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 machine’s cycle using the principles of mechanics then we 1design each element so that it will 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). 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 1
Transcript
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 machine’s cycle using the principles of mechanics then we 1design each element so that it will 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).
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.
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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 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).
3. For good design, equate the induced stress with the “allowable” or design strength of the 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%
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Medium carbon 0.25% - 0.50%
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.
Hooke’s 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: Ơ sigma (fiber stresses)
Ʈ 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.
Alloy - is a substance w/ metallic properties, composed to be the metallic 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 fracture 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 produce 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 tube by cold drawing.
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Homogeneous Materials - have the structure at all points.
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 original 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.
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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.
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.
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Hardening - is the heating of certain steels above the transformation range and then quenching, for the purpose of increasing the hardness.
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 iron is meant.
MALLEABLE IRON
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Malleable iron is a heat treated white cast iron.
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.
1. A journal bearing with of 76.2 mm is subjected to a load of 4900N 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 area= DxL
L=D (2.5) =76.2x 190.5
L= 76.2(2.5) = 190.5mm 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
E = 41.905x106 psi
3. The shaft whose torque varies from 2000 to 6000 in-lbs. has 1 1/2 inches in diameter and 60000psi yield strength. Compute for the shaft mean average stress.
Solution:
Tave = (2000+6000)/2
= 4000 in-lbs
Ss = (16T/π d3); d = 5 in.
= [(16)(4000 in – lb)]/ π(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 F = Ss x A
A = π dt = 50000 lb/in2 (.4295 in2)
= π(5/16 in)(7/16 in) 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, F2= 2400 lb is a repeated, reversed load. For N= 15, indefinite life and h= 3b, determine b and h (Ignore stress concentration)
GivenL = 30 in F1 = 30000 lb F.S. = 15a = 10 in d = 10 inh = 3b F2 = 2400 lb
AISI C1022 as rolled (Table AT 7; Faires p.576)
Oy= 52 ksi Ou= 72 ksiRequired
Dimension b and H
SolutionSince it is a ductile material use Soderberg equation
Therefore The dimension of the beam to satisfy the given condition are b= 0.8934 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 Materialr/d= 0.2 C1020 as rolledh= 1.5D Faires at 7F= 1000LB TO 3000 lb Ou= 65 ksi
Oy= 49 ksiBHN= 143
Required: Diameter D
Solution:Using Soderberg Equation for ductile material
1/FS = O m/Oy + kf (O a/On)
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O m = ((3000(16)/πD3 )+ (1000(16)/ πD3))/2O m = 32 kips/ πD3
O n = 0.8 O ‘n (surface factor)O ‘n = 0.5(65)O ‘n = 32.5 ksiO 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)+1Kf= (0.72/D+0.05)+1
Substituting the values to Spderberg Equation
1/FS = O m/Oy + kf (O a/On)1/1.5 = (32kips/πD3 /49ksi + 36kips/ πD3/32.5ksi)(0.72/(D+ 0.05)+1) 1/1.5= (0.6531/ πD3)+(1.1077/D3)(0.72/(D+0.05)(+1))D= 1.4084
Therefore diameter D= 1.4084in to satisfy the given condition
4. The same as Problem 3, except that the load Fos steady at F= 3000 lb and the beam rotates as a shaft.
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 1000ᵒF, and is subjected to a repeated, reversed axial load whose maximum value is 5 kips. For N=1.5, determine the diameter of the link at the hole a) for indefinite life b) for a life of 10 6 repetitions (no column action) c) in the link found in (a) what is the maximum tensile stress.
