finite element analysis of composite layered structures connor kaufmann – b. sc. ‘14

Finite Element AnalysisOf Composite Layered Structures

Connor Kaufmann – B. Sc. ‘14 Neola Putnam – M. Eng. ‘14 Ethan Seo – M. Eng. ‘14 Ju Hwan (Jay) Shin – B. Sc. ‘14

Cornell UniversitySibley School of Mechanical & Aerospace Engineering

Spring 2014 – Professor N. Zabaras

Objective Develop a linear 3D finite element analysis from scratch using MATLAB.

Consider a uniaxial loading of a symmetric laminate.

Verify the results with expected results, namely the state of out-of-plane stresses near the free edges.

Observe the effect of h-refinement (convergence of results through mesh refinements).

Perform a simple sanity check by doing a force reaction balance with the applied traction.

Compare the numerical results with a commercial FE software, or Ansys Composite PrepPost (ACP).

2

C. Kaufmann, N. Putnam, E. Seo, J. Shin

0 20 40 60 80 1000

2000

4000

6000

8000

10000

12000

14000

Stress_yz for [0 90]s laminate

Along y-direction [m] yz

[P

a]

𝜏 𝑦𝑧

𝑃0

3

Some stress contours! C. Kaufmann, N. Putnam, E. Seo, J. Shin

𝝈𝒙𝒙 𝝈𝒚𝒚 𝝈𝒛𝒛

𝝉𝒚𝒛 𝝉𝒙𝒛 𝝉𝒙𝒚

These results refer to a cross-ply laminate.

Overview of composite materials Composite materials are commonly used in aerospace structures to minimize mass.

We considered symmetric, unidirectional, fiber-reinforced composites.

Composite lamina (sheets) can be stacked to form high strength laminates.

Laminate stack-ups are characterized by the orientation angles of the fibers, and the materials used.

Anisotropy of the laminates allows one to tailor designs for stiffness and strength in specific directions.

4i.e.


Complexities of layered structures Composite materials can give a much more complicated mechanical response than monolithic materials.

Stress equilibrium must be satisfied in the laminate by way of interlaminar stresses.

Special care must be taken to consider the free-edge and free-corner effect in composite samples!

Non-intuitive effects, such as normal-shear coupling, can occur in anisotropic materials.

As a result, finite element analysis is often found useful for predicting the behavior of complicated composites.

5

Delamination


Normal-shear coupling

6

FEM Formulation Pre-processing:

Define the required size dimensions, material properties, and the laminate configuration. Discretize the model into finite elements. Consider a tri-linear hexahedron element. Apply any bias factor when discretizing (optional). Calculate the 3D elasticity matrix, . i.e. Compute the elemental matrices necessary in developing the stiffness equation. Specify the boundary conditions (includes the external load).

Processing: “Globalize” and assemble the local stiffness matrices and the local load vectors. Partition and rearrange the global stiffness equation. Solve for the nodal displacement field!

Post-processing: Compute the strain field by applying the kinematic equation (displacement ↔ strain). Compute the stress field by applying the constitutive equation (Hooke’s Law).


7

Pre-processing (1/7) Define the required size dimensions, material properties, and the laminate configuration. Discretize the model into finite elements. Consider a 8-noded hexahedron, or tri-linear element (3 translational DOF per node). Apply any bias factor when discretizing (optional). Calculate the 3D elasticity matrix, . i.e. Compute the elemental matrices necessary in developing the stiffness equation. Specify the boundary conditions (includes the external load).

Define the coordinate axes, based on fiber orientations. x-axis: longitudinal direction y-axis: transverse direction z-axis: normal direction (or thickness direction)

Specify the size dimensions. , , and

Specify the material properties. , , , , , , , , and

Specify the fiber orientations of the off-axis plies.


