Aect480 lecture 7

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Aect480 lecture 7

  1. 1. Lecture 7 – Two-Way SlabsTwo-way slabs have tension reinforcing spanning in BOTH directions, and may takethe general form of one of the following: Types of Two-Way Slab Systems Lecture 7 – Page 1 of 13
  2. 2. The following Table may be used to determine minimum thickness of various two-way slabs based on deflection: Minimum Suggested Thickness “h” for Two-Way Slabs Two-Way Slab System: Minimum Thickness h:Flat plate Ln/30Flat plate with spandrel beams Ln/33Flat slab Ln/33Flat slab with spandrel beams Ln/36Two-way beam-supported slab Ln/33 Ln = clear distance in long directionFlat 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 heightsDirect Design Method of Flat Plates per ACI 318-02 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 2 of 13
  3. 3. The ACI 318 code allows a direct design method that can be used in most typical situations. However, the following limitations apply: 1. Must have 3 or more continuous spans in each direction. 2. 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 2.0. 3. 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 2 times the unfactored dead load (L/D < 2.0).Design Strips a) If L1 > L2: L2 L2 Column (typ.) Exterior Column Strip Interior Column Strip Interior Column Strip Middle Strip Middle Strip L1 L2/4 L2/4 L2/4 Lecture 7 – Page 3 of 13
  4. 4. b) If L2 > L1: L2 L2 Exterior Column Strip Interior Column Strip Interior Column Strip Middle Strip Middle Strip L1 L1/4 L1/4 L1/4 Design Moment Coefficients for Flat Plate Supported Directly by ColumnsSlab End Span Interior SpanMoments 1 2 3 4 5 Exterior Positive First Positive Interior Negative Interior Negative NegativeTotal 0.26Mo 0.52Mo 0.70Mo 0.35Mo 0.65MoMomentColumn 0.26Mo 0.31Mo 0.53Mo 0.21Mo 0.49MoStripMiddle 0 0.21Mo 0.17Mo 0.14Mo 0.16MoStrip Mo = Total factored moment per span End Span Interior Span 1 2 3 4 5 2 wu L2 LnMo = where Ln = clear span (face-to-face of cols.) in the direction of analysis 8 Lecture 7 – Page 4 of 13
  5. 5. Bar Placement per ACI 318-02 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. 6. Example 1GIVEN: A two-way flat plate for an office building is shown below. Use the following: • Column dimensions = 20” x 20” • Superimposed service floor Dead load = 32 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 thedesign strip in the direction shown. L2 = 16’-0” L2 = 16’-0” L2 = 16’-0” 20’-0”Ln 20’-0” 20’-0” L2/4 L2/4½ 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. 7. Step 1 – Determine slab thickness h: Ln Since it is a flat plate, from Table above, use h = 30 where Ln = clear span in direction of analysis = (20’-0” x 12”/ft) – 20” Column size = 220” = 18.33’ 220" h= 30 = 7.333” Use 8” thick slabStep 2 – Determine factored uniform load, wu on the slab: wu = 1.2D + 1.6L Slab weight = 1.2[(32 PSF) + (8/12)(150 PCF)] + 1.6[(75 PSF)] = 278.4 PSF = 0.28 KSFStep 3 – Check applicability of “Direct Design Method”: 1) Must have 3 or more continuous spans in each direction. YES 2) 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 2.0. 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 2 times the unfactored dead load (L/D < 2.0). YES Lecture 7 – Page 7 of 13
  8. 8. Step 4 – Determine total factored moment per span, Mo: 2 wu L2 Ln Mo = 8 (0.28 KSF )(16 )(18.33 ) 2 = 8 Mo = 188 KIP-FT Step 5 – Determine distribution of total factored moment into col. & middle strips: Design Moment Coefficients for Flat Plate Supported Directly by ColumnsSlab End Span Interior SpanMoments 1 2 3 4 5 Exterior Positive First Positive Interior Negative Interior Negative NegativeTotal 0.26Mo = 48.9 0.52Mo = 97.8 0.70Mo = 131.6 0.35Mo = 65.8 0.65Mo = 122.2MomentColumn 0.26Mo = 48.9 0.31Mo = 58.3 0.53Mo = 99.6 0.21Mo = 39.5 0.49Mo = 92.1StripMiddle 0 0.21Mo = 39.5 0.17Mo = 32.0 0.14Mo = 26.3 0.16Mo = 30.1Strip Mo = 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) = 586.8 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. 9. Mu 586,800 LB − IN =φbd 2 (0.9)(96" )(7" ) 2 = 138.6 PSIFrom Lecture 4 → Table 2:Use ρmin = 0.0033 As ρ= bdSolve for As: As = ρbd = (0.0033)(96”)(7”) = 2.22 in2 As Number of bars required = As _ per _ bar 2.22in 2 = 0.20in 2 _ per _#4 _ bar = 11.1 → Use 12 - #4 TOP bars Lecture 7 – Page 9 of 13
  10. 10. b) Column strip for region 2 : 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” Mu 699,600 LB − IN = φbd 2 (0.9)(96" )(7" ) 2 = 165.2 PSI From Lecture 4 → Table 2: Use ρ = 0.0033 As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars Lecture 7 – Page 10 of 13
  11. 11. c) Middle strip for region 2 : 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” Mu 474,000 LB − IN = φbd 2 (0.9)(96" )(7" ) 2 = 112.0 PSI From Lecture 4 → Table 2: Use ρ = 0.0033 As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars Use 6 - #4 Bottom bars at each ½ Middle Strip Lecture 7 – Page 11 of 13
  12. 12. Step 7 – Draw “Summary Sketch” plan view of bars: 16’-0” 16’-0” 16’-0” Col. strip for region 1 12 - #4 TOP bars½ Middle strip forregion 2 20’-0”6 - #4 BOTTOM bars Col. strip for region 2 12 - #4 BOTTOM bars 20’-0”8” Thickconcrete slab 20’-0” 4’-0” 4’-0” ½ Middle strip = 4’-0” ½ Middle strip = 4’-0” Col. strip 16’ – 0” Lecture 7 – Page 12 of 13
  13. 13. Example 2GIVEN: The two-way slab system from Example 1.REQUIRED: Design the steel tension bars for design strip shown (perpendicular tothose in Example 1). 16’-0” 16’-0” 16’-0” 20’-0” ½ Middle strip = 6’-0”20’-0” Col. strip = 8’-0” ½ Middle strip = 6’-0” 20’-0” 20’-0” Solution → Similar to the procedure shown in Example 1, except: • Re-check slab thickness to verify that 8” is still acceptable • Re-calculate “M0” • Using new value of M0, determine “Design Moment Coefficients” • Design tension steel based on these moment coefficients Lecture 7 – Page 13 of 13

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