Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
Original Length = 235.6 mm
The maximum length of the specimen before deformation is; 235.6 mm
We are given;
Elastic modulus; E = 107 GPa = 107 × 10⁹ Pa
Tensile Load; P = 2000 N
original diameter; d = 3.7 mm = 0.0037 m
Allowable elongation; ΔL = 0.41 mm = 0.00041 m
σ = P/A
Where;
P = Tensile Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037)²/4 = 1.0752 × 10⁻⁵ m²
Thus;
σ = 2000/(1.0752 × 10⁻⁵)
σ = 186 × 10⁶ Pa
σ = 186 MPa
Formula for the strain is gotten from;
Elastic Modulus = Stress / Strain
E = σ/∈
Thus;
∈ = σ/E
∈ = 186 × 10⁶/(107 × 10⁹)
∈ = 1.74 × 10⁻³
∈ = ΔL/L
Where L is original length.
Thus;
L = ΔL/∈
L = 0.00041/(1.74 × 10⁻³)
L = 0.2356 m
L = 235.6 mm
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