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given the area to be sheared is 2 semi cirlces with diametre of 12 , and a rectangle 12 by 18  

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### Given:
- Thickness of the plate, \( t = 12 \, \text{mm} \)
- Ultimate shear strength, \( \tau_u = 320 \, \text{MPa} = 320 \times 10^6 \, \text{Pa} \)
- The area to be sheared includes:
  - 2 semicircles with a diameter of 12 mm
  - A rectangle of 12 mm by 18 mm

### (a) The Force Required to Punch the Hole

1. **Calculate the Perimeter of the Sheared Area:**

   - **Circumference of a full circle:**
     \[
     C = \pi \times d = \pi \times 12 \, \text{mm} = 37.7 \, \text{mm}
     \]
   - **Perimeter of 2 semicircles:**
     \[
     P_{\text{semi}} = 37.7 \, \text{mm}
     \]

   - **Perimeter of the rectangle (excluding the curved ends):**
     \[
     P_{\text{rect}} = 2 \times (12 \, \text{mm} + 18 \, \text{mm}) - 2 \times 12 \, \text{mm} = 36 \, \text{mm}
     \]

   - **Total Perimeter:**
     \[
     P_{\text{total}} = P_{\text{semi}} + P_{\text{rect}} = 37.7 \, \text{mm} + 36 \, \text{mm} = 73.7 \, \text{mm}
     \]

2. **Calculate the Area to be Sheared:**
   \[
   A_s = P_{\text{total}} \times t = 73.7 \, \text{mm} \times 12 \, \text{mm} = 884.4 \, \text{mm}^2
   \]

3. **Calculate the Force Required to Punch the Hole:**
   \[
   F = \tau_u \times A_s = 320 \times 10^6 \, \text{Pa} \times 884.4 \times 10^{-6} \, \text{m}^2
   \]
   \[
   F = 283.008 \, \text{kN}
   \]

So, the force required to punch the hole is **283.01 kN**.

### (b) The Maximum Compressive Stress in the Punch

To find the maximum compressive stress, we need the cross-sectional area of the punch.

1. **Cross-sectional area of the punch:**
   - The cross-sectional area will be the same as the area of the hole being punched out.

   - **Area of the rectangle:**
     \[
     A_{\text{rect}} = 12 \, \text{mm} \times 18 \, \text{mm} = 216 \, \text{mm}^2
     \]

   - **Area of the two semicircles:**
     \[
     A_{\text{semi}} = \frac{\pi \times (6 \, \text{mm})^2}{2} = 56.55 \, \text{mm}^2
     \]

   - **Total Area:**
     \[
     A_p = A_{\text{rect}} + A_{\text{semi}} = 216 \, \text{mm}^2 + 56.55 \, \text{mm}^2 = 272.55 \, \text{mm}^2
     \]

2. **Calculate the Maximum Compressive Stress:**
   \[
   \sigma_c = \frac{F}{A_p} = \frac{283.01 \times 10^3 \, \text{N}}{272.55 \times 10^{-6} \, \text{m}^2} = 1.039 \times 10^9 \, \text{Pa} = 1039 \, \text{MPa}
   \]

So, the maximum compressive stress in the punch is **1039 MPa**.

---

Would you like further clarification on any step or additional information?

Here are five related questions to extend your understanding:
1. How would changing the hole dimensions affect the punching force required?
2. What are the effects of using a material with a higher shear strength?
3. How do punch wear and tear influence the force required over time?
4. How does the compressive stress compare to the material limits of the punch tool?
5. What safety factors should be considered when designing such punching operations?

**Tip**: Always check if the dimensions are consistent with the units you are using to avoid calculation errors!

Learn how to calculate the shear force required and maximum compressive stress in punching operations. This problem involves calculating the force needed to punch out a shape consisting of two semicircles and a rectangle, with given dimensions and material properties. Suitable for advanced engineering and material science studies.

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