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 !
A circle with an 8-inch diameter has a perimeter of 25.1 inches.
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 weight of one foot of a three-quarter inch diameter steel bar will be approximately 1.42 pounds.
the 25-inch-diameter
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 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.
The surface area of a sphere with a 24 inch diameter is 1809.6 square inches = 12.6 sq feet