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Reinforced Concrete 2012 lecture 9/1

Budapest University of Technology and Economics

Department of Mechanics, Materials and Structures English courses Reinforced Concrete Structures Code: BMEEPSTK601 Lecture no. 9:

TWO-WAY SLABS


Content:

1. Definition, internal force components 2. Methods of analysis of two-way slabs 3. The method of Marcus 4. The effect of fixing the corners of the slab against lifting 5. The yield line theory 6. Design conditions set up for the parameter η to determine the

yield line pattern 7. Application of the yield line theory for more complex situations 8. Application of the yield line theory for different support condtions

and ground plan forms 9. Numerical example


1. Definition, internal force components specific force components in unit-length sections of two- way slabs If lll 5,0sh ≥ , the interaction of perpendicular strips of the slab through torsion (tx and ty (kNm/m)) is taken into account by determining the internal force distribution: the slab is regarded two-way slab


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, simply supported along the perimeter, EI= constant) -approximate analythical solutions by use of Fourier functions for limited no. of cases Numerical solutions can be computerized: -method of finite differences, finite element method (FEM) Results of both analythical and numerical solutions are available in tabulated form. Two manual approximate methods will be shown below.


3. The method of Marcus For rectangular slab, simply supported along the perimeter, loaded with uniformly distributed load. The load is distributed between series of two perpendicular strips by considering the condition, that at the intersection of two perpendicular strips the deflection is the same (effect of torsion is neglected):

EI

p

3845

wEI

p

3845

w4shsh

sh

4ll ll

l === →

4sh

shp

p

=

l

l

l

l (1)

p= psh+pℓ (2)

By introducing the parameters p

pshsh =α and

p

pll =α


4sh

sh

1

1

+

=α

ll

l

sh1 α−=αl

The load intensities can then be determined: psh= αshp and pℓ= αℓp For the two extreme cases, this yields in: 5,0sh =α=α l

9,0sh ≈α αℓ≈ 0,1


4. The effect of fixing the corners of the slab against lifting the corners of the slab lift, if they are not loaded by vertical forces of const- ructions above, resulting in torsional moments Moment distribution along a diagonal strip Moments’ equilibrium at corners

8

p3,0

82

)1,1(2

p

m2

2l

l

=⋅

=


Top reinforcement designed for negative moments bottom bars in practice: top mesh reinforcement parallel to sup- ports is used with the same intensity as that, designed for positive moments top bars


5. The yield line theory Yield lines: straight lines along primary cracks Along yield lines: m=mmax, v=0 ,

More exact model at corners:

y = ηηηη ll where ηηηη 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

lsh

msh

ml

ℓℓ


Equilibrium of triangular and trapezoidal panels of the slab Consider one of the triangular panels and the equilibrium of moments with respect to axis y :

ΣMy=0: shsh m

3

y

2

ypl

l

l=

lllll

l

lll

,o

22

22

22

2

m8

p

34

8

p

68

6

p

6py

m α=η=η=η==

where 2

34η=αl and

8

pm

2

,ol

l

l=


Consider now one of the trapezoidal panels and the equilibrium of moments with respect to axis x : ΣMx=0:

llll

lll

lllll

shshshshsh m22

p2

)2(32

p2

2 =⋅

η−+⋅

η⋅

Expressing msh, we get:

msh= sh,osh

2sh m

8

p)

34

1( α=η− l

where: η−=α34

1sh and mo,sh=8

p 2shl

The load p =pEd will be substituted in direction of the shorter and longer span respectively with modified values corresponding to the directions:

p3

41psh

η−= p3

4p 2η=l

( )ppp lsh ≤+


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 to the area of the distribution steel of that necessary in direction of the shorter span: as,ℓ=0,2as,sh (or approximately: mℓ=0,2msh) -let the reinforcement be same in the two perpendicular directions: as,ℓ=as,sh (or approximately: mℓ=msh) -let the triangle of the yield line pattern be rectangular: sh5,0y ll l =η= that is: lll /5,0 sh=η Limits of y determine limites of η, which should be checked:

2

y10

ll ll ≤≤ that is: 5,01,0 ≤η≤


7. Application of the yield-line theory for more komplex situations: -a system of continuous two-way slabs


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


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 maximum moment:

8/lpm 2xEd⋅α= ,


-rectangular slab with two free neighbouring edges and two restrained edges: the maximum moment:

8/lpm 2xEd⋅α=

-rectangular slab with two free neighbouring edges and two restrained edges Maximum moment:

2/lpm 2yEd⋅α=


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


9. Numerical example

Rectangular slab simply supported along the perimeter with optimal sidelength rate Problemt By what sidelength rate will the relationship – msh = 5 mℓ be true just at η = ηopt ? Lifting of the corners is impeded. Solution

Let be: γ=ll

lsh

ηopt = 2

2

1 γ (1)


(1-8

p

34

.58

p)

34 2

2opt

2sh

optlll η=η (2)

From (2) with the substitution γ=ll

lsh we get:

(1- 2opt

2opt 3

20)

3

4 η=γη

By expressing γ2 and substituting it in (1):

ηopt (1- 2optopt 3

20

2

1)

3

4 η⋅=η

We get the solution

ηopt = 0,214 = 22

h

r

2

1)(

2

1 γ=l

l


654,0214,02 =⋅=γ

That means: in case of ℓsh = 0,654ℓℓ , or ℓℓ = 1,528ℓsh that is at about

by the sidelength rate 1:1,5 will the rate of moments in the two directions be equal to 1:5 just by η = ηopt .

In the shorter direction will then be

715,0214,03

41

3

41sh =⋅−=η−=α

msh = 0,7158

p 2shl

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

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