l-04 analysis and design of two-way slab with beams (b&w)

1

Prof. Dr. Qaisar Ali Reinforced Concrete Design – II

Department of Civil Engineering, University of Engineering and Technology Peshawar

Lecture-04

Analysis and Design of Two-way

Slab System(Part-I: Two Way Slabs Supported on Stiff

Beams Or Walls)

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawarwww.drqaisarali.com

1



Topics

� Behavior

� Moment Coefficient Method

� Steps in Moment Coefficient Method

� Design Example 1 (Typical House with 2 Rooms and Verandah)

� Design Example 2 (100′ × 60′, 3-Storey Commercial Building)

� Practice Examples

� References

2

2



� Two-Way Slab System (Long span/short span < 2)

3

25′ 25′ 25′ 25′

20′

20′

20′

Behavior



4

One-Way Behavior Two-Way Behavior

Behavior

� Two-Way Bending of Two-Way Slabs

3



5

Behavior

� Two-Way Bending of Two-Way Slabs



6

Short Direction

Behavior

� Short Direction Moments in Two-Way Slab

4



7

Long Direction

Behavior

� Long Direction Moments in Two-Way Slab



� More Demand (Moment) in short direction due to size

of slab

� Δcentral Strip = (5/384)wl4/EI

� As these imaginary strips are part of monolithic slab, the deflection at

any point, of the two orthogonal slab strips must be same:

� Δa = Δb

(5/384)wala4/EI = (5/384)wblb4/EI

� wa/wb = lb4/la4 wa = wb (lb4/la4)

� Thus, larger share of load (Demand) is taken by the shorter direction.

8

Behavior

5



� The Moment Coefficient Method included for the first time in

1963 ACI Code is applicable to two-way slabs supported on

four sides of each slab panel by walls, steel beams relatively

deep, stiff, edge beams (h = 3hf).

� Although, not included in 1977 and later versions of ACI code,

its continued use is permissible under the ACI 318-11 code

provision (13.5.1). Visit ACI 13.5.1.

9

Moment Coefficient Method



� Moments:

Ma, neg = Ca, negwula2

Mb, neg = Cb, negwulb2

Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la2

Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb2

� Where Ca, Cb = Tabulated moment coefficients

wu = Ultimate uniform load, psf

la, lb = length of clear spans in short and long directions respectively.

10

Ma,neglb

laMa,neg

Ma,posMb,neg Mb,negMb,pos


6



� Cases

� Depending on the support conditions, several cases are possible:

11


4 spans @ 25′-0″

3 spans @ 20′-0″



12


� Cases


4 spans @ 25′-0″

3 spans @ 20′-0″

7



13


� Cases


4 spans @ 25′-0″

3 spans @ 20′-0″



14


� Cases


4 spans @ 25′-0″

3 spans @ 20′-0″

8



15


Note: Horizontal sides of the figure represents longer side while vertical side

represents shorter side.



16




9



17






18




10



19






20




11



21






� hmin = perimeter/ 180 = 2(la + lb)/180

� Calculate loads on slab

� Calculate m = la/ lb

� Decide about case of slab

� Use table to pick moment coefficients

� Calculate Moments and then design

� Apply reinforcement requirements (smax = 2hf, ACI 13.3.2)

22

Steps in Moment Coefficient Method

12



� 3D Model of the House

� In this building we will design a two-way slab (for rooms), a one-way

slab, beam and column (for verandah)

23

Design Example 1(Typical House with 2 Rooms and Verandah)



� Given Data:

� Service Dead Load

� 4″ thick mud

� 2″ thick brick tile

� Live Load = 40 psf

� f′c = 3 ksi

� fy = 40 ksi

24


13



� Sizes

� For two way slab system

� hmin = perimeter / 180 = 2(la + lb)/180

� hmin = 2 (12 +16) /180 = 0.311 ft = 3.73 inch

� Assume 5 inch slab

� For one way slab system

� For 5″ slab, span length l is min. of:

� l = ln+ hf = 8 + (5/12) = 8.42′

� c/c distance between supports = 8.875′

� Slab thickness (hf) = (8.42/24) × (0.4+��/100000)

= 3.367″ (min. by ACI)

