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
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Sommario
1 From the continuous to the discrete problem...
2 Isoparametric elements in elastic-crack problems
Nicola Cefis Quarter point element
From the continuous to the discrete problem...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?
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Two examples of Finite Elements
1 2
4 3
1 2
3
CONSTANT STRAIN
TRIANGLE
LINEAR ELEMENT
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
}
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Isoparametric elements
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Isoparametric elements
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).
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Isoparametric elements
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
}
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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)
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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)
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
]
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
From 8-node to 6-node element
1 2
6
4 7 3
5
8
1=4=8
5
7
3
6
2
Nicola Cefis Quarter point element
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
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
From the continuous to the discrete problem...Isoparametric elements in elastic-crack problems
Example
Nicola Cefis Quarter point element
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