Answer
$d_{0} =30 mm$
Work Step by Step
Given:
cylindrical rod of steel with $E = 207 GPa$ and yield strength of 310 MPa
load of 11,100 N
length of rod - 500 mm
Required:
rod diameter to allow elongation of 0.038 mm
Solution:
Using a combination of Equations 6.1, 6.2, and 6.5 and solving for $d_{0}$:
$σ = \frac{F}{A_{0}} = \frac{F}{π(\frac{d_{0}^{2}}{4})} = E \frac{Δl}{l_{0}}$
$d_{0} = \sqrt \frac{4l_{0}F}{πEΔl} =\sqrt \frac{(4)(500 \times 10^{-3} m(11,100 N)}{(π)(207 \times 10^{9} N/m^{2})(0.038 \times 10^{-3} m)} = 0.03 m = 30 mm$
