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Transcript of CHAPTER 3 SLAB - SKYSCRAPERS-CIVILIANS'...
CHAPTER 3
SLAB
3.1 INTRODUCTION
Reinforced concrete slabs are one of the most widely used structural elements. In many structures, in addition to providing a versatile and economical method of supporting gravity loads, the slab also forms an integral portion of the structural frame to resist lateral forces. Usually a slab is a broad, flat plate, with top and bottom surfaces parallel or nearly so. It may be supported by reinforced concrete beams, by masonry or reinforced concrete walls, by structural steel members, directly by columns, or continuously by the ground.
3.2 TYPES OF SLAB
• One-way slab : Independent of support condition. (Figure 3.1a 3.1b)
2
1
ll
> 2;
• Two-way slab : Depends on support condition. (Figure 3.1c)
2
1
ll
≤ 2
(a) One- way slab
Figure 3.1: Types of slab
2
1l
l >2
1l2l
2
1
ll > 2
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Two-way slabs are classified as:
• Two-way edge supported slab or slab with beams. ( Figure 3.1d ) • Two-way column supported slab or slab without beams. ( Figure 3.1e,3.1f, 3.1g )
(f) Flat slab (Column supported slab)
(d) Slab with beams (Edge supported slab)
(b) One- way slab
(g) Grid slab (Column supported slab)
(e) Flat plate (Column supported slab)
(c) Two- way slab
1l2l
1l2l
Figure 3.1: Types of slab (continued)
SLAB
16
3.3 DESIGN OF ONE-WAY SLAB
Step 1: Estimation of Slab Thickness (h)
Slab thickness is determined according to ACI Code 9.5.2 as given in Table 3.1
Table 3.1: Minimum thickness of non-prestressed one- way slab
Members wc = 145 pcf
fy = 60.000 psi
wc =90∼120 pcf fy <60.000 psi or,fy>60.000 psi
Rounding up the thickness
Simply supported l/20
One end continuous l/24
Both end continuous l/28
Cantilever l/10
Multiply by (1.65-0.005wc) but > 1.09
Multiply by
0.4 + 000,100yf
(1) h≤6 in next higher ¼ in
(2) h > 6 in next higher ½ in
# Span length l is in inches, as defined by ACI Code 8.7 given in Fig. 3.2(a), (b), & (c)
Step 2 : Calculation of Factored Load (wu)
wu = 1.4 D+ 1.7 L psf
Dead load, D = wc x 12h psf
wc = Unit weight of concrete (145 ~ 150 pcf for normal weight concrete )
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Step 3: Determination of Design Moment
Design moment is determined by using ACI Moment Coefficient (ACI Code 8.3.3) as given in Table 3.4.
Step 4 : Checking the Design Thickness
d =)59.01(
c
yy
u
ff
bf
M
′− ρφρ
putting ρ = ρmax = 0.75ρb.
Where, ρ b = 0.85*β 1y
c
ff /
*yf+87000
87000
Values of β 1 is given in Table 3.2
Table3.2 : Values of β 1 (ACI Code 10.2.7.3)
f c/ ≤ 4000 psi β 1 = 0.85
f c/ > 4000 psi β 1 shall be reduced at a rate of 0.05 for each 1000 psi of strength
in excess of 4000 psi.
0.65 ≤ β 1≤ 0.85
Table 3.3: Clear cover for slab ( ACI Code 7.7.1)
No 14 & No 18 bars .................................... .... 1 1/2 in No 11 bar & smaller .................................... ¾ in
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18
t
al
h
(a) Slabs not built integrally with the support (ACI Code 8.7.1)
l= la + h ≤ la + t t
al
h
l= la + t
(b) Slabs are continuous (ACI Code 8.7.2)
tal
Figure 3.2 : Span length
(c) Slabs built integrally with support
If (d + clear cover) < h; Design is ok.
Otherwise redesign the thickness.
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19
3. Discontinuous ends are
Table 3.4 : ACI moment coefficient
Support Condition
19
111
0 0
Moment Coefficent
For two span:
1. Discontinuous ends are
124
19
114
114
0 110
111
111
124
110
114
1 24
19
116
114
1 16
19
19
114
unrestrained
built integrally with support 2. Discontinuous ends are
(spandrel beam or girder)
116
1 11
1 11
116
111
1 11
1 11
114
116
116
110
111
1 11
1 11
built integrally with support (when support is a column only)
For continuous Span: 1. Discontinuous ends are unrestrained.
(spandrel beam or girder) built integrally with support 2. Discontinuous ends are
3. Discontinuous ends are built integrally with support
(when support is a column only)
w = Total factored load per unit length of beam or per unit area of slab
l = Clear span for positive moment and the average of two adjacent clear spans
for negative moment.
1. Shear in end members at first interior support
2. Shear at all other supports 1.15 w
2 w2
19
111
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Step 5 : Determination of Steal Area (As)
Reinforcement for 1 ft. strip towards shorter distance is calculated by Iteration. Details shown in Figure 3.3 (here b = 12 in)
Figure 3.3: Iteration process to determine the steel area
Calculate a for next trial with (As) corrected
(a) corrected = bf
fA
c
ys
′′′85.0
(As) trial = ( )2adf
M
y
u
−φ
(As) corrected
A corrs , ≈
As, trial
Yes
OK
Assume a (hints: a = 0.3d)
No
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Generally # 3 or # 4 bars are used for slab main reinforcement.
Spacing: ACI Code 7.6.5 specifies that
Spacing ≤ 3h or 18 in, whichever is smaller
but > 1.5 h
Finding out bar spacing: Let us chose # 3 bar (0.11 2in )
Spacing = sA
12*11.0 in c/c
Step 6 : Temperature and Shrinkage Reinforcement
Reinforcement is provided normal to main reinforcements. ACI Code 7.12.2.1 provides required area of temperature and shrinkage reinforcement as given in Table 3.5.
Table 3.5 : Minimum ratio of temperature and shrinkage reinforcement in slabs.
Slabs where grade 40 or 50 deformed bars are used
0.0020
Slabs where grade 60 deformed bars or welded wire fabric are used
0.0018
Slabs where reinforcement with yield strength exceeding 60,000 psi measured at a yield strain of 0.35 percent is used
yf000,600018.0 ×
But should be:
ρ > 0.0014
Required steel area, As = ρbh 2in per 1 ft. strip
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Spacing: ACI Code 7.12.2.2 specifies that
Spacing ≤ 5h
or ≤ 18 in , whichever is smaller
Using # 3 or # 4 bar required spacing can be obtained.
Step 7 : Shear Check
According to ACI shear coefficient given in Table 3.2
Shear at end members at first interior support is 2
15.1 nulw
Critical shear at a distance d from support, Vu = (1.15 122
dWlW unu − )
Design strength for shear, φ Vc = φ ′cf2 bd ; φ = 0.85
If φVc > Vu, slab design for shear is OK
Otherwise slab thickness should be revised.
