Answer
$d_{0} = 6.28 \times 10^{-3} m = 6.28 mm $
Work Step by Step
Given:
large tower to be supported by steel wire
load on each wire is 13,300 N
Required:
minimum wire diameter assuming factor of safety of 2.0 and yield strength of 860 MPa for steel.
Solution:
Using Equation 6.24, with N= 2.0:
$σ_{w} = \frac{σ_{y} }{2} = \frac{860 MPa }{2} = 430 MPa$
Using modified Equation 6.1 and solving for $d_{0}$:
$A_{0} = \frac{F}{σ_{w}} = π (\frac{d_{0}}{2})^{2} $
$d_{0} = \sqrt \frac{4F}{πσ_{w}} = \sqrt \frac{4 (13,300 N)}{π (430 \times 10^{6} N/m^{2})} = 6.28 \times 10^{-3} m = 6.28 mm $
