Fracture Mechanics_ANSYS

ANSYS Linear Elastic Fracture Mechanics Example

LEFM Example 

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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 K I at a crack tip.

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K I 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 

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 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 K I.

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Use special quarter-point crack elements and KCALC command to predict K I.

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

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Crack opening displacement approach to relate the relative displacement of the crack faces to K I.

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

LEFM Example 

Demonstration problem: —

Prediction and comparison of K I 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, K I for a compact specimen is given as:   a    K   B W   f    = I   P   W    +

4

a

  a     a     a     a     a    W  4  f    =  .444+ 4.44  − 44.44   + 44.44   − 4.44     W      a     W     W     W     W     4−     W    4

4

4

4 4

B = 1 in a = 1 in 1.25 W

W = 2 in P = 33.3 lb

KI = 227.7 psi-in1/2

LEFM Example 

Demonstration problem: ANSYS special crack tip elements. — —

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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 face

Crack tip Symmetry boundary

LEFM Example 

Demonstration problem: ANSYS special crack tip elements. —

KCALC command used with quarter-point elements to determine K I.

KI = 225.6 psi-in 1/2

LEFM Example 

Demonstration problem: ANSYS J-integral method. —

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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 face

Crack tip

LEFM Example 

Demonstration problem: ANSYS J-integral method. —

Printed J-integral values for 10 contours:

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Plotted J-integral values for 10 contours: J = 0.00154 lb/in

LEFM Example 

Demonstration problem: ANSYS J-integral method. —

Relating J and K I for plane strain, assuming no plasticity:

(− )

4

 J 

=

 K I  4 ν 

4

 E 

J = 0.00154 lb/in E = 30 x 10 psi 6

 ν

= 0.3

KI = 225.3 psi-in1/2

LEFM Example 

Demonstration problem: ANSYS direct method. —

In the direct method, elements near the crack tip are used to calculate K I by direct substitution of r, θ, and σyy: σ  yy

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=

 K  I  4 π r 

cos

θ     4+ sin θ  sin 4    4  4 4 

θ 

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

LEFM Example 

Demonstration problem: ANSYS direct method. An ANSYS macro was created to calculate K I at each element centroid. —  The distribution of K  I 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 K I versus radial distance from crack tip provides accurate result.

KI = 223 psi-in1/2