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From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems

Quarter point element

Computational fracture mechanics

Nicola CefisDipartimento di Ingegneria Civile e Ambientale

Politecnico di Milano

May 04, 2020Milano

Nicola Cefis Quarter point element


Sommario

1 From the continuous to the discrete problem...

2 Isoparametric elements in elastic-crack problems



Theoretical results

In elastic materials the stresses near the tip of crack tend to infinity.

The Finite Element Method can simulate this singularity in aproper way?



Two examples of Finite Elements

1 2

4 3

1 2

3

CONSTANT STRAIN

TRIANGLE

LINEAR ELEMENT



Isoparametric elements

1 5 2

8 6

4 7 3

s

t

3

5

2

6

1

8

7

4

x

y

11

1

1

PARENT

ELEMENT

(s,t)

EFFECTIVE

ELEMENT

(x,y)

T

(direct transformation)

T

-1

(inverse transformation)

{xy

}= T

{st

} {st

}= T−1

{xy

}




If an isoparametric formulation is adopted, the same model is usedfor the coordinates and for the displacements.

x = NX →

{x =

∑ni=1 xNi(s, t)xi

y =∑n

i=1 y Ni(s, t)yi

xi and yi are the nodal coordinates

u = NU →

{u =

∑ni=1 xNi(s, t)ui

v =∑n

i=1 y Ni(s, t)vi

ui and vi are the nodal displacements




Now, starting from the displacements model, the strain model can beobtained

u = NU → ε(x , y) = B(x , y)U

where B(x , y) is the strain-displacement matrix (derivatives of theshape functions).

B =

N1,x 0 N2,x 0 N3,x 0 N4,x 00 N1,y 0 N2,y 0 N3,y 0 N4,y

N1,y N1,x N2,y N2,x N3,y N3,x N4,y N4,x

In the isoparametric element the shape functions are defined in the(s, t) reference system while the derivatives have to be computedrespect to (x , y).




To calculate the matrix B it is possible to apply the chain rule:

Ni,s =∂Ni

∂x∂x∂s

+∂Ni

∂y∂y∂s

= Ni,xx,s + Ni,y y,s

Ni,t =∂Ni

∂x∂x∂t

+∂Ni

∂y∂y∂t

= Ni,xx,t + Ni,y y,t

In matrix form we can obtain{Ni,sNi,t

}=

[x,s y,sx,t y,t

]{Ni,xNi,y

}= JT

{Ni,xNi,y

}



If the mapping is regular, than J can be inverted, we obtain{Ni,xNi,y

}= J−1

{Ni,sNi,t

}where

J−1 =1

detJ

[y,t −y,s−x,t x,s

]=

1x,sy,t − y,sx,t

[y,t −y,s−x,t x,s

]

and it is possible to compute the stiffness matrix

K =

∫V

BT dBdV



A 8-node quadratic isoparametric element

Let’s now consider a 8-node quadratic isoparametric elementrepresented below. The node 1 is on the tip of crack and the edge1-5-2 is aligned whit the crack axis.

1 5 2

6

3

7

4

8

s

t

t=1

t=-1

s=1

s=-1

x

(crack axis)

y



8-node: shape functions for the mid-edge nodes

1 5 2

6

3

7

4

8

s

t

t=1

t=-1

s=1

s=-1

x

(crack axis)

y

N5(s, t) =12(1− s2)(1− t)

N6(s, t) =12(1− t2)(1 + s)

N7(s, t) =12(1− s2)(1 + t)

N8(s, t) =12(1− t2)(1− s)



8-node: shape functions for corner-edge nodes

N1(s, t) =14(1− s)(1− t)− 1

2N5(s, t)−

12

N8(s, t)

N2(s, t) =14(1 + s)(1− t)− 1

2N5(s, t)−

12

N6(s, t)

N3(s, t) =14(1 + s)(1 + t)− 1

2N6(s, t)−

12

N7(s, t)

N4(s, t) =14(1− s)(1 + t)− 1

2N7(s, t)−

12

N8(s, t)



Let focus on the edge 1-5-2 along the x axis (crack axis) in whichy = 0 and t = −1. Along this edge we have

y,s =∂y∂s

= 0 and: det(J) = x,sy,t − y,sx,t = x,sy,t

Thus,

J−1(t = −1) =1

x,sy,t

[y,t 0−x,t x,s

]=

[1

x,s0

− x,tx,sy,t

− 1y,t

]



8-node: the crack axis

In order to obtain a singularity of stress or of strain at x = y = 0 thederivative of displacement (and thus of the shape functions) has to besingular for x → 0+. {

Ni,xNi,y

}= J−1

{Ni,sNi,t

}Ni,x can became singular only if one of the therm of the first row ofJ−1(t = −1) becomes singular (Ni,s is finite for x → 0 because is apolymomil expression). The result is

1x,s→∞ ⇒ x,s → 0



8-node: the crack axis

Now let’s consider the shape function of nodes 1-5-2 in the particularcase of 8-node element

N5(s, t = −1) =12(1− s2)(1− t) = 1− s2

N1(s, t = −1) =14(1− s)(1− t)− 1

2N5(s, t)−

12

N8(s, t) =

=12(1− s)− 1

2(1− s2) =

12

s(s − 1)

N2(s, t) =14(1 + s)(1− t)− 1

2N5(s, t)−

12

N6(s, t) =

=12(1 + s)− 1

2(1− s2) =

12

s(s + 1)

and let’s write the model for x

x(t = −1) = N1x1 + N5x5 + N2x2 = 0 +12

s(s − 1) · x5 +12

s(s + 1) · x2



The quarter point elementNow we impose the condition of singularity

x,s =∂x∂s→ 0 →

(s +

12

x2 + (−2s)x5

)= 0

and in the node 1 (s → 1) we obtain

−12

x2 + 2x5 = 0 ⇒ x5 =14

x2

1 5 2

8 6

4 7 3

1 2

6

4 7 3

5

8

s

t

t

s

1

4

L

3

4

L

1

2

L

1

2

L

QUARTER

POINT

STANDARD

QUADRATIC Nicola Cefis Quarter point element


Power of the singularity

The power of singularity is proportional to√

r like the theoreticalresults?

Ler’s now consider the x position in the edge 1-5-2 in the quarterpoint element (x5 = 1

4 x2)

x(t = −1) = N1x1 + N5x5 + N2x2 = 0 +12

s(s − 1) · x5 +12

s(s + 1) · x2

=14(s + 1)2x2

s can be expressed as a function of x and x2:

(s + 1) = 2√

xx2

and the power of singularity is directly proportional to the derivative ofx along s

∂x∂s

(t = −1) =12(s + 1)x2 =

12

2√

xx2

x2 =√

xx2 ≈√

r



From 8-node to 6-node element

1 2

6

4 7 3

5

8

1=4=8

5

7

3

6

2



Spider web

- concentric rings of quadrilateral elements focused toward thecrack tip;

- the elements in the first ring are collapsed into triangles;- at least 10 element on a radial line to analyze the crack-tip

stresses;- easy transition from a fine discretion to crack tip to a coarse

mesh size far from the crack.Nicola Cefis Quarter point element


Example


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