Two Way Slab Analysis

Direct Design Method for Two-way Slab

Analysis

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar Direct Design Method

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk) Page 1 of 58

0.1l l

1. Introduction:-

The direct design method consists of a set of rules for distributing moments to slab

and beam sections in a two-way slab system.

2. Limitations on use of Direct Design method (ACI 13.6.1):-

(i) Minimum of 3 continuous spans in each direction (3 × 3 panel).

(ii) Rectangular panels with long span/short span ≤ 2.

(iii)Successive span in each direction shall not differ by more than 1/3 the longer

span.

(iv) Columns may be offset from the basic rectangular grid of the building by up to 0.1

times the span parallel to the offset (figure 01).

.

Figure 01: Column offset at a distance of 0.1l from the basic rectangular grid.

(v) All loads must be due to gravity only (N/A to un-braced laterally loaded frames,

mats or pre-stressed slabs).

(vi) Service (unfactored) live load ≤ 2 (service dead load).

(vii)For panels with beams between supports on all sides, relative stiffness of the

beams in the two perpendicular directions:

Shall not be less than 0.2 nor greater than 5.0.

212

221

l

l

α

α



slab of stiffness flexural

beam of stiffness flexural=α

scs

bcb

scs

bcb

4E

4E

/4E

/4E

I

I

lI

lI==α

slab uncracked of inertia ofMoment Ibeam uncracked of inertia ofMoment I

concrete slab of elasticity of Modulus Econcrete beam of elasticity of Modulus E

s

b

sb

cb

====

h < 4hw

hwhf

l /22

(a) Section for I (Edge beam)b

(b) Section for I (Edge beam)s

b + 2h < b + 8hw

hwhf

l /22

(c) Section for I (Interior beam)b

w f

l /22

bw

(d) Section for I (Interior beam)s

f w

Where α is the ratio of flexural stiffness of beam section to flexural stiffness of width

of slab bounded laterally by centerlines of adjacent panels (if any) on each side of the

beam.

The width of slab is bounded laterally by centerline of adjacent panels on each side of

the beam (figure 02).

Figure 02: Ib and Is in case of interior and exterior beams.

3. Definitions related to Direct Design Method:-

(i) Frames: Slab is considered to be a series of frames in two directions.

(ii) Panel (ACI 13.2.3): A panel is bounded by column, beam, or wall centerlines on

all sides. A panel includes all flexural elements between column centerlines.



ln

Exterior Frame EW1

Interior Frame EW2

0.25l or 0.25l(Whichever is less)

l2

l2

N

n

Exterior frame NS1

Interior frame NS2

Interior frame NS3

Exterior frame NS4

l

l2 l2 l2

Pane

lColumn strip

Middle strip

Half middle strip

Column strip

1 2

0.25l or 0.25l(Whichever is less)

1 2

Interior

Frame EW3

Exterior

frame EW4

N

(a)

(b)

l2

l1 l1 l1

1 l

1 l

1 l

Figure 03: Slab system divided into EW and NS frames.



(iii)Column Strip (ACI 13.2.1): Column strip is a design strip with a width on each

side of a column centerline equal to 0.25l2 or 0.25l1, whichever is less. Column

strip includes beams, if any.

(iv) l1: l1 is the length of span in direction that moments are being determined,

measured center-to-center of supports.

(v) l2: l2 is the length of span transverse to l1, measured center-to-center of supports.

(vi) Middle strips (ACI 13.2.2): Middle strip is a design strip bounded by two

column strips.

4. Distribution of Moments:-

(i) Total static Moment, Mo (ACI 13.6.2): The total static moment for a span length

ln and width l2 of a given frame is given by ACI equation 13-3 as:

Where,

wu = Factored load per unit area.

ln = length of clear span in direction that moments are being determined,

measured face-to-face of supports.

l2 = As defined above. However, two exceptional cases as defined by ACI

are given below:

ACI 13.6.2.3 — where the transverse span of panels on either side of the

centerline of supports varies, l2 in Eq. (13-3) shall be taken as the average of

adjacent transverse spans.

ACI 13.6.2.4 — when the span adjacent and parallel to an edge is being

considered, the distance from edge to panel centerline shall be substituted for l2 in

Eq. (13-3).

ln is defined by ACI as given below:

ACI 13.6.2.5 — clear span ln shall extend from face to face of columns, capitals,

brackets, or walls. Value of ln used in Eq. (13-3) shall not be less than 0.65l1.

Circular or regular polygon shaped supports shall be treated as square supports

with the same area.

( )3-13 ACI 8

2n2u

0llwM =



ACI R13.6.2.5 — if a supporting member does not have a rectangular cross

section or if the sides of the rectangle are not parallel to the spans, it is to be

treated as a square support having the same area, as illustrated in Fig. R13.6.2.5.

Figure 04: Equivalent square section for supporting members, ACI fig 13.6.2.5.

(ii) Longitudinal Distribution of Static Moment (ACI 13.6.3): For a typical interior

panel, the total static moment is divided into positive moment 0.35Mo and

negative moment of 0.65Mo.

For an exterior panel, the total static moment is dependent on the type of restraint

at the outside edge.

ACI table 13.6.3.3 (table 13.3 Nilson 13th Ed) as shown in figure 05 of this

document can be used for longitudinal distribution. Alternatively, figure 06 of this

document can also be used.

Figure 05: Table for longitudinal distribution of static moment in exterior

panel.



0

0.63

0.75 0.65 0.65

0.35

0.65 0.65

0.35

0.65 0.65

0.35

0.65 0.65

0.35

0.65 0.65

0.35

0.65

0.16

0.30

0.26

0.35

0.57

0.50

0.52

0.65

0.70

0.70

0.70

Exterior FirstInterior Interior

Support Support

1 No restraint

2 Full restraint

3 Slab withbeam between supports

4 Edge beamonly (no other beam)

5 No beams

Figure 06: Longitudinal moment distribution.

Note that the longitudinal moment values mentioned are for the entire width of

the equivalent building frame i.e., the width of two half column strips and two

half-middle strips of adjacent panels.



β = tE Ccb2E Ics s

y

y

x

x

2

2

1

1

Slab

Beam

(iii)Transverse or Lateral distribution of Longitudinal Moments:-

Tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI as given in figure 08 are used to

assign moments to column strip. The remaining moments are assigned to middle

strip in accordance with ACI 13.6.3. Beams between supports shall be

proportioned to resist 85 percent of column strip moments if α1l2/l1 {Where l2

shall be taken as full span length irrespective of frame location (exterior or

interior)} is equal to or greater than 1.0 (ACI 13.6.5.1). As an alternative the ACI

tables mentioned above for the assignment of moments to column strips, figure 09

(Graph A4, Nilson 13th Ed) can also be used.

Moreover, figure 10 of this document illustrates the summary of lateral

distribution of moments for slab system without beams.

Transverse distribution of the longitudinal moments to middle and column strips

is a function of the ratio of length (l2/l1), α1, and βt.

Where,

βt = the ratio of torsional stiffness of edge beam section to flexural stiffness

of a width of slab equal to span length of beam, center-to-center of

supports.

If there is no edge beam, βt is taken equal to zero. If there is edge beam, βt is

calculated as follows.

Where, C is the torsional constant of the edge beam. This is roughly equal to the

polar moment of inertia of edge beam and is given as:

C =∑[1–0.63{x/y}×x3y/3] = [1 – 0.63{x1/y1}×x13y1/3] + [1 – 0.63{x2/y2}×x2

3y2/3]

Where, “x” is the shorter side of the rectangle and “y” is the longer one.