� Taking 5 in. slab

25


9′ Wide Verandah

ln = 8′

lc/c = (8 + (9/12) / 2 + 0.5) = 8.875′

9″ Brick Wall

Slab

h

A

A

Section AA



� Loads

� 5 inch Slab = 5/12 x 0.15 = 0.0625 ksf

� 4 inch mud = 4/12 x 0.12 = 0.04 ksf

� 2 inch tile = 2/12 x 0.12 = 0.02 ksf

� Total DL = 0.1225 ksf

� Factored DL = 1.2 x 0.1225 = 0.147 ksf

� Factored LL = 1.6 x 0.04 = 0.064 ksf

� Total Factored Load = 0.211 ksf

26


14



� Analysis

� This system consist of both one way and two way slabs. Rooms

(two way slabs) are continuous with verandah (one way slab).

� A system where a two way slab is continuous with a one way slab

or vice versa is called a mixed slab system.

27




� Analysis

� The ACI approximate methods of analysis are not applicable to

such systems because:

� In case of two way slabs, the moment coefficient tables are applicable

to two way slab system where a two way slab is continuous with a two

way slab.

� In case of one ways slabs, the ACI approximate analysis is applicable

to one way slab system where a one way slab is continuous with a one

way slab.

28


15



� Analysis

� The best approach to analyze a mixed system is to use FE

software.

� However, such a system can also be analyzed manually by making

certain approximations.

� In the next slides, We will analyze this system using both of the

above mentioned methods.

29




� Analysis using FE Software (SAFE)

30


Moments in Longer Direction in Two Way Slabs

Moments in Shorter Direction in Two Way Slabs & Moment in Verandah One Way Slab

Two Way Slab Moments (ft-kip/ft)(Rooms)

One Way Slab Moment (ft-kip/ft)(Verandah)

Ma,pos Mb,pos Ma,neg Mb,neg Mver (+ve)

1.58 1.17 2.10 1.67 1.10

Mb,pos Mb,posMb,neg Ma,

pos

Ma,

pos

Ma,

neg

Ma,

neg

Mve

r(+

ve)

Mb,pos

Ma,

pos

Ma,

neg

Mb,negMb,pos

Ma,

neg

Ma,

pos

Mve

r(+

ve)

16



� Analysis using Manual Approach

� For approximate manual analysis of two way slab of rooms, we

know that two way slabs of rooms are not only continuous along the

long direction but also continuous along the short direction with the

verandah slab. We assume that the verandah slab is a two way slab

instead of one way slab in order to calculate Mb,neg

� Now, using Moment Coefficient Method for analysis.

31


Mb,pos

Ma,

pos

Ma,

neg

Mb,negMb,pos

Ma,

neg

Ma,

pos



� Two-Way Slab Analysis

Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.052

Cb,posLL = 0.016


32

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg

17




Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.052

Cb,posLL = 0.016


33

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg



Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.052

Cb,posLL = 0.016



34

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg

18



Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.052

Cb,posLL = 0.016



35

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg



Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.052

Cb,posLL = 0.016



36

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg

19



Case = 4

m = la/lb = 12/16 = 0.75

Ca,neg = 0.076

Cb,neg = 0.024

Ca,posDL = 0.043

Cb,posDL = 0.013

Ca,posLL = 0.055

Cb,posLL = 0.016



37

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg



38



� Calculating moments using ACI Coefficients:





� Using above relations the moments calculated are:

Ca,neg = 0.076 Cb,neg = 0.024

Ca,posLL = 0.052 Cb,posLL = 0.016

Ca,posDL = 0.043 Cb,posDL = 0.013

wu, d l = 0.147 ksf, wu, ll = 0.064 ksf, wu = 0.211 ksf

� Ma,neg = 2.31 ft-kip

� Mb,neg = 1.29 ft-kip

� Ma,pos = 1.39 ft-kip

� Mb,pos = 0.76 ft-kip

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4

Ma,

neg

20



� One Way Slab Analysis

� For calculation of Mver (+ve) along the short direction in the verandah

slab, we will pick the coefficients of continuous one way slabs having

two spans.

� Now, using ACI Approximate analysis procedure (ACI 8.33) for analysis.