𝑵 𝒙

8

Pre-processing (2/7) Define the required size dimensions, material properties, and the laminate configuration. Discretize the model into finite elements. Consider a 8-noded hexahedron, or tri-linear element (3 translational DOF per node). Apply any bias factor when discretizing (optional). Calculate the 3D elasticity matrix, . i.e. Compute the elemental matrices necessary in developing the stiffness equation. Specify the boundary conditions (includes the external load).

Store the global nodes as local nodes for each element.

Adhere to the given node-numbering scheme to ensure that the determinant of the Jacobian matrix is positive. Guarantee invertible mapping to natural coordinate system.


9

Pre-processing (3/7) Define the required size dimensions, material properties, and the laminate configuration. Discretize the model into finite elements. Consider a tri-linear hexahedron element. Apply any bias factor when discretizing (optional). Calculate the 3D elasticity matrix, . i.e. Compute the elemental matrices necessary in developing the stiffness equation. Specify the boundary conditions (includes the external load).

Use an eight-noded hexahedron element.

Each node has three translational degrees-of-freedom. , , and


10


Bias Factor: Allows us to have more concentrated mesh density near a particular region of interest.

The spacing between the node becomes a geometrical series.


0 100 200 300 400 5000

20

40

60

80

100Node Profile

x-axis

y-ax

isFiner m

esh

11


The three-dimensional elasticity (stiffness) matrix is defined by applying the generalized Hooke’s Law.

Take into account the anisotropy, assuming a transversely isotropic layer.

Material nonlinearity (plasticity) is neglected!


{𝜎 }=[𝑻 1 (−𝜃 ) ] [𝑪 ] [𝑻 2 (𝜃 ) ] {𝜀 }

{𝜎 }=[𝜎 𝑥𝑥

𝜎 𝑦𝑦𝜎 𝑧𝑧

𝜏 𝑦𝑧𝜏 𝑥𝑧𝜏𝑥𝑦

] {𝜀 }=[𝜀𝑥𝑥𝜀𝑦𝑦𝜀𝑧𝑧𝛾𝑦𝑧𝛾𝑥𝑧𝛾𝑥𝑦

][𝑪 ]=[

1−𝜈23𝜈32𝐸2𝐸3 Δ

𝜈21+𝜈23𝜈31𝐸2𝐸3 Δ

𝜈31+𝜈21𝜈32𝐸2𝐸3 Δ

0 0 0

𝜈21+𝜈23𝜈31𝐸2𝐸3 Δ

1−𝜈13𝜈31𝐸1𝐸3 Δ

𝜈32+𝜈12𝜈31𝐸1𝐸3 Δ

0 0 0

𝜈31+𝜈21𝜈32𝐸2𝐸3 Δ

𝜈32+𝜈12𝜈31𝐸1𝐸3 Δ

1−𝜈12𝜈21𝐸1𝐸2 Δ

0 0 0

0 0 0 𝐺23 0 00 0 0 0 𝐺13 00 0 0 0 0 𝐺12

]Δ≡

1−𝜈12𝜈21−𝜈23𝜈32−𝜈13𝜈31−2𝜈21𝜈32𝜈13𝐸1𝐸2𝐸3

12



[𝑵𝑒 ]=[𝑁 1𝑒 0 0 𝑁2

𝑒 0 0 ⋯ 𝑁 nen𝑒 0 0

0 𝑁 1𝑒 0 0 𝑁 2

𝑒 0 ⋯ 0 𝑁 nen𝑒 0

0 0 𝑁1𝑒 0 0 𝑁 2

𝑒 ⋯ 0 0 𝑁 nen𝑒 ]