Figure 07: Cross-section of torsional member (edge beam) for calculation of βt.



Figure 08: Lateral distribution.



Figure 09: Graph A4, (Nilson 13th Ed) for transverse distribution of longitudinal

moments.



12 middle strip

12 column stripA B C

l /22

l /22

12 column strip

12 middle strip

l /22

l /22

l /4 or l /4(whicheveris smaller)

1 2

Exterior panel Interior panel Interior panel

Center line of panel

Center line of panel

A B C

Panel momentM = 0.52Mp o

M = 0.26Mne o M = 0.70Mni o M = 0.65Mni o M = 0.65Mni o

M = 0.35Mp o

A B C

Column strip

(-0.26M ) 100%

o

(+0.312M ) 60%

o

(-0.525M ) 75%

o (-0.49M ) 75%

o (-0.49M ) 75%

o

(+0.21M ) 60%

o

A B C

Middle strip

0

(+0.208M ) 40%

o

(-0.175M ) 25%

o (-0.16M ) 25%

o (-0.16M ) 25%

o

(+0.14M ) 40%

o

Figure 10: Summary of longitudinal & lateral distribution for slabs without

beams.



( )2.0536200,000

8.0

m

yn

−+

+

=αβ

fl

h

β936200,000

8.0 yn

+

+

=

fl

h

5. Minimum Slab Thickness for two-way construction (ACI 9.5.3):-

The definitions of the terms are:-

h = Minimum slab thickness without interior beams.

ln = length of clear span in direction that moments are being determined, measured

face-to-face of supports.

β = ratio of clear spans in long to short direction of two-way slabs.

αm = average value of α for all beams on edges of a panel.

a. For 0.2 ≤ αm ≤ 2:

But not less than 5 in. fy in psi.

b. For αm > 2:

But not less than 3.5 in. fy in psi.

c. For αm < 0.2, use the ACI table 9.5 (c), reproduced in figure 11 of this document.

Figure 11: Table for minimum thickness of slabs without interior beams.



Additionally, slab systems with αm < 0.2 shall also fulfill the following requirements:

• For slabs without drop panels meeting ACI 13.3.7.1 and 13.3.7.2,

hmin = 5 in

• For slabs with drop panels meeting ACI 13.3.7.1 and 13.3.7.2,

hmin = 4 in

ACI 13.3.7.1 — Drop panel shall extend in each direction from centerline of support

a distance not less than one-sixth the span length measured from center-to-center of

supports in that direction.

ACI 13.3.7.2 — Projection of drop panel below the slab shall be at least one-quarter

the slab thickness beyond the drop.

6. Max Spacing and Min Reinforcement requirement of the ACI code:

• One way slab (ACI 7.6.5 & 7.12.2.2):

Main Reinforcement = 3 hf, or 18 in whichever is less (hf = slab thickness)

Temp reinforcement = 5 hf or 18 in whichever is less.

• Two way slab (ACI 13.3.2):

2 hf in each direction.

• Min Reinforcement in all cases (ACI 7.12.2.1):

0.0018 b hf for grade 60.

0.002 b hf for grade 40 and 50.

7. Detailing of flexural reinforcement for column supported two-way slabs:

• For protection of the steel against damage from fire or corrosion, at least 3/4 in.

concrete cover must be maintained.

• Because of the stacking that results when bars are placed in perpendicular layers,

the inner steel will have an effective depth 1 bar diameter less than the outer steel.

• In case of two way slabs supported on sufficiently stiff beams with α l2/l1 greater

than 1.0, curvatures and moments in the short direction are greater than in the

long direction of a rectangular panel, therefore short-direction bars are normally

placed closer to the top or bottom surface of the slab, with the larger effective

depth d, and long-direction bars are placed inside these, with the smaller d.



• In the case of flat plates/slabs, it is clear that the middle-strip positive moments

(for example) are larger in the long direction than the short direction, exactly the

opposite of the situation for the slab with stiff beams. In the column strips,

positive and negative moments are larger in the long than in the short direction.

On this basis, the designer is led to place the long-direction negative and positive

bars, in both middle and column strips, closer to the top or bottom surface of the

slab, respectively, with the larger effective depth.

• If column-line beams are added and if their stiffness is progressively increased for

comparative purposes, it will be found that the short-direction slab moments

gradually become dominant, although the long-direction beams carry larger

moments than the short-direction beams.

• The best guide in specifying steel placement order in areas where stacking occurs

is the relative magnitudes of design moments obtained from analysis for a

particular case, with maximum d provided for the bars resisting the largest

moment. No firm rules can be given. For square slab panels, many designers

calculate the required steel area based on the average effective depth, thus

obtaining the same bar size and spacing in each direction.

• In case of DDM standard bar cut off points from figure 13 of this document are

used as recommended in ACI code, figure 13.3.8.

• ACI 13.3.8.5 requires that all bottom bars within the column strip in each

direction be continuous or spliced with Class A splices (1.0 ld, For development

length see ACI 12.2.3 or Nelson 13th Ed, page 172 chapter 5) or mechanical or

welded splices. At least two of the column strip bars in each direction must pass

within the column core and must be anchored at exterior supports (ACI 13.3.8.5).

8. Reinforcement at exterior corners:

• Reinforcement should be provided at exterior corners in both the bottom and top

of the slab, for a distance in each direction from the corner equal to one-fifth the

longer span of the corner panel as shown in figure 12, below.

• The reinforcement at the top of the slab should be parallel to the diagonal from

the corner, while that at the bottom should be perpendicular to the diagonal.



• Alternatively, either layer of steel may be placed in two bands parallel to the sides

of the slab. The positive and negative reinforcement, in any case, should be of a

size and spacing equivalent to that required for the maximum positive moment in

the panel, according to ACI 13.3.6.

Figure 12: Reinforcement at exterior corners of slab



Figure 13: ACI fig 13.3.8.



l = 20'

20'-0"

2

l = 10'-7"2

7"

5'

5'

5'

5'

5'

5'

20'-0"

20'-0"

l = 25'-0"1 l = 25'-0"1 l = 25'-0"1

Half middle strip

Half middle strip

Half middle strip

Half column strip

Half column strip

Half column strip

All columns 14" square

All beams14" x 20"

A

Design Pb.1: Design the slab shown below (Follow the Direct Design Method for the

slab analysis).

Data Given:

A 75′ × 60′ building, divided into nine (9) panels using beams supported at their ends

on columns. Each panel is 20′ × 25′.

fc′ = 4 ksi

fy = 60 ksi

Height of building = 10′

Column dimensions = 14″ × 14″

Live load = 144 psf

Figure 14: Given slab beam system.



Solution: -

Check if the slab system satisfies all the limitations for Direct Design Method.

1) There must be a minimum of three continuous spans in each direction.

2) The panels must be rectangular, with the ratio of the longer to shorter spans

within a panel not greater than 2.

3) The successive span lengths in each direction must not differ by more than one

third of the longer span.

4) Loads must be due to gravity only and the service live load must not exceed 2

times the service dead load.

5) If beams are used on the column lines, the relative stiffness of the beam in the two

perpendicular directions, given by the ratio αl2/l1, must be in between 0.2 and 5.0.

6) Columns may be offset a maximum of 10 percent of the span in the direction of

the offset from either axis between centerlines of the successive columns.

Step No 1: Sizes for beams, slab and column.

• Beams: Let assume all beam sections equal to 14″ × 20″.

• Column: Let the column dimensions = 14″ × 14″.

• Slab Thickness: To find the minimum slab thickness, ACI equations (section

13.8, Nelson 13th Ed) will be used which utilizes αm (average value of α for all

beams on edges of a panel). For this purpose, the relevant calculation is given

in table 01.

