slab reinforcement

Slab reinforcement

0.Initital data

Dimensions

n 8:=

L1 4.3m 0.1 m⋅ n⋅+ 5.1m=:=

L2 6.4m 0.05 m⋅ n⋅− 6m=:=

T1 4m 0.05m n⋅+ 4.4m=:=

T2 6.7m 0.1m n⋅− 5.9m=:=

slab thickness

t 13cm:=

1.Loads

Permanent load

g t 25⋅kN

m3

3.25kN

m2

⋅=:=

Live load

p 1.3kN

m2

0.1nkN

m2

+ 2.1kN

m2

⋅=:=

Total load

F 1.35g 1.5p+( )1m 7.537kN

m⋅=:=

2.Moments on each type of panel

2.1. Panel with 2 adiacent simple supported contour lines

long- L.1short- T.1

L1

T1

1.159=

Negative moment on fixed contour lines

βsx1 0.0665:=

βsy1 0.045:=

MRx.1 βsx1 F⋅ T12

⋅ 9.704 kN m⋅⋅=:=

MRy.1 βsy1 F⋅ T12

⋅ 6.567 kN m⋅⋅=:=

Positive moment at the middle-span

βsx1 0.0495:=

βsy1 0.034:=

MCx.1 βsx1 F⋅ T12

⋅ 7.223 kN m⋅⋅=:=

MCy.1 βsy1 F⋅ T12

⋅ 4.961 kN m⋅⋅=:=

2.2. Panel with a higher contour line simple supported

long- T2

short- L.1

T2

L1

1.157=


βsx2 0.053:=

βsy2 0.037:=

MRx.2 βsx2 F⋅ L12

⋅ 10.391 kN m⋅⋅=:=

MRy.2 βsy2 F⋅ L12

⋅ 7.254 kN m⋅⋅=:=


βsx2 0.0395:=

βsy2 0.028:=

MCx.2 βsx2 F⋅ L12

⋅ 7.744 kN m⋅⋅=:=

MCy.2 βsy2 F⋅ L12

⋅ 5.489 kN m⋅⋅=:=

2.3. Panel with a higher contour line simple supported(2)

long- L2

short- T.1

L2

T1

1.364=


βsx3 0.06:=

βsy3 0.037:=


⋅ 8.756 kN m⋅⋅=:=


⋅ 5.399 kN m⋅⋅=:=


βsx3 0.0495:=

βsy3 0.028:=


⋅ 7.223 kN m⋅⋅=:=


⋅ 4.086 kN m⋅⋅=:=

2.4. Panel with all contour lines fixed

long- L2

short- T.2

L2

T2

1.017=


βsx4 0.031:=

βsy4 0.032:=


⋅ 8.134 kN m⋅⋅=:=


⋅ 8.396 kN m⋅⋅=:=


βsx4 0.024:=

βsy4 0.024:=


⋅ 6.297 kN m⋅⋅=:=


⋅ 6.297 kN m⋅⋅=:=

2.5. Panel with a higher contour line simple supported(3)

long- L1

short- T.1

L1

T1

1.159=


βsx5 0.053:=

βsy5 0.037:=


⋅ 7.734 kN m⋅⋅=:=


⋅ 5.399 kN m⋅⋅=:=


βsx5 0.039:=

βsy5 0.028:=


⋅ 5.691 kN m⋅⋅=:=


⋅ 4.086 kN m⋅⋅=:=

2.6. Panel with all contour lines fixed(2)

long- T2

short- L.1

T2

L1

1.157=


βsx6 0.0395:=

βsy6 0.032:=

MRx.6 βsx6 F⋅ L12

⋅ 7.744 kN m⋅⋅=:=

MRy.6 βsy6 F⋅ L12

⋅ 6.274 kN m⋅⋅=:=


βsx6 0.03:=

βsy6 0.024:=

MCx.6 βsx6 F⋅ L12

⋅ 5.882 kN m⋅⋅=:=

MCy.6 βsy6 F⋅ L12

⋅ 4.705 kN m⋅⋅=:=

3. Moment equilibration

3.1 Contour moment equilibration:

- on y direction:

from panel 1 (L.1/T.1) to panel 2 (L.1/T.2)

M12

MRy.1 MRy.2+

26.91 kN m⋅⋅=:=


M34

MRy.3 MRy.4+

26.898 kN m⋅⋅=:=

from panel 5 (L.1/T.1) to panel 6 L.1/T.2)

