Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL = [tex]\dfrac{FL}{AE}[/tex] ....................1
put here value and we get L
L = [tex]\dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}[/tex]
solve it we get
L = 0.312752 m
L= 312.75 mm