F
DGiven:
Fmax= 5 kipsLoad= repeated and reversed axial loadMaterial AISI 2330 WQT, 1000 ᵒF AT 7 (Faires, 576)
Ou= 105 ksi O y= 85 ks BHN=207
Required Diameter D a. for indefinite life
b. for a life of 106 repitition
Solution:Using Soderberg Equation for ductile material
Use AF 8 B/h flatKt=?d/D= o.125/1.25 = 0.1; assume D=1.25’’
Kt= 2.7Kf=0.9615(2.7-1)+1Kf=2.6346
For reversed load ; Om=0
Substituting the values; Solving for On
1/FS = O m/Oy + kf (O a/On) O ‘n= 0.4Ou=0.2(105)1/FS= Kf(O a/On) O ‘n= 42 ksi1/1.5= ((5kips/(πD4/4)-(dD/8))/28.56)(2.6346) O n=0.8 O ‘n (surface factor)28.56/(1.5(2.6346))= 5kips/(( πD4/4)-(D(1/8)) Surface factor = 0.85 at7.2263(πD4/4)-(D(1/8))=5 Table AF5 (Faire, p.583)5.676 D2-0.09034 D-5= 0 O n=0.8(42)(0.85)D=1.25in O u=28.56ksi
Therefore;Diameter D= 1.25 in for indefinite life
b) For definite life 106
Solving for O nO n=0.8 O n1(surface factor)O‘n1= O ‘n(106/nc) 0.09O n1= 42(1)0.09
O n1= 42Using the same value of Kf, FS, O a substitute the data to the same equation
Therefore;Daimeter D=0.8577 in for definite life of 106
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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 of rigidly is 13x106 psi.
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.
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Material SAE 1035 as rolled AT 7 Faires, p.576Ou=85 ksiOy= 55 ksiRepeated: Om= Oa
Substituting to the equivalent stress theory equation: 1/FS= √(Oe/On)2+ ¿e/Ҽn)2
1/1.5=√(
147647.9123πDᶟ
25510.03)
2
+(
94778.17529πDᶟ
19132.5225)
2
(1/1.5=√ (5.787837658 / πDᶟ)2+( 4.9537731πDᶟ
)2
)
1/2.25=(5.787837658 / πDᶟ)2+( 4.9537731πDᶟ
)2
)
1/ 2.25=(33.49906476/π2D6)+(24.53986793/π2D6)
1/2.25= 58.05893269
πDᶟπ2D6= 130.5875985 in6
D6= 13.23129005 in6
D= 1.537919835 in
Therefore;
Use standard size, D= 1 1116
∈¿
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8. A beam 2 ft long is made of AISI C1045 as rolled. The dimensions of the beam are 1’’x3’’. At the midpoint is repeated, reversed load of 4000 lb. What is the factor of safety?
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
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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)
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Om = 161.25/t
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 WQT, 1000oF. Determine the:
a. Diameter of the link for a FS = 1.5b. Diameter for a 105 cycle
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
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a. Determine these dimension for FS = 1.4 at a section without stress concentrationb. How much would these dimensions be decreased if the surfaces of the links were
mirror polished
Materials spec’s: Oy = 55ksi Ou = 85 BHN = 190
Given:
Fmin = 0kips
Fmax = 10kips
FS = 1.4
Solution:
a. Soderberg Equation1/FS = (Om/Oy) + (Oa/On)
Om = Omax + Omin
Omax = F/A = 10/bh = 10/1.5b2
Omin = 0
Om = Oa
On = 0.5(85)
On = 42.5
Om = (10/1.5b2 + 0)/2
Om = 3.33/b2
b. Mirror polishedS.F = 1
On = 0.8(47.5)(1)
On = 38
1/FS = (Om/Oy) + (Oa/On)
= (3.33/b2(55)) + (3.33/38b2)
1/1.4 = (0.061 + 0.0876)/b2
b = 0.456in
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h = 0.68in
BHN < 400
O`n = 0.5Ou
O`n = (0.25)(190)
= 47.5
On = 0.8O`n(SF)
= 0.8(47.5)(0.87)
On = 33.06
by Soderberg Equation
1/1.4 = (3.33/55b2) + (3.33/33.06b2)
1/1.4 = (0.061 + 0.1007)/b2
b = 0.48in
h = 0.714in
b; 0.48 – 0.456 = 0.024in diff
h; 0.714 – 0.68 = 0.034in diff
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 Spec’s:
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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 filletOa = OMmax/td2; Mmax=3000lb(22)
= 6(66)/0.5h2h
Oa = 792/h3
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
