Solved Examples on Flat Slab.pdf

Solved Examples on Flat Slab Example(1) The Given plan shows a general layout of flat slab . Using the empirical method of analysis , it is required to 1. Design and draw details of reinforcement for the flat slab. 2. Design the Col. And check punching for C1&C2&C3. 3. Choose the max. allowed dimension for the opening shows in plan and draw details of reinforcement. DATA:

F.C=1.5 kN/m 2 & No. of floor=4 Fcu=25 mpa

L.L= 6 kN/m 2 & Height of floor=4.50 m & steel used 360/520

.

Structural Eng. Dep. Ain Shams University

D. M. Nabil E. Reham Eltahawy

Reinforced Concrete II Third Civil Year Page 1 of 20

Solution 1- Design of flat slab 1. Concrete dimension 1. S lab: L.L≥ 6 kN/m 2

spacing≥6.0 m

use drop panel

(recommendation)

tsmin =15.0cm L e (6.5+70)/2 = = 18.75cm 36 36 L (6.5+7.5)/2 = I = =17.5 cm 40 40

=

take

t s = 20.0 cm=200 mm

2. Col. bmin = 30cm H 450 = =30 cm 15 15 L 750 = = =37.50 cm 20 20 =

take (400*400)cm 3. Drop panel t s 20 = = 10.0cm = 100mm 2 2 L 6.0 Ld = s = = 3.0m 2 2

td =




2. Loads D.L=F.c + o.w o.w=t save *25 10*3*3 =21.99mm 6.25*7.25 o.w=0.22*25=5.5 kN/m2 t save =20+

D.L=1.5+5.5=7.0 kN/m2 w u =D.L*1.4+L.L*1.6 w u =7*1.4+6*1.6=19.4 kN/m2




3. Calculation of B.M.D Long direction L1=7.50 m

L 2 =6.50 m

D=0.40 m

(w*L 2 ) *(L1 -2/3D)2 8 19.40*6.5 = *(7.5-(2/3)*0.4)2 8 =824.70 kN.m

M0 =

*actual width of field strip = 3.50 m

*theoretical width of field strip =6.5/2=3.25 m Correct the moment actual width theoretical width (L/2) 3.5 Mfieldcorrect=Mfieldcal* (B design =3.50m) 3.25

MFieldcorrect=Mfieldcal*




*then correct moment of the column strip (B design =3.0m) Mcolcorrect = Mcolcal + Mfieldcal - Mfieldcorrect

Short direction: McolcL1=6.50 m

L 2 =7.50 m

D=0.40 m

(w*L 2 ) *(L1 -2/3D)2 8 19.4*7.5 = *(6.5-(2/3)*0.4)2 8 = 706.70 kN.m

M0 =

Mcant.total =(w*L c 2 /2)*width of strip = (19.40*22 / 2) * 7.50 = 291.0kN.m Mcant.col. = 291* 0.75 = 218.25kN.m Mcant.field = 291* 0.25 = 72.75kN.m

Ss




Mfieldcorrect=Mfieldcal*

4.5 3.75

(B design = 4.50 m )

Mcol.correct = Mcol.cal + Mfieldcal - Mfieldcorrect




4. Design of sections 1. Long direction

Strip

C.S

F.S

Sec

1 2 3 4 5 6 7 8

D(mm) B(mm) Mult(kN.m) C1

280 180 280 180 180 180 180 180

3000 3000 3000 3000 3500 3500 3500 3500


323.7 234 361 196.7 88.8 176.4 133.2 133.2

4.26 3.22 4.03 3.51 5.72 4.01 4.61 4.61


J

0.81 0.76 0.80 0.78 0.826 0.80 0.821 0.821

As/strip(mm2

3954 4735 4456 3382 1622 3386 2503 2503

As/m (mm2 1318 1578 1485 1294 463 967 715 715

7 ࢶ16 8 ࢶ16 8 ࢶ16 7 ࢶ16 5 ࢶ12 9 ࢶ12 7 ࢶ12 7 ࢶ12


2. Short direction

strip

C.S

F.S

Sec D(mm)

B(mm) Mult(kN.m) C1

J

1 2 3 4 5 6 7 8

3000 3000 3000 3000 4000 4000 4000 4000

0.82 07 0.81 0.79 0826 0.81 0.824 0.824

270 170 270 170 170 170 170 170


266.4 183.4 296.9 155.8 87.3 170 126.5 126.5

4.53 3.44 4.29 3.73 5.75 4.12 4.78 4.78


As/strip(mm2 As/m’( mm2) 3340 1113 3855 1285 3757 1252 3211 1070 1727 432 3438 860 2506 626 2506 626

