Lecture-04€¦ · Prof. Dr. Qaisar Ali Reinforced Concrete Design –II Department of Civil Engineering, University of Engineering and Technology Peshawar Lecture-04 Analysis and

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 Peshawar

www.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)wblb

4/EI

wa/wb = lb4/la

4 wa = wb (lb4/la

4)

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




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 × la

2

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

2

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

la

Ma,neg

Ma,posMb,neg Mb,negMb,pos


6



Cases

Depending on the support conditions, several cases are possible:

11


4 spans @ 25′-0″

3 sp

ans @ 20

′-0″



12


Cases


4 spans @ 25′-0″

3 spans @ 20

′-0″

7



13


Cases


4 spans @ 25′-0″

3 sp

ans @ 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)

16' X 12' 16' X 12'

9' Wide Verandah

B1B1

RCC Column

9" brick masonry wall



Given Data:

Service Dead Load

4″ thick mud

2″ thick brick tile

Live Load = 40 psf

f′c = 3 ksi

fy = 40 ksi

24


16' X 12' 16' X 12'

9' Wide Verandah

B1B1

RCC Column


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) = (8.42/20) × (0.4+��/100000)

= 4.04″ (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

16' X 12' 16' X 12'

9' Wide Verandah

B1B1

RCC Column


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


16' X 12' 16' X 12'

9' Wide Verandah

B1B1

RCC Column




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

,po

s

Ma

,po

s

Ma

,ne

g

Ma

,ne

g

Mv

er

(+v

e)

Mb,pos

Ma,p

os

Ma,n

eg

Mb,negMb,pos

Ma,n

eg

Ma,p

os

Mver

(+v

e)

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,p

os

Ma,n

eg

Mb,negMb,pos

Ma,n

eg

Ma,p

os



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

,po

s

Case 4

Ma,n

eg

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

,po

s

Case 4

Ma,n

eg



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

,po

s

Case 4

Ma,n

eg

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

,po

s

Case 4

Ma,n

eg



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

,po

s

Case 4

Ma,n

eg

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

,po

s

Case 4

Ma,n

eg



38



Calculating moments using ACI Coefficients:




2


2

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, dl = 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

,po

s

Case 4

Ma,n

eg

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


Mver(

+v

e)



40


One-Way Slab Analysis

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

follows:

Mver (+ve) = wuln2/11

Mver (+ve) = 0.211 × (8)2/11

= 1.23 ft-k/ft

9′ Wide Verandah

ln = 8′

9″ Brick Wall

Wu = 0.211 ksf

h

1/11 1/11

1/91/9Simply Supported

Simply Supported

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 1.23

NOTE: All values are in ft-kip units

Mb,pos

Ma,p

os

Ma,n

eg

Mb,negMb,pos

Ma,n

eg

Ma,p

os

Mver

(+v

e)



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,p

os

Case 4

Ma,n

eg



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 in

2

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) = 15.87516.375

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



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 Vu (@ 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 reinforcement

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

S2 9' wide

M1 M1

M2M2

MT1 MT1 MT1

MT2 MT2

MT2MT2

M1

M2

C1

AA

B

S1 16' x 12' S1 16' x 12'

B MT2

B1 B1



54


Drafting Details for Slabs

SECTION A-A

L /4 = 4'-0" L /3 = 5'-3"

L = 16'-0"

#3 @ 10" c/c

#3 @ 10" c/c

#3 @ 18" c/c #3 @ 18" c/c

1 1

h = 5"

#3 @ 10" c/c

#3 @ 10" c/c

L/3=2'-8" L/4=2'-0"

#3 @ 10" c/c

#3 @ 10" c/c

#3 @ 18" c/c#3 @ 10" c/c

9" Brick Masonry Wall

12" x 18" Beam

#3 @ 10" c/c

#3 @ 10" c/c#3 @ 10" c/c

#3 @ 18" c/c #3 @ 6" c/c

2

2

SECTION B-B

L = 12'-0" L = 8'-0"

L /4 = 4'-0"2 L /3 = 4'-0"

28



55


Drafting Details for Verandah Beam

2 #5 Bars

2 #4 Bars

SECTION B-B

12"

18"

5"

3 #5 Bars

#3,2 legged stirrups @ 7.5" c/c

SECTION A-A

12"

18"

5"

3 #5 Bars

#3,2 legged stirrups @ 7.5" c/c

2 #4 Bars

A

A

A

A

2 #4 + 2 #5 Bars

BEAM (B1)

B

B

L = 11'-10.5" L = 11'-10.5"

0.33L = 4'-0" 0.33L = 4'-0"

2 #4 Bars

3 #5 Bars #3, 2 legged stirrups @ 7.5" c/c

s /2=3.75"

3 #5 Bars 3 #5 Bars

B

B



56


Drafting Details for Verandah Column

OR

12"

4 #6 Bars

#3 ties @ 9" c/c

12"

12"

#3 Ties @ 9" c/c

4 #6 Bars

8#4 bars

#3 ties @ 9" c/c

12"

12"

12"