Step 8 : Reinforcement Detailing
Shown in Figure 3.4
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23
Main positive reinforcement
'_zoom
Temperature & shrinkage reinforcement
(a) Plan of bottom reinforcement
A
A
2
l2/ 4
(b) Plan of Top reinforcement
'_zoom
negative reinforcement at discontinuous edge
negative reinforcement at continuous edge
(c) Cross section (A- A)
6"6"
l2/ 4
l2/ 4
l2/ 4
Temperature & shrinkage reinforcement
l2/ 4 l2/ 3 l2/ 3 l2/ 3
l2
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Figure 3.4: Reinforcement detailing in one-way slab (continued)
Temperature & shrinkage reinforcement
Main positive reinforcement
Negative reinforcement at continuous edge
Negative reinforcement at discontinuous edge
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3.4 TWO WAY SLAB
ACI Code 13.5.1 states that “a slab system shall be designed by any procedure satisfying conditions of equilibrium and geometric compatibility, if it is shown that the design strength at every section is at least equal to required strength, and that all serviceability conditions, including limits on deflections, are met.”
According to ACI Code 13. 5.1. 1, all Two-way slab system are to be analyzed and designed either by the Direct Design Method or the Equivalent Frame Method for gravity loads only.
For lateral loads, separate elastic analysis should be worked out. ACI Code 13. 5.1.3 permits the combining to the gravity load analysis with the result of lateral load analysis.
Adaptation of any one of the two methods demands fulfillment of certain requirements. However, when the requirements are not met, an old procedure is still followed by the Engineers as specified in 1963 ACI Code, named as Coefficient Method.
3.4.1 DIRECT DESIGN METHOD (DDM)
3.4.1.1 General
The design is based on equivalent rigid frame system as shown in Figure 3.5.
ACI Code13. 2.1 specifies:
Width of equivalent frame = l2
Width of column strip = 21 21
21 lorl whichever is less
The equivalent frames are considered in both longitudinal and transverse directions.
3.4.1.2 Limitations For DDM
ACI Code 13.6.1 specifies that the design of Two-way slab system by DDM shall be permitted within the following limitations:
• There shall be a minimum of three continuous spans in each direction. • The panels shall be rectangular, with the ratio of the longer to the shorter spans within a panel not greater than 2.
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26
• The successive span lengths in each direction shall not differ by more than one third of the longer span. • Column may be offset a maximum of 10% of the span in the direction of successive column. • Loads shall be due to gravity only and live load shall not exceed 2 times the dead load.
Figure 3.5 : Equivalent frame system for DDM
Interior Equivalent Frame
Exterior Equivalent Frame
H.M.S = Half Middle Strip; C.S= Column Strip
H.M.S
C.S
H.M.S
H.M.S
C.S
l2
l2
l2
l 1 l 1 l 1
l2
2l 2
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27
• If beams are used on the column lines, the relative stiffness of the beams in the two perpendicular direction shall be:
0.2< 1
22
2
21
l
l
α
α < 5.0
1α = α in direction of ℓ 1
2α = α in direction of ℓ 2
scs
bcb
IEIE
=α
ACI Code 13.6.1.8 allows the deviation from above limitations, if it can be shown that the requirements of ACI Code 13.5.1 (as stated in section 3.4) are satisfied.
3.4.1.3 DESIGN METHOD BY DDM
Step 1 : Determination of Slab Thickness
ACI Code 9.5.3 specifies the minimum thickness for two- way slab system
Table 3.6: Minimum thickness for slab without beams (ACI Table 9.5.c)
Without drops panels With drop panels Exterior panels Interior
panels Exterior panels Interior
panels
Yield strength fy psi Without
edge beams
Without edge beams *
Without edge beams
Without edge beams a
40,000 33ln
36ln
36ln
36ln
40ln
40ln
60,000 30ln
33ln
33ln
33ln
36ln
36ln
75,000 28ln
31ln
31ln
31ln
34ln
34ln
h > 5 in ( ACI Code 9.5.3.2.a) h > 4 in (ACI code 9.5.3.2.b)
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28
• For slabs with beams along exterior edges, the value of α for edge beam shall not be less than 0.80. • For f y between given values the minimum thickness should be obtained by linear interpolation.
Table 3.7 : Minimum thickness for slab with beams (ACI Code 9.5.3.3)
(1) αm < 0.2
(2) 0.2 < αm < 0.2*
(3) αm > 0.2*
h = )2.0(536
)000,200
8.0ln(
−+
+
m
yf
αβ h =
β936
)000,200
8.0ln(
+
+fy
h > 5 in h > 3.5 in
The provisions of Table 3.4 shall apply
For edge beam α > 0.80 , otherwise h min as provided by column (2), (3) must be increased by 10% in the panel with edge beam (3) (ACI Code 9.5.3.3. d)
Step 2: Determination of Total Factored Static Moment (Mo)
According to ACI Code 13.6.2.2
Mo = 8
22 nu llW
Mo, of an exterior equivalent frame =21 Mo, of an interior equivalent frame.
ln = face to face of columns or capitals or walls.
ln ≥ 0.65l1 and in determining ln, circular and polygon shaped supports shall be treated as square supports with the same area , shown in Fig 3.6. (ACI Code 13.6.2.5)
SLAB
29
Figure 3. 6 : Example of equivalent square section for supporting members
Step 3 : Longitudinal Distribution of Moments.
The total factored static moment, Mo is distributed to the negative and positive zone of a ‘equivalent frame’ according to ACI Code 13.6.3.2 and 13.6.3.3 as given in Table 3.8.
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30
0 .65 M o 0.65 M o
0 .35 M o
M o
0.75Mo
0.63Mo
Mo
0.65Mo 0.65Mo
0.35 Mo
0.26 Mo
0.70Mo
0.52Mo
Mo
0.30 Mo
0.70Mo
Mo
Table 3.8 : Distribution of total factored static moment 1. Interior Span (ACI Code 13.6.3.2) Completely fixed at both ends
Case -1: Exterior edge unrestrained
Case – 2 Beams on all Column lines
Case – 3 No beams ( Flat slab , Flat plate)
Case – 4 Edge beam only
Case - 5: Exterior edge fully restrained by monolithic concrete wall
0 .16 M o
0 .70M o
0 .57M o
M o
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31
Step 4 : Transverse Distribution of Longitudinal Moment
The longitudinal negative and positive moments are for the entire width of equivalent frame. Each of these moments is to be distributed proportionately among column strip and two half middle strips following ACI Code 13.6.4. Before distribution of moment the following 3 parameters are to be obtained:
• Aspect ratio = 1
2
ll
• Stiffness ratio, α = scs
bcb
IEIE
• Ratio βt = scs
cb
IECE
2
E cb = modulus of elasticity of beam concrete, psi
E cs = modulus of elasticity of slab concrete, psi
I b = moment of inertia of beam
I s = moment of inertia of slab
C = torsion constant
Evaluating the three parameters, distribute the percentage of longitudinal moment into column strip and the remainder into two half middle strips according to Table 3.9.