7 ࢶ16 7 ࢶ16 7 ࢶ16 6 ࢶ16 5 ࢶ12 8 ࢶ12 6 ࢶ12 6 ࢶ16


5. Details of reinforcement:




6. Design of column & check punching 1. Design of column A) Interior column (c 1) Assume thicknessof roof slab = 20.0 cm , L.L = 2kN / m2 , F.C = 2kN / m2 wuroof = (0.2 * 25 + 2) * 1.4 + 2 * 1.6 = 13.0kN / m2 − sec(1) : Case(1) : N = wuroof * L1 * L 2 + o.w o.w = 0.4 * 0.4 * 4.5 * 25 = 18kN Nu = 13 * 7.5 * 6.6 + 18 = 651.75kN M = 0.00( fromslab ) Calculation of M buckling K = 1.5 δ=

,

λ=

4.5 * 11.5 = 16.9 0.4

λ 2 * b (16.9)2 * 0.4 = = 0.057m 2000 2000

Madd (due tobuckling) = NU * δ = 651.75 * 0.057 = 37m.kN M = 0.057m N e 0.057 = = 0.1425 → (useint eraction diagram ) t 0.4 e=

µ(fromcurves ) = 0.5% µmin = 0.025 + 0.052 * λmax = 1.13 As =

1.13 * 40 * 40 = 18.08 cm 100 ( use8Φ18 )




Case (2): N = (D.L+0.5*L.L)*Area+o.w N= (9..8+0.5*3.2)*7.5*6.5 + 18 N = 573.75 kN M= 50%

M-VE of col. strip *ratio of stiffness

M = 0.5* 361.5*0.5 = 90.4 kN.m Madd (due to buckling ) = Nu *δ = 573.75*0.057 =32.70 m.kN M 123.1 = = 0.215 N 573.75 e 0.215 = = 0.538 (useint eraction diagram ) t 0.4 of uniform distributionof steelfor the sec tion

e=

NU = 0.14 Fcu * b * t Mu = 0.077 Fcu * b * t 2

∴ρ = 4 As = ρ * Fcu * 10 −4 * b * t As = 4 * 25 * 10 −4 * 400 * 400 = 1600mm2

Case (3): N= 573.75 kN M = 50% M− veof col.strip * ratio of stiffness M = 0.5 * 296.9 * 0.5 M = 74.23 m.kN * * * * * * fromcase 1,2,3 ∴ As = 1808 mm2 As = 8Φ18




Sec (2) Case (1): Nu = 651.75+19.40*7.5*6.5 =1597.5 kN M = 0.0 (from slab) +M add = 1597.5*0.057 = 91 m.kN e f 0.05 t ∴ use interaction diagram

ρ=6

Astotal =2400 mm2

Case (2): N = 573.75 +(7*1.4+05*6*1.6 )= 1285.5kN M= 90.4 (from slab) + Madd = 90.4 + 1285.5*0.057 =164.0 kN.m ∴ from interaction curves As total =3200 mm2

(critical)

(use 8Φ 25)

Sec (3) Case (1): N = 573.75 +(7*1.4+05*6*1. Nu = 1597.5 + 19.4*7.5*6.5 +18 = 2561.25 kN M = zero (from slab) + Madd calculation of M add. k = 1.5 4.5 * 1.5 = 13.5 0.5 λ 2 * b (13.5)2 * 0.5 δ= = = 0.046m 2000 2000 Madd (due tobuckling ) = Nu * δ = 118m.kN λ=

∴(Design using int eraction diagrams ) ∴ As total =2500mm2




Case (2) N = 1285.55 +(7*1.4+0.5*6*1.6)*7.5*6.5 Nu = 1997.25kN M =0.5*361.5*(

IL / HL ) IU / HU + IL / HL

504 ) = 128.20m.kN 504 + 404 M = 128.2m.kN (from slab ) + Madd . M =0.5*361.5*(

= 128.2 + 1997 * 0.046 = 220kN.m ∴designof int eractiondiagrams Astotal = 3125mm2

(critical )

∴ (use 8Φ 25 )

B) Edge Column (c2) Sec (4) N = 2*19.4*7/2*(3+3.250+13*7/2*(3+3.25) N u = 1133.1 kN M= 90% M-ve of col. strip *ratio of stiffness M = 0.9*268.4*(

50 4 ) =171.4 m.kN 50 4 + 40 4

M = 171.4 (from slab) + Madd =171.4 + 1133.1*0.046 = 223.5 m.kN ∴designusin gint eractiondiagrams Astotal =1875mm2 Asmin = µmin *b*t =

0.952 *500*5001=2380mm2 100

(use 8Φ 22)




Corner column (c3)

C)

Sec(4) Nu = 2*19.4*7/2*6/2+13*7/2*6/2 N u = 543.90 kN M= 90% M-veof half col. strip *ratio of stiffness 50 4 ) =103.4 m.kN 50 4 + 40 4 50 4 M2 = 0.9 * 0.5 * 238.4 * ( 4 ) = 85.8m.kN 50 + 40 4 Madd (due tobuckling) = Nu * δ = 0.046 * 543.9 = 25m.kN

