Slide 1

Report
Design Stress & Fatigue
MET 210W
E. Evans
P
Parts Fail When?
Crack initiation site
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
Failure Strength
Factor of Saf ety 
Applied Stress
Sy
Example :
N

• Design
Failure Strength
Allowable Stress 
Design Factor
Sy
Example :
 ALLOW 
N
Factors Effecting Design Factor
•
•
•
•
•
•
Application
Environment
Loads
Types of Stresses
Material
Confidence
Factors Effecting Design Factor
• Application
• How many will be produced?
•
•
•
•
•
• What manufacturing methods will
be used?
Environment
Loads
Types of Stresses
Material
Confidence
• 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
• Material
• Confidence
• What kind of stress?
• Direct tension or compression
• Direct shear
• Bending
Stresses • 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:
• Brittle:
%E  5%
%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
– Modified Mohr
- Uniaxial
- Biaxial
• Ductile Materials:
– Yield Strength
– Maximum Shear Strength
– Distortion Energy
- Uniaxial
- Biaxial
- 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
Modified Mohr
• Biaxial Static Stress on Brittle Materials
2
45° Shear Diagonal
Sut
Suc
Sut
2
1
1, 2
Suc
1
Stress concentrations
applied to stresses before
making the circle
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
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 tensiletest specimen when yielding begins.
Somewhat conservative – use Distortion
Energy for more precise failure estimate
Distortion Energy
• Static
Shear
Diagonal
Biaxial or Triaxial Stress on Ductile Materials
2
Best predictor of failure for
ductile materials under static
loads or completely
reversed normal, shear or
combined stresses.
Sy
Sy
Sy
1
'      12
2
1
2
2
’ = von Mises stress
Sy
Distortion Energy
Failure:
’ > Sy
Design:
’  d = Sy/N
von Mises Stress
• Alternate Form
'       x  y  3t
2
x
2
y
For uniaxial stress when y = 0,
• Triaxial Distortion Energy
2
xy
'    3t
2
x
2
xy
(1 > 2 > 3)
 2
 ( 2  1 )2  (3  1 )2  (3   2 )2
'  
 2 


Fluctuating Stress
• Varying stress with a nonzero mean.
alternating = a
Stress
max
max  min
mean 
2
max  min
a 
2
Stress Ratio,
mean
min
 min
R
 max
Time
-1  R  1
Fluctuating Stress Example
• Bending of Rocker Arm
Valve Spring Force
Valve Open
Valve
Closed
• Tension in Valve Stem
Valve
Closed
RBE
2/1/91
Valve Spring Force
Valve
Open
Adapted from R. B. Englund
Fatigue Testing
• Bending tests
– Spinning bending elements – most common
– Constant stress cantilever beams
Top View
Front View
Fixed Support
Applied Deformation
– Fully Reversed, R = -1
Test Data
Stress,  (ksi)
Fatigue Testing
Number of Cycles to Failure, N
Data from R. B. Englund, 2/5/93
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’
Sn
Cm
Cst
CR
CS
= actual endurance strength (ESTIMATE)
= endurance strength from Fig. 5-8
= material factor (pg. 174)
= stress type:
1.0 for bending
0.8 for axial tension
0.577 for shear
= reliability factor
= size factor
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
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
Axial Tension
Actual Sn’
Estimate
Reliability, Table 5-1
Size Factor, Fig. 5-9
99% Probability
Sn’ is at or above the
Guessing: diameter  .5”
calculated value
Goodman Diagram
a
Sy
Yield Line
Sn’
NO FATIGUE
FAILURE REGION
-Sy
0
FATIGUE
FAILURE REGION
Goodman Line
a m

1
Sn S u
Sy
Su
m
Goodman Diagram
Safe Stress Line
K t a
Sn
a
m
1


Su N
Sy
Yield Line
Sn’
FATIGUE
FAILURE REGION
Goodman Line
a m

1
Sn S u
Sn’/N
SAFE ZONE
-Sy
0
Su/N
Sy
Su
Safe Stress Line
m
Example: Problem 5-53.
Find a suitable titanium alloy. N = 3
1.5 mm Radius
42 mm DIA
30 mm
DIA
F varies from 20 to 30.3 kN
+
FORCE
MAX = 30.3
-
30.3  20
alt 
 5.15 kN
2
30.3  20
mean 
 25.15 kN
2
MIN = 20
TIME
Example: Problem 5-53 continued.
• Find the mean stress:
25,150 N
m 
 3 5 .6M P a

