Determine the new diameter (in mm) of a cylindrical Ti-6Al-4V alloy specimen tha

Question

Determine the new diameter (in mm) of a cylindrical Ti-6Al-4V alloy specimen that has an intial length of 0.217 m and an initial diameter of 79.656 mm under a compressive stress of 100.1 ksi. The elastic modulus and Poisson's ratio for brass are 16.5 x 106 psi and 0.34 respectively. Answer:79.820

A tensile test is performed on a cylindrical specimen originally 1.6 cm in diameter and 11.5 cm in length. Calculate the tensile stress (in MPa) if a force of 458.4 N is applied along the length of the specimen. Answer: 2.28

Explanation / Answer

as we know, increase in diameter=poisson's ration*(compressive stress/elastic modulous)*original dimater

note that here compressive stress is in units of ksi while elastic modulus is in psi

hence using 1 ksi=1000 psi, elastic modulus=16500 ksi

hence change in dimater=0.34*(100.1/16500)*79.656=0.1643 mm

hence new diamter=79.656+0.1643=79.8203 mm

part b:

tensile stress=force/cross sectional area

here cross sectional area=pi*(diameter/2)^2

=pi*(1.6*0.01/2)^2=2.01062 *10^(-4) m^2

then tensile stress=force/area

=458.4/(2.01062*10^(-4)) Pa=2.2798 MPa(rounding off to 2 digits after decimal will give the required result of 2.28 MPa)

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