Slab Design

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 l l > 2;

• Two-way slab : Depends on support condition. (Figure 3.1c) 2 1 l l ≤ 2

(a) One- way slab

2 1 l l

_>2

1 l 2 l 2 1 l l

_>

₂

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 1 l 2 l 1 l 2 l

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 , 100 y f (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 (w

u

)

wu = 1.4D+ 1.7L psf

Dead load, D = wc x

12

h

psf

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 . 0 1 ( c y y u f f b f M ′ − ρ φρ putting ρ = ρmax = 0.75ρb. Where, ρ_b= 0.85*β₁ y c f f / * y f + 87000 87000

Values of β1 is given in Table 3.2

Table3.2 : Values of β₁ (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 ≤ β₁≤ 0.85

Table 3.3: Clear cover for slab ( ACI Code 7.7.1)

No 14 & No 18 bars ... .... 1 1/2 in

t

a

l

h

(a) Slabs not built integrally with the support (ACI Code 8.7.1)

l= la + h ≤ la + t t a l h l= la + t

(b) Slabs are continuous (ACI Code 8.7.2)

t

a

l

(c) Slabs built integrally with support If (d + clear cover) < h; Design is ok.

Otherwise redesign the thickness.

3. Discontinuous ends are

Table 3.4 : ACI moment coefficient Support Condition 1 9 1 1 1 0 0 Moment Coefficent For two span:

1. Discontinuous ends are

1 2 4 1 9 1 1 4 1 1 4 0 1 1 0 1 1 1 1 1 1 1 2 4 1 1 0 1 1 4 1 2 4 1 9 1 1 6 1 1 4 1 1 6 1 9 1 9 1 1 4 unrestrained

built integrally with support 2. Discontinuous ends are (spandrel beam or girder)

1 1 6 1 1 1 1 1 1 1 1 6 1 1 1 1 1 1 1 1 1 1 1 4 1 1 6 1 1 6 1 1 0 1 1 1 1 1 1 1 1 1

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 w 2 1 9 1 1 1

Step 5 : Determination of Steal Area (A

s

)

Reinforcement for 1 ft. strip towards shorter distance is calculated by Iteration. Details shown in Figure 3.3 (here b = 12 in)

Calculate a for next trial

with

(As)corrected (a) corrected= b f f A c y s ′′ ′ 85 . 0 (As) trial =

(

)

2 a d f M y u − φ (As)corrected A_s,corr ≈ As,trial Yes OK

Assume a (hints: a =

0.3d

) No

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_in2₎

Spacing = s A 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 y f 000 , 60 0018 . 0 ×

But should be: ρ > 0.0014

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 wuln

Critical shear at a distance d from support, Vu = (1.15

12 2 d W l W_{u n u} − )

Design strength for shear, φ Vc = φ 2 fc′ 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

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

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

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 = _{1 2}

2 1 2 1 l or l 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

• 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. 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 l1 l1 l1 l2 2 l 2

• If beams are used on the column lines, the relative stiffness of the beams in the two perpendicular direction shall be:

0.2< 1 2 2 2 2 1 l l α α < 5.0 1 α = α in direction of ℓ₁ 2 α = α in direction of ℓ₂ s cs b cb I E I E = 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 33 ln 36 ln 36 ln 36 ln 40 ln 40 ln 60,000 30 ln 33 ln 33 ln 33 ln 36 ln 36 ln 75,000 28 ln 31 ln 31 ln 31 ln 34 ln 34 ln

• For slabs with beams along exterior edges, the value of α for edge beam shall not be less than 0.80.

• For fybetween 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 ( 5 36 ) 000 , 200 8 . 0 ln( − + + m y f α β h = 36 9β ) 000 , 200 8 . 0 ln( + + 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 (M

o

)

According to ACI Code 13.6.2.2 Mo = 8 2 2 n ul l W

Mo, of an exterior equivalent frame =

2 1

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

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

0 .6 5 M o 0 .6 5 M o 0 .3 5 M o M o 0.75Mo 0.63Mo Mo 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:

0 .1 6 M o

0 .7 0 M o 0 .5 7 M o

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 l l • Stiffness ratio, α = s cs b cb I E I E • Ratio βt = s cs cb I E C E 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

