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DESIGN OF REINFORCED CONCRETE STRUCTURES Desig n of Tow Way Design Way Slabs Sla bs Direct Design Method (DDM)
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DESIGN OF TWO-WAY FLOOR SLAB SYSTEM
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One-way and two-way slab Direct Design Method
Comparison of One-way and Two-way slab behavior 14
One-way slabs carry load in one direction.
Two-way slabs carry load in two directions.
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Comparison of One-way and Two-way slab behavior 15
One-way and twoway slab action carry load in two directions One-way slabs: Generally, long side/short side > 2
Comparison of One-way and Two-way slab behavior 16
Flat Plate
Waffle slab
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Comparison of One-way and Two-way slab behavior 17
Flat slab
Two-way slab with beams
Basic Steps in Two-way Slab Design 18
1. Choose layout and type of slab. 2. Choose slab thickness to control deflection. 3. Check if thickness is adequate to resist shear. 4. Choose Design method A. Equivalent Frame Method- use elastic frame analysis to compute positive and negative moments B. Direct Design Method - uses coefficients to compute positive and negative slab moments
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Basic Steps in Two-way Slab Design 19
5. Divide into column and middle strips … 6. Calculate positive and negative slab moments. 7. Determine distribution of moments across the width of the slab. 8. Assign a portion of moment to beams, if present. 9. Design reinforcement for moments (steps 5 and 6). 10. Distribute steel. 10. Repeat steps in the other direction.
Minimum Slab Thickness for two-way construction 20
Slabs without interior beams spanning between supports and ratio of long span to short span <2
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Minimum Slab Thickness for two-way construction 21
Slabs without drop panels meeting, t min = 5 in Slabs with drop panels meeting, t min = 4 in
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Minimum Slab Thickness for two-way construction 23
Maximum Spacing of Reinforcement At points of max. +/- M:
s 2t ACI 13.3.2 and s 18 in. ACI 7.12.3
Max. and Min Reinforcement Requirements
As min
As T &S
As max
from ACI 7.12 ACI 13.3.1
0.75 As bal
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DDM Definitions 27
DDM Definitions 28
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Effective beam sections
Shear in 2-way Slabs 30
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Definition of Beam-to-Slab Stiffness Ratio, 35
Accounts for stiffness effect of beams located along slab edge reduces deflections of panel adjacent to beams.
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flexural stiffness of beam flexural stiffness of slab
Beam-to-Slab Stiffness Ratio, 36
a
4E cb I b / l 4E cs I s / l
4E cb I b 4E cs I s
E cb
Modulus of elasticity of beam concrete
E sb
Modulus of elasticity of slab concrete
I b
Moment of inertia of uncracked beam
Is
Moment of inertia of uncracked slab
With width bounded laterally by centerline of adjacent panels on each side of the beam.
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Beam and Slab Sections for calculation of 38
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Beam and Slab Sections for calculation of 39
Beam and Slab Sections for calculation of 40
Definition of beam cross-section Charts may be used to calculate a Fig. 13-21
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Distribution of Moments 41
Total static Moment, M o
M 0
wu l 2l n2
ACI 13 - 3
8 where
wu
factored
l 2
l n
clear
load per unit area
transverse width of the strip span between columns
for circular columns, calc. l n using h 0.886dc
Column Strips and Middle Strips 42
Moments vary across width of slab panel Design moments are averaged over: 1. the width of column strips over the columns & 2. The middle strips between column strips.
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Critical Sections
Negative and positive design moments 46
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Distribution of M0 48
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Positive and Negative Moments in Panels 50
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Factored Moment in Column Strip 51
Ratio of flexural stiffness of beam to stiffness of slab in direction l 1. a1 =
b t = Ratio
of torsional stiffness of edge beam to flexural stiffness of slab (width = to beam length)
Factored Moment in Column Strip 52
a1 Ratio of flexural
stiffness of beam to stiffness of slab in direction l1.
