hollow core

PCIMANUAL FOR THE DESIGN

OFHOLLOW CORE SLABS

SECOND EDITION

by

Donald R. Buettner and Roger J. BeckerComputerized Structural Design, S.C.

Prepared for thePCI Hollow Core Slab Producers

Committee

John E. Saccoman, Chairperson

James Beerbower Ernest MarkleKevin Boyle James MarkleJeffrey Butler Milo J. NimmerLoris Collavino William C. Richardson, Jr.Edward J. Gregory Klaus RosensternPat Hynes Wes SchrootenPaul Kourajian Larry Stigler

PRECAST / PRESTRESSED CONCRETE INSTITUTE175 WEST JACKSON BOULEVARDCHICAGO, ILLINOIS 60604

312 786--0300 FAX 312 786--0353

Copyright 1998

By Precast/Prestressed Concrete Institute

First edition, 1985

Second edition, 1998

All rights reserved. This book or any part thereof may not be

reproduced in any form without the written permission of the

Precast/Prestressed Concrete Institute.

ISBN 0--937040--57--6

Printed in U.S.A.

INTRODUCTION

Purpose of Manual

The application and design of precast, prestressed hollow core slabs is similar to that of other pre-stressed members. However, there are situations which are unique to hollow core slabs either be-cause of the way the slabs are produced or because of the application of the slabs.

For special situations, hollow core producers have developed design criteria and conducted in-house testing to verify that their approaches are valid. In fact, there is consistency between the manytypes of hollow core slabs available. The purpose of this manual is to bring together those things thatare common, that are verified by test and that can be universally applied to hollow core slabs. Be-cause there are differences, some topics covered will also point to the differences where closer coor-dination with the local producer is required.

This manual was prepared by Computerized Structural Design, S.C., Milwaukee, Wisconsin withinput and direction from the PCI Hollow Core Slab Producers Committee. Additionally, the fire andacoustical sections were prepared by Armand Gustaferro of The Consulting Engineers Group, Inc.,Mt. Prospect, Illinois and Allen H. Shiner of Shiner and Associates, Inc., Skokie, Illinois, respective-ly. All reasonable care has been used to verify the accuracy of material contained in this manual.However, the manual should be used only by those experienced in structural design and should notreplace good structural engineering judgment.

Scope of Manual

This document is intended to cover the primary design requirements for hollow core floor androof systems. In instances where the design is no different than for other prestressed members, thePCI Design Handbook and the ACIBuilding Code should be consulted formore in-depth discussion.

For the architect or consulting engineer, this manual is intended as a guideline for working withhollow core slabs, a guide for the use and application of hollow core slabs and an indication of someof the limitations of hollow core slabs. For the plant engineer, the manual will hopefully presentsome backup and reference material for dealing with everyday design problems.

TABLE OF CONTENTS

Introduction

Notation

Chapter 1 -- Hollow Core Slab Systems1.1 Methods of Manufacturing 1--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.2 Materials 1--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.3 Advantages of Hollow Core Slabs 1--3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.4 Framing Concepts 1--3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.5 Wall Panel Applications 1--4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.6 Design Responsibilities 1--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.7 Cross-Sections and Load Tables 1--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.8 Tolerances 1--6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 2 -- Design of Hollow Core Slabs2.1 General Information 2--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2 Flexural Design 2--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2.1 ACI Requirements 2--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.2 Stresses at Transfer 2--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.3 Prestress Losses 2--3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.4 Service Load Stresses 2--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.5 Design Flexural Strength 2--6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3 Shear Design 2--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.3.1 ACI Requirements 2--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.4 Camber and Deflection 2--11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.4.1 Camber 2--12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.4.2 Deflections 2--14. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.5 Composite Design 2--15. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.6 Strand Development 2--19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.6.1 ACI Requirements 2--19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 3 -- Special Design Considerations3.1 General Information 3--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.2 Load Distribution 3--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.2.1 Load Distribution Mechanisms 3--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.2.2 Design Guidelines 3--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.3 Effect of Openings 3--8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.4 Continuity 3--10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.5 Cantilevers 3--10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.6 Horizontal Joints 3--12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 4 -- Diaphragm Action with Hollow Core Slabs4.1 General Information 4--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.2 Design Loads 4--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.3 Distribution of Lateral Forces 4--3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.4 Structural Integrity 4--4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.5 Elements of a Diaphragm 4--4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.6 Diaphragm Strength 4--6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.6.1 Longitudinal Joints 4--6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.6.2 Transverse Joints 4--8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.7 Collectors 4--8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.8 Topped vs. Untopped Diaphragms 4--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.9 Design Example 4--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 5 -- Connections in Hollow Core Slabs5.1 General 5--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.2 Details 5--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.3 Typical Details with Concrete Beams 5--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.4 Typical Details with Walls 5--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.5 Typical Details with Steel Beams 5--15. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.6 Typical Cantilever Details 5--20. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.7 Miscellaneous 5--23. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 6 -- Fire Resistance of Assemblies made with Hollow Core Slabs6.1 Introduction 6--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6.2 Heat Transmission through Floors or Roofs 6--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.2.1 Equivalent Thickness 6--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6.2.2 Toppings, Undercoatings, or Roof Insulation 6--2. . . . . . . . . . . . . . . . . . . . . . . . . .6.2.3 Ceilings 6--4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.3 Structural Fire Endurance of Floor or Roof Assemblies 6--4. . . . . . . . . . . . . . . . . . . . . . . .6.3.1 Simply Supported Slabs 6--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6.3.2 Effect of Spray Applied Coatings 6--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6.3.3 Structurally Continuous Slabs 6--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6.3.4 Detailing Precautions 6--11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.4 Restraint to Thermal Expansion 6--12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 7 -- Acoustical Properties of Hollow Core Slabs7.1 Glossary 7--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.2 General 7--1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.3 Approaching the Design Process 7--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7.3.1 Dealing with Sound Levels 7--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.4 Sound Transmission Loss 7--2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.5 Impact Noise Reduction 7--3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.6 Absorption of Sound 7--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.7 Acceptable Noise Criteria 7--5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.8 Establishment of Noise Insulation Objectives 7--8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.9 Leaks and Flanking 7--8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.10 Human Response to Building Vibrations 7--9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7.11 Vibration Isolation for Mechanical Equipment 7--10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 8 -- Guide Specification for Precast, Prestressed Hollow Core Slabs 8--1. . . . . . . . . .

References

Index

NOTATION

A = Cross-sectional areaa = Depth of equivalent compression stress

blockaθ = Depth of equivalent compression stress

block under fire conditionsAcr = Area of crack faceAe = Net effective slab bearing areaAps = Area of prestressed reinforcementAvf = Area of shear friction reinforcementb = Width of compression facebw = Net web width of hollow core slabC = Confinement factorC = Compressive forceC = Seismic factor dependent on site and

structure fundamental periodC = Factor for calculating steel relaxation

losses as given in Table 2.2.3.2c = Distance from extreme compression

fiber to neutral axisCR = Prestress loss due to concrete creepCs = Seismic coefficientD = Dead loadd = Distance from extreme compression fiber

to centroid of non-prestressedtension reinforcement

db = Nominal diameter of reinforcementdp = Distance from extreme compression fiber

to centroid of prestressedreinforcement

DW = Distribution widthe = Distance from neutral axis to centroid of

prestressed reinforcementEc = Modulus of elasticity of concreteEci = Modulus of elasticity of concrete at the

time of initial prestressES = Prestress loss due to elastic shortening of

concreteEs = Modulus of elasticity of steel

reinforcementf′c = Specified design compressive strength of

concretef′ci = Compressive strength of concrete at the

time of initial prestressfcir = Net compressive stress in concrete at

centroid of prestressed reinforcement attime of initial prestress

fcds = Stress in concrete at centroid ofprestressed reinforcement due tosuperimposed dead load

fd = Stress at extreme tension fiber due tounfactored member self weight

Fi = Portion of base shear applied at level ifpc = Compressive stress in concrete at the

centroid of the section due to effectiveprestress for non-composite sections ordue to effective prestress and momentsresisted by the precast section alone forcomposite sections

fpe = Compressive stress in concrete at extremefiber where external loads cause tensiondue to the effective prestress only

fps = Stress in prestressed reinforcement atnominal strength

fpsθ = Stress in prestressed reinforcement at firestrength

f′ps = Maximum steel stress in partiallydeveloped strand

fpu = Specified tensile strength ofprestressing steel

fpuθ = Tensile strength of prestressing steel atelevated temperatures

Fpx = Force applied to diaphragm at level underconsideration

fse = Effective stress in prestressing steel afterall losses

fsi = Stress in prestressing steel at initialprestress

Ft = Additional portionof base shear applied attop level

fu = Usable grout strength in a horizontal jointfy = Steel yield strengthh = Overall member depthhn = Net height of grout in keyway between

slab unitsI = Occupancy importance factorI = Cross-sectional moment of inertiaJ = Factor for calculating steel relaxation

losses as given in Table 2.2.3.1k = Fraction of total load in a horizontal joint

in a grout columnKcir = Factor for calculating elastic shortening

prestress lossesKcr = Factor for calculating prestress losses due

to concrete creepKes = Factor for calculating prestress losses due

to elastic shorteningKre = Factor for calculating prestress losses due

to steel relaxation as given in Table 2.2.3.1

Ksh = Factor for calculating prestress losses dueto concrete shrinkage

K′u = Factor from PCI Handbook Fig. 4.12.2 forcalculating flexural design strength

L = Live loadℓ = Span lengthℓd = Reinforcement development lengthℓe = Strand embedment length from member

end to point of maximum stressℓf = Flexural bond lengthℓt = Strand transfer lengthM = Service load momentMcr = Cracking momentMd = Unfactored dead load momentMg = Unfactored self-weight momentMn = Nominal flexural strengthMnθ = Flexural strength under fire conditionsMmax = Maximum factored moment due to

externally applied loads= Mu -- Md

Msd = Unfactored moment due tosuperimposed dead load

Mu = Factored design momentMθ = Applied fire momentP = Effective force in prestressing steel after

all lossesPo = Effective prestress force at release prior to

long term lossesPi = Initial prestress force after seating lossesQ = First moment of areaR = Fire endurance ratingRE = Prestress loss due to steel relaxationRe = Reduction factor for load eccentricity in

horizontal jointsRH = Ambient relative humidityRw = Seismic coefficient dependent on

structural system typeS = Section modulusSH = Prestress loss due to concrete shrinkageT = Tensile forcetg = Width of grout column in horizontal jointV = Seismic base shearVc = Nominal shear strength of concreteVci = Nominal shear strength of concrete in a

shear-flexure failure modeVcw = Nominal shear strength of concrete in a

web shear failure modeVd = Shear due to unfactored self weightVh = Horizontal beam shear

Vi = Factored shear force due to externallyapplied loads occurring simultaneouslywith Mmax

= Vu -- VdVn = Nominal shear strength of a memberVs = Nominal shear strength provided by shear

reinforcementVu = Design shear forceV/S = Volume to surface ratiow = Uniformly distributed loadw = Bearing area lengthW = Total dead load plus other applicable

loads for seismic designwi = Portion of W at level iwpx = Portion of W at level under

considerationyb = Distance from neutral axis to extreme

bottom fiberyt = Used as either distance to top fiber or

tension fiber from neutral axisZ = Seismic zone factorβ1 = Factor defined in ACI 318-95, Section

10.2.7.3γp = Factor for type of prestressing strandδall = Limiting free end slipδs = Actual free end slipεps = Strain in prestressed reinforcement at

nominal flexural strengthεs = Strain in prestressed reinforcementεse = Strain in prestressed reinforcement after

lossesµ = Shear friction coefficientµe = Effective shear friction coefficientρp = Ratio of prestressed reinforcementρ′ = Ratio of compression reinforcementφ = ACI strength reduction factorω = ρfy/f′cω′ = ρ′fy/f′cωp = ρpfps/f′cωw = Reinforcement index for flanged sectionsω′w = Reinforcement index for flanged sectionsωpw = Reinforcement index for flanged sectionsωpu = ρp fpu/f′cθ = Subscript denoting fire conditions

CHAPTER 1

1--1

HOLLOW CORE SLAB SYSTEMS

1.1 Methods of ManufacturingA hollow core slab is a precast, prestressed con-

crete member with continuous voids provided toreduce weight and, therefore, cost and, as a sidebenefit, to use for concealed electrical or mechan-ical runs. Primarily used as floor or roof deck sys-tems, hollow core slabs also have applications aswall panels, spandrel members and bridge deckunits.

An understanding of the methods used tomanufacture hollow core slabs will aid in the spe-cial considerations sometimes required in the useof hollow core slabs. Hollow core slabs are castusing various methods in the seven major systemsavailable today. Because each production systemis patented, producers are usually set up on a fran-chise or license basis using the background,knowledge and expertise provided with the ma-chine development. Each producer then has thetechnical support of a large network of associatedproducers.

Two basic manufacturing methods are current-ly in use for the production of hollow core slabs.One is a dry cast or extrusion system where a verylow slump concrete is forced through the ma-chine. The cores are formed with augers or tubeswith the concrete being compacted around thecores. The second system uses a higher slumpconcrete. Sides are formed either with stationary,fixed forms or with forms attached to the machinewith the sides being slip formed. The cores in thenormal slump, or wet cast, systems are formedwith either lightweight aggregate fed throughtubes attached to the casting machine, pneumatictubes anchored in a fixed form or long tubes at-tached to the casting machine which slip form thecores.

Table 1.1 lists the seven major hollow core sys-tems available today along with the basic in-formation on the casting technique. Variousnames may be used by local licensees to describethe same products. In most cases, the slabs arecast on long line beds, normally 300 ft to 600 ft

long. Slabs are then sawcut to the appropriatelength for the intended project.

The economy of the generalized hollow coresystem is in the quantity of slabs that can be pro-duced at a given time with a minimum of labor re-quired. Each slab on a given casting line will havethe same number of prestressing strands. There-fore, the greatest production efficiency is obtainedby mixing slabs with the same reinforcing re-quirements from several projects on a single pro-duction line. This implies that best efficiency for asingle project is obtained if slab requirements arerepetitive.1.2 Materials

As stated previously, hollow core slabs are pro-duced with two basic concrete mixes; low slumpand normal slump concrete. For the low slumpconcretes, water content is limited to slightlymore than that required for cement hydration.Water-cement ratios are typically about 0.3. Mix-ing is critical because the limited water availablemust be well dispersed in the mix. Water reducingadmixtures can be used to optimize a mix by re-ducing cement and water requirements while stillretaining adequate workability for proper com-paction of the concrete by the machine. Air en-trainment admixtures are not effective in the drymix concrete. With the low water-cement ratiosand compaction placing method, air is difficult todisperse well and maintain.

Table 1.1 Hollow Core Systems

Manufac-turer

MachineType

ConcreteType/Slump

Core Form

Dy-Core Extruder Dry/Low Tubes

Dynaspan Slip Form Wet/Normal Tubes

Elematic Extruder Dry/Low Auger/Tube

Flexicore Fixed Form Wet/Normal PneumaticTubes

Spancrete Slip Form Dry/Low Tubes

SpanDeck Slip Form Wet/Normal Filleraggregate

Ultra-Span Extruder Dry/Low Augers

1--2

Latex feathering ready for direct carpet application

Acoustical spray on exposed slab ceiling

Electrical and HVAC application

The wet cast products (those cast with normalslump concrete), have water-cement ratios in therange of 0.4 to 0.45. Depending on the slip form-ing system used, slumps of 2 to 5 inches (50 - 130mm) are used. The mix design and use of admix-tures is dependent on achieving a mix that willhold its shape consistent with the forming tech-nique used.

Aggregates vary in the various manufacturingprocesses depending on what type is locally avail-able. Maximum aggregate size larger than peagravel is rarely used because of the confined areasinto which concrete must be placed. Light weightaggregates are occasionally used to reduce theweight of the sections and to achieve a significantreduction in required equivalent thickness in a firerated application. Concrete unit weights rangingfrom 110 to 150 pcf (1760 - 2400 kg/m3) are usedin the industry.

Strand use in hollow core slabs includes aboutevery size and type of strand produced dependingon what is available to a particular producer. Thetrend is toward primary use of the larger 1/2 in (13mm) diameter, low relaxation strand. The philos-ophy of strand use varies from using many strandsizes to optimize cost for a given project to usingonly one or two strand sizes for simplicity of in-ventory and production.

Except for special situations, keyway grout isnormally a sand and Portland cement mixture inproportions of about 3:1. The amount of waterused is a function of the method used to place thegrout but will generally result in a wet mix so key-ways may be easily filled. Shrinkage cracks mayoccur in the keyways, but configuration of the keyis such that vertical load transfer can still occurwith the presence of a shrinkage crack. Rarely isgrout strength required in excess of 2000 psi (13.8MPa) for vertical load transfer.

Although it is discouraged, non-shrink, non-staining grout is occasionally specified for use inkeyways. In evaluating the potential benefits ofnon-shrink grout, the volume of grout must becompared to the overall volume of concrete in theslabs and support materials. Because the size ofthe keyway is small in relation to a floor or roof as-sembly of slabs, total shrinkage will be affectedonly to a minor degree. Shrinkage cracks can still

1--3

occur in the keyways and there is little benefit tobe gained in comparison with the additional cost.

1.3 Advantages of Hollow Core SlabsHollow core slabs are most widely known for

providing economical, efficient floor and roofsystems. The top surface can be prepared for theinstallation of a floor covering by feathering thejoints with a latex cement, installing non-structur-al fill concretes ranging from 1/2 in to 2 in (13 - 51mm) thick depending on the material used, or bycasting a composite structural concrete topping.The underside can be used as a finished ceiling asinstalled, by painting, or by applying an acousticalspray.

When properly coordinated for alignment, thevoids in a hollow core slab may be used for electri-cal or mechanical runs. For example, routing of alighting circuit through the cores can allow fix-tures in an exposed slab ceiling without unsightlysurface mounted conduit. Slabs used as the heatedmass in a passive solar application can be detailedto distribute the heated air through the cores.

Structurally, a hollow core slab provides the ef-ficiency of a prestressed member for load capac-ity, span range, and deflection control. In addi-tion, a basic diaphragm is provided for resistinglateral loads by the grouted slab assembly pro-vided proper connections and details exist. A de-tailed discussion of diaphragm capabilities ispresented in Chapter 4.