Let assume slab thickness (hf) equal to 7″. Then,

• Effective width for beam: We can now calculate the effective width (beff) for

interior and edge beams according to ACI R13.2.4:

Effective flange projection = minimum of 4hf and hw

4hf = 4 × 7 =28″

hw = h – hf = 20 – 7 =13″

Therefore, effective flange projection = 13″

beff = bw + 2(Effective flange projection) = 14 + 2 × 13 = 40″

And, for edge beams:

beff = bw + (Effective flange projection) = 14 + 13 = 27″



h < 4h = 13"w

h = 13"w

h = 7"f

b = 14"w

h = 13"w

b = 14"w

h = 7"f

b + 2h < b + 8h = 40"w w fwf

Figure 15: Interior and edge beams sections.

Note: -

IIS25 = Moment of inertia (MOI) of 25′ long interior slab.

IES25 = Moment of inertia (MOI) of 25′ long exterior slab.

IIS20 = Moment of inertia (MOI) of 20′ long interior slab.

IES20 = Moment of inertia (MOI) of 20′ long exterior slab.

IBINT = Moment of inertia of interior beam.

IBEXT = Moment of inertia of exterior beam.

α INT25 = Ratio of MOI of interior beam to MOI of 25′ long interior slab.

α EXT25 = Ratio of MOI of exterior beam to MOI of 25′ long exterior slab.

α INT20 = Ratio of MOI of interior beam to MOI of 20′ long interior slab.

α EXT20 = Ratio of MOI of exterior beam to MOI of 20′ long exterior slab

αm = (α INT25 + 2 × α INT20+ α EXT25)/4 {for panel A as shown in fig. 14}

= (2.7 + 2.2 + 2.2 + 3.9)/4 = 2.75

β = larger clear span / smaller clear span

= [{25 – (2 × (14/2)/12)}]/ [{20 – (2 × (14/2)/12)}] = 23.8 / 18.8 = 1.27

Table 01: Moment of inertia of beams, slab and corresponding values of α. 1 2 3 4 5 6 7 8 9

Moment of Inertia of slabs

Moment of Inertia of beams α

Panel Length

(ft)

Panel Width (b) (ft)

Slab height

(hf) (in) Notation Value

(in4) Notation Value (in4) Notation

Value (in4)

Col7/Col5

25 20 7 IIS25=bhf3/12 6860 α INT25 2.7

20 25 7 IIS20=bhf3/12 8575

IBINT = 2bwh3/12 18667 α INT20 2.2

25 10.6 7 IES25=bhf3/12 3636 α EXT25 3.9

20 13.1 7 IES20=bhf3/12 4493

IBEXT = 1.5bwh3/12 14000 α EXT20 3.1



• According to given conditions, Since αm > 2, following equation applies:

h = ln {0.8+ (fy/200000)}/ (36 + 9 β) [ACI 9.5.3]

h = (23.8 × 12) × {0.8+ (60000/200000)}/ (36 + 9 × 1.27) = 6.63″ < 7″

Therefore hf = 7″ is O.K. If not then revise assumed thickness.

Step No 2: Load on slab.

Service Dead Load (D.L) = γslabhf = 0.15 × (7/12) = 0.0875 ksf

Service Live Load (L.L) = 144 psf or 0.144 ksf

Factored Load (wu) = 1.2D.L + 1.6L.L = 1.2 × 0.0875 + 1.6 × 0.144 = 0.336 ksf

Step No 3: Analysis.

Though four frames are required to be analyzed for this specific slab system, only

two of the frames will be analyzed and designed for demonstration purpose. The

details are given in appendix A.

I. Analysis of E-W Interior Frame:

Step (A): Frame Data.

• Design Span of frame (c/c) = l1 = 25′

• Design Length of frame = ln = 25 – (2 × 14/2)/12 = 23.8′

• Width of frame = l2 = 20′

• Column strip width = (Shorter span)/ 4 = 20/4 = 5′

Step (B): Total static moment.

Mo = wul2ln2/8 (for Mo, l2 is the width of frame)

= 0.336 × 20 × 23.82/8 = 476 ft-k

Step (C): Longitudinal distribution of Total static moment (Mo).

Table 02: Longitudinal Distribution of Total Static Moment. Exterior span Interior span

25' 25'

Static Moment Mo (ft-k) 476 476

Section Exterior Negative Positive Interior

Negative Negative Positive Negative

Distribution Factor, ACI 13.6.3 (D.F) 0.16 0.57 0.7 0.65 0.35 0.65

Longitudinal Moment (L.M) = Mo x D.F 77 272 334 310 167 310



77 ft-k

272 ft-k167 ft-k

272 ft-k

334 ft-k

77 ft-k

310ft-k310 ft-k334 ft-k

Minimum of h or 4h =13"

h =7"

14"

20"

C = 11210

fw

f

Figure 16: Longitudinal Distribution of Total Static Moment (Mo)

Step (D): Lateral Distribution of Longitudinal moment (L.M).

βt calculation is as follows:

C = ∑ [(1- (0.63X/Y)X3Y/3]

C = {(1- (0.63 × 14/20)) x 143 × 20/3 + (1- (0.63 × 7/13)) × 73 × 13/3}= 11210

βt = C/ (2IIS25) = 11210/ (2 × 6860) = 0.81

Figure 17: βt calculation.

Other terms required are:

α INT25 =2.7

l2/l1 = 20/25 = 0.8{l2 shall be taken as full span length irrespective of frame

location (exterior or interior)}

α INT25l2/l1 = 2.2

The values of column strip and middle strip moments obtained from lateral

distribution of longitudinal moments are given in table 03.



Table 03: Lateral Distribution of Longitudinal moment.

Span Section Column Strip Moment (C.S.M) ft-k

Beam Moment (M.D.S) ft-k

Column Strip Slab Moment (C.S.S.M) ft-k

Middle Strip Moments

(M.S.M) ft-k Negative 0.81 x 310 = 252 0.85 x 252 = 214 0.15 x 252 = 38 0.19 x 310 = 58

Interior Positive 0.81 x 167 =135 0.85 x 135 = 115 0.15 x 135 = 21 0.19 x 167 = 32 Interior negative 0.81 x 334 =271 0.85 x 271 = 230 0.15 x 271 = 41 0.19 x 334 = 63

Positive 0.81 x 272 =220 0.85 x 220= 188 0.15 x 220 = 33 0.19 x 272 = 52 Exterior Exterior negative 0.93 x 77 =72 0.85 x 72 =61 0.15 x 72 = 11 0.07 x 77 = 5

Note: Coefficients for lateral distribution have been taken from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Step (E): Moment transferred to beam B1.

Self weight of beam = γbeambwhw

= 0.15 × (14/12) × (13/12) = 0.20 k/ft

Factored load (wu) = 1.2 × 0.20 = 0.24 k/ft

Moment due to self weight of beam B1 (M) = wuln2/8

= 0.24 × 23.82/8 = 17 ft-k

Table 04: Moment transferred to beam B1

Exterior span Interior span

25' 25' Static Moment M of beam (ft-k) 17 17



Distribution Factor (D.F) 0.16 0.57 0.7 0.65 0.35 0.65 Moment due to self weight (MDSB) = M x D.F (ft-k) 3 10 12 12 6 12

Beam moment from slab (MDS) (ft-k) 61 188 230 214 115 214

Total Moment (ft-k) 64 198 242 226 121 226



B2

B1

B1

B1

B1

B2

B2 B2

B3 B3

B3 B3

B4

B4

B4

B4

C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

Figure 18: Beam and Column plan.

