The weight of a steel rod can be calculated using the formula: weight = volume × density. The volume of a cylinder is given by the formula ( V = \pi r^2 h ), where ( r ) is the radius and ( h ) is the height. For an 8-foot long rod with a 3-inch diameter, the radius is 1.5 inches (0.125 feet). Using the density of steel, which is approximately 490 pounds per cubic foot, the rod weighs about 24.5 pounds.
The weight of a half-inch steel rod varies depending on its length and the specific type of steel. However, as a general estimate, a half-inch diameter steel rod weighs approximately 0.491 pounds per foot. To calculate the total weight, multiply the length of the rod in feet by this weight per foot. For example, a 10-foot length would weigh about 4.91 pounds.
To calculate the weight of a rolled steel with given dimensions, we need the density of the steel. Assuming a typical density of 7850 kg/m³, the weight can be calculated as follows: the volume of the rolled steel is (π * radius² * height) - (π * (radius - thickness)² * height). Substituting the values, the weight would be (π * (2 feet)² * 12 feet) - (π * (2 feet - 1/4 inch)² * 12 feet), which equals approximately 19,347 pounds.
8 "
The surface area of a sphere with a 24 inch diameter is 1809.6 square inches = 12.6 sq feet
If the radius is 18 feet - then the diameter is 36 feet. The circumference measurement is irrelevant !
You need to specify the wall thickness in order to calculate the weight.
A 56-inch diameter circle has an area of about 17.06 square feet.
A 72-inch diameter circle has an area of: 28.3 square feet.
A 48-inch diameter circle has an area of: 12.57 square feet.
To calculate the weight of a rolled steel with given dimensions, we need the density of the steel. Assuming a typical density of 7850 kg/m³, the weight can be calculated as follows: the volume of the rolled steel is (π * radius² * height) - (π * (radius - thickness)² * height). Substituting the values, the weight would be (π * (2 feet)² * 12 feet) - (π * (2 feet - 1/4 inch)² * 12 feet), which equals approximately 19,347 pounds.
8 "
the 25-inch-diameter
The weight of one foot of a three-quarter inch diameter steel bar will be approximately 1.42 pounds.
60 inch diameter = 2.5 ft radius.Area = pi*r2= 19.6 sq feet.
A 14-inch diameter by 20 foot long pipe has a volume of: 21.38 cubic feet.
The surface area of a sphere with a 24 inch diameter is 1809.6 square inches = 12.6 sq feet
The stakes should be made of rolled steel 1 inch in diameter. The should be 14-15 inches above the ground with a 3 inch lean. The stakes should be 40 feet apart.