Excellent fire resistance is another attribute ofthe hollow core slab. Depending on thickness andstrand cover, ratings up to a 4 hour endurance canbe achieved. A fire rating is dependent on equiva-lent thickness for heat transmission, concrete cov-er over the prestressing strands for strength in ahigh temperature condition, and end restraint.Underwriters Laboratories publishes fire ratingsfor various assemblies. However, many buildingcodes allow a rational design procedure forstrength in a fire. This procedure, described in de-tail in Chapter 6, considers strand temperature incalculating strength. Required fire ratings shouldbe clearly specified in the contract documents.Also, the fire rating should be considered in deter-mining the slab thickness to be used in prelimi-nary design.

Used as floor-ceiling assemblies, hollow coreslabs have the excellent sound transmission char-

acteristics associated with concrete. The SoundTransmission Class rating ranges from about 47 to57 without topping and the Impact InsulationClass rating starts at about 23 for a plain slab andmay be increased to over 70 with the addition ofcarpeting and padding. Detailed information onthe acoustical properties of hollow core slabs ispresented in Chapter 7.

1.4 Framing ConceptsThe primary consideration in developing a

framing scheme using hollow core slabs is thespan length. For a given loading and fire endur-ance rating, span length and slab thickness may beoptimized by consulting a producer’s publishedload tables. Section 1.7 presents sample loadtables and instructions for the use of the tables.The PCI Design Handbook1 recommends limitson span-depth ratios for the hollow core slabs. Forroof slabs, a span-depth ratio limit of 50 is sug-gested and for floor slabs, a limit of 40 is sug-gested. In practice, a span-depth ratio of 45 iscommon for floors and roofs when fire endurance,openings, or heavy or sustained live loads do notcontrol a design.

Consideration must be given to factors whichaffect slab thickness selection for a given span.Heavy superimposed loads, as required by thefunction of a system, would require a lower span-depth ratio. Similarly, heavy partitions or a largenumber of openings will result in higher load ca-pacity requirements. The fire resistance rating re-quired for the application will also affect the loadcapacity of a slab. As the code required fire ratingincreases, prestressing strands can be raised formore protection from the heat. The smaller effec-tive strand depth will result in a lower load capac-ity. Alternatively, a rational design procedure canbe used to consider the elevated strand tempera-tures during a fire. This fire design condition maycontrol a slab design and, again, result in a lowerload capacity.

Once slab thicknesses and spans are selected,the economics of layout become important.While ends cut at an angle can be designed andsupplied, it is most efficient to have the bearingperpendicular to the span so square cut ends canbe used.

It is also desirable to have the plan dimensionsfit the slab module. This is dependent upon the

1--4

slab systems available in the project area.Non-module plan dimensions can be accommo-dated using partial width slabs. Some producersintentionally cast narrow widths as filler pieceswhile others use a section split from a full slab.Such a split section might be created by a longitu-dinal sawcut or a break if the edge will not be ex-posed to view.

Construction tolerances must be accounted forin developing a plan layout. Tolerance on slablength may be taken up by allowing a gap at theslab ends in the bearing detail. On thenon-bearingsides, clearance may be provided by using a detailwhere the slabs lap over a wall or beam. If the slabedge butts a wall or beam, a gap should be pro-vided. Refer to local producers’ information forrecommendations of proper tolerances.

When a hollow core slab deck is exposed toweather for a long period of time during construc-tion, water can accumulate in the cores. The pri-mary source of water infiltration is at the buttjoints. In cold weather, this water can freeze andexpand causing localized damage. One remedyfor this situation is to drill weep holes at the slabends under each core. The need for such weepholes is generally known only after a constructionschedule is established. The specifier and the slabsupplier are not usually in a position to know ofsuch a need in advance.

Hollow core members will be cambered as withany other prestressed flexural member. In theplanning stages, consideration should be given tothe causes of differential camber. For two slabs ofidentical length and prestressing, the camber maybe different because of concrete and curing varia-tions. This factor is independent of a framingscheme. However, joints between slabs of un-equal spans or joints at which a change in the spandirection occurs, will cause a potential differentialcamber problem. This must be recognized anddealt with in the design layout. Wall locationsmay hide such a joint, but the door swing might bedirected to the least variable side.

Camber must also be accommodated when atopping is to be provided. The quantity of toppingrequired must consider the amount of camber andthe function of the floor. In occupancies whereflat floors are not a requirement, a constant top-ping thickness may be used to follow the curva-

ture of the slabs. At the other extreme, if a “flat”floor is required in a structure consisting of multi-ple bays of varying length and change in slabdirection, the highest point will determine the topelevation of the topping. A greater amount of top-ping will then be required in “low” areas. Theseconsiderations must be dealt with in the planningstages to both control costs and minimize ques-tions and potential for “extras” during construc-tion.

Camber, camber growth, and deflections mustbe considered when slabs run parallel to a stiff ver-tical element such as a wall (e.g. slabs runningparallel to the front wall of an elevator). The doorrough opening should allow for camber to pro-duce proper door installation. Alternatively, theslab span might be rearranged so the front wall is abearing wall. Then door problems would be alle-viated.

Camber, camber growth, and deflections mustbe taken into account in roofing details. Wherechanges in relative slab position can occur, coun-terflashings are suggested to accommodate suchchanges.

1.5 Wall Panel ApplicationsSome hollow core slab systems can also pro-

vide slabs to be used as walls. Long line manufac-turing can result in economical cladding or loadbearing panels used in manufacturing or commer-cial applications. The hollow core wall panels areprestressed with two layers of strands for accom-modating handling, structural loadings and bow-ing considerations. Some manufacturers can add2 in to 4 in (51 - 102 mm) of insulation to the hol-low core section with a 1 1/2 in thick to 3 in (38 - 76mm) thick concrete facing to create an insulatedsandwich panel.

A variety of architectural finishes are availablewith hollow core wall panels. While the finishescan be very good, the variety of finishes availableis different from those typically available withtrue architectural precast concrete panels. Injudging the quality of finish on hollow core wallpanels, consideration must be given to themanufacturing process.

1--5

1.6 Design ResponsibilitiesIt is customary in the hollow core industry for

the producer to perform the final engineering forthe product to be supplied to the job. This wouldinclude design for vertical loads and lateral loadsspecified by the Engineer of Record, embeddeditems for specified connection forces, and han-dling and shipping. However, the Engineer of Re-cord plays a very important role in the design pro-cess. Prior to selection of the hollow core produc-er, enough preliminary planning should be done toinsure that the specified floor and roof system isachievable. That is, the project should be one thatcan be engineered without requiring changes fromthe contract documents.

The contract documents must clearly indicatedesign criteria to which hollow core slabs willhave to conform. This is especially importantwhen the hollow core slabs must interface withother construction materials. When connectionsare required, the forces to be transmitted throughthe connections must be specified in the contractdocuments. The producer is best able to deter-mine the most efficient connection element to beembedded in the slab. However, the balance of aconnection which interfaces with another materi-al should be detailed in the contract documents.

The Engineer of Record also has a responsibil-ity in the review and approval of erection draw-ings prepared by the precast producer. Review ofthese drawings is the last opportunity to assurethat the producer’s understanding of the projectcoincides with the intent of design. Erectiondrawings should be checked for proper designloads, proper details and bearing conditions, con-formance with specified fire ratings, and the loca-tion of openings.

1.7 Cross-Sections and Load TablesEach of the major hollow core slab systems has

a standard set of cross-sections that can be pro-duced by their equipment. Available in thick-nesses ranging from 4 in to 15 in (102 - 380 mm),core configurations make each system unique.Each individual producer has additional produc-tion practices which may affect the capabilities oftheir product. Therefore, most producers prepareand distribute load tables in their market area.

Producer load tables define the allowable liveload that a given slab can safely support in addi-tion to the slab self weight. The load capacity willbe a function of the slab thickness, the amount ofprestressing provided, and the location of the pre-stressing strands. Fire rated slabs may requireadditional concrete cover below the strands whichwill affect the load capacity.

The design criteria used to develop these loadtables is defined by the ACI Building Code2 asoutlined in Chapter 2. Depending on the designcriteria controlling a slab’s load capacity, someadvantage may be gained by understanding that inmost applications, superimposed loads will con-sist of both dead and live loads. Where ultimatestrength controls, an equivalent live load can beused to enter a load table. It is calculated as:

wequivalent =1.41.7

superimposed Dead load

+ Live loadHowever, if bottom fiber tensile stresses con-

trol, no adjustment in superimposed loads may beused.

Similarly, many loading conditions consist ofloads other than uniform loads. For preliminarydesign only, an equivalent uniform load may becalculated from the maximum moment caused bythe actual loads.

wequivalent =8 Msuperimposed

ℓ2

Shear will not be properly addressed in this sit-uation. Thus, the final design must consider theactual load pattern.

Because of the uniqueness of each hollow coreslab system and the many possibilities of strandpatterns available from various producers, a ge-neric hollow core slab has been developed to dem-onstrate design procedures. Figure 1.7.1 depictsthe slab section and properties and illustrates atypical form for a producer’s load tables.Throughout this manual, this section will be usedto demonstrate various calculation procedureswhere any one of the proprietary cross-sectionscould be substituted. It must be emphasized thatthis cross-section is not available for use andshould not be specified.

Figures 1.7.2 through 1.7.8 present the propri-etary slab cross-sections currently available. Thesection properties are as provided by the manufac-

1--6

turers, but weights are based on 150 pcf (2400kg/m3) concrete. The actual weights may varyslightly from those given. The availability of anyparticular section in a given area must be verifiedwith the local producers. Figures 1.7.9 presentcharts of the general range of load capacitiesavailable in a given slab thickness. As with anychart of this nature, the chart should be carefullyapproached and verified with local producer loadtables, especially for the longest and shortest andlightest and heaviest conditions. Special care isalso required when fire rated slabs must be usedon a project. (See Chapter 6)

The following examples demonstrate the waysin which load tables may be used.

Example 1.7.1 Equivalent Uniform LoadFrom the load table in Figure 1.7.1 select a

strand pattern to carry a uniform superimposeddead load of 20 psf and a uniform live load of 60psf on a 24 foot span.

wtotal = 20 + 60 = 80 psf4-7/16 in dia. strands required: capacity = 118 psfflexural strength controls

wequivalent =1.41.7

20+ 60 = 77 psf

Use 4-3/8 in dia. strands: capacity = 79 psfflexural strength controls.

Example 1.7.2 Non-Uniform LoadsFrom the load table in Figure 1.7.1 select a

strand pattern to carry a superimposed uniformload of 20 psf dead plus 40 psf live and a continu-ous wall load of 600 plf located perpendicular tothe span and at midspan. The design span is 25feet.For preliminary design

Msuperimposed =252

820+ 40+ 25

4600

= 8438

wequivalent =88438

252

= 108 psf

ft-#/ft

Try 6-3/8 in dia. strands - capacity = 120 psf

For final design use the methods of Chapter 2particularly to check shear.

1.8 Tolerances3

Figure 1.8.1 shows the dimensional tolerancesfor precast hollow core slabs. These tolerancesare guidelines only and each project must be con-sidered individually to ensure that the tolerancesshown are applicable.

Figure 1.8.2 shows erection tolerances for hol-low core slabs. When establishing tolerances, thefunction of the slabs should be considered. Forexample, slabs covered by finish materials maynot need the close tolerances required for exposedslabs.

1--7

Section PropertiesA = 154 in2

I = 1224.5 in4

bw = 10.5 inyb = 3.89 inSb = 314.8 in3

St = 297.9 in3

wt = 53.5 psf

SAMPLE LOAD TABLE3

Allowable Superimposed Live Loads, psf

Strands, 270LR φMn, ft-k

Spans, ft

317 270 232 200 174 152 133 116 102 90356 311 272 240 212 188 168 150320 278 243 214 189 167 148 132

3431 3111 2831 258 231 208327 289 257 229 204 183

3171 2901 2671 2471

4-3/8″6-3/8″4-7/16″6-7/16″4-1/2″6-1/2″

Strands, 270LR φMn, ft-k

45.165.459.485.076.7

105.3

4-3/8″6-3/8″4-7/16″6-7/16″4-1/2″6-1/2″

45.165.459.485.076.7

105.3

79 79 69 61 53 46134 120 108 97 87 78 70118 105 94 84 75 67 59187 169 153 139 126 114 104165 148 134 121 109 99 902271 2101 1952 1782 1632 1492 1372

14 15 16 17 18 19 20 21 22 23

24 25 26 27 28 29 30

1 - Values are governed by shear strength.2 - Values are governed by allowable tension3 - Table based on 5000 psi concrete with allowable tension. Unless noted, values are

governed by strength design.

Note: This slab is for illustration purposes only. Do not specify this slab for a project.

6 f′c

Fig. 1.7.1 Generic hollow core slab

8"

5 1/

4"1

1/4"

1 1/

2"

4 1/4"1 1/2"1 1/2" 1"

36"

1--8

Fig. 1.7.2

Note: All sections not available from all producers. Check availability with local manufacturers.

Trade name: Dy-CoreEquipment Manufacturer: Mixer Systems, Inc., Pewaukee, Wisconsin

4′-0″ x 6″ 142 3.05 661 37 4.45 1475 624′-0″ x 8″ 193 3.97 1581 50 5.43 3017 754′-0″ x 10″ 215 5.40 2783 56 6.89 4614 814′-0″ x 12″ 264 6.37 4773 69 7.89 7313 944′-0″ x 15″ 289 7.37 8604 76 9.21 13225 101

Section Untopped with 2″ topping

widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

Fig. 1.7.3


Trade name: DynaspanEquipment Manufacturer: Dynamold Corporation, Salina, Kansas

4′-0″ x 4″ 133 2.00 235 35 3.08 689 604′-0″ x 6″ 165 3.02 706 43 4.25 1543 684′-0″ x 8″ 233 3.93 1731 61 5.16 3205 864′-0″ x 10″ 260 4.91 3145 68 6.26 5314 938′-0″ x 6″ 338 3.05 1445 44 4.26 3106 698′-0″ x 8″ 470 3.96 3525 61 5.17 6444 868′-0″ x 10″ 532 4.96 6422 69 6.28 10712 948′-0″ x 12″ 615 5.95 10505 80 7.32 16507 105


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

1--9

Fig. 1.7.4


Trade name: FlexicoreLicensing Organization: The Flexicore Co. Inc., Dayton, Ohio

1′-4″ x 6″ 55 3.00 243 43 4.23 523 682′-0″ x 6″ 86 3.00 366 45 4.20 793 701′-4″ x 8″ 73 4.00 560 57 5.26 1028 822′-0″ x 8″ 110 4.00 843 57 5.26 1547 821′-8″ x 10″ 98 5.00 1254 61 6.43 2109 862′-0″ x 10″ 138 5.00 1587 72 6.27 2651 972′-0″ x 12″ 141 6.00 2595 73 7.46 4049 98


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

Fig. 1.7.5

Trade name: ElematicEquipment Manufacturer: Mixer Systems, Inc., Pewaukee, Wisconsin

Note: Elematic is also availble in 96″ width. All sections not available from all producers. Check availability with local manufacturers.

4′-0″ x 6″ 157 3.00 694 41 4.33 1557 664′-0″ x 8″ 196 3.97 1580 51 5.41 3024 764′-0″ x 10″(5) 238 5.00 3042 62 6.49 5190 874′-0″ x 10″(6) 249 5.00 3108 65 6.44 5280 904′-0″ x 12″ 274 6.00 5121 71 7.56 8134 96


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

1--10

Fig. 1.7.6


Trade name: SpanDeckLicensing Organization: Fabcon, Incorporated, Savage, Minnesota

4′-0″ x 8″ 246 3.75 1615 62 5.55 2791 874′-0″ x 12″ 298 5.87 5452 75 8.01 7856 1008′-0″ x 8″ 477 3.73 3236 60 5.53 5643 858′-0″ x 12″ 578 5.86 10909 72 7.98 15709 97


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

Fig. 1.7.7

Note: Spancrete is also available in 40″ and 96″ widths. All sections are not available from all producers. Check availability withlocal manufacturer.

Trade name: SpancreteLicensing Organization: Spancrete Machinery Corp., Milwaukee, Wisconsin

4′-0″ x 4″ 138 2.00 238 34 3.14 739 594′-0″ x 6″ 189 2.93 762 46 4.19 1760 714′-0″ x 8″ 258 3.98 1806 63 5.22 3443 884′-0″ x 10″ 312 5.16 3484 76 6.41 5787 1014′-0″ x 12″ 355 6.28 5784 86 7.58 8904 1114′-0″ x 15″ 370 7.87 9765 90 9.39 14351 115

4′-0″ x 8″ 246 4.17 1730 60 5.41 3230 854′-0″ x 10″ 277 5.22 3178 67 6.58 5376 924′-0″ x 12″ 316 6.22 5311 77 7.66 8410 102


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

Ultralight Spancrete

1--11

Fig. 1.7.8

Fig. 1.7.9(a) Slab load ranges

Note: All sections are not available from all producers. Check availability with local manufacturers.

Trade name: Ultra-SpanLicensing Organization: Ultra-Span Technologies, Inc., Winnipeg, Manitoba, Canada

4′-0″ x 4″ 154 2.00 247 40 2.98 723 654′-0″ x 6″ 188 3.00 764 49 4.13 1641 744′-0″ x 8″ 214 4.00 1666 56 5.29 3070 814′-0″ x 10″ 259 5.00 3223 67 6.34 5328 924′-0″ x 12″ 289 6.00 5272 75 7.43 8195 100


widthx

depthA

in2ybin

Iin4

wtpsf

ybin

Iin4

wtpsf

Span, ft

10 20 30 40

50

100

150

200

250

300

~ = 45

Supe

rim

pose

d L

ive

Loa

d, p

sf

h

6" Hollow Core Slab

3/4" Concrete Cover

6" + 2" Topping

6"

1--12

Fig. 1.7.9 (b) Slab load ranges

Fig. 1.7.9(c) Slab load ranges

Span, ft

10 20 30 40

50

100

150

200

250

300

~ = 45

Supe

rim

pose

d L

ive

Loa

d, p

sf h

8" Hollow Core Slab

3/4" Concrete Cover

8" + 2" Topping

8"

Span, ft

10 20 30 40

50

100

150

200

250

300

~ = 45

Supe

rim

pose

d L

ive

Loa

d, p

sf

h

10" Hollow Core Slab

3/4" Concrete Cover

10"

10" + 2" Topping

1--13

Fig. 1.7.9 (d) Slab load ranges

Fig. 1.7.9(e) Slab load ranges

Span, ft

10 20 30 40

50

100

150

200

250

300

~ = 45

Supe

rim

pose

d L

ive

Loa

d, p

sf

h


3/4" Concrete Cover

12"

12" + 2" Topping

Span, ft

20 30 40 50

75

100

125

150

175

200

~ = 45

Supe

rim

pose

d L

ive

Loa

d, p

sf

h


3/4" Concrete Cover

5010 60

15" + 2 1/2" Topping

15"

1--14

Fig. 1.8.1 Product tolerances -- hollow core slabs

a = Length ±1/2 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .b = Width ±1/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .c = Depth ±1/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .dt = Top flange thickness

Top flange area defined by the actual measured values ofaverage dt x b shall not be less than 85% of the nominal areacalculated by dt nominal x b nominal.

db = Bottom flange thicknessBottom flange area defined by the actual measured valuesof average db x b shall not be less than 85% of the nominalarea calculated by db nominal x b nominal.

e = Web thicknessThe total cumulative web thickness defined by the actualmeasured valueΣe shall not be less than 85% of the nominalcumulative width calculated by Σe nominal.

f = Blockout location ±2 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .g = Flange angle 1/8 in per 12 in, 1/2 in max.. . . . . . . . . . . . . .h = Variation from specified end squareness

or skew ±1/2 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .i = Sweep (variation from straight line parallel to centerline of

member) ±3/8 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

j = Center of gravity of strand groupThe CG of the strand group relative to the top of the plankshall be within ±1/4 in of the nominal strand group CG. Theposition of any individual strand shall be within ±1/2 in ofnominal vertical position and ±3/4 in of nominal horizontalposition and shall have a minimum cover of 3/4 in.

k = Position of plates ±2 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .l = Tipping and flushness of plates ±1/4 in. . . . . . . . . . . . . . . .m = Local smoothness ±1/4 in in 10 ft. . . . . . . . . . . . . . . . . . . . .