39


Mve

r(+v

e)Mve

r(-v

e)



40


� One-Way Slab Analysis

� For two span one way slab system, positive moment at midspan is given as

follows:

� Mver (+ve) = wuln2/14

� Mver (+ve) = 0.211 × (8)2/14

= 0.96 ft-k/ft

� Mver (-ve) = wuln2/9 = 1.5 ft-k/ft

9′ Wide Verandah

ln = 8′

9″ Brick Wall

Wu = 0.211 ksf

h

1/14

1/9 1/9SpandrelSupport

Simply Supported

For negative moment above the long wall commonto rooms and veranda, maximum moment will bepicked from both analyses.

Moment of 2.3 from two way slab analysis is morethan 1.5, therefore we will design for 2.31.

1/24

21



41


� Design of Two-Way Slab

� Comparison of Analysis Results from FE Analysis and Manual Analysis

� Analysis results from both approaches are almost similar.

� Hence the intelligent use of manual analysis yields fairly reasonable results in most cases.

Analysis Type

Ma,neg Mb,neg Ma,pos Mb,pos Mver (+ve)

SAFE 2.10 1.67 1.58 1.17 1.10

Manual 2.31 1.3 1.39 0.76 0.96

NOTE: All values are in ft-kip units

Mb,pos

Ma,

pos

Ma,

neg

Mb,negMb,pos

Ma,

neg

Ma,

pos

Mve

r(+

ve)



42



� First determining capacity of min. reinforcement:

� As,min = 0.002bhf = 0.12 in2

� Using #3 bars: Spacing for As,min = 0.12 in2 = (0.11/0.12) × 12 = 11″ c/c

� However ACI max spacing for two way slab = 2h = 2(5) = 10″ or 18″ = 10″ c/c

� Hence using #3 bars @ 10″ c/c

� For #3 bars @ 10″ c/c: As,min = (0.11/10) × 12 = 0.132 in2

� Capacity for As,min: a = (0.132 × 40)/(0.85 × 3 × 12) = 0.17″

� ΦMn = ΦAsminfy(d – a/2) = 0.9 × 0.132 × 40(4 – (0.17/2)) = 18.60 in-kip

� Therefore, for Mu values ≤ 18.60 in-k/ft, use As,min (#3 @ 10″ c/c) & for Mu

values > 18.6 in-kip/ft, calculate steel area using trial & error procedure.

22



43



� For Ma,neg = 2.31 ft-kip = 27.71 in-kip > 18.60 in-kip: As = 0.20 in2 (#3 @ 6.6″ c/c)

� Using #3 @ 6″ c/c

� For Mb,neg = 1.29 ft-kip = 15.56 in-kip < 18.60 in-kip: Using #3 @ 10″ c/c

� For Ma,pos = 1.39 ft-kip = 16.67 in-kip < 18.60 in-kip: Using #3 @ 10″ c/c

� For Mb,pos = 0.76 ft-kip = 9.02 in-kip < 18.60 in-kip: Using #3 @ 10″ c/c

16′

12′

Mb,negMb,pos

Ma,

pos

Case 4M

a,ne

g



44



� Reinforcement at Discontinuous Ends

� Reinforcement at discontinuous ends in a two way slab is 1/3 of the positive

reinforcement.

� Positive reinforcement at midspan in this case is #3 @ 10″ c/c. Therefore

reinforcement at discontinuous end may be provided @ 30″ c/c.

� However, in field practice, the spacing of reinforcement at discontinuous ends

seldom exceeds 18″ c/c. The same is provided here as well.

� Supporting Bars

� Supporting bars are provided to support negative reinforcement.

� They are provided perpendicular to negative reinforcement, generally at spacing

of 18″ c/c.

