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


• 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


• 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


• 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 • 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


• 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


• 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


• 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


• 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


• 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


• 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


• 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

judy p. yang (楊子儀 · 2020. 4. 13. · 10.1 isoparametric formulation • isoparametric...

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