[𝑩𝑒 ]≡ [𝛁s𝑵𝑒 ]=[𝜕𝑁 1

𝑒

𝜕 𝑥 0 0𝜕𝑁2

𝑒

𝜕 𝑥 0 0 ⋯𝜕𝑁 nen

𝑒

𝜕 𝑥 0 0

0𝜕𝑁 1

𝑒

𝜕 𝑦 0 0𝜕𝑁2

𝑒

𝜕 𝑦 0 ⋯ 0𝜕𝑁 nen

𝑒

𝜕 𝑦 0

0 0𝜕𝑁1

𝑒

𝜕 𝑧 0 0𝜕𝑁2

𝑒

𝜕𝑧 ⋯ 0 0𝜕𝑁 nen

𝑒

𝜕 𝑧

0𝜕𝑁 1

𝑒

𝜕 𝑧𝜕𝑁1

𝑒

𝜕 𝑦 0𝜕𝑁2

𝑒

𝜕 𝑧𝜕𝑁2

𝑒

𝜕 𝑦 ⋯ 0𝜕𝑁 nen

𝑒

𝜕 𝑧𝜕𝑁 nen

𝑒

𝜕 𝑦𝜕𝑁 1

𝑒

𝜕 𝑧 0𝜕𝑁1

𝑒

𝜕 𝑥𝜕𝑁2

𝑒

𝜕 𝑧 0𝜕𝑁2

𝑒

𝜕𝑥 ⋯𝜕𝑁 nen

𝑒

𝜕 𝑧 0𝜕𝑁 nen

𝑒

𝜕 𝑥𝜕𝑁 1

𝑒

𝜕 𝑦𝜕𝑁 1

𝑒

𝜕 𝑥 0𝜕𝑁2

𝑒

𝜕 𝑦𝜕𝑁2

𝑒

𝜕𝑥 0 ⋯𝜕𝑁 nen

𝑒

𝜕 𝑦𝜕𝑁 nen

𝑒

𝜕 𝑥 0

]𝑁 𝑖

𝑒=𝐿𝐼𝑒 (𝜉 ) 𝐿 𝐽𝑒 (𝜂 )𝐿𝐾𝑒 (𝜁 )

𝐿𝑚𝑒 (𝜉 )=∏

𝑗 ≠𝑚

𝑝+1 𝜉−𝜉 𝑗𝑒

𝜉𝑚𝑒 −𝜉 𝑗

𝑒 𝐿𝑚𝑒 (𝜂 )=∏

𝑗 ≠𝑚

𝑝+1 𝜂−𝜂 𝑗𝑒

𝜂𝑚𝑒 −𝜂 𝑗

𝑒 𝐿𝑚𝑒 (𝜁 )=∏

𝑗≠𝑚

𝑝+1 𝜁 −𝜁 𝑗𝑒

𝜁𝑚𝑒 −𝜁 𝑗

𝑒

d 𝐿𝑚𝑒 (𝜉 )d 𝜉 =∑

h≠𝑚

𝑝+1 1𝜉𝑚𝑒 −𝜉 h𝑒 ( ∏

𝑗≠ h∧ 𝑗≠𝑚

𝑝+ 1 𝜉−𝜉 𝑗𝑒

𝜉𝑚𝑒 −𝜉 𝑗𝑒 )

d 𝐿𝑚𝑒 (𝜂 )d 𝜂 =∑

h≠𝑚

𝑝+1 1𝜂𝑚𝑒 −𝜂 h𝑒 ( ∏

𝑗 ≠h∧ 𝑗≠𝑚

𝑝+1 𝜂−𝜂 𝑗𝑒

𝜂𝑚𝑒 −𝜂 𝑗𝑒 )

d 𝐿𝑚𝑒 (𝜁 )d 𝜁 =∑

h≠𝑚

𝑝+1 1𝜁𝑚𝑒 −𝜁h𝑒 ( ∏

𝑗 ≠h∧ 𝑗≠𝑚

𝑝+1 𝜁 −𝜁 𝑗𝑒

𝜁𝑚𝑒 −𝜁 𝑗𝑒 )

13


Essential Boundary Condition

Natural Boundary Condition Pressure-based load, , at


14

Processing (1/3) “Globalize” and assemble the local stiffness matrices and the local load vectors. Partition and rearrange the global stiffness equation. Solve for the nodal displacement field!

The weak form of our finite element formulation is given below.

Use Gauss Quadrature rule to numerically evaluate the local integration.

