Answer:46.05 mm
Explanation:
Given
[tex]Power=260\ hp\approx 260\times 746=193.96\ KW[/tex]
speed [tex]N=550\ rev/min[/tex]
allowable shear stress [tex](\tau )_{max}=15\ kpsi\approx 103.421\ MPa[/tex]
Power is given by
[tex]P=\frac{2\pi NT}{60}[/tex]
[tex]193.96=\frac{2\pi 550\times T}{60}[/tex]
[tex]T=3367.6\ N-m[/tex]
From Torsion Formula
[tex]\frac{T}{J}=\frac{\tau }{r}-----1[/tex]
where J=Polar section modulus
T=Torque
[tex]\tau [/tex]=shear stress
For square cross section
[tex]r=\frac{a}{2}[/tex]
where a=side of square
[tex]J=\frac{a^4}{6}[/tex]
Substituting the values in equation 1
[tex]\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}[/tex]
[tex]a=0.04605\ m[/tex]
[tex]a=46.05\ mm[/tex]