(does not apply to top deck surface left rough to receive atopping or to visually concealed surfaces)

Plank weightExcess concrete material in the plank internal features iswithin tolerance as long as the measured weight of theindividual plank does not exceed 110% of the nominalpublished unit weight used in the load capacity calculation.

n = Applications requiring close control of differential camberbetween adjacent members of the same design should bediscussed in detail with the producer to determine applicabletolerances.

b

c

d b

td

e

CROSS SECTION

CGS

j

k

k

g

m

10 ft

fi

l

n

h

PLAN

ELEVATION

a

1--15

Fig. 1.8.2 Erection tolerances - hollow core floor and roof members

a = Plan location from building grid datum ± 1 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .a1 = Plan location from centerline of steel* ± 1 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .b = Top elevation from nominal top elevation at member ends

Covered with topping ± 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Untopped floor ± 1/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Untopped roof ± 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

c = Maximum jog in alignment of matching edges(both topped and untopped construction) 1 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

d = Joint width0 to 40 ft member length ± 1/2 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 to 60 ft member length ± 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61 ft plus ± 1 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

e = Differential top elevation as erectedCovered with topping 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Untopped floor 1/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Untopped roof** 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

f = Bearing length*** (span direction) ± 3/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .g = Differential bottom elevation of exposed hollow-core slabs**** 1/4 in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

* For precast concrete erected on a steel frame building, this tolerance takes precedence over tolerance on dimension “a”.** It may be necessary to feather the edges to ± 1/4 in to properly apply some roof membranes.

*** This is a setting tolerance and should not be confused with structural performance requirements set by the architect/engineer.**** Untopped installation will require a larger tolerance here.

e

gf

bldg. elevation datum

b

clearance

d

c

a

a bldg. Y grid datum

bldg. X grid datum

hollow corefloor or roof member

f

a1

ELEVATION

bldg. elevation datum

b

ELEVATION

Precast element to precast orcast-in-place concrete or masonry

floor or roof memberhollow core

precast or cast in placeconcrete support member

centerline of floor or roof memberhollow core

bldg. X grid datum

bldg. Y grid datum

d

c

a

f

a

Precast element to structural steel

g

e

PLAN PLANsteel structure

centerline ofsteel structure

CHAPTER 2

2--1

DESIGN OF HOLLOW CORE SLABS

2.1 GeneralThe design of hollow core slabs is governed by

the ACI (318-95) Building Code Requirementsfor Structural Concrete.2 As with prestressed con-crete members in general, hollow core slabs arechecked for prestress transfer stresses, handlingstresses, service load stresses, deflections and de-sign (ultimate) strength in shear and bending. Foruniform load cases, the manufacturer’s load tableswill take into account these various design consid-erations and print a load capacity based on thegoverning criteria. For loading conditions otherthan uniform, or for the development of loadtables, the design steps presented in this sectionare used.

An excellent reference for prestressed memberdesign exists in the PCI Design Handbook.1Charts and tables provide design aids to shortenthe calculation procedures. Another excellentsource for design information is the PCI StandardDesign Practice4 which reflects design practicesin the industry.

The generic slab presented in Section 1.7 willbe used for the calculations presented in this sec-tion. The cross-section was selected to provide ameans of demonstrating calculation proceduresand does not represent any slab currently in use.Therefore, this generic slab should never be speci-fied for use on a project. See Section 1.7 for theslabs currently available.

2.2 Flexural Design

2.2.1 ACI RequirementsChapter 18 of ACI (318-95) presents provi-

sions for the flexural design of prestressed con-crete members. The applicable limits from ACIare paraphrased as follows:

2.2.1.1 Permissible stresses at transfer(Section 18.4).a) Extreme fiber stress in compression

. 0.6 f′ci. . . . . . . . . . . . . . . . . . . . . . .

b) Extreme fiber stress in tension exceptas permitted in (c) 3 f′ci. . . . . . . . .

c) Extreme fiber stress in tension at endsof simply supported members . . . . . .

6 f′ci. . . . . . . . . . . . . . . . . . . . . . . .

2.2.1.2 Permissible stresses at serviceloads (Section 18.4)a) Extreme fiber stress in compression

due to prestress plus sustained loads0.45 f′c. . . . . . . . . . . . . . . . . . . . . . . .

b) Extreme fiber stress in compressiondue to prestress plus total load

0.60 f′c. . . . . . . . . . . . . . . . . . . . . . . .c) Extreme fiber stress in tension in pre-

compressed tensile zone 6 f′c. . . . .d) Extreme fiber stress in tension in pre-

compressed tensile zone where deflec-tions are calculated considering bili-near moment-deflection relationships

12 f′c. . . . . . . . . . . . . . . . . . . . . . . .

2.2.1.3 Loss of prestress (Section 18.6)Calculation of losses shall consider:a) Seating lossb) Elastic shortening of concretec) Creep of concreted) Shrinkage of concretee) Steel relaxation

2.2.1.4 Design (ultimate) strengtha) Load Factors (Section 9.2)

U = 1.4D + 1.7Lb) Strength Reduction Factors (Section

9.3)Flexure φ = 0.9

c) Flexural Strength (Section 18.7)

Mu ≤ φMn = ÔApsfpsdp − a2

a =Apsfps

0.85f′cbfps = value calculated by strain

compatibility

2--2

or

fps = fpu1− γp

β1pfpu

f′cρ

Mn > 1.2 Mcr

2.2.2 Stresses at TransferWhen the prestressing strands are cut to apply

the prestressing force to the concrete, only the slabself weight is present to counteract the effects ofeccentric prestress. A check of stresses is requiredat this point to determine the concrete strength re-quired to preclude cracking on the tension side orcrushing on the compression side. The concretestrength at the time of transfer may be only 50% to60% of the 28 day design strength.

Example 2.2.2.1 - Transfer StressesUsing the generic hollow core cross-section

defined in Section 1.7, check stresses at transfer ofprestress using the following criteria:

Prestressing steel: 4 - 1/2″ dia. 270 ksi, low relax-ation strands.Aps = 4(0.153) = 0.612 in2

assume 5% initial loss

dp = 7″

ℓ = 30′-6″

initial stress = 70% fpu

Solution:Stresses will be checked at the transfer point

and at midspanAt release prestress force

Po = (0.70)(0.95)(0.612)(270) = 109.9kPrestress effect

= PoA

PoeS

= 109.9154

109.92.89

297.9314.8

= --0.353 ksi top fiber

= +1.723 ksi bottom fiber

Self weight at transfer point

ℓt = 50db = 50(1/2) = 25 in

moment 25 in from slab end

Md = 30.52

2.08− 2.082

20.05353′

= 4.74 ft-k

MdS

=4.7412

279.9314.8

= +0.191 ksi top fiber= --0.181 ksi bottom fiber

Net concrete stress at transfer point= --0.162 ksi top fiber= +1.542 ksi bottom fiber

Self weight at midspan

Md = 30.52

8(0.0535)(3′) = 18.66 ft-k

MdS

=18.6612

279.9314.8

= +0.752 ksi top fiber= --0.711 ksi bottom fiber

Net concrete stress at midspan= +0.399 ksi top fiber= +1.012 ksi bottom fiber

Allowable stresses:

tension at end = 6 f′ci

f′ci = − 16262

= 729 psi

tension at midspan = 3 f′cidoes not control

compression = 0.6 f′ci

f′ci = 15420.6

= 2570 psi

Concrete strength required at release= 2570 psi

Note that if tension or compression in the endregion exceeds allowables based on a reasonableconcrete release strength, strands may be de-bonded in some manufacturing systems or, fortension, top mild reinforcement may be used insome manufacturing systems to resist the totaltension force.

2--3

If tension in the midspan region controls, eithera high release strength must be used or mild rein-forcement must be added to resist the total tensionforce. Mild reinforcement should only be used inthe wet cast manufacturing system.

2.2.3 Prestress LossesThe calculation of prestress losses affects the

service load behavior of a slab. The accuracy ofany calculation method is dependent on the pre-ciseness of concrete and prestressing steel materi-al properties as well as external factors such as hu-midity used in the calculation procedure. Theaccuracy of loss calculations has little effect onthe ultimate strength of a member.

Prestress loss calculations are required for pre-diction of camber and for service load stress cal-culations. Since the success of a project is judgedon service load performance rather than ultimatestrength, it behooves any slab producer to use aloss calculation procedure which best predicts thebehavior of the product as produced.

For low relaxation strand and for special cases(e.g., long spans or special loadings) using stressrelieved strand, the 1995 ACI Code referencesseveral sources for prestress loss calculations.The method presented here was developed by Zia,et al.5 and considers the following parameters:

1) Elastic Shortening

ES = KesEsEci

fcir

Kes = 1.0 for pretensioned members

fcir = Kcir PiA+ Pie2

I −MgeI

Kcir = 0.9 for pretensioned members

2) Concrete Creep

CR = KcrEsEc

(fcir -- fcds)

Kcr = 2.0 for normal weight pretensionedmembers

= 1.6 for sand lightweight pretensionedmembers

fcds =Msde

I3) Shrinkage of Concrete

SH = 8.2 x 10-6KshEs1− 0.06 VS

x (100 -- RH)

Fig. 2.2.3.1 Ambient relative humidity

8070

70

60

80

7060 40

30

40 50

4050 60

7075

7075

70

80

70

7570

2--4

Table 2.2.3.1Type of tendon Kre psi J

270 Grade stress-re-lieved strand or wire 20,000 0.15

250 Grade stress-re-lieved strand or wire 18,500 0.14

240 or 235 Grade stress-relieved wire 17,600 0.13

270 Grade low-relax-ation strand 5000 0.040

250 Grade low-relax-ation wire 4630 0.037

240 or 235 Grade low-re-laxation wire 4400 0.035

145 or 160 Grade stress-relieved bar 6000 0.05

Ksh = 1.0 for pretensioned members

RH = Ambient relative humidity from Fig-ure 2.2.3.1

4) Steel Relaxation

RE = [Kre -- J (SH + CR + ES)]C

Kre, J, C = factors from Tables 2.2.3.1and 2.2.3.2

5) Total Loss = ES + CR + SH + RE

Observations and experience in a plant mayprovide modifications to loss calculations to bet-ter predict slab performance.

Example 2.2.3.1 Loss of PrestressUsing the generic hollow core cross-section

defined in Section 1.7, calculate the loss of pre-stress based on the following information:

Prestressing steel: 4-1/2″ dia. 270 ksi, low re-laxation strandsApsfpu = 0.153(270) = 41.3k/strand

dp = 7″

initial stress = 70% fpu

ℓ = 30′-6″

Superimposed dead load = 20 psf

Table 2.2.3.2 Values of C

fsi/fpu

Stress-relieved

strand orwire

Stress-relievedbar or

low-relaxationstrand or wire

0.80 1.280.79 1.220.78 1.160.77 1.110.76 1.050.75 1.45 1.000.74 1.36 0.950.73 1.27 0.900.72 1.18 0.850.71 1.09 0.800.70 1.00 0.750.69 0.94 0.700.68 0.89 0.660.67 0.83 0.610.66 0.78 0.570.65 0.73 0.530.64 0.68 0.490.63 0.63 0.450.62 0.58 0.410.61 0.53 0.370.60 0.49 0.33

Solution:1) Elastic Shortening

Pi = 0.7(4)(41.3k) = 115.6k

Mg = 30.52

8(0.0535)(3′)

= 18.66 ft-k

= 224 in-k

fcir = 0.9115.6154

+ 115.6(2.89)2

1224.5

--2242.89

1224.5

= 0.857 ksi

using Es = 28,500 ksi and Eci = 3250 ksi

ES = KesEsEci

fcir

= (1.0)285003250

(0.857)

2--5

= 7.52 ksi2) Concrete Creep

fcds =Msde

I

=30.52

80.023122.89

1224.5

= 0.198 ksi

using Ec = 4300 ksi and normal weight concrete

CR = KcrEsEc

(fcir -- fcds)

= (2.0)285004300

(0.857 -- 0.198)

= 8.74 ksi3) Shrinkage of Concrete

VS

= AreaPerimeter

= 154236+ 8

= 1.75

use RH = 70%

SH = 8.2 x 10-6KshEs1− 0.06 VS

x (100 -- RH)

= 8.2 x 10-6(1.0)28500

x (1 -- 0.06 x 1.75)(100 -- 70)

= 6.27 ksi4) Steel Relaxation

From Table 2.2.3.1

Kre = 5000, J = 0.04

From Table 2.2.3.2

C = 0.75 for fsi/fpu = 0.7

RE = [Kre -- J(SH + CR + ES)]C

= [50001000

− 0.04x

(6.27 + 8.74 + 7.52)] 0.75

= 3.07 ksi5) Total Loss at Midspan

= 7.52 + 8.74 + 6.27 + 3.07

= 25.6 ksi

% = 25.60.7270

(100) = 13.5%

2.2.4 Service Load StressesService load concrete stresses are calculated as

a measure of performance or serviceability. Forthe in-service state when deflections must be cal-culated, a stress check must first be made to deter-mine whether gross section properties or cracked-transformed section properties are to be used.

In-service stresses are checked assuming thatall prestress losses have occurred. The calculatedstresses are compared to the permissible stressesnoted in Section 2.2.1. Hollow core slabs are nor-mally designed to be uncracked under full serviceloads. Tensile stress limits of between 6 f′c and7.5 f′c are commonly used. In special circum-stances where deflections will not be a problemand where cracking will not be of concern, the up-per limit of 12 f′c can be used.

Example 2.2.4.1 Service Load StressesUsing the generic hollow core cross-section

defined in Section 1.7, calculate the service loadstresses given the following criteria:

Prestressing steel:4-1/2″ dia. 270 ksi, low relaxation strands

Apsfpu = 0.153(270) = 41.3k/strand

dp = 7″

Initial stress = 70% fpu

f′c = 5000 psi

ℓ = 30′-6″

Clear Span = 30′-0″Superimposed Dead Load = 20 psf

Live Load = 50 psf

Solution:

Msustained = 302

8(0.0535 + 0.020)

= 8.27 ft-k/ft = 99.2 in-k/ft

Mservice = 302

8(0.0535 + 0.020 + 0.050)

= 13.89 ft-k/ft = 167 in-k/ftWith losses = 13.5% from Example 2.2.3.1

2--6

Apsfse = (0.7)(4)(41.3)(1 -- 0.135)

= 100.0k

Top fiber compression with sustained loads

ftop = 100.0154

− 100.02.89)297.9

+ 99.23297.9

= 0.649 -- 0.970 + 0.999

= + 0.679 ksi

Permissible compression

= 0.45f′c

= 0.45(5000)

= 2.25 ksi > 0.679 ksi OK

Top fiber compression with total load

ftop = 100.0154

− 100.02.89)297.9

+ 1673297.9

= 0.649 -- 0.970 + 1.679

= 1.358 ksi

Permissible compression

= 0.60f′c

= 0.60(5000)

= 3.00 ksi > 1.358 ksi OK

Bottom fiber tension

fbottom = 0.649 + (0.970 -- 1.679)297.9314.8

= --0.022 ksi (tension)

Permissible tension

= 7.5 f′c

= 7.5 5000

= 0.530 ksi > 0.022 ksi OK

2.2.5 Design Flexural StrengthThe moment capacity of a prestressed member

is a function of the ultimate stress developed in theprestressing strands. As with non-prestressedconcrete, upper and lower limits are placed on theamount of reinforcing to ensure that the stress inthe strands is compatible with concrete stressesfor ductile behavior.

The lower limit of reinforcing requires that:φMn ≥ 1.2 Mcr

Mcr = IybP

A+ Pe

Sb+ 7.5 f′c

This ensures that when the concrete developsflexural cracks, the prestressing steel will not havereached its full design stress. Violation of this cri-teria might result in strand fractures at the point offlexural cracking with a resulting brittle failure.However, ACI (318-95) Section 18.8.3 allowsviolation of this requirement for flexural memberswith shear and flexural strength at least twice thatrequired.

The upper limit of reinforcing requires that,ωp or,

ωp + ddp

ω−ω′ or

ωpw + ddp

ωw −ω′wbe not greater than 0.36β1

The need for an upper limit on reinforcing is re-lated to the assumptions of ultimate concrete com-pressive strain. Using a uniform compressionstress block forces more concrete to reach ulti-mate strain as reinforcing ratios increase. There-fore when the upper reinforcing limit is exceeded,the moment capacity must be based on the com-pression block. For this condition,

φMn = φf′cbd2p0.36β1 − 0.08β2

1for rectangular sections or for flanged sectionswith the neutral axis within the flange.

The stress in the prestressing steel at ultimatemay be calculated in several ways. The ACI equa-tion (18-3) may be used as an approximation,charts and tables from the PCI Design Handbookmay be used, or a strain compatibility analysismay be made.