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= 2.2037
1/Fs = Kf (Oa/On)
1/1.8 = 2.2037 (792h3/27.412)
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
1/Fs = 2.388(7922/h3)(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
1/1.8 = 2.7129(792/h3)(1/27.412)
h = 5.2059in
therefore use: h = 5.2059in, b=6.7677in, t = 2.60295in
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:
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Cold rolled, hot rolled, forge carbon steel
Relation of power, torque and speed:
P=2 πTN And T=Fr
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:
SS=TcJ And θ= TlJG (radians)
SS=16Tπ D3
(For solid circular shaft)
SS=16T DOπ [DO4−Di4 ] (For hollow circular shaft)
Where:SS = torsional shear stressT = torqueC = distance from the neutral axis to the outermost fibrer = radiusDO = outside diameterDi = inside diameterD = diameter of shaftJ = polar moment of inertia
=(π32
)D4 (for solid circular shaft)
=( π32 )(DO4−Di4) (for hollow circular shaft)
L = length of shaftθ= angular deformation in length, radiusG = 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 bending
SSmax=16π D3
√M 2+T2
STmax=16πD3
(M+√M 2+T 2)
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Where:SSmax = maximum shear stressSTmax = maximum tensile or compressive stressM = bending momentT = torsional moment
Strength of shaft with assumed allowable stress (PSME code p.18)For main power transmitting shafts (assumed stress = 4000 psi)
P=D3N80
Or D= 3√ 80 PNFor line shafts carrying pulleys: (assumed stress = 6000 psi)
P=D3N53.5
Or D= 3√ 53.5PNFor small, short shafts (assumed stress = 8500 psi)
P=D3N38
Or D= 3√ 38PNWhere:
P = power transmitted in HPD = diameter of shaft in inchesN = speed in rpm
Empirical formula from machinery’s handbookDiameter of shaft
1. For allowable twist not exceeding 0.08 deg per ft length
3. For short ,solid shaft subjected only to heavy transverse shear
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D=√ 1.7VSS
Where:V = max traverse shearing loads, lbsSS = max torsional shearing stress, lb/ in2
Linear deflection of shaftingFor 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 L=8.95 3√D2
2. For shafting subjected to bending action of pulleys, etc L=5.2 3√D2Where:
L = maximum distance between bearing, ftD = diameter of shafts, inches
Note:
1. Pulleys should be placed as closed to the bearings as possible2. In general, shafting up to three inches in diameter is almost always made
from cold – rolled steelSample Problems:
1. What power would a spindle 55 mm in diameter transmit at 480 rpm. Stress allowed for short shaft is 59 N/mm2
Solution:
Ss = 16T/πD3 P = 2πTN = 2π (1.92739)(480/60) = 96.88 kw59 = 16T/ π (55)3
T = 1,927,390 N-mm = 1.92739 kN-m
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.
3. A short 61 mm shaft transmits 120 HP. Compute the linear speed of a pulley 55 cm mounted on the shaft.
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Solution
For short shaft (PSME CODE):P = D3N/38where: D = 61 mm = 2.4 in 120 = (2.4)3N/38N = 330 rpmV = πDN = π0.55 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:
P=2 πTN
F=Tr= TP2
Where:P= power transmittedT torquer = radiusD = diameterN = speed
Crushing (compressive) stress:
SC=Fhs
(L)
Shearing stress;
SS=FWL
Generally, when the key and shaft are of the same material
W=D4 And L=1.2D
Where:W = width of keyh= thickness of keyL= length of key
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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.
2. A keyed sprocket delivers a torque of 778.8N-m thru the shaft of 54mm OD. The key thickness is 1.5875cm and the width is 1.11cm. Compute the length of the same key. The permissible stress value of 60MPa for shear and 90MPa for tension.