Step (F): Moment transferred to columns (ACI 13.6 9).

Exterior column (C3) moment = MDSB + LM…………………………..…… (A)

Where,

MDSB = moment due to self weight of beam as given in table 04.

LM = Longitudinal moment as given in table 02.

Exterior column (C3) moment = 3 + 77 = 80 ft-k

Interior column (C4) moment = MDSB(unbalanced)+(0.65/8){0.5wuLLl2ln2}…(B)

= (12–12)+(0.65/8) × {0.5 × (1.6 × 0.144) × 20 × 23.82}

= 106.3 ft-k

Step (G): Shear in beam B1 (ACI 13.6.8).

Tributary area (A) = 2 × (10 × 20/2) +5 × 20 =300 ft2

wub = wuslabA/l1 + Self weight of beam

= 0.336 × 300/ 25 + 0.24 = 4.27 k/ft

Vext = 46.17 k

Vcr, ext = 40.1 k

Vint = 60.59 k

Vcr, int = 54.52 k



Figure 19: Load on Beam (B1).

II. Analysis of E-W Exterior Frame:




• Width of frame = l2 = (20/2) + 14/(2 × 12) = 10.6′

• Column strip width = (shorter span)/ 4 = 20/4 = 5′

25'

4.27 k/ft

V = 60.59 k

V = 46.17 k

1.42'

10' 5' 10'

20'

B1 B1

45°

V =40.1 k

10.81'

14.19'

V =54.52 k

ext cr, ext

cr, int

int

64 k-ft 242 k-ft 226 k-ft

RDL

RDM Total

53.3 k 53.3 k

7.12 k 7.12 k

46.17 k 113.89 k

53.3 k



40 ft-k

144 ft-k88 ft-k

144 ft-k

176 ft-k

40 ft-k

164ft-k164 ft-k176 ft-k


Mo = wul2ln2/8 (for Mo, l2 is the width of frame)

= 0.336 × 10.6 × 23.82/8 = 252 ft-k


Table 05: Longitudinal Distribution of Total Static Moment.


25' 25'

Static Moment Mo (ft-k) 252 252



Distribution Factor(D.F) 0.16 0.57 0.7 0.65 0.35 0.65


Figure 20: Longitudinal Distribution of Total Static Moment (Mo)


α EXT25 = 3.9

l2/l1 = 20/25 = 0.8 {l2 shall be taken as full span length irrespective of frame


α EXT25l2/l1 = 3.12



Minimum of h or 4h =13"

h =7"

14"

20"

C = 11210

fw

f

Table 06: Lateral Distribution of Longitudinal moment.

Span Section Column Strip

Moment (C.S.M) ft-k

Beam Moment (M.D.S) ft-k

Column Strip Slab Moment (C.S.S.M) ft-k


(M.S.M) ft-k

Negative 0.81 x 164 = 133 0.85 x 133= 113 0.15 x 133 = 20 0.19 x 164 = 31 Interior

Positive 0.81 x 88 =71 0.85 x 71 = 61 0.15 x 71 = 11 0.19 x 88 = 17 Interior negative 0.81 x 176=143 0.85 x 144= 121 0.15 x 144 = 22 0.19 x 176 = 33

Positive 0.81 x 144 =117 0.85 x 117= 99 0.15 x 117 = 18 0.19 x 144= 27 Exterior Exterior negative 0.88 x 40 =35 0.85 x 35 =30 0.15 x 35 = 5 0.12 x 40 = 5

Note: Coefficients for lateral distribution have been taken from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Note: ACI 13.6.5.1 states that “Beams between supports shall be proportioned to

resist 85 percent of column strip moments if α1l2/l1 is equal to or greater than 1.0”.

Where, βt calculation for 10.06′ width of slab strip is given below:

C = ∑ [{1- (0.63X/Y)}X3Y/3]

C = {(1- (0.63× 14/20)) × 143 × 20/3 + (1- (0.63 × 7/13)) × 73 × 13/3}= 11210

βt = C/ (2IES25) = 11210/ (2 × 3636) = 1.54

Figure 21: βt calculation.

Step (E): Moment transferred to beam.

Self weight of beam = γbeambwhw = 0.15 × (14/12) × (13/12) = 0.20 k/ft

Factored load (wu) = 1.2 × 0.20 = 0.24 k/ft

Moment due to self weight of beam (M) = wuln2/8 = 0.24 × 23.82/8 = 17 ft-k



Table 07: Moment transferred to beam (B2)


25' 25' Static Moment M of beam (ft-k) 17 17



Distribution Factor (D.F) 0.16 0.57 0.7 0.65 0.35 0.65 Moment due to self weight

(MDSB) = M x D.F 3 10 12 12 6 12

Beam moment from slab (MDS) 30 99 121 113 61 113

Total T.M 33 109 132 125 67 125

Step (F): Moment transferred to columns

Exterior column (C1) moment = MDSB + L.M ……………………………… (A)

=3 + 40 = 43 ft-k

Interior column (C2) moment = MDSB(unbalanced)+(0.65/8){0.5wuLLl2ln2}…(B)

= (12 – 12)+ (0.65/8) × {0.5 × (1.6 × 0.144) ×10.6 × 23.82}

= 56.29 ft-k

Step (G): Shear in beam (B2).

Tributary area (A) = 2 × (10 × 10/2) + 5 × 10 =150 ft2

wub = wuslabA/l1 + Self weight of beam

= 0.336 × 150/25 + 0.24 = 2.256 k/ft

Vext = 24.24 k

Vcr, ext = 21 k

Vint = 32.16 k

Vcr, int = 28.9 k



10' 5' 10'

10'

B2 B2

25'

2.256 k/ft

V =32.16 k

V =24.24 k

1.42'

V =21.0 k

10.74'

14.26'

V =28.9k

ext

cr, ext

cr, int

int

33 k-ft 132 k-ft 125 k-ft

RDL

RDM

Total

28.2 k 28.2 k

3.96 k 3.96 k

24.24 k 60.36 k

28.2 k

Figure 22: Load on Beam (B2).



Direction along 25' length.

h = 7" d = 5.5"f l

Step No 4: Design.

(1) Design of slab strips.

A. E-W Interior slab strip:

Figure 23: Section of slab at E-W interior strip.

ds = hf – 1 = 7 – 1 = 6″

dl = ds – bar dia = 6 – (4/8) = 5.5″ (for # 4 bar)

Asmin = 0.0018bhf (for fy = 60 ksi) = 0.0018 × 12 × 7 = 0.151 in2 (ρmin = 0.0023, in

terms of actual effective depth)

Now, Equation used to calculate (ρ) in (table 1.4) is as follows:

Mu = Φfyρbdl2{1-0.59ρfy/fc′) = 0.9 × 60 × ρ × 12 × 5.52 × {1-0.59 x ρ x 60/4}

After solving the above equation for ρ, we get:

ρ = [19602 ± √{(196022) - (4 x 173477.7 x Mu′ x 12)}]/2(173477.7)…….(A)

Table 08: Slab Design.

