Dme Unit i Mech 30.6.12

Paavai Institutions Department of MECH

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UNIT I

STEADY AND VARIABLE

STRESSES


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CONTENTS

1.1 Introduction to Machine Design

1.1.1 Classification of Machine design

1.1.2 General consideration in Machine design

1.1.3 Factors influencing Machine design

1.2 Bending Stress in Straight Beams

1.3 Bending Stress in curved Beams

1.4 Stress Concentration

1.5 Theories of Failure

1.5.1 Rankine theory

1.5.2 Guests theory

1.5.3 Saint Venants theory

1.5.4 Maximum strain energy theory

1.5.5 Distortion energy theory

1.6 Design of Variable Loading

1.6.1 Notch Sensitivity

1.6.2 Endurance Limit

1.6.3 Factors affecting endurance strength

1.7 Solved Problems

1.8 Question Bank


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TECHNICAL TERMS

1. Factor of safety

The ratio between maximum stresses to working stress is known as factor of safety.

2. Endurance limit. Endurance limit is the maximum value of completely reversed stress that the standard

specimen can sustain an infinite number (106) of cycles without failure.

3. Impact load

If the time of load application is less than one third of the lowest natural period of

vibration of the part, it is called an impact load.

4. Stress concentration and stress concentration factor.

Stress concentration is the increase in local stresses at points of rapid change in cross

section or discontinuities.

Stress concentration factor is the ratio of maximum stress at critical section to the nominal

stress

5. Griffith theory. (Or) State the condition for crack growth.

A crack can propagate if the energy release rate of crack is greater than crack resistance

6. Notch sensitivity (q)

Notch sensitivity (q) is the degree to which the theoretical effect of stress

Concentration is actually reached.


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7. Machinability

It is the property of the material, which refers to a relative case with which a

material can be cut. It is measured in a number of ways such as comparing the tool

life for cutting different material.

8. Ductility

It is the property of the material enabling it to be drawn into wire, with the application of

tensile force. It must be both strong and plastic. It is usually measured in terms of

percentage elongation and reduction in area. (eg) Ni, Al, Cu

9. Fatigue When a material is subjected to repeated stress, it fails at stresses below the yield

point stress; such type of failure of the material is called fatigue.


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1. 1 INTRODUCTION TO MACHINE DESIGN

Machine design is the creation of new and better machines and improving the existing

one. A new or better machine is one which is more economical in the overall cost of production

and operation.

1.1.1 CLASSIFICATION OF MACHINE DESIGN

1. Adaptive design: The designers work is concerned with adaptation of existing design.

2. Development design: This type of design needs considerable scientific training and design

ability in order to modify the existing designs into a new idea.

3. New design: This type of design needs a lot of research technical ability and designers and

creative thinking.

1.1.2 GENERAL CONSIDERATION IN MACHINE DESIGN

(i) Type of load and stresses caused by the load.

(ii) Motion o parts

(iii) Selection of materials

(iv) Frictional resistance and lubrication

(v) Convenient and economical features

(vi) Safety of operation

1.1.3 FACTORS INFLUENCING MACHINE DESIGN

(i) Strength and stiffness

(ii) Surface finish and tolerances

(iii) Manufacturability

(iv) Ease of handling


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(v) Working atmosphere

(vi) Cooling and lubrication

(vii) Safety

(viii) Noise requirement

(ix) Cost

1.2 BENDING STRESS IN STRAIGHT BEAMS

Consider a straight beam subjected to a bending moment M as shown in figure. The

following assumption are usually made delivering the bending formula

(i) The material of the beam is perfectly homogeneous and isotropic.

(ii) The material of the beam obeys Hooks la

(iii) The Youngs modulus is same in tension and compression.

(iv) The loads are applied in the plane of bending.