Example 2.2.5.1 Design Flexural StrengthUsing the generic hollow core slab defined in

Section 1.7, check the design flexural strengthgiven the following criteria:

Prestressing steel: 4-1/2″ dia., 270 ksi, low re-laxation strandsdp = 7″initial stress = 70% fpuf′c = 5000 psi

ℓ = 30′-6″

2--7

Clear span = 30′-0″Superimposed Dead Load = 20 psf

Live Load = 50 psf

Solution:METHOD 1: ACI Equation (18-3)

φMn = φApsfps(dp -- a/2)

fps = fpu1− γp

β1 p

fpu

f′cρ

Use γp = 0.28 for low relaxation strands

β1 = 0.85 -- 5000− 40001000

0.05

= 0.80

ρp =Aps

bdp= 40.153

367= 0.0024

fps = 2701− 0.280.80

0.0024 2705

= 257.7 ksi

ωp = pfps

f′c= 0.0024257.7

5ρ

= 0.124 < 0.36 β1 = 0.288 OK

a =Apsfps

0.85f′cb= 40.153257.7

0.85536

= 1.03 in

Note: If “a” exceeds the top flange thickness, thecompression block will encroach on the core area.For this situation, multiple compression forces areused for the internal couple as is done with otherflanged members.

φMn = 0.9(4)(0.153)(257.7) 7− 1.032

= 920 in-k/slab = 76.7 ft-k/slab

wu = 1.4(0.0535 + 0.02) + 1.7(0.05)

= 0.188 ksf

Mu = 302

8(0.188)

= 21.14 ft-k/ft

= 63.4 ft-k/slab < 76.7 OK

Check minimum reinforcement

φMn ≥ 1.2 Mcr

From Example 2.2.3.1

Loss = 13.5%

Apsfse= 0.7(4)(41.3)(1 -- 0.135)

= 100.0 kBottom compression

= 100.0154

+ 100.02.89314.8

= 1.567 ksi

Mcr = 1224.53.891.567+ 7.5 5000

1000

= 660 in-k/slab

ÔMnMcr

= 920660

= 1.39 > 1.2 OK

METHOD 2: PCI Design HandbookUsing Figure 4.12.2 from the 5th Edition Hand-book.

ωpu =Aps

bdp

fpu

f′c

=441.33675

= 0.131K′u = 538

φMn = K′ubd2

p

12000

= 53836(7)2

12000= 79.0 ft-k/slab

METHOD 3: Strain CompatibilityThe stress-strain diagram from Figure 11.2.5 of

the PCI Design Handbook, shown in Fig. 2.2.5.1,will be used for this example. However, the actualstress-strain curves received with strand mill re-ports should be used when available.

The concrete ultimate strain is assumed to be0.003 in/in. The method involves a trial and errorprocedure to obtain equilibrium within the sectionwhere the force in the compression block equalsthe tensile force in the steel. The equations are de-veloped from the strain diagram shown.

a = β1c

2--8

Fig. 2.2.5.1 Stress-strain curves, prestressing strand

These curves can be approximated by the following equations:

250 ksi strand 270 ksi strand

εps ≤ 0.0076: fps = 28,500 εps (ksi)

εps > 0.0076: fps = 250 (ksi)

εps ≤ 0.0086: fps = 28,500 εps (ksi)

εps > 0.0086: fps = 270 (ksi)− 0.04Áps − 0.0064

− 0.04Áps − 0.007

1500 0.005 0.010 0.015 0.020 0.025 0.030

170

190

210

230

250

270

250 ksi strand

270 ksi strand

E = 28,500 ksis

MINIMUM YIELD STRENGTH AT1% ELONGATION FOR 270 ksi(ASTM A416)

(ASTM A416)1% ELONGATION FOR 250 ksiMINIMUM YIELD STRENGTH AT

Stre

ss -

f

(ks

i)ps

Strain- (in./in.)ps

dp

s

0.003

T

Cac

0.85 fc’

Using 13.5% loss from Example 2.2.3.1

fse = 0.7(270)(1 -- 0.135) = 163.4 ksi

εse = fseEs

= 163.428500

= 0.0057

Assume c = 1″ then a = 0.80(1) = 0.8″

εs =dpc (0.003) -- 0.003

= 71

(0.003) -- 0.003 = 0.018

2--9

εps = εse + εs

= 0.0057 + 0.018 = 0.0237

From stress-strain curve

fps = 268 ksi

T = 4(0.153)(268) = 163.8

C = 0.85(5)(0.8)(36)

= 122.4k < 163.8k

Try c = 1.3″ then a = 0.80(1.3) = 1.04″

εs = 71.30

(0.003) -- 0.003

= 0.0131

εps = 0.0131 + 0.0057 = 0.0188

From stress-strain curve

fps = 267 ksi

T = 4(0.153)(267) = 163

C = 0.85(5)(1.04)(36)

= 159k ≈ 163k

φMn = 0.9(4)(0.153)(267)7− 1.042

= 952 in-k/slab = 79.3 ft-k/slab

On occasion, conventional reinforcement isadded to a hollow core slab to locally provide add-ed flexural strength. When required, the bars areplaced in cores right after the slab is cast and con-crete is added to fill the cores with the bars. Thefollowing example illustrates the flexural strengthcalculation.

Example 2.2.5.2 Flexural Strength with BarsRepeat Example 2.2.5.1 but add 2 - #4 bars in

cores.Solution:

Use strain compatibility for strength calculationwith an effective depth of 5.5 in for the #4 bars.

Assume c = 1.53 in.

then a = 0.80(1.53) = 1.22 in

for strands

εs = 71.53

(0.003)− 0.003

= 0.0107 in/in

εps = 0.0057 + 0.0107

= 0.0164 in/in

fps = 266 ksi

for bars

εs = 5.51.53

(0.003)− 0.003

= 0.0078 in/in

yield strain = 6029000

= 0.002 in/in

T = 4(0.153)(266) + 2(0.2)(60)

= 162.8 + 24

= 186.8k

C = 0.85(5)(1.22)(36)

= 186.7k ≅ 186.8k ok

φMn = 0.9162.87− 1.222+ 245.5− 1.22

2

= 1042 in-k

= 86.8 ft-k

2.3 Shear Design

2.3.1 ACI RequirementsHollow core slabs are designed for shear ac-

cording to the same ACI Code provisions used ingeneral for prestressed members. In dry cast sys-tems, the normal practice is to not provide stirrupswhen the applied shear exceeds shear capacity be-cause of the difficulty encountered placing stir-rups in most production processes. The place-ment of stirrups in a wet cast system is certainlyeasier than in a dry cast extruded system and is aviable shear enhancement method. An alternativeused to increase shear capacity is to reduce thenumber of cores used in a given slab. This may bedone by either leaving out a core for the entirelength of a slab or by locally breaking into thecores and filling them solid while the concrete isstill in a somewhat plastic state.

The provisions for shear are found in Chapter11 of ACI 318-95. With some paraphrasing, therequirements are:Vu ≤ φVn

φ = 0.85 for shear

Vn = Vc + Vs

2--10

For the purpose of this discussion, Vs, the con-tribution of shear reinforcement, will be taken aszero. The nominal concrete shear strength may befound using equation (11-9),

Vc = 0.6 f′c + 700 VudMubwd (11-9)

when the effective prestress force is not less than40 percent of the tensile strength of the flexural re-inforcement. The term Vud/Mu shall not exceed1.0. The minimum value for Vc may be used as2 f′c bwd and the maximum value is the lesser of5 f′c bwd or the value obtained from Equation(11-12) considering reduced effective prestress inthe transfer zone.

Alternatively more refined shear calculationscan be made according to the lesser of Equations(11-10) or (11-12).

Vci = 0.6 f′c bwd+Vd +ViMcrMmax

(11-10)

Vcw = (3.5 f′c + 0.3fpc) bwd (11-12)Equation (11-10) predicts shear strength for an

inclined shear failure mode. For Equation(11-10), the following relationships are used:

Mcr = Iy(6 f′c + fpe -- fd) (11-11)

Vd = Unfactored self weight shear fornon-composite sections

Vi = Vu -- Vd

Mmax = Mu -- Md

Md = Unfactored self weight moment fornon-composite sections

The minimum value for Vci need not be lessthan 1.7 f′c bwd or 2 f′c bwd when the effectiveprestress force is not less than 40% of the tensilestrength of the flexural reinforcement. For equa-tions (11-10), (11-11) and (11-12), the reductionin prestressing force at the member end due totransfer must be considered. The ACI Code al-lows an assumption that prestressing force in-creases linearly from zero at the member end tofull effective prestress in a length equal to 50strand diameters.

Example 2.3.1.1 Shear Design

Using the generic hollow core cross-sectiondefined in Section 1.7, check the slab for sheargiven the following information:

Prestressing steel: 4-1/2″ dia., 270 ksi, lowrelaxation strands.

Initial stress = 70% fpu loss = 15%f′c = 5000 psi

ℓ = 25′-6″Clear span = 25′-0″

Superimposed Dead Load = 20 psfLive Load = 50 psf

Masonry dead load = 800 plf at 3′from one support

Solution:Uniform load: wu = 1.4(0.0535 + 0.020)

+ 1.7(0.05)= 0.188 ksf = 0.564 klf

Line Load: Pu = 1.4(0.800) = 1.12k/ft= (3′)(1.12) = 3.36k

Load, shear and moment diagrams for 3′ slabwidth:

1.12 x 3 = 3.36

0.188 x 3 = 0.564

k

kft

10k

k8.31

4.95k

k7.45

V

M

u

u

27.48 ft-k

49.25 ft-k

25

3

’

’

’

’

Using the more refined approach according toACI Equations (11-10) or (11-12), φVc is:

φVcw = 0.851000

3.5 5000 + 0.3fpcx (10.5)(7) (11-12)

= 15.46 + 0.0187fpc

fpc is calculated as a function of the transfer of pre-stress into the section along the span.

2--11

transfer length = 50 db = 50(1/2) = 25″with bearing length = 3″full prestress transfer is achieved 22″ fromthe face of support

Apsfse= 4(41,300)(0.70)(1 - 0.150)

x x+ 325 to x = 22″

fpc=Apsfse

A= 98294

154x+ 3

25

φVcw = 15.46 + 0.0187 98294154

x+ 325

= 15.46 + 11.96 x+ 325 to x = 22″

φVci = 0.6 50001000

10.57 +Vd +ViMcrMmax

x 0.85 (11-10)

Vd = Shear due to unfactored self weight(for non-composite section)

= 3(0.0535)252− x = 2.01 -- 0.16x

Vi = Shear due to factored loads minus Vd

Mcr = Iyb6 f′c + fpe − fd

fpe = Apsfse1A+ eyb

I

fpe = 98.294 x

1154

+3.89− 13.89

1224.5x+ 3

25

= 1.541x+ 325 ≤ 1.541 ksi

fd = flexural stress due to load used for Vd

=MdS

=

30.0535x2

25− x

314.8

= 2.01x− 0.08x2

314.8

Mcr = 314.812

x

0.424+ fpe −2.01x− 0.08x2

314.812

= 11.130 + 26.233fpe -- 2.01x + 0.8x2

Mmax = Moment due to factoredloads minus Md

Based on these definitions, φVcw, φVci, and Vu arecalculated at intervals across the span. A summa-ry is presented in Table 2.3.1.1. Figure 2.3.1.1presents the results graphically.

Table 2.3.1.1 Allowable Shear

x Vu φVcw φVcih/2 = 0.333′ 9.82k 18.81k 59.40k

0.5′ 9.72 19.76 45.741.0′ 9.44 22.64 31.921.5′ 9.16 25.51 27.152.0′ 8.88 27.42 23.342.5′ 8.59 27.42 18.933.0′ 8.31 27.42 15.983.0′ 4.95 27.42 10.023.5′ 4.67 27.42 9.114.0′ 4.39 27.42 8.34

Alternatively, the simplified equation (11-9)might be used.

φVc = 0.850.6 5000 + 700VuMu7

x10.571000

= 2.65 + 306.1 VuMu

(Mu in in-k).

The results of this equation are also shown on Fig-ure 2.3.1.1.

At all points, Vu < φVc so shear strength is ade-quate and stirrups are not required.

2.4 Camber and DeflectionCamber is the upward deflection of a pre-

stressed member and results from the prestressingforce being eccentric from the center of gravity ofthe cross-section. Since both prestressing forceand eccentricity are established by the requireddesign load and span length, camber is a result ofthe design rather than a design parameter. There-fore, camber requirements should not be speci-fied.

Deflections are also affected by the amount ofprestressing only because prestressing establishesthe load at which a member will crack. If tensile

2--12

Fig. 2.3.1.1 Shear for Example 2.3.1.1

Distance into Span, ft

00

Shea

r, k

ips

5

10

15

20

25

1 2 3 4

Eq. (11-10)

Eq. (11-9)

Eq. (11-12)

2 f’ b d

Vu

c w

5 f’ b dc w

stresses are kept below cracking, deflections willbe independent of the prestress level.

Cambers and deflections will change with timedue to concrete creep, prestress loss and other fac-tors. The sustained compression due to the pre-stressing will cause camber growth. Balancingthis is the effect of creep on deflections due to selfweight and other sustained loads. It is this timedependent movement which, in addition to instan-taneous deflections, must be considered in the de-velopment of framing schemes and detailing.

Instantaneous cambers and deflections are pre-dictable as long as the material properties areknown. The time dependent cambers and deflec-

tions are not predictable with any degree of accu-racy and any calculation of long term movementsmust be considered to be only estimates.

This section presents calculation proceduresfor determining long term deflections. From theproducer’s standpoint, history and experiencemust be used to modify the procedures to fit the lo-cal product. From the specifier’s standpoint, theseprocedures will allow only approximate estimatesof long term effects and should be complementedwith discussions with local producers.

2.4.1 Camber

2--13

1.85 1.85

1.80 1.80

2.70 2.40

2.45 2.20

3.00 3.00

------ 2.30

Table 2.4.1 Long term multipliers6

At Erection:1. Deflection (downward) component - apply to the elastic

deflection due to the member weight at release of prestress2. Camber (upward) component - apply to the elastic camber

due to the prestress at the time of release of prestressFinal:

3. Deflection (downward) component - apply to the elasticdeflection due to the member weight at release of prestress

4. Camber (upward) component - apply to the elastic camberdue to prestress at the time of release of prestress

5. Deflection (downward) - apply to elastic deflection due tosuperimposed dead load only

6. Deflection (downward) - apply to elastic deflection causedby the composite topping

ConditionWithout

CompositeTopping

WithCompositeTopping

Hollow core slabs are produced with straightstrand patterns rather than using draped or de-pressed strands. Using (+) to indicate upwardmovement and (--) to indicate downward move-ment, net camber can be calculated as:

camber = Peℓ2

8EI− 5wℓ4

384EI

To determine initial camber, the appropriatevalues for prestress force and modulus of elastic-ity of the concrete must be used. When ultimatemoment rather than tensile stresses govern a de-sign, the initial strand stress may be reduced tomodify the anticipated camber. Additionally, slabcamber is sensitive to support point locations dur-ing storage. Camber will increase as these supportpoints move in from the slab ends.

Example 2.4.1 Initial CamberUsing the generic hollow core slab defined in

section 1.7, calculate the initial camber given thefollowing:

Prestressing steel: 4-1/2″ dia., 270 ksi, low re-laxation strands

Apsfpu = 0.153(270) = 41.3k/strand

Initial stress: 70% fpu

dp = 7″

ℓ = 30′-6″

Solution:Estimate initial losses at 5% and use Eci = 3250

ksiPo = 0.95(0.7)(4)(41.3) = 109.9k

camber =109.93.89− 1[30.5(12)]2

832501224.5

--530.0535(30.5)41728

38432501224.5

= 1.34 -- 0.79

= 0.55″ Say 1/2″ to 3/4″ initial camberEstimating long term effects is complicated be-

cause, as time passes, the prestressing force de-creases due to losses and the modulus of elasticityof the concrete increases with concrete strengthgain. Traditionally, a creep factor of 2.0 has beenapplied to instantaneous deflections to estimatethe additional deflection due to creep. This hasbeen modified by Martin6 for prestressed con-crete. Table 2.4.1 presents suggested multipliersto determine both long term final deflections andposition at erection. It should be noted that in us-ing these multipliers, a total deflection is calcu-lated rather than the additional increment due tolong term effects.

Example 2.4.2 Long Term Camber

2--14

For the slab of Example 2.4.1, determine thenet camber at erection and the final camber.Table 2.4.2 Maximum Permissible Computed Deflections1

Type of member Deflection to be considered Deflection limitation

Immediate deflection due to live load L ℓ180

Floors not supporting or attached to non-structural elements likely to be damaged bylarge deflections

Immediate deflection due to live load L ℓ360

*

***

****

ℓ480

ℓ240

Roof or floor construction supporting orattached to nonstructural elements likely tobe damaged by large deflections

That part of the total deflection occurring afterattachment of nonstructural elements (sum ofthe long-term deflection due to all sustainedloads and the immediate deflection due to anyadditional live load)**Roof or floor construction supporting or at-

tached to nonstructural elements not likely tobe damaged by large deflections* Limit not intended to safeguard against ponding. Ponding should be checked by suitable calculations of deflection, including added deflections due to pondedwater, and considering long-term effects of all sustained loads, camber, construction tolerances, and reliability of provisions for drainage.

** Long-term deflection shall be determined in accordance with 9.5.2.5 or 9.5.4.2, but may be reduced by amount of deflection calculated to occur before attach-ment of nonstructural elements. This amount shall be determined on basis of accepted engineering data relating to time-deflection characteristics of memberssimilar to those being considered.

*** Limit may be exceeded if adequate measures are taken to prevent damage to supported or attached elements.

**** But not greater than tolerance provided for nonstructural elements. Limit may be exceeded if camber is provided so that total deflection minus camber doesnot exceed limit.

Flat roofs not supporting or attached to non-structural elements likely to be damaged bylarge deflections

Solution:At erection,initial camber = 1.34 -- 0.79

= 0.55″ from Example 2.4.1

Erection camber = 1.34(1.80) -- 0.79(1.85)

= 0.95″

Say 1″ erection camber

Final camber = 1.34(2.45) -- 0.79(2.70)

= 1.15″Say approximately 1 1/4″ final camber

2.4.2 DeflectionsAs with camber, concrete creep will also affect

deflections due to sustained superimposed loads.These long term effects must be considered forcomparison with Table 9.5(b) of the ACI Code todetermine acceptability. This table is reproducedhere as Table 2.4.2. Engineering judgementshould be used in comparing calculated deflec-tions to the ACI Code limits. Many building codespecified live loads exceed the actual loads in astructure. While it may be implied that the fulllive load be used for comparison to Table 9.5(b),situations may arise where it is more reasonable touse actual anticipated live loads for deflectioncomparisons. A further complication for super-imposed loads is that flexural cracking will reduce

the effective moment of inertia of the section.Calculations using bilinear moment-deflectionrelationships are required when tension exceeds6 f′c and are covered extensively in references 1and 2. By definition, cracking occurs at a tensilestress of 7.5 f′c . While the ACI Code requiressuch bilinear calculations when 6 f′c tension isexceeded, in effect bilinear behavior is meaning-less up to a tension of 7.5 f′c . Since hollow coreslabs are normally designed to be uncracked un-der service loads, the effects of cracking will notbe considered here.