Frame Span Location Section b (in) d (in)

Mu (ft-k)(Table

1.3)

Mu′ = Mu x 12/b

ρ (eqn A)

As = ρ x b x d (in2)

Bar taken

Spacing = b x Ab/As

Recommended

negative 106 5.5 38 4.30 0.0027 1.57 #4 13.48 9 Column Strip

Moment Positive 106 5.5 21 2.38 0.0023 1.34 #4 15.81 12

negative 120 5.5 58 5.90 0.0037 2.47 #4 9.73 9 Interior Middle

Strip Moment Positive 120 5.5 32 3.20 0.0023 1.52 #4 15.81 12

Interior negative 106 5.5 41 4.64 0.0029 1.70 #4 12.47 9

Positive 106 5.5 33 3.73 0.0023 1.36 #4 15.60 12 Column

Strip Moment Exterior

negative 106 5.5 11 1.24 0.0023 1.34 #4 15.81 12


Positive 120 5.5 52 5.20 0.0033 2.16 #4 11.09 9

Exterior

Middle Strip

Moment Exterior negative 120 5.5 5 0.50 0.0023 1.52 #4 15.81 12

Note: “b” is the width of strip.




h = 7" d = 5.5"f l

B. E-W Exterior slab strip:

Figure 24: Section of slab at E-W exterior strip.

ds = hf – 1 = 7 – 1 =6″

dl = ds – bar dia = 6 – (4/8) = 5.5″ (for # 4 bar)

Asmin = 0.0018 × 12 × 7 = 0.151 in2 (ρmin = 0.0023, in terms of actual effective

depth)

Now, Equation used to calculate (ρ) values is as follows:

Mu = Φfyρbdl2{1-0.59ρfy/fc′} = 0.9 × 60 × ρ × 12 × 5.52 × {1-0.59 × ρ × 60/4}

After solving the above equation for ρ

ρ = [19602 ± √ {(196022) – (4 × 173477.7 × Mu′ × 12)}]/2(173477.7)

Table 09: Slab Design.

Frame Span Location Section b

(in) d

(in) Mu (ft-k)

(Table 1.7) Mu′ =

Mu x 12/b ρ (eqn

A) As =ρ x b x d

(in2) Bar

taken Spacing = b x Ab/As

Recommended

negative 53 5.5 20 4.52 0.0028 0.83 #4 12.81 9 Column Strip

Moment Positive 53 5.5 11 2.48 0.0023 0.67 #4 15.81 12

negative 60 5.5 31 6.20 0.0039 1.30 #4 9.25 9 Interior

Middle Strip

Moment Positive 60 5.5 17 3.40 0.0023 0.76 #4 15.81 12 Interior negative 53 5.5 22 4.98 0.0031 0.91 #4 11.60 9

Positive 53 5.5 18 4.07 0.0025 0.74 #4 14.27 12 Column


negative 53 5.5 5 1.13 0.0023 0.67 #4 15.81 12


Positive 60 5.5 27 5.40 0.0034 1.12 #4 10.67 9

Exterior

Middle Strip

Moment Exterior negative 60 5.5 5 1.00 0.0023 0.76 #4 15.81 12




# 4 @ 9" c/c (B)

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (B)

# 4

@ 1

2" c

/c (B

)#

4 @

12"

c/c

(B)

# 4

@ 1

2" c

/c (B

)

# 4

@ 1

2" c

/c (B

)#

4 @

12"

c/c

(B)

# 4

@ 1

2" c

/c (B

)

# 4

@ 1

2" c

/c (B

)#

4 @

12"

c/c

(B)

# 4

@ 1

2" c

/c (B

)

# 4 @ 12" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)# 4 @ 9" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)

# 4 @ 12" c/c (T)

# 4 @ 12" c/c (T)# 4 @ 9" c/c (T) # 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)

# 4

@ 9

" c/

c (T

)#

4 @

9" c

/c (T

)#

4 @

9"

c/c

(T)

# 4

@ 9

" c/

c (T

)

# 4

@ 9

" c/

c (T

)

# 4

@ 9

" c/

c (T

)#

4 @

9"

c/c

(T)

# 4

@ 9

" c/

c (T

)

# 4

@ 9

" c/c

(T)

# 4

@ 9

" c/

c (T

)#

4 @

9" c

/c (T

)#

4 @

9" c

/c (T

)

25' 25' 25'

20'

20'

20'

B2

B1

B1

B2

B3 B3B4 B4

PLAN

B B

AA

Step No 5: Detailing.

All the frames may be analyzed and designed by the same procedure as given in steps

of analysis and design. However, the complete detail of reinforcement placement in

slab is given below.

Figure 25: Reinforcement details (plan view)



SECTION B-B

SECTION A-A

23.83' 23.83'

6'-0" 8'-0" 8'-0"

#4 @ 9" c/c#4 @ 12" c/c #4 @ 9" c/c#4 @ 9" c/c

#4 @ 12" c/c#4 @ 9" c/c #4 @ 12" c/c

#4 @ 12" c/c

18.83' 18.83'

5'-0" 6'-6"

#4 @ 9" c/c #4 @ 12" c/c

#4 @ 12" c/c

#4 @ 9" c/c

#4 @ 12" c/c

#4 @ 9" c/c

6'-6"

#4 @ 9" c/c #4 @ 12" c/c

Figure 26: Sectional views of slab.



9. Shear Strength of Slab without beams:-

There are two types of shear that needs to be addressed.

(a) One-way shear or beam shear at distance “d” from the column,

(b) Two-way or punch out shear which occurs along a truncated cone.

Figure 27: Beam shear and punching shear.

Shear design in flat plates & flat slabs.

Punching shear: For punching shear ΦVn = ΦVc + ΦVs

ΦVc is least of (a), (b), and (c).

(a) ΦVc = Φ4√ (fc′)bod

(b) ΦVc = (2 + 4/βc) √ (fc′)bod

(c) {(αsd/bo +2} √ (fc′)bod

βc = longer side of column/shorter side of column

αs = 40 for interior column

= 30 for edge column

= 20 for corner columns



c

c

d/2d/2

d/2

1

2Column

d = Depth of member

b = 2{(c + d) + (c + d)}b = 2(c + c + 2d)b = 2c + 2c + 4d)

o 1 2o 1 2

o 1 2

Figure 28: Critical perimeter for punching shear.

When ΦVc ≥ Vu (Φ = 0.75) O.K, Nothing required.

When ΦVc < Vu, then either increase ΦVc by:

(a) Increasing d ,depth of slab ( Drop Panel)

(b) Increasing bo, critical shear perimeter (Column capital)

(c) fc′ (high Strength Concrete)

Or provide shear reinforcement in the form of:

(a) Shear heads

(b) Bent Bars

(c) Integral beams

(d) Shear studs.

Drop Panels and Column Capitals:

(1) Drop panel: A drop panel with dimensions conforming to ACI 9.5.3.2 and

13.3.7.1 can be used for:

(i) Increasing the area of critical shear perimeter.

(ii) Increasing the depth of slab, reducing the amount of negative reinforcement.

(iii) Stiffening slab and reducing deflections.



Drop panel

l (length of panel)a l (length of panel)b

l /6a l /6b

hh/4

Figure 29: Drop Panel.

ACI 13.3.7.1 states that drop panel shall extend in each direction from and of the

support a distance not less than 1/6 of the span length measured from centre of

supports in that direction.

ACI 13.3.7.2 states that Projection of drop panel below the slab shall be at least

one-quarter the slab thickness beyond the drop.

(2) Column Capitals:

Occasionally, the top of the columns will be flared outward, as shown in

figure 34. This is done to provide a larger shear perimeter at the column and to

reduce the clear span, ln, used in computing moments.

ACI 6.4.6 requires that the capital concrete be placed at the same time as the

slab concrete. As a result, the floor forming becomes considerably more

complicated and expensive. For this reason, other alternatives, such as drop

panels or shear reinforcement in the slab, should be considered before capitals

are selected. If capital must be used, it is desirable to use the same size

throughout the project.