Figure1.1 Bending Stress in Straight Beams

When a beam is subjected to the bending moment the fibers on the upper side of the beam

will be compress and lower side elongate due to tension. The surface between the top and bottom

fibbers are neither neither shortens nor lengthened. Such a surface is called neutral surface. The

intersection of the neutral surface with any normal cross section of the beam is known is neutral

axis. The bending equation is given by


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M- Bending moment acting at the given section

- bending stress

I-moment of inertia of the cross section about the neutral axis

y- Distance from the neutral axis to the extreme fibber

E- Youngs modulus of the material of the beam

R- Radius of curvature of the beam

1.3BENDING STRESS IN CURVED BEAMS

In straight beams the neutral axis of the section coincides with its centroidal axis and the

stress distribution is linear. But in curved beams the neutral axis of the cross section is shifted

towards the centre of curvature of the beam causing a nonlinear distribution of stress. Application

of curved beam principle is used in crane hooks, chain links planers etc

Figure1.2 Bending Stress in Curved Beams

Consider a curved beam subjected to a bending moment M as shown in figure. The

general expression for bending stress b) in a curved beam at any fiber at a distance y from the

neutral axis is


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M- Bending moment acting at the given section about the centroidal axis

A- Area of cross-section e- Radius of curvature of the neutral axis

R- Radius of curvature of the centroidal axis

Rn- radius of curvature of the neutral axis

y- Distance from the neutral axis to the fiber under consideration.

It is positive for the distances towards the center of curvature and negative for the distance away

from the center of curvature.

1.4 STRESS CONCENTRATION

When every a machine component changes the shape of cross section the simple stress

distribution no longer holds good and the neighborhood of the discontinuity is different. This

irregularity in the stress distribution caused by abrupt changes of form is called stress

concentration. Consider a plate with transverse elliptical hole and subjected to a tensile load as

shown in figure. From the figure the stress at the point away from the hole is practically uniform

and the maximum stress will be induced at the edge of the hole.

Fig 1.3. kt = max stress/nominal stress


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Methods of Reducing Stress Concentration

Avoiding sharp corners

Providing fillets

Use of multiple holes instead of single hole.

Undercutting the shoulder part

1.5 THEORIES OF FAILURE

Stress produce in a body due to the application of the load is beyond the elastic limit the

permanent deformation occurs in the body. If the load is removed the body will not retain its

original shape. There are some permanent deformations in the body. Whenever permanent

deformation occurs in the body the body is said to be failed. The failure of a component due to

increase of tensile stress or due to other quantities such as shear stress and strain energy also

attain definite values and any one of these may be deciding factor of the failure have advanced to

explain the cause of failure. According to the important theories the failure takes place when a

certain limiting value is reached by one of the following

1.5.1 Maximum principal stress (or) maximum normal stress (or) Ranking theory

Failure occurs when the maximum normal stress is equal to the tensile yield strength.

This theory is based on failure in tensile or compression and ignores the possibility of

failure due to shearing stress, ductile material mostly fail by shearing. So this theory is used for

brittle material.

1.5.2 Maximum shear theory (or) Guests theory (or) Coloumb theory


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Failure occurs when the maximum shear stress developed in the machine member become

equal to the maximum shear stress at yielding in a tensile test.

This theory is mostly used for ductile materials.

1.5.3 Maximum strain theory (or) Venants theory

Failure occurs when the maximum strain in the member equal in the tensile yield strain.

1.5.4 Maximum strain energy theory

Failure is induced in the member when the strain energy stored per unit volume of the

member becomes equal to the strain energy per unit volume at the yield point.

1.5.5 Distortion energy theory (Vonmiseshenky theory)

1.6. DESIGN OF VARIABLE LOADING

Consider a rotating beam of circular cross section and carrying a load of W, this load

induces stresses in the beam which are cyclic in nature.

Upper fibres of beam(a) under compression and lower fibre (B) tensile after half revolution the

point B occupies the position of point A and point A occupies the point of B. thus point B is now

compression and point A is tensile.


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Fig 1.4 simply supported beam with center point load

The stresses which vary from one value of compressive to same value of tensile or vice

versa are known as completely reversed or cyclic stresses.

The stresses which vary from a minimum value to a maximum value of same nature is called

fluctuating stresses.

The stresses which vary from zero to a certain maximum value are called repeated

stresses.

The stresses which vary from a minimum value to a maximum value of the opposite

nature is called alternative stresses (from a certain minimum compressive to a maximum tensile

or minimum tensile to a certain maximum compressive)

fig 1.5 variable stresses

1.6.1 NOTCH SENSITIVITY (q)


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This is defined as the degree to which the actual stress concentration effect compares with

theoretical stress concentration effect.