Table 2.4.1 includes multipliers for determin-ing the long term effects for superimposed loads.Again, use of the multipliers gives an estimate oftotal deflection rather than an increment for theadditional long term deflection.

Example 2.4.3For the slab of Examples 2.4.1 and 2.4.2, deter-

mine the total deflection due to a superimposedload of 20 psf dead and 50 psf live on a clear spanof 30′-0″ including long term effects. Use Ec =4300 ksi.

Solution:

From Example 2.4.2

Final camber = 1.15″

superimposed dead load instantaneous deflection:

2--15

=50.023(30)4172838443001224.5

= 0.208″

Final deflection = 0.208 (3.0) = 0.62″

Instantaneous live load deflection:

=50.053(30)4172838443001224.5

= 0.52″

Final position

final camber = + 1.15″sustained dead load = -- 0.62

net camber + 0.53″live load increment = -- 0.52

+ 0.01″

For comparison to the provisions of Chapter 9of the ACI Code, when non-structural elementsare attached to the slabs, the portion of deflectionafter erection may be used for comparison.Change in camber = 1.15″ -- 0.95″ = + 0.20″Sustained dead load = -- 0.62″Instantaneous live loads = -- 0.52″

-- 0.94″When a composite topping is used, it will be

cast after a portion of the slab shrinkage has oc-curred. There will then be differential shrinkagebetween the topping and slab. This differentialcan cause additional deflection and bottom tensilestress. These effects will generally be negligible.

Example 2.4.4 Composite SlabGiven the slab of Example 2.4.3, add a 2″ com-

posite topping and recalculate deflections includ-ing the affects of differential shrinkage.Solution:

Final camber = 1.34 x 2.20 -- 0.79 x 2.40

= 1.05″

Instantaneous topping weight deflection:

=50.0253(30)4172838443001224.5

= 0.26″

Long term deflection due to topping weight

= 0.26″ (2.30) = 0.60″

Superimposed dead load deflection:

=50.023(30)4172838443002307

= 0.11″(Note: 2307 in.4 = composite moment of inertiausing a 3000 psi topping on a 5000 psi slab.)Long term dead load deflection

= 0.11(3.0) = 0.33″Instantaneous live load deflection:

= 5020

(0.11) = 0.28″

Final Position = +1.05 -- 0.60 -- 0.33 -- 0.26 =--0.14″ including instantaneous live load.Calculate increment due to differential shrinkageassuming shrinkage strain of 500 × 10--6 in/in inboth the topping and slab:

If total shrinkage = 500 × 10-6

and erection shrinkage = 250 × 10-6

differential shrinkage = 250 × 10-6

The differential shrinkage can be thought of asa prestress force from the topping where

P = Atopping (strain)(modulus)

= 36″(2″)(0.00025)(3320)

= 59.8kThe effect is lessened by concrete creep and,

using a factor of 2.30 from Table 2.4.1, reduces to:P = 59.8/2.30 = 26k

The eccentricity of this force is:e = 9″ -- 3.89″

= 5.11″

M = Pe = 26 x 5.11 = 133 in-k

downward deflection = Mℓ2

8EI

= 133(30x12)2

843002307

= 0.22″ ≅ 1/4″Considering the span used in this example and

the accuracy of the other camber and deflectioncalculations, it can be easily seen that differentialshrinkage will generally not be significant.

2.5 Composite DesignA composite, structural concrete topping is

commonly used in floor construction with hollow

2--16

core slabs. The composite action is desirable toadd stiffness and strength for gravity loads andmay also be required for load transfer within a dia-phragm. When a composite topping is used, con-sideration must be given to its strength, detailingand quality assurance.

The required compressive strength of the top-ping may be determined from the hollow core slabdesign requirements. Load tables provided by lo-cal producers will normally indicate that either a3000 psi (20.7 MPa) or 4000 psi (27.6 MPa) con-crete is required. Diaphragm requirements maynecessitate a higher strength topping concrete.

From a detailing standpoint, the primary con-sideration is that hollow core slabs will have cam-ber. If the topping is finished as a level surface, thecamber will reduce the topping thickness in themidspan region which will affect the load capacityof the slabs. With significant topping thicknessreduction, the integrity of the topping concretemay also be compromised. A preliminary slab de-sign can provide an estimate of camber and theminimum topping thickness necessary to supportthe design loads. The first option is to provide theminimum thickness topping at midspan and allowthe thickness to increase at the slab ends to main-tain a flat floor. Finish and bearing elevations canthen be set to this criteria.

A second option to minimize topping concretevolume is to allow the minimum topping thick-ness to follow the curvature of the slabs. This willresult in a finished floor with camber which maybe acceptable in some occupancies. In this option,it is important that all trades be made aware of thefinal camber as it may affect their work. Parti-tions, doorways and stairs will be particularly af-fected in this option.

When control joints are used in a structural top-ping, they should be located over the joints in theprecast units below where cracks would most nat-urally occur in the topping. At the ends of slabs,where movement will occur due to camberchanges, deflections, creep, shrinkage or elasticshortening, control joints are desirable.

Reinforcing of a topping may be required forstructural design. If not, consideration should begiven to using minimum shrinkage reinforcementfor crack control.

Since the composite topping and hollow coreslabs interact to create the final structural element,it is imperative that the topping bond well with theslabs. While the building designer may only be in-terested in the final product, the process of achiev-ing a well bonded, composite topping is very im-portant. The hollow core producer is dependenton a properly bonded topping, yet is not involvedin specifying, designing or installing the topping.The hollow core producer is responsible for sup-plying a slab that is capable of bonding with a top-ping. The installer of the topping is responsiblefor surface preparation, topping concrete mix de-sign and curing to assure proper bond.

At a minimum, the slab surface must be cleanand damp at the time of topping installation. It isrecommended that the surface be thoroughly satu-rated prior to topping placement, but all standingwater must be removed. ACI 301-967 specifiesthat a sand and cement grout be scrubbed into theslab surface ahead of topping placement. If thisprocedure is used, it is imperative that initial setnot be allowed prior to topping placement. If ini-tial set occurs, the grout can become a bond break-er. Similarly, bonding agents, which are rarelyspecified, will also act as a bond breaker if any ini-tial set occurs prior to topping placement.

The topping concrete mix and curing tech-niques will also affect bond of a composite top-ping. Curling at topping edges or joints will causelocal delamination. Curling is a result of differen-tial shrinkage between the top and bottom sur-faces of the topping. Generally, water is lost morequickly from the top surface causing additionaldrying shrinkage. This can be minimized by prop-er curing techniques and low shrinkage concrete.

Design of hollow core slabs for composite ac-tion is usually limited to a horizontal shearstrength of 80 psi (0.5 MPa) according to section17.5.2.1 of ACI 318-95. Through limited pub-lished8 and unpublished testing, the machine fin-ished surface has been found to meet the require-ments of that section. The horizontal shear checkshould be based on the shear diagram rather thanusing an average horizontal shear over the dis-tance from zero moment to maximum momentwhen checking compliance with the 80 psi limit.

Composite ties are not normally provided giv-en the difficulty and expense of installing the ties

2--17

in a machine casting operation. When the horizon-tal shear exceeds 80 psi (0.5 MPa) and compositeties are not used, the topping is considered to besuperimposed dead load on a non-composite slab.In a wet cast system, horizontal shear ties with 1/4in amplitude roughening may be used to take ad-vantage of the higher stresses allowed by ACI.

Design of a composite section is similar to thatpresented in Sections 2.2 and 2.3. The followingexample demonstrates the additional consider-ations with a composite section.

Example 2.5.1 Composite DesignUsing the generic hollow core cross-section de-fined in Section 1.7, add a 2 in structural toppingand check for the following conditions:Prestressing steel: 4-1/2″ dia., 270 ksi low relax-ation strands

Initial stress: 70% fpu

dp: 7 in

Slab: f′c = 5000 psi

Eci = 3250 ksi

Ec = 4300 ksi

Topping: f′c = 3000 psi

Ec = 3320 ksi

Slab length: 30′-6″

Slab span: 30′-0″

Loads: topping = 25 psfdead load = 20 psflive load = 50 psf

Calculate section properties:

Base section A = 154 in2

I = 1224.5 in4

yb = 3.89 in

Topping

n = 3320/4300 = 0.77

use width = 0.77(36)= 27.7 in

Composite

A = 154 + 2(27.7) = 209.4 in2

yb = 154(3.89)+ 2(27.7)(9)209.4

= 5.24 in.

I = 1224.5 + 154(5.24 -- 3.89)2

+ 23

12(27.7) + 2(27.7)(9 -- 5.24)2

= 2307 in4

Calculate prestress losses:

From Example 2.2.3.1

ES = 7.52 ksi

Concrete creep

Msd = 302

8(0.025 + 0.020)(3)

= 15.19 ft-k

fcds = 15.19(12)(2.89)1224.5

= 0.430 ksi

CR = (2.0) 285004300

(0.857 -- 0.430)

= 5.66 ksi

SH = 6.27 ksi

RE = 50001000

− 0.04(6.27+ 5.66+ 7.52)0.75

= 3.17 ksi

Loss = 7.52 + 5.66 + 6.27 + 3.17

= 22.62 ksi = 12%

Calculate service load stresses:

Apsfse = 0.7(4)(41.3)(1 -- 0.12)

= 101.8k

Mnon--comp = 302

8(0.0535 + 0.025)

= 8.83 ft-k/ft = 106 in-k/ft

Mcomp = 302

8(0.020 + 0.050)

= 7.88 ft-k/ft = 94.5 in-k/ft

At top of topping

ftop = 94.5(3)(10− 5.24)2307

(0.77)

= 0.450 ksi

At top of slab

ftop = 101.8154

− 101.8(2.89)(4.11)1224.5

+ 106(3)(4.11)1224.5

+ 94.5(3)(8− 5.24)2307

= 1.080 ksi

2--18

At bottom of slab

fbottom = 101.8154

+ 101.8(2.89)(3.89)1224.5

− 106(3)(3.89)1224.5

− 94.5(3)(5.24)2307

= --0.058 ksi

Calculate flexural strength

wu = 1.4(0.0535 + 0.025 + 0.020)+ 1.7(0.050)

= 0.223 ksf

Mu = 302

8(0.223)(3)

= 75.26 ft-k

Using ACI Eq. (18-3)

ρp = 4(0.153)36(9)

= 0.0019

fps = 2701− 0.280.85

0.0019 2703

= 254.8 ksi

a = 4(0.153)(254.8)0.85(3)(36)

= 1.7 in

φMn = 0.9(4)(0.153)(254.8)9− 1.72

= 1144 in-k = 95.3 ft-k

Check 1.2 Mcr

fbottom = 101.8154

+ 101.8(2.89)(3.89)1224.5

= 1.596 ksi

Mcr = 23075.241.596+ 7.5 5000

1000

= 936 in-k

ÔMnMcr

= 1144936

= 1.22 > 1.2 ok

Check horizontal shear:

φVnh = φ80bvd

= 0.85(80)(36)(9)

= 22030 lb

= 22 k

at h/2

Vu = 302− 10

2(12)(0.223)(3)

= 9.8 k < 22 k ok

Section is composite

Check web shear at h/2:

transfer length = 50(0.5) = 25 in

at h/2 plus 3 in bearing

Apsfse = 101.8 825 = 32.6 k

for composite section, fpc is calculated at centroidof composite section

fpc = 32.6154

− 32.6(2.89)(5.24− 3.89)1224.5

= 0.108 ksi

φVcw = 0.853.5 50001000

+ 0.3(0.108)(10.5)(9)

= 22.5k > 9.8 k ok

Check inclined shear at 4 ft

Vu = 302− 4(0.223)(3)

= 7.36k

Vd = 302− 4(0.0535+ 0.025+ 0.020)(3)

= 3.25k

Vi = 7.36 -- 3.25 = 4.11k

Mu = 0.223(3)(4)302− 4

2 = 34.8 ft-k

Md = (0.0535 + 0.025 + 0.020)(3)(4)302− 4

2

= 12.25 + 3.12 = 15.37 ft-k

Mmax = 34.8 -- 15.37 = 19.43 ft-k

fpe = 101.8154

+ 101.8(2.89)(3.89)1224.5

= 1.596 ksi

fd = 12.25(12)(3.89)1224.5

+ 3.12(12)(5.24)2307

= 0.552 ksi

Mcr = 23075.246 5000

1000+ 1.596− 0.552

= 646 in-k = 53.9 ft-k

2--19

φVci = 0.850.6 50001000

(10.5)(9)+ 0.853.25+ 4.11(53.9)

19.43

= 15.86k > 7.36k ok

2--20

Fig. 2.6.1.1St

eel S

tres

s fps

sef

t f

d

Length into span

Strand Development

2.6 Strand Development

2.6.1 ACI RequirementsSection 12.9 of the ACI Code covers develop-

ment length for prestressing strands. While thetopic has received considerable discussion9-16,the ACI Code expression currently remains:

ℓd = (fps -- 2/3fse)db

A further requirement is that the developmentlength shall be doubled when bonding of a stranddoes not extend to the end of the member and theprecompressed tensile zone is allowed to be intension at service loads.

The ACI Code expression for developmentlength describes two bond mechanisms. The firstis the transfer length which is the bond length re-quired to transfer the effective prestress afterlosses, fse, to the concrete. This portion of the de-velopment length is:

ℓt = fse3

db

With fse equal to 150 ksi (1034 MPa), the trans-fer length becomes 50db, the length used for shearcalculations.

The second mechanism is for bond length afterthe steel stress increases above fse. To develop thefull design strength of the strand, fps, a bond lengthin addition to the transfer length is required. Theflexural bond length is expressed as:

ℓf = (fps -- fse)db

Figure 2.6.1.1 depicts the increase in steelstress along the development length of the strand.

Fig. 2.6.1.2

fps

d

psf req’d.greater thanf availableps

Section 12.9.2 of the ACI Code limits inves-tigation of development length to the section near-est the end of the member where full designstrength is required. In conventionally reinforcedconcrete, the rate of moment increase must beconsidered in selecting reinforcing bar sizes. Thisconsideration is also valid in prestressed concretemembers. As shown in Figure 2.6.1.2, with asteep rate of moment increase, critical sectionsmay occur in the strand development length at lessthan maximum moment.

Demand on strand strength above fse does notoccur until after flexural cracking occurs. If flex-ural cracking occurs in the transfer length, thestrand cannot accept additional stress so bond fail-ure occurs. Therefore, the limit on member flexu-ral strength in the strand transfer length is thecracking moment.

In the flexural bond length, strand stress can in-crease above fse, but not to full fps. Therefore,there is additional flexural strength above thecracking moment, but less than full nominalstrength. If flexural cracking occurs at factoredload in the flexural bond length, the maximumvalue for fps can be calculated as:

f′ps= fse +x−ℓt

ℓf(fps -- fse)

where x = the distance from the end of themember to the section of interest

The nominal moment capacity is then calculatedon the basis of this maximum strand stress.

Martin and Korkosz17 suggest that with partial-ly developed strand, the full concrete compressive

2--21

failure strain will not be achieved. A strain com-patibility analysis can be performed to determinethe concrete strain that would be consistent withf′ps and nominal strength can then be calculatedusing that strain.

When debonded strands are mixed with fullybonded strands, a similar strain compatibilityanalysis may be required in the flexural bondlength for the debonded strands. In this case,nominal strength can be calculated in two ways:1. Analyze section with all strands at the f′ps for

the debonded strands.

2. Analyze section with only fully bonded strandsat their fps and ignore the debonded strands.

The greater of the two results would predict thenominal strength of the section.

For hollow core slabs, the strain compatibilityanalysis for partially developed strand will yieldvariable results as compared to a traditional ap-proach where f′ps is used with a full concrete strainof 0.003 in/in. If f′ps is close to fse, the strain com-patibility analysis will predict moment capacity ofabout 85% of the traditional analysis. When f′ps is10% greater than fse, the difference reduces to 5%or less. The additional complexity of the straincompatibility analysis would only seem war-ranted when flexural cracking is expected near thetransfer point or when debonded strands are used.

There are several aspects of a bond length dis-cussion that are significant to hollow core slab de-sign. In many framing schemes, there will be a re-quirement to use very short slabs to fill in an area.With fully developed strands, these slabs will nor-mally have very large load capacities. However,capacity may be reduced because the strandsmight only be partially developed. For example,for a slab prestressed with 1/2″ (12.7 mm) φ, 270ksi (1860 MPa) strands with fse = 150 ksi (1034MPa) and fps = 260 ksi (1790 MPa):

ℓd = fps − 23

fsedb

= 260− 231500.5

= 80″ = 6′-8″ (2030 mm)This slab would have to be two development

lengths, or 13′-4″ (4.1 m) long in order to developits full design strength. A shorter slab would havereduced capacity.

Hollow core slab systems are often required tocarry concentrated or wall loads which may affectthe rate of moment increase near the member end.While not required by ACI, it is suggested that thetransfer length and flexural bond length regionsbe investigated for reduced capacity when the mo-ment gradient is high.

The development length equations in the ACICode are based on testing conducted with mem-bers cast with concrete having normal water-ce-ment ratios. As noted in the Commentary to theACI Code, no slump concrete requires extra pre-cautions. Hollow core slabs produced with the ex-trusion process fall into this category. As original-ly presented by Anderson and Anderson10 andreinforced by Brooks, Gerstle and Logan18, ameasure of satisfactory bond is the free end slip ofa member after it is cut to length. A limit on freeend slip expressed as:

δall =fsefsi6Es

db

has been suggested as a maximum free end strandslip for using the ACI Code development lengths.This expression approximates the strand shorten-ing that would have to occur over the transferlength. For a 1/2″ (12.7 mm) dia. strand stressedinitially to 189 ksi (1300 MPa), the free end slipshould not exceed about 3/32″ (2.4 mm) if the ACICode transfer and development lengths are to beused.

When free end slip exceeds δall, the transferlength and the flexural bond length will increase.Shear strength in the transfer length and momentcapacity in the flexural bond length will be de-creased and the length into the span where fullmoment capacity is provided will be increased.

