 A conical pivot supports a load of 26 kN, the cone angle is 120° and the intensity of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 240 rpm and the coefficient of friction is 0.1. Find the power absorbed in friction. Assume uniform pressure.

### Given Data:

- Load \( W = 26 \, \text{kN} = 26 \times 10^3 \, \text{N} \)
- Cone angle \( 2\alpha = 120^\circ \), hence \( \alpha = 60^\circ \)
- Maximum normal pressure \( p = 0.3 \, \text{N/mm}^2 = 0.3 \times 10^6 \, \text{N/m}^2 \)
- External diameter \( D_{\text{ext}} = 2 \times D_{\text{int}} \)
- Rotational speed \( N = 240 \, \text{rpm} \)
- Coefficient of friction \( \mu = 0.1 \)
- Assumption: uniform pressure distribution

### Step 1: Relationship between External and Internal Radii
Let:
- \( r_{\text{int}} = r_1 \) (internal radius)
- \( r_{\text{ext}} = r_2 \) (external radius)

Since \( D_{\text{ext}} = 2 \times D_{\text{int}} \), we know:
\[
r_2 = 2r_1
\]

### Step 2: Load and Pressure Calculation
The load on the bearing surface \( W \) is given by the product of the pressure, area of the conical surface, and the angle adjustment:

\[
W = p \times \text{Area of the conical surface}
\]

The area of the conical surface (projected along the axis) is given by:
\[
A = \pi (r_2^2 - r_1^2) \sin \alpha
\]
Substituting \( \sin \alpha = \sin 60^\circ = \frac{\sqrt{3}}{2} \), we get:
\[
A = \pi (r_2^2 - r_1^2) \times \frac{\sqrt{3}}{2}
\]

The total load is:
\[
W = p \times A = 0.3 \times 10^6 \times \pi (r_2^2 - r_1^2) \times \frac{\sqrt{3}}{2}
\]
Substitute \( W = 26 \times 10^3 \, \text{N} \) and \( r_2 = 2r_1 \):
\[
26 \times 10^3 = 0.3 \times 10^6 \times \pi \times \frac{\sqrt{3}}{2} \times \left( (2r_1)^2 - r_1^2 \right)
\]
Simplifying:
\[
26 \times 10^3 = 0.3 \times 10^6 \times \pi \times \frac{\sqrt{3}}{2} \times (4r_1^2 - r_1^2)
\]
\[
26 \times 10^3 = 0.3 \times 10^6 \times \pi \times \frac{\sqrt{3}}{2} \times 3r_1^2
\]
\[
r_1^2 = \frac{26 \times 10^3}{0.3 \times 10^6 \times \pi \times \frac{\sqrt{3}}{2} \times 3}
\]
After solving for \( r_1 \):
\[
r_1 \approx 33.52 \, \text{mm}
\]
Thus:
\[
r_2 = 2r_1 \approx 67.04 \, \text{mm}
\]

### Step 3: Power Absorbed in Friction
The frictional force acting at the conical surface is given by:
\[
F_f = \mu \times W
\]
The frictional torque is:
\[
T_f = F_f \times r_{\text{avg}}
\]
where \( r_{\text{avg}} = \frac{r_1 + r_2}{2} = \frac{33.52 + 67.04}{2} \approx 50.28 \, \text{mm} = 0.05028 \, \text{m} \).

The frictional force is:
\[
F_f = \mu W = 0.1 \times 26 \times 10^3 = 2600 \, \text{N}
\]

So, the frictional torque is:
\[
T_f = 2600 \times 0.05028 \, \text{m} = 130.73 \, \text{Nm}
\]

The power absorbed in friction is:
\[
P = 2 \pi N T_f
\]
where \( N = \frac{240}{60} = 4 \, \text{rev/s} \).

Thus:
\[
P = 2 \pi \times 4 \times 130.73 = 3282.47 \, \text{W} = 3.28 \, \text{kW}
\]

### Final Results:
- Inner radius \( r_1 \approx 33.52 \, \text{mm} \)
- Outer radius \( r_2 \approx 67.04 \, \text{mm} \)
- Power absorbed in friction \( P \approx 3.28 \, \text{kW} \)

Would you like a detailed explanation on any part? Here are some related questions:

1. How is the area of the conical surface derived?
2. What are the units of power absorbed in friction and how are they converted?
3. Why is the coefficient of friction important in determining power losses?
4. How does the cone angle affect the bearing's load capacity?
5. What assumptions are typically made in uniform pressure bearing problems?

**Tip**: Always check the units consistency when solving mechanical problems involving loads and pressures.

A conical pivot supports a load of 26 kN, the cone angle is 120° and the intensity of normal pressure is not to exceed 0.3 N/mm². The external diameter is twice the internal diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 240 rpm and the coefficient of friction is 0.1. Find the power absorbed in friction. Assume uniform pressure.

Solve a conical pivot problem supporting a 26 kN load with a 120° cone angle and 0.3 N/mm² pressure. Learn how to calculate the outer and inner radii and the power absorbed due to friction at 240 rpm. The solution involves formulas for pressure-area relations, frictional torque, and power loss, suitable for college-level engineering students.

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