M56

MRy.5 MRy.6+

25.836 kN m⋅⋅=:=

- on x direction:


M13

MRx.1 MRx.3+

29.23 kN m⋅⋅=:=


M24

MRx.2 MRx.4+

29.262 kN m⋅⋅=:=


M35

MRx.3 MRx.5+

28.245 kN m⋅⋅=:=

from panel 4 (L.2/T.2) to panel 6 L.1/T.2)

M46

MRx.4 MRx.6+

27.939 kN m⋅⋅=:=

3.2 Central moment equilibration:

for interior panels,equilibration is made from both senses.

-on y:

Section 1 (panels 1-2-1)

panel 1

MCy1 MCy.1 M12− MRy.1+ 4.618 kN m⋅⋅=:=

panel 2(interior panel)

MCy2 MCy.2 2M12− 2MRy.2+ 6.177 kN m⋅⋅=:=


panel 3

MCy3 MCy.3 M34− MRy.3+ 2.587 kN m⋅⋅=:=


MCy4 MCy.4 2M34− 2MRy.4+ 9.294 kN m⋅⋅=:=


panel 5

MCy5 MCy.5 M56− MRy.5+ 3.649 kN m⋅⋅=:=


MCy6 MCy.6 2M56− 2MRy.6+ 5.58 kN m⋅⋅=:=

Verifications :


MCy.1 MRy.1+ MRy.2+ MCy.2+ MCy.1 MRy.1+ MRy.2+( )+ 43.053 kN m⋅⋅=

MCy1 M12+ M12+ MCy2+ MCy1 M12+ M12+( )+ 43.053 kN m⋅⋅=







-on x:

Section 1 (panels 1-3-5-3-1)

panel 1

MCx1 MCx.1 M13− MRx.1+ 7.698 kN m⋅⋅=:=

panel 3 (interior panel)

MCx3 MCx.3 M13− MRx.3+ M35+ MRx.5− 7.26 kN m⋅⋅=:=

panel 5 (interior panel)

MCx5 MCx.5 2M35− 2MRx.5+ 4.67 kN m⋅⋅=:=


panel 2

MCx2 MCx.2 M24− MRx.2+ 8.872 kN m⋅⋅=:=


MCx4 MCx.4 M24− MRx.4+ M46 MRx.6−+ 5.364 kN m⋅⋅=:=


MCx6 MCx.6 2M46− 2MRx.6+ 5.492 kN m⋅⋅=:=


2 MCx.1 MRx.1+ MRx.3+ MCx.3+ MRx.3+( ) MRx.5+ MCx.5+ MRx.5+ 104.483 kN m⋅⋅=

2 MCx1 M13+ M13+ MCx3+ M35+( ) M35+ MCx5+ M35+ 104.483 kN m⋅⋅=


2 MCx.2 MRx.2+ MRx.4+ MCx.4+ MRx.4+( ) MRx.6+ MCx.6+ MRx.6+ 102.768 kN m⋅⋅=

2 MCx2 M24+ M24+ MCx4+ M46+( ) M46+ MCx6+ M46+ 102.768 kN m⋅⋅=

So, after equilibration we have:

on y:

Section 1(1-2-1) Section 2(3-4-3) Section 1(5-6-5)

MCy1 4.618 kN m⋅⋅= MCy3 2.587 kN m⋅⋅= MCy5 3.649 kN m⋅⋅=

M12 6.91 kN m⋅⋅= M34 6.898 kN m⋅⋅= M56 5.836 kN m⋅⋅=


M12 6.91 kN m⋅⋅= M34 6.898 kN m⋅⋅= M56 5.836 kN m⋅⋅=


on x:

Section 1 (1-3-5-3-1) Section 2 (2-4-6-4-2)

MCx1 7.698 kN m⋅⋅= MCx2 8.872 kN m⋅⋅=

M13 9.23 kN m⋅⋅= M24 9.262 kN m⋅⋅=

MCx3 7.26 kN m⋅⋅= MCx4 5.364 kN m⋅⋅=

M35 8.245 kN m⋅⋅= M46 7.939 kN m⋅⋅=

MCx5 4.67 kN m⋅⋅= MCx6 5.492 kN m⋅⋅=

M35 8.245 kN m⋅⋅= M46 7.939 kN m⋅⋅=

MCx3 7.26 kN m⋅⋅= MCx4 5.364 kN m⋅⋅=

M13 9.23 kN m⋅⋅= M24 9.262 kN m⋅⋅=

MCx1 7.698 kN m⋅⋅= MCx2 8.872 kN m⋅⋅=

4. Reinforcements

4.1 Initital data

Choose C25/30 XC1 and S500 and on both direction we'll use:

