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تصميم المنشآت الخرسانية المسلحة

DESIGN OF REINFORCED

CONCRETE STRUCTURES

Design of Tow Way Slabs

Direct Design Method (DDM)

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DESIGN OF TWO-WAY FLOOR

SLAB SYSTEM

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One-way and two-way

slab

Direct Design Method

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Comparison of One-way and Two-way slab

behavior

One-way

slabs carry

load in one

direction.

Two-way

slabs carry

load in two

directions.

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Comparison of One-way and Two-way slab

behavior

One-way

and two-

way slab

action

carry

load in

two

directions

One-way slabs: Generally, long side/short side > 2

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Comparison of One-way and

Two-way slab behavior

Flat Plate Waffle slab

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Comparison of One-way and Two-way slab

behavior

Flat slab Two-way slab with beams

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Basic Steps in Two-way Slab Design

1. Choose layout and type of slab.

2. Choose slab thickness to control deflection.

3. Check if thickness is adequate to resist shear.

4. Choose Design method

A. Equivalent Frame Method- use elastic frame

analysis to compute positive and negative moments

B. Direct Design Method - uses coefficients to compute

positive and negative slab moments

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Basic Steps in Two-way Slab Design

5. Divide into column and middle strips …

6. Calculate positive and negative slab moments.

7. Determine distribution of moments across the width of the

slab.

8. Assign a portion of moment to beams, if present.

9. Design reinforcement for moments (steps 5 and 6).

10. Distribute steel.

10. Repeat steps in the other direction.

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Minimum Slab Thickness for two-way

construction

Slabs

without

interior

beams

spanning

between

supports

and ratio

of long

span to

short span

< 2

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Minimum Slab Thickness for two-way

construction

Slabs without drop panels meeting, tmin = 5 in

Slabs with drop panels meeting, tmin = 4 in

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Minimum Slab Thickness for two-way

construction

Maximum Spacing of Reinforcement

At points of max. +/- M:

Max. and Min Reinforcement Requirements

7.12.3 ACI in. 18 and

13.3.2 ACI 2

s

ts

balsmaxs

S&Tsmins

75.0

13.3.1 ACI 7.12 ACI from

AA

AA

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DDM Definitions27

DDM Definitions28

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Effective beam sections

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Shear in 2-way Slabs

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Definition of Beam-to-Slab Stiffness Ratio, a

Accounts for stiffness effect of beams located along

slab edge reduces deflections of panel adjacent to

beams.

slab of stiffness flexural

beam of stiffness flexurala

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Beam-to-Slab Stiffness Ratio, a

With width bounded laterally by centerline of

adjacent panels on each side of the beam.

scs

bcb

scs

bcb

4E

4E

/4E

/4E

I

I

lI

lIa

slab uncracked of inertia ofMoment I

beam uncracked of inertia ofMoment I

concrete slab of elasticity of Modulus E

concrete beam of elasticity of Modulus E

s

b

sb

cb

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Beam and Slab Sections for calculation of a

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Beam and Slab Sections for calculation of a

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Beam and Slab Sections for calculation of a

Definition of beam cross-section

Charts may be used to calculate a Fig. 13-21

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Distribution of Moments

Total static Moment, Mo

3-13 ACI

8

2

n2u0

llwM

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

wwhere

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Column Strips and Middle Strips

Moments vary across width of slab panel

Design moments are averaged over:

1. the width of column strips over the columns &

2. The middle strips between column strips.

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Critical

Sections

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Negative and positive design moments

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Distribution of M0

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Positive and Negative Moments in Panels

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Factored Moment

in Column Strip

a1 = Ratio of flexural

stiffness of beam to

stiffness of slab in direction

l1.

bt = Ratio of torsional

stiffness of edge beam to

flexural stiffness of slab

(width = to beam length)

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Factored Moment in Column Strip

Ratio of flexural

stiffness of beam to

stiffness of slab in

direction l1.

