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The A-36 hollow steel shaft is 2 m long and has an outer diameter of 40 mm. When it is rotating at 83 rad/s , it transmits 32 kW of power from the engine E to the generator G. The shear modulus of elasticity for A-36 steel is 75 GPaILIILI
Determine the smallest thickness of the shaft if the allowable shear stress is ?allow = 140 MPa and the shaft is restricted not to twist more than 0.05 rad. t=?

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beingteenowfmao

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Explanation:

${teks-lihat-gambar} beingteenowfmao
Lanuel

Based on the calculations, the smallest thickness of this shaft is equal to 0.00694 m or 69.4 mm.

Given the following data:

  • Length = 2 m.
  • Outer diameter = 40 mm to m = 0.04 m.
  • Angular speed = 83 rad/s.
  • Power transmitted = 32 kW.
  • Shear modulus of elasticity = 75 GPa.
  • Allowable shear stress = 140 MPa.
  • Allowable angular displacement = 0.05 rad.

How to calculate the smallest thickness?

First of all, we would determine the maximum torque applied on the shaft as follows:

T = P/ω

T = (32 × 10³)/83

T = 385.542 Nm.

Next, we would determine the inner radius by using the equation for shear stress of shaft:

[tex]\tau_{allowable}=\frac{T_C}{J_{BC}} \\\\\tau_{allowable}=\frac{T(\frac{d_o}{2} )}{\frac{\pi}{2}(0.02^4 - r_i^4) } \\\\140 \times 10^6 =\frac{385.542(\frac{0.04}{2} )}{\frac{\pi}{2}(0.02^4 - r_i^4) }\\\\[/tex]

Inner radius (ri) = 0.0188 meter.

Also, we would determine the inner radius by using an angle of twist limitation (0.05 rad):

[tex]\phi = \frac{TL}{JG} \\\\0.05 = \frac{385.542 \times 0.02}{\frac{\pi}{2}(0.02^4 - r_i^4) \times 75 \times 10^9}[/tex]

Inner radius (ri) = 0.01306 meter.

Now, we can calculate the smallest thickness of this shaft:

Smallest thickness = Outer radius - Inner radius

Smallest thickness = 0.02 - 0.01306

Smallest thickness, t = 0.00694 m or 69.4 mm.

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