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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 (hmin)
hmin =
𝑳
𝟏𝟖.𝟓
; 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 −
◊ Determine (Mn) : 𝐌𝐧 =
𝐚
𝟐
𝐌𝐮
𝟎.𝟗
◊ Find (As) : 𝐌𝐧 = 𝐀𝐬 ∗ 𝐟𝐲 (𝐝 −
◊ Determine (a) : 𝐚 =
𝐀𝐬∗𝐅𝐲
𝐚
𝟐
) = 0.9 d
)
𝟎.𝟖𝟓∗𝐟′𝐜∗𝐛
• a < hf … Rectangular section
• a > hf … T-section
◊ Find (As) in case and check of (ρ) :
𝐚
• 𝐌𝐧 = 𝐀𝐬 ∗ 𝐅𝐲 ∗ ( 𝐝 − )
• Actual ρ = (
𝐀𝐬
𝐛∗𝐝
𝟐
) > ρmin
◊ Check tension controlled section : 𝜺𝒕 = (
( 22 )
𝐝−𝐜
𝐜
) * 0.003 < 0.005
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
• Concrete above Hollow block = ( 24 * 0.08 ) = 1.92 KN\m2
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• Weight of hollow block =
• Weight of ribs =
5∗0.18
0.55
𝟎.𝟏𝟓+𝟎.𝟏𝟗
∗𝟐𝟒∗𝟎.𝟐𝟒
𝟐
𝟎.𝟓𝟓
= 1.64 KN\m2
= 1.78 KN\m2
• Weight of bleach = ( 22 * 0.02 ) = 0.44 KN\m2
∑ WD = 11 KN\m2
3.3.2.3 Live Load :
• For classes : LL = 3 KN\m2
• For corridors & Stairs : LL = 4 KN\m2
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 … ( As bar = 78.5 mm2 )
• Spacing =
∴ Use
𝟕𝟖.𝟓∗𝟏𝟎𝟎𝟎
𝟏𝟒𝟒
= 545 mm
Ø 10@550 mm in both direction.
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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 :
• 𝐀𝐬 =
=
∴
𝐌𝐮
𝟎.𝟗∗𝐟𝐲∗ 𝐣𝐝
; ( jd = 0.9 * d )
39.28∗10 6
420∗0.9∗0.9∗280
= 412.4 mm2
𝐔𝐬𝐞 𝟐∅𝟏𝟖 … (with As = 509 mm²)
• Check ρmin :
•𝛒=
509
280•150
= 0.012
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𝟏.𝟒
=
𝐟𝐲
• 𝛒𝐦𝐢𝐧 = 𝐦𝐚𝐱 𝐨𝐟
𝟎. 𝟐𝟓
𝐟𝐜′
𝐟𝐲
1.4
420
=
= 𝟎. 𝟎𝟎𝟑𝟑
0.25∗ 25
420
• 𝛒 = 0.012 > 𝛒𝐦𝐢𝐧 = 0.0033
∴
= 0.00298
That's OK
• Check tension controlled :
𝐀 𝐬 ∗𝐟𝐲
•𝐚=
𝟎.𝟖𝟓∗𝐟𝐜′ ∗𝐛𝐟
=
509∗420
0.85∗25∗550
= 18.3 mm
• 𝐚 < 𝐡𝐟 = 80 … rectangular action
•𝐜 =
𝐚
𝟎.𝟖𝟓
𝐝−𝐜
• 𝛆𝐭 =
𝐜
=
18.3
0.85
= 21.5 mm
∗ 𝟎. 𝟎𝟎𝟑 =
280−21.5
21.5
= 0.036 > 0.005
∗ 0.003 = 0.036
∴ Tension controlled
• Check section capacity ∶
• 𝐌𝐮 = 0.9 ∗ 509 ∗ 420 ∗ 280 −
18.3
2
= 52.11 KN. m > 39.28 𝑘𝑁. 𝑚
∗ 10−6
∴ 𝑇ℎ𝑎𝑡 ′ 𝑠 𝑂𝐾
3.4.2 For ( Negative Moment = - 41 KN.m ) :
• Find As :
• 𝐀𝐬 =
∴ 𝐔𝐬𝐞
𝐌𝐮
𝟎.𝟗∗𝐟𝐲∗ 𝐣𝐝
=
41∗10 6
420∗0.9∗0.9∗280
= 430.42 mm2
𝟐∅𝟏𝟖 … (with As = 509 mm²)
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• Check ρmin :
