Finite Element Methods Elastostatic Problems Finite Element Methods Two Dimensional Solid...

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Finite Element Methods

Two Dimensional Solid

Instructor: Mohamed Abdou Mahran Kasem, Ph.D.

Aerospace Engineering Department

Cairo University

Plane stress

Plane stress is a state of stress in which the normal stress and the shear stresses directed perpendicular to the plane are assumed to be zero.

๐‘–. ๐‘’. ๐œŽ๐‘ง , ๐œ๐‘ฅ๐‘ง , ๐œ๐‘ฆ๐‘ง = 0

Plane strain

Plane strain is a state of strain in which the normal strain and the shear strains directed perpendicular to the plane are assumed to be zero.

๐‘–. ๐‘’. ๐œ€๐‘ง, ๐›พ๐‘ฅ๐‘ง , ๐›พ๐‘ฆ๐‘ง = 0

Plane stress

As we mentioned before the governing equilibrium equation for elastic, static, linear analysis has the form

เถฑ๐‘ค ๐‘–,๐‘— ๐œŽ๐‘–๐‘— ๐‘‘ฮฉ = เถฑ๐‘ค๐‘–๐‘“๐‘– ๐‘‘ฮฉ + ๐ต. ๐‘‡. ๐œŽ๐‘–๐‘— = ๐ถ๐‘–๐‘—๐‘˜๐‘™๐œ€๐‘˜๐‘™ = ๐ถ๐‘–๐‘—๐‘˜๐‘™๐‘ข๐‘˜,๐‘™ + ๐‘ข๐‘™,๐‘˜

2= ๐ถ๐‘–๐‘—๐‘˜๐‘™๐‘ข ๐‘–,๐‘—

Plane stress

In this case the stress-strain relation is reduced to the form

๐œŽ11๐œŽ22๐œŽ12

=2๐œ‡ + ๐œ† ๐œ† 0

๐œ† 2๐œ‡ + ๐œ† 00 0 2๐œ‡

๐œ€11๐œ€22๐œ€12

=๐ธ

1 โˆ’ ๐‘ฃ2

1 ๐‘ฃ 0๐‘ฃ 1 0

0 01 โˆ’ ๐‘ฃ

2

๐œ€11๐œ€22๐œ€12

= ๐ƒ๐›†

ฮป and ฮผ are Lameยด constants. They are related to the well-known Youngโ€™s Modulus (E) and Poissonโ€™s ratio (ฯ…) by

the following relation

๐œ† =๐ธ ๐‘ฃ

1 + ๐‘ฃ 1 โˆ’ 2๐‘ฃ, ๐œ‡ =

๐ธ

2 1 + ๐‘ฃ

Plane stress

The strain-displacement relation takes the form

๐ฎ = ๐๐” โ†’ ๐›† = ๐››๐ ๐” = ๐๐”

Linear triangular element

Linear triangular element

3-node element

In matrix form:

Linear triangular element

Substitute by the BCโ€™s

I

II

By solving the two-set of equations together, one can obtain the shape functions for linear

triangular element

Linear triangular element

Linear triangular element

Linear triangular element

Element strains

Linear triangular element

Element strains

Similarly, we can obtain the other derivatives

Linear triangular element

By substitute in the weak form,

Where B is the strain displacement matrix and D in the material stiffness matrix

depends on the element either plane stress or strain.

เถฑ๐‘ค ๐‘–,๐‘— ๐œŽ๐‘–๐‘— ๐‘‘ฮฉ = เถฑ๐‘ค๐‘–๐‘“๐‘– ๐‘‘ฮฉ + ๐ต. ๐‘‡.

