Phillip Allen & Doug Wells NASA MSFC Damage Tolerance Team – EM20

1

Damage Tolerance Assessment BranchMSFC Engineering Directorate

Workshop on Life Prediction Methodology and Validation for Surface Cracks

Investigations into DeformationLimits for SSY and LSY forSurface Cracks in Tension

5/23/2007

Phillip Allen & Doug WellsNASA MSFC

Damage Tolerance Team – EM20

2


• Try to understand the current solutions to the 2-D problem• Compare with current length scale requirements in ASTM E 399 and E 1820

How do we determine the proper deformation limits for surface cracks?....

Our Proposed Method to Understand or Bound the Problem:Step 1: Revisit the 2-D Length Scale Problem

a

ra rb

2-D Plane Strain Boundary Layer Solution (Gives “exact” solution for crack tip stress field in infinite body)

Effect of constraint on crack tip plastic zone size and orientation

0

r rJ

0

yy

K, T

R>>rp

small strain analysis

3


• For valid KIC, and • Implicit requirement on B

Length Scale Requirements in Current ASTM E08 Standards

E399 – C(T) 2

0 5.2

ys

ICKb

WaW 55.045.0

a = crack lengthB = thicknessW = widthb0 = W-a

• For valid KIC, and

• For valid JC, (crack instability without stable tearing)

• For valid JIC, (crack instability proceeded by stable tearing)

• For J determination (ensures positive constraint)

E1820 – C(T), SEN(B)2

0 5.2

ys

ICKb

WaW 55.045.0

ys

QJBb

100,0

ys

QJBb

25,0

WaW 70.045.0

42 BW

4


• Can the 3-D surface crack front at some distance from the free surface in a finite body be approximated by a plane strain boundary layer solution?

• What is the influence of the stress tangential to the crack front, t? (Analogous to thickness requirements in E399 and E1820)

• What influence does the free surface behind the crack tip have for the shallow crack problem?

Step 2: Evaluate Finite Boundary 3-D Surface Crack Problem

ra

rb

ra rb

Stresses gradually decrease below SSY values as plasticity becomes uncontained

0

r rJ

0

yy

K, T

R>>rp

5


1/CJ(ys)

1/CK(ys)

A

D

C

B

E

A SSY, K or J dominance, 1 parameter

LSY, J dominance, 1 parameter

SSY, K or J with constraint, 2 parameters

LSY, J with constraint, 2 parameters

B

C

D

E Constraint Influenced Collapse, Alternative methods

Example: E399 KIc test

Example: E1820 JIc test

Examples: E740 KIe tests

Incr

easi

ng D

efor

mat

ion

Constraint measureConstraint condition equivalent to T = Q = 0

Loading trajectories

K- or J-

J-

J

K or J

K or J dominance, only1 parameter required

K or J dominance not achieved due to lack of constraint, 2 parameters required to describe fields

Collapse

At initiation of ductile tearing in a test sample or structure, the crack tip conditions will fall into one of the 5 regions A-E in the constraint/deformation diagram below. Evaluate the constraint () and the deformation limits (C) at the onset of ductile tearing to determine the applicable region for assessment of crack tip conditions.

1/C = J/(lys) Small Scale Yielding

Large Scale Yielding

6


Deformation Limit Study for E740•Determine reasonable deformation limits to compare to ra and rb to characterize test result•Proposed deformation limits based on

ys

KkbaJCrr ,

ysJba

JCrr

,

SSY Valid,

LSY Valid ,

ysba

JCrr ,

ysJba

JCrr

,

ysk

EfC

ysJ

EfC

Point (xe,B)m

Point (x, y)

ra

aB

Point (xint,0)2c

rb

Check at initiation of tearing

If prior to initiation

of tearing then classify as plastic collapse

7


Modified Boundary Layer FEMs•Plane strain boundary conditions•20 node bricks•WARP3D analysis•Linear Plus Power Law Mat’l Model•Apply displacement field as function of K, T•Vary T/ys,

K, T

R>>rp

0 1 2 3 4 5 6 7 8 9 101

1.5

2

2.5

3

3.5

4

r0/J

op

en/

0

0.10.1 ysT

T/ys = 0.0

T/ys = -0.9

T/ys = 0.9

-1 -0.5 0 0.5 11.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

T/0

op

en/

0

r0/J = 2

r0/J = 4

r0/J = 6

r0/J = 8

= r*

In this work 0 = ys

E/ys = 400, n = 10

E/ys = 400, n = 10

8


C(T) a/w = 0.5; E/ys = 400; n = 10•Plane strain boundary conditions•20 node bricks•WARP3D analysis•Linear Plus Power Law Mat’l Model

n

ys

ys

ys

for

for

ys

ys

9


C(T) a/w = 0.5, E/ys = 400, n = 10

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

T/0

op

en/

0

Reference Solution Comparison by T-stress at x/B = 0.000

101

102

103

104

0.85

0.9

0.95

1

1.05

1.1

(a0)/J

op

en/

ref


r* = 2 ar* = 2 br* = 4 ar* = 4 br* = 6 ar* = 6 br* = 8 ar* = 8 b

Reference Solution Comparison by T-Stress

CJ = 31

r* = 2

r* = 4r* = 6r* = 8

5% deviation curve (typ)

