Week 11 MECH3361 Recap 2D elements Displacement Shape function Strain T3, 3-node constant strain triangle (CST) Linear: u=b 1 + b 2 x +b 3 y v=b 4 +b 5 x +b 6 y (6 d.o.f.) N i = 1 2 A ( α i +β i x +γ i y ) (i =1,2,3) A = 1 2 | 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 | Constant strain ε xx = ∂ u ∂ x =b 2 ε yy = ∂ v ∂ y =b 6 2 ε xy = ( ∂ u ∂ y + ∂ v ∂ x ) ¿ b 3 + b 5 T6 quadratic triangular element linear strain triangle (LST) Quadratic: u=b 1 + b 2 x +b 3 y + b 4 x 2 +b 5 xy+b 6 y 2 v=b 7 +b 8 x +b 9 y+ b 10 x 2 +b 11 xy+b 12 y 2 (12 d.o.f.) N 1 =ξ ( 2 ξ−1 ) N 2 =η ( 2 η−1 ) N 3 =( 1−ξ−η ) [ 2 ( 1−ξ−η )−1] N 4 =4 ξη N 5 =4 η ( 1−ξ−η ) N 6 =4 ( 1−ξ− η ) ξ Fully-linear ε xx =b 2 + 2 b 4 x +b 5 y ε yy =b 9 + b 11 x +2 b 12 y 2 ε xy = ( b 3 + b 5) + ( b 3 +2 b 10 ) x+ ( 2 b 6 +2 b 11 ) y Linear Quadrilateral Element (Q4) Bilinear u=b 1 + b 2 x + b 3 y +b 4 xy v=b 5 +b 6 x + b 7 y +b 8 xy (8 d.o.f.) N 1 = 1 4 ( 1−ξ )( 1−η ) , N 2 = 1 4 ( 1+ξ )( 1−η ) , N 3 = 1 4 ( 1+ξ )( 1 +η ) , N 4 = 1 4 ( 1−ξ )( 1+η ) Half-linear ε xx = ∂ u ∂ x =b 2 +b 4 y ε yy = ∂ v ∂ y =b 7 +b 8 x 2 ε xy = ∂ v ∂ x + ∂ u ∂ y ¿ b 6 + b 8 y +b 3 +b 4 x x y x y 1 2 3 4 5 6 7 8 Quadratic Quadrilateral Element (Q8) Quadratic u=b 1 +b 2 x +b 3 y+ b 4 x 2 + b 5 xy+b 6 y 2 + b 7 x 2 y +b 8 xy 2 v=b 9 +b 10 x+ b 11 y + b 12 x 2 +b 13 xy+b 14 y 2 +b 15 x 2 y+b 16 xy 2 (16 d.o.f.) N 1 =− 1 4 ( 1−ξ )( 1−η )( 1+ξ+ η ) N 2 =− 1 4 ( 1+ξ )( 1−η )( 1−ξ+ η ) N 3 =− 1 4 ( 1+ξ )( 1 +η )( 1−ξ−η ) N 4 =− 1 4 ( 1−ξ )( 1+η )( 1 +ξ−η ) N 5 = 1 2 ( 1−ξ 2 )( 1−η ) N 6 = 1 2 ( 1+ξ )( 1−η 2 ) N 7 = 1 2 ( 1−ξ 2 )( 1+η ) N 8 = 1 2 ( 1−ξ )( 1−η 2 ) Half-quadratic ε xx =b 2 x +2 b 4 x+b 5 y+ 2 b 7 xy+b 8 y 2 ε yy =b 11 y+b 13 y +2 b 14 y +b 15 x 2 +2 b 16 xy ε xy =b 3 + b 5 x +2 b 6 y+ b 7 x 2 + 2 b 8 xy+ b 10 x+2 b 12 x+b 13 y +2 b 15 xy+b 16 y 2 1
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
8.7 Axisymmetric ProblemsEngineering problemsConditions: Both (1) Rotational structures and (2) Axisymmetric loading/boundary is applied.
Cylindrical coordinates ( x , y , z )⇒(r , θ , z )
Displacement field: u(r , z ) , w (r , z )Due to axisymmetry, there is no v – circumferential component of displacement vanishes.
Strain – displacement relation
ε rr=∂u(r , z )∂r
, εθϑ=u(r , z )
r, ε zz=
∂w (r , z )∂ z
, εrz=∂w(r , z )∂r
+∂u(r , z )∂ z
Note that ε rθ=εθz=0
Stress-strain relation {σrr
σθθ
σ zzσrz}= E(1+ν )(1−2 ν ) [1−ν ν ν 0
ν 1−ν ν 0ν ν 1−ν 0
0 0 0 1−2ν2
]{εrr
εθθ
εzzεrz}
Axisymmetric Elements:The element is in 2D but represents a ring swept around the axis.
