Chapter 5, Part B Failure Modes MET 210W E. Evans/ R.Michael.

Chapter 5, Part BFailure 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 andenergy absorption (toughness) before fracture

Brittle materials - little plastic deformation and low energyabsorption before failure

• Ductile fracture is desirable!

• Classification:

Ductile: warning before

fracture

Brittle: No

warning

DUCTILE VS BRITTLE FAILURE

(a) (b) (c)

• Resulting fracture surfaces (steel)

particles serve as voidnucleation sites.

50 µm

DUCTILE FAILURE

1 µm = 1 X 10-6 m = 0.001 mm

• Evolution to failure:

“cup and cone” fracture

Failure Prediction MethodsStatic Loads

• Brittle Materials - FT:– Maximum Normal Stress - Uniaxial stress– Modified Mohr - Biaxial stress

• Ductile Materials - FT:– Yield Strength - Uniaxial stress– Maximum Shear Strength - Biaxial stress– 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:

–In compression:

Static Load

Brittle Material

N

SK ut

dt max

N

SK uc

dt max

maxutSN

DESIGN:ANALYSIS:

maxutSN

DESIGN:ANALYSIS:

45° Shear Diagonal

Modified Mohr Method• Biaxial Static Stress on Brittle Materials

12

Stress concentrations applied to stresses before

making the circle

1

2

Sut

Sut

Suc

Suc

1, 2

Brittle materials often have a much larger compressive strength than tensile strength

Failure when outside of shaded area

Yield Strength Method• Uniaxial Static Stress on Ductile Materials

– In tension:

–In compression:

For most ductile materials, Syt = Syc

Static Load

Ductile Material

N

S ytd max

N

S ycd max

DESIGN:

ANALYSIS:

DESIGN:ANALYSIS:

maxytSN

maxycSN

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.

Somewhat conservative approach – use the Distortion Energy Method for a more precise failure estimate

N2

S

N

S yysdmax

avg, max

maxysSN

DESIGN:

ANALYSIS:

Distortion Energy

Best predictor of failure for ductile materials under static loads or under completely reversed normal, shear or combined stresses.

Shear Diagonal

1

2

2122

21 '

’ = von Mises stress

Failure: ’ > Sy

Design: ’ d = Sy/N

ANALYSIS:Sy’

• 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)()()('

Comparison of Static Failure Theories:

Maximum Shear – most conservative

Shows “no failure” zones

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

Brittle Ductile

AXIAL

Sn’

Goodman Method

m

a

-Sy

Sy

Sy Su

FATIGUE FAILURE REGION

NO FATIGUE FAILURE REGION

Goodman Line

0

Yield Line

1

u

m

n

a

SS

Good predictor of failure in ductile materials experiencing fluctuating stress

Sn’ = actual endurance strengtha = alternating stressm = mean stress

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

Sn’ = actual endurance strengtha = alternating stressm = mean stress

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 a valve stem made of AISI 4340 OQT 900°F steel.

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”

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

Design Factors, N (a.k.a. Factor of Safety)

•N = 1.25 to 2.0 Static loading, high level of confidence in all designdata

•N = 2.0 to 2.5 Dynamic loading, average confidence in all designdata

•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

FOR DUCTILE MATERIALS:

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

3. Maximum Shear Stress (Basis for MCH T 213)

Ductile Material/ Bi-axial Static Stress

4. Distortion Energy (von Mises)


5. Goodman Method Ductile Material/ Fluctuating Normal Stress (Fatigue Loading)

Ductile Material/ Fluctuating Shear Stress (Fatigue Loading)

Ductile Material/ Fluctuating Combined Stress (Fatigue Loading)

on)(for tensi max SutKt n)compressio(for max SucKt

on)(for tensi / NSutd n)compressio(for / NSucd

on)(for tensi max Syt

materialought ductile/wrfor SycSyt :Note

n)compressio(for max

Syc

on)(for tensi /d NSytn)compressio(for /d NSyc

Sy/2Sys wheremax Sys Sy/2Sys where/ NSysd

stress Misesvon ' where

' 2122

21

Sy

13-5 Figure

/'

see

NSyd

1'

u

m

n

at

SS

K

5.15 Figure

1'

see

NSS

K

u

m

n

at

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

NSS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

Failure Theories for STATIC LoadingFailure Theories for STATIC Loading

Uniaxial: Bi-axial:or

Failure Theory:




2. Yield Strength







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)






Syc




' 2122

21

Sy

13-5 Figure

/'

see

NSyd

1'

u

m

n

at

SS

K

5.15 Figure

1'

see

NSS

K

u

m

n

at

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

NSS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

Failure Theories for FATIGUE LoadingFailure Theories for FATIGUE Loading

Failure Theory:




2. Yield Strength







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)






Syc




' 2122

21

Sy

13-5 Figure

/'

see

NSyd

1'

u

m

n

at

SS

K

5.15 Figure

1'

see

NSS

K

u

m

n

at

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

NSS

K

75.0 and 577.0

here w1

''

'

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

usunsn

su

m

sn

at

SSSS

SS

K

75.0 and 577.0

re whe1)()(

''

max'

max

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:

Chapter 5, Part B Failure Modes MET 210W E. Evans/ R.Michael.

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Transcript of Chapter 5, Part B Failure Modes MET 210W E. Evans/ R.Michael.