Design Stress & Fatigue
MET 210W
E. Evans
Parts Fail When?
Crack initiation site
P
P
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.
Design Factor
• Analysis
• Design
ySN:Example
StressApplied
StrengthFailureSafetyofFactor
N
S:Example
FactorDesign
StrengthFailureStressAllowable
yALLOW
Factors Effecting Design Factor
• Application
• Environment
• Loads
• Types of Stresses
• Material
• Confidence
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?
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
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
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
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.
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?
Design Factor
Adapted from R. B. Englund
Design Factor
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
Predictions of Failure Fluctuating Loads
• Brittle Materials:– Not recommended
• Ductile Materials:– Goodman– Gerber– Soderberg
Maximum Normal Stress
•Uniaxial Static Loads on Brittle Material:
–In tension:
Kt d = Sut / N
–In compression:
Kt d = Suc / N
45° Shear Diagonal
Modified Mohr
• Biaxial Static Stress on Brittle Materials
12
Stress concentrations applied to stresses before
making the circle
1
2
Sut
Sut
Suc
Suc
1, 2
Often brittle materials have much larger compressive strength than tensile strength
Yield Strength Method
• Uniaxial Static Stress on Ductile Materials
– In tension:
d = Syt / N
– In compression:
d = Syc / N
For most ductile materials, Syt = Syc
Maximum Shear Stress
• Biaxial Static Stress on Ductile Materials
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.
max d = Sys / N = 0.5(Sy )/ N
Somewhat conservative – use Distortion Energy for more precise failure estimate
Distortion Energy
Best predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses.
Shear Diagonal
1
2
2122
21 '
’ = von Mises stress
Failure: ’ > Sy
Design: ’ d = Sy/N
• Static Biaxial or Triaxial Stress on Ductile Materials
Sy
Sy
Sy
Sy
Distortion Energy
von Mises Stress• Alternate Form
For uniaxial stress when y = 0,
• Triaxial Distortion Energy(1 > 2 > 3)
222 3 xyyxyx'
22 3 xyx'
223
213
2122
2)()()('
Fluctuating StressS
tres
s
Timemin
mean
max
• Varying stress with a nonzero mean.
alternating = a2
minmaxmean
2minmax
a
Stress Ratio,
max
minR
-1 R 1
Fluctuating Stress Example
Valve Spring Force
Valve Spring ForceValve Open
Valve Closed
Valve Closed
Valve Open
• Bending of Rocker Arm
• Tension in Valve Stem
Adapted from R. B. Englund
RBE 2/1/91
Fatigue Testing
• Bending tests– Spinning bending elements – most common– Constant stress cantilever beams
Fixed SupportApplied Deformation – Fully Reversed, R = -1
Top View
Front View
Fatigue Testing
Number of Cycles to Failure, N
Str
ess,
(
ksi)
Data from R. B. Englund, 2/5/93
Test Data
Endurance Strength
• 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.
Typical S-N Curve
Estimated Sn of Various Materials
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
Actual Sn Example
• Find the endurance strength for the valve stem. It is made of AISI 4340 OQT 900°F.
62 ksi
From Fig. A4-5.Su = 190 ksi
From Fig. 5-8.Sn = 62 ksi (machined)
Actual Sn Example Continued
Sn’ = Sn(Cm)(Cst)(CR)(CS)
= 62 ksi(1.0)(.8)(.81)(.94) = 37.8 ksi
Size Factor, Fig. 5-9
Wrought Steel
Axial Tension
Reliability, Table 5-1
99% Probability Sn’ is at or above the calculated value
Sn,Table 5-8
Actual Sn’ Estimate
Guessing: diameter .5”
Sn’
Goodman Diagram
m
a
-Sy
Sy
Sy Su
FATIGUE FAILURE REGION
NO FATIGUE FAILURE REGION
Goodman Line
0
Yield Line
1
u
m
n
a
SS
Sn’/N
Sn’
Goodman Diagram
m
a
-Sy
Sy
Sy Su
FATIGUE FAILURE REGION
Goodman Line
0
Yield Line
1
u
m
n
a
SS
Su/N
NSS
K
u
m
n
at 1
Safe Stress Line
Safe Stress Line
SAFE ZONE
MAX = 30.3
Example: Problem 5-53. Find a suitable titanium alloy. N = 3
42 mm DIA 30 mm DIA
1.5 mm Radius
F varies from 20 to 30.3 kN+
-
FO
RC
E
MIN = 20
TIME
kN..
mean 15252
20330
kN..
alt 1552
20330
Example: Problem 5-53 continued.
