twenty three continuous beams no beams concrete construction: - - PowerPoint PPT Presentation

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F2009abn

twenty three

concrete construction:

T-beams & slabs

Concrete Slabs 1 Lecture 23 Architectural Structures ARCH 331

lecture ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN

ARCH 331

• DR. ANNE NICHOLS

SPRING 2019

Concrete Slabs 2 Lecture 23 Foundations Structures ARCH 331 F2008abn

Systems

• beams separate from slab
• beams integral with slab

– close spaced

• continuous beams
• no beams

Concrete Slabs 3 Lecture 23 Foundations Structures ARCH 331 F2008abn

T sections

• two areas of compression in moment

possible

• one-way joists
• effective flange width

F2008abn Concrete Slabs 4 Lecture 23 Foundations Structures ARCH 331

T sections

• negative bending: min As, larger of:
• effective width (interior)

– L/4 – bw + 16t – center-to- center of beams

Cf Cw T

) ( 6 d b f f A

w y c s

  ) ( 3 d b f f A

f y c s

 

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F2008abn Concrete Slabs 5 Lecture 23 Foundations Structures ARCH 331

T sections

• usual analysis steps
• 1. assume no compression in web
• 2. design like a

rectangular beam

• 3. needs reinforcement

in slab too

• 4. also analyze for negative

moment, if any

a/2 T a=1c 0.85f’c C

Concrete Slabs 6 Lecture 23 Foundations Structures ARCH 331 F2008abn

One-Way

• Joists

– standard stems – 2.5” to 4.5” slab – ~30” widths – reusable forms

Concrete Slabs 7 Lecture 23 Foundations Structures ARCH 331 F2008abn

One-Way

• Joists

– wide pans – 5’, 6’ up – light loads & long spans – one-leg stirrups

Concrete Slabs 8 Lecture 23 Foundations Structures ARCH 331 F2008abn

Compression Reinforcement

• doubly reinforced
• negative bending
• two compression forces
• bigger Mn
• control deflection
• increase ductility
• needs ties because
• f buckling

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F2008abn Concrete Slabs 9 Lecture 23 Foundations Structures ARCH 331

Compression Reinforcement

• analysis

– As & As – T = Cc + Cs – T = Asfy – Cs = As (f s - 0.85f c) – Cc = 0.85f cba with – fs not known, so solve for c (n.a.) – fs < fy ? – Mn = T(d-a/2)+Cs(d-d )

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a c  

Concrete Slabs 10 Lecture 23 Foundations Structures ARCH 331 F2008abn

Slabs

• one way behavior – like beams
• two way behavior – more complex

Concrete Slabs 11 Lecture 23 Foundations Structures ARCH 331 F2008abn

Slab Design

• one unit wide “strip”
• with uniform loads

– like “wide” beams – moment / unit width – uniform curvature

• with point loads

– resisted by stiffness

• f adjacent strips

– more curvature in middle

F2008abn Concrete Slabs 12 Lecture 23 Foundations Structures ARCH 331

Slab Design

• min thickness by code
• reinforcement

– bars, welded wire mesh – cover – minimum by steel grade

• 40-50:
• 60:

0018 .   bt As  002 .   bt As 

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Concrete Slabs 13 Lecture 23 Foundations Structures ARCH 331 F2008abn

One-Way Slabs

• As tables
• max spacing

–  3(t) and 18” –  5(t) and 18” – temp & shrinkage steel

• no room for stirrups

Concrete Slabs 14 Lecture 23 Foundations Structures ARCH 331 F2008abn

Precast

• prestressed

– PCI Design Handbook – double T’s – hollow core – L’s

• topping
• load tables