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Construction Geometry. Cylinders Surface Area Volume. Cylinder. A cylinder is a solid figure with 2 congruent, parallel, circular bases. Campbells. Cylinders. Right cylinders have a vertical central axis while oblique cylinders do not. Application. circular base. Application.
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Construction Geometry Cylinders Surface Area Volume
Cylinder • A cylinder is a solid figure with 2 congruent, parallel, circular bases. Campbells
Cylinders • Right cylinders have a vertical central axis while oblique cylinders do not.
Application circular base
Application circular base circular base cylinder
Application • Cardboard tubes are used to form liquid concrete into solid cylinders.
Application • These cylinders can be used to support buildings and other structures.
Application • Concrete tubes come in all sizes. These are the most common sizes (8” to 12” in diameter).
Application • Larger tubes are used to support large structures such as bridges and skyscrapers.
Application • Reinforcement (rebar) is used to strengthen concrete pillars. frame of rebar
Application • The white sealant we see on bridge pillars is called parge. • Parge is a mixture of cement and waterproofing agents. parge
Application • Concrete cylinders are also used as parts for foundations.
Application • Here they are used as piers to support this building.
Application • To work correctly, piers must be spread evenly to support the weight of the structure.
Surface Area • The formula for the surface area of a cylinder is found on the Math Reference Sheet. • Surface Area = 2πr2 + 2πrh
Surface Area r • Surface Area = 2πr2 + 2πrh • 2πr2 = area of both circular bases. • 2πrh= area of rectangular side h r r 2πr h
Pi Circumference Review • The circumference is equal to: π * d (a little more than 3 diameters) or π * 2r (a little more than 6 radii). Pi ≈ 3.1416” d=1” 1” 1” 1”
Circumference Review • The circumference is the distance around a circle. • It’s the same length as the length of the rectangular face of the cylinder. • C = 2πr or πd
Circumference Review surface area h 2 π r
Practice #1 • Determine the surface area of the cylinder. • (2) ’s = 2πr2 = 2π(3)2 = 18πm2 • (1) = 2πrh= 2π(3)(15) = 90πm2 • Calculate the sum. • SA = 108π m2 • ≈ 339.3 m2 3m 15 m
Practice #2 • The tank needs to be painted. Calculate the surface area to determine the amount of paint needed. 10’ 8’
Practice #2 • Determine the surface area of the cylinder. • (2) ’s = 2πr2 = 2π(8)2 = 128πft2 • 1 = 2πrh= 2π(8)(10) = 160πft2 • Calculate the sum. • SA = 288π ft2 • ≈ 904.8 ft2 10’ 8’
Volume • The formula for the volume of a cylinder is found on the Math Reference Sheet. • V = πr2 h • πr2 = Area of base • h = Vertical height
Application • Carpenters often construct large tanks to hold oil and other liquids.
Application • This tank is actually a hollow cylinder designed to hold waste water.
Application • And the tank top is a poured concrete cylinder.
Practice #3 • How many cubic yards of concrete are needed to pour a concrete cylinder of this size? 8” thick 16 feet
Practice #3 • V = πr2h (d = 16’, so r = 8’) • Convert 8” to feet: ≈ .67’ • V ≈ π(82)(0.67) • V ≈ (42.88) π ft3 • V ≈ 134.7 ft3 • 1 cubic yard (yd3) = 27 ft3 • ≈ 4.98 yd3, so round up to 5 yd3 16’ 8”
Practice #4 • Determine the surface area and volume of this concrete tube. Diameter = 10” Height = 8’
Practice #4 • For the surface area: • d = 10” so r = 5”; • Convert feet to inches: h= 8’(12) = 96” • (2) ’s = 2πr2 = 2π(5)2 = 50πin2 • 1 = 2πrh = 2π(5)(96) = 960πin2 • Calculate the sum. • SA = 1010π in2 ≈ 3173 in2 Diameter = 10” Height = 8’
Practice #4 • V = πr2 h • d = 10” so r = 5”; h= 8’(12) = 96” • V = π(52)96 • V = 2400π • V ≈ 7539.8 in3 • If ft3 is more practical for the job: • 1 ft3 = 1728 in3 ≈ 4.4 ft3 Diameter = 10” Height = 8’
Practice & Assessment Materials • You are now ready for the practice problems for this lesson. • After completion and review, take the assessment for this lesson.