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Reinforced Concrete 2012 lecture 9/1

Budapest University of Technology and Economics

Department of Mechanics, Materials and Structures

English coursesReinforced Concrete StructuresCode: BMEEPSTK601

Lecture no. 9:

TWO-WAY SLABS

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Content:

1. Definition, internal force components2. Methods of analysis of two-way slabs3. The method of Marcus

4. The effect of fixing the corners of the slab against lifting5. The yield line theory6. Design conditions set up for the parameter η to determine the

yield line pattern

7. Application of the yield line theory for more complex situations8. Application of the yield line theory for different support condtions

and ground plan forms9. Numerical example

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1. Definition, internal force components

specific force componentsin unit-length sections of two-way slabs

If lll 5,0sh ≥ , the interaction of perpendicular strips of the slab throughtorsion (t x and t y (kNm/m)) is taken into account by determining theinternal force distribution: the slab is regarded two-way slab

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2. Methods of analysis of two-way slabs

The partial differential equation of slabs:

EI

p

yx

w

yx

w

x

w22

4

22

4

4

4

−=∂∂

∂+

∂∂

∂+

∂

∂ w : deflection p =p (x,y) load

Analythical solutions -exact: were only elaborated for very simple cases(for example: uniformly distributed load, rectangular slab, simplysupported along the perimeter, EI= constant)-approximate analythical solutions by use of Fourier functions forlimited no. of casesNumerical solutions can be computerized:-method of finite differences, finite element method (FEM)Results of both analythical and numerical solutions are available intabulated form .Two manual approximate methods will be shown below.

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3. The method of Marcus

For rectangular slab, simply supported along the perimeter, loaded withuniformly distributed load.The load is distributed between series of two perpendicular strips by

considering the condition, that at the intersection of two perpendicularstrips the deflection is the same (effect of torsion is neglected):

EI

p

384

5

wEI

p

384

5

w

4shsh

sh

4llll

l === →

4

sh

shp

p

=l

l

l

l

(1)

p = p sh +p ℓ (2)

By introducing the parametersp

pshsh =α andppl

l =α

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4sh

sh

1

1

+

=α

ll

l

sh1 α−=αl

The load intensities can then be determined:

p sh= αshp and p ℓ= αℓp

For the two extreme cases, this yields in:

5,0sh =α=α l

9,0sh ≈α αℓ≈ 0,1

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4. The effect of fixing the corners of the slab against lifting

the corners of the slab lift, if they arenot loaded by vertical forces of const-ructions above, resulting in torsional

moments

Moment distributionalong a diagonal strip

Moments’ equilibrium at corners

8

p3,0

82

)1,1(2p

m22

ll

=⋅

=

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Top reinforcement designed for negative moments

bottom bars

in practice:

top mesh reinforcement parallel to sup-ports is used with the same intensityas that, designed for positive moments

top bars

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5. The yield line theory

Yield lines: straight lines along primary cracksAlong yield lines: m =m max, v =0,

More exact model at corners:

y = η ηη η ll where η ηη η can be freely

adopted between 0,1 and 0,5 .

For example:

2

l

shopt

l

l

2

1

=η (results min.

quantity of reinforcement)

yield line

l sh

m

m

ℓℓ

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Equilibrium of triangular and trapezoidal panels of the slab

Consider one of the triangular panels and the equilibrium of momentswith respect to axis y :

ΣMy=0: shsh m

3

y

2

ypl

l

l=

ll

lll

l

lll

,o

22

22

22

2

m8

p

3

4

8

p

6

8

6

p

6

py

m α=η=η=η==

where 2

3

4η=αl and

8

pm

2

,ol

l

l=

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Consider now one of the trapezoidal panels and the equilibrium of

moments with respect to axis x :

ΣMx=0:

llll

lll

lllll

shshshshsh m

22

p

2

)2(

32

p

2

2 =

⋅

η−+

⋅

η⋅

Expressing msh, we get:

msh= sh,osh

2sh m

8

p)

3

41( α=η−

l

where: η−=α3

41sh and mo,sh=

8

p 2shl

The load p =p Ed will be substituted in direction of the shorter and longer

span respectively with modified values corresponding to the directions:p

3

41psh

η−= p

3

4p

2η=l ( )ppp lsh ≤+

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6. Design conditions set up for the parameter η to determine

the yield line pattern

Beside

2

shopt

2

1

=η=η

ll

l some further conditions that can be set up to

determine the value of the parameter η :-let the steel necessary in direction of the longer span be equal tothe area of the distribution steel of that necessary in direction of theshorter span: a s,ℓ=0,2a s,sh (or approximately: m ℓ=0,2m sh)

-let the reinforcement be same in the two perpendicular directions:a s,ℓ=a s,sh (or approximately: m ℓ=m sh)-let the triangle of the yield line pattern be rectangular:

sh5,0y lll =η= that is: lll / 5,0 sh=η

Limits of y determine limites of η, which should be checked:

2y

10

ll ll≤≤ that is: 5,01,0 ≤η≤

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7. Application of the yield-line theory for more komplex situations:

-a system of continuous two-way slabs

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-irregular ground plan forms can be completed to rectangle and handled

like that

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8. Application of the yield line theory for different support

condtions of rectangualr slabs and different ground plan forms

-rectangular slab panel simply supported along three sides

and free along the forth side:the maximummoment:

8 / lpm 2xEd⋅α= ,

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-rectangular slab with two free neighbouring edges and two

restrained edges:

the maximummoment:

8 / lpm2xEd⋅α=

-rectangular slab with two free neighbouring edges and tworestrained edges

Maximum

moment:2 / lpm 2

yEd⋅α=

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Circular and other ground plan forms that can be characterized

with inscribed circle, simply supported along the perimeter

8

Dpm

2Edα=

b

a

≤ 1

≤ 1

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9. Numerical example

Rectangular slab simply supported along the perimeter withoptimal sidelength rate

ProblemtBy what sidelength rate will the relationship – msh = 5 mℓ be true just atη = ηopt ? Lifting of the corners is impeded.

Solution Let be: γ =

ll

lsh

ηopt =

2

2

1

γ (1)

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(1-8

p

3

4.5

8

p)

3

42

2

opt

2sh

opt

lllη=η (2)

From (2) with the substitution γ =ll

lsh we get:

(1- 2opt2opt3

20)34 η=γ η

By expressing γ2 and substituting it in (1):

ηopt (1- 2optopt

3

20

2

1)

3

4η⋅=η

We get the solution

ηopt = 0,214 = 22

h

r

2

1)(

2

1γ =

l

l

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654,0214,02 =⋅=γ That means: in case of ℓsh = 0,654ℓℓ , or ℓℓ = 1,528ℓsh that is at aboutby the sidelength rate 1:1,5 will the rate of moments in the twodirections be equal to 1:5 just by η = ηopt .

In the shorter direction will then be

715,0214,03

41

3

41sh =⋅−=η−=α

m sh = 0,7158

p 2shl

that is by 28,5 % smaller, then in case of not taken into account theeffect of two-way ection. It is reasonable not to forget this rate, and tryto apply in the practice when possible!

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