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Definition Of Slab Definition Of One-Way Slab Deflection Design Concept Minimum Thickness Of One-Way Slab Load Assigned To Slabs Temperature And Shrinkage Reinforcement Temperature And Shrinkage Reinforcement Ratio Design Procedure
PRESENTATION OUTLINE
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Layer of reinforce-concrete with uniform/variable thickness supported by restraints
Figure:
DEFINITION OF SLAB
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The ratio of the longer to the shorter side (L/B) of the slab is at least equal to 2.0
Support conditions is an important impact. Deflection is another important factorLoad is distributed to the direction perpendicular of the
supporting restraints.
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DEFINITION OF ONE WAY SLAB
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PICTURE OF ONE WAY SLAB
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PICTURE OF ONE WAY SLAB
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PICTURE OF ONE WAY SLAB
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DEFLECTION
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DEFLECTION
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For the purpose of analysis and design, a unit strip of a slab is cut out at right to the supported beams.
The method here is used for rectangular beams. Figure
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DESIGN CONCEPT
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To control deflection, ACI Code 9.5.2.1 specifies minimum thickness values for one-way solid slabs, shown in Table.
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MINIMUM THICKNESS OF SLAB
Element Simply Supported
One-End Continuous
Both-End Continuous
Cantilever
One-way Slab
L/20 L/24 L/28 L/10
L is the span length direction of bending.If Wc ≈ 90 – 120 psf factor will be applied by ( 1.65-0.005Wc)If Fy ≠ 60 ksi factor will be ( 0.4 + Fy/100000)
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LOAD ASSIGNED OF SLAB
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LOAD ASSIGNED OF SLAB
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LOAD ASSIGNED OF SLAB
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Concrete gets shrinks and results crackingCracks known as Hairline cracks.provide special reinforcement for shrinkage and
temperature contraction in the direction perpendicular to the main reinforcement.
Maximum lateral spacing 5h or 8″
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TEMPERATURE AND SHRINKAGE
REINFORCEMENT
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Based on the Gross Concrete AreaSlabs with 40 or 50 grade reinforcement-0.002Slabs with 60 grade reinforcement-0.0018Slabs where reinforcement with yield stress exceeding 60 KSI
at 35% yield-(0.0018*60000)/Fy
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TEMPERATURE AND SHRINKAGE
REINFORCEMENT RATIO
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Step-1: From given panel find out the Tmin.
Step-2: Find out the Factored Load.(Wu = 1.2*DL + 1.6*LL)
Step-3: Find out Moments of the panel.
(M = Moment Coefficients*Wu*L²)
Step-4: Find out maximum steel ratio
(ρmax = 0.85*β1*(f´c/fy)*(εu/ εu + 0.004)
Step-5: Find dreq. ( )
Step-6: As and a. (
)
Step-7: Temperature and shrinkage reinforcement
Step-7: Reinforcement Detailing.
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DESIGN PROCEDURE
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REINFORCEMENT DETAILING
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REINFORCEMENT DETAILING
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THANK YOU
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