1.6.2 ENDURANCE LIMIT

It is defined as maximum value of completely reversed bending stress which a polished

specimen can withstand without failure for infinite number of cycles.

1.6.3 FACTORS AFFECTING ENDURANCE STRENGTH

1. Surface finish factor (KSF)

2. Size factor (KSZ)

3. Reliability factor (KR)

4. Miscellaneous factors (K)

5. Load factor (KL)

1.7 SOLVED PROBLEMS

1. A 20KN tensile load acts on the following members. Considering stress concentration,

calculate the maximum stress induced in each member.

(a)A stepped shaft of diameter 50mm and 25mm with fillet radius 5mm. (b) a rectangle

plate 60mm wide and 10mm thick with a transverse hole of 12mm diameter.

Solution:

Case1: stepped shaft:

Tensile load , p=20 KN =20,000N

Maximum diameter, D=50mm

Minimum diameter, d=25mm

Fillet radius=5mm.


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In stepped shaft, due to minimum cross-section more stress is induced in small shaft than the

bigger shaft.

Hence,

Nominal stress induced in small shaft, o=P/A=P/(/4x )

= (4X20, 000)/x

=40.7N/mm2

Also, due to change of diameters, stress concentration factor should be considered for obtaining

maximum induced stress.

Now, r/d=5/25=0.2 and D/d=50/25=2

Hence, the stress concentration factor for the above ratio conditions is, Kt=1.5Maximum induced

stress =KtXo=1.5X40.7=61N/mm2

Case2: Rectangular plate

Tensile load, P=20KN=20,000N

Width of the plate, b=60mm

Thickness of plate, t=10mm

Diameter of hole, d=12mm

Here also, more stress is induced in the minimum cross section of the plate ,where hole is present.


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Now, nominal stress induced, o =P/A=P/(b-d)t

=20,000/(60-12)10 =41.7N/mm2

Also, ratio of diameter of hole to width of plate,

D/b=12/60 =0.2

Stress concentration factor, Kt=2.5 for d/b=0.2(from graph 4.99)

Maximum induced stress =KtXo =2.5X41.7 =104N/mm2

2. A tie-bar has to carry a load of 100KN. What must be the thickness of bar of 110mm

Width, if there is the rivet hole of 22mm diameter on its centre line? Working

stress for the tie bar is 75Mpa.

Solution:

Load =100KN =100X103N

Width of plate, b =110mm

Diameter of hole, d=22mm

Thickness 0f plate =t mm

Working stress, t=75Mpa=75x106N/m2=75N/mm2

Nominal stress induced in the cross-section passing across hole

o=P/A=P/(b-d)t=(100x10 3

)/(110-22)t=1136/t

Now, d/b=22/110=0.2, for which Kt=2.5

Hence, maximum stress=K t xo=2.5x(1136/t)=(2840/t)

This maximum stress should be less than working stress for safe design.

i.e., (2840/t)


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(or) t >(2840/75) > 37.8mm= 40mm

Thickness of plate=40mm.

3. A round bar of dia40mm has a smooth ground surface and is made of steel for which

ultimate strength is 500N/mm2

yield strength is 400N/mm2 and endurance limit is

250N/mm2 .

Determine the factor of safety of the bar when it is subjected to alternating bending stress

in a symmetric cycle if the maximum bending moment in the cycle is 640 N-m.

Solution.

Given

Diameter of the shaft d=40mm

u=500N/mm 2

y=400N/mm2

e=250N/mm2

Maximum bending moment Mama=640N-m=640x103N-mm

Maximum bending stress,


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bmax=32 Mmax/d3

= (32x640x103)/x403

=102N/mm2

Since this is a symmetric cycle,

bmin=-102N/mm2

then, mean stress m=bmax+bmin/2 =(102-102)/2 =0

And variable stress v=bmax-bmin/2

=(102-(-102))/2 =204/2 =102N/mm2

According to soderberg equation, we have

(Kfv/e)+(m/y) =1/fs

WhereKf=1

e=exKsr

=250x0.9 (assume Ksr=0.9,ground surface)

=225N/mm2

Sub the in the soderberg equation,

We get,

(102/225)+(0/400)=(1/fs)

1/fs=(102/225)

Fs=225/102=2.2

Factor of safety =2.2

5.A stepped shaft of diameters D and d is subjected to a variable axial load P which

cyclically varies between 0 and 10KN. The shaft is made of C20 steel, mirror polished with

Su=500N/mm2 and Sy=260N/mm

2


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Determine the diameters D and d with D/d=1.5, FOS=2, notch sensitivity factor=0.8 and

r/d=0.2 where r is the shoulder radius.

