Answer
$l_{f} = 76.25 mm$
Work Step by Step
Given:
brass alloy- plastic deformation begins at 345 MPa stress
modulus of elasticity - 103 GPa = $103 \times 10^{3} MPa$
Required:
maximum length stretched without deformation if original length is 76 mm
Solution:
Using a combination of Equation 6.2 and 6.5, it follows:
$l_{f} = l_{0}(1 + \frac{σ}{E}) = (76 mm)(1 + \frac{345 MPa}{103 \times 10^{3} MPa} )= 76.25 mm$
