MACHINE DESIGN(PART3,design principles)
Ø Design Principles
A machine may be defined as a combination of stationary and moving parts constructed for the useful purpose of generating, transforming or utilizing mechanical energy. Machines can be classified in to:
1. Machines for generating mechanical energy: Converts some form of energy (electrical, heat, hydraulics etc.) into mechanical work. eg: steam engines, IC engines, water turbines etc.
2. Machines for transforming mechanical energy: known as converting machines. These types of machines transform mechanical energy into another form of energy. eg: Electric generators, hydraulic pumps etc.
3. Machines for utilizing mechanical energy: These machines receive mechanical energy and deliver and utilize it as such in the performance of useful work. eg: lathe, m/c tools etc.
A m/c element or part is a separate part of machine, either integral or consist of several small pieces which are rigidly joined together by riveting, welding etc. The m/c element can be classified into
1. General purpose elements. eg: nuts, bolts, key, axles, shafts.
2. Special purpose elements: These m/c elements are employed only with a particular type of machine. eg: Piston, Connecting rods, Cam shafts etc. Further sub divided into.
3. Fasteners: which connects or join the parts of a machine.
4. Elements of Rotary motion drive: These elements transmit power: Eg: belt, rope, chain, gears, shafts etc.
“It is engineers tasks to define, calculate, motions, forces and changes in energy in order to determine the sizes, shapes, and materials needed for each the interrelated parts in the machine.”
Basic Requirements for Machine Elements and Machine
v Cost
v High o/p and efficiency
v Strength
v Stiffness or rigidity
v Wear resistant
v Light weight and Mini dimensions
v Reliability
v Durability
v Economy of performance
v Accessibility
v Processability
General Steps in Design
Market survey
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Define specification of product
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Feasibility study
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Creative Design synthesis
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Preliminary Design and Development
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Detailed Design
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Prototype
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Design
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Design for product
Material Properties
Mechanical properties of a material generally determined through destructive testing samples under controlled loading conditions.
Tensile test
This is one of the simplest and be basic test and determines values of number of parameters concerned with mechanical properties. A materials like strength, ductility and toughness. The information which can be obtained from the tests are:
i) Proportional Limit
ii) Elastic limit
iii) Modulus of Elasticity
iv) Yield strength
A typical tensile specimen is shown in fig. 1.1.
The tensile bar is mechanical from the material to be tested in one of the several standard diameters ‘do’ and gauge lengths ‘l0’. The gauge length is an arbitrary length defined along the small-diameter portion of the specimen by two indentations so that its increase can be measured during the test. The larger dia sections are threaded for insertion into a tensile test machine which is capable of applying either controlled loads or controlled deflections to the end of the bars; and gauge length portion in mirror polished to eliminate stress concentration from surface defects. The bar is stretched slowly in tension until it breaks, while the load and distance across the gauge length are monitored.
STRESS-STRAIN DIAGRAMS
Stress (s) is defined as the load per unit area
p – Applied load at any instant.
A0 – Original cross-section area of the specimen.
Strain is the change in length per unit length and is calculated from
The results of tensile test are expressed by means of stress-strain relationships and plotted in the form of a graph.
I. It is observed from the diagram that stress-strain relationship is linear from O to P. OP is a straight line and after P, the curve begins to deviate from straight line. Point ‘P’ is the proportional limit below which stress is proportional to strain; as expressed by Hooke’s law.
Where E defines the slope of stress-strain curve up to proportional limit called Young’s Modulus OR Modulus of Elasticity of the material.
E=tan q =
E is the measure of the stiffness of the material in its elastic range and has the units of the of stress.
For most ductile materials, the modulus of elasticity in compression is the same as in tension.
II. ELASTIC LIMIT:
Even if the specimen is stressed beyond point P and up to E, it will regain its initial size and shape when load is removed. This shows that the material is the elastic stage up to point E. Therefore E is called elastic limit. Elastic limit can be defined as the maximum stress without any permanent deformation. Point P and E are typically close together that they are often considered as the same.
