Judy P. Yang (楊子儀 · 2020. 4. 13. · 10.1 Isoparametric Formulation • Isoparametric...
Transcript of Judy P. Yang (楊子儀 · 2020. 4. 13. · 10.1 Isoparametric Formulation • Isoparametric...
Introduction to Finite Element Method
Judy P. Yang (楊子儀)
Nov. 22, 2016
Department of Civil Engineering National Chiao Tung University (NCTU)
Chapter 10 Isoparametric Formulation and Numerical Integration
• 10.1 Isoparametric Formulation – Element stiffness matrix
• 10.2 Gauss Quadrature
10.1 Isoparametric Formulation
• Isoparametric elements – Non-rectangular quadrilateral elements with curved
boundary • Useful for modelling structures with curved edges and for
grading a mesh from coarse to fine • Effective in 2D and 3D elasticity problems, shell analysis,
non-structural application
10.1 Isoparametric Formulation
• Isoparametric elements – If the same shape function defines both
• the displacements of a point in the element
𝑢 = 𝑁1 𝑁2𝑢1𝑢2 = 𝑵𝑵
– 𝑵: nodal displacements
• the global coordinates of the point in the element
𝑥 = 𝑁1 𝑁2𝑥1𝑥2 = 𝑵𝒄
– 𝒄: nodal coordinates
10.1 Isoparametric Formulation
• Isoparametric elements – Computation of stiffness matrix is done on element
defined in natural coordinates (η, ξ, ζ = < −1, 1 >): line, rectangle, cube
– Element in natural coordinates (η, ξ, ζ) is transformed to real finite in (x, y, z) coordinates with use of shape functions
– Shape functions are used also as approximation functions of unknown displacements
• A two-node bar element – Consider a straight bar
– 𝜉: natural coordinate • 𝜉: the axial coordinate of the bar regardless of the bar’s
orientation in global coordinates
10.1 Isoparametric Formulation
• A two-node bar element – The assumed displacement function
• 𝑢 = 𝑎1 + 𝑎2𝜉 = 1 𝜉𝑎1𝑎2
– Nodal conditions • 𝑢 𝜉 = −1 = 𝑢1 • 𝑢 𝜉 = 1 = 𝑢2
– 𝑢 = 1−𝜉2
1+𝜉2
𝑢1𝑢2 = 𝑵𝑵
• The displacement of a point in the element is expressed in terms of nodal displacements
10.1 Isoparametric Formulation
