Lecture-04 - Dr. Qaisar Ali · Lecture-04 Analysis and Design of Two-way Slab System (Part-I: Two...

1

Prof. Dr. Qaisar Ali CE-416: 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



Objectives

At the end of this lecture, students will be able to

Classify one-way and two-way slab systems

Make use of ACI coefficient method for two-way slabs

analysis

Analyze and design two-way slab system for flexure

Compare manual and computer results

Recall basic design steps for beam shear and flexure

design

3



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

4

25′ 25′ 25′ 25′

20′

20′

20′

Behavior

3



5

One-Way Behavior Two-Way Behavior

Behavior

Two-Way Bending of Two-Way Slabs



6

Behavior

Two-Way Bending of Two-Way Slabs

4



7

Short

Direction

Behavior

Short Direction Moments in Two-Way Slab



8

Long

Direction

Behavior

Long Direction Moments in Two-Way Slab

5



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.

9

Behavior



More Demand (Moment) in short direction due to size

of slab

Ma = Ca*wala2

As from the previous slide wa = wb (lb4/la

4)

Ma = Ca*wala2 = Ca*wb(lb

4/la2)………………i

Mb = Cb*wblb2 ………………………………. ii

Comparing (i) and (ii) , we get

Ma = Mb*(lb /la)^2 Provided that Ca = Cb

ΔThus, larger share of load (Demand) is taken by the shorter

direction.

10

Behavior

6



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 318-14 code

provision (8.2.1). Visit ACI 8.2.1.

11




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.

12

Ma,neglb

la

Ma,neg

Ma,posMb,neg Mb,negMb,pos


7



Cases

Depending on the support conditions, several cases are possible:

13


4 spans @ 25′-0″

3 sp

ans @

20

′-0″



14


Cases


4 spans @ 25′-0″

3 sp

ans @

20

′-0″

8



15


Cases


4 spans @ 25′-0″

3 sp

ans @

20

′-0″



16


Cases


4 spans @ 25′-0″

3 sp

ans @

20

′-0″

9



17


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

short side.



18



short side.

10



19



short side.



20



short side.

11



21



short side.



22



short side.

12



23



short side.



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

Calculate loads on slab

Calculate m = la/ lb

Decide about case of slab

Use tables to pick moment coefficients

Calculate Moments and then design

Apply reinforcement requirements (smax = 2h, ACI 8.7.2.2)

24

Steps in Moment Coefficient Method

13



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)

25

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

fc′= 3 ksi

fy = 40 ksi

26

Design Example 1


16' X 12' 16' X 12'

9' Wide Verandah

B1B1

RCC Column


14



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+ h = 8 + (5/12) = 8.42′

c/c distance between supports = 8.875′

Slab thickness (h) = (8.42/24) × (0.4+𝑓𝑦/100000)

= 3.367″ (min. by ACI)

Taking 5 in. slab

27

Design Example 1


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 A-A



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, wu, dl = 1.2 x 0.1225 = 0.147 ksf

Factored LL, wu, ll = 1.6 x 0.04 = 0.064 ksf

Total Factored Load, wu= 0.211 ksf

28

Design Example 1


15



Analysis

This system consists 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 can be called as a mixed slab system.

29

Design Example 1


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.

30

Design Example 1


16



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.

31

Design Example 1




Analysis using FE Software (SAFE)

32

Design Example 1


Moments in Long Direction in Two

Way SlabsMoments in Short 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

Mve

r(+

ve

)

Mb,pos

Ma,p

os

Ma,n

eg

Mb,negMb,pos

Ma,n

eg

Ma,p

os

Mv

er

(+v

e)

17



Analysis using Manual Approach

Moment Coefficient Method for analysis.

33

Design Example 1


Mb,pos

Ma,p

os

Ma,n

eg

Mb,negMb,posM

a,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

Design Example 1


34

16′

12′

Mb,negMb,pos

Ma,p

os

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

Design Example 1


35

16′

12′

Mb,negMb,pos

Ma,p

os

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


Design Example 1


36

16′

12′

Mb,negMb,pos

Ma,p

os

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 = 0.052

Cb,posLL = 0.016


Design Example 1


37

16′

12′

Mb,negMb,pos

Ma,p

os

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


Design Example 1


38

16′

12′

Mb,negMb,pos

Ma,p

os

Case 4

Ma,n

eg

20



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


Design Example 1


39

16′

12′

Mb,negMb,pos

Ma,p

os

Case 4

Ma,n

eg



40

Design Example 1



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 = 0.75 ft-kip

16′

12′

Mb,negMb,pos

Ma,p

os

Ma,n

eg

21



One Way Slab Analysis

ACI Approximate analysis procedure (ACI 6.5.2).

41

Design Example 1


Mv

er(

+v

e)

Mv

er(

-ve)



42

Design Example 1


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 1/9Spandrel

SupportSimply

SupportedFor negative moment above the long wall common

to rooms and veranda, maximum moment will be

picked from both analyses.

Moment of 2.31 from two way slab analysis is more

than 1.5, therefore we will design for 2.31.

1/24

22



43

Design Example 1


Design of Two-Way Slab

Comparison of 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.

Mb,pos

Ma,p

os

Ma,n

eg

Mb,negMb,pos

Ma,n

eg

Ma,p

os

Mv

er

(+v

e)



44

Design Example 1



First determining Moment capacity of min. reinforcement:

As,min = 0.002bh = 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.

23



45

Design Example 1



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

Ma,n

eg



46

Design Example 1



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.

24



47

Design Example 1


Design of One-Way Slab

Main Reinforcement:

Mver (+ve) = 11.52 in-kip

As,min = 0.002bh = 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



48

Design Example 1


Design of One-Way Slab

Shrinkage Reinforcement:

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

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

25



49

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 9.3.1.1).

Take h = 1.5′ = 18″

d = h – 3 = 15″

b = 12″

Design Example 1


Verandah Beam

16.375 16.375

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



50

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′

Design Example 1


ln = 8′

9″ Brick Wall

8/2 = 4′

12″

Column

4′ + 1′ = 5′

26



51

Verandah Beam

Design

Step 03: Analysis

Using ACI Moment

Coefficients for analysis

of verandah beam

Design Example 1


16.375

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

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: (Class Activity)

Shear Design: (Class Activity)

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″

52

Design Example 1


Mu

(in-kip)

d

(in.)

b

(in.)

As

(in2)

Asmin

(in2)

Asmax

(in2)

As

(governing)

Bar

used

# of

bars

343.66 (+) 15 27.875 (beff) 0.64 0.90 3.645 0.90#4 5

#5 3

420.03 (-) 15 12 0.81 0.90 3.645 0.90#4 5

#4 + #5 2 + 2

Location Vu (@ d)

(kip)ΦVc = Φ2 𝐟′𝐜bwd

(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″

beff for spandrel beam is minimum of:

• 6h + bw = 6×6 + 12 =48″

• clear length of beam/12 + bw =(15.875′/12) × 12 + 12 =27.875″

• clear spacing between beams/2 + bw = not applicable

So beff = 27.875″

27



53

Design Example 1


Column Design

Sizes:

Column size = 12″ × 12″

Loads:

Pu = 11.41 × 2 = 22.82 kip

Verandah Column



54

Design Example 1


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 22.4.2.2}

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

28



55

Design Example 1


Drafting Details for

Slabs

Panel Depth (in) Mark Bottom Reinforcement Mark Top reinforcement

S1 5"

M1 #3 @ 10" c/cMT1 #3 @ 10" c/c Continuous End

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

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

MT2 #3 @ 18" c/c Non Continuous End

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

M2 #3 @ 10" c/c MT2 #3 @ 18" c/c Non Continuous End



56

Design Example 1


Drafting Details for Slabs

29



57

Design Example 1


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



58

Design Example 1


Drafting Details for Verandah Column

OR

COLUMN (C1)

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

30



59

Practical Example

(Crack in slab in village house due to

absence of negative reinforcement)



60

Practical Example

(Crack in slab in village house due to

absence of negative reinforcement)

31



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.

61

Design Example 2

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



Plan view of floor

62

4 spans @ 25′-0″

3 s

pans @

20′-0

″

Design Example 2


32



Given Data:

Material Properties:

fc′ = 3 ksi, fy = 40 ksi

Sizes:

Slab thickness, h = 7″ ; using #04 bar, davg = 7-0.75-0.5= 5.75″

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, dl = 0.105 ksf

wu, ll= 0.2304 ksf

wu = 1.2 (0.0875)+1.6 (0.144) = 0.336 ksf

63

Design Example 2


davgh



Complete analysis of the slab is done by analyzing four panels

64

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 s

pans @

20′-0

″

Design Example 2


33



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

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Design Example 2


65




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

Design Example 2


66

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

34




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

Design Example 2


67

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,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

Design Example 2


68

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

35




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

Design Example 2


69

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,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

Design Example 2


70

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

36



71






2


2

Using above relations the moments calculated are:

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)

Design Example 2


la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg



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

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg

Design Example 2


72

37




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

Design Example 2


73

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg




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

Design Example 2


74

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg

38




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

Design Example 2


75

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg




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

Design Example 2


76

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg

39




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

Design Example 2


77

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg



78






2


2

Using above relations the moments calculated are: (Class Activity)

Ca,neg = 0.075 Cb,neg = 0.017




Ma,neg = 8.93 ft-k (107.2 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)

Design Example 2


la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,negMb,neg

Ma,n

eg

40



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

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

Design Example 2


79




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

Design Example 2


80

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

41




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

Design Example 2


81

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg




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

Design Example 2


82

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

42




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

Design Example 2


83

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg




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

Design Example 2


84

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

43



85






2


2

Using above relations the moments calculated are: (Class Activity)

Ca,neg = 0.055 Cb,neg = 0.041




Ma,neg = 6.55 ft-k (78.6 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)

Design Example 2


la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg



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

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

Design Example 2


86

44




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

Design Example 2


87

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg




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

Design Example 2


88

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

45




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

Design Example 2


89

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg




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

Design Example 2


90

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

46




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

Design Example 2


91

la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg



92






2


2

Using above relations the moments calculated are:(Class Activity)

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)

Design Example 2


la = 18.83′

lb = 23.83′

Ma,n

eg

Ma,p

os

Mb,posMb,neg

Mb,neg

Ma,n

eg

47



93

Analysis Results (All values are in ft-kip)

Design Example 2


8.42

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 1.60Mneg at Non-

Continuous End =

1/3 of Mpos

NOTE: White: Long Direction Moments, Yellow: Short Direction Moments

4 spans @ 25′-0″

3 s

pans @

20′-0

″



94

Design Example 2


Slab Design

First determining the capacity of min. reinforcement:

As,min = 0.002bh = 0.002 × 12 × 7 = 0.168 in2

For #4 bars, spacing = (0.20/0.168) × 12 = 14.28 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(davg– a/2) = {0.9 × 0.20 × 40 × (5.75 – (0.26/2))}/12 = 3.37 ft-kip

48



95

Design Example 2


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

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

Longer direction.



96

Design Example 2


Slab Design (Class Activity)

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 = 0.2davg= 0.2(5.75) = 1.15, As = (5.37 × 12)/{0.9 × 40(5.75 – (1.15/2))} = 0.35 in2

Now, a = (0.35 × 40)/(0.85 × 3 × 12) = 0.45, after trails As = 0.32 in2

For #4 bars, spacing = (0.20/0.32) × 12 = 7.5

As all the above moments are almost same, we will use #4 @ 6 c/c for all above moments.

49



97

Design Example 2


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 3.37 ft-kip/ft, therefore using

#4 @ 12 c/c



98

Design Example 2


Slab Design (Class Activity)

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 = 0.55 in2

Using #4 bars, spacing = 4.5 c/c

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

50



99

Design Example 2


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 @ 6 c/c

Designing for 7.80 ft-kip/ft

Designing for 5.14 ft-kip/ft



Slab Reinforcement Details

100

A

BBB

C

C

A

C CB

A

A

BA

C

C

A

A B

A

A= #4 @ 12″

B = #4 @ 6″

C = #4 @ 4.5″

4 spans @ 25′-0″

3 s

pans @

20′-0

″

Design Example 2


NOTE: White: Long Direction Moments, Yellow: Short Direction Moments

51



101

Design Example 2


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



102

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 )

• Load transfer in longer direction = (wu × l / 2 × Wb ) + (wu × l / 2 × Wb )

Longer Direction

4 spans @ 25′-0″

3 s

pans @

20′-0

″

Shorte

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

Wb = 1 - Wa

ACI Table for Wa values

Design Example 2


52



103



4 spans @ 25′-0″

3 s

pans @

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-II

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

= 5.17 k/ft

Shorte

r Dire

ctio

n

20′

Wu = 0.336 ksfShorte

r Dire

ctio

n

20′

Wu = 0.336 ksf

• Wu = 0.336 ksf

• Wa,Panel-I = 0.71

• Wa,Panel-II = 0.83

Design Example 2




104



4 spans @ 25′-0″

3 s

pans @

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-III

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

(1 - 0.55)

