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3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 1 -)

Flat slab with drop panel For the shown floor plan given that:

- fcu= 300 kg/cm2 - fy= 4000 kg/cm2

- Live load = 800 kg/m2 -Covering = 150 kg/m2

- walls= 150 kg/m2

- Column dimensions (30*60)cm -no of stories =4

-story height = 3.0m

Required to: (a) Make complete design of the reinforced concrete flat

slab in both directions and draw reinforcement details?

(b) Check punching shear for interior column (C1) and exterior column (C2).

(C) Design the column (C1) and draw its details of reinforcement to scale 1 :10.

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 2 -)

3rd Civil year Reinforced Concrete - Notes No (13)

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Solution 1- slab theicknes (ts) 1 - - ts = flat slab with drop panel

= =20.84 cm take ts = 22.0cm -Dimension of drop panel:

-take td =ts/2= 22/2=11.0cm

for slab : l1=7.50 m & l2=6.0 m

-take ld= l2/2 = 6.0/2 =3.0m 2- laods (wsu): 2-

-Wdl = ts*rc +( )*rc ( )+ covering + walls Wdl = 0.22 * 2.50 +( ..... )*2.50 + 0.15 +0.15 = 0.90 t/m 2

-Wll = 800 kg/m2 = 0.80 t/m2 . (given)

- ultimate slab loads : > 0.75

Wsu = (1.40 wdl + 1.60 wll ) = 1.40 *0.90 +1.60 *0.80

= 2.54 t/m 2 .

3rd Civil year Reinforced Concrete - Notes No (13)

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3- Width of column strip & field strip: l1= 7.5m ( ) & l2= 6.0m ( ) -width of column strip (C.s) = =

. =3.0 m

-width of field strip (F.s) : -in(X) direction =l2- =6. 0 -

. =3.0 m

-in (y) direction =l1- =7.5 - . =4.5 m

3rd Civil year Reinforced Concrete - Notes No (13)

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4-total bending moment in (X & y) direction: -in(X) direction

Mx=

(lx- D)

2

D=b =0.30m for slab without column head Mx= .. (6.0 -

*0.30)

2 =80.10 t.m -in(y) direction

My=

(ly- D)

2 = .. (7.50 - *0.30)

2 =101.52 t.m 5-distribution of total moment between column and field strip:

)6-48: ( - for slab without marginal beam:

-in(X) direction : Mx=80.10 t.m

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 6 -)

-in(y) direction : My=101.52 t.m

-Mc=

* l where: lc : & l : Mc= .. * 6.0 = 30.48 t.m

: (Mc)c.s =0.75 *Mc = 0.75 *30.48 =22.86 t.m

(Mc)f.s =0.25 *Mc = 0.25 *30.48 =7.62 t.m

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 7 -)

6-Design critical sections in (X & y) direction: :

Section (1): Mu=50.76 t.m at support

: = +

t = ts+td=22+11=33cm

1- d = C1 . ....(1) & d = t - cover (2.0 cm) d = 33.0 - 2.0 = 31 cm

B (width of column strip)=300cm

1 ..... C1 31 = C1. ^ get C1 = 4.13

from table (shaker 401) using C1 = 4.13 get j = 0.808

2- As = = .^

. = 50.66 cm2 / strip (3.0 m) .

As / m = As /3.0=50.66 /3.0=16.89 cm2 / m

Take As = 7 18 / m .

)31( oN setoN - etercnoC decrofnieR raey liviC dr3

)- 8 -(oN egap

)2( noitceS- naps dim ta m.t 64.03=uM

= : : mc22=st = t

)1(.... . 1C = d -1 mc02= 0.2 - 22 = d &

mc003=)pirts nmuloc fo htdiw( B

1C ..... 1

44.3 = 1C teg ^ .1C = 02 777.0 = j teg 44.3 = 1C gnisu )104 rekahs( elbat morf

= = sA -2^.

mc 0.94 = . . )m 0.3( pirts / 2

m / 2mc 43.61=0.3/ 0.94=0.3/ sA = m / sA

. m / 81 7 = sA ekaT

: (mc33=t) - 1 - 2

(.mc22=t)

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 9 -)

1 : )x(: )Mu=40.05 t.m( Ex: Mu = 50.76 t.m use As= 16.89 cm2 / m

Mu = 40.05t.m need As=?? cm2 / m As=16.89 *40.05/50.76 =13.33 cm2/m =7 16 / m

2 : )x(: )Mu=24.03 t.m( Ex: Mu = 30.46 t.m use As= 16.34 cm2 / m

Mu = 24.03t.m need As=?? cm2 / m As=16.34 *24.03t /30.46 =12.89 cm2/m =7 16 / m

(y) Direction

(X) Direction

3rd Civil year Reinforced Concrete - Notes No (13)

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7-Draw reinforcement details in (X & y) direction:

3rd Civil year Reinforced Concrete - Notes No (13)

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3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 12 -)

2-Check punching shear stresses:

: : - )d/2( )d/2( .

Column (C1): interior column:

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 13 -)

Section (1-1)

-d=t-2.0=33.0-2.0=31.0cm

-lx=t+d=0.60+0.31=0.91m

-ly=b+d=0.30+0.31=0.61m

- = 1.15 internal column -Qu=wsu*(l1*l2 -lx*ly)=2.54(7.50*6.0 - 0.91*0.61)=112.89 t

-bo = 2(lx+ly)=2(0.91 +0.61)=3.04m =304.0cm

-qu = . =1.15 * . ^

.. =12.32 kg/cm2

- qcup= =.. =14.14 kg/cm2 qu < qcup Safe

-Section (2-2)

-d=t-2.0=22.0-2.0=20.0cm

-lx= ly = ld+d=3.0+0.20=3.20m

-Qu=wsu*(l1*l2 -lx*ly) =2.54(7.50*6.0 - 3.20*3.20)=88.29 t

-bo = 2(lx+ly) =2(3.20+3.20)=12.80m =1280 cm

-qu = . =1.15 * . ^

. =3.96 kg/cm2

qu < qcup Safe

3rd Civil year Reinforced Concrete - Notes No (13)

page No(- 14 -)

3- Design of column Column (C1): -No of story =4 - story height =3.0m -Column section (30*60)cm pu=(wsu*Area +Column weight) *No. of stories. -Column weight =1.40(b*t*h*rc)

=1.40 *0.30*0.60*3.0*2.5=1.89 t Internal column: pu =(2.54*7.0*6.0 +1.89) *4 =428 ton Moment transfer from slab to column (C1): -in(X) direction : (internal col.): Mc=0.50 *M-ve Col. strip in (x) direction Mc= 0.50 *40.05 * 0.50 ( )= 10.0 t.m -in(y) direction : (internal col.): Mc=0.50 *M-ve Col. strip in (y) direction =0.50 *50.76 * 0.50 ( )=12.69 t.m Design column on: pu=428 ton & Mx=12.69 t.m & My=10.0 t.m

: 1-Check column buckling. 2-get final moment. 3-Design section.

covere.pdf13-design of flat slab-slab with drop panel