Apply the transformation rule to and using the connectivity matrix.

Sum individual matrices and vectors for global assembly.


∑𝑒=1

nel (𝑳𝑒⊤∫−1

+1

∫−1

+1

∫− 1

+1

𝑩𝑒⊤𝑫𝑒𝑩𝑒| 𝑱𝑒|d 𝜉 d𝜂 d 𝜁 𝑳𝑒)⏟𝑲

𝑑=∑𝑒=1

nel

(𝑳𝑒⊤∫Γ t𝑒

𝑵 𝑒⊤ 𝑡 𝑒dΓ )⏟𝑓

15


Partition the global stiffness equation into the known and unknown components.

Apply the transformation rule to rearrange them as shown below.


[𝑲 ] {𝑑 }= { 𝑓 }

[ 𝑲 𝐸 𝑲 𝐸𝐹

𝑲 𝐸𝐹⊤ 𝑲 𝐹 ] [𝑑𝐸𝑑𝐹 ]=[ 𝑓 𝐸𝑓 𝐹]

16


Solve the system of equations efficiently by using the Gaussian elimination method. In MATLAB, a built-in function, d=K\f can be employed.


𝑲 𝐸𝐹⊤ 𝑑𝐸+𝑲 𝐹 𝑑𝐹= 𝑓 𝐹

𝑑𝐹=𝑲 𝐹−1 𝑓 𝐹 , ∀𝑑𝐸= 0⃑

𝑓 𝐸=𝑲 𝐸𝐹 𝑑𝐹=𝑲 𝐸𝐹 𝑲 𝐹−1 𝑓 𝐹

17

Post-processing (1/2) Compute the strain field by applying the kinematic equation (displacement ↔ strain). Compute the stress field by applying the constitutive equation (Hooke’s Law).

The elemental strain vector can be computed as shown below. Weighted average of the strain values, evaluated the Gauss points.


𝜀=𝛁s𝑢

[𝜀𝑥𝑥𝜀𝑦𝑦𝜀𝑧𝑧𝛾𝑦𝑧𝛾𝑥𝑧𝛾𝑥𝑦

]=[𝜕𝜕 𝑥 0 0

0 𝜕𝜕 𝑦 0

0 0 𝜕𝜕 𝑧

0 𝜕𝜕 𝑧

𝜕𝜕 𝑦

𝜕𝜕 𝑧 0 𝜕

𝜕 𝑥𝜕𝜕 𝑦

𝜕𝜕 𝑥 0

] [𝑢𝑥𝑢𝑦𝑢𝑧 ]

18

Post-processing (2/2) Compute the strain field by applying the kinematic equation (displacement ↔ strain). Compute the stress field by applying the constitutive equation (Hooke’s Law).

The elemental stress vector can be computed as shown below. Weighted average of the stress values, evaluated the Gauss points.


{𝜎 }=[𝑫 ] {𝜀 }=[𝑻 1 (−𝜃 ) ] [𝑪 ] [𝑻 2 (𝜃 ) ] {𝜀 }

𝜎 vm=1√2 √ (𝜎𝑥𝑥−𝜎 𝑦𝑦 )2+(𝜎 𝑦𝑦−𝜎 𝑧𝑧 )2+(𝜎 𝑥𝑥−𝜎 𝑧𝑧 )2+6 (𝜏 𝑦𝑧2 +𝜏𝑥𝑧2 +𝜏 𝑥𝑦2 )

