For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 * 106 psi).

(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2 (0.5 in. 2) without

plastic deformation?

(b) If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be stretched without

causing plastic deformation?

(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

(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