bt Ratio of torsional stiffness of edge beam to flexural stiffness of slab(width= to beam length)
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Example: Using the ACI Code, determine the required thickness for slabs in Panels 3 and 2 . Edge beams are used around the building perimeter.(300 mm wide x 200mm drop), f y =414 Mpa (G60)
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Solution 54
For interior Panels 3: ln = 6 - 0.4 = 5.6 m, h = ln/33 = 5.6/33 = 0.17 m) > 125 mm
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h = 7 in
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Flat plate without edge beams
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Example 57
Design an interior flat plate… LL =80 psf = 3.83 kN/m2, DL +Own wt. =110psf = 5.27 kN/m2, fy = 60ksi = 414 MPa, f’c = 3 ksi = 21 MPa Column height = 12ft =3.6m Slab h same as before the required h = 7 in =0.17m (Use: h = 7.5 “ = 0.19m 1 Psf = 47.9 N/m2 Wu = 1.2*5.27+1.6*3.8 3= 13 kN/m2
Shear check for h 58
d = 6.25*25.4 = 159 mm say 0.16m Wu = 13.0 kPa One Way shear Ln1 = 6.1- 0.4 = 5.7 m Ln1@d = 0.5*5.7-0.16 =2.69 m Vu = 13 *1* (2.69) = 35 kN < F Vc F Vc = 0.75*0.16 * 210.5 * 160 = 88 kN OK
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bo = 2 (0.4+0.16+0.3+0.16) = 2.04 m
A = 6.1*4.88- (0.56*.46) = 29.8 m2 Vu2 = 13*(29.8) =387 kN F Vc = 0.75*.33*(21)0.5*2.04*160 = 387 kN > 355 kN OK
Mo = ? (in short and long directions 60
Mol =13*4.88*5.72/8 = 245 kN.m Mos =13* 6.1 *4.582/8 = 208 kN.m
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Interior -Ve, Exterior – Ve and +Ve moments in Long direction
Moment Distribution in long direction (between col. And middle strips
Column Strip M 0l wu .l 2
l n21 8
181.2k . ft
Middle strip
(0.26)16 * (20 16 / 12) 2 / 8
234kN .m
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Moment Distribution in long direction: col. And middle strips
M 0l wu .l 2
l n21 8
0.26 *16 * (20 16 / 12) 2 / 8 181.2k . ft 245kN .m
M-ext = 0.65 M0 =75% for col. Strip + 25% for Middle strip M+ = 0.35 M0 =60% for col. Strip + 40% for Middle strip 66
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Similarly the distribution in the short direction
l n21 M 0 s wu .l 2 8
(0.26) * 20 * (16 12 / 12) 2 / 8 146.2k . ft
Summary and design (short span) 68
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Slab with Interior beams Steps: 1.Find slab thickness (Eqns). 2.Find: Wu = 1.2*WD+1.6*WL, and Find Mol, Mos 3.Distribute to –ve int Mom, -ve ext mom, +Ve Mom. 4.Distribute to the moments in step 3 into beam and column strip moments 5.Distribute column Strip moments into Beam and slab moments 6.Find As ….. And distribue steel
Beam and slab contributions 70
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Slabs with Interior Beams:
Thickness in SI units:
USC Units:
b= l n long / l n shor
Example: Slabs with interior beams 72
Given:
2-Way Slab with beams as shown, h = 7”, F’c = 3 ksi, fy = 60 ksi Req’d: Check the ACI requirements for int panel
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a2 and am 74
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Minimum Slab thickness: 75
Exmple: Slabs With Beams (DDM) Determine the slab thickness and the +ve and the – ve moments required for the design of the exterior panel of the shown slab. LL = 120psf, DL = 100 psf (including own weight . 15 ”x15” and 12’ long. The slabs are supported by beams. F’c = 3ksi, fy = 60ksi.
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Check the code limitations 1. more than 3 spans 2. equal spans 3. no offsets 4. rectangular shape with long/short spans < 2 5. ….
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Direction of the 18’ slab width
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Slab Thickness:
Moments/ Interior Panel:
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Interpolation: Col strip will resist from M-int =0.75 – (0.75-0.45)*(1.22-1) = 0.68 of 0.65 M0 = - 0.68*065*241 = -107 k.ft (for col strip + beam) As the beam stiffness 85% of this value will be resisted by the beam = (0.85* 107)= 90.95 k.ft and 16.05 k.ft should be resisted by remaining col. strip
similarly:
0.68(+84)= 57.1 k.ft (48 For beam and 9.1 for slab)
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Short span on the edge beam
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18ft strip
107 +50 = 157 k.ft +57+ 30 =87 k.ft -57 -27 +31+16 k.ft
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