Design B1, C1, F1Design one way Ribbed slab

225 mmNfc

4.50

4.50

2.00

B1 C1F1F1

C1

F2C2

F2C2

2420 mmNf y

LoadsMinimum thickness of Beams & one-way ribbed slabs

ElementSimply supported

One end continuous

Both ends continuous

Cantilever

One-way ribbed slabs

l/16l/18.5l/21l/8

cmhcmhcmh

258/2002518/4502816/450

min

min

min

Dead Load• Total volume (hatched) = 0.52 x

0.25 x 0.28 = 0.035 m3

• Volume of one hollow block = 0.4 x 0.20 x 0.25 = 0.02 m3

• Net concrete volume = 0.035 - 0.02 = 0.015 m3

• Weight of concrete = 0.015 x 2.5 (10)= 0.375 kN

• Weight of concrete /m2 = 0.375/(0.52)(0.25) = 2.88 kN/ m2

• Weight of hollow locks /m2 = 20(10)/(0.52)(0.25)(1000) = 1.54 kN/ m2

• Covering Materials = 2.5 kN/m2

• Total dead load = 2.88 + 1.54 + 2.5 = 7 kN/ m2

• Live Load = 2.5 kN/m2

Beam Analysis & design

mkNM u . 6.1688

5.46.66 2

2/6.665.45.260.195.4720.1 mtonw wallu

kNVu 1502

5.46.66max

Flexural Design

mmcmdmmcmhmmcmbmkNM u

240242/4.18.05.2282802880080

. 6.168

14cover

0108.0

240008 25 9.0 85.06.16810211

4202585.0

9.0,

/,.

2

6

lim

assumemmbd

MpaormmNf'fmkNM

w

yc

u

dbfM

ff

wc

u

y

c

85.01021185.0 6

0184.042025)85.0)(85.0(

73)85.0(

73

1max

y

c

ff

0161.042025)85.0)(85.0(

83)85.0(

83

1lim

y

c

ff

16111414

73.206.20732408000108.0

0033.0420

4.1,420425max4.1,

4max

22

lim

min

usecmmmdbA

ok

fff

ws

yy

c

9.0limmax

if

General Note

21

21.0420100042060025.065.0

1000600

25.065.01

y

y

ff

011.04201000

60042025)85.0)(85.0(25.0

1000600)85.0(25.0 12

yy

c

fff

old 011.021.0

dbfM

ff

wc

u

y

cnew 85.0

1021185.0 6

You can check using newIf there is no clear difference yoursolve is OK

Shear Design

kNVu 150kNV

kNVV

C

CCd

60 5.0

1201000/80024062575.0

No shear reinforcement

minimumShear R.

600,120,315min600,2

240,800

4202003min600,2

,3

min

5.0

max

db

fAs

VVV

w

yv

cducd

cdu VV 5.0

ends twoat the12@8

504100030

24042020075.0

cmuse

mmV

dfAs

sd

yv

kNVVV cdusd 30120150

cdu VV

2401000/24080032575.030

300,4

,3

min3

mmd

bfA

sdbf

Vw

yvw

csd

Checks

4801000/240*800253275.030

32

ifadequat issection the

c

wcsd dbfV

Column Design• For 5 storiesC1 Factored Load is 150kN for 1 floorFor 5 floors P for C1 = 750kN

cygcgu fffAP 85.085.08.065.0

148

10205001.0

20cm50cm200mm500mm

1.48983

3085.042001.03085.08.065.0000,750

2

2

usecmA

use

mmA

A

s

g

g

Footing design تصميم لك سأوضحبالتفصيل F1 Isolated Footing القواعد

Ps = 580 kNPu = 750kNqall =200kpa

Area required

kPaPaAPq

mq

PA

kPamtq

uu

netall

sg

netall

230102308.1*8.1

10750

8.1*8.19.21020010580

200/20

33

23

3

)(

2)(

Check for punching Sheard = 43.5 cm

6.608935.0*635.08.1*230

32191000/31404351225

314043540275.0

12'

2

4.17071000/314043532575.0

3'