ϕ 8mm:=

Resistances

fcd 16.67N

mm2

:=

fyd 435N

mm2

:=

Coverings :

For XC1:

cmin 15mm:=

cnom cmin 15 mm⋅=:=

Design height

dx t cnom−ϕ

2− 0.111m=:=

dy t cnom− ϕ−ϕ

2− 0.103m=:=

The reinforcement is computed on strips of 1m.

b 1m:=

The supporting resistance structure is considered of confined masonry having

the thickness 0.3m.The columns cross section have the dimension 0.3m x 0.3m.

4.2 Inferior reinforcement(in the field)

Note. For the greater moment we put the reinforcement down and for the lower

moment above the other one.Maximum distance between reinforcements is

0.3-0.4m.

All the central moments have values below 10kNm and they are close to each other.

So, we will find the reinforcement for maximum central moment on each

direction.The obtained reinforcement will be used in all panels.

( )

MCymax max MCy1 MCy2, MCy3, MCy4, MCy5, MCy6, ( ) 9.294 kN m⋅⋅=:=

MCxmax max MCx1 MCx2, MCx3, MCx4, MCx5, MCx6, ( ) 8.872 kN m⋅⋅=:=

µy

MCymax

b dy2

⋅ fcd⋅

0.053=:=

ωy 0.0545:=

Ay ωy b⋅ dy⋅fcd

fyd

⋅ 2.151 cm2

⋅=:=

Choose 5ϕ8 over 1m. So the distances between them will be 0.2m .

As.eff 2.51cm2

:=

µx

MCxmax

b dx2

⋅ fcd⋅

0.043=:=

ωx 0.0438:=

Ax ωx b⋅ dx⋅fcd

fyd

⋅ 1.863 cm2

⋅=:=


As.eff 2.51cm2

:=

4.3 Superior reinforcements(in the supports)

In the same manner,we will find the reinforcement for maximum support moment

on each direction

MCymax max M12 M34, M56, ( ) 6.91 kN m⋅⋅=:=

MCxmax max M13 M24, M35, M46, ( ) 9.262 kN m⋅⋅=:=

µy

MCymax

b dy2

⋅ fcd⋅

0.039=:=

ωy 0.0398:=

Ay ωy b⋅ dy⋅fcd

fyd

⋅ 1.571 cm2

⋅=:=

Because hf<30cm,maximum distance between reinforcements is 0.2m.


As.eff 2.51cm2

:=

µx

MCxmax

b dx2

⋅ fcd⋅

0.045=:=

ωx 0.0461:=

Ax ωx b⋅ dx⋅fcd

fyd

⋅ 1.961 cm2

⋅=:=


As.eff 2.51cm2

:=

Besides inferior and superior reinforcement we will have repartition reinforcement of ϕ6/25 on

both directions.

4.4 Anchorage length

for C25/30 we have:

fbd 2.7N

mm2

:=

lbrqd 0.25 ϕ⋅fyd

fbd

⋅ 0.322m=:=

α1 1:=

α2 1 0.15cnom ϕ−( )

ϕ− 0.869=:=

take

α2 0.7:=

α3 1:=

α4 0.7:=

α5 1:=

lbd α1 α2⋅ α3⋅ α4⋅ α5⋅ lbrqd⋅ 0.158m=:=

Take

lbd 16cm:=

4.5 Elevation lengths over supports

Elevation lenths is given by the smallest distance(on x or y) from one wall face to the next

divided with 5 .

The ends bend at t - 2 x cnom = 10cm

Panels 1,3,5

ln1 T1 0.15m− 0.15m− 4.1m=:=

ln1

50.82m=

Panels 2,6

ln2 L1 0.15m− 0.15m− 4.8m=:=

ln2

50.96m=

Panel 4

ln3 T2 0.15m− 0.15m− 5.6m=:=

ln3

51.12m=

slab reinforcement

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