Explanation:
The given data is as follows.
Load on each wire = 11,100 N
Factor of safety (N) = 2
[tex]\sigma_{yield}[/tex] = 1030 MPa (150,000 Psi)
Now,
[tex]\sigma_{working} = \frac{\sigma_{yield}}{2}[/tex]
= [tex]\frac{1030 MPa}{2}[/tex]
= 515 MPa (75,000 Psi)
So, we will calculate the diameter as follows.
[tex]\sigma_{w} = \frac{F}{A_{o}}[/tex]
or, [tex]A_{o} = \frac{F}{\sigma_{w}}[/tex]
[tex]\frac{\pi}{4} \times d^{2} = \frac{11,100}{515} \times 10^{6} N/m^{2}}[/tex]
[tex]d^{2}[/tex] = 27.45
d = 5.24 mm
Thus, we can conclude that the minimum required wire diameter is 5.24 mm.