(30 mm)2
4
• Find the alternating stress:
a 
5,150 N

(30 mm)2
4
 7 .3 M P a
• Stress concentration from App. A15-1:
D 42 mm

 1.4;
d 30 mm
r 1.5 mm

 .05
d 30 mm
 K t  2 .3
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
K t a
Sn
m
1


Su
N
(Eqn 5-20)
2.3(7.3 MPa )
35.6 MPa 1

  .297
.8(.86)(448 MPa / 4) 448 MPa N
Size
Tension
Reliability 50%
1
N
 3 .3 6
.297
3.36 is good, need further information on Sn for titanium.
Example:
Find a suitable steel for N = 3 & 90% reliable.
3 mm Radius
50 mm DIA
30 mm
DIA
T varies from 848 N-m to 1272 N-m
TORQUE
+
-
MAX = 1272 N-m
MIN = 848 N-m
TIME
1272  848
alt 
 212 N  m
2
1272  848
mean 
 1060 N  m
2
T = 1060 ± 212 N-m
Example: continued.
• Stress concentration from App. A15-1:
D 50 mm

 1.667;
d 30 mm
r
3 mm

 .1  K t  1 .3 8
d 30 mm
• Find the mean shear stress:
)
Tm 1060 N  m(1000 mm
m
tm 

 2 0 0M P a

Zp
(30 mm)3
16
• Find the alternating shear stress:
Ta
212000 N  mm
ta 

 40M Pa
3
Zp
5301 mm
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:
Ductile
tmax = 200 + 40 = 240 MPa  Sys  600/2 = 300MPa
• Find the ultimate shear stress:
Sus = .75Su = .75(779 MPa) = 584 MPa
Example: continued.
• Assume machined surface, Sn  295 MPa
(Fig. 5-8)
• Find actual endurance strength:
S’sn = Sn(Cm)(Cst)(CR)(CS)
= 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa
Sn
Wrought steel
Shear Stress
Size – 30 mm
90% Reliability
Example: continued.
• Goodman:
K t ta
Ssn
tm
1


S su N
(Eqn. 5-28)
1.38( 40 MPa ) 200 MPa 1

  .7606
132 MPa
584 MPa N
1
N
 1 .3 1
.7606
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%
Ductile
• 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
Sn
shear
wrought
size
reliable
Example: continued.
• Goodman:
K t ta
Ssn
tm
1


S su N
(Eqn. 5-28)
1.38( 40 MPa ) 200 MPa
1

  .378
272 MPa
1140 MPa N
1
N
 2 .6 4
.378
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
PIPE: TS4 x .237 WALL
MATERIAL: ASTM A242
Equivalent
Reversed,
Repeated
DEAD WEIGHT:
SIGN + ARM + POST = 1000#
(Compression)
45°
Bending
RBE
2/11/97
Repeated one direction
Example: Combined Stress Fatigue Cont’d
Stress Analysis:
Dead Weight:

P
1000 #

 315.5 psi
2
A 3.17 in
(Static)
Vertical from Wind:

P
200 #

 63.09 psi
2
A 3.17 in
(Cyclic)
Bending:
M 500# (60 in)


 9345.8 psi
3
Z
3.21 in
(Static)
Example: Combined Stress Fatigue Cont’d
Stress Analysis:
Torsion:
T
200# (100 in)
t

 3115.3 psi
3
ZP
2(3.21 in )
Stress Elements:
STATIC:
315.5 psi
(Cyclic)
(Viewed from +y)
CYCLIC:
9345.8 psi
z
x
z
x
63.09 psi – Repeated
One Direction
t = 3115.3 psi
Fully Reversed
Example: Combined Stress Fatigue Cont’d
Mean Stress:
Static
Repeated / 2
-
8998.8 psi
t (CW)
TIME
Stress
9345.8
-315.5
-31.5
+
Alternating Stress:
m
MIN = -63.09 psi
t (CW)
a
tmax
tmax
(0,-3115.3)

1 
(-31.5,-3115.3)
tmax
8998.8 psi

 4499.4 psi
2
tmax  3115.34 psi
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”) = 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 steel
Combined or Shear Stress
Size – 4.50” dia
90% Reliability
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
Alternating Stress, ta
K t ta
Ssn
10
S’sn

tm
1

S su N
1.0(3115.3 psi) 4499.4 psi 1

  .426
8900 psi
52500 psi N
N
5
S’sn/N
Su
Kttalt
3115.3
0
0
5
tmean = 4499.4
10
15
Mean Stress, tm
1
 2 .2 9
.426
Su/N
20

similar documents