A cylindrical metal specimen 10.7000 mm in diameter and 95.000 mm long is to be subjected to a tensile force of 6300 N; at this force level, the resulting deformation will be totally elastic. (a) If the final length must be less than 95.040 mm, which of the metals in Table 6.1 are suitable candidates? Why? (b) If, in addition, the diameter must be no greater than 10.698 mm while the tensile force of 6300 N is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why?

a) Y = 4.16 10¹⁰ N / m², b) B = 37.4 10¹⁰ Pa

Explanation:

a) For this exercise we must use the Young's modulus formula

Y = P / DL / L

We look for the magnitude salt system SI

d₁ = 10.7000 mm = 10.7000 10⁻³ m

l₁ = 95,000 mm = 95,000 10⁻³ m

l₂ = 95.040 mm = 95.040 10⁻³ m

d₂ = 10.698 mm = 10.698 10⁻³ m

The pressure is

P = F / A

A = π r²

P = F / pi r²

P = 6300 / π (10.7 10⁻³) ²

P = 17,5155 10⁶ Pa.

Change in length

ΔL = l₂-l₁

ΔL = (95.04000 - 95.000) 10⁻³

ΔL = 0.040 10⁻³ m

Let's calculate

Y = 17,515 10⁶ 95,000 10⁻³ / 0.040 10⁻³

Y = 4.16 10¹⁰ N / m²

This Young or elasticity module is very close to the medium density fiberboard or solid poly styrene module

b) If in addition to the length the diameter also changes, the whole body volume changes, so we can use the ratio of the volume module

B = - P / DV / V

Let's look for the initial volume of the cylinder

V₁ = π r₁² l₁

r₁ = d₁ / 2 = 10.7000 10⁻³ / 2

r₁ = 5.3500 10⁻³ m

r₂ = 10.698 / 2 10⁻³

r₂ = 5.349 10⁻³ m

V₁ = π (5.35 10⁻³) ² 95.00 10⁻³

V₁ = 8.5424 10⁻⁶ m³

Let's calculate the final volume

V₂ = π r₂² L₂

V₂ = π 5.349² 10⁻⁶ 95.040 10⁻³

V₂ = 8.54282 10⁻⁶ m³

Let's look for the volume change

ΔV = V₂ - V₁

ΔV = (8.54280 - 8.5424) 10⁻⁶

ΔV = 0.0004 10⁻⁶ m³

Let's calculate the Bulk module

B = 17,515 10⁶ 8.5424 10⁻⁶ / (0,0004 10⁻⁶)

B = 3.74 10¹¹ Pa

B = 37.4 10¹⁰ Pa

This volume module corresponds to some type of glass made with poly styrene