Linear triangular element

๐พ

=๐ธ๐‘ก

แˆป4๐ด(1 โˆ’ ๐œˆ2

๐›ฝ๐‘–2 โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘–

2 1

2(1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘– ๐›ฝ๐‘–๐›ฝ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘–๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘– + ๐œˆ๐›ฝ๐‘–๐›พ๐‘— ๐›ฝ๐‘–๐›ฝ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘–๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘– + ๐œˆ๐›ฝ๐‘–๐›พ๐‘š

1

2(1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘– โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–

2 + ๐›พ๐‘–2 ๐œˆ๐›ฝ๐‘—๐›พ๐‘– โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›ฝ๐‘— + ๐›พ๐‘–๐›พ๐‘— ๐œˆ๐›ฝ๐‘š๐›พ๐‘– โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›ฝ๐‘š + ๐›พ๐‘–๐›พ๐‘š

๐›ฝ๐‘–๐›ฝ๐‘— โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘–๐›พ๐‘— ๐œˆ๐›ฝ๐‘—๐›พ๐‘– โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘— ๐›ฝ๐‘—

2 โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘—

2 1

2(1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘— ๐›ฝ๐‘—๐›ฝ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘—๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘— + ๐œˆ๐›ฝ๐‘—๐›พ๐‘š

โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘– + ๐œˆ๐›ฝ๐‘–๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›ฝ๐‘— + ๐›พ๐‘–๐›พ๐‘—

1

2(1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—

2 + ๐›พ๐‘—2 ๐œˆ๐›ฝ๐‘š๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›ฝ๐‘š + ๐›พ๐‘—๐›พ๐‘š

๐›ฝ๐‘–๐›ฝ๐‘š โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘–๐›พ๐‘š ๐œˆ๐›ฝ๐‘š๐›พ๐‘– โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›พ๐‘š ๐›ฝ๐‘—๐›ฝ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘—๐›พ๐‘š ๐œˆ๐›ฝ๐‘š๐›พ๐‘— โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›พ๐‘š ๐›ฝ๐‘š

2 โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›พ๐‘š

21

2(1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘š

โˆ’1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘– + ๐œˆ๐›ฝ๐‘–๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘–๐›ฝ๐‘š + ๐›พ๐‘–๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘— + ๐œˆ๐›ฝ๐‘—๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘—๐›ฝ๐‘š + ๐›พ๐‘—๐›พ๐‘š

1

2(1 + ๐œˆแˆป๐›ฝ๐‘š๐›พ๐‘š โˆ’

1

2(โˆ’1 + ๐œˆแˆป๐›ฝ๐‘š

2 + ๐›พ๐‘š2

Example

Evaluate the stiffness matrix for the element shown in Figure. The coordinates

are shown in units of inches. Assume plane stress conditions. Let ๐ธ = 30๐‘ฅ106psi,

๐œ = 0.25, and thickness t = 1 in. Assume the element nodal displacements have been

determined to be ๐‘ข1 = 0, ๐‘ฃ1 = 0.0025 ๐‘–๐‘›, ๐‘ข2 = 0.0012 ๐‘–๐‘›, ๐‘ฃ2 = 0, ๐‘ข3 = 0, ๐‘ฃ3 = 0.0025 ๐‘–๐‘›

Determine the element stresses.

Example

Example

Example

Surface Forces

Surface Forces

Surface Forces

Equivalent nodal forces

Example

For a thin plate subjected to the surface traction shown in Figure, determine the

nodal displacements and the element stresses.

The plate thickness t = 1 in., ๐ธ = 30๐‘ฅ106psi, and ๐œˆ = 0.3.

Example

Plate mesh

Example

Calculate element stiffnesses

Example

Calculate element stiffnesses

Example

Calculate element stiffnesses

Example

Calculate element stiffnesses

Example

Calculate element stiffnesses

Example

Calculate element stiffnesses

Example

The global stiffness matrix

Example

After applying the BCโ€™s

Example

Determine the unknown displacements

Example

Comparing to analytical solution

- The analytical solution represents 1-D approximation, while the FE solution represents 2-

D approximation.

- We used a coarse mesh in the FE solution, which results in an inaccurate solution.

Example

Element stresses

Example

Stresses for element 1

Example

Stresses for element 2

This lecture is prepared from: Logan โ€œA first course in the finite element methodโ€

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