CIVL311 CIVL911 2020 Week 6 Design of one way slab S4 PDF

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Week 6 lecture ...

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Examination Part 1 (Mid-session Examination) (Duration: 2 hours)

CIVL311: STRUCTURAL DESIGN 1 CIVL911: STRUCTURAL DESIGN BASED ON AUSTRALIAN STANDARDS AUTUMN 2020 (WEEK 6): DESIGN OF ONE-WAY SLAB

28 April 2020 (Week 7) TUESDAY

A/Prof. Neaz Sheikh Room 4.127

10:30- 12:30

Email: [email protected]

OPEN BOOK EXAMINATION Examination Books should be uploaded to the Moodle Submission box by 13:00 (half-hour is allowed for scanning and uploading) Consultation time: Tuesday 11:30 – 13:30 Friday 11:30 – 13:30 (Moodle Discussion Forum Only) 1

A suspended slab in Nowra with edge beams and one band beam through the middle. The slab is supported on six columns and four blade walls.

2

Slab reinforcement is N16‐250 each way top and bottom 8N24 at the top and bottom for edge beams Photos by Nick Idziak in 2011

Photos by Nick Idziak in 2011

Motivation for Week 6 Lecture

Important Notes These lecture notes do not substitute recommended textbooks.

 What is the difference between one-way slab and twoway slab?

These lecture notes are not peer-reviewed. They may contain errors.

 How does one-way slab behave under loading?

The lecturer is not responsible for the learning of students who rely exclusively on the lecture notes

 How can I design one-way slab? What is the deemed-to-comply span-depth ratio for deflection for one-way slab?

It is highly recommended that students review the lecture materials using the recommended textbooks.

What about detailing?

Lecture notes are based on Foster et al. (2010) 5

Motivation for Week 6 Lecture

Agenda for today

two-way General description and difference between one-way and two-way slab

Feedback on week 5 Tutorial Assessment Topics for Today: Slabs and floor system Behaviour of slab under load Simplified methods for slab analysis Design requirements for slabs

one-way slab Behaviour and design of one-way slab

Reinforcement details of one-way slab

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Week 6 Essential Reading

RECAP ON WEEK 5

Foster, Kilpatrick and Warner (2010) Chapter 4: Slabs and Floor Systems

Structural concrete columns

Loo and Chowdhury (2013)

Behaviour and load capacity of short column

Chapter 9: Slabs

Behaviour and load capacity of slender column

AS 3600 Section 6: Methods of Structural Analysis

Reinforcement details

Section 9: Design of Slabs for Strength and Serviceability

Design of structural walls 10

9

ASSIGNMENT FEEDBACK: WEEK 5

ASSIGNMENT FEEDBACK: WEEK 5 y

y

Bending takes place about the x-axis. Consider: f’c= 40 MPa; fsy=500 MPa; Es=200 GPa; concrete cover= 50 mm; transverse reinforcement (stirrup)= N10.

4N28

x

x

4N28 y

500 mm

d= 600-50-10-(28/2)= 526 mm

4N28 y

500 mm

74 mm

C. Determine Nub and Mub (Point C of N-M interaction diagram)

x

74 mm

N10 x

B. Determine Nu and Mu at decompression point(Point B of N-M interaction diagram)

B. Determine Nu and Mu at decompression point(Point B of N-M interaction diagram)

dsc= 50+10+(28/2)= 74 mm

12

600 mm

4N28

600 mm

A. Determine the squash load (Nuo) and position of the plastic centroid for the column section (500 mmx600 mm) shown in figure that contains 8-N28 bars.

ASSIGNMENT FEEDBACK: WEEK 5

Slabs and floor systems C. Determine Nub and Mub (Point C of N-M interaction diagram)

Slabs are thin, planar, horizontal, flexural member RC floor slabs may be supported on a rectangular grid of beams or walls, or directly on columns Prime stress resultants for design are bending Depending on the arrangement of supports, a slab may be subjected to one way bending: one-way slab One way Slab: Ly/Lx ≥ 2 two way bending: two-way slab Two way slab: Ly/Lx ≤ 2 13