Assume 5% deviation from MBL open as limit of LSY validity

“a” in deformation scale can be ra or rb. The minimum dimension is the limiting case. ra = rb for this geometry.

@ r* = 2

10


101

102

103

104

0.85

0.9

0.95

1

1.05

1.1

(a0)/J

op

en/

ref

Reference Solution Comparison by Q at x/B = 0.000


C(T) a/w = 0.5, E/ys = 400, n = 10

Reference Solution Comparison by Q

CJ = 49@ r* = 4

-1 -0.8 -0.6 -0.4 -0.2 0 0.21

1.5

2

2.5

3

3.5

4

Q at r*=2

op

en/

0

Reference Solution Comparison by Q at x/B = 0.000

11


102

103

104

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

1.18

1.2

(a0)/J

J tota

l/Jel

astic

Ratio of J/Je vs Deformation at x/B = 0.000

CrackLigament

101

102

103

104

1

1.5

2

2.5

3

3.5

4

4.5

(a0)/J

J tota

l/Jel

astic

Ratio of J/Je vs Deformation at x/B = 0.000

CrackLigament

C(T) a/w = 0.5, E/ys = 400, n = 10

Jtotal vs. Jelastic Comparison

Ck = 110

1.12.1 KK

JJ J

elastic

totalAssume 10% deviation from elastic K prediction as limit of SSY validity

12


C(T) a/w = 0.5, E/ys = 400, n = 10

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

T/0

op

en/

0


101

102

103

104

0.85

0.9

0.95

1

1.05

1.1

(a0)/J

op

en/

ref



Reference Solution Comparison by T-Stress – Another look at Deform. Limits

SSY, K, JelLSY, JPlastic Collapse

CJ = 31CK = 110

CK-E399 = 1100

E399, KIC,

Note: this value is a function of E/ys

Traditional definition of SSY, at T = 0, r* = 2

LSYSSY Plastic Collapse

13


Point (xe,B)m

Point (x, y)

ra

aB

Point (xint,0)2c

rb

•20 node bricks•WARP3D analysis•Linear Plus Power Law Mat’l Model

a/B = 0.50, a/c = 1.0

SC(T) FEMs

14


SC(T) Test conducted at NASA MSFC

2219-T87, E/ys = 190, n = 10

Sample description:• W = 3.00 in.• B = 0.375 in.• 2c = 0.494 in.• a = 0.229 in.• a/c = 0.92• a/B = 0.61

Test conditions, results:• 70F• Monotonic load to crack

initiation• Initiation force = 54.95 kip

Tearing present 180 degGeneral tear length = 0.006 in.Maximum tear length = 0.013 in.

15


= 18 degrees or 2 / = 0.2

SC(T) Test conducted at NASA MSFC

Location of Tearing Initiation

16


Reference Solution Comparison by T-Stress

2/ = 0.19

101

102

103

104

0.85

0.9

0.95

1

1.05

1.1

(a0)/J

op

en/

ref

Reference Solution Comparison by T-stress at 2/ = 0.188


-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

T/0

op

en/

0

Reference Solution Comparison by T-stress at 2/ = 0.188

SC(T) a/B = 0.61, a/c = 0.92, 2219-T87, E/ys = 190, n = 10

CJ ≈ 50

Initiation of ductile tearing in SC(T) test

17


Reference Solution Comparison by Q

2/ = 0.19

SC(T) a/B = 0.61, a/c = 0.92, 2219-T87, E/ys = 190, n = 10

101

102

103

104

0.85

0.9

0.95

1

1.05

1.1

(a0)/J

op

en/

ref

Reference Solution Comparison by Q at 2/ = 0.188


-1 -0.8 -0.6 -0.4 -0.2 0 0.21

1.5

2

2.5

3

3.5

4

Q at r*=2

op

en/

0

Reference Solution Comparison by Q at 2/ = 0.188

CJ ≈ 50


18


Jtotal vs. Jelastic Comparison

2/ = 0.19

SC(T) a/B = 0.61, a/c = 0.92, 2219-T87, E/ys = 190, n = 10

101

102

103

104

105

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

(a0)/J

J tota

l/Jel

astic

Ratio of J/Je vs Deformation at 2/ = 0.188

CrackLigament

Ck = 110


19


SSY Deformation Limit Determination

0

100

200

300

400

500

600

700

800

0 200 400 600 800 1000

SSY Deformation Limit

C(T), a/W = 0.5SE(B), a/W = 0.5SE(B), a/W = 0.7SC(T), a/c = 1, a/B = 0.5SC(T), a/c = 1, a/B = 0.9SC(T), a/c = 92, a/B = 0.61SC(T), a/c = .24, a/B = 0.6Proposed LimitE399 Limit

