Answer:
a) 0.347 mm
b) - 0.0126 mm
Step-by-step 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