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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
Reinforced Concrete 2012 lecture 9/2
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
Reinforced Concrete 2012 lecture 9/3
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
Reinforced Concrete 2012 lecture 9/4
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.
Reinforced Concrete 2012 lecture 9/5
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
Reinforced Concrete 2012 lecture 9/6
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
Reinforced Concrete 2012 lecture 9/7
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
=⋅
=
Reinforced Concrete 2012 lecture 9/8
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
Reinforced Concrete 2012 lecture 9/9
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
Reinforced Concrete 2012 lecture 9/10
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=
Reinforced Concrete 2012 lecture 9/11
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 ≤+
Reinforced Concrete 2012 lecture 9/12
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≤
Reinforced Concrete 2012 lecture 9/13
7. Application of the yield-line theory for more komplex situations: -a system of continuous two-way slabs
Reinforced Concrete 2012 lecture 9/14
-irregular ground plan forms can be completed to rectangle and handled like that
Reinforced Concrete 2012 lecture 9/15
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= ,
Reinforced Concrete 2012 lecture 9/16
-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=
Reinforced Concrete 2012 lecture 9/17
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
Reinforced Concrete 2012 lecture 9/18
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)
Reinforced Concrete 2012 lecture 9/19
(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
Reinforced Concrete 2012 lecture 9/20
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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