If the free end strand slip is known from qualitycontrol measurements, the member capacity canbe evaluated with consideration of extendedtransfer and flexural bond lengths. As a functionof measured end slip, the transfer length and flex-ural bond length can be calculated for each strandas follows:

ℓt = 2δsEs/fsi

ℓf = 6δsEs(fps -- fse)/(fsifse)

Shear strength can be evaluated by substitutingthe extended transfer length for 50 db in evaluat-ing the rate of increase of prestress. Flexural

2--22

Fig. 2.6.1.3 Effect of End Slip

0.0 5.0 10.0 15.0 20.0 25.0 30.0

80.00

60.00

40.00

20.00

0.00

Mom

ent (

ft-k

)

Distance Along Member Length (ft)

Uniform LoadMoment Diagram

Moment Capacitywith 5/32" End Slip

Moment Capacitywith Normal End Slip

Distance Along Member Length (ft)

Mom

ent (

ft-k

)

(b)

Moment Capacitywith 5/32" End Slip40.00

0.0

20.00

0.00

Uniform LoadMoment Diagram

5.0 10.0

60.00

80.00

15.0

Moment Capacitywith Normal End Slip

20.0 25.0

(a)

strength calculations are affected only by the ex-tension of the strand development length and po-tential reduction of f′ps. The strain compatibilityanalysis suggested by Martin and Korkosz forsections with partially developed strand becomesmore complex as there can be variation in devel-opment lengths within a given member.

Figure 2.6.1.3 illustrates the change in momentcapacity for the generic slab of Section 1.7 fromnormal slip to 5/32 in (4 mm) slip on all strands. In(a), the span length is 30 ft (9.1 m) and there wouldbe no change in slab capacity for uniform load. In(b), the span is reduced to 25 ft (7.6 m) and it isclear that the extended development length wouldresult in reduced capacity even with uniform load.End slip in excess of normal slip has a more signif-icant effect in shorter slabs.

The following example demonstrates the use ofthe Martin and Korkosz strain compatibility anal-ysis for partially developed strand and the use offree end slip for evaluating strength. The proce-dure illustrated is also valid with normal end slipby using the appropriate transfer and bondlengths.

Example 2.6.1.1 Initial Strand SlipGiven the generic hollow core slab defined in

Section 1.7, calculate the design flexural strengthgiven the following:Prestressing steel: 4-1/2″ dia., 270 ksi low

relaxation strands.Es = 28500 ksidp = 7″f′c = 5000 psi

2--23

fsi = 185 ksifse = 163.4 ksifps = 267 ksiδs = 3/16 in. all strandsSolution:

ℓt = 2(3/16)(28500)/185= 57.8″

ℓf = 6(3/16)(28500)(267 -- 163.4)/185/163.4= 109.9″

ℓd = 57.8 + 109.9= 167.7″The minimum slab length required to achieve

full flexural capacity is 2(167.7)/12 or 28 ft. Cal-culate flexural capacity at 10 ft.

f′ps = 163.4 +10x12− 57.8

109.9(267 -- 163.4)

= 222 ksiApsf′ps = 4(0.153)(222)

= 135.9kTraditional analysis

a = 135.9.85536

= 0.89 in.

Mn = 135.9(7 -- 0.89/2)/12= 74.24 ft-k

Strain compatibility analysis

dp

s

CC

T = Apsf’ps

c21

cE c

εps = εse + εs

εse = 163.4/28500= 0.00573 in/in

εps = 222/28500= 0.00779 in/in

εs = 0.00779 -- 0.00573= 0.00206 in/in

Using trial and error for

T = C

Find

c = 2.18″

εc = 0.000929 in/in

Concrete stress at top

= 4300(0.000929)

= 3.995 ksi

Concrete stress at top of core

=2.18− 1.25

2.18(3.995) = 1.704 ksi

C1 =3.984+ 1.704

2(1.25)(36)

= 128k

C2 = 1.7042

(10.5)(2.18 -- 1.25)

= 8.3k

Mn = (135.9(7 -- 0.54) -- 8.3(1.56 -- 0.54))/12

= 72.45 ft-k

CHAPTER 3

3--1

SPECIAL DESIGN CONSIDERATIONS

3.1 General InformationThe application of hollow core slabs as roof

and floor deck members creates several situationsfor consideration in design which are either notcompletely covered by ACI Code provisions orwhich involve consideration of production pro-cesses. This section presents information whichmay be used as a guideline for the situations de-scribed but not as hard and fast rules. The criteriapresented represent conservative practices andshould be verified with local producers. Pub-lished data relative to each situation is referenced.However, extensive in plant testing has been con-ducted by hollow core producers which may al-low less conservative criteria to be used becauseof the unique characteristics of a particular slab.

3.2 Load DistributionAs demonstrated in Chapter 2 of this manual,

hollow core slabs are designed as individual, oneway, simple span slabs. When the slabs areinstalled and grouted together at the keyways, theindividual slabs become a system that behavessimilarly to a monolithic slab. A major benefit ofthe slabs acting together is the ability to transferforces from one slab to another. In most hollowcore slab deck applications, non-uniform loadingoccurs in the form of line loads, concentratedloads, or load concentrations at openings. Theability of individual slabs to interact allows theseload concentrations to be shared by several slabs.The ability to distribute loads among several slabshas been demonstrated in several published tests19-25 and many unpublished tests.

In many cases, load concentrations do not haveto be carried by the slabs. For example, a header ata large opening may be supported directly to afoundation or vertical support element; a beammight be installed to directly carry a heavy con-centrated load; or a heavy wall parallel to a slabspan might be designed to carry its own weight orany load superimposed on the wall as a deep beamspanning between vertical supports. However,when such loads must be supported by the slabsystem, a method is required to decide how many

slabs will contribute in carrying a given load in agiven location. This section presents a designmethod that may be used when the slabs do actual-ly have to support non-uniform loads.

3.2.1 Load Distribution MechanismsAs load is applied to one slab in a system, the

response of the slab system is to deflect and alsotwist if the load is not on the longitudinal center-line of the system. As the loaded slab edges try tomove down, the interlock of the grout in the jointswith the keyways formed in the slab edges forcesadjacent slabs to deflect a similar amount. Theflexural and torsional stiffness of the adjacentslabs reduce the deflection of the loaded slab fromwhat might be expected if the slab were alone.Shear forces are developed along the keyways andthe loaded slab then gets some support from theadjacent slabs. As this effect trickles through thesystem, the keyways between slabs force equaldeflections for slab edges at any given keyway.

Many times shrinkage cracks will occur in thegrouted joints at the interface between the groutand slab edge. This cracking does not impair themechanism described above because the configu-ration of the keyways in the slab edges still pro-vides mechanical interlock even with the presenceof a crack.

Shear forces transferred along keyways causetwo sets of forces that are normally not consideredin hollow core slab design. The first is torsionwhich develops because the shear on one edge of agiven slab is different in magnitude from the shearon the opposite edge. As depicted in Figure 3.2.1,the keyway shears reduce as the distance from theload increases. These torsions cause shear stressin the slabs in addition to the direct shear stress.

The second set of forces is induced because thesystem is tending to behave as a two way slab.Transverse bending moments occur because ofthe edge support provided by adjacent slabs. Theresult is transverse tensile stress developed in thebottom of the slab and compressive stress in thetop. Hollow core slabs are not provided withtransverse reinforcement. Transverse tensile

3--2

Fig. 3.2.1

V1 > V2 > V3

V1 V1 V2 V2 3V

stresses must then be resisted by plain concrete.The magnitude of load concentration causing thetransverse tension must be limited to preclude asplitting failure. See Section 3.2.2.

Several factors affect the ability of a slab sys-tem to distribute loads to adjacent slabs. As thewidth of an assembly of slabs gets narrower thanthe span length, a reduction in the number of slabscontributing to the support of a concentration ofload occurs. This occurs because the freedom ofthe free edges of the system to deflect and twist be-comes more significant. A second factor is thespacing of the slab joints. With slabs available inwidths ranging from 2 feet to 8 feet (0.6 to 2.4 m),some differences in load distribution behavior canbe expected. Finally, the span length affects thenumber of contributing slabs. As span lengthchanges for a wide system, the interaction of flex-ural and torsional stiffnesses changes. For longerspans, flexural stiffness reduces relative to tor-sional stiffness. This results in relatively less slabrotation and less transverse curvature. The resultis that more slabs can contribute to distribution onlonger spans as long as the system is wide relativeto its length.

3.2.2 Design GuidelinesACI 318-95 recognizes the load transfer capa-

bilities of hollow core slabs in Section 16.3.1.That section specifies that distribution of forcesbe established by analysis or test. The guidelinespresented here were based on extensive, full scaletesting of a specific slab system. Additionally, acomparison of these guidelines to an analyticalstudy has been done. Therefore, the guidelinespresented here should satisfy the requirement ofACI (318-95) 16.3.1.

The two basic design parameters consideredfor hollow core slabs are flexure and shear. De-sign for flexure is straightforward with the effec-tive load resisting width being a function of thespan length. Conversely, shear design is compli-cated by torsions developed in the system. If tor-sion is not to be used as a design parameter, directshear must then be modified to reflect the increasein shear stress from the torsion.

Figure 3.2.2 depicts a method of establishingan effective resisting section for any type of loadto be distributed between slabs. In the midspan re-gions, the effective width is defined as a functionof span length. At the supports, the effectivewidth is defined as an absolute width. The widthat the support is restricted to account for shearstresses due to torsion. Use of these resisting sec-tions will result in prediction of peak values ofmoment and shear. That is, the effective widthconcept is simply a mechanism used to determinethe maximum design moments and shears ratherthan a depiction of the actual load path through thesystem.

The performance of slab systems indicates thatshear and moment might affect additional slabs.For example, for a load located some distancefrom a free edge, the peak moment due to that loadcan be predicted by assuming the load is resistedby a width equal to 0.50ℓ. In reality, in flexure, atotal width equal to 85% to 90% of the span lengthmight have some moment attributable to that load.In shear, the 1′-0″ (0.3m) effective section at thesupport at a free edge may be used to predict thepeak shear but, because of torsion, the total reac-tion due to an edge load will not actually be con-centrated in the 1′-0″ (0.3m).

Several limitations should be recognized forFigure 3.2.2.

3--3

Fig. 3.2.2 Effective resisting width of slab for load anywhere along span

0.50

1 -0

Interpolate

0.25

4 -0

Interpolate

0.50

0.25

0.25

’ " ’ "

1) As the width of the system becomes narrow-er than the span length, the effective resist-ing widths will become narrower.

2) For extremely high span-depth ratios (in ex-cess of approximately 50), the effective sec-tion at midspan may be reduced by 10 to 20percent.

3) For spans less than about 10 ft, the effectivewidth at the support may become narrower.

4) Local load concentrations can cause longi-tudinal splitting failures due to transversebending in the system. Punching shear typefailures can also occur. The magnitude ofconcentrated loads must be limited to pre-clude such failures. These limits are best es-

tablished by test for each slab system. Ref-erence 23 provides guidance for edge loads.

The concept of using an effective resisting sec-tion is subtlely different from the traditional con-cept of load distribution width. Traditionally,loads have been divided by distribution widths fordesign. Using an effective resisting section meansthat a given load is resisted by a varying width de-pending on the location of the section being inves-tigated in the span. This is best illustrated by ex-ample.

Example 3.2.1 General CaseGiven an untopped hollow core system using

36″ wide slabs as shown in Figure 3.2.3, deter-mine the slab design loads. Slab weight = 53.5 psf

3--4

Fig. 3.2.3.

DL = 10 psfLL = 40 psf

w1D = 650 #/ftw1L = 1040 #/ft

P1D = 500 #P1L = 1000 #

P2D = 1000 #P2L = 3000 #

25 -

0P

P

w

w

P

P

5 -6

5 -6

9 -6

9 -6

6 -0

1

2

2

1

1

1

’"

’"

"’

"’

"’

’"

Solution

Step 1: Evaluate the shear and moment diagramsfor the non-distributable loads.

wu = 1.4(53.5 + 10) + 1.7(40) = 157 psf

Vx = w (ℓ/2 -- x) = 0.157(25/2 -- x)

Mx = w x2

(ℓ− x) = 0.157x2

(25− x)

See Table 3.2.1

Step 2: Evaluate the shear and moment diagramsfor the distributable loads.

wu = 1.4(650) + 1.7(1040) = 2678#/ft

Pu = 1.4(500) + 1.7(1000) = 2400#

Pu = 1.4(1000) + 1.7(3000) = 6500#

See Table 3.2.1

Step 3: Evaluate the effective width along thespan.

At supportDW = 4.0 ftAt 0.25ℓ = 0.25(25) = 6.25 ft

DW = 0.5ℓ = 0.5(25) = 12.5 ft

Between x = 0 and x = 6.25 ftDW = 4 + x

6.25(12.5 -- 4)

= 4 + 1.36xSee Table 3.2.1

Step 4: Divide the shears and moments from Step2 by the effective width from Step 3 andadd to the shears and moments in Step 1.

See Table 3.2.1

Step 5: Design the slabs for the web shear, in-clined shear, and moments obtained fromStep 4.

The solution for the general case where theshears and moments are calculated at intervalsalong the span is best suited for use with a comput-er. The information could then also be used to cal-culate shear strength at the same time.

For many cases, a general solution is not neces-sary. Simplifying shortcuts can be used to shortenthe design process. Consider the case where shearis known not to be a problem.

3--5

Table 3.2.1 Shears and Moments for Example 3.2.1

0 1.96 0 34.34 0 4.0 10.55 0h/2 1.91 0.65 33.45 11.28 4.45 9.43 3.18

1 1.81 1.88 31.66 33.0 5.36 7.72 8.042 1.65 3.61 28.98 63.32 6.72 5.96 13.033 1.49 5.18 26.31 90.98 8.08 4.75 16.444 1.33 6.59 23.63 115.94 9.44 3.83 18.875 1.18 7.85 20.95 138.23 10.80 3.12 20.656 1.02 8.95 15.87 156.64 12.16 2.32 21.837 0.86 9.89 13.19 171.17 12.5 1.92 23.58

10 0.39 11.78 0 195.78 12.5 0.39 27.4411 0.24 12.09 0 195.78 12.5 0.24 27.7512.5 0 12.27 0 195.78 12.5 0 27.93

Non-distributedLoads

DistributableLoads

EffectiveWidthDWx

Final

x Vux Mux Vux Mux Vu(k/ft) Mu(ft-k/ft)

Example 3.2.2

D = 250 plf

D = 10 psfL = 40 psf

Effective widthSlab wt. = 53.5 psf

25 -

0’

"

Given the system shown, determine the designload.

Solution:

Shear is judged to be not criticalFrom Figure 3.2.2 the effective width resisting

the line load is 0.50ℓ = 0.50(25) = 12.5 ft

Determine the design superimposed load:

w = 40 + 10 + 250/12.5

= 70 psf

Using the generic slab load table in Figure 1.7.1select an 8″ slab with 4 - 7/16″ diameter strands.

If it is not known whether shear is critical, sim-ple iterative checks may be made.

Example 3.2.3

D = 250 plf


Effective widthSlab wt. = 53.5 psf

L = 200 plf

25 -

0’

"

Given the system shown select a generic slabfrom Figure 1.7.1 to support the loads shown.

Solution:

Make preliminary selection based on flexure:

Superimposed w = 10+ 40+250+ 2000.5025

= 86 psf

Select 4 - 7/16″, 270 ksi low relaxation strandsfrom Figure 1.7.1

First shear checkeffective width at support = 4′-0″ = DW

wu = 1.4(10 + 53.5) + 1.7(40)+ (1.4 x 250 + 1.7 x 200)/DW

= 157 + 690/DW

Using DW = 4.0

3--6

Example 3.2.4

Typ. wall loadD = 300 plfL = 400 plf

Effectivewidth Typ. point load

D = 1800#L = 2880#

10 10

25 -

0

9 -6

6 -0

9 -6

Uniform loads: slab wt = 53.5 psfD = 10 psfL = 40 psf

’"

’ ’

’"

’"

’"

wu = 157 + 690/4.0 = 330 psf

Check shear based on this load and find@ h/2 Vu = 4.02 k/ft and φVcw = 6.04 k OK

@ 3.0 ft Vu = 3.13 k/ft and φVci = 3.05 k NG

Second Shear CheckInclined shear did not check at 3.0 ft so deter-

mine effective width at 3.0 ft, recalculate distrib-uted load and recheck shear.

At ℓ/4, DW = 0.5ℓ = 0.5(25) = 12.5 ftAt support, DW = 4.0 ft

Interpolate at 3 ft, DW = 325∕4

(12.5 -- 4) + 4

= 8.08 ftwu = 157 + 690/DW

= 157 + 690/8.08

= 242 psf

Again check shear at 3.0 ft and beyond and find

φVci > Vu at all points.

Therefore, shear check is complete and slab is ad-equate.

To summarize the steps taken to check shear inExample 3.2.3, distributable loads were dividedby the effective width at the support to make aconservative shear check. If shear along the spanis found to be satisfactory, no further steps are re-quired and the shear check is complete. If shear inthe span at some point is found to be inadequate,the effective width at that point is used to calcu-lated a new load which will then be conservativefor points further into the span. Shear is re-checked. This iterative approach is used until allpoints further into the span check for shear. Ifshear works for a given situation, generally nomore than three cycles will be required.

A combination of loads will be used to furtherdemonstrate this method in the following exam-ple.

Example 3.2.4Given the center bay of an apartment building

as shown, design for the applied loads using thegeneric slab shown in Figure 1.7.1

3--7

Solution:

Select preliminary slab based on flexure:

Use DW = 0.50ℓ = 0.5(25) = 12.5 ftBecause wall and point loads are spaced closer

than 12.5 ft, conservatively use spacing of loads asDW.At design strip:

1758# 1758#

50

468#

70

468#

70

Point loads =1800+ 2880

10= 468 plf

Parallel walls =300+ 400

10= 70 psf

Uniform load = 10 + 40 = 50 psf

M = 11,511 ft-#/ft

equivalent uniform load = 8 x 11511/252

= 147 psf

Select 4-1/2″, 270 ksi, low relaxation strandscapacity = 148 psf at 25 ft.

