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E-04 (Brick Patch Test)

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Problem Description

 

This is a patch test for a unit cube [Ref 1].  The cube is modeled with 7 eight-node brick elements.  Nodal coordinates, element connectivity, and boundary conditions are given in the following tables.  Boundary conditions are given as forced displacements.  No additional loads are prescribed.

 

Material:  E = 1.e6 psi; ν = 0.25

 

Find stresses for each element.

E-04 (Brick Patch Test)1

 

Nodal coordinates (inch)

 

Node

X

Y

Z

1

0.249

0.342

0.192

2

0.826

0.288

0.288

3

0.85

0.649

0.263

4

0.273

0.75

0.23

5

0.32

0.186

0.643

6

0.677

0.305

0.683

7

0.788

0.693

0.644

8

0.165

0.745

0.702

9

0

0

0

10

1

0

0

11

1

1

0

12

0

1

0

13

0

0

1

14

1

0

1

15

1

1

1

16

0

1

1

 

 

Displacement field

u = 0.001 * (2x + y + z) / 2

v = 0.001 * (x + 2y + z) / 2

w = 0.001 * (x + y + 2z) / 2

Forced displacements (inch) on boundary

 

Node

Dx

Dy

Dz

9

0

0

0

10

0.001

0.0005

0.0005

11

0.0015

0.0015

0.001

12

0.0005

0.001

0.0005

13

0.0005

0.0005

0.001

14

0.0015

0.001

0.0015

15

0.002

0.002

0.002

16

0.001

0.0015

0.0015

 

All strains are constant. For example E-04 (Brick Patch Test)4

E-04 (Brick Patch Test)5

 

 

Element Connectivity

Element

Node1

Node2

Node3

Node4

Node5

Node6

Node7

Node8

1

1

2

3

4

5

6

7

8

2

4

3

11

12

8

7

15

16

3

9

10

2

1

13

14

6

5

4

2

10

11

3

6

14

15

7

5

9

1

4

12

13

5

8

16

6

9

10

11

12

1

2

3

4

7

5

6

7

8

13

14

15

16

 

Results

The displacements of internal nodes can be calculated based on the boundary conditions.  The constant stresses are also given by [Ref 1].

Units: displacement – in

Nodes

ENERCALC 3D

(compatible and incompatible)

Theoretical

Dx

Dy

Dz

Dx

Dy

Dz

1

5.16E-04

5.63E-04

4.88E-04

5.16E-04

5.63E-04

4.88E-04

2

1.11E-03

8.45E-04

8.45E-04

1.11E-03

8.45E-04

8.45E-04

3

1.31E-03

1.21E-03

1.01E-03

1.31E-03

1.21E-03

1.01E-03

4

7.63E-04

1.00E-03

7.42E-04

7.63E-04

1.00E-03

7.42E-04

5

7.35E-04

6.68E-04

8.96E-04

7.35E-04

6.68E-04

8.96E-04

6

1.17E-03

9.85E-04

1.17E-03

1.17E-03

9.85E-04

1.17E-03

7

1.46E-03

1.41E-03

1.38E-03

1.46E-03

1.41E-03

1.38E-03

8

8.89E-04

1.18E-03

1.16E-03

8.89E-04

1.18E-03

1.16E-03

Units: stress - psi

 

Sxx

Syy

Szz

Sxy

Syz

Sxz

ENERCALC 3D

(compatible)

1999.982

1999.982

1999.982

399.999

399.999

399.999

ENERCALC 3D

(incompatible)

1999.978

1999.978

1999.978

399.998

399.998

399.998

[Ref. 1]

2000

2000

2000

400

400

400

 

Comments

Both compatible and incompatible brick elements pass the patch test.  Therefore, “the results for any problem solved with the element will converge toward the correct solution as the elements are subdivided.” [Ref. 1]  The tiny differences in stresses are due to the penalty approach employed in support enforcement during solution.

 

Reference

[1]. MacNeal & Harder, “A Proposed Standard Set of Problems to Test Finite Element Accuracy”, Finite Elements in Analysis and Design, 1 (1985) 3-20