Ck fo

r Jto

tal/J

el =

1.2

E/ys

Ck = (1/2)(E/

ys)

20


LSY Deformation Limit Determination

0

20

40

60

80

100

120

0 200 400 600 800 1000

C(T), a/W = 0.5SE(B), a/W = 0.5SE(B), a/W = 0.7SC(T), a/c = 1, a/B = 0.5SC(T), a/c = 1, a/B = 0.9SC(T), a/c = 92, a/B = 0.61SC(T), a/c = .24, a/B = 0.6Proposed Limit

CJ

E/ys

CJ = (1/14)(E/

ys)+30

E 1820 JIC

E 1820 JC

21


Deformation Limit Study for E740•Determine reasonable deformation limits to compare to ra and rb to characterize test result•Proposed deformation limits based on

ys

KkbaJCrr ,

ysJba

JCrr

,

SSY Valid,

LSY Valid ,

ysba

JCrr ,

ysk

EC2

1

30141

ys

JEC

E/ys CK CJ

80 40 36100 50 37200 100 44300 150 51400 200 59600 300 73800 400 871000 500 101

ysJba

JCrr

,If prior to initiation

of tearing then classify as plastic collapse

22


SC(T) Test Evaluation per E740

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.50

0.2

0.4

0.6

0.8

1

1.2

Phi=18 deg; Xphi=0.23491; Yphi=0.070765; Xint=0.03299; Lfront=0.21396; Lback=0.91987; Ltotal=1.1338

Plots on pp 16-18 also indicate that SSY should be valid for initiation of ductile tearing. Likely need to increase value for CK, to ensure that J/JK < 1.2, especially for materials with low E/ys.

2219-T87 Matl. Props. E740 EvaluationYield Stress, Ksi = 55 E/Sys = 191Elastic Modulus, Ksi = 1.05E+04Poissons Rato = 0.3

Spe

cim

en

Nam

e / I

D

B (i

n.)

W (i

n.)

A (i

n²)

a c a/c

a/2c

a/B

P (k

ip)

σ1 (K

si)

Init

Loc'

n (2

*phi

/pi)

SC(T) - 1 0.375 3.00 1.13 0.229 0.247 0.927 0.464 0.611 54.95 48.84 0.200

CK Cj CK/Cj

95 44 2.19

Init

Loc'

n -

Par

amet

ric

Ang

le

(deg

)

Kin

it (K

si*in

^½)

Km

ax

(Ksi

*in^½

)

Jint

from

K

(in-lb

/in^2

)

J (in

-lb

/in^2

)

r ,a

(in)

r ,b

(in)

CK*J

k/ys

(in)

Cj*J

/ys

(in)

SS

Y

Sat

isfie

d

LSY

S

atis

fied

far/yield J

/JK

18.00 33.6 36 97.55 130.60 0.214 0.920 0.169 0.104 YES YES 0.89 1.34

23


Deformation Limit Comparison

0

100

200

300

400

500

600

0 200 400 600 800 1000

Proposed CK Limit

Proposed CJ Limit

Ck o

r CJ

E/ys

Ck = (1/2)(E/

ys)

CJ = (1/14)(E/

ys)+30

SSY

LSY

Plastic Collapse

Incr

easi

ng

Load

May need to modify CK limit for materials with low E/ys.

24


Deformation Limit Study for E740 - Questions• What are reasonable deformation limits to compare to specimen

dimensions to characterize test results?• Can we use deviation from Jel solution to determine limits for SSY (K or

Jel valid solution)?• Is a 5% deviation from J-T MBL solution a valid cut off point for LSY

validity? Is this just “in the noise” in test data? • Should our deformation limits be a function of E/ys, n, or other? Which

material variables have the strongest influence on deformation limits?• Should we use different deformation limits to compare to crack size (ra)

and ligament length (rb)?• How do r* distances compare to process zone sizes for ductile tearing?

Is r* = 2 the right place to focus or other?

Phillip Allen & Doug Wells NASA MSFC Damage Tolerance Team – EM20

Documents

Transcript of Phillip Allen & Doug Wells NASA MSFC Damage Tolerance Team – EM20