Ratio of torsional

stiffness of edge

beam to flexural

stiffness of

slab(width= to

beam length)

bt

a1

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Example: Using the ACI Code, determine the

required thickness for slabs in Panels 3 and 2 . Edge

beams are used around the building perimeter.(300

mm wide x 200mm drop), fy =414 Mpa (G60)

Solution54

For interior Panels 3: ln = 6 - 0.4 = 5.6 m, h = ln/33 = 5.6/33 = 0.17 m) > 125 mm

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h = 7 in

Flat

plate

without

edge

beams

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Example

Design an interior flat plate…

LL =80 psf = 3.83 kN/m2,

DL +Own wt. =110psf = 5.27 kN/m2,

fy = 60ksi

= 414 MPa,

f’c = 3 ksi = 21 MPa

Column height = 12ft =3.6m

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Slab h same as before the required h = 7 in =0.17m(Use: h = 7.5 “

= 0.19m1 Psf = 47.9 N/m2

Wu = 1.2*5.27+1.6*3.8 3= 13 kN/m2

Shear check for h58

d = 6.25*25.4

= 159 mmsay 0.16m

Wu = 13.0 kPaOne Way shear

Ln1 = 6.1- 0.4 = 5.7 m

Ln1@d = 0.5*5.7-0.16 =2.69 m

Vu = 13 *1* (2.69)

= 35 kN < F Vc

F Vc = 0.75*0.16 * 210.5 * 160

= 88 kN OK

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bo = 2 (0.4+0.16+0.3+0.16)

= 2.04 m

A = 6.1*4.88- (0.56*.46)

= 29.8 m2

Vu2 = 13*(29.8)

=387 kN

F Vc = 0.75*.33*(21)0.5*2.04*160

= 387 kN > 355 kN OK

Mo = ? (in short and long directions60

Mol =13*4.88*5.72/8 = 245 kN.m

Mos =13* 6.1 *4.582/8 = 208 kN.m

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Interior -Ve, Exterior –Ve and +Ve

moments in Long direction

Moment Distribution in long direction (between

col. And middle strips

mkNftk

llwM n

ul

.234.2.181

8/)12/1620(*16)26.0(8

. 22

120

Column Strip Middle strip

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Moment Distribution in long direction: col. And middle strips

mkNftkl

lwM nul .245.2.1818/)12/1620(*16*26.0

8. 2

2

120

M-ext = 0.65 M0 =75% for col. Strip + 25% for Middle strip

M+ = 0.35 M0 =60% for col. Strip + 40% for Middle strip

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Similarly the distribution in

the short direction

ftkl

lwM nus .2.1468/)12/1216(*20*)26.0(

8. 2

2

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Summary and design (short span)68

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Slab with Interior beams

Steps:1.Find slab thickness (Eqns).2.Find: Wu = 1.2*WD+1.6*WL, and Find Mol, Mos

3.Distribute to –ve int Mom, -ve ext mom, +VeMom.

4.Distribute to the moments in step 3 into beam and column strip moments

5.Distribute column Strip moments into Beam and slab moments

6.Find As ….. And distribue steel

Beam and slab contributions70

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Slabs with

Interior Beams:

b= ln long / ln shor

USC Units:

Thickness in SI units:

Example: Slabs with interior beams72

Given: 2-Way Slab with beams as shown,

h = 7”, F’c = 3 ksi, fy = 60 ksi

Req’d: Check the ACI requirements for int panel

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a2 and am

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Minimum Slab thickness:75

Exmple:

Slabs With Beams

(DDM)Determine the slab thickness and

the +ve and the –ve moments

required for the design of the

exterior panel of the shown slab.

LL = 120psf, DL = 100 psf

(including own weight . 15”x15”

and 12’ long. The slabs are

supported by beams. F’c = 3ksi,

fy = 60ksi.

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Check the code limitations1. more than 3 spans 2. equal spans 3. no offsets

4. rectangular shape with long/short spans < 2

5. ….

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Direction of the 18’ slab width

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Slab Thickness:

Moments/ Interior Panel:

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Interpolation:

Col strip will resist from M-int

=0.75 – (0.75-0.45)*(1.22-1)

= 0.68 of 0.65 M0

= - 0.68*065*241

= -107 k.ft (for col strip + beam)

As the beam stiffness 85% of this value will be resisted

by the beam = (0.85* 107)= 90.95 k.ft

and 16.05 k.ft should be resisted by remaining col. strip

similarly: 0.68(+84)= 57.1 k.ft

(48 For beam and 9.1 for slab)

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Short span on the edge beam

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18ft strip

107 +50

= 157 k.ft

+57+ 30

=87 k.ft

-57 -27

+31+16 k.ft

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