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The design specifications of a 1.2-m-long solid transmission shaft require that the angle of twist of the shaft not exceed 48 when a torque of 750 N?m is applied. Determine the required diameter of the shaft, knowing that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77.2 GPa

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Olajidey

Answer:

c = 18.0569 mm  

Explanation:

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given that,

Applied Torque,T = 750 N.m

Length of shaft, L = 1.2 m

Modulus of Rigidity,G = 77.2 GPa

Allowable Stress, г = 90 MPa

Maximum Angle of twist  

∅ = 4°

∅ = 4π/180

∅ = 69.813 *10⁻³ rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅ = [tex]TL/G*\pi/2*c^4[/tex]

∅ =  [tex]2TL/G*\pi*c^4[/tex]

[tex]c=\sqrt[4]{2TL/\pi G }[/tex]∅

c = 18.0869 *10⁻³ rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*/2*c⁴)]*c

г =[2T/(J**c^⁴)]*c

c = 17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10⁻³ m  

c = 18.0569 mm  

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