Chapter 5, Part B
Failure Modes
MET 210W
E. Evans/ R.Michael
Ductility and Percent Elongation
• Ductility is the degree to which a material
will deform before ultimate fracture.
• Percent elongation is used as a measure
of ductility.
• Ductile Materials have %E  5%
• Brittle Materials have %E < 5%
• For machine members subject to repeated
or shock or impact loads, materials with
%E > 12% are recommended.
Ductile materials - extensive plastic deformation and
energy absorption (toughness) before fracture
Brittle materials - little plastic deformation and low energy
absorption before failure
DUCTILE VS BRITTLE FAILURE
• Classification:
(a)
• Ductile
fracture is
desirable!
Ductile:
warning before
fracture
(b)
(c)
Brittle:
No
warning
DUCTILE FAILURE
• Evolution to failure:
“cup and cone” fracture
• Resulting
fracture
surfaces
(steel)
particles serve as void
nucleation sites.
50 µm
1 µm = 1 X 10-6 m = 0.001 mm
Failure Prediction Methods
Static Loads
• Brittle Materials - FT:
– Maximum Normal Stress
– Modified Mohr
- Uniaxial stress
- Biaxial stress
• Ductile Materials - FT:
– Yield Strength
– Maximum Shear Strength
– Distortion Energy
- Uniaxial stress
- Biaxial stress
- 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:
Brittle Material
Static
Load
–In tension:
DESIGN:
 max
S ut
 K t   d 
N
ANALYSIS:
N
Sut
 max
–In compression:
DESIGN:
 max
ANALYSIS:
Suc
 K t   d 
N
N
Sut
 max
Modified Mohr Method
• Biaxial Static Stress on Brittle Materials
2
45° Shear Diagonal
Sut
Suc
Sut
2
1
1, 2
Failure when outside of shaded area
Suc
1
Stress concentrations
applied to stresses before
making the circle
Brittle materials often have a
much larger compressive
strength than tensile strength
Yield Strength Method
• Uniaxial Static Stress on Ductile Materials
Static
Load
Ductile Material
– In tension:
ANALYSIS:
DESIGN:
 max   d 
S yt
N
–In compression:
S yc
 max   d 
N
N
 max
ANALYSIS:
DESIGN:
For most ductile materials, Syt = Syc
S yt
N
S yc
 max
Maximum Shear Stress
• Biaxial Static Stress on Ductile Materials
avg, max
DESIGN:
max  d 
ANALYSIS:
N
Sy s
N

Sy
2N
S ys
 max
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 approach – use the Distortion Energy Method
for a more precise failure estimate
Distortion Energy
• Static
Biaxial or Triaxial Stress on Ductile Materials
Shear
Diagonal
2
Best predictor of failure for
ductile materials under static
loads or under completely
reversed normal, shear or
combined stresses.
Sy
Sy
Sy
1
'  12  22  12
’ = von Mises stress
Sy
Distortion Energy
Failure:
’ > Sy
Design:
’  d = Sy/N
ANALYSIS:
N  Sy/’
von Mises Stress
• Alternate Form
'   2x   2y   x  y  3 2xy
For uniaxial stress when y = 0,
• Triaxial Distortion Energy
'    3
2
x
2
xy
(1 > 2 > 3)
 2
 ( 2  1 )2  (3  1 )2  (3   2 )2
'  
 2 


Comparison of Static Failure
Theories:
Shows “no failure” zones
Maximum Shear – most conservative
Summary Static Failure Theories:
• Brittle materials fail on planes of max
normal stress:
– Max Normal Stress Theory
– Modified Mohr Theory
• Ductile materials fail on planes of max
shear stress:
– Max shear stress theory
– Distortion energy theory
• See summary table!
• Do example problems for static loading!
Brittle failure or ductile failure? Key: is the fracture surface
on a plane of max shear or max normal stress.
TORQUE:
DUCTILE
BRITTLE
AXIAL
Brittle
Ductile
Goodman Method
Good predictor of failure in ductile materials
experiencing fluctuating stress
a
Sn’ = actual endurance
strength
a = alternating stress
m = mean stress
Sy
Yield Line
Sn’
FATIGUE
FAILURE REGION
Goodman Line
a m

1
Sn S u
NO FATIGUE
FAILURE REGION
-Sy
0
Sy
Su
m
Goodman Diagram
Safe Stress Line
K t a
Sn
Sn’ = actual endurance strength
a = alternating stress
m = mean stress
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
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 a valve stem
made of AISI 4340 OQT 900°F steel.
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
Example: Problem 5-53.
Find a suitable titanium alloy. N = 3
1.5 mm Radius
30 mm
DIA
42 mm DIA
F varies from 20 to 30.3 kN
+
FORCE
MAX = 30.3
-
30 .3  20
 5.15 kN
2
30 .3  20
mean 
 25 .15 kN
2
alt 
MIN = 20
TIME
Example: Problem 5-53 continued.
• Find the mean stress:
25,150 N
m 
 35.6 MPa

(30 mm )2
4
• Find the alternating stress:
a 
5,150 N

(30 mm )2
4
 7.3 MPa
• 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
 3.36
.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
1272  848
 212 N  m
2
1272  848
mean 
 1060 N  m
2
alt 
MIN = 848 N-m
TIME
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  1.38
d 30 mm
• Find the mean shear stress:
)
Tm 1060 N  m(1000 mm
m
m 