Table 3.9 : Percentage of longitudinal moment in column strip (ACI code 13.6.4.2.1, 13.6.4.2, 13.6.4.4)
Aspect ratio = l2/l1 0.5 1.0 2.0 βt = o 100 100 100 α1 l 2 /l11 = o βt > 2.5 75 75 75 βt = o 100 100 100
Negative moment at exterior support α1 l2 /l11 > 1.o
βt > 2.5 90 75 45
α1
l2
/l11 = o 75 75 75
Negative moment at interior support α1 l2 /l11 > 1.o 90 75 45
α1 l2 /l11 = o 60 60 60 Positive moment α1 l2 /l11 > 1.o 90 75 45
# Linear interpolations shall be made between values shown.
SLAB
32
Calculation of Three Parameters
• Calculation of Aspect Ratio = 1
2
ll
• Calculation of Stiffness Ratio (α)
Ratio of flexural stiffness is related to slab with beams either on all sides or on edge only. For slab without beams e.g. flat plate or flat slab, α = 0.
α = scs
bcb
IEIE
Determination of Moment of Inertia of the Beam
Ib = k12
3hbw
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33
Definition of be is shown in Figure 3.7 ( ACI code 13.2.4)
Figure 3.7 : Example of portion of slab to be included with beam or definition of b e
The Moment of Inertia of Slab Section
Is = 12
3bh
h = slab thickness
b = l2 for interior equivalent frame, or
= l2/2 for exterior equivalent frame
It should be determined in both directions.
• Calculation of Ratio β t
βt = scs
cb
IECE
2
h w fh4≤ b w +2h w fw hb 8+≤
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34
Torsion Constant
ACI Code 13.7.5.1 specifies that torsional members shall be assumed to have a constant cross section throughout their length consisting of the largest of (a), (b) and (c) as shown in Figure 3.8.
ACI Code 13.0 defines torsion constant as:-
C = ∑ ⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
363.01
3 yxyx
x = shorter dimension of a component rectangle
y = longer dimension of a component rectangle
The component rectangle should be taken in such a way that the largest value of C is obtained.
Example for slab with beam & flat plate shown in Figure 3.9.
Step 5 : Determination of Effective Depth (d)
Where two steel layers (along two directions, perpendicular with each other) are in contact, the larger d is assigned to the steel of greater moment ( i.e. steel for greater moment shall be placed near to either top or bottom face ).
Large d = h – clear cover (min ¾ in) - d b + ½ d b in
Short d = h – clear cover - ½ d b in
SLAB
35
directionof
momentFlat plate
(a) Torsional member (ACI Code 13.7.5.1a)
t
Torsional member
Slab with beam t
beam
Column
Slab
t
(b) Torsional member (ACI Code 13.7.5.1b)
t
h f
(c) Torsional member ( ACI Code 13.7.5.1c)
be=(bw+hw) ≤ (bw+4hf)
h
bw
hf
hw≤4hfhw≤4h f
hw
bw
hw≤4hf
be=(bw+2hw) ≤ (bw+8hf)
Edge beam Interior beam
t
Figure 3.8 : Torsional member
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36
y1
C1
1
2
1
2 C2
C=Larger of C1 and C2
(a) Slab with beam (Edge section)
y1
1
2 C1
2
1 1
C2
t1
x h f
t2
t1
t2
Short direction
Imaginary beam
Long direction
(c) Flat plate
y1
y2
y2
y1
y2
C=Larger of C1 and C2
(b) Slab with beam (Interior section)
X2
X1X1
X2
X1
X2
Y2
x1
X2
x
YY
Figure 3.9 : Determination of C
SLAB
37
Step 6 : Determination of Steel Area (A s )
Total steel area for the strip is obtained by iteration process as shown in Figure 3.3. b = width of strip or, b= width of drop panel in the direction of moment (For slab with drop panel in negative moment zone)
Step 7 : Check for Minimum Steel
A min,s = ρbh
ρ = minimum steel ratio for temperature and shrinkage as shown in Table 3.5
A provideds , > A min,s (OK)
Otherwise provide A min,s
Step 8 : Total Number of Bar
b
s
AAN = , A b = cross section area of bar used
Spacing = Nb ≤ 2h; b = width of strip
Step 9 : Check for Deflection Control
According to ACI Code 9.5.3 control of deflection is achieved by providing the slab thickness in accordance to Table 3.6 and Table 3.7. For details see section 3.5.
SLAB
38
Step 10 : Reinforcement Detailing
• Placing: Flexural reinforcement in two- way slab system is placed in an orthogonal grid, with bars parallel to the sides of the panels. • Straight Bars: Straight bars are generally used throughout, although in some cases positive moment steel is bent up where no longer needed, in order to provide for part or the entire negative moment requirements. • Spacing: Maximum spacing ≤ 2h. Figure 3.10.a (ACI Code 13.3.2). • Concrete Cover: Minimum concrete cover = ¾ in. Figure 2.10.a. (ACI Code 7.7.1). • Effective Depth: When bars are placed in perpendicular layers either on top or bottom together, stacking problem arises. The inner steel will have an effective depth 1-bar diameter less than the outer steel. For relatively larger moment bars in one direction are provided with greater d. Details in Figure 3.10.a & Figure 3.10.b. • Embedment for Positive Moment: Positive moment reinforcement perpendicular to a discontinuous edge shall extend to the edge of slab and have embedment, straight or hooked, at least 6 in. in spandrel beams, columns or walls. Details in Figure 3.10.c. (ACI Code 13.3.3). • Embedment for Negative Moment: Negative moment reinforcement perpendicular to a discontinuous edge shall be bent, hooked or anchored in spandrel beams, columns or walls. Details in Figure 3.10.c. (ACI Code 13.3.4). • Cantilever Slab: Where slab is not supported by a spandrel beam or wall at a discontinuous edge or where a slab cantilevers beyond the support, anchorage of reinforcement shall be permitted within the slab. Details in Figure 3.10.d. (ACI Code 13.3.5). • Corner Reinforcement: In slabs with beams between supports, with value α > 1.0, special top and bottom reinforcement shall be provided at exterior corners. Details in Figure 3.11. (ACI Code 13.3.6). • Slab with Drop Panel: Detail dimensions are shown in Figure 3.12. (ACI Code 13.3.7). • Details of Reinforcement in Slabs without Beams: In addition to other requirements as mentioned through paragraph 1- 10, detailing shown in Figure 3.13 should be observed. (ACI Code 13.3.8)
SLAB
39
Figure 3.10 : Details of reinforcement in two-way slab
(a) Section along short span ( 2)
Maximum spacing < 2h ( ACI Code 13.3.2)
34" min
h
(b) Section along long span ( 1)
34" min
SLAB
40
Figure 3.10 : Details of reinforcement in two-way slab (continued)
(c) Embedment of positive moment reinforcement (ACI Code 13.3.3)
6"
(d) Anchorage of reinforcement in cantilever slab (ACI Code 13.3.5)
Anchorage
SLAB
41
Figure 3.11 : Spiral reinforcement at exterior corner
G ride type
l5
l 5
l5
l 5
Top bar
Bottom bar
(Provided in tw o layers)
(Providedin band)D iagonal type
L/5
L/5
L/5
L/5
SLAB
42
Figure 3.12 : Details of drop panels
b
ht
b ≥ 16 l1
t
t min ≥ 14 h
SLAB
43
Figure 3.13 : Minimum extension for reinforcement in slabs without beams (ACI Figure 13.3.8)
SLAB
44
3.4.2 EQUIVALENT FRAME METHOD (EFM)
3.4.2.1 General
The EFM is an alternate method to the DDM for computing longitudinal moments and shear for gravity loads in slabs, supported on column or walls. ACI Code Commentary R13.7 states that EFM involves the representation of the three dimensional slab systems by a series of two-dimensional frames that are then analyzed for loads acting in the plane of the frames.
ACI code 13.7.2 defines the equivalent frame as in Figure 3.14
Figure 3.14 : Elements of equivalent frame system
SLAB
45
3.4.2.2 Moment Of Inertia of Slab Beam (Is)
• Considering gross area of concrete (ACI Code 13.7.3.1)
Is = 12
32hl
• Variation in Is along axis of slab-beam shall be taken into account. The first change from midspan Is occurs at the edge of drop panels, the next occurs at the edge of the column or capital. (ACI code 13.7.3.2)
• Is from center of column to face column = 2
2
21 ⎟⎠⎞⎜
⎝⎛ − l
CI s ; I s at face of column
(ACI Code 13.7.3.3).
3.4.2.3 The Equivalent Column
ACI Code Commentary R13.7.4 establishes a concept of an “equivalent column” that combines the stiffness of the slab- beam and torsional member into a composite element. The column flexibility is modified to account for the torsional flexibility of the slab- to- column connection that reduces its efficiency for transmission of moment. The equivalent column is shown in Figure 3.17.
Figure3.15 : Equivalent column
SLAB
46
3.4.2. 4 Moment of Inertia of Column (Ic )
ACI code 13.7.4 defines the moment of inertia of column as shown in Figure 3.16.
Figure 3.16 : Column area for moment of inertia
3.4.2.5 Design Method by EFM
Step 1 : Determination of Factored Load
Step 2: Determination of Slab Thickness
Minimum required slab thickness s obtained from Table 3.6 & Table 3.7.
Step 3 : Flexural Stiffness of Actual Column (Kc)
Kc = c
ccccc
lIEκ
κc = Column stiffness coefficient ( to be obtained from Appendix C-1)
E cc = modulus of elasticity of column concrete
c1
I = ∞
I = ∞
I c = 12
312cc
Variable
SLAB
47
I c = moment of inertia of column = 12
312CC
l c = length of column (c/c)
Step 3 : Torsional Stiffness of Transverse Torsional Member (Kt)
Kt = 3
2
22 1
9
⎟⎠⎞⎜
⎝⎛ −
∑
lCl
CEcs
• C can be determined as mentioned in step 4 (c) of section 3.4.1.3. • `∑’ Sign implies that Kt of the transverse member in each side of interior column is computed separately and added. For exterior columns, there is only one transverse member. • For beam along center line of column Kt should be corrected.(ACI Code 13.7.5.2 )
K correctedt , = Kt * s
sb
II
Isb = Moment of inertia of slab with a beam Is = Moment of inertia of slab without such beam
Step 5 : Flexural Stiffness of Equivalent Column (Kec)
tcec KKK111
+=∑
∑ Kc = Kc1 + Kc2
Step 6 : Flexural Stiffness of Slab (Ks)
1lIEK scss
sκ
=
κs = coefficient of slab stiffness ( to be obtained from Appendix C-2 and C-3)
Is = 12
32hl
SLAB
48
Step 7 : Distribution Factor (D.F)
11
1
ecs
s
kkk+
232
2
ecSS
S
KKKK
++
232
3
ecSS
s
KKKK
++
11
1
ecs
ec
kkk+
232
2
ecss
ec
kkkk
++
Figure 3.17 : Distribution factors for slab-column joints
Step 8 : Carry Over Factors and Moment Coefficient (M)
Carry over factors (C.O.F) and moment coefficient (M) for slab beam are obtained from Appendix C-1, 2, 3)
Step 9 : Moment Analysis
The longitudinal moments of equivalent frames are obtained by Moment Distribution Method.
• For different loading conditions distributed negative and positive moments are computed. Maximum moments are taken as design moment.
Live loading pattern is known, frame shall be analyzed for that load. ( ACI Code 13.7.6.1)
Variable LL, but LL < 43 DL, then maximum factored moment occur at all sections
with full LL on entire slab system. ( ACI Code 13.7.6.2)
k 2s
K 2eck 1ec
k 1s k 3s
DF
DF
DF DF
DF
SLAB
49
Variable LL, but LL > 43 DL, three loading case to be considered : (ACI Code
13.7.6.3)
Total load (w u ) on all panels
DL on all panels and 43 LL on midspan of a panel.
DL on all panels and 43 LL on adjacent panels.
• Total panel moment (Mp) is computed using equation:
M p = 8
22 nu llw
• F.E.M are computed using equation:
F.E.M = M wu l2l12
• Analyzing by Moment Distribution Method final negative moments at the supports are computed.
• Positive moments at midspan is obtained by
M(+) = M p - 21 [sum of M(-) in a panel after distribution]
• Reduction in Negative Moments: The negative moment as obtained is applicable for centerlines of support. Since the support is not a knife edge but rather a broad band, ACI Code 13.7.7 specifies a reduction in negative moment at critical section.
• When a slab system satisfy the six imitations of DDM, but are analyzed by EFM,
further reduction in computed moments are permitted to the proportions of TM
Mo as
such, ∑ Design moments < Mo (ACI code 13.7.7.4)
MT = total panel moment
Mo = 8
22 nlWul
SLAB
50
Step 10 : Transverse Distribution of Longitudinal Moment
According to ACI Code 13.7.7.5 the distribution of longitudinal moments to column strip and half middle strips to be done as mentioned in step 4 of section 3.4.1.3.
For Step 11 to Step 16 follow Step5 to Step10 of section 3.4.1.3.
Step 11: Determination of Effective Depth (d)
Step 12: Determination of Steel Area (A s )
Step 13: Check for Minimum Steel
Step 14: Total Number of Bar
Step 15: Check for Deflection Control
Step 16: Reinforcement Detailing
SLAB
51
3.4.3 COEFFICIENT METHOD
3.4.3.1 General
The method makes use of tables of moment coefficient for a variety of conditions. These coefficients are based on elastic analysis but also account for inelastic redistribution. This method was recommended in 1963 ACI Code for the special case of two-way slabs supported on four sides by relatively deep, stiff, edge beams.
C.S = column strip; M.S = middle strip
Figure 3.18 : Elements of two- way slab with beam by coefficient method
l a = length of clear span (face of support to support) in short direction
l b = length of clear span (face of support to support) in long direction
4bl
;C 4bl
;C S2bl
;
4al
; C.
2al
;M
4al
; C S
bl
l a
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52
The moments in the middle strips in two directions are:
Ma = Ca wu la2
Mb = Cb wu lb2
Ca, Cb = tabulated moment coefficients
3.4.3.2 DESIGN BY COEFFICIENT METHOD
Step 1 : Selection of Stab Thickness
h = 5.3180
≥P in, P = panel perimeter
Step 2 : Calculation of Factored Load
wu = 1.4 D + 1.7L
D = dead load = wc *12h psf ; wc = 150 lb
Step 3 : Determination of Moment Coefficient
m = lbla
Case type is identified from end conditions. Using the value of ‘m’ corresponding moment coefficients are obtained for respective ‘case type’:
• Ca, neg and Cb, neg are obtained from Appendix D-1. • Ca, dl, pos and Cb, dl, pos are obtained from Appendix D-2. • Ca, ll, pos and Cb, ll, pos are obtained from Appendix D-3.
SLAB
53
Step 4 : Calculation of Moment
Middle strip moment
Positive moment
Ma, pos = Ca, dl wu la2 + Ca, ll wu la
2
Mb, pos = Cb,dl wu lb2 + Cb, ll wu lb
2
Negative Moments for continuous Edge
Ma, neg,cont = Ca, neg wu la2
Mb, neg,cont = Cb, neg wu lb2
Negative Moments for Discontinuous Edge
Ma, neg, discont = 31 Ma, pos
Mb, neg, discont = 31 Mb, pos
Column strip moment
The moments in column strips should be taken as 2/3rd of middle strip’s moment in respective directions.
Step 5 : Check the Design Thickness
d =)59.01(
c
yy
u
ff
bf
M
′− ρφρ
If (d + clear cover) < h ; design is ok.
Otherwise redesign the thickness. (For details see step 4 of section 3.3)
SLAB
54
Step 6 : Reinforcement for Middle Strip
Required reinforcement can be determined by Iteration process as given in Figure 3.3. Reinforcement shall be determined for short direction and long direction separately as follows:
• Short Direction
Midspan Continuous Edge Discontinuous Edge
• Long Direction
Midspan Continuous Edge Discontinuous Edge
Check for Minimum Reinforcement: According to ACI Code 13.3.1 the minimum reinforcement in each direction shall be as mentioned in Table 3.5.
Spacing: Using # 3 or # 4 bar required spacing is determined.
Maximum spacing < 2h (ACI Code 13.3.2)
Step 7 : Reinforcement for Column Strip
Bars selected for middle strip are used in column strips, with the spacing 3/2 times that in the middle strip, but spacing < 2h.
Step 8 : Check for Shear
Percent of total load as transmitted in each direction is obtained from Appendix D-4 Load per foot on the beams are determined.
The shear to be transmitted by the slab to these beams is = beam loads
Shear at critical section at a distance d from beam face = V u
SLAB
55
Shear strength of the slab, φ V c = bdcf ′2
φ VuVc > design is ok, otherwise thickness should be redesigned.
3.5 CONTROL OF DEFLECTION
3.5.1 GENERAL
ACI Code Commentary R 9.5.1 establishes two methods for controlling deflections:
• For non-prestressed two-way construction, minimum thickness as required by Table 3.4 & Table 3.5 will satisfy the requirements of the code.
• When there is need to use member depths shallower than are permitted by Table 3.4 & Table 3.5 or when members support construction is likely to be damaged by large deflections, deflections should be calculated and compared with ACI Code limiting values as given in Table 3.13.
3.5.2 IMMEDIATE DEFLECTION
Immediate deflection is also termed as Short-Term deflection and calculated using the formula given in Table 3.11.
SLAB
56
Table 3.11 : Calculation of immediate deflection
1. Live load deflection
Δ l = ec
bb
IElM
323 2
# Both ends continuous or equally restrained or,
One or both ends discontinuous, but monolithic with beam. # Mb = live load +ve moment
2. Live load deflection
Δ l = ec
bb
IElM
485 2
# Slab supported by masonry walls # Mb = live load +ve moment
3. Dead load deflection
Δ d = ec
bb
IElM
161 2
# Both ends continuous and fully fixed # Mb = maximum dead load +ve moment
4. Dead load deflection
Δ d = ec
bb
IElM
485 2
# Both ends free of restraint (Supported on masonry wall) # Mb = maximum dead load +ve moment
lb = Clear span in long direction
M b = Unfoctored moment = 7.14.1,,,, llposbdlposb M
orM
in long direction Deflection can be calculated in short direction also in the same way. Ie = Effective moment of inertia for computation of deflection
l b = Clear span in long direction; I e = effective moment of inertia
M b = Unfoctored moment = 7.14.1,,,, llposbdlposb M
orM
in long direction
Determination of Ie
Where, M cr = t
gr
yIf
(ACI Code 9.5.2.3)
SLAB
57
y t = distance from centroidal axis of gross section, neglecting reinforcement, to extreme fibre in tension, in.
f r = modulus of rupture of concrete, psi.
For normal weight concrete:
f r = 7.5 /cf
For light weight concrete one of the following modifications shall apply:
• When average tensile strength, f ct is specified
f r = 7.57.6ctf = 1.12 f ct , 7.6
ctf ≤ /cf
• When f ct is not specified
f r = 0.75 * 7.57.6ctf ; for all lightweight concrete
f r = 0.85 * 7.57.6ctf ; for sand- lightweight concrete
• I e for Continuous Spans ( ACI Code 9.5.2.4)
I e = 0.50 I em + 0.25 ( I 1e + I 2e )
I em = effective moment of inertia for the midspan section I 1e , I 2e = negative moment sections at the respective bean ends
3.5.3 LONG TERM DEFLECTION
Initial deflections increase significantly if dead loads sustain over a long period of time, due to the effects of shrinkage and creep According to ACI Code 9.5.2.5
SLAB
58
Δlong = Δd,short * λ
Where, λ= /501 ρξ
+
/ρ = value at midspan for simple and continuous span
= at support for cantilever
ξ = time- dependent factor (Table 3.12 or Figure 3.19)
Table 3.12 : Values of ξ ( ACI Code 9.5.2.5 )
5 years or more 2.0
12 months 1.4 6 months 1.2 3 months 1.0
Figure 3.19 : Values of ξ
TOTAL DEFLECTION
totalΔ = shortllong ,Δ+Δ
Deflection should be calculated along both direction and maximum values will be considered. (ACI Code 9.5.2.6)
SLAB
59
Desired value : totalΔ < limiting value given in Table 3.13
Table 3.13 : Maximum permissible computed deflection (ACI Table 9.5.b)
Type of member Deflection to be considered Deflection limitation
Flat roofs not supporting or attached to nonstructural elements likely to be damaged by large deflection
Immediate deflection due to live load L 180
l
Floors not supporting or attached to nonstructural elements likely to be damaged by large deflection
Immediate deflection due to live load L 360
l
Roof or floor construction supporting or attached to nonstructural elements likely to be damaged by large deflection
480l
Roof or floor construction supporting or attached to nonstructural elements not likely to be damaged by large deflection
That part of the total deflection which occurs after attachment of the nonstructural elements, the sum of the long- time deflection due to all sustained loads, and the immediate deflection due to live load L.
240l
3.6 STRIP METHOD FOR SLABS
Introduction:
The strip method is a lower bound approach, based on satisfaction of equilibrium requirements everywhere in the slab. By the strip method a moment field is first determined that fulfills equilibrium requirements, after which the reinforcement of the slab at each point is designed for this moment field. The strip method gives results on the safe side, which is certainly preferable in practice, and differences from the true carrying capacity will never impair safety. The strip method is a design method, by which the needed reinforcement can be calculated. It encourages the designer to vary the reinforcement in a logical way, leading to an economical arrangement of steel as well as a safe design.
