A linear elastic, isotropic epoxy resin with elastic modulus, E=2GPa, and Poisson’s ratio, ν=0.3, is loaded by the following stress state:
σ=⎡⎣10 5 0
5 −20 0
0 0 0⎤⎦MPa
Give numerical values for all strain components in the corresponding strain matrix.
Give a numerical value for the elastic strain energy, U, in the resin in terms of J/m3.
U in ( J/m3):
εx=
εy=
εz=
γyz=
γxz=
γxy
εx=0.008
εy=-0.0115
εz=0.0015
γyz=0
γxz=0
γxy=0.0065
U=171250
To find the strain components in the strain matrix, we can use Hooke's Law for linear elasticity. Hooke's Law relates stress and strain through the elastic modulus and Poisson's ratio.
The strain matrix, ε, can be determined using the stress matrix, σ, as follows:
ε = (1/E) * [σ - ν * Tr(σ) * I]
Where E is the elastic modulus, ν is the Poisson's ratio, Tr(σ) is the trace of the stress matrix, and I is the identity matrix.
Let's calculate the strain matrix:
E = 2 GPa = 2 * 10^9 Pa
ν = 0.3
σ = [10 5 0
5 -20 0
0 0 0] MPa = [10 5 0
5 -20 0
0 0 0] * 10^6 Pa
Tr(σ) = 10 - 20 + 0 = -10 MPa = -10 * 10^6 Pa
I = [1 0 0
0 1 0
0 0 1]
ε = (1/E) * [σ - ν * Tr(σ) * I]
= (1/2 * 10^9) * [ [10 5 0
5 -20 0
0 0 0] * 10^6 - 0.3 * (-10 * 10^6) * [1 0 0
0 1 0
0 0 1] ]
Calculating ε, we get:
ε = 5 * 10^-5 * [ 20 5 0
5 -10 0
0 0 0] - 3 * 10^-6 * [1 0 0
0 1 0
0 0 1]
= [1 * 10^-3 2 * 10^-5 0
2 * 10^-5 -5 * 10^-6 0
0 0 0]
Thus, the numerical values for all strain components in the strain matrix are:
ε = [1 * 10^-3 2 * 10^-5 0
2 * 10^-5 -5 * 10^-6 0
0 0 0]
Next, to find the elastic strain energy, U, in the resin, we can use the following formula:
U = (1/2) * σ * ε
Where U is the elastic strain energy, σ is the stress matrix, and ε is the strain matrix.
Considering the given stress and strain matrices:
σ = [10 5 0
5 -20 0
0 0 0] MPa = [10 5 0
5 -20 0
0 0 0] * 10^6 Pa
ε = [1 * 10^-3 2 * 10^-5 0
2 * 10^-5 -5 * 10^-6 0
0 0 0]
Calculating U, we get:
U = (1/2) * σ * ε, where * denotes element-wise multiplication
= (1/2) * [10 5 0
5 -20 0
0 0 0] * [1 * 10^-3 2 * 10^-5 0
2 * 10^-5 -5 * 10^-6 0
0 0 0]
= (1/2) * [10 * 1 * 10^-3 + 5 * 2 * 10^-5 + 0 * 0
5 * 1 * 10^-3 + (-20) * 2 * 10^-5 + 0 * 0
0 * 1 * 10^-3 + 0 * 2 * 10^-5 + 0 * 0]
= (1/2) * [0.000102 + 0.0002 + 0
0.00005 - 0.00004 + 0
0 + 0 + 0]
= 0.500152 J/m^3 (approximately)
Therefore, the numerical value for the elastic strain energy, U, in the resin is approximately 0.500152 J/m^3.
To find the strain components in the strain matrix, we can use Hooke's law which relates stress and strain in a linear elastic material:
ε = D * σ
Where ε is the strain, D is the elastic modulus matrix, and σ is the stress matrix.
The elastic modulus matrix, D, can be calculated using the elastic modulus, E, and Poisson's ratio, ν, as follows:
D = 1/(1-ν^2) * ⎡⎣1 ν 0
ν 1 0
0 0 (1-ν)/2⎤⎦ * E
Substituting the given values, we have:
D = 1/(1-0.3^2) * ⎡⎣1 0.3 0
0.3 1 0
0 0 (1-0.3)/2⎤⎦ * 2GPa
D = 1/(1-0.09) * ⎡⎣1 0.3 0
0.3 1 0
0 0 0.35⎤⎦ * 2GPa
D = 1/0.91 * ⎡⎣1 0.3 0
0.3 1 0
0 0 0.35⎤⎦ * 2GPa
D = 1.0989 * ⎡⎣1 0.3 0
0.3 1 0
0 0 0.35⎤⎦ * 2GPa
D = ⎡⎣1.0989 0.329 0
0.329 1.0989 0
0 0 0.7694⎤⎦ GPa
Now we can substitute the given stress matrix, σ, into the strain equation:
ε = D * σ
ε = ⎡⎣1.0989 0.329 0
0.329 1.0989 0
0 0 0.7694⎤⎦ * ⎡⎣10 5 0
5 -20 0
0 0 0⎤⎦
Calculating this, we find:
ε = ⎡⎣14.9315 4.3945 0
4.3945 -0.7945 0
0 0 0.7694⎤⎦
Therefore, the numerical values for all strain components in the corresponding strain matrix are:
ε11 = 14.9315
ε22 = -0.7945
ε33 = 0.7694
ε12 = 4.3945
ε21 = 4.3945
ε32 = ε31 = ε23 = ε13 = 0
To calculate the elastic strain energy, U, in the resin, we can use the formula:
U = (1/2) * σ * ε
Where U is the elastic strain energy, σ is the stress matrix, and ε is the strain matrix.
Substituting the given stress and strain matrices, we have:
U = (1/2) * ⎡⎣10 5 0
5 -20 0
0 0 0⎤⎦ * ⎡⎣14.9315 4.3945 0
4.3945 -0.7945 0
0 0 0.7694⎤⎦
Calculating this, we find:
U = (1/2) * ⎡⎣149.315 58.84475 0
58.84475 79.45 0
0 0 0⎤⎦
U = ⎡⎣74.6575 29.422375 0
29.422375 39.725 0
0 0 0⎤⎦
Therefore, the numerical value for the elastic strain energy, U, in the resin is:
U = 74.6575 J/m3