10.1 Isoparametric Formulation • A two-node bar element
– Consider the interpolation
• 𝑥 = 𝑵𝑥1𝑥2 (*)
• The 𝑥 coordinate of a point in the element is expressed in terms of nodal coordinates 𝑥𝑖
– Recall the displacement
• 𝑢 = 𝑵𝑢1𝑢2
– Since both 𝑢 and 𝑥 are defined by the same nodes and by the same 𝑵, the element is isoparametric
• 𝑵 = 1−𝜉2
1+𝜉2
10.1 Isoparametric Formulation • A two-node bar element
– Strain-displacement relationship
• 𝜀𝑥 = 𝑑𝑑𝑑𝑥
= 𝑑𝑵 𝜉𝑑𝑥
𝑢1𝑢2 = 𝑩
𝑢1𝑢2
• 𝑑𝑑𝑥
= 𝑑𝜉𝑑𝑥
𝑑𝑑𝜉
• From (*), 𝑑𝑥𝑑𝜉
= −12
12
𝑥1𝑥2 = 𝐿
2
– In calculus, the scale factor between two coordinate systems is called the Jacobian (denoted by 𝐽)
• In this case, 𝐽 = 𝑑𝑥𝑑𝜉
= 𝐿2
⇒𝑑𝜉𝑑𝑥 =
2𝐿
10.1 Isoparametric Formulation
• A two-node bar element – Strain-displacement matrix
• 𝑩 = 𝑑𝑵 𝜉𝑑𝑥
= 𝑑𝜉𝑑𝑥
𝑑𝑵𝑑𝜉
= 1𝐽𝑑𝑵𝑑𝜉
= 2𝐿
−12
12
= −1𝐿
1𝐿
– Stiffness matrix
• 𝒌 = ∫ 𝑩T𝐴𝐴𝑩𝐿0 𝑑𝑥 = ∫ 𝑩T𝐴𝐴𝑩𝐽1
−1 𝑑𝜉
10.1 Isoparametric Formulation • A three-node bar element
– The assumed displacement function
• 𝑢 = 1 𝜉 𝜉2𝑎1𝑎2𝑎3
– Nodal conditions • 𝑢 𝜉 = −1 = 𝑢1, 𝑢 𝜉 = 1 = 𝑢2, 𝑢 𝜉 = 0 = 𝑢3 • Find 𝑎𝑖
– 𝑢 = 𝑵𝑵
• 𝑵 = −𝜉+𝜉2
2𝜉+𝜉2
21 − 𝜉2
10.1 Isoparametric Formulation
• A three-node bar element
– 𝑢 = 𝑵 𝜉𝑢1𝑢2𝑢3
and 𝑥 = 𝑵 𝜉𝑥1𝑥2𝑥3
– Strain-displacement relationship
• 𝜀𝑥 = 𝑑𝑑𝑑𝑥
= 𝑑𝑵 𝜉𝑑𝑥
𝑵 = 𝑩𝑵
– Jacobian
• 𝐽 = 𝑑𝑥𝑑𝜉
= −1+2𝜉2
1+2𝜉2
−2𝜉𝑥1𝑥2𝑥3
• Only if 𝑥3 represents the midpoint of the bar, then
𝐽 = −1+2𝜉2
0 + 1+2𝜉2
𝐿 − 2𝜉 𝐿2
= 𝐿2
10.1 Isoparametric Formulation • A three-node bar element
– Strain-displacement matrix
• 𝑩 = 𝑑𝑵 𝜉
𝑑𝑥= 1
𝐽 −1+2𝜉
21+2𝜉2
−2𝜉
– Stiffness matrix • 𝒌 = ∫ 𝑩T𝐴𝐴𝑩𝐿
0 𝑑𝑥 = ∫ 𝑩T𝐴𝐴𝑩𝐽1−1 𝑑𝜉
– Comment • 𝐽 and 𝑩 are functions of 𝜉, especially for polynomials in 𝜉 in
denominator of 𝑩, which infers that 𝒌 cannot be integrated in closed form in general
10.1 Isoparametric Formulation
• The isoparametric formulation can be applied to other isoparametric elements – e.g. rectangular plane element
10.1 Isoparametric Formulation
• The isoparametric formulation can be applied to other isoparametric elements – e.g. rectangular plane element
