Problem 2.8. Stiffness matrix

Based on laboratory measurements the stiffness matrix of a material is the following (values are in kN/mm2):

101011000010
Prove that the material is stable, and not isotropic.

Solve Problem

Solve

Young modulus, E [GPa]=

Poisson ratio, ν=

Shear modulus [GPa]=

Check proof

The material is stable, because both the Young and shear moduli are positive. The stiffness matrix is positive definite.

G=10kNmm2E2(1+ν)=9.92(1+0.1)=4.5kNmm2

thus the material is not isotropic.

Do you need help?

Steps

Step by step

Step 1. Write parametrically the stiffness matrix and determine the material properties. Assume orthotropic material.

Check material properties

Stiffness matrix of an orthotropic material is given by the following engineering constants:

Eq.(2-63) and Footnote k in Section 2.1.4

11νxyνyxExνyxEx0νxyEyEy000G1νxyνyx=101011000010

From the above equation the unknown material properties can be unambiously expressed:

νxy=νyx=0.1, Ex=Ey=9.9kNmm2, G=10kNmm2

Step 2. Check the Young modulus and the shear modulus to prove whether the material is stable.

Check criterion of stable material

The material is expected to be stable, because both the Young and the shear moduli are positive.

See Eq.(2-80)

E=9.9 kNmm2>0, G=10 kNmm2>0

For isotropic material the Poisson ratio must be smaller than 0.5, for anisotropic material the material is stable if the compliance (or the stiffness) matrix is positive definite. Symmetric, real matrices are positive definite when all of their eigenvalues are positive.

In Matlab eig101011000010 results in the following eigenvalues: 9.00, 10.00, 11.00, which are positive, hence the material is stable.

Step 3. Check isotropy.

Check isotropy

The material is isotropic when the following relation between the engineering constants is true

Eq.(2-62)

G=E2(1+ν)

Here

G=10kNmm2E2(1+ν)=9.92(1+0.1)=4.5kNmm2

thus the material is not isotropic.

Results

Worked out results

Material properties can be determined from the parametric form of the stiffness matrix. Assume orthotropic material. Stiffness matrix of an orthotropic material is given by the following engineering constants:

Eq.(2-63) and Footnote k in Section 2.1.4

11νxyνyxExνyxEx0νxyEyEy000G1νxyνyx=101011000010

From the above equation the unknown material properties can be unambiously expressed:

νxy=νyx=0.1, Ex=Ey=9.9kNmm2, G=10kNmm2

The material is expected to be stable, because both the Young and the shear moduli are positive.

See Eq.(2-80)

E=9.9 kNmm2>0, G=10 kNmm2>0

For isotropic material the Poisson ratio must be smaller than 0.5, for anisotropic material the material is stable if the compliance (or the stiffness) matrix is positive definite. Symmetric, real matrices are positive definite when all of their eigenvalues are positive.

In Matlab eig101011000010  results in the following eigenvalues: 9.00, 10.00, 11.00, which are positive, hence the material is stable.

The material is isotropic when the following relation between the engineering constants is true

Eq.(2-62)

G=E2(1+ν)

Here

G=10kNmm2E2(1+ν)=9.92(1+0.1)=4.5kNmm2

thus the material is not isotropic.