Respuesta :
The diameter d of shaft is 0.319m
The actual maximum shear stress of the shaft is 44.95N/m^2
The angle of twist of the 3-m length shaft is 0.01409 rad
The question is incomplete, here is the complete question:
A solid steel shaft 3 m long is transmitting 12 MW at 400 rev/min. The working conditions
to be satisfied by the shaft are:
• The shaft must not twist more than 0.015 radian on a length of 10 times the diameter, i.e., 10D
• The maximum shear stress must not exceed 70 MN/m2.
If the modulus of rigidity of steel is 60 GN/m2 determine:
(a) The diameter of the shaft required
(b) The actual maximum shear stress and the angle of twist of the 3-m length shaft.
Part a
The diameter of the shaft
We use Torque equation to find diameter d:
[tex]\frac{T}{J} =\frac{G\theta }{L}[/tex]
[tex]\frac{12*10^{6}*60 }{25600d^{4} } =\frac{60*10^{9}*0.015 }{10d}[/tex]
[tex]d^{3} =0.032414[/tex]
d=0.319m
Hence, diameter of shaft d=0.319m
Part b
Actual maximum shear stress Y, and the angle of twist of the 3m length shaft
[tex]Y=\frac{16T}{\pi d^{3} }[/tex]
[tex]Y=\frac{16*286478.90}{\pi *0.319^{3} }[/tex]
[tex]Y=44.95N/m^{2}[/tex]
Angle of the twist
Let angle of twist = [tex]\theta[/tex]
[tex]\theta=\frac{TL}{GJ}[/tex]
[tex]=\frac{286478.90*3}{60*10^{9}(\frac{\pi }{32}(0.019)^{9} }[/tex]
=0.01409rad
Angle of twist=0.01409 rad
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