6.15361000/3140435625

2050

2175.03

'21

3140)200435()500435(2

2 kNV

kNdbf

bdV

kNdbf

V

kNdbf

V

cmb

U

ocs

C

oc

C

oc

cC

o

O.K CU VV

Check for beam shearb = 1800mm, d = 435mm

11.1518.1*435.02

2.08.1*230

facecolumn from at

5.4891000/180043562575.0

CU

U

C

C

VV

kNV

dV

kNV

O.K

Bending moment & Area of steel

direction two/128 990010005000018.0

0.001 1000*435*250.859.0

73.6*102-1-1 42025*85.0

435d ,0001b

6.735.0*2

2.08.1*0.230

facecolumn from dat

22

min2

6

2

musecmmmA

mmmm

kNM

M

S

U

U

Slab design بالتفصيل الحقا ذلك لك سأوضح

DL L

D

Wallwu

Case of Loading

DL L

D

Wall

Wall

LD D

D DL

Two way slab

Two way slabs Types

Flat Plates

When the ratio (L/S ) is less than 2.0, it is called two - way slab

A flat plate floor is a two-way slab with no supporting beams, only columns.Flat Plate suitable span 7.5 m with LL= 500 kg/m2

AdvantagesLow cost formworkExposed flat ceilings Fast

DisadvantagesLow shear capacityLow Stiffness (notable deflection)

A two-way slab with column capitals or drop panels, or both.Flat Slab Max. suitable span 10m with LL= 700kg\m2

Flat Slabs

AdvantagesLow cost formworkExposed flat ceilings Fast

DisadvantagesNeed more formwork for capital and panels

Edge supported solid Slabs

The system can be used economically for spans up to 7.0 meters .

Waffle Slab• The waffle slab is capable of providing the largest spans of the conventional concrete floor systems, and can be economically used for spans up to 14.0 meters. • The ribs are formed with fiberglass or metal dome forms. The ribs are usually 0.60 to 0.90 meter on center. Shear is transferred to the columns by using beams or shear heads.

AdvantagesCarries heavy loadsAttractive exposed ceilings Fast

DisadvantagesFormwork with panels is expensive

Edge supported Ribbed SlabsThis system can be economically used for spans up to 7.0 meters. It is similar to the waffle slab, but the voids between ribs are filled with hollow blocks

Design MethodsSimplified Design Methods

Empirical formulae and approximate theories have been formulated which give bending moments in slabs supported on all four edges

ls www

Grashoff Method

The central defection of the strip in L direction

EILwl

l 3845 4

The central defection of the strip in S direction

EISws

s 3845 4

EISw

EILw sl

3845

3845 44

4

4

SL

ww

l

s

44

4

SLL

www

ls

s

Since

ls www

Grashoff load coefficientsr=L/S1.01.11.21.31.41.51.61.71.81.92.0

0.5000.5940.6750.7410.7930.8350.8680.8930.9130.9290.941

0.5000.4060.3250.2590.2070.1650.1320.1070.0870.0710.059

44

4

SLLwws

wr

rwws

4

4

1

wr

wwl

41

1

let r = L/S

Marcus MethodMarcus has given an approximate method for determining bending moments in slabs simply supported on four edges with corners prevented from being lifted, and considering torsion in the slab.

Marcus load coefficientsr=L/

S1.01.11.21.31.41.51.61.71.81.92.0

0.2920.3550.4110.4700.5260.5770.6230.6630.6990.7300.757

0.2920.2400.1980.1650.1370.1140.0950.0790.0670.0560.047

Egyptian Code MethodThe Egyptian code method has extended the last methods for analysis of continuous slabs, where individual panels have different supporting conditions. To account for the various supporting conditions, the rectangularity ratio r =L/S is modified to be

s

l

mSmL

r

For simply supported spans, m = 1.0, for spans continuous from one side, m = 0.87, and for spans

continuous from both sides, m = 0.76.