Calculation of Three Parameters

• Calculation of Aspect Ratio =

1 2

l l

• 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 = s cs b cb I E I E

Determination of Moment of Inertia of the Beam

Ib = k 12

3

h b_w

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 3 bh 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 = s cs cb I E C E 2 h_w ≤4h_f b_w+2h_w ≤b_w+8h_f

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 =

∑

_⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − 3 63 . 0 1 x3y y x

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

direction of moment Flat 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

hf

(c) Torsional member ( ACI Code 13.7.5.1c) be=(bw+hw) ≤ (bw+4hf) h bw hf hw≤4hf hw≤4hf hw bw hw≤4hf be=(bw+2hw) ≤ (bw+8hf)

Edge beam Interior beam

t

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 hf 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 X1 X1 X2 X1 X2 Y2 x1 X2 x Y Y Figure 3.9 : Determination of C

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_s,min = ρbh

ρ = minimum steel ratio for temperature and shrinkage as shown in Table 3.5 A_s,provided > A_s,min (OK)

Otherwise provide A_s,min

Step 8 : Total Number of Bar

b s

A A

N = , A_b = cross section area of bar used

Spacing =

N b

≤ 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.

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 1-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).

Figure 3.10 : Details of reinforcement in two-way slab (a) Section along short span (2)

Maximum spacing < 2h ( ACI Code 13.3.2)

3 4" min h

(b) Section along long span ( 1)

3 4" min

(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

Figure 3.11 : Spiral reinforcement at exterior corner G ride type l 5 l 5 l 5 l 5 T o p b a r B o tto m b a r (Provided in tw o layers) (Providedin band) D iagonal type

L/5

L/5

L/5

L/5

Figure 3.12 : Details of drop panels

b

h

t

b

≥ 1₆

_l

1 t

t

min ≥ 1₄

h

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.

3.4.2.2 Moment Of Inertia of Slab Beam (I

s

)

• Considering gross area of concrete (ACI Code 13.7.3.1)

Is = 12

3 2h

l

• 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 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −C _l I_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.

3.4.2. 4 Moment of Inertia of Column (I

c

)

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 (K

c

)

Kc = c cc cc c l I E κ

c

1

I = ∞

I = ∞

I

c

=

12 3 1 2c c

Variab

le

I_c= moment of inertia of column = 12 3 1 2C C l_c = length of column (c/c)

Step 3 : Torsional Stiffness of Transverse Torsional Member (K

t

)

K

t

=

₃ 2 2 2 1 9 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −

∑

l C l C E_cs

• 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_t,corrected= Kt * s sb

I I

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 (K

ec

)

t c ec K K K 1 1 1 _{= +}

∑

∑ Kc = Kc1 + Kc2

Step 6 : Flexural Stiffness of Slab (K

s

)

1 l I E K s cs s s κ =

κs = coefficient of slab stiffness ( to be obtained from Appendix C-2 and C-3)

Is = 12

3 2h

Step 7 : Distribution Factor (D.F)

1 1 1 ec s s k k k + K_S2 K_SS₃2 K_ec2 K + + 2 3 2 3 ec S S s K K K K + + 1 1 1 ec s ec k k k + k_s2 k_sec₃2 k_ec2 k + +

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.

¾ Live loading pattern is known, frame shall be analyzed for that load. ( ACI Code

13.7.6.1)

k

s2

K

ec2

k

ec1

k

s1

k

s3

DF DF DF DF

DF

¾ Variable LL, but LL >

4 3

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

4 3

LL on midspan of a panel.

™ DL on all panels and

4 3

LL on adjacent panels.

• Total panel moment (Mp) is computed using equation:

Mp = 8 2 2 n ul l w

• 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 -

2

1 _{[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 T

M Mo

as such,

∑ Design moments < Mo (ACI code 13.7.7.4)

MT = total panel moment

Mo = 8

2 2 nl

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

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

4 b l

;C

4 b l

;C S

2 b l

;

4 a l

; C.

2 a l

;M

4 a l

; C S

b l

l

a

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 = 3.5

180≥

P

in, P = panel perimeter

Step 2 : Calculation of Factored Load

wu = 1.4 D + 1.7L

D = dead load = wc *

12

h

psf ; wc = 150 lb

Step 3 : Determination of Moment Coefficient

m =

lb la

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.

Step 4 : Calculation of Moment

Middle strip moment

¾ Positive moment

Ma, pos = Ca, dl wu la2 + Ca, ll wu la2 Mb, pos = Cb,dl wu lb2 + Cb, ll wu lb2

¾ Negative Moments for continuous Edge Ma, neg,cont = Ca, neg wu la2

Mb, neg,cont = Cb, neg wu lb2

¾ Negative Moments for Discontinuous Edge Ma, neg, discont = 3 1 Ma, pos Mb, neg, discont = 3 1 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 . 0 1 ( c y y u f f b f M ′ − ρ φρ

If (d + clear cover) < h ; design is ok.

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 ¾ Midspan ¾ Continuous Edge ¾ Discontinuous Edge • Long Direction ¾ Midspan ¾ Continuous Edge ¾ 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.

Shear strength of the slab, φ V_c = 2 f ′cbd

φV_c >Vu 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.

Table 3.11 : Calculation of immediate deflection

1. Live load deflection

Δ_l = e c b b I E l M 32 3 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 = e c b b I E l M 48 5 2

# Slab supported by masonry walls

# Mb = live load +ve moment

3. Dead load deflection

Δ_d = e c b b I E l M 16 1 2

# Both ends continuous and fully fixed

# Mb = maximum dead load +ve moment

4. Dead load deflection

Δ_d = e c b b I E l M 48 5 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 = 1.4 1.7 , , , ,posdl bposll b M or M 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 = 1.4 1.7 , , , ,posdl bposll b _or M M in long direction

Determination of I

e

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 /

c

f

For light weight concrete one of the following modifications shall apply:

• When average tensile strength, f_ct is specified

f_r= 7.5 7 . 6 ct f = 1.12 f_ct, 7 . 6 ct f ≤ / c f

• When f_ct is not specified

f_r= 0.75 * 7.5 7 . 6 ct f

; for all lightweight concrete

f_r= 0.85 * 7.5 7 . 6 ct f

; for sand- lightweight concrete

• I_e for Continuous Spans ( ACI Code 9.5.2.4)

I_e = 0.50 I_em+ 0.25 ( I_e1+ I_e2)

I_em= effective moment of inertia for the midspan section

I_e1, I_e2= 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

Δlong = Δd,short * λ Where, λ= _/ 50 1 ρ ξ + /

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

total Δ = Δ_long +Δ_l,short

Desired value : Δ_total< 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 ₁₈₀l

Floors not supporting or attached to nonstructural elements likely to be damaged by large deflection

Immediate deflection due to

live load L ₃₆₀l

Roof or floor construction supporting or attached to nonstructural elements likely to

be damaged by large deflection 480

l

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.

240

l

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.

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

m_x = my =

16

2

wa

Simple supports 4 sides

16 2 wa

(d) m

x

across

X=a/2

Y

X

A A

a

a

w/2

(a) Plan view

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

m_x = 2

2

wy

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

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 m_x = 2 w x 4 a x 8 a = 64 2 wa

And for a strip along section B-B, the maximum moment is

m_x = w x 4 a x 8 a + 2 w x 4 a x 8 3a = 64 5_wa2

(d) m

x

across

x=a/2

Y

X

B

A

B

a

w/2

(a) Plan view

(b) w

x

and

m

x

along A-A

w/2

a/2 a/4

a/4

a/4

5Wa

2

/6

4

a/2

A

a/4

a

w

w/2

w/2

w/2

w/2

w/2

w

Wa

2

/64

(c) w

x

and

m

x

along B-B

5Wa

2

/6

4

w/2

w w

Condition-4:

The preferred arrangement, shown in figure below, gives design moment as follows: In the X direction: Side strips: m_x = 2 w x 4 b x 8 b = 64 2 wb Middle strips: m_x = w x 4 b x 8 b = 32 2 wb In the Y direction: Side strips: m_x = 2 w x 4 b x 8 b = 64 2 wb Middle strips: m_x = w x b x 8 b = 8 2 wb

2 b a b ×

Wa

2

/64

b

2 b a b × a b a− 2

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

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

xf m = 2 wb α x 4 b α = 8 2 2_wb α

Accordingly, the distance from the right support, figure “c”, to the maximum positive moment

section is chosen as α . It follows that the maximum positive moment is b

yf m = αwb x 2 b α = 2 2 2_wb α

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.