23



45


� Design of One-Way Slab

� Main Reinforcement:

� Mver (+ve) = 14.73 in-kip

� As,min = 0.002bhf = 0.002(12)(5) = 0.12 in2

� Using #3 bars, spacing = (0.11/0.12) × 12 = 11″ c/c

� For one-way slabs, max spacing by ACI = 3h = 3(5) = 15″ or 18″ = 15″ c/c

� For #3 bars @ 15″ c/c, As = (0.11/15) × 12 = 0.09 in2. Hence using As,min = 0.12 in2

� a = (0.12 × 40)/(0.85 × 3 × 12) = 0.16″

� ΦMn = ΦAsminfy(d – a/2) = 0.9 × 0.12 × 40(4 – (0.16/2)) = 16.94 in-kip > Mver (+ve)

� Therefore, using #3 @ 11″ c/c

� However, for facilitating field work, we will use #3 @ 10″ c/c



46


� Design of One-Way Slab

� Shrinkage Reinforcement:

� Ast = 0.002bhf = 0.12 in2 (#3 @ 11″ c/c)

� However, for facilitating field work, we will use #3 @ 10″ c/c

24



47

� Verandah Beam Design

� Step 01: Sizes

� Let depth of beam = 18″

� ln + depth of beam = 15.875′ + (18/12) = 17.375′

� c/c distance between beam supports

= 16.375 + (4.5/12) = 16.75′

� Therefore l = 16.75′

� Depth (h) = (16.75/18.5) × (0.4 + 40000/100000) × 12

= 8.69″ (Minimum requirement of ACI 9.5.2.2).

� Take h = 1.5′ = 18″

� d = h – 3 = 15″

� b = 12″


Verandah Beam

16.375′ 16.375′

ln = 16.375 – 0.5(12/12) = 15.875′ ln = 16.375 – 0.5(12/12) = 15.875′



48

� Verandah Beam Design

� Step 02: Loads

� Load on beam will be equal to

� Factored load on beam from slab + factored

self weight of beam web

� Factored load on slab = 0. 211 ksf

� Load on beam from slab = 0. 211 ksf x 5 =

1.055 k/ft

� Factored Self load of beam web =

� = 1.2 x (13 × 12/144) × 0.15 = 0.195 k/ft

� Total load on beam = 1.055 + 0.195

= 1.25 k/ft

5′


ln = 8′

9″ Brick Wall

8/2 = 4′

12″ Column

4′ + 1′ = 5′

25



49

� Verandah Beam

Design

� Step 03: Analysis

� Using ACI Moment

Coefficients for analysis

of verandah beam


16.375′ln = 16.375 – 0.5(12/12) = 15.875′ ln = 16.375 – 0.5(12/12) = 15.875′16.375′

16.375′

ln = 15.875′ ln = 15.875′

9.92 k

11.41 k

Vu(ext) = 8.34 k

Vu(int) = 9.61 k

343.66 in-kip 343.66 in-kip

420.03 in-kip

Verenda beam is l beam,be



� Beam Design

� Flexure Design:

� Shear Design:

� Smax is min. of: (1) Avfy/(50bw) = 14.67″ (2) d/2 =7.5″ (3) 24″ c/c (4) Avfy/ 0.75√(fc′)bw = 17.85″

50


Mu(in-kip)

d (in.)

b (in.)

As(in2)

Asmin(in2)

Asmax(in2)

As(governing)

Bar used

# of bars

343.66 (+) 15 28.75 (beff) 0.64 0.90 3.654 0.90 #4 5

#5 3

420.03 (-) 15 12 0.81 0.90 3.654 0.90 #4 5

#4 + #5 2 + 2

Location V u (@ d)(kip)

ΦVc = Φ2 �′�bwd (kips) ΦVc > Vu, Hence

providing minimum reinforcement.

smax, ACI

S taken(#3 2-legged)

Exterior 8.34 14.78 7.5″ 7.5″

Interior 9.61 14.78 7.5″ 7.5″

26



51


� Column Design

� Sizes:

� Column size = 12″ × 12″

� Loads:

� Pu = 11.41 × 2 = 22.82 kip

Verandah Column



52


� Column Design

� Main Reinforcement Design:

� Nominal strength (ΦPn) of axially loaded column is:

ΦPn = 0.80Φ {0.85fc′ (Ag – Ast) + Astfy} {for tied column, ACI 10.3.6}

� Let Ast = 1% of Ag (Ast is the main steel reinforcement area)

� ΦPn = 0.80 × 0.65 × {0.85 × 3 × (144 – 0.01 × 144) + 0.01 × 144 × 40}

= 218.98 kip > Pu = 22.82 kip, O.K.