 200 MPa

Zp
(30 mm)3
16
• Find the alternating shear stress:
Ta
212000 N  mm
a 

 40 MPa
3
Zp
5301 mm
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:
Ductile
max = 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 a
Ssn
m
1


S su N
(Eqn. 5-28)
1.38( 40 MPa) 200 MPa 1

  .7606
132 MPa
584 MPa N
1
N
 1.31
.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 a
Ssn
m
1


S su N
(Eqn. 5-28)
1.38( 40 MPa) 200 MPa
1

  .378
272 MPa
1140 MPa N
1
N
 2.64
.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)


 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
 = 3115.3 psi
Fully Reversed
Example: Combined Stress Fatigue Cont’d
Mean Stress:
Static
Repeated / 2
-
8998.8 psi
 (CW)
TIME
Stress
9345.8
-315.5
-31.5
+
Alternating Stress:
m
MIN = -63.09 psi
 (CW)
a
max
max
(0,-3115.3)

1 
(-31.5,-3115.3)
max
8998 .8 psi

 4499 .4 psi
2
max  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:
a = S’sn / N = 8.9 ksi / 3 = 2.97 ksi
m = Ssu / N = 52.5 ksi / 3 = 17.5 ksi
Alternating Stress, a
K t a
Ssn
10
S’sn

m
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
Ktalt
3115.3
0
0
5
mean = 4499.4
10
15
Mean Stress, m
1
 2.29
.426
Su/N
20
Design Factors, N
(a.k.a. Factor of Safety)
FOR DUCTILE
MATERIALS:
•N = 1.25 to 2.0
Static loading, high level of confidence in all design
data
•N = 2.0 to 2.5
Dynamic loading, average confidence in all design
data
•N = 2.5 to 4.0
Static or dynamic with uncertainty about loads,
material properties, complex stress state, etc…
•N = 4.0 or higher Above + desire to provide extra safety
Failure
Theory:
When Use?
Failure When:
Design Stress:
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tensi on)
 max  Syc (for compressio n)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
 max  Kt   Sut (for tensi on)
 max  Kt   Suc (for compressio n)
Uniaxial:
Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
 d  Syt / N (for tensi on)
 d  Syc / N (for compressio n)
Note : Syt  Syc for ductile/wr ought material
 d  Sys / N where Sys  Sy/2
 'd  Sy / N
where  '  von Mises stress
see Figure 5 - 13
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
see Figure 5.15
Ductile Material/


Failure
Theories Kfor
 STATIC
1
where Loading
Fluctuating Shear Stress
(Fatigue Loading)
 d  Sut / N (for tensi on)
 d  Suc / N (for compressio n)
t a
'
sn
S
m
S su
S sn'  0.577 S n' and S su  0.75Su
Bi-axial:
K t ( a ) max ( m ) max

1
S sn'
S su
or
whe re
S sn'  0.577 S n' and S su  0.75Su
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
whe re
S sn'  0.577 S n' and S su  0.75Su
Failure
Theory:
When Use?
Failure When:
Design Stress:
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tensi on)
 max  Syc (for compressio n)
 d  Syt / N (for tensi on)
 d  Syc / N (for compressio n)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
 d  Sys / N where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
a. Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
 max  Kt   Sut (for tensi on)
 max  Kt   Suc (for compressio n)
: Syt FATIGUE
Syc for ductile/wr ought
material
Failure TheoriesNotefor
Loading
b. Ductile Material/
Fluctuating Shear Stress
(Fatigue Loading)
c. Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
 'd  Sy / N
where  '  von Mises stress
see Figure 5 - 13
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
K t a  m

1
S sn'
S su
 d  Sut / N (for tensi on)
 d  Suc / N (for compressio n)
see Figure 5.15
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
whe re
S sn'  0.577 S n' and S su  0.75Su
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
whe re
S sn'  0.577 S n' and S su  0.75Su
Failure
Theory:
When Use?
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tensi on)
 max  Syc (for compressio n)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
a. Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
b. Ductile Material/
Fluctuating Shear Stress
(Fatigue Loading)
c. Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
Failure When:
Design Stress:
 max  Kt   Sut (for tensi on)
 max  Kt   Suc (for compressio n)
 d  Sut / N (for tensi on)
 d  Suc / N (for compressio n)
 d  Syt / N (for tensi on)
 d  Syc / N (for compressio n)
Note : Syt  Syc for ductile/wr ought material
 d  Sys / N where Sys  Sy/2
 'd  Sy / N
where  '  von Mises stress
see Figure 5 - 13
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
K t a  m

1
S sn'
S su
see Figure 5.15
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
whe re
S sn'  0.577 S n' and S su  0.75Su
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
whe re
S sn'  0.577 S n' and S su  0.75Su
General Comments:
1.
Failure theory to use depends on material (ductile vs. brittle) and type of loading (static or
dynamic). Note, ductile if elongation > 5%.
2.
Ductile material static loads – ok to neglect Kt (stress concentrations)
3.
Brittle material static loads – must use Kt
4.
Terminology:
•
Su (or Sut) = ultimate strength in tension
•
Suc = ultimate strength in compression
•
Sy = yield strength in tension
•
Sys = 0.5*Sy = yield strength in shear
•
Sus = 0.75*Su = ultimate strength in shear
•
Sn = endurance strength = 0.5*Su or get from Fig 5-8 or S-N curve
•
S’n = estimated actual endurance strength = Sn(Cm) (Cst) (CR) (Cs)
•
S’sn = 0.577* S’n = estimated actual endurance strength in shear
5.9 What Failure Theory to Use: