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10/1/2014
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تصميم المنشآت الخرسانية المسلحة
DESIGN OF REINFORCED
CONCRETE STRUCTURES
Design of Tow Way Slabs
Direct Design Method (DDM)
2
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3
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DESIGN OF TWO-WAY FLOOR
SLAB SYSTEM
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7
8
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9
10
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11
12
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One-way and two-way
slab
Direct Design Method
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Comparison of One-way and Two-way slab
behavior
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
One-way
and two-
way slab
action
carry
load in
two
directions
One-way slabs: Generally, long side/short side > 2
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Comparison of One-way and
Two-way slab behavior
Flat Plate Waffle slab
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Comparison of One-way and Two-way slab
behavior
Flat slab Two-way slab with beams
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Basic Steps in Two-way Slab Design
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
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.
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Minimum Slab Thickness for two-way
construction
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
Slabs without drop panels meeting, tmin = 5 in
Slabs with drop panels meeting, tmin = 4 in
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Minimum Slab Thickness for two-way
construction
Maximum Spacing of Reinforcement
At points of max. +/- M:
Max. and Min Reinforcement Requirements
7.12.3 ACI in. 18 and
13.3.2 ACI 2
s
ts
balsmaxs
S&Tsmins
75.0
13.3.1 ACI 7.12 ACI from
AA
AA
24
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25
26
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DDM Definitions27
DDM Definitions28
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Effective beam sections
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Shear in 2-way Slabs
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31
32
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33
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Definition of Beam-to-Slab Stiffness Ratio, a
Accounts for stiffness effect of beams located along
slab edge reduces deflections of panel adjacent to
beams.
slab of stiffness flexural
beam of stiffness flexurala
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Beam-to-Slab Stiffness Ratio, a
With width bounded laterally by centerline of
adjacent panels on each side of the beam.
scs
bcb
scs
bcb
4E
4E
/4E
/4E
I
I
lI
lIa
slab uncracked of inertia ofMoment I
beam uncracked of inertia ofMoment I
concrete slab of elasticity of Modulus E
concrete beam of elasticity of Modulus E
s
b
sb
cb
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Beam and Slab Sections for calculation of a
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Beam and Slab Sections for calculation of a
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Beam and Slab Sections for calculation of a
Definition of beam cross-section
Charts may be used to calculate a Fig. 13-21
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Distribution of Moments
Total static Moment, Mo
3-13 ACI
8
2
n2u0
llwM
cn
n
2
u
0.886d h using calc. columns,circular for
columnsbetween span clear
strip theof width e transvers
areaunit per load factored
l
l
l
wwhere
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Column Strips and Middle Strips
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
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Negative and positive design moments
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Distribution of M0
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Positive and Negative Moments in Panels
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Factored Moment
in Column Strip
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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Factored Moment in Column Strip
Ratio of flexural
stiffness of beam to
stiffness of slab in
direction l1.
Ratio of torsional
stiffness of edge
beam to flexural
stiffness of
slab(width= to
beam length)
bt
a1
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53
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), fy =414 Mpa (G60)
Solution54
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
Flat
plate
without
edge
beams
56
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Example
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
57
Slab h same as before the required h = 7 in =0.17m(Use: h = 7.5 “
= 0.19m1 Psf = 47.9 N/m2
Wu = 1.2*5.27+1.6*3.8 3= 13 kN/m2
Shear check for h58
d = 6.25*25.4
= 159 mmsay 0.16m
Wu = 13.0 kPaOne 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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59
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 directions60
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
mkNftk
llwM n
ul
.234.2.181
8/)12/1620(*16)26.0(8
. 22
120
Column Strip Middle strip
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Moment Distribution in long direction: col. And middle strips
mkNftkl
lwM nul .245.2.1818/)12/1620(*16*26.0
8. 2
2
120
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
ftkl
lwM nus .2.1468/)12/1216(*20*)26.0(
8. 2
2
120
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, +VeMom.
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 contributions70
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Slabs with
Interior Beams:
b= ln long / ln shor
USC Units:
Thickness in SI units:
Example: Slabs with interior beams72
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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73
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 limitations1. 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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