Elemental stiffness matrix
2
Week 11 MECH3361
Applications: Rotating Flywheel:
Axisymmetric Condition:(1) Structure is rotational (2) The loading of body force due to angular velocity is axisymmetric.So such a problem can be modelled as a 2D axisymmetric problem.
Examples for 2D finite element method:Example 8.9 Calculate stiffness matrix of the CST element (plane stress). Soln: To install the elemental stiffness matrix for CST element, we can use Matlab as follows:Step 1: Material matrix:
Example 8.10 Sketch the distribution of v-displacement and y-strain in these two connected CST elements for the global displacement vector {u} as shown.
x
y
v
D (1,0)
C (1,1)
B (0,1)
A (0,0)
x
y
D (1,0)
C (1,1)
B (0,1)
A (0,0)
1 2
132
21
401
D
D
C
C
B
B
A
A
vuvuvuvu
u1 2
Soln
Step 1: Select v-displacement in each node: v={v A v B vC v D }T= {0 −1 −2 1 }T .Step 2: Draw the displacement in each corresponding node and connect them as shown.Step 3: Calculate y-strains in each of the CST elements according to the follow equation.
}Note that node numbering of 1, 2, 3 should be in count-clockwise sequence. For element #1, we can choose: Node (1,2,3) = Node (A,D,C) (we cannot use Node (A,C,D)). For element #2, we can choose Node (1,2,3) = Node (C,B,A) (we cannot use Node (A,B,C)). For element #1: nodal coordinates: For element #2: nodal coordinates:
4
Week 11 MECH3361
x(1)={x1
y1
x2
y2
x3
y3
}={x A
y A
x D
yD
xC
yC
}={001011 } x(2)={x1
y1
x2
y2
x3
y3
}={xC
yC
x B
yB
x A
y A
}={110100 }Strain in element #1 (Node 1 = Node A, Node 2 = Node D, Node 3 = Node C)
Strain in element #2 (Node 1 = Node C, Node 2 = Node B, Node 3 = Node A)
ε yy(2)=1
2 A [ (x3−x2)v1+ (x1−x3 )v2+( x2−x1) v3]¿1
2 A [ (x A−xB )vC+(xC−x A )v B+ (xB−xC )v A ]¿1
2×0.5 [ (0−0 )× (−2 )+ (1−0 )×(−1 )+(0−1 )×(0 ) ]=−1Step 4: plot the y-strain yy for
each constant strain element (as shown on the right). Note that the strain distributions are constant!
Example 8.11 Determine coordinate G1, G2, G3, G4 of Gaussian points in Cartesian coordinate system of iso-parametric element Q4 as shown (the nodal coordinates have been give in the figure). Compute Jacobian matrix [J] and det[J] for the Q4 element.
A(1,0) B(2,0)
C(2.25,1.5)
D(1.25,1)
x
y
A’(-1, -1)
B’(1, -1)
C’(1, 1)D’(-1, 1)
* *
**
G1G2
G3 G4
0.5774
-0.5774
Soln: Q4 and Q8 are “iso-parametric” element in which displacement and coordinate share the same “shape functions” for mapping and interpolation, as follows:
Coordinate mapping: x=∑i=1
4
N i xi , y=∑i=1
4
N i y i
Displacement interpolation: u=∑
i=1
4
N i ui , v=∑i=1
4
N i vi
Step 1: Node numbering: Node (1,2,3,4) = Node (A,B,C,D)Step 2: Coordinate in terms of Shape functionsShape functions
5
x
y
yy
D (1,0)
C (1,1)
B (0,1)
A (0,0)
1 2
-1
-3
-1-3
Week 11 MECH3361
N1=14(1−ξ )(1−η )
, N2=
14(1+ξ )(1−η)
, N3=
14(1+ξ )(1+η)
, N4=
14(1−ξ )(1+η )
Cartersian coordinate (x,y) in terms of natural coordinate (, )
where Ni are shape functions for 3D element and d= {u1 v1 w1 u2 v2 w2 ⋯}T is the nodal displacement vector.
Using strain – displacement relation, we can have the following similar formula:
7
Week 11 MECH3361
ε(6×1 )=B(6×3 N )d(3 N×1)
Similarly, the Stiffness Matrix:
k(3 N×3 N )=∫V e
BT(3 N×6)E(6×6 )B(6×3 N ) dV e
Numerical quadratures are often needed to calculate the above integration.