10.2 Gauss Quadrature
• Gauss quadrature – A technique for numerical integration used to
approximate a function
– The Gauss quadrature locates the sampling points (Gauss points) so that the greatest accuracy is achieved
10.2 Gauss Quadrature
• Consider the integral 𝐼
𝐼 = ∫ 𝜙 𝜉 𝑑𝜉1−1
– The integration can be approximated by
𝐼 ≈ 𝑊1𝜙 𝜉1 + 𝑊2𝜙 𝜉2 + ⋯+ 𝑊𝑛𝜙 𝜉𝑛
• 𝜉𝑖 : Gauss point • 𝑊𝑖: Gauss weight
10.2 Gauss Quadrature • Integration table
– The Gauss points and weights for Gauss quadrature up to order 𝑛 = 4 are shown below
– The Gauss points are located symmetrically w.r.t. the
center of the interval
10.2 Gauss Quadrature
• 1D example • Consider 𝑛 = 1 𝐼 ≈ 𝑊1𝜙 𝜉1 = 2.0𝜙 0 • Consider 𝑛 = 2 𝐼 ≈ 𝑊1𝜙 𝜉1 + 𝑊2𝜙 𝜉2 = 1.0𝜙 0.57735 + 1.0𝜙 −0.57735 • In general, a polynomial of degree 2𝑚− 1 is integrated
exactly by an 𝑚-point Gauss quadrature
10.2 Gauss Quadrature
• 1D example – Use Gaussian quadrature to obtain an exact value for
the integral
– For a polynomial of degree 𝑛 = 2 • Choose Gauss points 𝑚 = 2, then 2𝑚− 1 = 3 > 𝑛 = 2
( )1 2
13 7I r r dr
−= − +∫
( )2 2
1 2
1
1 1 1 13 7 1 3 7 1 3 73 3 3 3
14.6667
r r dr−
− + = ⋅ − + + ⋅ − − − +
=
∫
10.2 Gauss Quadrature • 2D example
• 𝐼 = ∫ ∫ 𝜙 𝜉, 𝜂 𝑑𝜉1−1 𝑑𝜂1
−1 ≈ ∑ ∑ 𝑊𝑖𝑊𝑗
𝑛𝑗=1
𝑚𝑖=1 𝜙 𝜉𝑖 , 𝜂𝑗
• For 𝑚 = 𝑛 = 3, – 9 Gauss points – The integration is approximated by order of 5
𝐼 ≈ ∑ ∑ 𝑊𝑖𝑊𝑗3𝑗=1
3𝑖=1 𝜙 𝜉𝑖 , 𝜂𝑗
= ∑ 𝑊1𝑊𝑗𝜙 𝜉1, 𝜂𝑗3𝑗=1 + ∑ 𝑊2𝑊𝑗𝜙 𝜉2, 𝜂𝑗3
𝑗=1 + ∑ 𝑊3𝑊𝑗𝜙 𝜉3, 𝜂𝑗3𝑗=1
= 𝑊1𝑊1𝜙 𝜉1, 𝜂1 + 𝑊1𝑊2𝜙 𝜉1, 𝜂2 + 𝑊1𝑊3𝜙 𝜉1, 𝜂3 +𝑊2𝑊1𝜙 𝜉2, 𝜂1 + 𝑊2𝑊2𝜙 𝜉2, 𝜂2 + 𝑊2𝑊3𝜙 𝜉2, 𝜂3 +𝑊3𝑊1𝜙 𝜉3, 𝜂1 + 𝑊3𝑊2𝜙 𝜉3, 𝜂2 + 𝑊3𝑊3𝜙 𝜉3, 𝜂3
10.2 Gauss Quadrature
• 2D example 𝐼 ≈ ∑ ∑ 𝑊𝑖𝑊𝑗3
𝑗=13𝑖=1 𝜙 𝜉𝑖 , 𝜂𝑗
= ∑ 𝑊1𝑊𝑗𝜙 𝜉1, 𝜂𝑗3𝑗=1 + ∑ 𝑊2𝑊𝑗𝜙 𝜉2, 𝜂𝑗3
𝑗=1 + ∑ 𝑊3𝑊𝑗𝜙 𝜉3, 𝜂𝑗3𝑗=1
= 𝑊1𝑊1𝜙 𝜉1, 𝜂1 + 𝑊1𝑊2𝜙 𝜉1, 𝜂2 + 𝑊1𝑊3𝜙 𝜉1, 𝜂3 +𝑊2𝑊1𝜙 𝜉2, 𝜂1 + 𝑊2𝑊2𝜙 𝜉2, 𝜂2 + 𝑊2𝑊3𝜙 𝜉2, 𝜂3 +𝑊3𝑊1𝜙 𝜉3, 𝜂1 + 𝑊3𝑊2𝜙 𝜉3, 𝜂2 + 𝑊3𝑊3𝜙 𝜉3, 𝜂3
𝜉1 = 0.7746 𝜉2 = 0 𝜉3 = −0.7746 𝜂1 = 0.7746 𝜂2 = 0 𝜂3 = −0.7746
𝑊1 = 0.5556 𝑊2 = 0.8889 𝑊3 = 0.5556