� Ast =0.01 × 144 =1.44 in2

� Using 3/4″ Φ (#6) with bar area Ab = 0.44 in2

� No. of bars = 1.44/0.44 = 3.27 ≈ 4 bars

� Use 4 #6 bars (or 8 #4 bars) and #3 ties @ 9″ c/c

27



53


� Drafting Details for

Slabs

Panel Depth (in) Mark Bottom Reinforcement Mark Top rei nforcement

S1 5"M1 #3 @ 10" c/c

MT1 #3 @ 10" c/c Continuous EndMT1 #3 @ 18" c/c Non Continuous End

M2 #3 @ 10" c/cMT2 #3 @ 6" c/c Continuous EndMT2 #3 @ 18" c/c Non Continuous End

S2 5"M1 #3 @ 10" c/c MT2 #3 @ 6" c/c Continuous EndM2 #3 @ 10" c/c MT2 #3 @ 18" c/c Non Continuous End



54


� Drafting Details for Slabs

28



55


� Drafting Details for Verandah Beam



56


� Drafting Details for Verandah Column

OR

29



� A 100′ × 60′, 3-storey commercial building is to be designed. The grids of

column plan are fixed by the architect.

� In this example, the slab of one of the floors of this 3-storey building will

be designed.

57

Design Example 2(100′ × 60′, 3-Storey Commercial Building)



� Given Data:

� Material Properties:

� f′c = 3 ksi, fy = 40 ksi

� Sizes:

� Slab thickness = 7″

� Columns = 14″ × 14″

� Beams = 14″ × 20″

� Loads:

� S.D.L = Nil

� Self Weight = 0.15 x (7/12) = 0.0875 ksf

� L.L = 144 psf = 0.144 ksf ; wu = 0.336 ksf

58


30



� Complete analysis of the slab is done by analyzing four panels

59

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

4 spans @ 25′-0″

3 spans @ 20′-0″




� Two-Way Slab Analysis (Panel-I)

Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg


60

31




Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020


61

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg




Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020


62

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg

32




Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020


63

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg




Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020


64

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg

33




Case = 4

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.071

Cb,neg = 0.029

Ca,posDL = 0.039

Cb,posDL = 0.016

Ca,posLL = 0.048

Cb,posLL = 0.020


65

Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg



66








Ca,neg = 0.071 Cb,neg = 0.029



wu, dl = 0.105 ksf, wu, ll = 0.2304 ksf, wu = 0.336 ksf

� Ma,neg = 8.44 ft-k (101.2 in-k)

� Mb,neg = 5.52 ft-k (66.2 in-k)

� Ma,pos = 5.37 ft-k (64.4 in-k)

� Mb,pos = 3.57 ft-k (42.8 in-k)


Case 4 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,neg

34



� Two-Way Slab Analysis (Panel-II)

Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg


67




Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017


68

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

35




Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017


69

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017


70

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

36




Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017


71

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 9

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.075

Cb,neg = 0.017

Ca,posDL = 0.029

Cb,posDL = 0.010

Ca,posLL = 0.042

Cb,posLL = 0.017


72

Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

37



73








Ca,neg = 0.075 Cb,neg = 0.017




� Ma,neg = 8.91 ft-k (106.9 in-k)

� Mb,neg = 3.24 ft-k (38.8 in-k)

� Ma,pos = 4.51 ft-k (54.1 in-k)

� Mb,pos = 2.82 ft-k (33.8 in-k)


Case 9 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg



� Two-Way Slab Analysis (Panel-III)

Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg


74

38




Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019


75

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019


76

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

39




Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019


77

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019


78

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

40




Case = 8

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.055

Cb,neg = 0.041

Ca,posDL = 0.032

Cb,posDL = 0.015

Ca,posLL = 0.044

Cb,posLL = 0.019


79

Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg



80








Ca,neg = 0.055 Cb,neg = 0.041




� Ma,neg = 6.54 ft-k (78.4 in-k)

� Mb,neg = 7.80 ft-k (93.6 in-k)

� Ma,pos = 4.78 ft-k (57.4 in-k)

� Mb,pos = 3.38 ft-k (40.5 in-k)


Case 8 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

41



� Two-Way Slab Analysis (Panel-IV)

Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg


81




Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017


82

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

42




Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017


83

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017


84

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

43




Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017


85

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg




Case = 2

m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80

Ca,neg = 0.065

Cb,neg = 0.027

Ca,posDL = 0.026

Cb,posDL = 0.011

Ca,posLL = 0.041

Cb,posLL = 0.017


86

Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg

44



87








Ca,neg = 0.065 Cb,neg = 0.027




� Ma,neg = 7.72 ft-k (92.7 in-k)

� Mb,neg = 5.14 ft-k (61.6 in-k)

� Ma,pos = 4.31 ft-k (51.8 in-k)

� Mb,pos = 2.88 ft-k (34.5 in-k)


Case 2 la = 18.83′

lb = 23.83′

Ma,

neg

Ma,

pos

Mb,posMb,negMb,neg

Ma,

neg



88

� Analysis Results (All values are in ft-kip)


8.42

4 spans @ 25′-0″

3 spans @ 20′-0″

5.375.523.57

8.91

8.91

3.244.51

2.82

6.54

7.80 7.804.783.38

7.72

7.72

5.145.144.31

2.880.94

1.19

1.79 1.60 Mneg at Non-

Continuous End =

1/3 of Mpos

NOTE: White: Longer Direction Moments, Yellow: Short er Direction Moments

45



89


� Slab Design

� First determining the capacity of min. reinforcement:

� As,min = 0.002bhf = 0.002 × 12 × 7 = 0.17 in2

� For #4 bars, spacing = (0.20/0.17) × 12 = 14.2″ c/c.

� ACI max spacing for two-ways slabs = 2h = 2(7) = 14″ or 18″

� Using #4 @ 12″ c/c

� For the 12″ spacing: As = (0.20/12) × 12 = 0.20 in2. Hence As,min = 0.20 in2

� Capacity for As,min: a = (0.20 × 40)/(0.85 × 3 × 12) = 0.26″

� ΦMn = ΦAsminfy(d – a/2) = {0.9 × 0.20 × 40(6 – (0.26/2))}/12 = 3.52 ft-kip



90


� Slab Design

� Positive Moments in Long Direction:

� 3.57 ft-kip/ft

� 3.38 ft-kip/ft

� 2.82 ft-kip/ft

� 2.88 ft-kip/ft

� Since all the above moments values are almost equal to or less than

3.52 ft-kip/ft. Therefore using #4 @ 12″ c/c for all positive moments in

Longer direction.

46



91


� Slab Design

� Positive Moments in Short Direction:

� 5.37 ft-kip/ft

� 4.51 ft-kip/ft

� 4.78 ft-kip/ft

� 4.31 ft-kip/ft

� Using trial and success method for determining As for 5.37 ft-kip/ft:

� Assume a = 0.2d= 0.2(6) = 1.2″, As = (5.37 × 12)/{0.9 × 40(6 – (1.2/2))} = 0.33 in2

� Now, a = (0.33 × 40)/(0.85 × 3 × 12) = 0.43″, As = 0.31 in2

� For #4 bars, spacing = (0.20/0.31) × 12 = 7.7″

� As all the above moments are almost same, we will use #4 @ 7″ c/c for all above

moments.



92


� Slab Design

� Negative Moments at Non-Continuous Ends in Short & Long Directions:

� 1.19 ft-kip/ft

� 1.79 ft-kip/ft

� 1.60 ft-kip/ft

� 0.94 ft-kip/ft

� Since, all the above moments are

less than 3.52 ft-kip/ft, therefore using

#4 @ 12″ c/c

47



93


� Slab Design

� Negative Moments at Continuous Ends in Short Direction:

� 8.42 ft-kip/ft

� 8.91 ft-kip/ft

� 7.72 ft-kip/ft

� 6.54 ft-kip/ft

� Using trial and success method:

� For 8.91 ft-kip/ft: As = 0.52 in2

� Using #4 bars, spacing = 4.5″ c/c

� As all above moments are almost same, we will use #4 @ 4.5″ c/c



94


� Slab Design

� Negative Moments at Continuous Ends in Long Direction:

� 5.52 ft-kip/ft

� 7.80 ft-kip/ft

� 3.24 ft-kip/ft

� 5.14 ft-kip/ft

� Reinforcement:

� 7. 80 ft-kip/ft is almost equal to 8.91 ft-kip/ft:

Therefore using #4 bars @ 4.5″ c/c

� 5.14 ft-kip/ft is almost equal to 5.37 ft-kip/ft:

Therefore using #4 bars @ 7″ c/c

Designing for 7.80 ft-kip/ft

Designing for 5.14 ft-kip/ft

48



� Slab Reinforcement Details

95

A

BBB

C

C

A

C CB

A

A

BA

C

C

A

A B

A

A= #4 @ 12″B = #4 @ 7″C = #4 @ 4.5″

4 spans @ 25′-0″

3 spans @ 20′-0″





96


� Load Transfer from Slab to the Beam

� Review of Load transfer to Beam from One-Way Slabs

• In case of one-way slab system the entire slab load is transferred in short direction.