M1 = 0.9*0.5*323.7*(

* * * *it willbe de signedas sec tion subjected to double(M) Nu = 0.09 b*t*fcu Mu = 0.06 b*t*fcu ∴ ρ=3 ∴µ = ρ * fcu * 10 −4 = 0.75% µmin = 0.25 + 0.052 λmax = 0..25 + 0.052 * 13.5

As total

= 0.952% 0.952 * 500 * 500 = 2380m m2 = 100




2. check punching A) Interior column Case (1)

QPunching =Ws *(L1 *L 2 -(c1+d)(c 2 +d)) Qp = 19.4*(6.25*7.25-(0.4+0.27)(0.4+0.27)) = 870 kN

qpunching =

Qpunching

Area Area= 2d*((c1+d)+(c 2 +d)) Area=2*0.27*((0.4+0.27)+(0.4+0.27))=0.724 m2 870*1000 =0.72 m2 0.724*106 qcu p = the least of qpunching =

qcu p =0.8(

f α*d +0.2)* cu γc b0

N/mm2

f a qcu p = 0.3116(0.5+ )* cu b γc qcu p ≤ 0.316

where

fcu γc

N/mm2

N/mm2

α=4

int erior column

=3

edgecolumn

=2

corner column

b0 =perimeter of criticalsectionof punching a,b= the short & thelong dimensionof column section b0 = 4*0.87 =3.48m

→

qcu p =0.8(

f 4*270 +0.2)* cu 3480 γc

qcu p = 0.316 (0.5 +1) * qcu p = ∴ Case (2)

fcu γc

=0.408 * =0.474* = 0.316*

qcu p = 0.316*

fcu γc fcu γc fcu γc

fcu = 1.29 N/mm2 γc

qpunching p qcu p ( safe )




QPunching =(D.L+0.5 L.L)*(L1.L 2 -(c1+d)(c 2 +d)) = (9.8+3.2*0.5)*(6.25*7.2-(0.4+0.27)(0.4+0.27)) = 511 kN Q qpunching = punching *1.15 Area 511*1000*1.15 qpunching = = 0.812 N/mm2 6 0.724*10 qcu p = 0.316 *

fcu = 1.26 N / mm2 γc

qpunching p qcup ∴ (safe)

B) Edge column (c2) Qp = 19.4*(3.5*6.25-(0.4+0.27/2)(0.4+0.27)) = 416.9 kN Area=d(c1 + d) + 2d(c 2 + d / 2) = 0.47m2 qpunching =

Qpunching

* 1.3 Area 416.9*1000 qpunching = *1.3=1.153 N/mm2 6 0.47*10 → b0 = 2(c 2 + d / 20 + (c1 + d) = 1.74m qcu p = the least of qcu p =0.8(

f 3*270 +0.2)* cu γc 1740

qcu p = 0.316(0.5+1)*

fcu γc

N/mm2 = 0.532 N/mm2

qcu p = qcu p ≤ 0.316 qpunching

fcu γc p

fcu γc

= 0.474

fcu γc

= 0.316

fcu γc

= 1.29 N/mm2 qcup ( safe )




C) corner column (c3) Qpunching = w s *(L1.L 2 /4-(c1+d)(c 2 +d)) = 19.4*(3.0*3.5-(0.535*0.535) = 198 kN Area=d((c1 + d / 2) + (c 2 + d / 2)) = = 0.27 * (0.535 + 0.535) = 0.29m2 qpunching =

Qpunching

* 1.5 Area 198*1000 qpunching = *1.5=1.024 N/mm2 0.29*10 6 → b0 = (c 2 + d / 2) + (c1 + d / 2) = 1.10m

qcu p =0.8(

f 2*270 +0.2)* cu 1100 γc

qcu p = 0.316(0.5+1)*

fcu γc

N/mm2 = 0.553 N/mm2

qcu p = qcu p = 0.316 qpunching

fcu γc p

fcu γc

= 0.474

fcu γc

= 0.316

fcu γc

= 1.29 N/mm2 qcup ( safe )




7. Opening & Details of RFT Max. Allowed dimension : This opening is in the intersection of 2 field stirps So we can choose max. dimension 0.4*L

Opening Dimension is

(3.0 *2.6 ) m

X direction

7Φ12 / m'* 2.60 = 18.2 Φ 12 = 18.2 * 1.12 = 2038mm2 = 9Φ18 Y direction

5Φ12 / m'* 3.0 = 15 Φ 12 = 15 * 1.12 = 1680mm2 = 9Φ16










Solved Examples on Flat Slab.pdf

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