Typical 3D elements
Tetrahedron (Tet):
Hexahedron (brick):
Penta:
Remarks: Each node has 3 degrees of freedom (u, v, w) in these 3D elements 4-node tet element is a “constant strain element” and do not use 4-node tet elements for
capturing high stress/strain gradients.
Linear Hexahedron Element (8-node brick element)Mapping:
Coordinate transformation (from natural coordinate to Cartesian coordinate)
N7 (ξ ,η , ζ )=18(1+ξ )(1+η)(1+ζ ) , N 8(ξ , η ,ζ )=1
8(1−ξ )(1+η)(1+ζ ) ,
Isoparamatric Elements: The same shape functions are used as for the displacement field.
Coordinate system:x=∑
i=1
N
N i x i , y=∑i=1
N
N i y i , z=∑i=1
N
N i zi
Displacement field:u=∑
i=1
N
N i ui , v=∑i=1
N
N i v i , w=∑i=1
N
N i wi
The name of “isoparamatric element” is applied to 2D elements as well.Jacobian Matrix: (to apply strain – displacement relation, derivatives w.r.t. different coordinate systems are needed). Take u as example, one should have:
{∂ u∂ξ∂ u∂ η∂ u∂ ζ}=[∂ x∂ξ
∂ y∂ξ
∂ z∂ξ
∂ x∂η
∂ y∂η
∂ z∂ η
∂ x∂ ζ
∂ y∂ ζ
∂ z∂ ζ]{∂ u∂ x∂ u∂ y∂ u∂ z}=J {
∂ u∂ x∂ u∂ y∂u∂ z} or
{∂ u∂ x∂ u∂ y∂u∂ z}=[∂ x∂ ξ
∂ y∂ ξ
∂ z∂ ξ
∂ x∂ η
∂ y∂ η
∂ z∂ η
∂ x∂ ζ
∂ y∂ ζ
∂ z∂ ζ]−1
{∂u∂ ξ∂u∂η∂u∂ ζ}=J−1{
∂ u∂ ξ∂ u∂ η∂u∂ζ}
where J is named Jacobian matrix. Similarly for v and w, one can extend as follows:
{∂ v∂ x∂ v∂ y∂ v∂ z}=[∂ x∂ ξ
∂ y∂ ξ
∂ z∂ ξ
∂ x∂ η
∂ y∂ η
∂ z∂ η
∂ x∂ ζ
∂ y∂ ζ
∂ z∂ ζ]−1
{∂ v∂ ξ∂ v∂η∂v∂ζ}=J−1{
∂ v∂ξ∂ v∂ η∂ v∂ ζ},
{∂w∂ x∂w∂ y∂w∂ z}=[∂ x∂ ξ
∂ y∂ ξ
∂ z∂ ξ
∂ x∂ η
∂ y∂ η
∂ z∂η
∂ x∂ζ
∂ y∂ζ
∂ z∂ζ]−1
{∂w∂ ξ∂w∂ η∂w∂ζ}=J−1{
∂w∂ ξ∂w∂η∂w∂ζ}
in which, for example, the calculation of ∂u /∂ξ can be done as:∂u∂ξ=∑
i=1
8 ∂N i
∂ ξui=
∂N1
∂ξu1+
∂N2
∂ ξu2+∂N 3
∂ ξu3+
∂N4
∂ ξu4+
∂ N5
∂ξu5+
∂N6
∂ ξu6+
∂N7
∂ ξu7+
∂ N8
∂ξu8
To calculate Jacobian matrix, the isoparamatric relations of
x=∑i=1
8
N i (ξ ,η , ζ )xi , y=∑i=1
8
N i( ξ , η , ζ ) y i , z=∑i=1
8
N i(ξ ,η ,ζ ) zi can be used as:
9