• Load transfer in short direction = (Wu × l / 2 × 1) + (Wu × l / 2 × 1)

• Load transfer in long direction = Wu × l / 2 × 0

l/2

l/2

l/2 l/2

49



97

� Load transfer from Slab to Beam

� Load Transfer to Beam from Two-Way Slab

• In case of two way slab system, entire slab load is NOT transferred in shorter direction.

• Load transfer in shorter direction = (Wu × l / 2 × Wa ) + (Wu × l / 2 × Wa )

Longer Direction

4 spans @ 25′-0″

3 spans @ 20′-0″

Shorter D

irection

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

This value will NOT be 1 in this case, It is specified by ACI Table• Load transfer in longer direction = (Wu × l / 2 × Wb ) + (Wu × l / 2 × Wb )

Wb = 1 - Wa

ACI Table for W a values




98



4 spans @ 25′-0″

3 spans @ 20′-0″

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

Load Transfer from Slab to Beam B1

• Load is transferred to B1 from Panel-I and Panel-II in

short direction

= Wu × l / 2 × Wa,Panel-I + Wu × l / 2 × Wa,Panel-I I

= 0.336 × 20/2 × 0.71 + 0.336 × 20/2 × 0.83

= 5.17 k/ft

Shorter D

irection

20′

Wu = 0.336 ksfShorter D

irection

20′

Wu = 0.336 ksf

• Wu = 0.336 ksf

• Wa,Panel-I = 0.71

• Wa,Panel-II = 0.83


50



99



4 spans @ 25′-0″

3 spans @ 20′-0″

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

Load Transfer from Slab to Beam B2

• Load is transferred to B2 from Panel-I and Panel-III in

long direction

= Wu × l / 2 × Wb,Panel-I + Wu × l / 2 × Wb,Panel-I II

= 0.336 × 25/2 × (1 - 0.71) + 0.336 × 25/2 ×

(1 - 0.55)

= 3.11 k/ft

Shorter D

irection

20′

Wu = 0.336 ksf

Shorter D

irection

20′

Wu = 0.336 ksf

• Wu = 0.336 ksf

• Wb,Panel-I = (1 – Wa,Panel-I) = 1 - 0.71 = 0.29

• Wb,Panel-III = (1 – Wa,Panel-III) = 1 - 0.55 = 0.45




� Moment Coefficient Method: Example 2

� Load On Beams from coefficient tables

100

Table: Load on beam in Panel I, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)

Width (bs) of slab panel

supported by beam

Wa Wb

Load due to slab, Wwubs

(k/ft)

B1 25 10 0.71 - 2.39B2 25 10 0.71 - 2.39B3 20 12.5 - 0.29 1.22B4 20 12.5 - 0.29 1.22

Panel I

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4


51



Table: Load on beam in Panel I, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.71 - 2.39B2 25 10 0.71 - 2.39B3 20 12.5 - 0.29 1.22B4 20 12.5 - 0.29 1.22



B1

B1

B2

B2

B3 B3 B3 B4B4

4 spans @ 25′-0″

3 spans @ 20′-0″

Panel I

101


Panel I





102

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel II, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.83 - 2.78B3 20 12.5 - 0.17 0.714B4 20 12.5 - 0.17 0.714


Panel II

52



Table: Load on beam in Panel II, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.83 - 2.78B3 20 12.5 - 0.17 0.714B4 20 12.5 - 0.17 0.714



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel II

4 spans @ 25′-0″

3 spans @ 20′-0″

103


Panel II





104

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel III, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.55 - 1.84B2 25 10 0.55 - 1.84B3 20 12.5 - 0.45 1.89


