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ENGINEERING DESIGN

PRESENTATION

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INDEX

INTRODUCTION

PROCEDURE

EXAMPLE

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AXIAL FLUCTUATING LOAD

It is stress induced due to change of

magnitude with respect to time.

It is time delayed fracture under cyclic

loading.

So it is called as FATIGUE FAILURE.

80% of mechanical components fails due to

this only.

Eg: Crank,Connecting rod ,Gearsetc

This is represented by endurance limit.

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ENDURANCE LIMIT

It is maximum amplitude of stress a material

can handle for unlimited number of cycles

without failure.

Endurance limit for a component

.

Where Se=endurance limit for a material.

Ka=surface finish factor.

Kb=size factor

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Kc=reliability factor.

Kd=modifying factor.

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COMPLETELY REVERSED STRESS

It means that stress variation occurs equally

on both positive and negative side.

In other words stress which varies from one

value of tensile to same value of

compressive is called as COMPLETELY

REVERSED STRESS.

sm=0

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STRESS CONCENTRATION

Whenever a mechanical component changes

shape of its cross section the simple stress

distribution no longer holds good as stress at

neighbourhood of discontinuity varies.

This stress discontinuity due to abrupt

changes in form is called as STRESS

CONCENTRATION.

This occurs due to presence of

fillets,notches,holes,

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Stress concentration factor, .

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FORMS OF STRESS CONCENTRATION

1)Rectangular plate with transverse hole:

Where w=width of plate t=thickness of plate

d=diameter of hole

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2)Plate with shoulder fillet:

so=P/dt

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PROCEDURE

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STEP 1:CALCULATION OF ENDURANCE LIMIT

STRESS

Se=(KaKbKcKd)Se Where

Se=e.l.s for whole material

Se=.5(Sut)

Ka=surface finish factorKa=a(Sut)^b

Kb=size factor

Kc=reliability factor

Kd= stress concentration factorKd=Endurance limit of notch free specimen/Endurance limitof notched specimen

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Kd=1/ Kf

Kd=1/1+q(Kt-1)

Kd=1 if there is no stress concentration q=notch sensitivity factor

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STEP 2:CALCULATION OF STRESS AMPLITUDE

(Se)axial=.8(Se)

sa=(Se)axial/FOS

where FOS=factor of safety

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STEP 3:CALCULATION OF GEOMETRIC

CHARACTERISTICS:

(s)a = P/A

here A= (w-d)t for transverse hole

P= dt for filletso , by using the above formula we find the

geometric characteristic of system.

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EXAMPLE :

A rectangle plate in a chassis has transverse

hole in it .It is made steel alloy whose

Sut=650N/mm^2 and FOS is 2.It is subjected

to completely reversed axial load of 15 KN.The plate is in the hot rolled condition .The

notch sensitivity factor is 0.8. The expected

reliability is 90%.if the diameter of hole is10mm and width of plate is 70mm.find the

thickness of the plate .

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GIVEN

Sut=650 N/mm^2

P =15KN

q =0.8R =90%

d =10mm

w =70mm FOS =2

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STEP 1:CALAULATION OF ENDURANCE LIMIT

STRESS:

Se= (Ka Kb Kc Kd) Se Se= 0.5 (Sut)

=0.5(650)=325N/mm^2

Ka = a(Sut)^b

mode of surface finish :hot rolled conditionso a =57.7

b =-0.718

By subs above values in formula we get ,

Ka =0.5514Kb =0.85[d

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CONTD..

Kd =1/Kf

Kd =1/1+q(Kt-1)

d/w =10/70 =0.143

Kt =2.61 (from DD book 7.10)

Kd =1/1+0.8[2.61-1]

=0.437

Se =(0.5514*.85*.897*.437)325

= 59.718N/mm^2

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STEP 2: CALCULATION OF STRESS AMPLITUDE:

(Se) axial =0.8 times Se

=47.77

sa =(Se)axial/FOS

= 47.77/2 => 23.885

sa =P/A

=P/(w-d)t

23.885 =15000/(70-10)t

t =10.467mm

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By

10a207

10a24310a251

10a254

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