A new mold has four cavities and a hot runner. Each part is round with a diameter of 6 inches and has a hole with a diameter of 0.500 inch in the center. With a tonnage factor of t... A new mold has four cavities and a hot runner. Each part is round with a diameter of 6 inches and has a hole with a diameter of 0.500 inch in the center. With a tonnage factor of three tons per square inch, what is the minimum required clamp force needed to support this mold if using a 400-ton machine? Read more

Steps to Solve

1. Calculate the area of the entire part The part is round with a diameter of 6 inches. The radius is half the diameter, so the radius is 3 inches. The area of a circle is $A = \pi r^2$. $$A = \pi * (3 \text{ inches})^2 = 9\pi \text{ inches}^2$$

2. Calculate the area of the hole The hole has a diameter of 0.500 inches, so the radius is 0.250 inches. $$A = \pi * (0.250 \text{ inches})^2 = 0.0625\pi \text{ inches}^2$$

3. Calculate the projected area of one part Subtract the area of the hole from the area of the entire part. $$9\pi - 0.0625\pi = 8.9375\pi \text{ inches}^2$$

4. Calculate the total projected area Since there are four cavities, multiply the projected area of one part by 4. $$4 * 8.9375\pi = 35.75\pi \text{ inches}^2$$

5. Calculate the clamp force With a tonnage factor of 3 tons per square inch, multiply the total projected area by 3. $$35.75\pi * 3 \approx 336.84 \text{ tons}$$