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2 years ago

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The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat

ions require that the diameter of the shaft be uniform from A to D and that the angle of twist between A and D not exceed 1.45°. Knowing that τmax ≤ 60 MPa and G = 77.2 GPa, determine the minimum diameter shaft that can be used. (Round the final answer to one decimal place.)

2 answers:

Tamiku [17]2 years ago

8 0

The image of this electric motor with its shaft is missing. I have attached it.

Answer:

Minimum Diameter = 40.8 mm

Explanation:

From the question,

Torque(T) = 800 N·m

φ = 1.45° = 1.45π/180 = 0.0253 rads

τ = 60 MPa = 60 x 10^(6) Pa

G = 77.2 GPa = 77.2 x 10^(9) Pa

From the image attached, length of shaft (L) = 0.4 + 0.6 + 0.3 = 1.3m

Now, the polar moment of inertia is calculated from,

J = πc⁴/2 where c is the radius of shaft

To solve this, we will find the diameter based on the angle of twist and also based on the shear stress and we will choose the smaller one.

Based on angle of twist;

Formula for angle twist is given as;

φ = TL/GJ

Now J = πc⁴/2

Thus, φ = 2TL/G(πc⁴)

c⁴ = 2TL/φGπ

c⁴ = [2 x 800 x 1.3]/(0.0253) x 77.2 x 10^(9) x π)

c⁴ = 0.00000033898

c = ∜0.00000033898

c = 0.024m

Now based on shear stress;

Formula for shear stress is given as;

τ = Tc/J

Putting πc⁴/2 for J, we have;

τ = 2T/πc³

c = ∛(2T/πτ)

c = ∛(2 x 800)/(π x 60 x 10^(6))

c = ∛0.00000848826

c = 0.0204m

So, comparing the two values of radius gotten, the one based on the shear stress is bigger and it's c = 0.0204m

Diameter = 2 x radius = 2 x 0.0204m = 0.0408m which is 40.8 mm

kodGreya [7K]2 years ago

3 0

Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.

Maximum Torque, ζ= τ × \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 × \frac{ π}{16} × d³

800= 11.78 × d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

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