6. A flat plate specimen of a soft metal alloy is to be tested by pulling on round pins inserted through the ends of the plate, as illustrated below

(a) Calculate the required diameter of the pins, given that the plate is 8.0 mm thick, the maximum tensile load is 4.0 kN and the maximum allowable average bearing stress is 50 MPa.
(b) What stresses are apparent in the pins?

To calculate the required diameter of the pins, we need to consider the maximum allowable average bearing stress and the applied load. Let's go step by step to find the answer:

(a) Calculate the required diameter of the pins:

1. Determine the area of the pins required to withstand the applied load.
- The applied load is 4.0 kN, which is equivalent to 4000 N.
- Since the load is being distributed over two pins, each pin will need to support half of the load, or 2000 N.

2. Calculate the maximum allowable average bearing stress.
- The maximum allowable average bearing stress is given as 50 MPa.

3. Use the formula to relate the bearing stress, load, and pin area.
- The formula is: Bearing stress = Load / Area

4. Rearrange the formula to solve for the pin area.
- Area = Load / Bearing stress

5. Substitute the values into the formula to calculate the required area.
- Area = 2000 N / 50 MPa = 2000 N / (50 * 10^6 N/m^2) = 0.00004 m^2

6. Since the pins are round, the area can be related to the diameter.
- Area = π * (diameter/2)^2

7. Rearrange the formula to solve for the diameter.
- diameter = 2 * √(Area / π) = 2 * √(0.00004 m^2 / π) = 2 * 0.004 m = 0.008 m = 8 mm

Therefore, the required diameter of the pins is 8 mm.

(b) To determine the stresses apparent in the pins, we need to consider the load distribution.

1. Calculate the load per pin.
- Since the load is evenly distributed among two pins, each pin will experience half of the total load, or 2000 N.

2. Calculate the apparent stress in the pins.
- The apparent stress can be obtained by dividing the load by the cross-sectional area of each pin.

3. The cross-sectional area of each pin can be obtained using the formula: Area = π * (diameter/2)^2.

4. Substituting the diameter of 8 mm into the area formula:
- Area = π * (0.008 m/2)^2 = π * (0.004 m)^2 = 0.0000503 m^2

5. Calculate the stresses in the pins:
- Stress = Load / Area = 2000 N / 0.0000503 m^2 = 39,713.8 N/m^2 or 39.7 MPa

Therefore, the stresses apparent in each pin are approximately 39.7 MPa.