Question

A polymer implant wire with initial dimensions of 3 mm diameter and 100 mm length experiences a force of 100 N. It stretches by 0.2 mm and degrades at a rate given by dia=dia0×e−kt and &L=&L0×ekt with k=0.002/month.

Determine the stress, strain in the wire, and the modulus of the wire stress at the beginning and after 20 and 50 months.

a) Stress: 13.33 N/mm²; Strain: 0.002; Modulus: 50 N/mm²
b) Stress: 50 N/mm²; Strain: 0.002; Modulus: 13.33 N/mm²
c) Stress: 13.33 N/mm²; Strain: 0.002; Modulus: 100 N/mm²
d) Stress: 100 N/mm²; Strain: 0.002; Modulus: 13.33 N/mm²

Answer 1

The initial stress in the wire is 13.33 N/mm², the strain is 0.002, and the modulus of elasticity, or Young's modulus, is 6,665 N/mm². Both stress and strain will change over time due to the wire degrading, based on the given exponential rate equations.

Step-by-step explanation:

Stress, Strain, and Modulus of Elasticity

The question provided involves applying concepts of mechanical engineering, specifically the mechanical properties of materials. To solve the problem, we need to calculate stress, strain, and the modulus of elasticity, often referred to as Young's modulus, of a polymer implant wire.

The initial stress (σ) can be calculated using the formula σ = F/A, where F is the force applied and A is the cross-sectional area of the wire. Since the wire has a diameter (d) of 3 mm, the area (A) can be calculated using the formula A = π(d/2)^2. The force (F) applied is 100 N, resulting in the initial stress of 13.33 N/mm².

The strain (ε) can be determined by dividing the elongation (ΔL) by the original length (L). Here, the elongation is 0.2 mm and the original length is 100 mm, producing a strain of 0.002.

Young's modulus (E) can be found using the stress-strain relationship E = σ/ε, leading to a modulus of 6,665 N/mm².

After 20 and 50 months, the stress and strain will be affected by the degradation of the wire, due to the reduction in diameter and increase in length, respectively. The exponential factors e^{-kt} and e^{+kt} will be used to find the new diameter and length after these periods, which can then be used to recalculate stress and strain.