The diameter or effective dimension of the capital, “dc” or “c”, is defined in

ACI 13.1.2 as that part of the capital lying within the largest circular cone or

pyramid with a 90o vertex that can be induced within the outlines of the

supporting column. The diameter is measured at the bottom of the slab or drop

panel, as illustrated in figure 30.



Figure 30: Column Capital.

Figure 31: Circular column with column capital and drop panel.

45°

Slab

Drop Panel

"dc" or "c"

Capital



Shear Reinforcements:

(1) Shear Heads:

The shear heads shown in Fig. 32a consist of standard structural steel

shapes embedded in the slab and projecting beyond the column. They

serve to increase the effective perimeter bo of the critical section for shear.

In addition, they may contribute to the negative bending resistance of the

slab. It consists of short lengths of I or wide-flange beams, cut and welded

at the crossing point so that the arms are continuous through the column.

Normal negative slab reinforcement passes over the top of the structural

steel, while bottom bars are stopped short of the shear head. Column bars

pass vertically at the corners of the column.

(2) Bent-bar:

The bent-bar arrangement in Fig. 32b is suited for use with concrete

columns. The bars are usually bent at 45° across the potential diagonal

tension crack, and extend along the bottom of the slab a distance sufficient

to develop their strength by bond.

(3) Integral Beams:

Another type of shear reinforcement is illustrated in Fig. 32c, where

vertical stirrups have been used in conjunction with supplementary

horizontal bars radiating outward in two perpendicular directions from the

support to form what are termed integral beams contained entirely within

the slab thickness. These beams act in the same general way as the shear

heads shown in Fig. 32a. Adequate anchorage of the stirrups is difficult in

slabs thinner than about 10 in. ACI 11.12.3 requires the slab effective

depth d to be at least 6 in., but not less than 16 times the diameter of the

shear reinforcement. In all cases, closed hoop stirrups should be used with

a large diameter horizontal bar at each bend point, and the stirrups must be

terminated with a standard hook (Ref. 13.18).

(4) Shear Stud:

A more recent development is the shear stud reinforcement shown in Fig.

32d. This consists of large-head studs welded to steel strips.



(c) (d)

Figure 32: Shear reinforcement for flat plates.

The strips are supported on wire chairs during construction to maintain the

required concrete cover to the bottom of the slab below the strip and the

usual cover is maintained over the top of the head. Because of the positive

anchorage provided by the stud head and the steel strip, these devices are



24"

24"

20'

20'24"

24"

d/2 = 3"

18"

18"

d/2 = 3"

more effective, according to tests, than either the bent bar or integral beam

reinforcement. In addition, they can be placed more easily, with less

interference with other reinforcement, than other types of shear steel.

Design for punching shear:

Design Problem 02 (pg 455, Nelson 13th Ed): A flat plate has thickness h = 7 ½ in

and is supported by 18″ square columns spaced 20 ft on centers each way. The floor

will carry a total factored load of 300 psf. Check the adequacy of the slab in resisting

punching shear at a typical interior column, and provide shear reinforcement, if

needed. d = 6 in, fc′ = 4000 psi, fy = 60000 psi.

Solution:

(a) Vu = 300{(20)2 – (2)2}

= 118800 lb = 118.8 k

Figure 33: Critical perimeter calculation.

(b) Shear capacity of concrete in punching shear:

ΦVc = 0.75 × 4 √ (fc′) bod

bo = 2(c1 + d) +2(c2 + d)

bo = 4(18 + 6) = 96″



C

1.5" 1.5"18"

θ

1.5"

y

ΦVc = 0.75 × 4 √ (4000) × 96 × 6/1000 = 109 k

ΦVc < Vu

Options available:

(i) Capital: Determine minimum (bo). Equating the applied critical shear to shear

capacity.

Vu = ΦVc

118.8 = 0.75 × 4 √ (fc′) × bo × 6

bo = 104.5″

Now bo = 4 (c + d) = 4(c + 6) =104.5

c = 20.13 ≈ 21″

Figure 34: Column capital.

According to ACI code, θ < 45o

Let θ = 30o, then y = 2.6″

For θ = 20o, then y = 4″

(ii) Drop panel: To determine the minimum “d” required.

Vu = ΦVc

118.8 = 0.75 × 4 × √ (4000) × 96 × d/1000

d = 6.5″ and h = 6.5 + 1.5 = 8″

Thickness of drop panel = 2″ (say) {h/4, ACI recommendation}

h = 2 + 7.5 = 9.5″



α

VV sinαs

s

1 2

Figure 35: Drop panel.

(iii) Increasing column size or slab thickness (overall) would not be economical.

(iv) Shear reinforcement:

(a) Bent bar reinforcement: When bent bars are to be used, ACI 11.12.3 reduces

ΦVc by 2.

ΦVc = Φ × 4 × √ (fc′) × bo × d

Therefore, ΦVc = 109/2 = 54.5 kip

Reinforcement required, Av = (Vu – ΦVc)/ Φfysinα

Figure 36:

Av = (118.8 – 54.5)/ (0.75 × 60 × sin45o) = 2.03 in2

Using total 4 bars (two in each direction), providing 8 legs crossing the

critical section, the area per bar = 2.03/8 = 0.25, using # 5 bar.

Figure 37: Bent bar.

20 /6=3'-4" 20 /6=3'-4"

7.5"

2″

6'-8"



5" 15"

No 5 bar

Development Length

According to ACI, Vs = Vu – ΦVc = Avfysinα is not to exceed 3√ (fc′)bod.

3√ (fc′)bod = 3 × √ (4000) × 96 × 6/1000 = 109 kip

Vs = 2.4 × 60 × sin45 = 100 O.K.

Figure 38: Bent bar reinforcement.

(b) Stirrup reinforcement:

Using 3/8″ Φ, 2 legged (0.22 in2), 4 (side) = 4 × 0.22 = 0.88 in2

Spacing (s) = ΦAvfyd/ (Vu – ΦVc)

s = 0.75 × 0.88 × 60 × 6/ (118.8 – 54.5) = 3.68 ≈ 3.5″

Maximum spacing allowed d/2 = 6/2 = 3″ controls.

Four #5 bars are to be provided in each direction to hold the stirrups. The

beams such formed should be extended to a length at which bo can provide

shear capacity. We know minimum bo = 104.5″

bo = 4R + 4c1

R = √ (x2 + x2)

x = (3/4)(lv – c1/2)

R = √ (2) x

bo = 4√ (2) x + 4c1

bo = 4√ (2){(3/4)(lv – c1/2)} + 4c1

Or bo = 4.24lv – 2.12c1 + 4c1= 4.24lv + 1.88c1

Therefore lv = 16.67 ≈ 17″



x = 34 (l - c /2)1v

x

R

45°

Figure 39:

Figure 40: Stirrups Reinforcement (ACI R11.12.3).



l = 15'

15'-0"

2

l = 8.08'2

7"

3.75'

3.75'

15'-0"

15'-0"

l = 20'-0"1 l = 20'-0"1 l = 20'-0"1

Half middle strip

Half middle strip

Half middle strip

Half column strip

Half column strip

Half column strip


3.75'

3.75'

3.75'

3.75'

Design Pb. 03: Design the slab for the building shown below. (Follow the Direct Design

Method).

Data Given:

A 60′ × 45′ building, divided into nine (9) panels, supported at their ends on

columns. Each panel is 20′ × 15′.

fc′ = 4 ksi

fy = 60 ksi

Height of building = 10′

Column dimensions = 14″ × 14″

Live load = 144 psf

Figure 41: Given flat plate.



