3/12 O u/FS= (FL(12)(h))/(4(bh3)(2)) M=FL/4 O u= 3FL
C= h/2
2
FS= 2bh 2 FS= ((2bh )( O u))/(3FL) 2 FS = (2 (1in)(3in) ()96000)psi)/(3(4000)(2)(12)in-lb) FS= 6
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1. A thin walled cylindrical pressure vessel is subjected to internal pressure which varies from 750kPa
to 3550kPa continuously. The diameter of a shell is 150 cm. find the required
thickness of the cylinder wall based on yield point of 480MPa. Net endurance of 205MPa and factor of safety equal to 2.5. Given: Pmin = 750kPa Pmax = 3550kPa d = 150 cm = 1.5m Oy = 480MPa On = 205MPa Fs = 2.5 Solution: Soderberg Equation 1/Fs = (Om/Oy)+ (Oa/On) Om = (Omax + Omin)/2 Omin = PD/2t = (750)(1.5)/2t Omin = 56.25/t Omax = 3550(1.5)/2t = 266.25/t Om = (56.25/t) + (266.25/t) 2 Om = 161.25/t
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2. A connecting link shown is under a maximum load of 10kips, repeated and reversed. The link has a radial hole drilled through it at the center of the section. It is machined from AISI 2330 o
WQT, 1000 F. Determine the: a. Diameter of the link for a FS = 1.5 5 b. Diameter for a 10 cycle Given: Fmax = 10kips D = 1/8 in Material Specs: o
AISI 2330 WQT, 1000 F Ou = 105ksi Oy = 85ksi BHN = 207 FS = 1.5 Solution: 1/Fs = (Om/Oy) + Kf (Oa/On) Solve for Kf Kf = a (Kt - 1) +1 a = 1/(1 + (a/t)) ; a = 0.0025 a = 1/(1 + (0.0025)/(1/16)) a = 0.96154 for Kt assume D=1.35in d/h = d/D = 0.125/1.25 = 0.1 use solid curve Kt = 2.7 Solve for Kf Kf = 0. 96154 (2.7-1)+1 = 2.634618
26
For Om and Oa repeated and reversed wll prevail Omax = Omin Om = 0 Oa = Omax Oa = Fmax/A 2
= 10kips/(d /4)-d 2
Om = 10/(d /4)-0.125D 2
1/1.5 = 2.65385 [10/(0.785D -0.125D)][1/35.7] 2
0.7854D 0.124D 1.11506 = 0 Quadratic formula D = 1.2737 in b. Nc = 105 cylce Fs = 1.5 Kf = 2.634615 ; D=1.25 Oa=
10
`
2
0.7854D 0.125D On = 0.8On(Surf) 2
5]
On1 = antilog[log 0.8(105) /52.5 1/3 log 0.9(105)/52.5 log 10 On = 63.074 6
5 0.09
On1 = 52.5 (10 /10 )
= 64.589
On = 0.8(63.074)(0.85) 2
1/1.5 = 2.65 [10/(0.785D -0.125D)][1/12.89] D = 1.1656 in Oa = (Omax Omin)/2 = (266.25-56.25)/t 2 Oa = 1050/t
27
1 /2.5 = (161.25/t) + (1050/t) 480000
205000
-3
-3
1 /2.5 = (1.36x10 )/t + (5.12x10 )/t 1 /2.5 = 8.48/t x10
-3
t = 0.0212m
3. The link shown is machined from AISI 1035 steel as rolled and subjected to a repeated tensile load that varies to 0 to 10 kips, h = 1.5b a. Determine these dimension for FS = 1.4 at a section without stress concentration b. How much would these dimensions be decreased if the surfaces of the links were mirror polished
Materials specs: Oy = 55ksi
Ou = 85 BHN = 190
Given: Fmin = 0kips Fmax = 10kips FS = 1.4 Solution: a. Soderberg Equation 1/FS = (Om/Oy) + (Oa/On) Om = Omax + Omin Omax = F/A = 10/bh = 10/1.5b
2
Omin = 0 Om = Oa On = 0.5(85) On = 42.5 2
Om = (10/1.5b + 0)/2 2
Om = 3.33/b
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b. Mirror polished S.F = 1 On = 0.8(47.5)(1) On = 38 1/FS = (Om/Oy) + (Oa/On) 2
2
= (3.33/b (55)) + (3.33/38b ) 2
1/1.4 = (0.061 + 0.0876)/b b = 0.456in h = 0.68in BHN < 400 O`n = 0.5Ou O`n = (0.25)(1 90) = 47.5 On = 0.8O`n(SF) = 0.8(47.5)(0.87) On = 33.06 by Soderberg Equation 2
2