SLAB
60
Choice of Load Distribution
Condition-1:
The simplest load distribution is obtained by setting k = 0.5 over the entire slab, as shown in figure below. The load on all strips in each direction is then w/2, as illustrated by load dispersion arrows in figure. This gives maximum design moments
xm = ym = 16
2wa
Simple supports 4 sides
16
2wa
(d) mx across X=a/2
Y
X
A A
a
a
w/2
(a) Plan view
(b) wx along A-A
(c) mx along A-A
Figure 3.20: Square slab with load shared equally in two directions
SLAB
61
Condition-2:
An alternative, more reasonable distribution is shown in figure below. Here the regions of different load dispersion, separated by dash-dotted “discontinuity lines,” follow the diagonals, and all of the load on any region is carried in the direction giving the shortest distance to the nearest support. The solution proceeds, giving k values of either 0 or 1, depending on the region, with load transmitted in the direction indicated by the arrows in figure. For a strip A-A at a distance y ≤ a/2 from the X-axis, the design moment is
xm = 2
2wy
Simple supports 4 sides
(d) mx across X=a/2
Y
X
A A
a
a
w
(a) Plan view
(b) wx along A-A
(c) mx along
Wa2/2
w
y
Wy2/2
y
Figure 3.21: Square slab with load dispersion lines following diagonals
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62
Condition-3:
A third alternative distribution is shown in figure below. Here the division is made so that the load is carried to the nearest support, as before, but load near the diagonals has been divided, with one-half taken in each direction. Thus k is given values of 0 or 1 along the middle edges and value of 0,5 in the corner and center of the slab. For an X direction strip along section A-A, the maximum moment is
xm = 2w x
4a x
8a =
64
2wa
And for a strip along section B-B, the maximum moment is
xm = w x 4a x
8a +
2w x
4a x
83a =
645 2wa
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63
(d) mx across x=a/2
Y
X
B
A
B
a
w/2
(a) Plan view
(b) wx and mx along A-A
Wa2/64
w/2
a/2 a/4 a/4
a/4
5Wa2/64
a/2
Aa/4
a w
w/2 w/2
w/2w/2
w/2 w
Wa2/64
(c) wx and mx along B-B
5Wa2/64
w/2 w w
Figure 2.22: Square slab with load near diagonals shared equally in two directions
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64
Condition-4:
The preferred arrangement, shown in figure below, gives design moment as follows:
In the X direction:
Side strips: xm = 2w x
4b x
8b =
64
2wb
Middle strips: xm = w x 4b x
8b =
32
2wb
In the Y direction:
Side strips: xm = 2w x
4b x
8b =
64
2wb
Middle strips: xm = w x b x 8b =
8
2wb
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65
2b
ab
×
Wa2/64
b
2b
ab
× a
ba2
−
b
b/2
b/2
a
b/4 b/4 A-b/2
w/2
w/2
w/2
w/2
a
b/4
b/2
b/4
Figure 2.23: Rectangular slab with discontinuity lines originating at the corners
Figure 2.24: Discontinuity lines parallel to the sides for a rectangular slab
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66
Condition-5:
For slab strips with one end fixed and one end simply supported, the duel goals of constant moment in the unloaded central region and a suitable ratio of negative to positive moments govern the location to be chosen for the discontinuity lines. Figure ‘a’ shows a uniformly loaded rectangular slab having two adjacent edges fixed and the other two edges simply supported. The moment curve of figure “b” is chosen so that moment is constant over the unloaded part, i.e., shearing force is zero. The maximum positive moment in the X direction middle strip is then
xfm = 2wbα x
4bα =
8
22wbα
Accordingly, the distance from the right support, figure “c”, to the maximum positive moment section is chosen as bα . It follows that the maximum positive moment is
yfm = αwb x 2bα =
2
22wbα
With the above expressions, all the design moments for the slab can be found once a suitable value for α is chosen. The values of α from 0.35 to 0.39 give corresponding ratios of negative to positive moments from 2.45 to 1.45.
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67
( )2
1 bα− 2bα
2ba −
b
a
w/2 w/2
w
( )8
212wbα−
8
22 wbα
(a) Plan view
Figure 2.25: wx and mx along A-A
2bα
b/2
( )2
1 bα−
w/2 w/2
A A
B
B
w
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68
3.6.1 DESIGN BY STRIP METHOD:
Step-1: Selection of Slab Thickness
From table-3.7 (ACI Code-9.5.3.3)
180
p≥ 3.5
2
22 wbα
( )2
212wbα−
Figure 2.26:-wy and my along B-B
SLAB
69
Step-2: Calculation of Factored Load
Wu = 1.4 D+ 1.7 L (ACI Code-00)
Wu = 1.2 D+ 1.6 L (ACI Code-02)
Where, D =150 x 12h (psf)
Step-3: Selection of Load Distribution
From choice of load distribution.
Step-4: Calculation of Moment
From the equations of loading condition moments are calculated.
Step-5: Check For Design Thickness
d = )59.01(
'c
yy
u
ff
f
M
ρφρ −
If (d + clear cover) ≤ h; design is OK.
Step-6: Reinforcement Calculation
Reinforcement calculation is done by iteration method from figure-3.3 but compare of moment should be done with minimum steel requirement.
φMn=φρfybd2 ('
59.01(c
y
ff
ρ− )
If φMn<M; then only minimum reinforcement.
If φMn>M; then iteration from figure-3.3.
SLAB
70
ρmin from Table-3.3 (ACI Code-7.12)
Spacing:
Using #3 and #4 bar.
Maximum spacing ≤ 2h (ACI Code-13.3.2)
Cut-off points can be calculated from moment diagrams and development length should be provided.
SLAB
71
Table 3.14: Choice of Load Distribution
Case-1 xm = ym =
16
2wa
Case-2 xm =
2
2wy
Case-3
For an X direction strip along section A-A, the maximum moment is
xm = 2w x
4a x
8a =
64
2wa
And for a strip along section B-B, the maximum moment is
xm = w x 4a x
8a +
2w x
4a x
83a =
645 2wa
Case-4
In the X direction:
Side strips: xm = 2w x
4b x
8b =
64
2wb
Middle strips: xm = w x 4b x
8b =
32
2wb
In the Y direction:
Side strips: xm = 2w x
4b x
8b =
64
2wb
Middle strips: xm = w x b x 8b =
8
2wb
Case-5
The maximum positive moment in the X direction middle strip is then
xfm = 2wbα x
4bα =
8
22wbα
The maximum positive moment is
yfm = αwb x 2bα =
2
22wbα
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72
3.7 EXAMPLE FOR DESIGN OF SLAB
3.7.1 EXAMPLE: DESIGN OF SLAB BY DDM
Problem:
A plan of a market building is given in Figure 3.20. Necessary data are furnished below:
Live load = 60 psf f′c = 4.000 psi
Story height = 9 ft fy = 50,000 psi
Slab thickness = 521 in. No edge beam.
Design the slab as Flat plate by DDM.
All Column 5 @ 12’ = 60 ‘ = 12 in.x10 in. 5 @ 15’ = 75
Figure 3.20 : Floor plan of the building of Example 3.6.1
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73
Solution:
The problem is solved with reference to section 3.4.1.3.