III YIELD STRENGTH
When the specimen is stressed beyond point E, plastic deformation occurs and material starts yielding. During this stage, it is not possible to recover the initial size and shape of the specimen on the removal of the load. From the diagram, beyond point E, the strain increases at a faster rate up to point y1. In the case of mild steel, it is observed that there is small reduction in load and the curve drops down to point y2, immediately after yielding starts. The points y1 and y2 are called upper and lower yield points respectively. For many materials, y1 and y2 are close to each other and in such a case, two points are considered as same and denoted by y. The stress corresponding to yield point y is called yield strength. Yield strength is defined as the maximum stress at which a marked increase in elongation occurs without increase in load.
Many varieties of steel, especially heat-treated steels, aluminium and cold-drawn steels do not have a well defined yield point on the stress – strain diagram. This type of material yields gradually after passing through elastic limit E. If the loading is stopped at point Y, at a stress level slightly higher than elastic limit E, and specimen is unloaded and readings taken, the curve would follow the dotted line and a permanent at a plastic deformation will exist.
The strain corresponding to this permanent deformation is indicated by OA. For such materials which do not exhibit a well defined yield point, the yield strength is defined as the stress corresponding to a permanent set of 0.2% of gauge length. In such a case the yield strength is determined by offset method. A distance OA equal to 0.002 mm/mm strain is marked on X-axis. A line is constructed from pt A parallel to st. line portion OP of the stress-strain curve. The pt of intersection of this line and the stress-strain curve is called Y. The corresponding stress is called 0.2% yield strength. The term proof load G proof strength are frequently used in the design of fasteners. The proof strength is similar to yield strength. It is determined by offset method, however offset in this case is 0.001 mm/mm. 0.1% proof strength denoted by the symbol Rp 0.1.
Ultimate tensile strength
After the yield point Y2, plastic deformation of the specimen increases. The material become stronger due to strain hardening and higher and higher load is required to deform the material. Finally the load and corresponding stress reaches a maximum value, given by pt u. The stress corresponding to pt U is called ultimate strength. The ultimate tensile strength is the maximum stress that can be reached in tension test. For ductile material the diameter of the specimen begins to decrease rapidly beyond maximum load point U. C.A.
Ultimate tensile strength is considered as failure criterion in brittle material.
vi) Percentage Elongation:
Ductility is measured by percentage elongation and is given by
vii) Percentage reduction in area.
Ratio of decrease in C.A. C.A.
Percentage reduction in area =
A0 – Original C.A.
A – Final C.A.
SIMPLE STRESSES IN MACHINE PARTS
Stress: Force per unit area
Stress (s) = P/A
P = Force or load acting on the body
A = Cross sectional area
Units:
In S.I. unit; stress is usually expressed as (Pa)
1 Pa = 1 N/m2
Strain: Deformation per unit length
Strain
Young’s Modulus:
Stress is directly proportional to strain (with in elastic limit)
s a e
s = Ee
E = s /e =
E – Constant of proportionality: Young’s Modulus unit : GPa : GN/m2 or kN/mm2.
PROBLEMS
A coil chain of a crane required to carry a max. load of 50 KN is shown in fig. A
Find diameter of link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa?
p = 50 KN = 50 × 103 N
st = 75 MPa = 75 × 106 N/m2 = 75 N/mm2
d = ?
A =
st =
2. A M.S. bar of 12 mm dia: is subjected to an axial load of 50 KN in tension. Find magnitude of induced stress?
3. If the length of bar in 1 m (dia 12 mm ) and the modulus of elasticity of material of bar is 2 × 105 MPa, find elongation of bar.
SHEAR STRESS AND SHEAR STRAIN
When the external force acting on a component tends to slide the adjacent planes w.r.t each other, the resulting stress on these planes are called direct shear stresses.
t – shear stress
A – Cross sectional area (mm2)
r – Shear strain (radians)
t = G.r.
