Question

An isotropic epoxy resin (𝐸=2𝐺𝑃𝑎, 𝜈=0.3) is reinforced by unidirectional glass fibers (𝐸𝑔𝑙𝑎𝑠𝑠=70𝐺𝑃𝑎), aligned in the 2 direction, such that the fiber composite is transversely isotropic, with the 1-3 plane being the plane of isotropy. The elastic constants of the fiber composite are:

𝐸1=3.3𝐺𝑃𝑎,
𝜈13=0.25
𝐸2=29.2𝐺𝑃𝑎,
𝜈12=0.3
𝐸3=3.3𝐺𝑃𝑎,
𝐺12=1.27𝐺𝑃𝑎
(Note that 𝜈𝑖𝑗=−𝜀𝑗/𝜀𝑖) Give numerical values for all strain components in the corresponding strain matrix for the stress state below.
𝜎=⎡10 5 0 ⎤
⎥5 -20 0 ⎥ 𝑀𝑃𝑎
⎣ 0 0 0⎦

Value for 𝜀1 ?? (It isn’t 3.235*10^(-3))

Answer 1

[AI-generated answer — Explain Bot]

To find the numerical value for 𝜀1 (strain along the 1-direction), we can start by using Hooke's law to relate stress and strain:

𝜎 = [ C ] * 𝜀

where 𝜎 is the stress matrix, [ C ] is the stiffness matrix, and 𝜀 is the strain matrix.

Given:
𝜎 = ⎡10 5 0 ⎤
⎥5 -20 0 ⎥
⎣ 0 0 0⎦

To find 𝜀1, we need to determine the first row of the strain matrix.

Let 𝜀1 = 𝜀, 𝜀2 = 0, 𝜀3 = 0, 𝜀23 = 0.

Using the relationship 𝜎 = [ C ] * 𝜀, we can write the stress-strain equations for the first component (𝜎11):

10 = 𝑪11 * 𝜀1 + 𝑪12 * 𝜀2 + 𝑪13 * 𝜀3 + 𝑪14 * 𝜀23

Since 𝜀2 = 𝜀3 = 𝜀23 = 0, the equation simplifies to:

10 = 𝑪11 * 𝜀1

Now, we need to find the value of 𝑪11 from the given elastic constants.

From the expression 𝐸 = 2𝐺(1 + 𝜈):

𝐺 = 𝐸/(2(1 + 𝜈))

For the epoxy resin, 𝐸 = 2𝐺𝑃𝑎 = 2𝑥10^9 𝑃𝑎.

𝐺 = (2𝑥10^9 𝑃𝑎)/(2(1 + 0.3)) = 1.111111111 x 10^9 𝑃𝑎

Now, using the relationship 𝐸 = 𝑪11/(1 - 𝜈12*𝜈21):

𝐸 = 𝑪11/(1 - 𝜈12*𝜈21)

Solving for 𝑪11:

𝑪11 = 𝐸(1-𝜈𝜈21)

Substituting the given values, we get:

𝑪11 = (3.3𝐺𝑃𝑎)(1-0.3*0.3)

𝑪11 = 2.739 𝐺𝑃𝑎

Plugging 𝑪11 back into the stress-strain equation:

10 = (2.739 𝐺𝑃𝑎) * 𝜀1

Solving for 𝜀1:

𝜀1 = 10/(2.739 𝐺𝑃𝑎)

𝜀1 ≈ 3.65 x 10^(-9)

Therefore, the numerical value for 𝜀1 is approximately 3.65 x 10^(-9) (Note: this value is different from 3.235 x 10^(-3) which is not the correct answer).