Step 1 : Calculation of Factored Load
Thickness of the slab = 5½″
DL = 150 x 12
5.5 = 69 psf and LL = 60 psf
W= 1.4D + 1.7L = (1.4 * 69) + (1.7 * 60) = 198 psf = 0.198 ksf
Step 2 : Check for Slab Thickness
• Clear span
ln.long = (15-1) = 14′ and ln,short = (12-10/12) = 11.17′
For α = o; ln = 15-1 = 14′
Using Table 3.6 by interpolation for fy = 50 ksi
• For Exterior Panel
tmin = 34.5301
331168*
21
303321
=⎟⎠⎞
⎜⎝⎛ +=⎟
⎠⎞
⎜⎝⎛ + nn ll
in
• For Interior Panel
tmin = 88.4331
361168*
21
333621
=⎟⎠⎞
⎜⎝⎛ +=⎟
⎠⎞
⎜⎝⎛ + nn ll
in
According to ACI Code 9.5.3.2 (a) the minm thickness for flat plate is 5″
So, given thickness of slab = 5½″ (Ok)
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74
Step 3 : Determination of Total Factored Static Moment
Mo = 8
22 nu llW
Mo,long = 81 (0.198)* 12 *142 = 58.2 ft – kips
Mo,short = 81 (0.198) *15 * 11.182 = 46.3 ft – kips
Step 4 : Longitudinal Distribution of Moment
From Table 3.8 for Flat Plate (Case 3 and Case 5)
Mo for A = 58.2 ft – kips
Mo for B = 21 (58.2) = 29.1 ft - kips
Mo for C = 43.3 ft – kips
Mo for D = 23.1 ft – kips
0.20
0.52
0.70
0.35
0.65 0.60
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75
Step 5 : Transverse Distribution of Longitudinal Moment
• Calculation of Aspect Ratio
For A & B : 80.01512
1
2 ==ll
For C & D : 25.11215:
1
2 ==ll
• Calculation of α
Since no edge beam α = o for all
• Calculation of βt
Is in βt
Is = 12
3bh
For A & B : Is = 43
200012
)5.5)(1212( inx=
For C & D : 43
250012
)5.5)(1215( inx=
Torsional Constant, C
Since no actual edge beam, use Figure 2.8 (b) for calculation of the torsional member
C = ∑ ⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
363.01
3 yxyx
For long direction
C = 43
3633
105.510
5.5*63.01 inx=⎥
⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
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76
For short direction:
C = 43
4743
125.512
5.5*63.01 inx=⎥
⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
s
t IC
2=β
For A & B : 118.020002
474==
xtβ
For C & D : 073.025002
363==
xtβ
• Finding out α 1
2
ll
A B C D βt 0.118 0.118 0.073 0.073 α 0 0 0 0
1
2
ll 0.80 0.80 0.80 0.80
α1
2
ll 0 0 0 0
Percentage of Longitudinal Moment in Column Strip
For Exterior Negative Moment
A B C D
98.8% 98.8% 99.3% 99.3%
Explanation for A and B
1
2
ll
= 0.80: 1
2
ll
α = 0; βt = 0.118
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77
From Table 3.9:
1
2
ll 0.5 0.80 1.0 2.0
βt = 0 100 100 100 100
1
2
ll
α = 0 βt = 0.118 98.8
βt > 2.5 75 75 75 75
Interpolation in both directions
For βt = 0.118 % of moment = 118.005.275100 x
−− = 1.2% decrease
∴ % of moment in column strip = (100 – 1.2) = 98.8%
Explanation for C and D
1
2
ll
= 1.25: 1
2
ll
α = 0 ; βt = 0.073
From above Table, by interpolation in both direction
For, βt = 0.073, % of moment decrease = %70.0073.0*)050.2
75100( =−
−
∴ % of moment in column strip = (100 – 0.70) = 99.3%
For Positive Moment
For 1
2
ll
α = 0
A B C D 60% 60% 60% 60%
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78
For Interior Negative Moment
For 1
2
ll
α = 0 and 1
2
ll
= 0.80 and 1
2
ll
= 1.25
A B C D
75% 75% 75% 75%
Table 3.15 : Summary of Calculation
Transverse Distribution of Longitudinal Moment
Ser. Equivalent Rigid Frame A B C D 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
Transverse width (in) Column strip width (in) Half middle strip width (in) C (in4) Is (in4) in βt βt α
1
2
ll
1
2
ll
α
Exterior –ve moment, percent to column strip.
Positive moment percent to column strip Interior negative moment, percent to column strip.
144 72 2 @ 36 474 2000 0.188 0 0.80 0 98.8% 60% 75%
72 36 36 474 2000 0.188 0 0.80 0 98.8% 60% 75%
180 72 2 @ 54 363 2500 0.073 0 1.25 0 99.3% 60% 75%
90 36 54 363 2500 0.073 0 1025 0 99.3% 60% 75%
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79
Table 3.16 : Distribution of factored moment in Column Strip & Middle Strip
For equivalent Rigid Frame A
Exterior span Interior span Ser. Moments at Vritical section (ft.- kips) -ve
moment +ve moment
-ve moment
-ve moment
+ve moment
-ve moment
1. Total moment in equivalent rigid frame ‘A’
-15.1 +30.3 -40.7 -37.8 +20.4 -37.8
2. Percentage to column strip
98.8% 60% 75% 75% 60% 75%
3. Moment in column strip
-14.92 +18.20 -30.53 -28.53 +12.24 -28.35
4. Moment in middle strip
-0.18 +12.10 -10.17 -9.45 +8.16 -9.45
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80
3.7.2 EXAMPLE: DESIGN OF SLAB BY EFM
Problem:
A multi-story market building is planned using a flat plate floor system as shown in Figure 3.21. Necessary data are given below:
Live Load = 100 psf fc′ = 4000 psi
Floor finish = 20 psf fy = 60,000 psi
Floor to floor height = 12 ft. Column size = 18 in. x 18 in.
Design the Interior Panel C by EFM.
Figure 3.21 : Floor plan for Example 3.7.2
A
C
B
B
22ft
22ft
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81
Solution:
The problem is solved with reference to section 2.4.2.5. The EFM is used to determine the longitudinal moments only. As mentioned earlier the transverse distribution process of longitudinal moments and reinforcement calculation are similar to DDM (Section 3.4.1.3, step 4). Hence, this problem is solved upto determination of longitudinal moments. The structure is identical in each direction, permitting the design for one direction to be used for both.
Step 1 : Determination of Slab Thickness
Minimum thickness h for a flat plate is obtained from Table 3.6.For an exterior panel:
h= 30
nl = 30
12*5.20 = 8.20 in.≈ 8.50 in.
Step 2 : Determination of Factored Load
Slab DL = wc * h = (150*8.50)/12 = 106 psf ; wc = weight of concrete= 150 pcf
Super imposed DL = 20 psf
Total DL = (106=20) 126 psf
Factored Load:
DL= 1.4D = 1.4* 126= 176psf
LL= 1.7L = 1.7*100= 170 psf
Step 3 : Determination Flexural Stiffness of Actual Column
Kc = c
ccccc
lIEk
Obtain ck from Appendix C-1
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82
For flat plate structure it is assumed all members are prismatic, neglecting the increase in stiffness within the joint region. Take ck = 4
Consider Ec = constant
Step 4 : Determination Torsional Stiffness of Transverse Torsional Member
• C = ∑ ⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
363.01
3 yxyx
hf = 8.5 in.
