• Author / Uploaded
• Suhail Ahamed

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Design of Flat Slabs

Dr. Ayman Hussein Hosny Professor of RC Structures Ain Shams University

Lecture Layout | |

Types Concrete dimensions z Slab thickness z Drop thickness z Columns z Column heads z Beams (if any)

|

Analysis z z z

| | |

Empirical analysis Frame Analysis Finite element

Check p punching g Examples Details of RFT

Types of Flat Slabs |

Regular flat slabs

Slab

column l

Ordinary flat slabs

Flat plate slab

Flat slab with Drop Parts

Slab

drop column

Flat slab with Drop Parts

Failures of Flat Slab

Punching failure

Punching failure

Flat slab with Column Heads

Flat slab

Column head or column capital

column

Flat slab with Column Heads

Flat Slab with Drop Part and C l Column h head d

Flat slab

REAL BUILDING

Concrete Dimensions Minimum thickness (ts)= 150 mm | Thickness depends on L |

z |

L= larger of L1 or L2

Without drop z ts

≥ Lext/32 z ts ≥ Lint/36 |

With drop z ts

≥ Lext/36 z ts ≥ Lint/40

Example 1: |

Without drops z

ts is bigger of • 150 mm • 6000/32=187.5 mm • 7000/36 7000/36= 194.4 mm

z

Take ts = 200 mm

Example 2: |

With drops z

ts is bigger of • 150 mm • 6000/36=166.6 mm • 7000/40= 7000/40 175 mm

z |

Take ts = 180 mm

Note slight effect of drop pp panel here

When to use drop? If ts is large | If heavy live load (> 5-8 kN/m2) | If architect permits | In general, drop parts are not common in residential buildings | They y are common in factories and buildings g with false ceilings |

Drop panels |

Drop panels below slab l b

|

Additionall panels Additi l above slab

Drop Part Dimensions |

Thickness (td) z td≥ts/4 z Common, Common td=ts/2

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Drop panel width (X) z z z

|

X ≥Lmin/3 X≤ Lmin/2 Common X= X Lmin i /2

Example z z

td=80mm X=2750 mm

Notes

Columns |

Bmin=bigger of z z z

300 mm h/15 Spacing/20

Column Head Mostly used to safeguard against p punching g | αmax= 45 degrees | Dmax=L Lmin/4 |

What if α>45 degrees?

Marginal Beams |

Might be used z z z

Stiffen slab edge Carry walls Resist lateral loads

Analysis of Flat Slabs Empirical method 3rd civil | Frame analysis 3rd civil | Finite element (computer) | Yield line analysis |

Empirical method conditions |

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Columns lay on straight lines (or maximum eccentricity=10%) t i it 10%) Number of spans at least three in each direction Diff Difference b between t ttwo adjacent dj t spans within ithi 10% Maximum difference between largest and smallest span in the same direction within ithin 20% Inner spans larger or equal outer spans M i Maximum span/minimum / i i span≤1.30 ≤1 30 Constant slab thickness, ts Li lload< Live d 2* Dead D d lload d

Example |

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Columns lay on straight lines (or maximum eccentricity=10%) eccentricity 10%) Number of spans at least three in each direction Difference between two adjacent spans within 10% Maximum difference between largest and smallest span in the same direction within 20% Inner spans larger or equal outer spans Maximum span/minimum span≤1.30 Constant slab thickness, ts Live load< 2* Dead load

Analysis Procedure |

Obtain loads, Ws z

Consider wall loads

Check punching around columns | Get straining actions in long and short di ti directions |

z

Divide into strips

Design of sections | Details of reinforcement |

SOLUTION |

Divide slab into column and field strips in both directions

In case of drop panels

Calculation in Long Direction In long direction, L1 | Mo=WsL2(L1-2/3D)2/8 | Mo is the moment in a span length L1 having a width of L2 | Distribute Mo between column and field strips as per the following p g table |

