Question

A circular aluminum tube 40 cm in length is subjected to a tensile load of 2kN. The outside and inside diameters of the tube are 4.2 cm and 4.0 cm respectively. Assume that Aluminum has an elastic modulus of 73.1 GPa.
a. What is the tensile stress applied to the tube? (10 points)
b. Determine the axial strain and increase in length of the tube due to the applied stress (15 points)
c. If a typical elastic modulus for an artery is 0.1 MPa, what tensile load will produce the same axial strain as that of the aluminum tube? (15 points)

Answer 1

Answer:

a) σ=15.527 MPa; b) ε=2.12*10^(-4); Δl=84.96*10^(-6) m; c) Fa=2.736*10^(-3) N

Explanation:

a) To find the tensile stress, we should divide applied force by the area. In the given case, area should be calculated using the following equation:

A=π*(do^2-di^2)/4, where do- outer radius, di- inner radius, π=3.14, then:

σ=F/A=4F/(π*(do^2-di^2))=15.527 MPa

b) To find the strain and elongation, we should use the Hooks's law:

σ=Eε , where E- elastic module of the material. Knowing stress, from the question a), we can find the strain:

ε=σ/E=2.12*10^(-4) -note, that stress was calculated in MPa and E is given in GPa! In addition, ε is a unit-less variable.

Knowing strain, we can calculate tube's elongation: Δl=l0*ε , where l0 is tube's initial length. As a result: Δl=l0*ε=84.96*10^(-6) m. You can convert given answer in the recommended units upon multiplication factors.

c) To find the applied force, we can use Hooke's law again:

σ=Eε, from where, we can add equation for the strain from section b):

Fa/A=Ea*F/(A*E), where Fa- force, applied to the artery and Ea- elastic module of an artery, F- force applied to the tube, E- elasticity of the aluminum. Solving this equation, we can get: Fa=F*Ea/E=2.736*10^(-3) N. Please, note, that elastic modulus for artery is in MPa (10^6) and for aluminum it is given in GPa (10^9).