Check shearFor design stripincluding slab wt.

wu = 1.4(10 + 53.5)) + 1.7(40)+ (1.4 x 300 + 1.7 x 400)/DW

= 157 + 1100/DW

Pu = (1.4 x 1800 + 1.7 x 2880)/DW

= 7416/DW

Start at support where effective width = 4.0 ftwu = 157 + 1100/4 = 432 psf

Pu = 7416/4 = 1854 plf

Obtain the following results:

x h/2 1.09′ 1.85′ 2.61′ 3.37′

Vu k/ft 7.11 6.93 6.45 6.12 5.79

φVn k/ft 6.30 7.79 8.44 6.09 4.79

Note that web shear at h/2 does not work. Nofurther modifications can be made to adjust theshear calculation. Shear enhancement is requiredin the form of stirrups, solid cores, higher concretestrength or using a deeper section.

Proceed to check inclined shear which was notadequate at 2.61 ft.

Recalculate effective width at 2.61 ft as:

= 2.610.25ℓ

0.5ℓ− 4+ 4

= 2.616.25

(12.5 -- 4) + 4 = 7.54′

wu = 157 + 1100/7.54 = 303 psf

Pu = 7416/7.54 = 984 plf

Obtain the following results:

x 2.61 3.37 4.14 4.90 5.66

Vu k/ft 3.97 3.74 3.51 3.28 3.05

φVn k/ft 6.07 4.77 3.94 3.36 2.94

Inclined shear is now adequate to a distance of5.66 ft into the span. Recalculate the effectivewidth at 5.66 ft.

5.660.25ℓ

0.5ℓ− 4+ 4

= 11.7 ft

Note that loads are located only 10 ft apartwhich means that design strips would start tooverlap. For this case, the maximum effectivewidth might be used as the distance betweenloads, or 10 feet, rather than 0.5ℓ.

wu = 157 + 1100/10 = 267 psf

Pu = 7416/10 = 742 plf

With these loads, it is found that Vu < φVn forthe balance of the span. Therefore, the slab se-lected is adequate except for the shear enhance-ment required for web shear as previously noted.

3--8

3.3 Effect of OpeningsOpenings may be provided in hollow core sys-

tems by saw cutting after a deck is installed andgrouted, by shoring and saw cutting, by formingor sawing the openings in the plant or by installingshort slabs with steel headers. Some typical head-er configurations are shown in Section 5.7. In lay-ing out openings for a project, the least structuraleffect will be obtained by orienting the longest di-mension of an opening parallel to a span, or bycoring small holes to cut the fewest prestressingstrands, or when several openings must be pro-vided, aligning the openings parallel to the span toagain cut the least number of prestressing strands.

For slab design, openings cause load con-centrations which may be distributed over the slabsystem as discussed in Section 3.2. As with non-uniform loads, openings cause torsion in the slabs.Therefore, the method of determining shear ade-quacy must also consider the effects of torsion onthe shear stresses. In flexure, the primary consid-erations are the length of the opening parallel tothe span and the length of strand embedmentavailable from the end of an opening to the pointof maximum moment.

Figure 3.3.1 shows some general opening loca-tions with suggested interpretations of the effec-tive resisting slab width described in Section 3.2.Local slab producers may have information whichwould allow different design approaches for theirparticular slab.

Figure 3.3.1(a) depicts a relatively small open-ing located at midspan. In flexure, the load fromthe short slabs can be resisted by slabs within0.25ℓon each side of the opening. As a guideline,if an end of the opening shown is not closer to thesupport than 3/8ℓ, there will be no special consid-erations for shear design with only uniform loads.When non-uniform loads are superimposed nearthe opening, the effective resisting section shownin Figure 3.2.2 would then be used for those non-uniform loads.

Figure 3.3.1(b) shows a similar conditionwhere an opening is located with an end closer tothe support than 3/8ℓ. In this case, shear is con-sidered as though the opening created a free edge.That is, load from the short slabs or opening willbe transmitted as an edge load to the adjacent

Fig. 3.3.1 Effects of openings

< 3/8

0.25

< 3/8

0.25

0.25

0.25

1 -0

0.25 0.25> 3/8

> 3/8

0.5

(a)

(b)

(c)

0.25

’ "1 -0’ "

1 -0’ " 1 -0’ "

slabs. The resulting torsion on the adjacent slabsrequires that a reduced effective width at the sup-port be used if torsional shear stresses are not di-rectly calculated.

Figure 3.3.1(c) depicts the extreme where anopening is located right at the end of a span.Again, the reduced shear width adjacent to theopening is required to reflect torsional shearstresses. An end opening extending less than thelesser of 0.125ℓor 4 ft (1.2m) into the span may be

3--9

neglected when considering flexure. However,some capacity reduction might be required for theslab with the opening when strand embedmentlength is less than full required development.When non-uniform loads are superimposed in thearea of an end opening, these loads should be con-sidered as being at a free edge for shear calcula-tions.

Example 3.3.1

25 -

0

11 -

611

-6

2

2


Slab wt = 53.5 psf

’"

’

’’

"

’"

Given the slab system shown, select a genericslab from Figure 1.7.1 to carry the loads givenconsidering the opening.

Solution:

Check proximity of opening to support

3/8ℓ = 0.375(25) = 9.38′11.5′ > 9.38′ no special shearconsiderations

Distribute load from strip with opening:

superimposed w = 10 + 40 = 50 psfload on strip with opening =

2(10 + 40 + 53.5) = 207 plf

distributing 1/2 of strip load each side

w = 50 +207∕20.25ℓ

= 50 + 1040.2525

= 67 psf

Select, from Figure 1.7.1, 4 - 3/8″ dia., 270 ksi,low relaxation strands

Example 3.3.2

Given the floor system shown, select a genericslab from Figure 1.7.1 to carry the loads given.

25 -

0


Slab wt = 53.5 psf3

19

’

"

’6’

’

Solution:

The ends of the opening are closer than 3/8ℓ tothe support on both ends. Therefore, consider theopening as though it were a free edge.

load on strip with openingw = 3(10 + 40 + 53.5)

= 311 plffor flexure and preliminary slab selection use ef-fective width = 0.25ℓ to each side

w = 10 + 40 +311∕2

0.2525

= 75 psf

Try 4 - 7/16″ dia., 270 ksi, low relaxation strands

Check Shear

effective width at support = 1′-0″ each sidewu = 1.4(10 + 53.5) + 1.7(40) +

3[1.4(10+ 53.5)+ 1.740)]∕2DW

= 157 + 235/DW

where DW = effective width on each sidewu = 157 + 235/1 = 392 psf

Using this load, obtain:

x h/2 1.1 1.85 2.61 3.37 4.14

Vu k/ft 4.77 4.47 4.17 3.87 3.58 3.28

φVn k/ft 6.08 7.29 6.62 4.80 3.80 3.15

Shear is adequate to 4.14 ft into span.Modify effective width at 4.14 ft

DW = 4.140.25ℓ (0.25ℓ− 1)+ 1

= 4.146.25

(6.25 -- 1) + 1

3--10

= 4.48 ft

wu = 157 + 235/DW

= 157 + 235/4.48

= 209 psf

Find shear is adequate at 4.14 ft and all points fur-ther into span. Use 4 - 7/16″ dia., 270 ksi, low re-laxation strands.

3.4 ContinuityHollow core slabs are normally designed as

part of a simple span system. However, continuityover supports can be achieved by placing rein-forcing steel in the grouted keyways, in a compos-ite structural topping, or by concreting bars intocores. Within limits, the result will be better con-trol of superimposed load deflections and a lowerrequirement for positive moment capacity.

With reinforcing steel in either a compositetopping or in cores, elastic moments with allow-ance for negative moment redistribution deter-mine the amount of reinforcing required. Becauseof the relative efficiencies of positive prestressingsteel and negative mild reinforcing, it is difficultto economically justify a continuous system de-sign.

When reinforcing is required at supports forreasons such as structural integrity ties or dia-phragm connections, the reinforcing ratios aregenerally quite low, and therefore, develop littlemoment capacity. While this reinforcing may beconsidered in calculating service load deflections,it is recommended that full simple span positivemoment capacity be provided for strength designunless moment-curvature relationships existing atthe supports at ultimate loads are known.

One situation that seems reasonable for consid-ering a reduction in the positive moment require-ments is where the slabs are required to have a firerating developed using the rational design proce-dure. In this case, a limit analysis approach wouldbe reasonable.

The negative moment reinforcing, which is un-affected by fire loads, can develop full yield mo-ment potential and effectively provide a plastichinge at the support. As a result, the positive mo-ment at midspan may be correspondingly re-

duced. A detailed discussion of this is presentedin Section 6.3.3.

3.5 CantileversCantilever design in hollow core slabs differs

from design with conventional precast membersbecause of the production procedures used forhollow core slabs. Guidelines noted here are con-servative and may be exceeded depending on thespecific product used.

Because long line beds are used for the produc-tion of hollow core slabs, top prestressing strandsmay be economical only when full bed capacity isutilized. Even then, substantial amounts of pre-stressing strand may be used inefficiently becauseof debonding requirements. The economics of us-ing top strand must, therefore, be determined bythe local producer.

When top strands are used, the length of thecantilever is usually not sufficient to fully developa strand. A reduced value for fps is required andthe design procedures given in Section 2.6 shouldbe used. In dry cast systems, the bond of topstrands may be less than desired so a further re-duction in fps is required. This reduction may besubstantial and each producer should be consultedon top strand bond performance.

When top strands are not economical,non-prestressed reinforcement may be placed inthe cores or directly in the unit in the case of a wetcast product. This is generally done while the slabconcrete is still plastic so bond of the fill concretewith the slab may be achieved. The reinforcementis selected based on conventional design with dueconsideration given to bar development length.

With either top strands or reinforcing bars, itmay be necessary to debond portions of the bot-tom prestressing strand in the cantilever zone tohelp minimize the top tension under service loads.Not all producers have the ability to debond bot-tom strands which could potentially limit cantile-ver length or load capacity.

It is desirable to limit service level tensions incantilevers so that uncracked section propertiesmay be used to more accurately predict deflec-tions. The tensile stress limit may vary for differ-ent systems used. For example, the practice withsome dry cast systems is to limit tensile stresses to100 psi (0.7 MPa). In other dry cast systems and in

3--11

wet cast systems, the limit may be raised to 6 f′c .The tension limit will basically be a function of aproducer’s past experience.

As a rule of thumb, cantilever lengths falling inthe range of 6 to 12 times the slab thickness will beworkable depending on the superimposed loadand individual producer’s capabilities.

Example 3.5.1 Cantilever DesignUsing the generic hollow core slab section de-

fined in Section 1.7, design for the followingconditions shown in Figure 3.5.1.Fig. 3.5.1

26 7

15 psf D.L.40 psf L.L.

200 plf D.L.

7.25 ft-k/ft

2.87

4.06 ft-k/ft

D.L. + L.L.

Full (D.L. + L.L.)

11.02 ft-k/ft

5.98 ft-k/ft

D.L. on backspan

5.39 ft-k/ft

5.98 ft-k/ft

(D.L. + L.L.) oncantilever

u

u

u

’

’

’2.82

4.79’

’

Solution:

From the load table in Figure 1.7.1, select 4 -3/8″ dia., 270 ksi strands as the primary reinforce-ment. Try 2 - 3/8″ dia., 270 ksi, low relaxationstrands as cantilever reinforcement. Assume 15%losses and 70% initial stress.

Check stresses at cantilever:

Bottom strands:

ftop = 0.7(0.85)(4)(23) 1154

− 2.89× 4.111224.5


Top strands:

ftop = 0.7(0.85)(2)(23) 1154

− 3.11× 4.111224.5

= + 0.463 ksi

Applied moment:

ftop = --4.063(12)(4.11)

1224.5


Net tension with fully bonded bottom strands:ft = --0.176 + 0.463 -- 0.491

= --0.204 ksi

Allow 6 5000 = 0.424 ksi OK

Note that some of the bottom strands couldhave been debonded for the length of the cantile-ver if top tensile stresses had exceeded a desirablelevel.

Stresses in backspan:

Because the backspan is long in this example,stresses will not be critical in the backspan. Whenthe backspan is short relative to the cantileverlength, stresses may require a check in the back-span to determine the length of bonding of the topstrands.

Ultimate Strength

At the cantilever, strain compatibility will gen-erally show that the bottom strands may be ig-nored in determining the nominal moment capac-ity. When the bottom prestress is very heavy orthe bottom strands are high in the slab, a straincompatibility analysis should be performed con-sidering both strand layers.

For this example, assume the bottom strandsmay be ignored.

fps = 270 1−0.280.82(0.085)(270)

(36)(7)(5)

= 265 ksi

a = 2(0.085)(265)(0.85)(5)(36)

= 0.294 in

φMn = 0.912

(2)(0.085)(265)7− 0.2942

= 23.15 ft-k/slab

Mu = 3(5.98) = 17.94 ft-k/slab

Mcr = 1224.54.110.463− 0.176+ 7.5 5000

1000 1

12

= 20.29 ft-k/slab

ÔMnMcr

= 23.1520.29

= 1.14 < 1.2 NG

Add 1 - #4 bar top per slab.

3--12

Design summary

2- " strands bonded 13’-5" at cantilever endDebond 1- #4 x 12’-5" at cantilever end

38

4- " strands bonded full length38

Check length of top strand to be bonded:

lavailable = 7(12) = 84 in

ℓd = (fps -- 2/3fse)db

= (265 -- 2(0.7)(0.85)(270)/3)(0.375)

= 59.2 in < 84 in

Therefore, the strand is fully effective in the canti-lever. The moment capacity would have to be re-calculated by the procedures of Section 2.6 if thedevelopment length were found to be greater thanthe length available.

Bond of the top strands in the backspan must belong enough to develop the fps required in the can-tilever design. The top strands should also bebonded for a distance of one transfer length (50 di-ameters) past the inflection point under the worstload condition. For this example a bonded lengthof 77 in. would be required.

Alternate DesignProvide mild reinforcement in lieu of top pre-

stressing strandsTry 2 - #5 Gr 60 bars at d = 7″

a = 2(0.31)(60)(0.85)(5)(36)

= 0.243″

φMn = 0.912

(2)(0.31)(60)7− 0.2432

= 19.19 ft-k/slab

= 6.4 ft-k/ft > 5.98 OK

Top stress = --0.176 -- 0.491

= --0.667 ksi with fully bondedbottom strands

Note that a cracked section must be consideredin calculating cantilever deflections because thetop stress exceeds a tension of 6 f′c .

3.6 Horizontal JointsFigure 3.6.1 depicts three conditions typically

used in a multistory wall bearing building wherehollow core slabs are used in a platform detail.Several expressions 27-31 have been proposed todescribe the transfer of axial load through this hor-izontal joint.

With hollow core slabs used for floors, the mostefficient detail is to build the slab ends into thewall. Depending on the butt joint size, thestrength of the joint for transfer of vertical loadscan be enhanced with the addition of grout in thebutt joint, Fig. 3.6.1(b), and in both the joint andcores, Fig. 3.6.1(c). Grout fill in the cores in-creases the net slab width and provides confine-ment for a grout column.

The strength of the joint for vertical load trans-fer can be predicted using Eq. 3.6.1 for an un-grouted joint, Fig. 3.6.1(a). For a grouted joint,Fig. 3.6.1(b) or (c), the greater of Eq. 3.6.1 and Eq.3.6.2 can be used. Both grouted and ungroutedjoints can have the slab cores either filled or notfilled. Both equations include a capacity reduc-tion term for load eccentric from the centerline ofthe joint. With single story walls braced at the topand bottom, this eccentricity will be negligible.φPn =φ0.85Aef′cRe (Eq. 3.6.1)

φPn =φtgℓfuCRe/k (Eq. 3.6.2)wherePn = nominal strength of the jointAe = effective bearing area of slab in joint = 2wbw

w = bearing strip widthbw = net web width of slab when cores are not

filled= unit width as solid slab when cores are filled

f′c = design compressive strength of slab con-crete or grout whichever is less

tg = grout column thickness

3--13

Fig. 3.6.1 Common platform details

w w

(a)

w wtg

(c)

gt ww

(b)

ℓ = width of slab being considered

fu = design compressive strength of wall or groutwhichever is less when walls are reinforcedagainst splitting and slab cores are filled

= 80% of design compressive strength of wallor design compressive strength of grout,whichever is less when walls are not rein-forced against splitting or slab cores are notfilled

C = 1.0 when cores are not filled

= 1.4 2500∕f′c(grout) ≥ 1.0 when cores arefilled

k = 0.65 + (f′c(grout) -- 2500)/50,000

Re = reduction factor for eccentricity of load

= 1 -- 2e/h

e = eccentricity of applied load measured fromjoint centerline

h = wall thickness

φ = 0.7Where bearing strips with a modulus of elastic-

ity other than 50,000 psi (345 MPa) are used, theamount of force in the grout column will be al-tered. A theoretical approach presented in Refer-ence 30 considers pad stiffness, grout columnstrength as compared to grout strength, and con-finement of the grout column. A comparison ofthis theoretical procedure with the Johal proce-dure indicates that conservative capacity will bepredicted by substituting the actual pad modulusof elasticity for 50,000 when calculating k.

The bearing strips need to also be checkedagainst manufacturer’s recommended stress lim-its. Figure 3.6.2 summarizes the forces in thejoint and the recommended effective bearingstrip width.

Another set of forces acting on the horizontaljoint develop from the negative moments inducedin the floor slabs due to the clamping effect of abearing wall on the slab ends. Two consequences

3--14

Fig. 3.6.2. Force distribution in horizontal joint

V1 V2

2/3w

kPu1 - k 2

Pu + V1

2/3w

Pu + V1 - k 2

2

Pu

result. The splitting strength of the bearing wall isreduced when the normal force restraining slabend rotation is considered. Secondly, the joint orslab may crack to relieve the frictional restraint.This condition is undesirable from either thestandpoint of joint or slab integrity. Reinforcingnormal to the slab butt joint is most efficient forcontrolling this condition. To date, there are nopublished studies to evaluate effects of this rota-tional restraint. No adverse effects have beencited when nominal diaphragm or structural integ-rity reinforcement has been provided across thejoint.

