What is the total weight of a structure that has a 6-inch outer diameter with a 5/8 inch wall thickness and is 4 feet long?

To determine the total weight of the structure, we first need to calculate its volume using its dimensions and then convert that volume into weight.

The outer diameter of the structure is 6 inches, which means it has a radius of 3 inches (since radius is half the diameter). The wall thickness is 5/8 inch, so the inner diameter can be found by subtracting twice the wall thickness from the outer diameter. This gives us an inner diameter of 6 - 2(5/8) = 6 - 1.25 = 4.75 inches, which gives an inner radius of 2.375 inches.

The formula for the volume of a cylinder is given by:

[

\text{Volume} = \pi h (R^2 - r^2)

]

where (h) is the height (or length) of the cylinder, (R) is the outer radius, and (r) is the inner radius.

In our case:

• Height (h = 4 \text{ feet} = 48 \text{ inches})

• Outer radius (R = 3 \text{ inches})

• Inner radius (r = 2.375 \text{