G – Modulus of rigidity (N/mm2)
The relationship between the modulus of elasticity, the modulus of rigidity and Poisson’s ratio is given by
E = 2G [1+ m]
m - Poisson’s ratio
m = strain in lateral dim.
strain in axial dim.
The permissible shear stress is given by
STRESSES DUE TO BENDING MOMENT
A st. beam is subjected to bending moment Mb as in fig.
For an irregular cross section.
Mb - Applied Bending Moment
I – Moment of inertia of C.O. abt NA (mm4)
I × g =
The parallel – axis theorem:
STRESSES DUE TO TORSIONAL MOMENT
The internal stresses, which are induced to resist the action of twist, are called torsional shear stress.
Mt – applied torque (N-mm)
J – Polar moment of inertia (mm4)
The angle of twist is given by
l – length of shaft (mm)
Power transmitted
1. Two plates subjected to a tensile force of 50 kN are fixed together by means of three rivets (as in fig.). The plates and rivets are made of plain carbon steed 10C4 with a tensile strength of 250 N/mm2. The yield strength in shear is 50% of tensile yield strength, and factor of safety is 2.5. Neglecting stress concentration determines:
(i) The diameter of rivets
(ii) Thickness of plate
yield strength in shear is 50% of yield strength in tension.
To find diameter of Rivets
d = 20.68 = 22 mm
To find plate thickness:
2. A suspension link, used in bridge is shown in fig. The plates and the pins are made of plain carbon steel (Syt = 400 N/mm2) and factor of safety is 5. The maximum load in link is 100 kN. The ratio of width of the link plate to its thickness (b/t) can be taken as 5. Calculate.
(i) Thickness and width of link plate
(ii) Dia. of knuckle pin
(iii) Width of the link plate at centre line of pin
(iv) Crushing stress on pin.
LINK PLATE
t = 11.18 = 12 mm
b = 5 t = 60 mm
1. A link shown in fig. is made of gray cast iron FG 150. It transmits a pull p of 10 KN. Assume that the link has square cross section (b = h) and using for 5, determine the dimensions of the cross section of the link?
Solution:
Given
Made of Grey cast iron (FG 150)
According to Indian Standard Specification (IS: 210-1993) the grey cast iron is designated by alphabets FG, followed by a figure indicating the minimum tensile strength in MPa or N/mm2. ‘FG150’ means C.I. with 150 MPa or 150 N/mm2.
[C.I. is a brittle material]
- Tensile strength will be 100 – 200 MPa.
- Compressive strength = 400 to 1000 MPa.
- Shear strength = 120 MPa
Load P = 10 KN = 10 × 103 N
Square cross section \ A = b × h
[allowable stress]
b = h
\ h = 18.26 mm = 20 mm
\ The cross section of link is 20 × 20 mm
Points to be remembered:
- The dimensions of simple m/c parts are determined in the basis of pure tensile stress, pure compressive stress, direct shear stress, bending stress or torsional shear stress. The analysis is simple but approximate, because number of factors such as principal stresses, stress concentration reversal of stresses is neglected. Therefore a higher FOS up to 5 can be taken into....
- It is incorrect to take allowable stress as data are design.
- The max. shear stress theory; proposes the yield strength in shear is 50% of yield strength is tension.
\ Sys = 0.5 Syt
2. The forces exerted by the levers of the pump on a rocking shaft are shown in fig. The rocking shaft does not transmit torque. It is made of plain carbon steel 30 C8 [Syt – 400 N/mm2] and factor of safety is 5. Calculate diameter of the shaft.
Taking Moment about A:
20 × 200 + 30 × 800 – RB × 1050 = 0
RB × 1050 = 28000
RB = 26.67 KN
ey = 0
FACTOR OF SAFETY - FOS
While designing a component, it is necessary to ensure sufficient reserve strength in the case of an accident. It is ensured by taking a suitable factor of safety (fs)
FOS can be defined as: [sall = allowable stress]
The allowable stress is the stress value which is used in design to determine the dimensions of the component. It is considered as a stress which the designer, expects will not exceed under normal operating conditions.
