ribbed slab design

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3.1 Introduction :

• A concrete slab is a common structural element of modern buildings,

horizontal slabs of steel reinforced concrete, typically between 10 and 50 centimeters thick, are most often used to construct floors and ceilings, while thinner slabs are also used for exterior paving.

• It is essentially a two-dimensional reinforced or non-reinforced cement concrete structural element of the modern building design concept that serves the purpose of a floor and/or a ceiling and/or a landing base. It is termed a two-dimensional structural element as it spans across the length and width (or trigonometric components of horizontal and/or vertical planes thereof like that in case of inclined roofs, stair waist slabs, etc.). • The most common examples of slabs being the floors, roofs, ramps, concrete staircases, etc. The slabs may or may not have composite

beam(s) network associated with them, which provide inherent support and rigidity to the slabs per se.

• Structural slabs are often supported by beams and are named one or two way slabs depending on their shape, or they may be supported by

columns and these are termed flat slabs

• Slabs also may be supported directly by the natural ground surface, or by a prepared and compacted sub base and sub grade. Highway pavements, airport runways, are common examples.

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3.2 Types of one-way slab :

3.2.1 One-way Beam and slab, One-way flat slab :

• These slabs are supported on two opposite sides and all bending moment and deflections are resisted in the short direction. A slab supported on Four sides with length to width ratio greater than two, should be designed as one-way slab.

3.2.2 One-way joist floor system :

• This type of slab, also called ribbed slab, is supported by reinforced concrete ribs or joists .The ribs are usually tapered and uniformly spaced and supported on girders that rest on columns.

Figure 3.1 : Typical types of slabs

3.2.3 One way Ribbed Slabs :

• Ribbed slabs are widely used in many countries such as Jordan. This is attributed to the rapid shattering, ease of construction, and the reduction in the time of erection. This type of slabs or flooring system consists of series of small closed spaced reinforced concrete T-beams. These floors are suitable for building with light live loads.

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3.2.3.1 One-way Ribbed Slab :

• In one-way ribbed slab, loads are transferred in one direction, and the main reinforcement is distributed in the same direction of the load. With accurate to temperature and shrinkage, minimum of Φ33 bars diameter will be used in both direction and crossing each other over the blocks ( practically ).

Figure 3.2 : one way ribbed slab

3.2.4 Solid and voided slabs :

• It consists of a thin topping slab supported by closely spaced small beams called ribs. The space between successive ribs is made by temporary

removable forms, or by introducing permanent fillers made of light weight hollow blocks of standard dimensions.

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3.3 Design of one–way ribbed slab :

Determination of the minimum thickness of slab for deflection control, according to table9-5(a) from the ACI318_M-2008 code for the case of “one end continuous one way ribbed slab “the minimum thickness is:

Table 3.1 : Min. Thickness (h

_min

)

h

_min

=

_{𝟏𝟖.𝟓}𝑳

;

L = span length of beam or one-way slab ; clear projection of cantilever, ( in mm ).

Figure 3.2 :

Local Example

for one way

ribbed slab

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3.3.1 Design steps :

◊ Assume section dimensions ◊ determine the load for rib : • DL ( rib ) = total DL * 0.55 • LL ( rib) = total LL * 0.55 ◊ From PROKON 2.4 we get ( Mu ) * If ( Mu ) is Negative :

◊ Find ( ρ ) in the case : 𝛒 = 𝟎.𝟖𝟓∗𝐟′𝐜

𝐅𝐲 (𝟏 − 𝟏 − 𝟐𝐑𝐧 𝟎.𝟖𝟓∗𝐟𝐜 ) ◊ Find As : 𝐀𝐬 = 𝛒 ∗ 𝐛 ∗ 𝐝 ◊ Check of ρmin ; 𝛒 𝐦𝐢𝐧 = 𝟏.𝟒 _𝐅𝐲 * If ( Mu ) is Positive :

◊ Check (a) … • Assume ( 𝐝 − 𝐚 _𝟐 ) = 0.9 d ◊ Determine (Mn): 𝐌𝐧 = 𝐌𝐮_𝟎.𝟗 ◊ Find (As) : 𝐌𝐧 = 𝐀𝐬 ∗ 𝐟𝐲 (𝐝 − 𝐚_𝟐 ) ◊ Determine (a) : 𝐚 = 𝐀𝐬∗𝐅𝐲 𝟎.𝟖𝟓∗𝐟′𝐜∗𝐛 • a < hf … Rectangular section • a > hf … T-section