1/1.4 = (3.33/55b ) + (3.33/33.06b ) 2
1/1.4 = (0.061 + 0.1007)/b b = 0.48in h = 0. 714in
b; 0.48 0.456 = 0.024in diff h; 0.714 0.68 = 0.034in diff
29
4. A cantilever beam as shown is to be subjected to a reversing load of 3000lb. Let the radius of the fillet bar = 1/8inch and the material cold rolled SAE1015. Determine the dimensions t, h (b=1.3h) for a design factor of 1.8 based on variable stresses. Consider section A and B, indefinite life. Given: F = 3000lb r = 1/8 in b = 1.3h t = 0.5h Material Specs: SAE 1015 Ou = 77ksi Oy = 63ksi BHN = 140 FS = 1.8 Solution: Soderberg Equation: 1/Fs = (Om/Oy) + Kf (Oa/Ou) reversed Om = 0 Oa = Omax On = 0.8h O`n = 0.5Ou = 38.5 On = 0.8(38.5)(0.89) = 27.412ksi a. Consider fillet 2 Oa = OMmax/td ; Mmax=3000lb(22)
30
2
= 6(66)/0.5h h 3
Oa = 792/h
Kf = 9(kt - 1) + 1 9
= 1/(1+ ( 9/r)) = 1/(1+(0.01/0.125)) = 0.9259
r/d = 0.125/h ; assume h = 2.8 h/d = 1.3h/h = 1.3 Kf = 0. 9259 (2.3-1)+1 = 2.2037 1/Fs = Kf (Oa/On) 3
1/1.8 = 2.2037 (792h /27.412) h = 4.85467 : wrong assumption h = 4.85 r/d = 0.125/4.85 = 0.02577 h/d = 1.3h/h = 1.3 Kf = 0. 9259 (25-1)+1 = 2.388 2
3
1/Fs = 2.388(792 /h )(1/27.412) h = 4. 9898 assuming h = 5.2 r/d = 0.125/5.2 = 0.0240 Kf = 0. 9259(285-1)+1 = 2.7129 3
1/1.8 = 2.7129(792/h )(1/27.412) h = 5.2059in therefore use: h = 5.205 9in, b=6.7677in, t = 2.60295in
31
Shafts Shafts a rotating member transmitting power Axle a stationary member carrying rotating wheels, pulleys, etc. Spindle a short shaft or axle on machines Machine shaft a shaft which is an integral part of the machine Transmission shaft which is used to transmit between the source and the machine
absorbing the power Line shaft or main shaft transmission shaft driven by the prime mover Counter shaft, jack shaft, head shaft, short shaft transmission shaft intermediate
between the line shaft and driven machine Materials for transmission shaft:
Cold rolled, hot rolled, forge carbon steel Relation of power, torque and speed:
And Where: P = power transmitted (KW) T = torque (KN-m) N = speed (rev/sec) F = transmitted load or force (KN) r = radius (m) Stresses in shafts, subject to torsion only:
And (radians) (For solid circular shaft) (For hollow circular shaft)
Where:
SS = torsional shear stress T = torque C = distance from the neutral axis to the outermost fibre r = radius DO = outside diameter Di = inside diameter 32
D = diameter of shaft J = polar moment of inertia = (for solid circular shaft)
= (for hollow circular shaft)
L = length of shaft = angular deformation in length, radius G = modulus of rigidity in year = 11,500,000 psi to 12, 000,000 psi for steel Stresses in solid circular shaft subject to torsion and be nding
Where:
= maximum shear stress = maximum tensile or compressive stress M = bending moment T = torsional moment Strength of shaft with assumed allowable stress (PSME code p.18) For main power transmitting shafts (assumed stress = 4000 psi)
Or
For line shafts carrying pulleys: (assumed stress = 6000 psi) Or For small, short shafts (assumed stress = 8500 psi)
Or Where:
P = power transmitted in HP D = diameter of shaft in inches N = speed in rpm Empirical formula from machinerys handbook Diameter of shaft