For ductile material, the allowable stress (sall) is given by,
For brittle material:
The FOS ensures against
- Uncertainty in the magnitude of external force acting on the component.
- Variations in the properties of materials like yield strength or ultimate strength.
- Variations in the dimensions of the component due to imperfect workmanship.
The magnitude of FOS depends upon following conditions:
1. Effect of failure:
Failure of the ball bearing in gear box.
Failure of valve in pressure vessel
2. Types of Load
- When external force acting on the m/c element is static - FOS is low.
- Impact load – FOS is high
3. Degree of Accuracy in force analysis.
When the force acting on the m/c element is precisely determined low FOS can be selected. Where as higher FOS is considered when the m/c component is subjected to a force whose magnitude or direction is uncertain and unpredictable.
4. Material of Component
When the component is made of homogenous ductile material, like steel, yield strength is the criterion of feature. FOS is small in such cases. Cast Iron component has non-homogenous structure and a higher FOS based on ultimate strength is chosen.
5. Reliability of the component:
FOS & Reliability
- Defense
- Power stations
6. Cost of Components:
FOS & Cost
7. Testing of Machine element
Low FOS – when m/c comp. are tested under actual condition of service and operation.
8. Service conditions:
Higher FOS – when m/c element is likely to operate in corrosive atmosphere or high temp. environment.
9. Quality of Manufacture:
Quality is inversely proportional to FOS.
Lecture - 5
TYPES OF CYCLIC LOADING
Machine components are subjected to external force or load. The external load acting on the component is either static or dynamic. The dynamic load is further classified into cyclic and impact loads. Static load is one: as a load which does not vary in magnitude or direction with respect to time, after it has been applied.
Dynamic load is a load which varies in magnitude and direction w.r.t time, after it has been applied.
There are two types of dynamic loads.
Cyclic load
Impact load
Cyclic Loads:
This is a load, which when applied, varies in magnitude in a repetitive cyclic manner; either completely reversing itself from tension to compression or oscillating about some mean value. In this case, the pattern of load variation w.r.to time is repeated again and again.
Examples of cyclic loads are
- Force induced in gear teeth
- Loads induced in a rotating shaft subjected to B.M.
There are three types of mathematical models are of cyclic loads.
- Fluctuating OR alternating load
- Repeated loads
- Reversed loads
Stress – time relationship corresponding to these three types of loads are given below.
(a) Fluctuating stresses
(b) Repeated stresses
(c) Reversed stresses
The fluctuating or alternative load varies in a sinusoidal manner with respect to time. It has some mean value as well as amplitude value. It fluctuates between two limits – maximum and minimum load. The load can be tensile compressive or partially tensile.
The repeated load varies in a sinusoidal manner w.r.to time but varies from zero to some maximum value. The minimum value (load) is zero in this case and therefore, amplitude load and mean loads are equal.
The reversed load varies in a sinusoidal manner w.r.to time, but it has zero mean load. In this case, half portion of the cycle consists of tensile load and remaining half of compressive load. There is complete reversal from tension to compression between these two halves and therefore, mean load is zero.
smax - Maximum stress
smin - Minimum stress
sm - Mean stress
sa - Stress Amplitude
sm - ½ (smax + smin)
sa - ½ (smax – smin)
In the analysis of fluctuating stresses, tensile stress is considered as +ve, while compressive stresses are –ve.
STRESS CONCENTRATION
In design of m/c elements, following fundamental equations are used.
These equations are called ‘elementary’ equations. These equations are based on a number of assumptions:
1. Stress is proportional to strain.
2. Modulus of elasticity is same in tension and in compression.
3. No high intensity contact pressure at the regions of the contact of support and loading.
4. Machine members have constant section.
5. No abrupt change in section.
However, in practice, discontinuities and abrupt changes in cross-section are unavoidable due to certain features of the component such as oil holes and grooves, keyways and splines, screw threads etc. Hence it cannot be assumed that cross-section of the machine component is uniform. Under these circumstances, the elementary equations do not give correct result.