◊ Find (As) in case and check of (ρ) : • 𝐌𝐧 = 𝐀𝐬 ∗ 𝐅𝐲 ∗ ( 𝐝 − 𝐚_𝟐 )

• Actual ρ = ( _𝐛∗𝐝𝐀𝐬 ) > ρmin

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3.3.2 Select slab thickness :

3.3.2.1 Thickness of one way ribbed slab :

Hollow Block = 24 cm • Use thickness of slab

32 cm

Slab Concrete = 8 cm

Figure 3.3 : Dead Load Calculations

3.3.2.2 Dead load calculations :

• Tile = ( 22 * 0.025 ) = 0.55 KN\m2 • Mortar = ( 22 * 0.025 ) = 0.55 KN\m2 • Fill = ( 17 * 0.1 ) = 1.7 KN\m2

• Partitions = 2.38 KN\m2

• Asphalt = ( 14 * 0.01 ) = 0.14 KN\m2

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• Weight of hollow block = 5∗0.18

0.55 = 1.64 KN\m2 • Weight of ribs = 𝟎.𝟏𝟓+𝟎.𝟏𝟗 𝟐 ∗𝟐𝟒∗𝟎.𝟐𝟒 𝟎.𝟓𝟓 = 1.78 KN\m2 • Weight of bleach = ( 22 * 0.02 ) = 0.44 KN\m2

∑ W

_D

= 11 KN\m

2

3.3.2.3 Live Load :

• For classes :

L

_L

= 3 KN\m

2

• For corridors & Stairs :

L

_L

= 4 KN\m

2

3.3.2.4 Combination loads :

• WD = 11* 0.55 = 6 KN\rib • WLc = 3 * 0.55=1.65 KN\rib • Wu = 1.2 * 6 +1.6 * 1.65 = 9.84 KN\m • Fy =420 Mpa … fc’ = 25 Mpa • bw for ribs = 15 cm

• Effective depth for slab = 320 – 40 = 280 mm

3.3.2.5 Shrinkage or thermal reinforcement ( secondary steel ) :

• As min = ρmin * b * t = 0.0018 * 1000 * 80 = 144 mm2

∴ Use Ø10 … ( Asbar = 78.5 mm2 ) • Spacing = 𝟕𝟖.𝟓∗𝟏𝟎𝟎𝟎

𝟏𝟒𝟒 = 545 mm

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3.4 Design of All Ribs :

• A Designed example for ribbed slab ( 1st _{floor ) :} • Given : ◊ f'c = 25 Mpa ◊ fy = 420 Mpa ◊ h =320 mm ◊ d =280 mm ◊ bw = 150 mm ◊ bf = 550 mm

• According to PROKON analysis : * For rib #1 :

3.4.1 For ( Positive Moment = 39.28 KN.m ) :

• Find As : • 𝐀_𝐬 = 𝐌𝐮 𝟎.𝟗∗𝐟_𝐲∗𝐣𝐝 ; ( jd = 0.9 * d ) = 39.28∗106 420∗0.9∗0.9∗280 = 412.4 mm 2 ∴ 𝐔𝐬𝐞 𝟐∅𝟏𝟖 … (with As = 509 mm²) • Check ρmin : • 𝛒 = 509 280•150 = 0.012

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• 𝛒_𝐦𝐢𝐧 = 𝐦𝐚𝐱 𝐨𝐟 𝟏.𝟒 𝐟_𝐲 = 1.4 420 = 𝟎. 𝟎𝟎𝟑𝟑 𝟎. 𝟐𝟓 𝐟𝐜 ′ 𝐟_𝐲 = 0.25∗ 25 420 = 0.00298 • 𝛒 = 0.012 > 𝛒_𝐦𝐢𝐧 = 0.0033 ∴ That's OK

• Check tension controlled : • 𝐚 = 𝐀𝐬∗𝐟𝐲 𝟎.𝟖𝟓∗𝐟_𝐜′∗𝐛_𝐟 = 509∗420 0.85∗25∗550 = 18.3 mm • 𝐚 < 𝐡_𝐟 = 80 … rectangular action • 𝐜 = 𝐚 𝟎.𝟖𝟓 = 18.3 0.85 = 21.5 mm • 𝛆_𝐭 = 𝐝−𝐜_𝐜 ∗ 𝟎. 𝟎𝟎𝟑 = 280−21.5_21.5 ∗ 0.003 = 0.036 = 0.036 > 0.005 ∴ Tension controlled • Check section capacity ∶

