Lecture 7 Two-Way Slabs
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1 Lecture 7 Two-Way Slabs Two-way slabs have tension reinforcing spanning in BOTH directions, and may take the general form of one of the following: Types of Two-Way Slab Systems Lecture 7 Page 1 of 13
2 The following Table may be used to determine minimum thickness of various twoway slabs based on deflection: Minimum Suggested Thickness h for Two-Way Slabs Two-Way Slab System: Minimum Thickness h: Flat plate L n /30 Flat plate with spandrel beams L n /33 Flat slab L n /33 Flat slab with spandrel beams L n /36 Two-way beam-supported slab L n /33 L n = clear distance in long direction Flat Plates Flat plates are the most common type of two-way slab system. It is commonly used in multi-story construction such as hotels, hospitals, offices and apartment buildings. It has several advantages: Easy formwork Simple bar placement Low floor-to-floor heights Direct Design Method of Flat Plates per ACI Two-way slabs are inherently difficult to analyze by conventional methods of statics because of the two-way bending occurring. Accurately determining the moments on a two-way slab is typically accomplished by finite element computer analysis. Computer analysis of two-way slab Lecture 7 Page of 13
3 The ACI 318 code allows a direct design method that can be used in most typical situations. However, the following limitations apply: Design Strips 1. Must have 3 or more continuous spans in each direction.. Slab panels must be rectangular with a ratio of the longer span to shorter span(measured as centerline-to-centerline of support) not greater than Successive span lengths in each direction must not differ by more than 1/3 of the longer span. 4. Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns. 5. Loads must be uniformly distributed, with the unfactored live load not more than times the unfactored dead load (L/D <.0). a) If L 1 > L : L L Column (typ.) Interior Column Strip Middle Strip Interior Column Strip Middle Strip Exterior Column Strip L 1 L /4 L /4 L /4 Lecture 7 Page 3 of 13
4 b) If L > L 1 : L L Interior Column Strip Middle Strip Interior Column Strip Middle Strip Exterior Column Strip L 1 L 1 /4 L 1 /4 L 1 /4 Design Moment Coefficients for Flat Plate Supported Directly by Columns Slab End Span Interior Span Moments Exterior Negative Positive First Interior Positive Interior Negative Negative Total 0.6M o 0.5M o 0.70M o 0.35M o 0.65M o Moment Column 0.6M o 0.31M o 0.53M o 0.1M o 0.49M o Strip Middle Strip 0 0.1M o 0.17M o 0.14M o 0.16M o M o = Total factored moment per span End Span Interior Span M o = w u L Ln 8 where L n = clear span (face-to-face of cols.) in the direction of analysis Lecture 7 Page 4 of 13
5 Bar Placement per ACI The actual quantity of bars required is determined by analysis (see Example below). However, usage of the Direct Design Method prescribes bar placement as shown below: Lecture 7 Page 5 of 13
6 Example 1 GIVEN: A two-way flat plate for an office building is shown below. Use the following: Column dimensions = 0 x 0 Superimposed service floor Dead load = 3 PSF (not including slab weight) Superimposed service floor Live load = 75 PSF Concrete f c = 4000 PSI #4 Grade 60 main tension bars Concrete cover = ¾ REQUIRED: Use the Direct Design Method to design the two-way slab for the design strip in the direction shown. L = 16-0 L = 16-0 L = L n L /4 L /4 ½ Middle strip = ½(16 Col. strip) ½ Middle strip Col. strip = ½(16 Col. strip) Design Strip = 16 Lecture 7 Page 6 of 13
