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To determine the modulus of elasticity of the material, we can use Hooke's Law, which states that the stress (σ) is directly proportional to the strain (ε) for a linear-elastic material.

Hooke's Law can be expressed as:
σ = E * ε

Where:
σ = stress (force per unit area)
E = modulus of elasticity (also known as Young's modulus)
ε = strain (change in length divided by original length)

In this case, we are given the following information:
Force (F) = 60 kN = 60,000 N
Original length (L) = 150 mm = 0.15 m
Change in length (ΔL) = 0.08 mm = 0.00008 m
Cross-sectional area (A) = 550 mm^3 = 550 * 10^-6 m^2 (since 1 mm^3 = 10^-6 m^3)

First, let's calculate the stress (σ):
σ = F / A
σ = 60,000 N / (550 * 10^-6 m^2)
σ ≈ 109,091,000 N/m^2

Next, let's calculate the strain (ε):
ε = ΔL / L
ε = 0.00008 m / 0.15 m
ε ≈ 0.000533

Now, we can rearrange Hooke's Law to solve for the modulus of elasticity (E):
E = σ / ε
E = 109,091,000 N/m^2 / 0.000533
E ≈ 204,500,000 N/m^2

Therefore, the modulus of elasticity of the material is approximately 204,500,000 N/m^2.