Sudi Jibril
Oct 30, 2023
Guy?
A bar having a length of 150mm and cross-sectional area of 550mm^2 is subjected to an axial force of 60KN. if the bar stretches 0.08mm, determine the modulus of elasticity of a material. The material has linear-elastic behavior
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ChatGPT
Oct 30, 2023
To determine the modulus of elasticity of the material, we can use Hooke's Law, which states that the stress (σ) in a material is directly proportional to the strain (ε) it experiences.
The stress (σ) is given by the formula:
σ = F / A
Where F is the axial force applied to the bar (60 kN) and A is the cross-sectional area of the bar (550 mm^2).
The strain (ε) is given by the formula:
ε = ΔL / L
Where ΔL is the change in length of the bar (0.08 mm) and L is the original length of the bar (150 mm).
Using Hooke's Law, we can write:
σ = E * ε
Where E is the modulus of elasticity we want to determine.
Rearranging the equation, we get:
E = σ / ε
Substituting the values we have:
E = (60 kN / 550 mm^2) / (0.08 mm / 150 mm)
Converting kN to N and mm to m:
E = (60,000 N / 0.00055 m^2) / (0.00008 m / 0.15 m)
Simplifying the equation:
E = 109,090,909 N/m^2
Therefore, the modulus of elasticity of the material is approximately 109,090,909 N/m^2.
The stress (σ) is given by the formula:
σ = F / A
Where F is the axial force applied to the bar (60 kN) and A is the cross-sectional area of the bar (550 mm^2).
The strain (ε) is given by the formula:
ε = ΔL / L
Where ΔL is the change in length of the bar (0.08 mm) and L is the original length of the bar (150 mm).
Using Hooke's Law, we can write:
σ = E * ε
Where E is the modulus of elasticity we want to determine.
Rearranging the equation, we get:
E = σ / ε
Substituting the values we have:
E = (60 kN / 550 mm^2) / (0.08 mm / 150 mm)
Converting kN to N and mm to m:
E = (60,000 N / 0.00055 m^2) / (0.00008 m / 0.15 m)
Simplifying the equation:
E = 109,090,909 N/m^2
Therefore, the modulus of elasticity of the material is approximately 109,090,909 N/m^2.
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