509
•𝛒=
280∗150
= 0.012
𝟏.𝟒
𝐟𝐲
• 𝛒𝐦𝐢𝐧 = 𝐦𝐚𝐱 𝐨𝐟
𝟎. 𝟐𝟓
=
𝐟𝐜′
𝐟𝐲
1.4
420
=
= 𝟎. 𝟎𝟎𝟑𝟑
0.25∗5
420
• 𝛒 = 0.012 > 𝛒𝐦𝐢𝐧 = 0.0033
= 0.00298
∴ That's
Ok
• Check tension controlled :
𝐀 𝐬 ∗𝐟𝐲
•𝐚=
𝟎.𝟖𝟓∗𝐟𝐜′ ∗𝐛𝐟
=
509∗420
0.85∗25∗550
= 18.3 mm
• 𝐚 < 𝐡𝐟 = 80 … rectangular action
•𝐜 =
𝐚
𝟎.𝟖𝟓
𝐝−𝐜
• 𝛆𝐭 =
𝐜
=
18.3
0.85
= 21.5 mm
∗ 𝟎. 𝟎𝟎𝟑 =
= 0.036 > 0.005
280−21.5
21.5
∗ 0.003 = 0.036
∴ Tension controlled
• Check section capacity ∶
• 𝐌𝐮 = 0.9 ∗ 509 ∗ 420 ∗ 270 −
18.3
= 52.11 KN. m > 41 KN. m
2
∗ 10−6
∴ That ′ s OK
3.4.3 Reinforcement :
• 𝐕𝐮 @𝐝 = 36.72 KN
• 𝐕𝐜 = 𝟎. 𝟏𝟕 ∗ 𝛌 ∗ 𝐟𝐜′ ∗ 𝐛𝐰 ∗ 𝐝
= 0.17 ∗ 1 ∗ 25 ∗ 150 ∗ 280 ∗ 10−3 = 35.7 KN
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• 𝐕𝐮 > 𝟎. 𝟓 ∗ ∅𝐕𝐜 … ??
• 39.49 KN > 0.5 ∗ 0.75 ∗ 35.7 = 13.4 KN
∴ That ′ s OK … We need shear reinforcement
• 𝐕𝐮 < ∅ 𝐕𝐜 … ??
• 39.49 > 26.8
∴ That ′ s OK … We need shear reinforcement
• 𝐕𝐬 =
𝐕𝐮 @𝐝
∅
36.72
− 𝐕𝐜 =
− 35.7 = 13.26 KN
0.75
• 𝐕𝐬 < 𝟎. 𝟑𝟑 ∗ 𝐟𝐜′ ∗ 𝐛𝐰 ∗ 𝐝 … ??
• 15.1 KN < 69.3 𝐾𝑁
∴ That ′ s OK . . We′ ll use 𝐀𝐯 𝐦𝐢𝐧
• Assume ∅10mm stirrups diameter :
• 𝐀𝐯 = 𝟐𝐀𝐛 = 157 mm2 …. 𝐅𝐲𝐯 = 280 KN
𝐝
• 𝐒𝐦𝐚𝐱 = 𝐦𝐢𝐧 𝐨𝐟
280
= 𝟏𝟒𝟎 𝐦𝐦
or
𝟔𝟎𝟎 𝐦𝐦
or
𝐀𝐯 ∗ 𝐅𝐲𝐯
= 945.4 mm
𝟐
𝟎.𝟎𝟔𝟐∗
=
2
𝐟𝐜′ (𝐛𝐰 )
𝐀𝐯 ∗𝐅𝐲𝐯
𝟎.𝟑𝟓∗ (𝐛𝐰
or
= 810.7 mm
)
∴ We Should Use ∅𝟏𝟎@𝟏𝟓𝟎 𝐦𝐦
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1) For Rib #1 :
Figure 3.4 : Design of Rib #1 according to PROKON
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2) For Rib #2 :
Figure 3.5 : Design of Rib #2 according to PROKON
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3) For Rib #3 :
Figure 3.6 : Design of Rib #3 according to PROKON
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4) For Rib #4 :
Figure 3.7 : Design of Rib #4 according to PROKON
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5) For Rib #5 :
Figure 3.8 : Design of Rib #5 according to PROKON
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6) For Rib #6 :
Figure 3.9 : Design of Rib #6 according to PROKON
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7) For Rib #7 :
Figure 3.10 : Design of Rib #7 according to PROKON
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