Design Compute moment per meter | Choose rft (same regulations as solid slabs) |

5-10 bars/meter z Minimum Mi i di diameter=10 t 10 mm z Preferable to have constant number of bars per meter z

Calculation in Short Direction In short direction, L2 | Mo=WsL1(L2-2/3D)2/8 | Mo is the moment in a span length L2 having a width of L1 | Distribute Mo between column and field strips as per same table | Note that larger moment results in longerspan direction |

Notes Table ratios are based on column strip width equals field strip width | In case field strip width is larger larger, increase field strip moment by the following ratio, R | R=real R real width of field strip/(half C C.L. L to C C.L.) L) | Reduce column strip moments accordingly |

Example

F.S.=4.25 m C.S.=2.75 m

Details of RFT

Columns Resist normal force and bending moments |

Internal columns resist 50% of column strip p negative moment z

Moment is distributed b t between lower l and d upper columns

|

Exterior columns resist 90% of column strip p negative moment z

Moment is distributed b t between lower l and d upper columns

Effect of transferred moment on slab |

Moment transferred to column partially by flexure and partially ti ll b by ttorsion i

M flexure = γ f M M torison = γ q M

γf =

|

1 2 b1 1+ 3 b2

Put As to resist Mflexure in column width+1 width+1.5 5 ts per side

Punching

q=

Qup bo d

Qup = β Q bo= 2(A+B) β = 1.15 bo=2A+B β = 1.30

bo=A+B

β = 1 . 50

Punching check

q ≤ q cup q cup = 0.80[ q cup

αd b0

+ 0.20]

f cu

γc

α = 4......inner.column α = 3......exterior.column α = 2......corner.column

a f cu = 0.316[0.50 + ] b γc

q cup = 0.316

f cu

γc

q cup = 1.6 N / mm 2

The least shall be considered

EXAMPLE Data Walls=3 kN/m2 Live load=3 load 3 kN/m2 Flooring=2 kN/m2 Fcu=30MPa 30MP

Example ws = t s γ c + FC + Walls + LL ws = 0.24 x 25 + 2 + 3 + 3 = 14kN / m 2 ws u = 14 x1.5 = 21kN / m

2

d = 210mm Q = 21x[3.5 x3.5] = 257 kN b0 = 2 x 455 = 910mm 1.50 x 257000 q= = 2.03 N / mm 2 910 x 210

EXAMPLE

q > q cup USE COLUMN HEAD

qcupp = 0.80[

αd b0

+ 0.20]

f cu

γc

qcup

2 x210 30 = 0.80[ + 0.20] = 2.37 N / mm2 910 1.5

qcup

a f cu = 0.316[0.50 + ] b γc

qcup

350 30 = 0.316[0.50 + ] = 2.11N / mm2 350 1.5

qcup = 0.316 qcup qcup

f cu

γc

30 = 0.316 = 1.41N / mm2 1.5 = 1.6 N / mm2

EXAMPLE q cupp = 0.80[ q cup = 0.80[

αd b0

+ 0.20]

f cu

γc

1.50 x 257000 q= = 1.14 N / mm 2 1610 x 210 1610x

2 x 210 30 + 0.20] = 1.64 N / mm 2 1610 1.5

q cup

a f cu = 0.316[0.50 + ] b γc

q cup

805 30 = 0.316[0.50 + ] = 2.11N / mm 2 805 1.5

q cup = 0.3316 6

f cu

γc

q cup = 0.316

q < q cup

q cup

SAFE& OK

30 = 1.41N / mm 2 1 .5 = 1.6 N / mm 2

Column Head Rft

OPENINGS

FRAME ANALYSIS

FRAME ANALYSIS |

Consider frames from columns and slabs C.L. to C.L.

FRAME ANALYSIS

FRAME ANALYSIS

FRAME ANALYSIS

Quiz |

Flat slab LL=10 kN/m2 z Draw to scale 1:100 concrete dimensions for the roof z