• 𝐌_𝐮 = 0.9 ∗ 509 ∗ 420 ∗ 280 −18.3₂ ∗ 10−6

= 52.11 KN. m > 39.28 𝑘𝑁. 𝑚 ∴ 𝑇ℎ𝑎𝑡′_{𝑠 𝑂𝐾}

3.4.2 For ( Negative Moment = - 41 KN.m ) :

• Find As : • 𝐀_𝐬 = 𝐌𝐮 𝟎.𝟗∗𝐟_𝐲∗𝐣𝐝 = 41∗106 420∗0.9∗0.9∗280 = 430.42 mm 2 ∴ 𝐔𝐬𝐞 𝟐∅𝟏𝟖 … (with As = 509 mm²)

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• Check ρmin : • 𝛒 = 509 280∗150 = 0.012 • 𝛒_𝐦𝐢𝐧 = 𝐦𝐚𝐱 𝐨𝐟 𝟏.𝟒 𝐟_𝐲 = 1.4 420 = 𝟎. 𝟎𝟎𝟑𝟑 𝟎. 𝟐𝟓 𝐟𝐜 ′ 𝐟_𝐲 = 0.25∗5 420 = 0.00298 • 𝛒 = 0.012 > 𝛒_𝐦𝐢𝐧 = 0.0033 ∴ That's Ok

• Check tension controlled : • 𝐚 = 𝐀𝐬∗𝐟𝐲 𝟎.𝟖𝟓∗𝐟_𝐜′∗𝐛_𝐟 = 509∗420 0.85∗25∗550 = 18.3 mm • 𝐚 < 𝐡_𝐟 = 80 … rectangular action • 𝐜 = 𝐚 𝟎.𝟖𝟓 = 18.3 0.85 = 21.5 mm • 𝛆_𝐭 = 𝐝−𝐜 𝐜 ∗ 𝟎. 𝟎𝟎𝟑 = 280−21.5 21.5 ∗ 0.003 = 0.036 = 0.036 > 0.005 ∴ Tension controlled • Check section capacity ∶

• 𝐌_𝐮 = 0.9 ∗ 509 ∗ 420 ∗ 270 −18.3 2 ∗ 10 −6 = 52.11 KN. m > 41 KN. m ∴ That′s OK 3.4.3 Reinforcement : • 𝐕_𝐮@𝐝 = 36.72 KN • 𝐕_𝐜 = 𝟎. 𝟏𝟕 ∗ 𝛌 ∗ 𝐟_𝐜′ _{∗ 𝐛} 𝐰 ∗ 𝐝 = 0.17 ∗ 1 ∗ 25 ∗ 150 ∗ 280 ∗ 10−3 = 35.7 KN

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• 𝐕_𝐮 > 𝟎. 𝟓 ∗ ∅𝐕_𝐜 … ??

• 39.49 KN > 0.5 ∗ 0.75 ∗ 35.7 = 13.4 KN

∴ That′_{s OK … We need shear reinforcement}

• 𝐕_𝐮 < ∅ 𝐕_𝐜 … ?? • 39.49 > 26.8

∴ That′s OK … We need shear reinforcement • 𝐕_𝐬 = 𝐕𝐮@𝐝 ∅ − 𝐕𝐜 = 36.72 0.75 − 35.7 = 13.26 KN • 𝐕_𝐬 < 𝟎. 𝟑𝟑 ∗ 𝐟𝐜′ ∗ 𝐛𝐰 ∗ 𝐝 … ?? • 15.1 KN < 69.3 𝐾𝑁 ∴ That′s OK . . We′ll use 𝐀_{𝐯 𝐦𝐢𝐧}

• Assume ∅10mm stirrups diameter : • 𝐀_𝐯 = 𝟐𝐀_𝐛 = 157 mm2 …. 𝐅_𝐲𝐯 = 280 KN • 𝐒_𝐦𝐚𝐱 = 𝐦𝐢𝐧 𝐨𝐟 𝐝 𝟐 = 280 2 = 𝟏𝟒𝟎 𝐦𝐦 or 𝟔𝟎𝟎 𝐦𝐦 or 𝐀_𝐯∗ 𝐅_𝐲𝐯 𝟎.𝟎𝟔𝟐∗ 𝐟_𝐜′ (𝐛_𝐰) = 945.4 mm or 𝐀_𝐯∗𝐅_𝐲𝐯 𝟎.𝟑𝟓∗ (𝐛_𝐰) = 810.7 mm ∴ We Should Use ∅𝟏𝟎@𝟏𝟓𝟎 𝐦𝐦

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1) For Rib #1 :

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2) For Rib #2 :

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3) For Rib #3 :

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4) For Rib #4 :

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5) For Rib #5 :

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6) For Rib #6 :

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7) For Rib #7 :