Egyptian Code load coefficientsr'1.01.11.21.31.41.51.61.71.81.92.0

0.350.400.450.500.550.600.650.700.750.800.85

0.350.290.250.210.180.160.140.120.110.090.08

Beams for Two-way Slabs Designed by Approximate Methods

L

S

Shear and moment equivalent load coefficients for trapezoidal load distribution

r =L/S1.01.11.21.31.41.51.61.71.81.92.0

Cs0.5000.5450.5830.6150.6430.6670.6880.7060.7220.7370.750

Cb0.6670.7250.7690.8030.8300.8520.8700.8850.8970.9080.917

Minimum Slab Thickness of Two-way Edge-supported

SlabsThe following values give some indication for the minimum thickness of two way slab (solid or ribbed slab)

Suitable for Gaza

Using the Grashoff coefficients design the simply supported solid slab shown in Figure . Assume that the beam webs are 25 cm wide. Weight of flooring materials is 2 kN/m2, and the live load is 4 kN/m2. use h=140mm

Example 1

Mpafc 25

Mpaf y 420

Solution

Minimum Slab Thickness for two-way construction (ACI

code)The ACI Code 9.5.3 specifies a minimum slab thickness to control deflection. There are three empirical limitations for calculating the slab thickness (h), which are based on experimental research. If these limitations are not met, it will be necessary to compute deflection.

22.0 m (a) For

2.05361400

8.0

m

yn

fl

h

fy in N/mm2.

scs

bcb

scs

bcb

EE

/4E/4E

II

lIlI

slab uncracked of inertia ofMoment Ibeam uncracked of inertia ofMoment I

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

s

b

sb

cb

13

4

2

44321 m

2m (b) For

9361.1

/420 9361400

8.0

n

2y

yn

lh

mmNffor

fl

h

2.0m (c) ForUse the following table

The definitions of the terms are:

h = Minimum slab thickness without interior beamsln =

m=

Clear span in the long direction measured face to face of column

the ratio of the long to short clear span

The average value of a for all beams on the sides of the panel.

Example 1 Ref 2Design Sec. A-A & B-B in the two-way solid slab. The

covering materials weigh is 2kN/m2, and the live load is 3kN/m2. Also,

All beams are 25 cm wide.

Mpafc 25

Mpaf y 420

B

A

B

A

3 - Evaluate load distribution in both directions:A

A

BB

Sec. A-A

Sec. B-B

Example 2Design the two-way ribbed slab

The covering materials weigh 2.00 N/mm2, live load is 3.00 N/mm2.

Mpafc 25

Mpaf y 420

Ref 2

2533.2333

77033min

cmcm

lh n

(assumption)

2/8.12

360.1205.164.320.1

mkN

wu

Analysis

bentstraight

use

161141

Two way slabDirect Design Method DDM

Direct Design Method for Two-way Slab

1 - Minimum of 3 continuous spans in each direction. (3 x 3 panel)

2 - Rectangular panels with long span/short span

Method of dividing total static moment Mo into positive and negative moments.

Limitations on use of Direct Design method

2

Limitations on use of Direct Design method

3- Successive span in each direction shall not differ by more than 1/3 the longer span.4 - Columns may be offset from the basic rectangular grid of the building by up to 0.1 times the span parallel to the offset.

5- All loads must be due to gravity only (N/A to unbraced laterally loaded frames, from mats or pre-stressed slabs)

6- Service (unfactored) live load service dead load

2

7- For panels with beams between supports on all sides, relative stiffness of the beams in the 2 perpendicular directions.

Shall not be less than 0.2 nor greater than 5.0

212

221

l

l

Definition of Beam-to-Slab Stiffness Ratio,

Accounts for stiffness effect of beams located along slab edge reduces deflections of panel adjacent to beams.

slab of stiffness flexural

beam of stiffness flexural

With width bounded laterally by centerline of adjacent panels on each side of the beam.

scs

bcb

scs

bcb

IEIE

/lIE/lIEα

44

44

slab uncracked of inertia ofMoment Ibeam uncracked of inertia ofMoment I

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

s

b

sb

cb

Basic Steps in Two-way Slab Design DDM

1- Choose slab thickness to control deflection. Also, check if thickness is adequate for shear.

2- Calculate positive and negative moments in the slab.

3- Determine distribution of moments across the width of the slab. - Based on geometry and beam stiffness.

4- Assign a portion of moment to beams

Minimum slab thickness (see lecture 6)

Maximum Spacing of Reinforcement

At points of max. +/- M:

Min Reinforcement Requirements

7.12.3 ACI in. 18 and

13.3.2 ACI 2

s

ts

s min s T&S from ACI 7.12 ACI 13.3.1A A

Distribution of Moments

Slab is considered to be a series of frames in two directions:

Total static Moment, Mo

3-13 ACI 8

2n2u

0llwM

cn

n

2

u

0.886d h using calc. columns,circular for

columnsbetween span clear

strip theof width e transvers

areaunit per load factored

l

l

l

w

where

Column Strips and Middle Strips

Moments vary across width of slab panel

Design moments are averaged over the width of column strips over the columns & middle strips between column strips.

