ed batch 5
TRANSCRIPT
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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
11a44311a444