FLUCTUATING FLUCTUATING STRESSESSTRESSES

SUBJECT: Machine Design

Mechanical Engineering

Department Presented by: Flt Lt Dinesh Gupta

Associate Professor

Deptt. Of Mech. Engg.

NIET, Alwar (Raj.)

Fatigue loadFatigue load Reduced material resistance

under fluctuating stresses or reversals, which may culminate in cracks or failure after a number of cycles.

Fatigue is the tendency of a member to fail at stress levels below yield stress when subject to cyclical loading.

Therefore failure criterion is not yield stress or ultimate tensile stress.Then what is that?

Completely reversed cycle of stress:

Illustrates the type of fatigue loading where a member is subjected to opposite loads alternately with a means of zero.

For example bending of steel wire continuously in either direction leads to alternate tensile and compressive stresses on its surface layers and failure fatigue.

Repeated & Reversed Repeated & Reversed StressStress

an element subjected to a repeated and alternating tensile and compressive stresses.

Continuous total load reversal over time

Repeated and Reversed Repeated and Reversed StressStress

The average or mean stress is zero.

1max

min R

Definitions:Definitions:

22minmax

a

2minmax

m

max

min

RR = 0, repeated and one direction, i.e. stress cycles from 0 to max value.R =-1, Fully reversed

= Alternating stress

= Mean stress

Fluctuating StressFluctuating StressWhen an element experiences alternating

stress, but the mean stress is NOT zero.

Load varies between P and Q over time

Fluctuating StressFluctuating Stress

2minmax

a

2

minmaxm

The average or mean stress is not zero.

Fatigue properties : Fatigue life (N): it is total

number of cycles are required to bring about final fracture in a specimen at a given stress.

Fatigue life for a given condition is a property of the individual specimen and is arrived at after testing a number of specimens at the same stress.

Fatigue strength (σn) It is the maximum stress at

which a material can withstand repeatedly N number of cycles before failure.

OR it is the strength of a material for a particular fatigue life.

Fatigue limit or Endurance limit (Se ’):

it is stress below which a material will not fail for any unlimited number of cycles.

For ferrous materials it is approximately half of the ultimate tensile strength.

This value is obtained with the help of standard rotating beam test.

Endurance limit The cyclic stress level that the material

can sustain for 10 million cycles is called the Endurance Limit (EL).Denoted by Se’.

Rotating-beam specimen (Bending) For Steels, Se’ = 0.5 Sut, Sut ≤ 1400 MPa For Cast iron and cast steels, Se’ = 0.45 Sut MPa, Sut ≤ 600 MPa

The endurance limit of a real machine component is lower than the estimated of experimentally determined endurance limit because the size, shape, surface finish, etc of a components are generally quite different from those of the specimens used in testing. As such the following relationship is generally used to determine endurance strength of real component.

Se = Ka Kb Kc Kd Ke Se’

Se = endurance strength of a component, MPa

Se’ = endurance limit of a standard specimen of the respective material in a rotating beam machine.

Ka = surface finish factor Kb = size factorKc = reliability factor Kd = modifying factor for stress

concentration Ke = miscellaneous factor like temperature,

environmental etc

Fatigue Failure, S-N Fatigue Failure, S-N CurveCurve

Finite life

Infinite life

N < 103 N > 103

S′e

= endurance limit of the specimenSe′

Generally, the region upto 1000 cycles is low cycle fatigue .Some defense equipment is designed for life in this region.

Mean stress

Alternating stress

m

a

Se

SySoderberg line Sut

Goodman line

Gerber curve

Sy Yield line

Design of components subjected to fluctuating stresses for infinite life

The Effect of Mean Stress on The Effect of Mean Stress on Fatigue Life Modified Goodman Fatigue Life Modified Goodman DiagramDiagram

Mean stress

Alternating stress

m

a

Sut

Goodman line

Sy Yield line

Sy

Se

Safe zone C

Example on reversed Example on reversed stressstress

1. A plate made of steel 20C8 Sut=440N/mm2. It is subjected to a completely reversed axial load of 30KN. Assume ka=0.67, kb=0.85, kc=0.89, kd=0.45 and factor of safety as 2. Determine the plate thickness for infinite life?

Sol. Se’=0.5 Sut=0.5x440=220N/mm2

Se=KaxKbxKcxKdxSe’=0.67x0.85x0.89x0.45x220=50.9N/mm2

Allowable stress= Se/fs=50.9/2=25.45N/mm2 Also allowable stress= P/(w-d)t=(30x1000)/(50-

10)t=25.45 t=29.47mm

Example on fluctuating Example on fluctuating stressstress1. A circular rod made of ductile material has endurance

strength of 280N/mm2 and ultimate strength of 350N/mm2. The member is subjected to variable axial load varying from 300KN(tensile) and 70KN(compressive). Assume Ka=0.85, Kb=1.0, Kc=1,Kd=0.55 and factor of safety as 2. Find suitable diameter of the rod?

Sol. Se’=280N/mm2Sut=350N/mm2Using Goodman line equation:Sm/Sut + Sa/Se=1/fs……………………(1)Se=KaxKbxKcxKcxKdxSe’=0.85x1x1x0.55x280=132.3MPaSm=(Pmax+Pmin)/2A=[300+(-70)]/2A=115/A N/mm2Sa=(Pmax-Pmin)/2A=[300-(-70)]/2A=185/A N/mm2Put values in eqn. 1 we get: A= 3456mm2Now A= π/4 d2=3456 or d=66.3mm

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