Solution:

Given

Pmax=10KN=10000N

Pmin=0

u=500N/mm2

y=260N/mm2

Endurance limit for C20 steel is

e =0.5 Su=0.5x500=250N/mm2 which is less thenybecause e=0.85y.

Now considering mirror polished surface condition , and axial loading system, assume

Ka=load correction factor for axial loading =0.8

Ksr=surface finish factor =1

Now e=250xKaxKsr

i.e.,e=250x0.8x1=200N/mm2

in this problem ,since the ultimate stress (u) and the yield stress (y) are given the diameters may

be determined using Goodman equation or soderberg equation.

According to soderberg equation.

Kfv=e[1/fs-m/y]

Now

Kf=fatigue stress concentration factor

=1+q(Kf-1)

=1+0.8(1.45-1)=1.36

v =variable stress amplitude

=Pmax-Pmin/2 =10000-0/2A =5000/A N/mm2


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=Pmax+Pmin/2A =10000+0/2A =5000/A N/mm2

Where A=area of critical section=/4xd2

Now using soderberg equation

1.36x5000/(Ax200) +(5000/(Ax260)) =1/2

By solving, we get A=106.5mm2

I.e., /4xd2=106.5 d={106.5x4/}0.5

=12mm (D=1.5d)

D=18mm

1.8 QUESTION BANK

PART-A (2 Marks)

1. Define Design and explain the design process

2. What is innovative design?

3. Explain a method of reducing cost of the final product from the design Perspective.

4. What is optimization? What are the methods for optimization?

5. Define factor of safety. What factors dictate the selection of factor of safety?

6. Differentiate between hardness and toughness of materials.

7. Explain creep, resilience.

8. List the various types of loads and explain.

9. Distinguish between different types of variable stresses.

10. Explain endurance limit. What factors influence endurance strength?

11. Explain the effect of product reliability on endurance strength.

12. State the significance of S-N curve.


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13. Define stress intensity factor, notch sensitivity and fatigue stress concentration factor.

14. Explain Goodman and Seidenberg diagrams

15. Differentiate between the use of Goodman diagram and Seidenberg diagram for designing.

16. Comment on the statement In curved beams maximum stress always occurs at the inner fibre .

17. What is stress concentration? What are the methods to determine it? What are the methods to

reduce it?

18. Explain how the maximum shear stress theory is used for biaxial and triaxail Stress cases.

PART-B (16 Marks)

EXCERCISE PROBLEMS

1. A bar of circular cross section is subject to alternating tensile forces varying from 200kN to

500kN. Materials ultimate tensile strength is 900Mpa, endurance limit is 700Mpa. Determine the

diameter of the bar using safety factors of 3.5 related to ultimate strength and 4 related to ultimate

strength and 4 related to endurance limit. Stress concentration factor us 1.65 use Goodman

criteria.

2. A steel bar is subjected to a reversed axial load of 180kN. Find the diameter of the bar for a design

factor of 2. Ultimate tensile strength 1070N/mm2 yield strength 910N/mm2 . Endurance limit in

bending is half of ultimate tensile strength. Use the following data. Load factor 0.7, Surface finish

factor 0.8, Size factor 0.85, and stress concentration factor 1.

3. A steel cantilever bean 180mm long has a diameterd for a length of 125mm from the free end

and 2d for the remaining length. A fillet of radius 0.2d is provided at the junction of the two

sections. A transverse load varying from 4N up and 135N down is acting in combination with an

axial load that varies from- 110N to + 450N . Using a design factor of 2, calculate the diameter at


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the fillet section for infinite life. Ultimate strength 550Mpa, yield strength 470Mpa, Endurance limit

275Mpa. Size factor 0.85 surface factor 0.9 stress concentration factor for bending 1.44, for axial

load 1.63

4. A C Clamp is acted upon by a load of 20kN as shown in figure. The clamp has a square cross

section throughout the length And is made of a materials With an allowable tensile stress of 150Mpa.