Solution: -

The given slab system satisfies all the necessary limitations for Direct Design Method

to be applicable.

Step No 1: Sizes for slab and columns.

Slab: To find minimum slab thickness (hf), ACI 9.5.3.2 {ACI Table 9.5 (c) or table

13.50, Nelson 13th Ed} will be used.

Table 10: Minimum thickness of slabs without interior beams. Without drop panels With drop panels

Exterior panels Exterior panels Yield strength, fy

(psi) Without edge

beams

With edge beams

Interior panels

Without

edge beams

With edge

beams

Interior panels

40000 ln/33 ln/36 ln/36 ln/36 ln/40 ln/40 60000 ln/30 ln/33 ln/33 ln/33 ln/36 ln/36 75000 ln/28 ln/31 ln/31 ln/31 ln/34 ln/34

For our case (Slab without drop panels, interior and edge beams)

hf = ln/30 (Exterior panel)

hf = ln/33 (Interior panels)

Exterior panel governs. Therefore,

hf = ln/30

= [{20 – (2 × 14/2)/12}/30] × 12 = 7.53″ (Minimum requirement)

Take hf = 8″; as there are no beams; α = 0

Columns: Let the column dimensions = 14″ × 14″.

Step No 2: Load on slab.

Service Dead Load (D.L) = γslabhf = 0.15 × (8/12) = 0.1 ksf

Service Live Load (L.L) = 144 psf or 0.144 ksf

Factored Load (wu) = 1.2D.L + 1.6L.L

= 1.2 × 0.1 + 1.6 × 0.144 = 0.3504 ksf



l = 15'

15'-0"

2

3.75'

15'-0"

15'-0"

l = 20'-0"1 l = 20'-0"1 l = 20'-0"1

Half middle strip

Half middle strip

Half column strip

Half column strip

3.75'

3.75'

3.75'

Step No 3: Analysis.

Though four frames are required to be analyzed for this specific slab system, only two

of the frames will be analyzed and designed for demonstration purpose. The details

are given in appendix A.

I. Analysis of E-W Interior Frame.




• Width of frame = l2 = 15′

• Column strip width = (Shorter span)/ 4 = 15/4 = 3.75′

Figure 42: East West Interior Frame.


Mo = wul2ln2/8 (for Mo, l2 is the width of the frame)

= 0.3504 × 15 × 18.832/8 = 233 ft-k




Table 11. Longitudinal distribution of Total static moment (Mo)


20' 20' Static Moment Mo (ft-k) 233 233 Distribution Factor (D.F)

ACI 13.6.3 (D.F) 0.26 0.52 0.7 0.65 0.35 0.65



α = 0

l2/l1 = 15/20 = 0.75 {l2 shall be taken as full span length irrespective of frame


αl2/l1 = 0

βt = 0

Table 12: Lateral Distribution of Longitudinal moment (L.M).


Moment (C.S.M) ft-k


(M.S.M) ft-k Negative 0.75 x 151=113 0.25 x 151=38

Interior Positive 0.6 x 82=49 0.4 x 82=33

Interior negative 0.75 x 163=122 0.25 x 163=41 Positive 0.6 x 121=73 0.4 x 121=48 Exterior

Exterior negative 1.00 x 61=61 0 x 61=0

Note: Coefficients for lateral distribution have been selected from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.



C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

Figure 43: Column plan.

Step (E): Moment transferred to columns.

Exterior column (C3) moment = L.M……………………..……….……..……… (A)

= 61 ft-k

Interior column (C4) moment = (0.65/8)x{0.5xwuLLx l2xln2}…………………….. (B)

= (0.65/8) x {0.5x (1.6x0.144) x 15 x 18.832}

= 49.78 ft-k

II. Analysis of E-W Exterior Frame.



• Design Length of frame = ln = 20 – (2 x 14/2)/12 = 18.83′

• Width of frame = l2 = (15/2) + 14/(2 x 12) = 8.08′

• Column strip width = (shorter span)/ 4 = 15/4 = 3.75′



15'-0"

l = 8.08'2

7"

3.75'

15'-0"

15'-0"

l = 20'-0"1 l = 20'-0"1 l = 20'-0"1

Half middle strip

Half column strip


3.75'

Figure 44: East west exterior frame.


Mo = wul2ln2/8

= 0.3504 x 8.08 x 18.832/8 = 126 ft-k


Table 13: Longitudinal Distribution of Total Static Moment (Mo)


20' 20' Static Moment Mo (ft-k) 126 126

Distribution Factor (D.F) ACI 13.6.3 (D.F) 0.26 0.52 0.7 0.65 0.35 0.65





α = 0

l2/l1 = 15/20 = 0.75

αl2/l1 = 0

βt = 0 (no edge beam)

Table 14: Lateral Distribution of Longitudinal moment (L.M).


Moment (C.S.M) ft-k

Middle Strip Moments (M.S.M) ft-k

Negative 0.75 x 82 = 62 0.25 x 82 = 21 Interior

Positive 0.60 x 45 =27 0.40 x 45 = 18 Interior negative 0.75 x 89=67 0.25 x 89 = 22

Positive 0.60 x 66 =40 0.40 x 66= 26 Exterior Exterior negative 1.00 x 33 =33 0 x 33 = 0

Note: Coefficients for lateral distribution have been selected from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Step (E): Moment transferred to columns.

Exterior column (C1) moment = L.M ………………………….…………… (A)

=33 ft-k

Interior column (C2) moment = (0.65/8){0.5wuLLl2ln2}………..………….… (B)

= (0.65/8) x {0.5 x (1.6 x 0.144) x 8.08 x 18.832}

= 27 ft-k



h = 8" d = 7"f


s

Step No 4: Design.

(1) Design of Slab strips:

(A) E-W Interior slab strip:

ds = hf – ¾ - (4/8)/2 = 8 – 1 =7″

Asmin = 0.0018x 12 x 8 = 0.1728 in2 (ρmin = 0.002, in terms of actual effective depth)


Mu = Φfyρbds2{1-0.59ρfy/fc′) = 0.9 x 60 x ρ x 12 x 72 x {1-0.59 x ρ x 60/4)

281005 ρ2 - 31752 ρ + Mu = 0


ρ = [31752 ± √{(317522) -(4 x 281005 x Mu′ x 12)}]/2(281005)…………(A)

Figure 45: Slab section at E-W interior strip.

Table 15: Slab Design (E-W, Interior Frame).

Frame Span Location Section b (in) d

(in) Mu (ft-

k) Mu′ =

Mu x12/b

ρ (equation

A)

As = ρ*b*d (in2)

BAR taken

Spacing =

b*Ab/As Recommended

negative 90 7 113 15.07 0.0060 3.79 #4 4.75 3 Column Strip

Moment Positive 90 7 49 6.53 0.0025 1.59 #4 11.31 9

negative 90 7 38 5.07 0.0020 1.26 #4 14.29 9 Interior

Middle Strip

Moment Positive 90 7 33 4.40 0.0020 1.26 #4 14.29 9

Interior negative 90 7 122 16.27 0.0065 4.11 #4 4.38 3

Positive 90 7 73 9.73 0.0038 2.40 #4 7.51 3 Column


negative 90 7 61 8.13 0.0032 1.99 #4 9.03 9


Positive 90 7 48 6.40 0.0025 1.56 #4 11.55 9

Exterior

Middle Strip

Moment Exterior negative 90 7 0 0.00 0.0020 1.26 #4 14.29 9



h = 8" d = 7"f


s

(B) E-W Exterior slab strip:

ds = hf – 1 = 8 – 1 =7″

Asmin = 0.0018x 12 x 8 = 0.1728 in2 (ρmin = 0.002, in terms of actual effective depth)


Mu = Φfyρbds2{1-0.59ρfy/fc′) = 0.9 x 60 x ρ x 12 x 72 x {1-0.59 x ρ x 60/4)

281005 ρ2 - 31752 ρ + Mu = 0


ρ = [31752 ± √{(317522) -(4 x 281005 x Mu′ x 12)}]/2(281005)…………(A)

Figure 46: Slab section at E-W exterior strip.