In order to visualize the effect of discontinuities and abrupt changes in cross section on distribution of stresses, a concept called “FLOW ANALOGY” is used. Fig. (a) shows a bar of uniform cross section subjected to axial tensile force. In flow analogy concept, the force is visualized as flowing through the bar. Each flow line in the figure represents a certain amount of force. These flow lines are called Force flow lines. Since bar has uniform cross section, flow lines are uniformly spaced. Fig. (b) shows an identical bar but with notch cut on its circumference. If we consider a cross section of this bar away from the notch, the flow lines are uniformly spaced showing normal distribution of tensile stress. As the line approaches the notch they are bent in order to pass through restricted openings. The bending of force flow lines indicates weakening of the material. The effect of stress concentration is proportional to bending of flow lines. When force flow lines are bent, the load carrying capacity is reduced and material becomes weak. Let us consider a flat plate Fig. (c) of uniform thickness subjected to axial tensile force. At the section xx, there is sudden change in height.
At both ends flow lines are parallel indicating uniform distribution of stresses. At the right end, they are close together indicating higher magnitude of stresses. At the left end they are spaced comparatively away from each other, indicating lower magnitude of stress. When these lines from left and right side join at section at xx, they are bent indicating weakening of material.
There are two terms – normal stresses and localized stresses. Normal stresses are shown at two ends with uniform distribution. The localized stresses are restricted to local regions of the component such as sections of discontinuity.
(i) Normal stresses: are stresses in the machine component at a section away from discontinuity or abrupt change of cross section. Localized stresses are stresses in local regions of the component such as the sections of discontinuity or sections of change of cross section.
(ii) Normal stresses are determined by elementary equations. It is not possible to use these formulae for localized stresses.
(iii) The normal stresses are comparatively of small magnitude. The localized stresses in the vicinity of discontinuity are frequently of large magnitude. This may give rise to a crack.
(iv) Failure rarely occurs in region of normal stresses. The region of localized stresses is more vulnerable to fatigue failure.
Let us consider a plate with a small circular hole as in figure (d).
The distribution of stresses near the hole can be observed by keeping a model of the plate made of epoxy resin in circular polariscope. The localized stresses in the neighbourhood of the hole are far greater than the stress obtained by elementary equation.
“Stress concentration is defined as the localization of high stresses due to irregularities or abrupt changes of cross section.” In order to consider the effect of stress concentration and find out localized stresses ‘stress concentration factor’ is used. It is denoted by Kt.
The causes of stress concentration are:
(i) Variation of properties of materials
(ii) Local application
(iii) Abrupt change in section
(iv) Discontinuities in the component
(iv) Machining scratches
Methods to reduce stress concentration
(i) Additional notches and holes in tension member.
Eg. It is observed that a single notch results in a high degree of stress concentration. The severity of stress concentration can be reduced by three methods:
(a) Use of multiple notches
(b) Drilling additional holes
(c) Removal of undesired material
The method of removal of undesired material is called “Principle of Minimization of Material”.
STRESS CONCENTRATION FACTORS
Case – 1
Stress concentration factor for a rectangular plate with a transverse hole loaded in tension:
Case – 2
The values of stress concentration factor for a flat plate with a shoulder fillet subjected to tensile or compressive force are determined from,
Case – 3
Stress concentration factor for a round shaft with shoulder fillet subjected to tensile force, bending moment and torsional moment are taken from Data book; The nominal stresses in these three cases are as follows:
(a) Tensile force,
(b) Bending Moment,
(c) Torsional Moment,
There are a number of geometric shapes and conditions of loading. A separate chart for the stress concentration factor should be used for each combination.
The effect of stress concentration depends upon the material of component.
References
- Mechanical Engg. Design – Joseph Shigley
- Machine Design – Mubeen
- Machine Design – Black
- Machine Design – R. K. Jain
- Machine Design an integral approach – Norton, Pearson
- Machine Design data hand book – Lingayah Vol I.
- Elements of Machine Design – Pandya & Shah
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