Answer:
a) The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm
b) The change in diameter of the specimen is - 0.015 mm
Explanation:
Given the data in the question;
(a) The amount by which this specimen will elongate in the direction of the applied stress.
First we find the area of the cross section of the specimen
A = [tex]\frac{\pi }{4}[/tex] d²
our given diameter is 20.5 mm so we substitute
A = [tex]\frac{\pi }{4}[/tex] ( 20.5 mm )²
A = 330.06 mm²
Next, we find the change in length of the specimen using young's modulus formula
E = σ/∈
E = P/A × L/ΔL
ΔL = PL/AE
P is force ( 46300 N), L is length ( 201 mm ), A is area ( 330.06 mm² ) and E is elastic modulus (60.5 GPa) = 60.5 × 10⁹ N/m² = 60500 N/mm²
so we substitute
ΔL = (46300 N × 201 mm) / ( 330.06 mm² × 60500 N/mm² )
ΔL = 0.466 mm
Therefore, The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm
(b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Using the following relation for Poisson ratio
μ = - Δd/d / ΔL/L
given that Poisson's ratio of the metal is 0.33
so we substitute
0.33 = - Δd/20.5 / 0.466/201
0.33 = - Δd201 / 20.5 × 0.466
0.33 = - Δd201 / 9.143
0.33 × 9.143 = - Δd201
3.01719 = -Δd201
Δd = 3.01719 / - 201
Δd = - 0.015 mm
Therefore, The change in diameter of the specimen is - 0.015 mm
