Answer to Question #128734 in Mechanics | Relativity for Ariyo Emmanuel

Question #128734
solid shaft of 10cm diameter transmits 74 kW at 150 rev/min. Calculate (a) the torque on the shaft, (b) the maximum shear stress developed, (c) the angle of twist in a length of 1.50m, and (d) the shear stress at a radius of 3cm. Take G = 8 MN/cm2.
1
Expert's answer
2020-08-17T09:03:27-0400

a)The torque on the shaft


"P=Tw"

Where P is the power transmitted, T is the torque, and ω is the angular velocity (units are in terms of rad/s).

We can get the angular velocity from the rotation speed of the shaft (150 rev/min) knowing that "1 rev=2\\pi rad"

Solving for the torque,

"74 kW= 74 (\\frac {kJ}{S}) =T *( 150\\frac{rev}{min}) (\\frac{2\\pi rad}{1 rev}) (\\frac{1 min}{60 sec})"


"T= 4.711 kJ"


b)The maximum shear stress developed

"T_{max} =\\frac {Tc}{J}"

Where "T_{max}" is the maximum shear stress, T is the torque applied, c is the radius of the circular shaft (half of the shaft diameter, 5 cm = 0.05 m), and J is the polar moment of inertia of the shaft.

J is calculated for a circular solid shaft via the following:

"J= \\frac{\\pi c4}{2}"

"T_{max} =\\frac{Tc}{J} =\\frac{2T}{\\pi(c^{3})} =\\frac{(2)(4.711 kJ)}{\\pi(0.05 m)^{3}}"


"T_{max} =239992.9 kPa"


c) The angle of twist in a length of 1.50 m

 "\\theta= \\frac{TL}{GJ}"


Where "\\theta" is the angle of twist, L is the length of the shaft, and G is the Modulus of Rigidity (provided as"G= 8mN\/cm^{2} =8*10^{10} Nm^{2}=80 GPa)"


With "T=4.711kJ =4711 Nm", solving for the angle o twist:


"J= \\frac{\\pi c^{4}}{2} =\\frac{\\pi(0.05 m)^{4}}{2} =9.817*10^{-6} m^{4}"


"\\theta=\\frac{TL}{GJ}=\\frac{(4711Nm)(1.50 m)}{(8*10^{10} \\frac{N}{m^{2})} (9.817*10^{-6}m^{4})}"


"\\theta=8.997*10^{-3} rad"


d)The shear stress at a radius of 3 cm

"T=\\frac{Tp}{J}"

Where p is the radial distance considered for the analysis. Solving,


"T=\\frac{TP}{J}=\\frac{(4.711kJ)(0.03 m)}{9.817*10^{-6}m{^4}}"


"T=14395.7 kPa"



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