C1= 18 in
C = ∑ ⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
318*5.8
185.863.01
3
= 2590 in4
• Kt = 3
2
22 1
9
⎟⎟⎠
⎞⎜⎜⎝
⎛−
∑
lC
l
CEcs
Kt = 3
2212/181264
2590*9
⎟⎠⎞
⎜⎝⎛ −
∑ cE = 109 Ec
Step 5 : Determination of Flexural Stiffness of Equivalent Column
tcec KKK111
+=∑
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83
∑ cK = 2 * 243 Ec
Kt = 2 * 109 Ec
ccec EEK 2181
48611
+= ⇒ ecK = 151 Ec
Step 6 : Determination of Flexural Stiffness of Slab
1lIE
K scsss
κ=
Obtain value of κs from Appendix C-2. For flat plate assume κs = 4
12
32hlI s = =
125.8*264 3
=13510
26413510*4 c
s
EK = = 205 Ec
Step 7 : Calculation of Distribution Factor
Distribution factors at each joint are calculated according to step 7 of section 3.4.2.5.
Step 8 : Determination of Carry Over Factors and Moment Coefficient
C.O.F and Moment coefficient for slab- beam are obtained from Appendix C-2.Moment coefficient 0.083.COF=.503(for both cases).
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Step 9 : Moment Analysis
Since LL = 170 psf > 43 DL=132 psf three loading cases should be considered:
• Total load on all panel • DL load on al panels and ¾ LL on midspan of a panel • DL on all panels and ¾ LL on adjacent panel
Table 3.17 : Longitudinal moment in flat plate floor
Panel B C B
Joint 1 2 2 3 3 4 DF .424 .576 .447 .22 - -
Fixed end moments +307 -307 +307 -307 +307 -307 Final moments +139 -359 +328 -328 +359 -139 Span moment in C 132 (b) 176 psf panels B 304 psf panel C Fixed end moments +156 -156 +270 -270 +156 -156 Final moments +59 -229 +253 -253 +229 -59 Span moment in C 152 (c) 304 psf panels B(left) & C &176 psf panel B (right) Fixed end moments +270 -270 +270 -270 +156 -156 Fixed end moments +120 -325 +306 -235 +220 -62 Span moment in C 134
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3.7.3 EXAMPLE: DESIGN BY COEFFICIENT METHOD
Problem:
A plan of a residential building is given in Figure 3.22. Necessary data are given below:
Live Load = 140 psf
fc′ = 3000 psi
fy = 60,000 psi
Column size = 12” x 12”
Design the corner panel A as two-way slab with beam by coefficient Method.
16’
Figure 3.22 : Floor plan for example 3.7.3
Solution:
The problem is solved with reference to section 3.4.3.2 and Appendix D-1, 2, 3, 4.
AColumn size: 12
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Step1: Determination of Minimum Thickness
h = 18064
180=
p = 0.36 ft. = 4.27 in.
whwre, P = 2 (16+16) = 6
Select h = 5 in. as trial depth
Step 2 : Calculation of Factored Load
DL = wc * 12h psf
DL = 150 x psf635.62125
≈= ; where wc = 150 pcf
LL = 140 psf
W = 1.4D + 1.7 L = (1.4 * 63 + 1.7 * 140) = 326 psf
Step 3 : Determination of Moment Coefficient
Length ratio, m = 11616
==b
a
ll
From the end condition case type is ‘Case 4’
• From Appendix D-1
Ca, neg = 0.05;Cb, neg = 0.05 • From Appendix D-2
Ca,dl,pos = 0.027; Cb,dl,pos = 0.027 • From Appendix D-3
Ca,ll,pos= 0.032; Cb,ll,pos= 0.032
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Step 4 : Calculation of Moment
Middle Strip Moment:
• Positive Moments at Midspan
Ma,pos = Ca,dl Wla2 + Ca, ll Wla
2
Ma,pos = 0.027 * 326 * 162 + 0.032 * 326 * 162
= 4924 ft-lb
Mb,pos = Cb, dl Wlb2 + Ca, ll Wlb
2
Mb,pos = 0.027 * 326 * 162 + 0.032 * 326 * 162
= 4924 ft-lb
• Negative Moments at Continuous Edge
Ma,neg = Ca,neg W la2 = 0.05 * 326 * 162 = 4173 ft – lb
Mb,neg = Cb,neg W lb2 = 0.05 * 326 * 162 = 4173 ft – lb
• Negative Moment at Discontinuous Edge
Ma,neg, discontinuous = 31 * Ma,pos =
31 * 4924 = 1642 ft – lb
Mb,neg, discontinuous = 31 * Mb,pos =
31 * 4924 = 1642 ft – lb
Column Strip Moment:
Column strip moments are 2/3 of corresponding middle strip’s moments in respective direction.
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Step 5 : Check the Design Thickness
d = )59.01(
'c
yy
u
ff
f
M
ρφρ −
Here ρ = ρmax = 0.75ρ b = 0.75 * 0.85*β 1y
c
ff /
*yf+87000
87000 = 0.016
d = )
360*016.0*059.01(12*000,60*016.0*90.0
12*4924
− = 2.41 in.
hrequired = (d + clear cover= 1 in ) = 3.41 in.
hrequired < hdesign, design is OK
Step 6 : Calculation for Reinforcement for Middle Strip
In Short Direction:
• Midspan
Mu = 4924 *12 lb- in.
By Iteration process as given in Step 5 of section 3.3 find:
As = 0.30 in.2/ft
Using # 3 bar required spacing:
Spacing = ccx /44.430.0
11.012 ′′≈=
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89
• Continuous Edge
Mu = 4273 *12 lb- in.
By Iteration process as given in Step 5 of section 3.3 find:
As = 0.25 in2/ft.
Using # 3 bar required spacing:
Spacing = ccx /528.525.0
11.012 ′′≈=
• Discontinuous Edge
The negative moment at discontinuous edge is one third of positive moment in the span. It would be adequate to bend up every third bar from the bottom to provide negative moment steel at discontinuous edge.
However the spacing would be = 12″
But maximum allowable spacing = 2h = 10 in.
So, using # 3 @ 10 in c/c.
In Long Direction:
Being equal moments, the reinforcement in long direction will be equal to short direction in this case.
Step 7 : Calculation for Reinforcement for Column Strip
The average moments in columns being two-third of the corresponding moments in the middle
strips, adequate steel will be furnished if the spacing of this steel is 23 times that in the middle
strip.
Using # 3 bar spacing for column strip
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• Midspan = 4 x 23 = 6″ c/c
• Continuous edge = 5x23 = 7.5″ c/c
• Discontinuous edge = 8x23 = 12″ c/c
But maximum allowable spacing = 2h = 10 in.
Use 10″ c/c.
Step 8: Detailing
A
A
B
B
Figure 3.23 : Detailing for example 3.7.3
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91
Sect
ion
A-A
Sec
tion
B-B
# 3
@ 5
in c
/c
# 3
@ 1
0 in
c/c
#
3 @
4 in
c/c
Figure 3.23 : Detailing for example 3.7 .3 (continued)