(

)

2 1−α b 2 b α 2 b a−

b

a

w/2

w/2

w

(

)

8 2 1− α wb2 8 2 2 wb α

(a) Plan view

Figure 2.25: wx and mx along A-A

2 b α

b/2

(

)

2 1−α b

w/2

w/2

A A

B

B

w

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 2 2 wb α

(

)

2 2 1 2 wb α −

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 12 h (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 . 0 1 ( ' c y y u f f f M ρ φρ −

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 . 0 1 ( c y f f ρ − )

If φMn<M; then only minimum reinforcement. If φMn>M; then iteration from figure-3.3.

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.

Table 3.14: Choice of Load Distribution Case-1 m_x = my = 16 2 wa Case-2 m_x = 2 2 wy Case-3

For an X direction strip along section A-A, the maximum moment is m_x = 2 w x 4 a x 8 a = 64 2 wa

And for a strip along section B-B, the maximum moment is

m_x = w x 4 a x 8 a + 2 w x 4 a x 8 3a = 64 5_wa2 Case-4 In the X direction: Side strips: m_x = 2 w x 4 b x 8 b = 64 2 wb Middle strips: m_x = w x 4 b x 8 b = 32 2 wb In the Y direction: Side strips: m_x = 2 w x 4 b x 8 b = 64 2 wb Middle strips: m_x = w x b x 8 b = 8 2 wb Case-5

The maximum positive moment in the X direction middle strip is then xf m = 2 wb α x 4 b α = 8 2 2_wb α The maximum positive moment is

yf m = αwb x 2 b α = 2 2 2_wb α

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 = 5

2 1

in. No edge beam. Design the slab as Flat plate by DDM.

All

Column

5 @ 12

’

= 60

‘

= 12 in.x10 in.

Solution:

The problem issolved 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 = 5.34 30 1 33 1 168 * 2 1 30 33 2 1 ₌ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ln ₊ ln _in

• For Interior Panel

tmin = 4.88 33 1 36 1 168 * 2 1 33 36 2 1 ₌ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ln ₊ ln _in

According to ACI Code 9.5.3.2 (a) the minm thickness for flat plate is 5″

Step 3 : Determination of Total Factored Static Moment

Mo = 8 2 2 n ul l W Mo,long = 8 1 (0.198)* 12 *142 = 58.2 ft – kips Mo,short = 8 1 (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 = 2 1 (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

Step 5 : Transverse Distribution of Longitudinal Moment

• Calculation of Aspect Ratio

For A & B : 0.80 15 12 1 2 _{= =} l l For C & D : 1.25 12 15 : 1 2 _{= =} l l • Calculation of α

Since no edge beam α = o for all

• Calculation of βt ¾ Is in βt Is = 12 3 bh For A & B : Is = 4 3 2000 12 ) 5 . 5 )( 12 12 ( in x = For C & D : 3 ₂₅₀₀ 4 12 ) 5 . 5 )( 12 15 ( in x ₌

¾

Torsional Constant, C

Since no actual edge beam, use Figure 2.8 (b) for calculation of the torsional member C =

∑

_⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − 3 63 . 0 1 3_y x y x For long direction

C = 3 ₃₆₃ 4 3 10 5 . 5 10 5 . 5 * 63 . 0 1 x _⎥= in ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −

For short direction: C = 3 ₄₇₄ 4 3 12 5 . 5 12 5 . 5 * 63 . 0 1 x _⎥= in ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −

¾

s t I C 2 = β For A & B : 0.118 2000 2 474 ₌ = x t β For C & D : 0.073 2500 2 363 ₌ = x t β • Finding out α 1 2 l l

A B C D

β

t

0.118 0.118 0.073 0.073

α

0 0 0 0

1 2 l l

0.80 0.80 0.80 0.80

α

1 2 l l

0 0 0 0

Percentage of Longitudinal Moment in Column Strip

¾ For Exterior Negative Moment

A B C D

98.8% 98.8% 99.3% 99.3%

From Table 3.9: 1 2 l l

0.5 0.80 1.0 2.0

β

t

= 0

100 100 100 100

1 2 l l α

= 0

β

t

= 0.118

98.8

β

t

> 2.5

75 75 75 75

Interpolation in both directions

For βt = 0.118 % of moment = 0.118 0 5 . 2 75 100 x − − = 1.2% decrease ∴ % of moment in column strip = (100 – 1.2) = 98.8%