14

By definition Ly>Lx

FLOOR SYSTEM CURRENTLY USED IN PRACTICE

two-way

flat plate one-way slab

One-way slab i) acts like a wide beam ii) significant bending moment only in the direction of the span iii) main reinforcement in the direction of the span iv) secondary reinforcement in the orthogonal direction to control cracking and transfer of load across the slab during local events 15 Foster et al. (2010)

flat slab

one-way slab

two-way slab

16 Foster et al. (2010)

BEHAVIOUR OF SLAB AND FLOOR SYSTEM UNDER LOAD Stress Resultant and Slab Behaviour

one-way joists

waffle slab Theoretical equation for moments and shears are based on assumption that the slab is a medium thick plate Membrane forces are ignored Shear deformation are ignored As an approximation span-to-slab thickness ratio from 10 to 40 may be regarded as medium-thick plates Because complementary shear stresses are equal and opposite

composite slab band-beam/slab 17

mxy= -myx

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FKW(2010)

Foster et al. (2010)

Behaviour of one-way slab under load Under increasing applied load, one-way slab behaviour is similar to beam Transverse strain induced due to longitudinal strain-poission’s effect Transverse expansion in the compressive fibres and transverse contraction in the tensile fibres- anticlastic bending Usually Slab cross-section is partially restrained from distortion and hence produces transverse stresses in the slab. Minor effect on the behaviour and hence ignored. Load curvature relationship has longer plateau- ductile behaviourredistribution of moments. AS 3600 allows an adjustment of negative moments by 30% if ductile reinforcement is used The applied load on the element is carried partly by bending in the beam strip in the X-direction and partly by bending in the beam strip in the Y-direction , and remaining by interaction between X and Y strip, i.e. by torsion

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FKW(2010)

The minimum reinforcement required in the transverse direction is usually adequate to accommodate any local bending resulting from usual effects at edge or from non-uniform loads

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Method of Analysis for One-way Slab

One way slab: beam of unit width

One way slab is less efficient than two-way slab Simple to construct and simple arrangement of reinforcement Advanced analyses not necessary for one-way slab - maybe useful for two-way slab 1m

AS 3600 Simplified method of flexural analysis: determine design moments in slab and floor system (AS3600 Clause 6.10) - not for buildings that carry horizontal loads -takes advantage of the slab capacity for redistribution to achieve simple numerical values -assumes no torsional moment

Beams

Design slab as a beam of unit width (1m)

Slab as continuous beam

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Simplified Method AS 3600: Restrictions

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Simplified Method AS 3600: Moment Coefficients

Ductile cross sections (N class steel) Uniform cross section Gravity loads only Not for horizontal loading Adjacent spans must be roughly equal

Negative design moment: two spans Fd = uniformly distributed design load i.e. w* kN/m for Ultimate limit state FdLn²/9

Ratio must not exceed 1.2

Uniformly distributed loads

take larger Ln

Ductility Class N

UDL Imposed load < 2 x Permanent action

AS3600 Clause 6.10.2

Reinforcement according to AS 3600 Clause 9.1.3.2 23

Ln1

Ln2

24

Simplified Method AS 3600: Moment Coefficients

Simplified Method AS 3600: Moment Coefficient

Negative design moment: More than two spans

Positive design moment: two or more spans

Fd = uniformly distributed design load i.e. w* kN/m for Ultimate limit state

Fd = uniformly distributed design load i.e. w* kN/m for Ultimate limit state

Ductility Class N May be non zero

FdLn²/10

Ln1

FdLn²/11

FdLn²/11

AS3600 Clause 6.10.2

Slab continues to right

Ln2

Ln1 FdLn1²/11

Note: at interior faces of exterior supports if built integrally with: beam:

Ln2 FdLn2²/16

Ln3 FdLn3²/16

Slab continues to right AS3600 Clause 6.10.2

FdLn2/24 26

25

Column: FdLn2/16

Have you really understood?

Simplified Method AS 3600: Shear Coefficient End Span

(i) At the face of interior support (ii) At mid-span (iii) At the face of end span

1.15FdLn/2 FdLn/7 FdLn/2

The slab is supported by 150 mm wide parallel walls with a clear distance between them of 5000 mm (see Figure). Calculate moments and shears in the slab segment. Consider Fd=10.2 kN/m

Interior Span

(i) At the face of supports (ii) At mid-span FdLn/2

FdLn/7

1.15x FdLn/2 FdLn/2

FdLn/2 FdLn/8

Mid-span positive moment MM*= + wLn2/11= 10.2x 5.02/11= 23.18 kN-m At the face of interior support MRS*= - wLn2/10 = 10.2x 5.02/10=-25.5 kN-m At the face of exterior support MLS*= - ? kN-m

