A solid steel shaft is to transmit a torque of 10kNm. If the shearing stress is not to exceed 45MPa, find the minimum diameter of the shaft.?

Question

Answer

To determine the minimum diameter of a solid steel shaft that can safely transmit a torque of 10 kNm without exceeding a shearing stress of 45 MPa, we can use the following formula:

1. Torque Formula
The relationship between torque (T), shearing stress (τ), and the diameter (d) of the shaft is given by the equation:

T = (π/16) × τ × d^3

Where:
- T = Torque in Nm
- τ = Shearing stress in MPa
- d = Diameter in m

2. Convert Units
First, we need to convert the torque from kNm to Nm:

T = 10 kNm = 10,000 Nm

Next, convert the shearing stress from MPa to Pa (1 MPa = 1,000,000 Pa):

τ = 45 MPa = 45 × 1,000,000 Pa = 45,000,000 Pa

3. Rearranging the Formula
Now, we can rearrange the torque formula to solve for d:

d^3 = (16 × T) / (π × τ)

4. Substituting Values
Now substitute the values of T and τ into the equation:

d^3 = (16 × 10,000) / (π × 45,000,000)

5. Calculation
Calculating the above expression:

d^3 = (160,000) / (141,372.6) ≈ 1.131 m^3

Now take the cube root to find d:

d ≈ (1.131)^(1/3) ≈ 0.105 m
d ≈ 105 mm

6. Conclusion
Thus, the minimum diameter of the shaft needed to safely transmit a torque of 10 kNm without exceeding a shearing stress of 45 MPa is approximately:

105 mm

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