Explanation for C and D

1 2 l l = 1.25: 1 2 l l α = 0 ; βt = 0.073

From above Table, by interpolation in both direction

For, βt = 0.073, % of moment decrease = )*0.073 0.70%

0 50 . 2 75 100 ( = − −

∴ % of moment in column strip = (100 – 0.70) = 99.3%

¾ For Positive Moment

For

1 2 l l α

= 0

A

B

C

D

60%

60%

60%

60%

¾ For Interior Negative Moment For 1 2 l l α = 0 and 1 2 l l = 0.80 and 1 2 l l = 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 l l 1 2 l l 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%

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

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.

A

C

B

B

22ft

22ft

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 n l = 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 cc cc c l I E k

For flat plate structure it is assumed all members are prismatic, neglecting the increase in

stiffness within the joint region. Take k = 4_c

Consider Ec = constant

Step 4 : Determination Torsional Stiffness of Transverse Torsional Member

• C =

_∑

_⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − 3 63 . 0 1 3_y x y x

h

f

= 8.5 in.

C

1

= 18 in

C =

∑

_⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 3 18 * 5 . 8 18 5 . 8 63 . 0 1 3

= 2590 in

4

• K

t = ₃ 2 2 2 1 9 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −

∑

l C l C E_cs

K

t = ₃ 22 12 / 18 1 264 2590 * 9 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −

∑

Ec =

109 E

c

Step 5 : Determination of Flexural Stiffness of Equivalent Column

1 1

∑

K = 2 * 243 Ec c Kt = 2 * 109 Ec c c ec E E K 218 1 486 1 1 _{= + ⇒} ec K = 151 Ec

Step 6 : Determination of Flexural Stiffness of Slab

1 l I E K s cs s s κ =

Obtain value of κs from Appendix C-2. For flat plate assume κs = 4

12 3 2h l I_s = = 12 5 . 8 * 264 3 =13510 264 13510 * 4 _c s E K = = 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).

Step 9 : Moment Analysis

Since LL = 170 psf >

4 3

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

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 issolved with reference to section 3.4.3.2 and Appendix D-1, 2, 3, 4.

A

Column

size: 12

Step1: Determination of Minimum Thickness

h = 180 64 180 = 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 * 12 h psf DL = 150 x 62.5 63psf 12 5 _{= ≈} ; 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 = 1 16 16 = = b a l l

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 C = 0.032; C = 0.032

Step 4 : Calculation of Moment

Middle Strip Moment:

• Positive Moments at Midspan Ma,pos = Ca,dl Wla2 + Ca, ll Wla2 Ma,pos = 0.027 * 326 * 162 + 0.032 * 326 * 162 = 4924 ft-lb Mb,pos = Cb, dl Wlb2 + Ca, ll Wlb2 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 = 3 1 * Ma,pos = 3 1 * 4924 = 1642 ft – lb Mb,neg, discontinuous = 3 1 * Mb,pos = 3 1 * 4924 = 1642 ft – lb

Column Strip Moment:

Step 5 : Check the Design Thickness

d = ) 59 . 0 1 ( ' c y y u f f f M ρ φρ − Here ρ = ρmax = 0.75ρ_b = 0.75 * 0.85*β₁ y c f f / * y f + 87000 87000 = 0.016 d = ) 3 60 * 016 . 0 * 059 . 0 1 ( 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 = 12x0.11=4.4≈4′′c/c

• 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 = x 5.28 5 c/c 25 . 0 11 . 0 12 _{= ≈ ′′} • 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

2 3

times that in the middle strip.

• Midspan = 4 x 2 3 _{= 6″ c/c} • Continuous edge = 5x 2 3 _{= 7.5″ c/c} • Discontinuous edge = 8x 2 3 _{= 12″ c/c}

But maximum allowable spacing = 2h = 10 in. Use 10″ c/c.

Step 8: Detailing

A

A

B

B

Section A-A

Section B-B

# 3 @ 5 in c/c

# 3 @ 10 in c/c

# 3 @ 4 in c/c