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

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