8 #4 Bars

#3 Ties @ 9" c/c

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 sp

ans @

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,neg



66






2


2


Ca,neg = 0.071 Cb,neg = 0.029




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 4la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g


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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

37



73






2


2


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 9la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g



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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g


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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g



80






2


2


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 8la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g


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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g




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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g

44



87






2


2


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 2la = 18.83′

lb = 23.83′

Ma

,ne

gM

a,p

os

Mb,posMb,negMb,neg

Ma

,ne

g



88

Analysis Results (All values are in ft-kip)


8.42

4 spans @ 25′-0″

3 spans @ 2

0′-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

2.881.50

1.79

1.19 1.19 Mneg at Non-

Continuous End =

1/3 of Mpos

NOTE: White: Longer Direction Moments, Yellow: Shorter 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 in

2

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

short 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

1.50 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

3.24 ft-kip/ft

7.80 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

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

Therefore using #4 bars @ 7 c/c

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 sp

ans @

20

′-0″


NOTE: White: Longer Direction Moments, Yellow: Shorter Direction Moments



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 sp

ans @

20′-0

″

Sh

orte

r Dire

ctio

n

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 Wa values




98



4 spans @ 25′-0″

3 sp

an

s @ 2

0′-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-II

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

= 5.17 k/ft

(For Wa Values, refer to the tables on next slides)• Wu = 0.336 ksf

• Wa,Panel-I = 0.71

• Wa,Panel-II = 0.83


50



99


Load Transfer to Beam from Two-Way Slab (For Beam B1)

Sh

orte

r Dire

ction

20′

Wu = 0.336 ksf

Sh

orte

r Dire

ctio

n

20′

Wu = 0.336 ksf




100



4 spans @ 25′-0″

3 sp

an

s @ 2

0′-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-III

= 0.336 × 25/2 × (1 - 0.71) + 0.336 × 25/2 × (1 - 0.55)

= 3.11 k/ft

(For Wa values, refer to the tables on next slide)• 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


51



101

Load transfer from Slab to Beam (For Beam B2)


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

(1 - 0.55)

= 3.11 k/ft

Sh

orte

r Dire

ction

20′

Wu = 0.336 ksf

Sh

orte

r Dire

ctio

n

20′

Wu = 0.336 ksf




Moment Coefficient Method: Example 2

Load On Beams from coefficient tables

102

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 - 2.39

B2 25 10 0.71 - 2.39

B3 20 12.5 - 0.29 1.22

B4 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


52



Table: Load on beam in Panel I, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


B1 25 10 0.71 - 2.39

B2 25 10 0.71 - 2.39

B3 20 12.5 - 0.29 1.22

B4 20 12.5 - 0.29 1.22



B1

B1

B2

B2

B3 B3 B3 B4B4

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

Panel I

103


Panel I





104

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


B1 25 10 0.83 - 2.78

B3 20 12.5 - 0.17 0.714

B4 20 12.5 - 0.17 0.714


Panel II

53



Table: Load on beam in Panel II, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


B1 25 10 0.83 - 2.78

B3 20 12.5 - 0.17 0.714

B4 20 12.5 - 0.17 0.714



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel II

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

105


Panel II





106

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


B1 25 10 0.55 - 1.84

B2 25 10 0.55 - 1.84

B3 20 12.5 - 0.45 1.89


Panel III

54



Table: Load on beam in Panel III, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


B1 25 10 0.55 - 1.84

B2 25 10 0.55 - 1.84

B3 20 12.5 - 0.45 1.89



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel III

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

107


Panel III





108

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


B1 25 10 0.71 - 2.39

B3 20 12.5 - 0.29 1.22


Panel IV

55



Table: Load on beam in Panel IV, usingCoefficients

(wu = 0.336 ksf)

BeamLength

(ft)


supported by beam

Wa Wb


B1 25 10 0.71 - 2.39

B3 20 12.5 - 0.29 1.22



B1

B1

B2

B2

B3 B3 B3 B4B4

Panel IV

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

109


Panel IV




Load On Beams

110

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


56




Load On Beams

111

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 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4





Load On Beams

112

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

57




Load On Beams (including self-weight)

113

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 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4




114

lnln ln

Simplesupport

Integral withsupport

wu

Spandrelsupport

NegativeMoment

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)


58



115


Analysis of Beams

Interior Beam B1



116


Analysis of Beams

Exterior Beam B2

59



117


Analysis of Beams

Interior Beam B3



Analysis of Beams

Exterior Beam B4

118


60



119

Pictures of a Multi-StoreyCommercial Building



120

Pictures of a Multi-StoreyCommercial Building

61



Different Stages of Building Construction

121

Phases of Construction



Moment Coefficient Method: Home Work

Design the given slab system

122

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″


SDL = 40 psf

LL = 60 psf

fc′ =3 ksi

fy = 40 ksi

Practice Examples

62



Moment Coefficient Method: Home Work

Design the given slab system

123

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 sp

ans @

15′-0″


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.

124

References

63



The End

125

Documents

Lecture-04€¦ · Prof. Dr. Qaisar Ali Reinforced Concrete Design –II Department of Civil Engineering, University of Engineering and Technology Peshawar Lecture-04 Analysis and