ANSYS Linear Elastic

Fracture Mechanics Example

LEFM Example

ANSYS has the ability to perform linear elastic fracture mechanics (LEFM) using several approaches.

— LEFM uses derived elasticity solutions to determine the stress intensity factor KI at a crack tip.

— KI can be compared to the material fracture toughness to determine if the crack will propagate.

ANSYS also has the ability with the J-integral feature to predict crack behavior in the presence of plasticity.

— This feature is not presented here.

LEFM Example

The approaches that can be performed in ANSYS include:— Direct method – Substitute near-crack element stress and

locations directly into the near-crack elasticity solution to estimate KI.

— Use special quarter-point crack elements and KCALC command to predict KI.

— Use J-integral method to predict J from path integration around crack tip, then use relationship between J and KI to determine KI.

— Crack opening displacement approach to relate the relative displacement of the crack faces to KI.

Several of these methods are demonstrated and compared to theoretical results.

LEFM Example

Demonstration problem: — Prediction and comparison of KI of compact specimen using

the following methods:• Hand calculation.• ANSYS special crack tip elements.• ANSYS J-integral method.• ANSYS direct method.

LEFM Example

Demonstration problem: Hand calculation. — From fracture mechanics text, KI for a compact specimen is

given as:

432

2

3 60.572.1432.1364.4886.0

1

2

W

a

W

a

W

a

W

a

Wa

W

a

W

af

P

WBK

W

af I

1.25 W

B = 1 in

a = 1 in

W = 2 in

P = 33.3 lb

KI = 227.7 psi-in1/2

LEFM Example

Demonstration problem: ANSYS special crack tip elements.

— 2D plane strain mesh.— KSCON command used to automatically create local crack tip

mesh with quarter-point nodes.— Half specimen modeled using symmetry boundary conditions.

Crack tipCrack face

Symmetry boundary

LEFM Example

Demonstration problem: ANSYS special crack tip elements.

— KCALC command used with quarter-point elements to determine KI.

KI = 225.6 psi-in1/2

LEFM Example

Demonstration problem: ANSYS J-integral method. — CINT commands used to define crack tip node and request

number of contours to use (10).— Same model as before, but special crack tip elements are not

required.— Paths are created automatically around the crack tip, using

the next available row of elements.

Path 7 of 10

Crack tipCrack face

LEFM Example

Demonstration problem: ANSYS J-integral method. — Printed J-integral values for 10 contours:

— Plotted J-integral values for 10 contours:

J = 0.00154 lb/in

LEFM Example

Demonstration problem: ANSYS J-integral method. — Relating J and KI for plane strain, assuming no plasticity:

KI = 225.3 psi-in1/2J = 0.00154 lb/in

E = 30 x 106 psi

= 0.3

E

KJ I

22 1

LEFM Example

Demonstration problem: ANSYS direct method. — In the direct method, elements near the crack tip are used to

calculate KI by direct substitution of r, , and yy:

— Using the same mesh used in the J-integral calculation, the values were determined at the centroid of the elements shown below.

2

3sin2

sin12

cos2

r

K Iyy

LEFM Example

Demonstration problem: ANSYS direct method. — An ANSYS macro was created to calculate KI at each element

centroid.— The distribution of KI shows a wide range of values right near

the crack tip.

LEFM Example

Demonstration problem: ANSYS direct method. — To provide a more accurate result, consider only the first row

of elements directly in front of the crack tip.• Extrapolating KI versus radial distance from crack tip provides

accurate result.

KI = 223 psi-in1/2