Week 11 MECH3361
J=[∂ x∂ξ
∂ y∂ξ
∂ z∂ξ
∂ x∂η
∂ y∂η
∂ z∂η
∂ x∂ζ
∂ y∂ζ
∂ z∂ ζ]=[∑i=1
8 ∂N i
∂ ξx i ∑
i=1
8 ∂N i
∂ ξy i ∑
i=1
8 ∂N i
∂ξzi
∑i=1
8 ∂N i
∂ ηx i ∑
i=1
8 ∂N i
∂ηy i ∑
i=1
8 ∂N i
∂ηzi
∑i=1
8 ∂N i
∂ ζ x i ∑i=1
8 ∂N i
∂ζ y i ∑i=1
8 ∂N i
∂ζ zi]
J=[∂ N1
∂ξ∂N2
∂ξ∂N3
∂ ξ∂N 4
∂ ξ∂N5
∂ ξ∂N6
∂ ξ∂N 7
∂ ξ∂ N8
∂ξ∂ N1
∂η∂N2
∂η∂N3
∂η∂N 4
∂η∂N5
∂η∂N6
∂ η∂N 7
∂ η∂ N8
∂η∂ N1
∂ ζ∂N2
∂ζ∂N3
∂ζ∂N 4
∂ ζ∂N5
∂ζ∂N6
∂ ζ∂N 7
∂ ζ∂ N8
∂ζ][
x1 y1 z1
x2 y2 z2
x3 y3 z3
x4 y 4 z 4
x5 y5 z5
x6 y6 z6
x7 y7 z7
x8 y8 z8
]Strain – Displacement relation:
ε(6×1 )={ε xx
ε yy
εzz
ε xy
ε yz
ε zx
}={∂ u/∂ x∂v /∂ y∂w /∂ z
(∂ v /∂ x+∂ u/∂ y )/2(∂w /∂ y+∂ v /∂ z )/2(∂ u/∂ z+∂w /∂ x )/2
}=B(6×24 )d (24×1 )
in which B(6×24 )= [B1 B2 B3 B4 B5 B6 B7 B8 ]
where:
Bi=[∂ N i/∂ x 0 0
0 ∂N i /∂ y 00 0 ∂N i/∂ z0 ∂N i/∂ z ∂N i/∂ y
∂N i /∂ z 0 ∂N i /∂ x∂N i /∂ y ∂N i/∂ x 0
](6×3)
and Jacobian transformation for chain rule of differentiation is needed here.
{∂ N i
∂ x∂ N i
∂ y∂N i
∂ z}=[∂ x∂ξ
∂ y∂ξ
∂ z∂ξ
∂ x∂η
∂ y∂η
∂ z∂ η
∂ x∂ ζ
∂ y∂ζ
∂ z∂ ζ]−1
{∂N i
∂ ξ∂N i
∂ η∂N i
∂ζ}=J−1{
∂ N i
∂ξ∂ N i
∂ η∂ N i
∂ ζ}
So Jacobian matrix needs to be positive definite.
Nodal coordinates are d(24×1 )= {u1 v1 w1 u2 v2 w2 ⋯ u8 v8 w8 }T
Strain energy
U=12∫V
σΤ ε dV=12∫V(Eε )Τ ε dV=1
2∫VεΤ Eε dV=1
2∫V(Β )Τ E(Β)dV=dT [ 12∫V ΒΤ EΒ dV ]d
10
Week 11 MECH3361
Elemental Stiffness matrix: k(24×24 )=
12∫V
ΒΤ(24×6)E(6×6 )Β(6×24 )dV
In natural coordinate system: dV=(det J )dξdηdζ
k(24×24 )=12∫−1
1
∫−1
1
∫−1
1
ΒΤ(24×6 )E(6×6)Β(6×24 ) (det J )dξdηdζ
which can be calculated using numerical integration.
Example 8.12 The 8-node iso-paramatric brick element as shown in Cartesian coordinate system (x,y,z) below is mapped to the natural coordinate (,,). Use shape function to determine (1) Cartesian coordinate of centroid (2) displacement u, v, w at the centroid if the
Similarly, Displacement at the centroid: u0=0 , v0=0 . 625 , w0=0 . 125 .