Determine the side of the square cross section. Determine the stress developed at section BB.

UNIVERSITY QUESTIONS

1. Determine the thickness of a 120mm wide uniform plate for safe operation if plate is to

subjected to a maximum tensile load of 250 KN and minimum value of 100 KN. Properties of

plate materials are as follows. Endurance limit stress is 225 MPa and yield point stress is 300

MPa .FOS at yield point 1.5. (AU- Chennai Nov/Dec 2006)

2. A hollow shaft of 40mm outer diameter and 25mm inner diameter is subjected to a

twisting moment of 120 N-m also it is subjected to axial thrust of 10 KN and a bending moment

of 80 N-m. Calculate the maximum compressive and shear stress.

(AU- Chennai Nov/Dec 2006)

3. A simply supported beam has a concentrated load at the centre which fluctuates from

value P to 4P, span of beam is 500mm and circular cross section of diameter 60mm, take

ultimate tensile stress of 70 MPa, yield stress of 500 MPa, endurance limit 330 MPa for reversed

bending and FOS 1.3. calculate value of P taking size factor of 0.85 and surface finish factor of

0.9. (AU- Chennai Nov/Dec 2007)

4. A shaft of 760mm length is simply supported at its ends. It is subjected a central

concentrated cyclic load varies from 12 to 36 KN. Find diameter of the shaft assuming FOS 2,

size correction factor 0.8, surface correction factor 0.85. the material properties are ultimate


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tensile strength of 500 MPa, yield strength of 280 MPa and endurance limit of 250 MPa. Fatique

stress concentration factor of 1.5. (AU-Chennai Apr/May 2008)

5. A cast iron pulley transmits 12 KW at 330 rpm, diameter of the pulley is 1.3m and it has

4 straight arms of elliptical cross section in which major axis is twice the minor axis. Determine

dimensions of the arm where allowable bending stress is 18MPa.

(AU- Chennai Apr/May 2008)

6. A flat bar 32mm wide and 12mm thick is loaded by steady tensile load of 85 KN. The

material is mild steel with yield point stress of 315 N/mm2. Find factor of safety based on yield

point. (AU- Chennai Nov/Dec 2008)

7. A flat plate of width 60mm has a central hole of 10mm diameter; plate is subjected to

axial tensile load of 10 KN, Find thickness of the plate. Assume yield point stress 300 MPa and

FOS 2.5. (AU- Chennai Nov/Dec 2008)

8. A hot rolled steel shaft of 40mm diameter is subjected to a torsional moment varies from

330N to 100 Nm and an applied bending moment varies from 440 Nm to 220 Nm. Material has

ultimate tensile strength of 550 MN/m2 and yield strength of 410 MN/m

2. Find FOS using

soderberg equation allowing endurance limit to be half the ultimate strength and size factor and

surface finish factor to be 0.85 and 0.62 respectively. (AU-ChennaiNov/Dec 2008)

9. Determine the diameter of the circular rod made of ductile material with an endurance

limit 25MPa and tensile yield strength 350 MPa. The member is subjected to varying axial load

from 300 KN to700KN and has stress concentration factor 1.8. take FOS as 2.

(AU-Chennai May/June 2009)

10. Two rods made up of plain carbon steel 40 C8 with yield strength 380N/mm2 are

connected by means of cotter joint. Diameter of each rod is 50mm and cotter is made of steel

plate of 15mm thickness. Calculate dimensions of the socket and make following assumptions:

(a) Yield strength in compression is twice tensile yield strength


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(b) Yield strength in shear is 50% tensile yield strength.

(c) FOS is 4. (AU- Chennai Nov/Dec 2009)

11. A rod of linkage mechanism made of steel 40Cr1 subjected to completely reverse axial

load of 100 KN. The rod is machined on lathe and expected reliability is 95%. There is no stress

concentration. Determine the diameter of the rod using a factor of safety of 2 for an infinite life

condition. (AU- Chennai Nov/Dec 2009)

Documents

Dme Unit i Mech 30.6.12