Answer
given,
Stress for plastic deformation = 267 MPa
modulus of elasticity = 115 GPa
cross sectional area = 377 mm²
a) maximum load (in N) that may be applied to a specimen
= σ x A
= 267 x 10⁶ x 377 x 10⁻⁶
= 100659 N
b) modulus of elasticity = stress/strain
115 x 10⁹ =[tex]\dfrac{267 \times 10^6}{\dfrac{\Delta l}{L}}[/tex]
L = 127 mm
115 x 10⁹ =[tex]\dfrac{267 \times 10^6}{\dfrac{\Delta l}{127}}[/tex]
[tex]\dfrac{\Delta l}{127}=\dfrac{267 \times 10^6}{115\times 10^9}[/tex]
Δ l = 0.295 mm
maximum length after the stretched = 127 mm + 0.295 mm
= 127.295 mm