Matlab code: the following simple Matlab code can be used for the calculations:xi=0; nu=0; zt=0;x1=0; x2=2; x3=2; x4=0; x5=0; x6=3; x7=3; x8=0;y1=0; y2=0; y3=1; y4=3; y5=0; y6=0; y7=3; y8=4;z1=0; z2=0; z3=0; z4=0; z5=4; z6=3; z7=3; z8=4;u1=0; u2=0; u3=1; u4=-2; u5=0; u6=0; u7=2; u8=-1;v1=0; v2=0; v3=2; v4=0; v5=0; v6=0; v7=2; v8=1;w1=0; w2=0; w3=-1;w4=-2; w5=0; w6=0; w7=2; w8=2;N1=(1-xi)*(1-nu)*(1-zt)/8; N2=(1+xi)*(1-nu)*(1-zt)/8;N3=(1+xi)*(1+nu)*(1-zt)/8; N4=(1-xi)*(1+nu)*(1-zt)/8;N5=(1-xi)*(1-nu)*(1+zt)/8; N6=(1+xi)*(1-nu)*(1+zt)/8;N7=(1+xi)*(1+nu)*(1+zt)/8; N8=(1-xi)*(1+nu)*(1+zt)/8;x=N1*x1+N2*x2+N3*x3+N4*x4+N5*x5+N6*x6+N7*x7+N8*x8y=N1*y1+N2*y2+N3*y3+N4*y4+N5*y5+N6*y6+N7*y7+N8*y8z=N1*z1+N2*z2+N3*z3+N4*z4+N5*z5+N6*z6+N7*z7+N8*z8u=N1*u1+N2*u2+N3*u3+N4*u4+N5*u5+N6*u6+N7*u7+N8*u8v=N1*v1+N2*v2+N3*v3+N4*v4+N5*v5+N6*v6+N7*v7+N8*v8w=N1*w1+N2*w2+N3*w3+N4*w4+N5*w5+N6*w6+N7*w7+N8*w8
Example 8.13 For the axisymmetric problem, if a 3-node triangular element is used, please derive the formula for B matrix (or namely strain-displacement matrix).Soln:Shape function: After replacing x by using r, y by z (in axisymmetric problem, two coordinates governing the mesh are r and z).
N1=12 A [( x2 y3−x3 y2)+( y2− y3) x+( x3−x2) y ]→1
2 A [(r2 z3−r3 z2 )+( z2−z3 )r+(r3−r2 )z ]
N2=12 A [( x3 y1−x1 y3)+( y3− y1) x+( x1−x3) y ]→1
2 A [(r3 z1−r1 z3 )+( z3−z1 )r+(r1−r3) z ]
N3=12 A [( x1 y2−x2 y1)+( y1− y2 ) x+( x2−x1) y ]→
Nature of Finite Element Solutions FE Model – A mathematical model of the
real structure, based on many approximations.
Real Structure -- Infinite number of nodes (physical points or particles), thus infinite number of DOF’s.
FE Model – finite number of nodes, thus limited number of DOF’s.
Displacement field is controlled (or constrained) by the values at a limited number of nodes
Stiffening Effect: FE Model is stiffer than the real structure. In general, displacement results are smaller in
magnitudes than the exact values. Hence, exact solution provides an upper
bound for the FEM solution. In the convergence test, it usually provides the following diagram: The FEM solution approaches the exact solution from below. This is true for displacement based FEA only!
Numerical Error Numerical Error ( in solving FE equations) Modeling Error (approximation to beam, plate … theories) Discretization Error (finite, piecewise … )
Ill-conditionLet’s look at an example:FE equilibrium equation (after apply B.C.)
[k1 −k1
−k1 k1+k2 ]{u1
u2}={P0 }
14
Week 11 MECH3361
det[k1 −k1
−k1 k1+k2 ]=k1k2
The system will be singular if k 2 is much smaller than k1 (or vice verse)
Convergence of FE SolutionsAs the mesh in an FE model is “refined” repeatedly, the FE solution will converge to the exact solution of the mathematical model of the problem (the model based on bar, beam, plane stress/strain, plate, shell, or 3-D elasticity theories or assumptions).
Types of Refinement: h-refinement: reduce the size of the element (“h” refers to the typical size of the
elements), e.g. let smart size down from 5 to 1; p-refinement: Increase the order of the polynomials on an element, i.e. use a higher
order of elements (linear to quadratic, etc.; “h” refers to the higher order in a polynomial);
r-refinement: re-arrange the nodes in the mesh; hp-refinement: Combination of the h- and p-refinements (better results!).
Adaptivity (h-, p-, and hp-Methods)pi
Stress concentration, needs dense mesh to capture the stress gradient d) t = 0.0150 (fracture of bridge) Fracture pathFracture path
Mesh quality Know the behaviours of each type of elements:
T3 and Q4: linear displacement, constant strain and stress;T6 and Q8: quadratic displacement, linear strain and stress.
Choose the right type of elements for a given problem:When in doubt, use higher order elements or a finer mesh.
Avoid elements with large aspect ratios and corner angles:Aspect ratio = Lmax / Lminwhere Lmax and Lmin are the largest and smallest characteristic lengths of an element, respectively.
15
Week 11 MECH3361
Avoid singular elements
Example 8.14: Write elemental and global equilibrium equations of the following two CST finite element model. Go on to determine the nodal displacements and strain in both elements (assume: E=1, v=0.3, thickness = 1 for a plane stress problem).
Soln: Step 1: Connectivity of FE modelNumber of elements = 2; Number of nodes = 4; d.o.f.=8
Step 2: Elemental Stiffness matrix calculation for plane stress problem[K e ]=∫