4. Determine the maximum torque that can be applied to a hollow circular steel shaft of 100-mm outside diameter and an 80-mm inside diameter without exceeding a shearing stress of 60 MPa or a twist of 0.5 deg/m. Use G-83 GPa.

To determine the maximum torque \( T \) that can be applied to a hollow circular steel shaft, we need to consider two conditions:

1. **Shearing stress** \( \tau \) should not exceed 60 MPa.
2. **Twist per unit length** \( \theta \) should not exceed 0.5 degrees per meter.

Given dataTo determine the maximum torque \( T \) that can be applied to a hollow circular steel shaft, we need to consider two conditions:

1. **Shearing stress** \( \tau \) should not exceed 60 MPa.
2. **Twist per unit length** \( \theta \) should not exceed 0.5 degrees per meter.

Given 
- Outside diameter \( D_o = 100 \) mm
- Inside diameter \( D_i = 80 \) mm
- Maximum shearing stress \( \tau_{\text{max}} = 60 \) MPa
- Maximum twist \( \theta_{\text{max}} = 0.5 \) degrees/m
- Shear modulus \( G = 83 \) GPa

### 1. Maximum Torque Based on Shearing Stress

The formula for the maximum shearing stress \( \tau_{\text{max}} \) in a hollow circular shaft is:

\[
\tau_{\text{max}} = \frac{T \cdot r_o}{J}
\]

where:
- \( T \) is the torque,
- \( r_o \) is the outer radius,
- \( J \) is the polar moment of inertia.

The polar moment of inertia \( J \) for a hollow circular shaft is given by:

\[
J = \frac{\pi \cdot (r_o^4 - r_i^4)}{2}
\]

where:
- \( r_o = \frac{D_o}{2} = 50 \) mm,
- \( r_i = \frac{D_i}{2} = 40 \) mm.

Substituting the values:

\[
J = \frac{\pi \cdot ((50 \, \text{mm})^4 - (40 \, \text{mm})^4)}{2}
\]

Let's calculate \( J \) first:

\[
J = \frac{\pi}{2} \times (50^4 - 40^4) \, \text{mm}^4
\]

\[
J = \frac{\pi}{2} \times (62500000 - 25600000) \, \text{mm}^4
\]

\[
J = \frac{\pi}{2} \times 36900000 \, \text{mm}^4
\]

\[
J \approx 579366.44 \, \text{mm}^4
\]

Now, substituting \( J \) into the equation for \( \tau_{\text{max}} \):

\[
T = \frac{\tau_{\text{max}} \cdot J}{r_o} = \frac{60 \times 10^6 \times 579366.44 \times 10^{-12}}{0.05} \, \text{N}\cdot\text{m}
\]

\[
T \approx 695.24 \, \text{N}\cdot\text{m}
\]

### 2. Maximum Torque Based on Twist

The angle of twist per unit length is given by:

\[
\theta = \frac{T \cdot L}{J \cdot G}
\]

Rearranging to solve for \( T \):

\[
T = \frac{\theta \cdot J \cdot G}{L}
\]

Convert the twist to radians:

\[
\theta = 0.5^\circ/m = \frac{0.5 \times \pi}{180} \, \text{rad/m} \approx 0.008727 \, \text{rad/m}
\]

Substituting the values:

\[
T = \frac{0.008727 \times 579366.44 \times 10^{-12} \times 83 \times 10^9}{1} \, \text{N}\cdot\text{m}
\]

\[
T \approx 419.72 \, \text{N}\cdot\text{m}
\]

### Conclusion

The maximum torque that can be applied to the shaft without exceeding either the shearing stress or the twist limit is determined by the smaller of the two calculated values:

\[
\boxed{T_{\text{max}} = 419.72 \, \text{N}\cdot\text{m}}
\]

This is the torque governed by the twist condition.

Would you like further details or explanations? Here are some related questions you might find helpful:

1. How does the polar moment of inertia affect the strength of the shaft?
2. What would happen if the torque exceeded the calculated maximum value?
3. Can this shaft be used in high-torque applications? Why or why not?
4. How would changing the material of the shaft affect the maximum torque?
5. What are the implications of reducing the shaft's inner diameter?

**Tip:** Always check both stress and deformation criteria when determining the maximum load a structure can handle.

Calculate the maximum torque that can be applied to a hollow circular steel shaft without exceeding shearing stress or twist limits. Detailed solution with formulas and data for shaft dimensions, material properties, and torque constraints.

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