1. FLUCTUATING STRESSES SUBJECT: Design of machine elements
Submitted to- Dhaval Darji Sir
Mechanical Department By

2. Fatigue load
The loads, which vary in magnitude and/or direction with respect to time, are known as fatigue, fluctuating or alternating loads.
It has been observed that, when the mechanical component is subjected to fluctuating loads, it fails at a stress considerably below the ultimate strength and quite frequently even below the yield strength. Such failure is Fatigue failure.

4. B
Instantaneuos Fast Fracture! A
Crack nucleation and Growth

6. Fluctuating stresses
When the mechanical component is subjected to the fatigue or fluctuating load, the stress induced is known as fluctuating stress.

7. Repeated & Reversed Stress
an element subjected to a repeated and alternating tensile and compressive stresses.
Continuous total load reversal over time

8. Definitions: = Alternating stress = Mean stress = R value:
R = 0, repeated and one direction, i.e. stress cycles from 0 to max value.
R =-1, Fully reversed (R-R Moore)

9. 1.Repeated and Reversed Stress
The average or mean stress is zero.

10. Fluctuating Stress
When an element experiences alternating stress, but the mean stress is NOT zero.
Load varies between P and Q over time

11. 2.Fluctuating Stress Example
Bending of Rocker Arm
Valve Spring Force
Valve Open
Valve Closed
Tension in Valve Stem
Valve Closed
Valve Spring Force
Valve Open

12. Tensile Stress w/ Tensile Mean

13. Partially Reversed w/ Tensile Mean
smax is tensile and smin is compressive

14. Partially Reversed w/ Compressive Mean
smax is tensile and smin is compressive

15. Compressive Stress w/ Compressive Mean
smax and smin are both compressive

16. Repeated – One Direction Stress

17. Fatigue Failure, S-N Curve
Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied.
Motor
Load
Rotating beam machine – applies fully reverse bending stress
Typical testing apparatus, pure bending

18. Fatigue Failure, S-N Curve
Finite life
Infinite life
S′e
= endurance limit of the specimen Se ′

19. Design for Finite Life Sn = a (N)b equation of the fatigue lineSe
106
103 A B N S Sf
5x108
103 A B
Point A
Sn = .9Sut
N = 103
Point A
Sn = .9Sut
N = 103
Point B
Sn = Se
N = 106
Point B
Sn = Sf
N = 5x108

20. Sn = a (N)b Sn Se ( ) ⅓ Sn n a log .9Sut = log a + b log 103
log Sn = log a + b log N a =
(.9Sut)2 Se b
.9Sut 1 3
log
log .9Sut = log a + b log 103
log Se = log a + b log 106 Sn = Se ( N
106 ) ⅓
.9Sut
log Sn
Kf a = n
Design equation
Calculate Sn and replace Se in the design equation

21. Design of components subjected to fluctuating stresses for infinite life
Mean stress
Alternating stress m a Sy
Yield line
Gerber curve Se Sy
Soderberg line
Sut
Goodman line

23. The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
Alternating stress m a Sy
Yield line Se
Safe zone C
Goodman line Sy
Sut
Mean stress

24. a Sy Yield line - Syc Se Safe zone Sut Goodman line Safe zone - m Sy

25. a a m a m a = a + m = a + m = m > 0 m ≤ 0 nf Se Sut Sn
Fatigue,
m ≤ 0
Fatigue,
m > 0 a nf Se 1 =
Sut a m +
Infinite life
a = Se nf
Finite life Sn 1 =
Sut a m +
a + m = Sy ny
Yield Se
a + m = Sy ny
Yield C
Safe zone
Safe zone
- m
- Syc Sy
Sut
+m

26. Combined loading xya alternating component of shear stress
All four components of stress exist,
xa alternating component of normal stress
xm mean component of normal stress
xya alternating component of shear stress
xym mean component of shear stress
Calculate the alternating and mean principal stresses,
1a, 2a = (xa /2) ± (xa /2)2 + (xya)2
1m, 2m = (xm /2) ± (xm /2)2 + (xym)2

27. a′ = (1a + 2a - 1a2a)1/2 m′ = (1m + 2m - 1m2m)1/2 ′a ′m nf
Calculate the alternating and mean von Mises stresses,
a′ = (1a + 2a - 1a2a)1/2 2
m′ = (1m + 2m - 1m2m)1/2 2
Fatigue design equation nf Se 1 =
Sut
′a
′m +
Infinite life

28. THANK YOU