7 Step 1 Determine slab thickness h: Ln Since it is a flat plate, from Table above, use h = 30 where L n = clear span in direction of analysis = (0-0 x 1 /ft) 0 Column size = 0 = h = 0" 30 = Use 8 thick slab Step Determine factored uniform load, w u on the slab: w u = 1.D + 1.6L Slab weight = 1.[(3 PSF) + (8/1)(150 PCF)] + 1.6[(75 PSF)] = 78.4 PSF = 0.8 KSF Step 3 Check applicability of Direct Design Method : 1) Must have 3 or more continuous spans in each direction. YES ) Slab panels must be rectangular with a ratio of the longer span to shorter span(measured as centerline-to-centerline of support) not greater than.0. YES 3) Successive span lengths in each direction must not differ by more than 1/3 of the longer span. YES 4) Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns. YES 5) Loads must be uniformly distributed, with the unfactored live load not more than times the unfactored dead load (L/D <.0). YES Lecture 7 Page 7 of 13
8 Step 4 Determine total factored moment per span, M o : M o = = w u L Ln 8 (0.8KSF)(16')(18.33') 8 M o = 188 KIP-FT Step 5 Determine distribution of total factored moment into col. & middle strips: Design Moment Coefficients for Flat Plate Supported Directly by Columns Slab End Span Interior Span Moments Exterior Negative Positive First Interior Positive Interior Negative Negative Total 0.6M o = M o = M o = M o = M o = 1. Moment Column 0.6M o = M o = M o = M o = M o = 9.1 Strip Middle Strip 0 0.1M o = M o = M o = M o = 30.1 M o = Total factored moment per span = 188 KIP-FT Step 6 Determine tension steel bars for col. & middle strips: a) Column strip for region 1 : Factored NEGATIVE moment = 48.9 KIP-FT (see Table above) = KIP-IN = 586,800 LB-IN b = 96 8 d d = 8 conc. cover ½(bar dia.) = 8 ¾ ½(4/8 ) = 7 Lecture 7 Page 8 of 13
9 M u φbd 586,800LB IN = (0.9)(96")(7") = PSI From Lecture 4 Table : Use ρ min = A ρ = s bd Solve for A s : A s = ρbd = (0.0033)(96 )(7 ) =. in Number of bars required = = A A _ per _ bar s 0.0in s.in _ per _#4 _ bar = 11.1 Use 1 - #4 TOP bars Lecture 7 Page 9 of 13
10 b) Column strip for region : Factored POSITIVE moment = 58.3 KIP-FT (see Table above) = 699,600 LB-IN b = 96 8 d d = 8 conc. cover ½(bar dia.) = 8 ¾ ½(4/8 ) = 7 M u φbd 699,600LB IN = (0.9)(96")(7") = 165. PSI From Lecture 4 Table : Use ρ = A s =. in (see calcs. above) Use 1 - #4 BOTTOM bars Lecture 7 Page 10 of 13
11 c) Middle strip for region : Factored POSITIVE moment = 39.5 KIP-FT (see Table above) = 474,000 LB-IN b = 96 8 d d = 8 conc. cover ½(bar dia.) = 8 ¾ ½(4/8 ) = 7 M u φbd = 474,000LB IN (0.9)(96")(7") = 11.0 PSI From Lecture 4 Table : Use ρ = A s =. in (see calcs. above) Use 1 - #4 BOTTOM bars Use 6 - #4 Bottom bars at each ½ Middle Strip Lecture 7 Page 11 of 13
12 Step 7 Draw Summary Sketch plan view of bars: Col. strip for region #4 TOP bars ½ Middle strip for region 6 - #4 BOTTOM bars 0-0 Col. strip for region 1 - #4 BOTTOM bars Thick concrete slab ½ Middle strip = 4-0 Col. strip 16 0 ½ Middle strip = 4-0 Lecture 7 Page 1 of 13
13 Example GIVEN: The two-way slab system from Example 1. REQUIRED: Design the steel tension bars for design strip shown (perpendicular to those in Example 1) ½ Middle strip = Col. strip = 8-0 ½ Middle strip = Solution Similar to the procedure shown in Example 1, except: Re-check slab thickness to verify that 8 is still acceptable Re-calculate M 0 Using new value of M 0, determine Design Moment Coefficients Design tension steel based on these moment coefficients Lecture 7 Page 13 of 13
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