Question: Knowing that a force $P$ of magnitude 44 kN is required to punch a hole of diameter $d = 20$ mm in an aluminum sheet of thickness $t = 5$ mm, determine the average shearing stress in the aluminum at failure.
1. **State the problem:**
We need to find the average shearing stress in the aluminum sheet when a force $P = 44$ kN punches a hole of diameter $d = 20$ mm through a sheet of thickness $t = 5$ mm.
2. **Formula used:**
The average shearing stress $\tau$ is given by the formula:
$$\tau = \frac{P}{A}$$
where $A$ is the shear area.
3. **Calculate the shear area $A$:**
The shear area is the lateral surface area of the cylindrical hole punched, which is the circumference of the hole times the thickness:
$$A = \pi d t$$
4. **Substitute the values:**
$$A = \pi \times 20 \text{ mm} \times 5 \text{ mm} = 100\pi \text{ mm}^2$$
5. **Convert force to Newtons:**
$$P = 44 \text{ kN} = 44,000 \text{ N}$$
6. **Calculate the average shearing stress:**
$$\tau = \frac{44,000}{100\pi} = \frac{44,000}{314.159}$$
7. **Simplify the fraction:**
$$\tau = \frac{\cancel{44,000}}{\cancel{314.159}} \approx 140.0 \text{ MPa}$$
8. **Final answer:**
The average shearing stress in the aluminum at failure is approximately **140 MPa**.