= 3.11 k/ft

Shorte

r Dire

ctio

n

20′

Wu = 0.336 ksf

Shorte

r Dire

ctio

n

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

Design Example 2


53



Moment Coefficient Method: Example 2

Load On Beams from coefficient tables

105

Table: Load on beam in Panel I, using

Coefficients

(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 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Design Example 2




Table: Load on beam in Panel I, using

Coefficients

(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



B1

B1

B2

B2

B3 B3 B3 B4B4

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

Panel I

106

Design Example 2


Panel I

54





107

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel II, using

Coefficients

(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 0.83 - 2.78

B3 20 12.5 - 0.17 0.714

B4 20 12.5 - 0.17 0.714

Design Example 2


Panel II



Table: Load on beam in Panel II, using

Coefficients

(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 0.83 - 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″

108

Design Example 2


Panel II

55





109

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel III, using

Coefficients

(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 0.55 - 1.84

B2 25 10 0.55 - 1.84

B3 20 12.5 - 0.45 1.89

Design Example 2


Panel III



Table: Load on beam in Panel III, using

Coefficients

(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 0.55 - 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″

110

Design Example 2


Panel III

56





111

4 spans @ 25′-0″

3 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Table: Load on beam in Panel IV, using

Coefficients

(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

B3 20 12.5 - 0.29 1.22

Design Example 2


Panel IV



Table: Load on beam in Panel IV, using

Coefficients

(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

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″

112

Design Example 2


Panel IV

57




Load On Beams

113

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 sp

ans @

20

′-0″

B1

B1

B2

B2

B3 B3 B3 B4B4

Design Example 2





Load On Beams

114

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

Design Example 2


58




Load On Beams

115

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

Design Example 2


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 (including self-weight)

116

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

Design Example 2


59



117

nn n

Simple

supportIntegral with

support

wu

Spandrel

supportNegative

Moment

x wun2

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

*1/9 (2 spans)

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

Column

support1/16

Positive

Moment

x

1/14 1/16 1/11

wun2

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

(ii) To calculate negative moments, ℓn shall be the average of the adjacent clear span lengths.

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

Design Example 2




Analysis of Beams

Interior Beam B1

118

Design Example 2


60



Analysis of Beams

Exterior Beam B2

119

Design Example 2




Analysis of Beams

Interior Beam B3

120

Design Example 2


61



Analysis of Beams

Exterior Beam B4

121

Design Example 2


1.4

5 k

/ft



Design of Beams

Do design of beams yourself.

122

Design Example 2


62



123

Pictures of a Multi-Storey

Commercial Building



124

Pictures of a Multi-Storey

Commercial Building

63



Different Stages of Building Construction

125

Phases of Construction



Moment Coefficient Method: Home Work

Design the given slab system

126

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″

Slab thickness = 6″

SDL = 40 psf

LL = 60 psf

fc′ =3 ksi

fy = 40 ksi

Practice Examples

64



Moment Coefficient Method: Home Work


127

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″

Slab thickness = 6″

SDL = 40 psf

LL = 60 psf

fc′ =3 ksi

fy = 40 ksi

Practice Examples



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

For Two-way slab system

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

Select ACI moment coefficients based on la/ lb ratio

Calculate moments by using equations

● 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


128

Summary

65



As,min = 0.002bh

Max spacing requirements

smax = 2hf (ACI 8.7.2.2)

Load transfer in shorter direction beam

(Wu × l / 2 × Wa ) + (Wu × l / 2 × Wa )

Load transfer in longer direction beam

(Wu × l / 2 × Wb ) + (Wu × l / 2 × Wb )

Where Wb = 1 – Wa & Wa values selected from ACI tables

129

Summary



CRSI Design Handbook

ACI 318-14

Design of Concrete Structures 14th Ed. by Nilson, Darwin and

Dolan.

130

References

66



The End

131

Lecture-04 - Dr. Qaisar Ali · Lecture-04 Analysis and Design of Two-way Slab System (Part-I: Two...

Documents

Transcript of Lecture-04 - Dr. Qaisar Ali · Lecture-04 Analysis and Design of Two-way Slab System (Part-I: Two...