1. Design Stress & Fatigue
MET 210W
E. Evans

2. Parts Fail When? P P Crack initiation site
This crack in the part is very small. If the level of stress in the part is SMALL, the crack will remain stable and not expand. If the level of stress in the part is HIGH enough, the crack will get bigger (propagate) and the part will eventually fail. P

3. Design Factor
Analysis
Design

4. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence

5. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
How many will be produced?
What manufacturing methods will be used?
What are the consequences of failure?
Danger to people
Cost
Size and weight important?
What is the life of the component?
Justify design expense?

6. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
Temperature range.
Exposure to electrical voltage or current.
Susceptible to corrosion
Is noise control important?
Is vibration control important?
Will the component be protected?
Guard
Housing

7. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
Nature of the load considering all modes of operation:
Startup, shutdown, normal operation, any foreseeable overloads
Load characteristic
Static, repeated & reversed, fluctuating, shock or impact
Variations of loads over time.
Magnitudes
Maximum, minimum, mean

8. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
What kind of stress?
Direct tension or compression
Direct shear
Bending
Torsional shear
Application
Uniaxial
Biaxial
Triaxial

9. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
Material properties
Ultimate strength, yield strength, endurance strength,
Ductility
Ductile: %E  5%
Brittle: %E < 5%
Ductile materials are preferred for fatigue, shock or impact loads.

10. Factors Effecting Design Factor
Application
Environment
Loads
Types of Stresses
Material
Confidence
Reliability of data for
Loads
Material properties
Stress calculations
How good is manufacturing quality control
Will subsequent handling, use and environmental conditions affect the safety or life of the component?

11. Design Factor
Adapted from R. B. Englund

12. Design Factor

13. Predictions of Failure Static Loads
Brittle Materials:
Maximum Normal Stress - Uniaxial
Modified Mohr - Biaxial
Ductile Materials:
Yield Strength - Uniaxial
Maximum Shear Strength - Biaxial
Distortion Energy - Biaxial or Triaxial

14. Predictions of Failure Fluctuating Loads
Brittle Materials:
Not recommended
Ductile Materials:
Goodman
Gerber
Soderberg

15. Maximum Normal Stress Uniaxial Static Loads on Brittle Material:
In tension:
Kt s  sd = Sut / N
In compression:
Kt s  sd = Suc / N

16. Stress concentrations applied to stresses before making the circle
Modified Mohr
Biaxial Static Stress on Brittle Materials s2
Sut
45° Shear Diagonal s2 s1
Suc
Sut s1
Stress concentrations applied to stresses before making the circle
s1, s2
Often brittle materials have much larger compressive strength than tensile strength
Suc

17. For most ductile materials, Syt = Syc
Yield Strength Method
Uniaxial Static Stress on Ductile Materials
In tension:
s  sd = Syt / N
In compression:
 sd = Syc / N
For most ductile materials, Syt = Syc

18. tmax  td = Sys / N = 0.5(Sy )/ N
Maximum Shear Stress
Biaxial Static Stress on Ductile Materials
tmax  td = Sys / N = 0.5(Sy )/ N
Ductile materials begin to yield when the maximum shear stress in a load-carrying component exceeds that in a tensile-test specimen when yielding begins.
Somewhat conservative – use Distortion Energy for more precise failure estimate

19. Distortion Energy s2 s1 Distortion Energy
Static Biaxial or Triaxial Stress on Ductile Materials
Shear Diagonal s2
Best predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses. Sy Sy s1 Sy
s’ = von Mises stress
Failure: s’ > Sy
Design: s’  sd = Sy/N Sy
Distortion Energy

20. von Mises Stress Alternate Form
For uniaxial stress when sy = 0,
Triaxial Distortion Energy (s1 > s2 > s3)

21. Fluctuating Stress Varying stress with a nonzero mean.
salternating = sa
smax
Stress
Stress Ratio,
smean
Time
-1  R  1
smin

22. 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
RBE 2/1/91
Adapted from R. B. Englund

