Answer :
Approximately 31.4159 Newtons of axial force would be required to extend the pin by 0.1 mm.
To calculate the axial force required to extend the pin by a certain amount, we can use Hooke's law, which states that the stress (σ) in a material is proportional to the strain (ε) it undergoes. In mathematical terms, this relationship is expressed as:
σ = E * ε
where σ is the stress, E is the Young's modulus, and ε is the strain.
In this case, we want to find the force (f) required to produce a specific strain of 0.1 mm (or 0.0001 m) in the pin. We know the diameter (d) of the pin is 2 mm (or 0.002 m), and its length (L) is 2 cm (or 0.02 m).
The strain can be calculated using the following formula:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
Given that ΔL = 0.1 mm (or 0.0001 m) and L = 0.02 m, we can calculate the strain:
ε = 0.0001 m / 0.02 m
= 0.005
Now, we can substitute this strain value into Hooke's law equation to find the stress:
σ = E * ε
= 2 GPa * 0.005
= 10 MPa
Finally, we need to convert the stress into force. The stress is defined as force (f) divided by the cross-sectional area (A) of the pin:
σ = f / A
Rearranging the equation, we have:
f = σ * A
The cross-sectional area of a cylindrical pin can be calculated using the formula:
A = π * (d/2)^2
Substituting the given values:
A = π * (0.002 m / 2)^2
= π * (0.001 m)^2
= 3.14159 * 0.000001 m^2
= 3.14159e-6 m^2
Now, we can calculate the force:
f = σ * A
= 10 MPa * 3.14159e-6 m^2
≈ 31.4159 N
Therefore, approximately 31.4159 Newtons of axial force would be required to extend the pin by 0.1 mm.
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