𝜆3− 𝐼1 𝜆2− 𝐼 2 𝜆− 𝐼3=0det (𝝈− 𝜆 𝑰 )=0

𝐼 1=𝜎𝑥𝑥+𝜎 𝑦𝑦+𝜎 𝑧𝑧

𝐼 2=𝜎 𝑦𝑦𝜎 𝑧𝑧+𝜎𝑥𝑥 𝜎𝑧𝑧+𝜎𝑥𝑥 𝜎 𝑦𝑦−𝜏 𝑦𝑧2 −𝜏𝑥𝑧2 −𝜏 𝑥𝑦2

𝐼 3=𝜎 𝑥𝑥𝜎 𝑦𝑦 𝜎𝑧𝑧−𝜎𝑥𝑥𝜏 𝑦𝑧2 −𝜎 𝑦𝑦 𝜏𝑥𝑧2 −𝜎 𝑧𝑧 𝜏𝑥𝑦2 +2𝜏 𝑦𝑧 𝜏𝑥𝑧𝜏𝑥𝑦

𝜏max=MAX (|𝜎 p2−𝜎 p 3

2 |,|𝜎 p1−𝜎 p 3

2 |,|𝜎 p1−𝜎 p 2

2 |)

𝜆1=𝜎 p1 𝜆2=𝜎 p2 𝜆3=𝜎 p3

19

Pathwise-results! C. Kaufmann, N. Putnam, E. Seo, J. Shin

These results refer to a angle-ply laminate.

16800 seconds ≈ 4.5 hours!

0 5 10 15 20 25 30 35

-1

-0.5

0

0.5

1x 106 Stress_yz for [-35 35]

s laminate

Along z-direction [m]

yz [

Pa]

Ply interface

𝜏 𝑦𝑧0 5 10 15 20 25 30 35

-3

-2.5

-2

-1.5

-1

-0.5

x 107 Stress_xz for [-35 35]s laminate

Along z-direction [m]

xz [

Pa]

Stress singularity!

𝜏𝑥𝑧0 20 40 60 80 100

-18000

-16000

-14000

-12000

-10000

-8000

-6000

-4000

-2000

Stress_yz for [-35 35]s laminate

Along y-direction [m]

yz [

Pa]

𝜏 𝑦𝑧

20

Contour Plots!

𝝈𝒙𝒙 𝝈𝒚𝒚 𝝈𝒛𝒛

𝝉𝒚𝒛 𝝉𝒙𝒛 𝝉𝒙𝒚



21

More contours… C. Kaufmann, N. Putnam, E. Seo, J. Shin

𝝈𝐩𝟑𝝈𝐩𝟐𝝈𝐩𝟏

𝝈𝐯𝐦 𝝉𝐦𝐚𝐱


22

Sanity Check! A force reaction balance check would indicate that our analysis was modeled correctly!

External load Reaction load≟• Reaction load is extracted from nodes (DOF to be more specific), where an essential BC is specified.


InputLaminate width, Laminate thickness, Pressure Load,

𝑓 𝐸=𝑲 𝐸𝐹 𝑑𝐹

𝑲 𝐸𝑑𝐸+𝑲 𝐸𝐹 𝑑𝐹= 𝑓 𝐸

Nodal Reaction Forces,

External Pressure Load,

∑𝐹 𝑥= 𝑓 ext+ 𝑓 r=0

23

Error Analysis• and Energy error norms are considered.

• In the below formula, is equal to 1, since a linear element is considered.


||𝑒||L2=||𝑢 (𝑥 )−𝑢h (𝑥 )||L2=(∑∫𝑥1𝑒

𝑥2𝑒

(𝑢 (𝑥 )−𝑢h (𝑥 ) )2d 𝑥)12

||𝑒||en=||𝑢 (𝑥 )−𝑢h (𝑥 )||en=(∑ 12∫𝑥1𝑒𝑥2𝑒

𝐸𝑒 (𝜀 (𝑥 )−𝜀h (𝑥 ) )2d 𝑥)12

||𝑒||L2≤𝐶 h𝑝+1 ||𝑒||en ≤𝐶h𝑝

h=√h𝑥2+h𝑦2+h𝑧2

24

Comparison to Ansys? C. Kaufmann, N. Putnam, E. Seo, J. Shin

5.3539e7 Pa(Ansys ACP)

5.358e7 Pa(MATLAB)

𝜎 𝑥𝑥

Averaged over an element

finite element analysis of composite layered structures connor kaufmann – b. sc. ‘14

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