Answer:
(a) The maximum load that may be applied to a specimen with a cross-sectional area of 325 mm² without plastic deformation = 89,375 Newton.
(b) The maximum length to which the specimen may be stretched without causing plastic deformation = 115.275 mm.
Explanation:
(a) Given that the applied stress at which plastic deformation begins = 275 MPa = 275,000,000 N/m²,
and the cross-sectional area of the specimen = 325 mm² = 0.000325 m²,
the maximum load can be calculated from the formula: maximum load = applied stress × area of specimen = 275,000,000 N/m² × 0.000325 m² = 89,375 Newton.
(b) To calculate the maximum length, we will use the formula:
L = L1 (1 + [stress at which plastic deformation begins ÷ modulus of elasticity])
where L = maximum length to which sample may be stretched/deformed without causing plastic deformation.
L1 = original length of specimen = 115 mm
stress at which plastic deformation begins = 275,000,000 N/m²
modulus of elasticity = 115 GPa = 115,000,000,000 N/m²
Therefore, L = 115 mm (1 + [275,000,000 ÷ 115,000,000,000]) = 115.275 mm
