Answer
$F_{y} = 44,850 N$
Work Step by Step
Given:
brass alloy- plastic deformation begins at 345 MPa stress
modulus of elasticity - 103 GPa = $103 \times 10^{9} N/m^{2}$
Required:
maximum load to be applied to specimen with cross sectional area of $130 mm^{2}$ without plastic deformation ($F_{y}$)
Solution:
Using Equation 6.1, it follows:
$F_{y} = σ_{y}A_{0} = (345 \times 10^{6} N/m^{2})(130 \times 10^{-6} m^{2}) = 44,850 N$