• Find the mean stress:
• Find the alternating stress:
• Stress concentration from App. A15-1:
MPa35.6
2
m
)mm30(4
N150,25
MPa7.3
2
a
)mm30(4
N150,5
2.3Kt 05.mm30
mm5.1
d
r;4.1
mm30
mm42
d
D
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
3.36
297.
1N
297.N
1
MPa448
MPa6.35
)4/MPa448)(86(.8.
)MPa3.7(3.2
N
1
SS
K
u
m
n
at(Eqn 5-20)
Tension
Size
Reliability 50%
3.36 is good, need further information on Sn for titanium.
MAX = 1272 N-m
Example: Find a suitable steel for N = 3 & 90% reliable.
T varies from 848 N-m to 1272 N-m+
-
TO
RQ
UE
MIN = 848 N-m
TIME
mN10602
8481272mean
mN2122
8481272alt
50 mm DIA30 mm
DIA
3 mm Radius
T = 1060 ± 212 N-m
Example: continued. • Stress concentration from App. A15-1:
• Find the mean shear stress:
• Find the alternating shear stress:
MPa200
3
mmm
p
mm
)mm30(16
)1000(mN1060
Z
T
MPa40
3
p
aa
mm5301
mmN212000
Z
T
1.38Kt 1.mm30
mm3
d
r;667.1
mm30
mm50
d
D
Example: continued. • So, = 200 ± 40 MPa. Guess a material.
TRY: AISI 1040 OQT 400°F Su = 779 MPa, Sy = 600 MPa, %E = 19%
• Verify that max Sys:
max = 200 + 40 = 240 MPa Sys 600/2 =
300MPa
• Find the ultimate shear stress:
Sus = .75Su = .75(779 MPa) = 584 MPa
Ductile
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
Sn
Wrought steelShear Stress
90% ReliabilitySize – 30 mm
(Fig. 5-8)
Example: continued.
• Goodman:
1.31
7606.
1N
7606.N
1
MPa584
MPa200
MPa132
)MPa40(38.1
N
1
SS
K
su
m
sn
at(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.
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
wrought
shear
reliable
size
Example: continued.
• Goodman:
2.64
378.
1N
378.N
1
MPa1140
MPa200
MPa272
)MPa40(38.1
N
1
SS
K
su
m
sn
at(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
Example: Combined Stress Fatigue
RBE 2/11/97
Example: Combined Stress Fatigue Cont’d
RBE 2/11/97 Repeated one direction
PIPE: TS4 x .237 WALL
MATERIAL: ASTM A242 Equivalent
DEAD WEIGHT:SIGN + ARM + POST = 1000#
(Compression)
Reversed, Repeated
45°
Bending
Example: Combined Stress Fatigue Cont’dStress Analysis:
psi5.315in17.3
#1000
A
P2
Dead Weight:
psi09.63in17.3
#200
A
P2
Vertical from Wind:
(Static)
(Cyclic)
psi8.9345in21.3
)in60(#500
Z
M3
Bending:
(Static)
Example: Combined Stress Fatigue Cont’dStress Analysis:
Torsion:
(Cyclic)psi3.3115)in21.3(2
)in100(#200
Z
T3
P
Stress Elements: (Viewed from +y)
CYCLIC:STATIC: 315.5 psi
9345.8 psi
63.09 psi – Repeated One Direction
= 3115.3 psi Fully Reversedx
z
x
z
Example: Combined Stress Fatigue Cont’dMean Stress:
9345.8-315.5
-31.5
8998.8 psi
Static
Repeated / 2
Alternating Stress:
(CW)
1
max
psi4.44992
psi8.8998max
+TIME
-
Str
ess
MIN = -63.09 psi
m
a (CW)
(-31.5,-3115.3)
(0,-3115.3)max
psi34.3115max
Example: Combined Stress Fatigue Cont’dDetermine Strength:
Try for N = 3 some uncertainty
Size Factor? OD = 4.50 in, Wall thickness = .237 inID = 4.50” – 2(.237”) = 4.026 in
Max. stress at OD. The stress declines to 95% at 95% of the OD = .95(4.50”) = 4.275 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
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
Wrought steelCombined or Shear Stress
90% ReliabilitySize – 4.50” dia
Example: Combined Stress Fatigue Cont’d“Safe” Line for Goodman Diagram:
a = S’sn / N = 8.9 ksi / 3 = 2.97 ksi
m = Ssu / N = 52.5 ksi / 3 = 17.5 ksi
Mean Stress, m
0 15105 200
5
10
Alte
rnat
ing
Str
ess,
a
2.29
426.
1N
426.N
1
psi52500
psi4.4499
psi8900
)psi3.3115(0.1
N
1
SS
K
su
m
sn
at
Su/N
S’sn/N
N = 1 FailNot Fail
Su
S’sn
mean = 4499.4
3115.3 Ktalt
N = 3Safe