A cylindrical bar of metal having a diameter of 19.9 mm and a length of 186 mm is deformed elastically in tension with a force of 42600 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. (b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answer:

a) 0.347 mm

b)  - 0.0126 mm

Explanation:

Diameter of metal bar = 19.9 mm

length = 186 mm = 0.186 m

Tension force = 42600 N

elastic modulus = 67.1 GPa

Poisson's ratio = 0.34

a) calculate the amount by which the specimen will elongate

first calculate the area of metal bar

Area = πd^2 / 4 = π/4 ( 19.9 )^2  = 3.11 * 10^-4 m^2

Elongation ( E ) = б / e = P/A * L / ΔL

and ΔL = PL / AE

hence the elongation ( ΔL) = [ (42600 * 0.186 ) / ( 3.4*10^-4 *  67.1 * 10^9 ) ]

                                            = 3.47 * 10^-4 m ≈ 0.347 mm

b) Calculate the change in diameter of specimen

μ = -( Δd / d) / ( ΔL/L )

0.34 = - (Δd / 19.9 ) / ( 0.347 / 186 )

∴ Δd = 0.34 * 0.00186 * 19.9  = - 0.0126 mm