Question

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is

Answer 1

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

Answer 2

Answer:

(a) The maximum load that may be applied to a specimen without causing plastic deformation should be less than 89,375 N

(b) The maximum length to which a specimen may be stretched without causing plastic deformation should be less than 0.115275 m

Explanation:

(a) Load = stress × area

Stress = 275 MPa = 275×10^6 = 2.75×10^8 Pa

Area = 325 mm^2 = 325 mm × (1 m/1000 mm)^2 = 3.25×10^-4 m^2

Load = 2.75×10^8 × 3.25×10^-4 = 89,375 N (This is the load at which plastic deformation begins)

Maximum load without plastic deformation should be less than 89,375 N

(b) Strain = Stress ÷ Young's modulus

Young's modulus = 115 GPa = 115×10^9 = 1.15×10^11 Pa

Strain = 2.75×10^8 ÷ 1.15×10^11 = 0.00239

Extension = strain × original length

Original length = 115 mm = 115/1000 = 0.115 m

Extension = 0.00239 × 0.115 = 0.000275 m

Maximum length = original length + extension = 0.115 + 0.000275 = 0.115275 m (This is the maximum length at which plastic deformation begins)

Maximum length without causing plastic deformation should be less than 0.115275 m