For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}\) (50,000 psi), and the modulus of elasticity is \(103 \mathrm{GPa}\left(15.0 \times 10^{6}\right.\) psi). (a) What is the maximum load that can be applied to a specimen with a cross- sectional area of \(130 \mathrm{~mm}^{2}\left(0.2 \mathrm{in}{ }^{2}\right.\) ) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{~mm}\) (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?

The stress at which plastic deformation begins is given to be 345 MPa. This is the maximum stress that can be applied without causing plastic deformation. We have the cross-sectional area as 130 mm². To find the maximum load that can be applied without plastic deformation, let's use the formula for stress, which is given by: Stress = Load / Cross-sectional Area We can rearrange the formula to solve for the maximum load: Maximum Load = Stress × Cross-sectional Area Now let's plug in the values: Maximum Load = 345 MPa × 130 mm² Before performing the calculation, convert MPa to N/mm²: 1 MPa = 1 N/mm² So, Maximum Load = 345 N/mm² × 130 mm² Calculating the result, Maximum Load = 44850 N

To find the maximum length to which the specimen can be stretched without causing plastic deformation, we first need to determine the maximum strain (change in length per unit length) that can be applied. We know that the modulus of elasticity is 103 GPa, the maximum stress is 345 MPa, and the original length is 76 mm. The modulus of elasticity (E) is defined as the ratio of stress to strain: E = Stress / Strain Rearranging to find the strain: Strain = Stress / E First, convert the modulus of elasticity to the same units as stress (MPa): 103 GPa × (1000 MPa / 1 GPa) = 103000 MPa Now, we can find the strain: Strain = (345 MPa) / (103000 MPa) Strain ≈ 0.00335 Now, we will use the definition of strain: Strain = (Change in Length) / (Original Length) Rearranging to find the change in length: Change in Length = Strain × Original Length Change in Length ≈ 0.00335 × 76 mm Change in Length ≈ 0.2546 mm The maximum length the specimen can be stretched to without causing plastic deformation is: Maximum Length = Original Length + Change in Length Maximum Length = 76 mm + 0.2546 mm Maximum Length ≈ 76.2546 mm So, the answers to the exercise are (a) The maximum load that can be applied without plastic deformation is 44850 N. (b) The maximum length to which the specimen can be stretched without causing plastic deformation is approximately 76.2546 mm.