Let's break down the problem and solve it step-by-step. **Given Data:** - Load on bearing (\(W\)): \(20\) kN \(= 20,000\) N - Shaft speed (\(N\)): \(1000\) rpm - Maximum bearing pressure (\(P_{max}\)): \(1.6\) N/mm² - Type: Full journal bearing (i.e., 360°) You are to **design the bearing**, i.e., find suitable diameter (\(d\)) and length (\(L\)), based on the pressure criterion. --- ## Step 1: **Bearing Pressure Formula** The **bearing pressure** is defined as: \[ P = \frac{W}{L \times d} \] where: - \(W\) = Load (N) - \(L\) = Length of bearing (mm) - \(d\) = Diameter of journal (mm) --- ## Step 2: **Pressure Constraint** Given: \[ P_{max} = 1.6 \text{ N/mm}^2 \] So, \[ \frac{W}{L \times d} \leq 1.6 \] \[ L \times d \geq \frac{W}{1.6} \] Plug in the value: \[ L \times d \geq \frac{20,000}{1.6} = 12,500 \text{ mm}^2 \] --- ## Step 3: **Assume Ratio L/d** It's common to assume a length-to-diameter ratio \((L/d)\) for journal bearings. For full journal bearings, typical values are between 0.5 to 2. Let's assume \(L = d\) (i.e., \(L/d = 1\)), which is a common starting point for design. Thus: \[ L = d \] So, \[ L \times d = d^2 \] \[ d^2 = 12,500 \] \[ d = \sqrt{12,500} = 111.8 \text{ mm} \] So, \[ d = 112 \text{ mm (rounded up)} \] \[ L = 112 \text{ mm} \] --- ## Step 4: **Check the Bearing Pressure** Let's check if our values satisfy the pressure limit: \[ P = \frac{W}{L \times d} = \frac{20,000}{112 \times 112} = \frac{20,000}{12,544} = 1.595 \text{ N/mm}^2 \] Which is just below the maximum allowed pressure of 1.6 N/mm². --- ## Step 5: **Final Answer** ### **Design Dimensions:** - **Diameter of journal (\(d\))**: **112 mm** - **Length of bearing (\(L\))**: **112 mm** **(assuming \(L = d\) for initial design; can be adjusted based on further design constraints such as heat dissipation, lubrication, etc.)** --- ## **Summary Table:** | Parameter | Value | |---------------------|----------| | Load (\(W\)) | 20,000 N | | Max Pressure (\(P\))| 1.6 N/mm²| | Shaft speed (\(N\)) | 1000 rpm | | Diameter (\(d\)) | 112 mm | | Length (\(L\)) | 112 mm | --- ### **Step-by-step summary:** 1. Used pressure formula to relate load, length and diameter. 2. Set \(L \times d \geq 12,500 \text{ mm}^2\). 3. Assumed \(L = d\) for initial design, giving \(d = 112 \text{ mm}\). 4. Checked pressure is within limit. --- **Final Answer:** \[ \boxed{ \text{The required diameter and length for the full journal bearing are:} \\ \text{Diameter } (d) = 112\,\text{mm} \\ \text{Length } (L) = 112\,\text{mm} } \] *(If you want a different L/d ratio, let me know!)*