23. Fatigue Testing Bending tests Spinning bending elements – most common
Constant stress cantilever beams
Top View
Front View
Applied Deformation – Fully Reversed, R = -1
Fixed Support

24. Fatigue Testing Test Data Stress, s (ksi)
Number of Cycles to Failure, N
Data from R. B. Englund, 2/5/93

25. Endurance Strength = 0.50(Su)
The stress level that a material can survive for a given number of load cycles.
For infinite number of cycles, the stress level is called the endurance limit.
Estimate for Wrought Steel:
Endurance Strength = 0.50(Su)
Most nonferrous metals (aluminum) do not have an endurance limit.

26. Typical S-N Curve

27. Estimated Sn of Various Materials

28. Actual Endurance Strength
Sn’ = Sn(Cm)(Cst)(CR)(CS)
Sn’ = actual endurance strength (ESTIMATE)
Sn = endurance strength from Fig. 5-8
Cm = material factor (pg. 174)
Cst = stress type: 1.0 for bending
0.8 for axial tension
0.577 for shear
CR = reliability factor
CS = size factor

29. Actual Sn Example
Find the endurance strength for the valve stem. It is made of AISI 4340 OQT 900°F.
From Fig. A4-5.
Su = 190 ksi
From Fig. 5-8.
Sn = 62 ksi (machined)
62 ksi

30. Actual Sn Example Continued
Sn’ = Sn(Cm)(Cst)(CR)(CS)
= 62 ksi(1.0)(.8)(.81)(.94) = 37.8 ksi
Sn,Table 5-8
Wrought Steel
Actual Sn’ Estimate
Axial Tension
Reliability, Table 5-1
Size Factor, Fig. 5-9
99% Probability
Sn’ is at or above the calculated value
Guessing: diameter  .5”

31. FATIGUE FAILURE REGION NO FATIGUE FAILURE REGION
Goodman Diagram sa
Yield Line Sy
FATIGUE FAILURE REGION
Sn’
Goodman Line
NO FATIGUE FAILURE REGION sm
-Sy Sy Su

32. FATIGUE FAILURE REGION
Goodman Diagram
Safe Stress Line sa
Yield Line Sy
FATIGUE FAILURE REGION
Sn’
Goodman Line
Sn’/N
SAFE ZONE sm
-Sy
Su/N Sy Su
Safe Stress Line

33. Example: Problem 5-53. Find a suitable titanium alloy. N = 3
1.5 mm Radius F
30 mm DIA
42 mm DIA
F varies from 20 to 30.3 kN +
MAX = 30.3
FORCE
MIN = 20 -
TIME

34. Example: Problem 5-53 continued.
Find the mean stress:
Find the alternating stress:
Stress concentration from App. A15-1:

35. Example: Problem 5-53 continued.
Sn data not available for titanium so we will guess! Assume Sn = Su/4 for extra safety factor.
TRY T2-65A, Su = 448 MPa, Sy = 379 MPa
(Eqn 5-20)
Size
Tension
Reliability 50%
3.36 is good, need further information on Sn for titanium.

36. Example: Find a suitable steel for N = 3 & 90% reliable.
3 mm Radius
50 mm DIA
30 mm DIA T T
T varies from 848 N-m to 1272 N-m +
MAX = 1272 N-m
TORQUE
MIN = 848 N-m -
TIME
T = 1060 ± 212 N-m

37. Example: continued. Stress concentration from App. A15-1:
Find the mean shear stress:
Find the alternating shear stress:

38. Example: continued. So, t = 200 ± 40 MPa. Guess a material.
TRY: AISI 1040 OQT 400°F
Su = 779 MPa, Sy = 600 MPa, %E = 19%
Verify that tmax  Sys:
tmax = = 240 MPa  Sys  600/2 = 300MPa
Find the ultimate shear stress:
Sus = .75Su = .75(779 MPa) = 584 MPa
Ductile

39. Assume machined surface, Sn  295 MPa Find actual endurance strength:
Example: continued.
Assume machined surface, Sn  295 MPa
Find actual endurance strength:
S’sn = Sn(Cm)(Cst)(CR)(CS)
= 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa
(Fig. 5-8) Sn
Wrought steel
Shear Stress
90% Reliability
Size – 30 mm

40. Goodman: Example: continued. No Good!!! We wanted N  3
(Eqn. 5-28)
No Good!!! We wanted N  3
Need a material with Su about 3 times bigger than this guess or/and a better surface finish on the part.

