TB-Lecture19-Example-of-PT-Slab

TB-Lecture19-Example-of-PT-Slab - EGN-5439 The Design of...

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EGN EGN -5439 The Design of Tall Buildings 5439 The Design of Tall Buildings Lecture #19 Lecture #19 Example of a post-tensioned plate for a residential high-rise building © L. A. Prieto-Portar - 2008
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This example illustrates the design method presented in ACI 318-05 and IBC 2003 for the design of a simple two-way post-tensioned plate for a residential high-rise building. Step 1. Define the Loads. 1. Framing Dead Load = self-weight of the structure 2. Super-imposed Dead Load = 25 psf for partitions, MEP and miscellaneous 3. Live Load = 40 psf (residential) 4. Provide a 2 hour fire-rating Step 2. Define the Materials. 1. Normal weight concrete = 150 pcf 2. 28-day compressive strength f‘ c = 5,000 psi 3. Compressive strength when post-tensioned (typically after 24-hours) f’ ci = 3,000 psi 4. Mild reinforcing steel f y = 60,000 psi 5. Post-tensioning steel = un-bonded tendons, 0.5 inch-diameter, 7-wire strands, A = 0.153 in 2 f pu = 270 ksi 6. Estimated pre-stress losses = 15 ksi (as per ACI 318-05 Section 18.6) f se = 0.7 f pu = 0.7(270 ksi) - 15 ksi = 174 ksi (as per ACI 318-05 Section 18.5.1) P eff = A * f se = (0.153 in 2 )(174 ksi) = 26.6 kips/tendon
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Layout of a typical residential floor plate.
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Step 3. Determine the preliminary slab thickness h. Start with L / h = 45 where the longest span = 30 feet = 30 x 12 = 360 inches h = (360 inches) / 45 = 8.0 inches is the preliminary plate (slab) thickness . Step 4. Loads. DL = Self weight = (8 in)(150 pcf) = 100 psf SIDL = 25 psf LL o = 40 psf IBC-2003 Section 1607.9.1 allows for a LL reduction: For an exterior bay: A T = (25 feet)(27 feet) = 675 ft 2 K LL = 1 LL = 0.83 LL o = 33 psf For an interior bay: A T = (25 feet)(30 feet) = 750 ft 2 K LL = 1 LL = 0.80 LL o = 32 psf
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Step 5. Design the East-West interior frame. Use the Equivalent Frame Method of ACI 318-05 Section 13.7 (but excluding Sections 13.7.7.4-5). Total bay width between centerlines = 25 feet. Ignore the column stiffness in the equations for simplicity of hand calculations. No pattern loading required, since LL / DL = 40/125 < 3/4 (as per ACI 318-05 Section 13.7.6) Step 6. Calculate the section properties. The two-way plate must be designed as Class U (ACI 318-05 Section18.3.3). The gross cross- sectional properties are allowed (ACI 318-05 Section 18.3.4), A = bh = (300 in)(8 in) = 2,400 in 2 S = bh 2 /6 = (300 in)(8 in) 2 / 6 = 3,200 in 3
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Step 7. Set the design parameters. Allowable stresses: Class U (ACI 318-05 Section 18.3.3) Strength at time of jacking (ACI 318-05 Section 18.4.1) f’ ci = 3,000 psi Compression = 0.60 f' ci = 0.6(3,000 psi) = 1,800 psi Tension = 3 f' ci = 3 3,000 = 164 psi At service loads (ACI 318-05 Sections 18.4.2(a) and 18.3.3), f‘ c = 5,000 psi Compression = 0.45 f' c = 0.45(5,000 psi) = 2,250 psi Tension = 6 f' c = 6 5,000 = 424 psi Average pre-compression limits (ACI 318-05 Section 18.12.4) , P / A = 125 psi minimum to 300 psi maximum. Target load balances, use 60% - 80% of DL (self-weight) for plate (good approximation for
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This note was uploaded on 08/29/2011 for the course CES 4600 taught by Professor Staff during the Fall '08 term at FIU.

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TB-Lecture19-Example-of-PT-Slab - EGN-5439 The Design of...

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