Column Strips and Middle Strips

Column strips Design w/width on either side of a column centerline equal to smaller of

1

2

25.0 25.0

ll

l1= length of span in direction moments are being determined.

l2= length of span transverse to l1

Column Strips and Middle Strips

Middle strips: Design strip bounded by two column strips.

Positive and Negative Moments in Panels

M0 is divided into + M and -M Rules given in ACI sec. 13.6.3

Longitudinal Distribution of Moments in Slabs

For a typical interior panel, the total static moment is divided into positive moment 0.35 Mo and negative moment of 0.65 Mo.

For an exterior panel, the total static moment is dependent on the type of reinforcement at the outside edge.

Distribution of M0

The factored components of the moment for the beam.

Transverse Distribution of Moments

Transverse distribution of the longitudinal moments to middle and column strips is a function of the ratio of length l2/l1,1, and t.

363.01

22

3

sscs

cbt

scs

bcb1

yxy

xC

IC

IECE

IEIE

torsional constant

363.01

3 yxy

xC

Take largest value of C from the following

Factored Moment in Column Strip

Factored Moment in an Exterior Panel

Minimum extension for reinforcement in slabs

without beams(Fig. 13.3.8)

Example 1 Design the long direction of an interior panel of the two-way slab for the floor system.The floor consists of six panels at each direction, with a panel size 7.5 m x 6 m. All panels are supported by 40 cm square columns. The slabs are supported by beams along the column line with cross sections. The service live load is to be taken as 4kN/m2 and the service dead load consists of 6.5kN/m2 of floor finishing in addition to the self-weight.

h=18cm

The cross-sections are:h = 18cm

Example 1- Loading The weight of the slab is given as.

16_4.146.1218

/2.14)4(6.15.62.1 2

usecmdmkNwu

Example 1 – Strip SizeCalculate the strip sizes

4.5m

Example 1, Static Moment Computation

Moment Mo for the two directions.

long direction

short direction

mkNlwlM

cmL

nol

n

.5378

1.762.148

7104075022

2

mkNlwlM

cmL

nol

n

.4188

6.55.72.148

5604060022

2

The factored components of the moment for the beam (long).

Negative - Moment

Positive + Moment

mkN

mkN.18853735.0.34953765.0

Example 1 – Moments (long)

Example 1- - Moment (long) Coefficients

The moments of inertia about beam, Ib = 0.01 m4 and Is = 0.0027 m4 (long direction) are need to determine the distribution of the moments between the column and middle strip.

96.2)8.0(7.3

7.30027.0

01.0

8.05.7

6

1

21

1

1

2

ll

IIll

s

b

Example 1- Column Strip Factors (negative)

Need to interpolate to determine how the negative moment is distributed.

81.0)5.08.0(1.0-0.50.75-0.90.9

factor strip col.

Need to interpolate to determine how the positive moment is distributed

81.0)5.08.0(1.0-0.50.75-0.90.9

factor strip col.

Example 1 - Column Strip Factors (positive)

Example 1 - Moment (long) column/middle strips

Components on the beam (long).

Negative – Moment

Positive + Moment

Column Strip

Negative – Moment

Positive + Moment

Middle Strip

mkNmkN

.152)188(81.0.7.282)349(81.0

mkNmkN

.36)188(19.0.66)349(19.0

Example 1 - Moment (long)-beam/slab distribution

When 1 (l2/l1) > 1.0, ACI Code Section 13.6.5 indicates that 85 % of the moment in the column strip is assigned to the beam and balance of 15 % is assigned to the slab in the column strip.

Beam Moment

Slab Moment

Column Strip - Negative Moment

mkNmkN

.4.42)320(15.0.3.240)7.282(85.0

Beam Moment

Slab Moment

Column Strip - Positive Moment

mkNmkN

.23)152(15.0.129)152(85.0

Results

shearcheckAmmcmb

mmcmd

s _40040

520528.08.0458

shearcheckAmmmb

mmcmd

s _30003

1444.14