Panel III

53



Table: Load on beam in Panel III, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.55 - 1.84B2 25 10 0.55 - 1.84B3 20 12.5 - 0.45 1.89



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel III

4 spans @ 25′-0″

3 spans @ 20′-0″

105


Panel III





106

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel IV, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.71 - 2.39B3 20 12.5 - 0.29 1.22


Panel IV

54



Table: Load on beam in Panel IV, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


(k/ft)

B1 25 10 0.71 - 2.39B3 20 12.5 - 0.29 1.22



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel IV

4 spans @ 25′-0″

3 spans @ 20′-0″

107


Panel IV




� Load On Beams

108

2.39 k/ft

1.22 k/ft

2.39 k/ft

1.22 k/ft

2.77 k/ft

2.77 k/ft

0.71 k/ft 0.71 k/ft

1.84 k/ft

1.89 k/ft

1.84 k/ft

1.89 k/ft

2.39 k/ft

1.22 k/ft

2.39 k/ft

1.22 k/ft

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4


55




� Load On Beams

109

5.16 k/ft

1.22 k/ft

2.39 k/ft

3.11 k/ft

5.16 k/ft

0.71 k/ft 1.93 k/ft

4.23 k/ft

1.84 k/ft

3.11 k/ft

4.23 k/ft

1.93 k/ft

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4





� Load On Beams

110

Self weight of beam

= 1.2 × (14 ×

13/144) × 0.15

= 0.23 kip/ft

Adding this with the

calculated loads on

beams:

14″

20″


5.16 k/ft

1.22 k/ft

2.39 k/ft

3.11 k/ft

5.16 k/ft

0.71 k/ft 1.93 k/ft

4.23 k/ft

1.84 k/ft

3.11 k/ft

4.23 k/ft

1.93 k/ft

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

56




� Load On Beams (including self-weight)

111

1.45 k/ft

2.62 k/ft

5.39 k/ft

0.94 k/ft

3.34 k/ft

2.07 k/ft

2.16 k/ft

4.46 k/ft

4 spans @ 25′-0″

3 spans @ 20′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4




112

lnln ln

Simplesupport

Integral withsupport

wu

Spandrelsupport Negative

Moment

x wuln2

1/24 1/10* 1/11 1/11 1/10* 0

*1/9 (2 spans)

* 1/11(on both faces of other interior supports)

Columnsupport 1/16

PositiveMomentx

1/14 1/16 1/11

wuln2

Note: For simply supported slab, M = wul2/8, where l = span length (ACI 8.9).

* 1/12 (for all spans with ln < 10 ft)


57



� Analysis of Beams

� Interior Beam B1

113





� Exterior Beam B2

114


58




� Interior Beam B3

115





� Exterior Beam B4

116


1.45 k/ft

59



117

Pictures of a Multi-Storey

Commercial Building



118

Pictures of a Multi-Storey

Commercial Building

60



� Different Stages of Building Construction

119

Phases of Construction



� Moment Coefficient Method: Home Work

� Design the given slab system

120

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

4 spans @ 25′-0″

3 spans @ 20′-0″

Slab thickness = 6″

SDL = 40 psf

LL = 60 psf

fc′ =3 ksi

fy = 40 ksi

Practice Examples

61



� Moment Coefficient Method: Home Work

� Design the given slab system

121

Panel I Panel I

Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel III Panel III

Panel IV Panel IV

4 spans @ 20′-0″3 spans @

15′-0″

Slab thickness = 6″

SDL = 40 psf

LL = 60 psf

fc′ =3 ksi

fy = 40 ksi

Practice Examples



� CRSI Design Handbook

� ACI 318

� Design of Concrete Structures 13th Ed. by Nilson, Darwin and

Dolan.

122

References

62



The End

123



124

63



125

� Analysis Results (All values are in ft-kip)


8.42

4 spans @ 25′-0″3 spans @

20′-0″

5.375.523.57

8.91

8.91

3.244.51

2.82

6.54

7.80 7.804.783.38

7.72

7.72

5.145.144.31

2.880.94

1.19

1.79 1.60 Mneg at Non-

Continuous End =

1/3 of Mpos


l-04 analysis and design of two-way slab with beams (b&w)

Documents