1. For allowable twist not exceeding 0.08 deg per ft length
Where:
Or
D = shaft diameter; inches T = torque; in-lb HP = horse power N = speed, rpm In S.I. units (allowable twist 0.26 deg per meter length)
Where:
Or D = shaft diameter; mm T = torque; N mm 33
P = power; KN N = speed, rpm 2. For allowable twist not exceeding 1 deg per 20D length
Where:
Or
D = shaft diameter; inches T = torque; in-lb HP = horse power N = speed, rpm 3. For short ,solid shaft subjected only to heavy transverse shear Where:
V = max traverse shearing loads, lbs 2 SS = max torsional shearing stress, lb/ in Linear deflection of shafting
For steel line shafting, it is considered good practice to limit the linear deflection to a maximum of 0.01 inch per foot of length. Maximum distance: 1. For shafting subject to no bending action except its own weight
For shafting subjected to bending action of pulleys, etc
2.
Where:
L = maximum distance between bearing, ft D = diameter of shafts, inches Note: 1. Pulleys should be placed as closed to the bearings as possible 2. In general, shafting up to three inches in diameter is almost always made from cold rolled steel Sample Problems:
1. What power would a spindle 55 mm in diameter transmit at 480 rpm. Stress allowed for 2 short shaft is 59 N/mm Solution: Ss = 16T/ D 3 59 = 16T/ (55) T = 1,927,390 N-mm = 1.92739 kN-m 3
P = 2 TN = 2 (1.92739)(480/60) = 96.88 kw
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2. A hollow shaft has a inner diameter of 0.035 m and an outer diameter of 0.06 m. Compute for the torque if the shear stress is not to exceed 120 Mpa in N-m. Solution: Ss = 16TDo/ (Do - Di ) 4 4 120,000,000 = 16T(0.06)/ [( 0.06) (0.035) ] T = 4500 N-m 4
4
3. A short 61 mm shaft transmits 120 HP. Compute the linear speed of a pulley 55 cm mounted on the shaft. Solution For short shaft (PSME CODE): 3 P = D N/38 where: D = 61 mm = 2.4 in 3 120 = (2.4) N/38 N = 330 rpm V = DN = 0.55 x 3.28)(330) = 1870 ft/min Keys Definitions: 1. Key a machine member employed at the interface of a fair of mating female and male
circular cross sectional members to prevent relative angular motion between these mating members. 2. Key way a groove in the shaft and mating member to which the key fits 3. Splines permanent keys made integral with the shaft and fitting into keys ways broached into the mating hub Stresses in keys:
Where:
P= power transmitted T torque r = radius D = diameter N = speed
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Crushing (compressive) stress:
Shearing stress;
Generally, when the key and shaft are of the same material And
Where:
W = width of key h= thickness of key L= length of key
Sample Problems:
1. A rectangular key was used in a pulley connected to a line shaft with a power of 125 KW at a speed of 900 rpm. If the shearing stress of the shaft is 40 N/mm² and the key to be 22 N/mm². Determine the length of the rectangular key if the width is one fourth that of the shaft diameter. Solution: P=2 TN 125=2T(900/60) T=1.326291 kN-m=1,326,291 N-mm Ss = 16T/D
3
40=16(1326291)/D
3
D= shaft diameter = 55.2 7mm F=T/(D/2) = 13262 91/(55.27/2)= 47,993N 36
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