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We are tasked with determining the smallest diameters \( d_1 \), \( d_2 \), and \( d_3 \) in the assembly of disks under a load of 140 kN, with specific allowable stresses provided.

Here's how we'll approach this:

1. **For \( d_1 \)**:  
   \( d_1 \) is the diameter of the top disk (A). The bearing stress on the disk \( \sigma_b \) is defined as:

   \[
   \sigma_b = \frac{P}{A} = \frac{P}{\frac{\pi}{4}d_1^2}
   \]

   Given the allowable bearing stress \( \sigma_{\text{allow}} = 350 \, \text{MPa} \) and the load \( P = 140 \, \text{kN} = 140,000 \, \text{N} \), we solve for \( d_1 \).

2. **For \( d_2 \)**:  
   \( d_2 \) is the diameter of the disk in the support space (B). The stress here is also a bearing stress, but the calculation is based on the area that supports the disk. We will use the same method as for \( d_1 \), considering the area of the disk B.

3. **For \( d_3 \)**:  
   \( d_3 \) is the diameter of the hole in the bottom disk (C), and it is affected by the shear stress \( \tau_{\text{allow}} = 125 \, \text{MPa} \). The relation for shear stress is:

   \[
   \tau = \frac{P}{A} = \frac{P}{\pi d_3 t}
   \]

   Here, \( t = 10 \, \text{mm} \) (the thickness of the disk) and \( P = 140 \, \text{kN} \), so we can solve for \( d_3 \).

Let's start by calculating these diameters step by step. I will proceed with the calculations now.The diameters of the disks are calculated as follows:

- The smallest diameter \( d_1 \) of the top disk A: **22.57 mm**
- The diameter \( d_2 \) within the support space (disk B): **22.57 mm** (same as \( d_1 \))
- The diameter \( d_3 \) of the hole in the bottom disk C: **188.81 mm**

These diameters ensure that the assembly can safely support the 140 kN load while respecting the given allowable bearing and shear stresses.

Would you like more details on any of these steps or explanations? Here are a few related questions to deepen your understanding:

1. What is the relationship between load, area, and stress in a material?
2. How does the thickness of the material influence the shear stress calculation for \( d_3 \)?
3. Why are \( d_1 \) and \( d_2 \) the same in this problem?
4. What happens if the actual stress exceeds the allowable stress?
5. How would changing the material properties (allowable stresses) affect the diameters?

**Tip:** Always double-check the units when performing calculations involving stress, as stresses are typically given in MPa (MegaPascals), which is \( \text{N}/\text{mm}^2 \).

The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is 350 MPa, and the allowable shear stress is 125 MPa.

Determine the diameters of three disks supporting a 140 kN load while considering the allowable bearing and shear stresses. Includes step-by-step calculations for d1, d2, and d3 based on the given stress limits.

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