FdLn/8

AS3600 Clause 6.10.2

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Shear at the face of interior support Shear At mid-span Shear at the face of end span

1.15FdLn/2= 1.15*10.2*5/2= 29.3 KN FdLn/7 = 10.2*5/7=7.3 kN FdLn/2 = 10.2*5/2= 25.5 kN 28

Strength Design

Design Requirements

Strength Design Vu V*

Flexural Shear

Mu= Tz=Cz

MuM*

T= Ast. fsy (Tensile steel at yield]

(=0.85)

Less reinforcement than beam

MuM*

Bending

Design slab as a rectangular beam, b=1000 mm

=Ast/bd; M*=Mu=0.85xMu

Mu=Astxfsyx0.925d

Mu= Ast.fsy.z= b.d.fsyx0.925xd

 (approx)= (2.54xM*)/d2

Serviceability Design

This  is per metre width of the slab

Deflection control

 (approx)= (2.54xM*)/d2

min=0.20 (Ds/d)2(f’ct.f/fsy)

Strength Design

Serviceability Design

Flexural shear is unlikely to be critical VucV* 𝑣

b=1000 mm fsy= 500 MPa

Minimum reinforcement for slab, not to fail upon cracking: 29

𝑣

= M*/ (0.85xbxfsyx0.925xd2)

M* is in KN-m and d in mm

Crack control

𝑢𝑐 = 𝑣

Mu= Ast.fsy.z

Conservative assumption z=0.925d

Deflection Control

𝑐

The design concrete strength does not exceed 65 MPa and the size of the maximum aggregate particle is not less than 10 mm, the value of kv and angle of the strut v may be determined by the simplified method (AS 3600 Clause 8.2.4.3)

Deflection do not exceed acceptable value under service load Condition AS 3600: Three tire approach •

Limiting span-to-depth ratio: deemed to comply requirements

•

Simplified method

•

Rigorous analysis: nonlinear behaviour + load-time history 32

Limiting Span-to-Depth Ratio Deemed to comply span-to-depth ratio (AS 3600 Clause 9.4.4)

Lef= effective span (mm)

  / Lef 1000 Ec  Lef  k3 k4   Fd .ef d  

Deemed to comply span-to-depth ratio

1/ 3

  / Lef 1000 Ec   k 3k 4   Fd .ef d  

1/ 3

Lef

= Lesser of L (centre-to-centre of supports) and (Ln+Ds)

tot= s + Kcs sus ws=FG+ sFQ wl=FG+lFQ

Ln= the clear span face-to-face supports d= effective depth K3= 1.0 for one-way slab K4=1.4 for a simply supported slab =1.75 for the end span of a continuous slab =2.1 for the interior span of a continuous slab /Lef= deflection ratio limit Ec=modulus of elasticity of concrete (in MPa)

Fd.ef= the effective design load (in kN/m) for either

L Ln Lef Lo

total deflection (1.0+kcs)FG+(s+kcsl)FQ incremental deflection kcsFG+(s+Kcsl)FQ s, l

short term and long term live load factor (AS/NZS 1170.0:2002; Table 4.1) When Asc= 0 Kcs= 2.0

Kcs= 2-1.2 (Asc/Ast)0.8 33

Live Load Factors for Serviceability Design (AS/NZS 1170.0) AS/NZS 1170.0 Table 4.1 Type of Load

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Simplified Method of Deflection Calculations AS 3600 Simple Formula for Deflection of a beam supporting uniformly distributed load (Foster et al. 2010)

Short-term factor, s Long-term factor, l



Floors Domestic Offices Parking area Retail store Storage

0.7 0.7 0.7 0.7 1.0

0.4 0.4 0.4 0.4 0.6

Roofs Trafficable Non-trafficable

0.7 0.7

0.4 0.0

L2 [ M L 10 M M  M R] 96E c I ef .av

Short-term Deflection Short term deflections are due to external loads and prestressing, which occur immediately on their application. Should be calculated based on appropriate values of Elastic Modulus (Ecj) and effective second moment of area (Ief) of the member

tot= s + Kcs sus

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ws=FG+sFQ wl=FG+lFQ

36

Simplified Method of Deflection Calculations AS 3600 Effective second moment of area: tension stiffening effect

Simple Formula for Deflection of a beam supporting uniformly distributed load (Foster et al. 2010)



[Section 8.5.3.1, AS 3600]

L2 [M L  10M M  M R ] 96E c I ef .av

At service load Icr...