41. Example: continued. Guess another material. TRY: AISI 1340 OQT 700°F
Su = 1520 MPa, Sy = 1360 MPa, %E = 10%
Find the ultimate shear stress:
Sus = .75Su = .75(779 MPa) = 584 MPa
Find actual endurance strength:
S’sn = Sn(Cm)(Cst)(CR)(CS)
= 610 MPa(1.0)(.577)(.9)(.86) = 272 MPa
Ductile Sn
shear
size
wrought
reliable

42. Goodman: Example: continued. No Good!!! We wanted N  3
(Eqn. 5-28)
No Good!!! We wanted N  3
Decision Point:
Accept 2.64 as close enough to 3.0?
Go to polished surface?
Change dimensions? Material? (Can’t do much better in steel since Sn does not improve much for Su > 1500 MPa

43. Example: Combined Stress Fatigue
RBE 2/11/97

44. Example: Combined Stress Fatigue Cont’d
PIPE: TS4 x .237 WALL
MATERIAL: ASTM A242 Equivalent
DEAD WEIGHT:
SIGN + ARM + POST = 1000#
(Compression)
Reversed, Repeated
45°
Bending
RBE 2/11/97
Repeated one direction

45. Example: Combined Stress Fatigue Cont’d Stress Analysis:
Dead Weight:
(Static)
Vertical from Wind:
(Cyclic)
Bending:
(Static)

46. Example: Combined Stress Fatigue Cont’d Stress Analysis:
Torsion:
(Cyclic)
Stress Elements: (Viewed from +y)
STATIC:
CYCLIC:
315.5 psi
63.09 psi – Repeated One Direction
psi x z x z
t = psi Fully Reversed

47. Example: Combined Stress Fatigue Cont’d
Mean Stress:
Alternating Stress: +
TIME -
Stress
MIN = psi sm sa
9345.8
-315.5
-31.5
8998.8
psi
Static
Repeated / 2
t (CW) s
(-31.5, )
(0, )
tmax
t (CW) s1 s
tmax

48. Example: Combined Stress Fatigue Cont’d Determine Strength:
Try for N = 3  some uncertainty
Size Factor? OD = 4.50 in, Wall thickness = .237 in
ID = 4.50” – 2(.237”) = in
Max. stress at OD. The stress declines to 95% at 95% of the OD = .95(4.50”) = in. Therefore, amount of steel at or above 95% stress is the same as in 4.50” solid.
ASTM A242: Su = 70 ksi, Sy = 50 ksi, %E = 21%
t  3/4”
Ductile

49. Example: Combined Stress Fatigue Cont’d
We must use Ssu and S’sn since this is a combined stress situation. (Case I1, page 197)
Sus = .75Su = .75(70 ksi) = 52.5 ksi
S’sn = Sn(Cm)(Cst)(CR)(CS)
= 23 ksi(1.0)(.577)(.9)(.745) = 8.9 ksi
Hot Rolled Surface
Size – 4.50” dia
Wrought steel
90% Reliability
Combined or Shear Stress

50. Example: Combined Stress Fatigue Cont’d
“Safe” Line for Goodman Diagram:
ta = S’sn / N = 8.9 ksi / 3 = 2.97 ksi
tm = Ssu / N = 52.5 ksi / 3 = 17.5 ksi 10
N = 1 Fail
Not Fail Su
S’sn
Alternating Stress, ta 5
tmean =
3115.3
Kttalt
S’sn/N
N = 3
Safe 5 10 15
Su/N 20
Mean Stress, tm