Table 16: Slab Design (E-W, Exterior Frame).

Frame Span Location Section b (in) d

(in) Mu (ft-k)

(Table 1.6) Mu′ =

Mu x 12/b ρ (eqn A) As = ρ x

b x d (in2)

Bar taken

Spacing = b x Ab/As

Recommended

negative 45 7 62 16.53 0.0066 2.09 #4 4.30 3 Column Strip

Moment Positive 45 7 27 7.20 0.0028 0.88 #4 10.24 9

negative 45 7 21 5.60 0.0022 0.68 #4 13.24 9

Interior Middle Strip

Moment Positive 45 7 18 4.80 0.0020 0.63 #4 14.29 9


Positive 45 7 40 10.67 0.0042 1.32 #4 6.82 3 Column

Strip Moment

Exterior negative 45 7 33 8.80 0.0034 1.08 #4 8.33 9


Positive 52 7 26 6.00 0.0023 0.84 #4 12.34 9

Exterior

Middle Strip

Moment Exterior negative 52 7 0 0.00 0.0020 0.73 #4 14.29 9




d/2 = 3.5"

d/2 =3.5"

14"

14"

C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

(2) Shear Design for Flat-plate:

(i) At Column C1:

hf = 8″

d = 8 – 1 = 7″

Figure 47: Critical perimeter for column (C1).

(a) Vu = 0.3504{(7.5 x 10) – (17.5/12)2}

= 25.53 k


ΦVc = 0.75 x 4 √ (fc′) bod

bo = (c1 + d/2) +(c2 + d/2)

bo = 2(14 + 7/2) = 35″

ΦVc = 0.75 x 4 √ (4000) x 35 x 7/1000 = 46.48 k

ΦVc > Vu, O.K.



d/2 = 3.5"

d/2 = 3.5"

14"

14"

d/2 = 3.5"

C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

(ii) At Column C2:


(a) Vu = 0.3504{(7.5 x 20) – (21 x 17.5/144)}

= 51.66 k


ΦVc = 0.75 x 4 √ (fc′) bod

bo = (c1 + d) +2(c2 + d/2)

bo = (14 + 7) +2(14 + 7/2) = 56″

ΦVc = 0.75 x 4 √ (4000) x 56 x 7/1000 = 74.37 k

ΦVc > Vu, O.K.



d/2 = 3.5"

d/2 = 3.5"

14"

14" d/2 = 3.5"

C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

(iii) At Column C3:


(a) Vu = 0.3504{(10 x 15) – (21 x 17.5/144)}

= 51.66 k


ΦVc = 0.75 x 4 √ (fc′) bod

bo = 2(c1 + d/2) +(c2 + d)

bo = 2(14 + 7/2) + (14 + 7) = 56″

ΦVc = 0.75 x 4 √ (4000) x 56 x 7/1000 = 74.37 k

ΦVc > Vu, O.K.



d/2 = 3.5"

d/2 = 3.5"

14"

14" d/2 = 3.5"

d/2 = 3.5"

C1

C1 C1

C1C2 C2

C2 C2

C3

C3 C3

C3C4 C4

C4 C4

C3

C2C1

(iv) At Column C4:


(a) Vu = 0.3504{(20 x 15) – (21 x 21/144)}

= 104 k


ΦVc = 0.75 x 4 √ (fc′) bod

bo = 2(c1 + d) +2(c2 + d)

bo = 4(14 + 7) = 84″

ΦVc = 0.75 x 4 √ (4000) x 84 x 7/1000 = 111 k

ΦVc > Vu, O.K.



20'-0"

15'-0"D D

C C

A

A

B

B

15'-0"

20'-0"

#4 @ 9" c/c (B)

#4 @ 3" c/c (T)

3.17' 12'-6" 7'-6" 12'-6" 3.17'

3.17'

7'-6"

7'-6"

7'-6"

3.17'

#4 @

9"

c/c

(B)

#4 @ 9" c/c (B)

#4 @ 9" c/c (B)#4 @ 9" c/c (B)

#4 @

5"

c/c

(B)

#4 @

5"

c/c

(B)

#4 @

9"

c/c

(B)

#4 @

9"

c/c

(B)

#4 @

9"

c/c

(B)

#4 @

5"

c/c

(B)

#4 @ 5" c/c (B) #4 @ 5" c/c (B)

#4 @ 5" c/c (B)#4 @ 5" c/c (B)

#4 @ 5" c/c (B) #4 @ 5" c/c (B)

#4 @

5"

c/c

(B)

#4 @

9"

c/c

(B)

#4 @

9"

c/c

(B)

#4 @ 9" c/c (T) #4 @ 9" c/c (T)#4 @ 9" c/c (T)

#4 @ 9" c/c (T)#4 @ 9" c/c (T)#4 @ 9" c/c (T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @

9"

c/c

(T)

#4 @ 5" c/c (T)

#4 @ 5" c/c (T) #4 @ 3" c/c (T)

#4 @ 3" c/c (T)#4 @ 5" c/c (T)

#4 @ 3" c/c (T)#4 @ 5" c/c (T)

#4 @ 3" c/c (T)

#4 @

5"

c/c

(T)

#4 @

5"

c/c

(T)

#4 @

5"

c/c

(T)

#4 @

5"

c/c

(T)

#4 @

5"

c/c

(T)

#4 @

5"

c/c

(T)

#4 @ 3" c/c (T)

#4 @ 3" c/c (T)

C1 C2 C2

C3

C3

C4C4

C4 C4

N

S

EW

PLAN According to ACI 13.3.8.5,At least two of thecolumn strip bottom barsor wires in each directionshall pass within thecolumn core and shall beanchored at exterior supports.

Step No 5: Detailing.

All the frames may be analyzed and designed by the same procedure as given in steps

of analysis and design. However, the complete detail of reinforcement placement in

slab is given below.

Figure 51: Reinforcement details (Plan view)



SECTION A-A (N-S exterior column strip)L = 13.83' L = 13.83'

0.3L = 4'-3" 0.3L = 4'-3"

#4 @ 9" c/c

0.3L = 4'-3"

#4 @ 5" c/c

SECTION B-B (N-S middle strip)L = 13.83' L = 13.83'

0.3L = 4'-3" 0.3L = 4'-3"

#4 @ 9" c/c

0.3L = 4'-3"

#4 @ 9" c/c

n n

nnn

n n

nnn

#4 @ 5" c/c

#4 @ 9" c/c

SECTION D-D (E-W middle strip)

L = 18.83' L = 18.83'

#4 @ 9" c/c

#4 @ 9" c/c #4 @ 9" c/c

0.3L = 6'-0" 0.3L = 6'-0" 0.3L = 6'-0"

SECTION C-C (E-W exterior column strip)L = 18.83' L = 18.83'

#4 @ 5" c/c #4 @ 3" c/c

0.3L = 6'-0" 0.3L = 6'-0" 0.3L = 6'-0"

n n

nnn

n n

nnn

#4 @ 5" c/c

Figure 52: Sectional view.

Two Way Slab Analysis

Documents

Transcript of Two Way Slab Analysis