Example 3.6.1Using the generic hollow core slab section de-

fined in Section 1.7, determine the grouting re-quirements for an interior butt joint as depicted inFigure 3.6.1(a) given the following criteria:

Slab span: 28 feet18 story building with:

8″ concrete bearing wallsf′c wall = 5000 psi

Loads: Roof -- DL = 15 psf LL = 30 psf

Floors -- DL = 10 psf LL = 40 psf

Walls -- DL = 800 plf/story

LL Reduction: None for example

Solution:Loads

Roof: wu = 28 (1.4(53.5+15)+1.7(30))= 4.ll klf

Floors: wu = 28 (1.4(53.5+10)+1.7(40))= 4.39 klf

Walls: wu = 1.4(800)= 1.12 klf/story

Accumulate loads above floor noted

Floor wu Σwu

18 4.11 + 1.12 5.23

17 4.39 + 1.12 10.74

16 5.51 16.25

15 5.51 21.76

14 5.51 27.27

13 5.51 32.78

12 5.51 38.29

11 5.51 43.80

10 5.51 49.31

9 5.51 54.82

8 5.51 60.33

7 5.51 65.84

6 5.51 71.35

5 5.51 76.86

4 5.51 82.37

3 5.51 87.88

2 5.51 93.39

a) Evaluate capacity of ungrouted joint (Fig.3.6.1(a))

bw = 10.5″ for generic slab = 3.5 in/ft ofwidth

f′c (slab) = 5000 psi

3 in bearing strips

φPn = φ0.85 Ae f′c Re (Eq. 3.6.1)

φPn = 0.7(0.85)(2)(3)(3.5)(5)1− 2(0)8

= 62.48 k/ft

Adequate for Floors 8 through roof

3--15

b) Evaluate strength of grouted joint using3000 psi grout for1) 2 in butt joint with no filled cores (Fig.

3.6.1(b))φPn = φ 0.85Aef′cRe (Eq. 3.6.1)

= 0.7(0.85)(2)(3)(3.5)(5)1− 2(0)8

= 62.48 k/ftor

φPn = φtgℓfuCRe/k (Eq. 3.6.2)fu = 3000 psiC = 1.0k = 0.65 + (3000 -- 2500)/5000

= 0.66φPn = 0.7(2)(12)(3)(1.0)

x 1− 2(0)8∕0.66

= 76.36 k/ft > 62.48Therefore φPn = 76.36 k/ft

2) 1/2 in butt joint with cores filled (Fig.3.6.1(c))φPn = φ 0.85Aef′cRe (Eq. 3.6.1)

= 0.7(0.85)(2)(3)(12)(3)1− 2(0)8

= 128.5 k/ftor

φPn = φtgℓfuCRe/k (Eq. 3.6.2)fu = 3000 psi

C = 1.4 2500∕3000 = 1.28k = 0.65 + (3000 -- 2500)/50,000 = 0.66φPn = 0.7(0.5)(12)(3)(1.28)

x 1− 2(0)8∕0.66

= 24.4 k/ft < 128.5Therefore φPn = 128.5 k/ft

Use 1/2 in butt joint with cores filled below 8thfloor.

It should be noted that this example may over-state the height of building that can be supportedon an ungrouted joint. Concentrated loads due tocorridor lintels, wall openings, or exterior span-drels must also be considered in most buildings re-sulting in an increase in load to be transferredthrough the horizontal joint.

CHAPTER 4

4--1

DIAPHRAGM ACTION WITH HOLLOW CORE SLABS

4.1 General InformationWhen hollow core slabs are used as floor or

roof decks to support vertical loads, the naturalextension is to use the slabs as a diaphragm to re-sist and transmit lateral loads. Lateral loads willbe applied to building structures in the form of lat-eral earth pressures, wind loads or seismic loads.The function of a diaphragm is to receive theseloads from the building elements to which theyhave been applied and transmit the loads to the lat-eral-resisting elements which carry the lateralloads to the foundation. The design issues in ahollow core diaphragm are the design of connec-tions to get loads into the diaphragm, the strengthand ductility of the slab system to transmit theseloads to the lateral-resisting elements and the de-sign of the connections required to unload the lat-eral forces from the diaphragm to the lateral-res-isting elements.

Clear communication is required between thebuilding designer and the hollow core slab suppli-er when the hollow core system is to be used as adiaphragm. Some elements of the diaphragm de-sign may be delegated to the hollow core slab sup-plier. However, only the building designer is inthe position to know all theparameters involved ingenerating the applied lateral loads. Because ofmany design issues, only the building designercan determine the location and relative stiffnessesof the lateral-resisting elements. These parame-ters dictate the distribution of forces in the dia-phragm. If any design responsibility will be dele-gated to the hollow core supplier, the location andmagnitude of the lateral loads applied to the dia-phragm and the location and magnitude of forcesto be transmitted to lateral-resisting elementsmust be specified. Where hollow core slabs mustconnect to other building materials, or where de-mands on connections go beyond simple strengthdemands, the connection details should be shownin the contract documents.

An additional consideration in detailing dia-phragms is the need for structural integrity. ACISection 16.5 provides minimum requirements to

satisfy Section 7.13 in precast concrete structures.For large panel bearing wall structures, minimumforces are specified to provide ties throughout thestructure. For other types of precast structures,only general detailing philosophies are specified.In either case, the fundamental requirement is toprovide a complete load path from any point in astructure to the foundation. Clearly, a diaphragmis a significant element in this load path. A tie sys-tem that satisfies the strength and force transferdemands on a diaphragm will generally satisfy thedetailing requirements for structural integrity.

4.2 Design LoadsLateral loads imposed on hollow core dia-

phragms can include lateral earth pressures, windloads or seismic loads. Lateral earth pressureswill be established by the characteristics of thesoil being retained. Wind and seismic loads willbe dictated by the applicable building code for thestructure. Soil and wind loads are forces actuallyapplied to the structure. Seismic forces are gener-ated from within the structure as inertial forcesdue to lateral displacement from ground motions.While soil and wind loads can be safely treated asstatic loads, seismic loads must be considered asdynamic loads. In all cases, the same elementswill comprise a complete diaphragm, but the duc-tility demands on a seismic resistant system aresignificantly more important.

The balance of the discussion in this chapterwill be concerned with lateral loads from windand seismic. This is not intended to slight the im-portance of considering unbalanced soil pressureswhich can commonly be a significant consider-ation in many projects using hollow core slabs.The basic principles of hollow core diaphragmswhich will be discussed are equally applicable tolateral soil pressures.

There are many documents which cover designfor wind and seismic loads. The references usedfor this chapter are the 1994 UBC code32 and the1996 BOCA code33. For wind load, both codesare similar in that a basic wind speed is selected

4--2

based on the building location, an exposure cate-gory is selected based on the surrounding terrain,an importance factor is selected based on the oc-cupancy of the building, modifying factors are de-termined for the geometry of the building and thedesign positive and negative wind pressures arecalculated.

For seismic loads, the two codes take differentapproaches. The UBC code allows an equivalentstatic load approach for many building types. Forothers, where certain heights or irregularities arepresent, a dynamic lateral force procedure is re-quired. The static force procedure allows designfor a base shear of:

V = ZICRw

W (Eq. 4.2.1)

where

Z = seismic zone factor

I = importance factor

C = factor dependent on site and structure fun-damental period

Rw = coefficient dependent on structural systemtype

W = total dead load plus other applicable loadsThis base shear is then distributed over the

height of the structure in proportion to the dis-tribution of weights over the height. Additionally,a minimum eccentricity of 5% of the building di-mension perpendicular to the direction being con-sidered shall be included when determining thedistribution of forces to the lateral-resisting ele-ments when the diaphragm is not flexible. Specif-ic to diaphragms, for Zones 2, 3, and 4, the UBCrequires that a floor or roof diaphragm resist aforce equal to:

Fpx =Ft + Σ

n

i=xFi

Σn

i=xwi

wpx (Eq. 4.2.2)

where

Fpx = force applied to diaphragm at level underconsideration

Ft = additional portion of base shear applied attop level

Fi = portion of base shear applied at level i

wi = portion of W at level i

wpx = portion of W at level under considerationThe magnitude of Fpx need not exceed

0.75ZIwpx but shall not be less than 0.35 ZIwpx.Many other requirements are included in the UBCcode which are not restated in this summary.

The BOCA code prescribes a seismic designprocedure. A very important note is that theBOCA provisions result in forces that are alreadyfactored and are intended to be used with ultimatestrength design methods with no additional loadfactors. The base shear is calculated as:V = CsW (Eq. 4.2.3.)

where

Cs = coefficient related to peak velocity-relatedacceleration, soil profile, structural systemtype and building fundamental period

W = total dead load plus other applicable loadsThe base shear is distributed over the height of

the building in proportion to the distribution of thebuilding mass with consideration of the buildingperiod. A minimum eccentricity of 5% of the per-pendicular building dimension is also required byBOCA when distributing forces to the lateral-re-sisting elements. For Seismic Performance Cate-gories B and greater, each floor or roof diaphragmshall be designed for a minimum load equal to50% of the effective peak velocity-related accel-eration times the weight attributable to the levelunder consideration. The peak velocity-relatedacceleration is determined by the project location.Again, there are many provisions in the BOCAcode which are not covered in this summary.

In light of the performance of some diaphragmsin recent earthquakes, the seismic demand on dia-phragms is an area of new focus. Preliminary in-dications are that diaphragms should remain elas-tic during a seismic event to ensure that post-elas-tic behavior can be achieved in thelateral-resisting elements. By designing a dia-phragm to remain elastic, several things are ac-complished. Diaphragm flexibility, discussed inSection 4.3 will be less significant. The ductilityrequirements for connection details will be of lessconcern. The horizontal distribution of forces tolateral-resisting elements can be maintained.

The building code provisions summarizedabove are based on achieving post-elastic perfor-mance. To keep a diaphragm compatible with

4--3

Fig. 4.3.1 Diaphragm bending momentsw

Flexible diaphram onRigid supports

(a)

Moment

Rigid diaphram onflexible supports

Moment

(b)

w

M + +M

M -

M + M +

post-elastic performance in the lateral-resistingsystem system, an analysis can bedone to evaluatethe total potential post-elastic capacity of the lat-eral-resisting elements. Providing a diaphragmwith strength beyond this capacity will achievecompatibility, but will involve significant analy-sis. Alternatively, the diaphragm design forcesprescribed by the building codes can be increasedby a factor of 2R/5 to keep the diaphragm elasticand minimize required analysis. Whether build-ing code provisions are based on service or fac-tored load levels, use of 2R/5 loads will result infactored loads for design.

4.3 Distribution of Lateral ForcesOnce the lateral forces to be applied to the dia-

phragm have been determined, the next problemis to determine the distribution of those lateralforces to the lateral-resisting elements which willcarry the forces to the foundation. This problem isusually structurally indeterminate which meansthat deformation compatibilities must be consid-ered for establishing equilibrium. The stiffnessesto be considered are those of the diaphragm andthe lateral-resisting elements. Concrete dia-phragms are normally considered to be rigid whencompared to the lateral-resisting elements. De-pending on the type and magnitude of lateralforces applied, a hollow core diaphragm may need

to be considered as a flexible diaphragm. Analy-sis considering flexible diaphragms is much morecomplex than for rigid diaphragms and should beconsidered in light of the project complexity andseismicity for the project location. For most lowand mid-rise structures in low seismic risk areas,an assumption of a rigid diaphragm will be rea-sonable.

The difference in behavior of flexible and rigiddiaphragms is illustrated in Figure 4.3.1. In (a),the flexible diaphragm with rigid supports be-haves as a continuous beam. Shears and momentsin the diaphragm are a function of the plan geome-try. In (b), the deflections of the flexible supportsmust be the same because of the rigid diaphragm.The diaphragm shears and moments will be afunction of the relative stiffnesses of the supports.The differences between (a) and (b) can be consid-erable. Actual behavior will fall between the twocases tending toward one or the other as a functionof the diaphragm stiffness.

In seismic areas, the topic of diaphragm flexi-bility has become a more significant issue. UBCrequires consideration of the diaphragm flexibil-ity for the horizontal distribution of forces. Aflexible diaphragm is defined by the UBC as onehaving a maximum lateral deformation more thantwice the average story drift for the level underconsideration. It may be inferred from UBC Sec-

4--4

tion 1631.1 that this consideration would only ap-ply in seismic zones 2, 3, and 4. The BOCA codesimply states that the horizontal distribution offorces consider the relative stiffnesses of the later-al-resisting system and the diaphragm. This pro-vision would apply to Seismic Performance Cate-gories B and greater. By code then, diaphragmflexibility need not be considered when designingfor wind or for seismic loads in Zones 0 and 1 un-der the UBC or Seismic Performance Category Aunder BOCA.

When diaphragm flexibility must be consid-ered, a cracked moment of inertia calculation issuggested in Reference 34 and a Virendeel trussmodel in suggested in Reference 35. Since theanalysis of a structure with a flexible diaphragm isdependent on so many factors beyond the dia-phragm itself, such analysis is beyond the scope ofthis manual.

4.4 Structural IntegrityAs noted in the introduction to this chapter, the

ACI code requires consideration of structural in-tegrity for all precast concrete structures. Whileproper detailing for lateral loads will satisfy thecomplete load path philosophy of structural integ-rity, there are some minimum provisions in ACISection 16.5 which must be met. With specific re-gard to diaphragms, provisions to be aware of in-clude:

1. For buildings other than large panel bearingwall buildings, the connection to the dia-phragm of members being laterally braced bythe diaphragm shall have a minimum nominaltensile strength of 300 lb per lin ft (4.4KN/m).

2. For large panel bearing wall structures, a sum-mary of the tie forces is given in Figure 4.4.1and are required to have the following mini-mum nominal strengths:

T1 = nominal strength of 1500 lb per lin ft(21.9 KN/m) of floor or roof span

T2 = nominal strength of 16,000 lb (71 KN)

T3 = nominal strength of 1500 lb per lin ft(21.9 KN/m) of wall

These minimum strengths shall not control ifthe actual forces in the diaphragm are greater.

Fig. 4.4.1 Tie forces in bearing wall buildings

T3

b 2T

1

T3

1T

2T

22

2T

T1

3

For seismic loading, it is preferable to use con-ventional reinforcing steel for these types of ties tolimit the elongations and deformations. Whenstructural integrity requirements control in non-seismic areas, untensioned prestressing strandsmay be used to satisfy the strength requirements.

4.5 Elements of a DiaphragmFigure 4.5.1 illustrates the various elements

which comprise a complete diaphragm. The fol-lowing definitions will be used to describe thevar-ious elements:Boundary Element:Edge member around the pe-

rimeter of a diaphragm orthe perimeter of an openingin a diaphragm which tiesthe diaphragm together. Theboundary element may func-tion as a chord or a dragstrut.

Collector: Elements which transfershear from the diaphragm toa lateral-resisting element.

Chord: Tension or compression ele-ment creating a flange forthe diaphragm to developflexural integrity in the dia-phragm.

Drag Strut: Element used to “drag” lat-eral loads into the lateral-re-sisting elements and to dis-tribute shears over a greaterlength of the diaphragm

4--5

web. (Also called dia-phragm strut.)

Longitudinal joint: Joint oriented parallel to theslab span.

Transverse joint: Joint oriented perpendicularto the slab span.

To satisfy structural integrity, all diaphragmsshould have boundary elements of some type.The boundary elements are essential to ensure thata diaphragm will have the strength to transfer lat-eral loads to the lateral-resisting system. As achord, tension reinforcement is placed in theboundary element to allow the diaphragm to act asa deep horizontal beam or tied arch. This rein-forcement can also provide shear friction steel forshear transfer along the longitudinal joints.

Collectors are required in all diaphragms totransfer forces from the diaphragm to the lateral-resisting elements. Such connectors are also re-quired for structural integrity to provide a com-plete load path for lateral forces to the foundation.Collectors may also function to get forces into adiaphragm.

Drag struts act to engage a longer length of dia-phragm web for transferring diaphragm shearsinto the lateral-resisting elements. A drag strut isparallel to the applied load, receives load from thediaphragm and transfers load to the lateral-resist-

ing element as an axial tension or compression.Drag struts are not required for structural integrityas long as the diaphragm is connected directly tothe lateral-resisting elements. Drag struts simplyspread out shears that might otherwise be highlylocalized. Under the UBC code, it is implied thatdrag struts are required elements in Zones 2, 3,and 4. The BOCA code is silent on the use of dragstruts, but it can be implied that they are requiredfor Seismic Performance Categories B and great-er.

When a bonded structural topping is used witha hollow core slab diaphragm, these elements canbe provided directly by reinforcement in the top-ping. When no topping is provided, these ele-ments are developed as grouted or concrete ele-ments external to the hollow core slabs. As a sim-ple example, Figure 4.5.2 depicts two commonboundary conditions. In (a), the boundary rein-forcement is placed in a masonry bond beam andthe collector reinforcement is placed in the key-ways between slabs. In (b), the boundary rein-forcement is placed in a grouted or concrete filledspace at the end of the slabs. The collector rein-forcement is again placed in the keyways betweenslabs. The primary difference between the detailsis that the boundary reinforcement in (a) is eccen-tric from the diaphragm web while it is concentric

Fig. 4.5.1 Diaphragm elements

Lateral - ResistingElement

Boundary Element(chord)

Boundary Element(chord)

Collector

Drag Strut

LateralResistingElement

Load

DragStrut

Lateral - ResistingElement

LongitudinalJoint

TransverseJoint

4--6

Fig. 4.5.2 Boundary elements

(a) (b)

in (b). The concentric boundary element will ex-hibit better performance in a seismic situation andshould be used in Zones 3 and 4 under the UBC orSeismic Performance Categories C, D and E un-der the BOCA code.

4.6 Diaphragm StrengthThe diaphragm must have the strength to trans-

fer imposed lateral loads from the point of ap-plication to the point of resistance. The dia-phragm spans between lateral-resisting elementsas a deep beam or tied arch. Shears and tensionswill develop and must be resisted in the dia-phragm to have a complete system.

4.6.1 Longitudinal JointsThe grouted keyways between slabs do have

capacity to transfer longitudinal shear from oneslab to the next. Using a shear stress of 80 psi(0.55 MPa), the useable ultimate strength for lon-gitudinal shear is:

φVn =φ(0.08)hnℓ (Eq. 4.6.1)

where

ℓ = length of joint under consideration (in)

hn = net height of grout key (in)

φ = 0.85

Fig. 4.6.1 Shear friction steel in butt joint

When the grout strength is exceeded or ductilebehavior is required, shear friction principles maybe used to design reinforcement to be placed per-pendicular to the longitudinal joints.36 This rein-forcement may be placed in the transverse joints atthe slab ends rather than being distributed alongthe length of the joints. Placed as shown in Figure4.6.1, the area of steel is calculated as:

Avf = VuÔfyμ

(Eq. 4.6.2)

where

Vu = factored applied shear

4--7

Fig. 4.6.2 Alternate longitudinal shearconnections

(a) (b)

Reinforcement acrossgrout keyway

Welded connection

µ = 1.0 for shear parallel to longitudinal joints

= 1.4 for shear parallel to transverse jointswhere concrete can flow into cores

φ = 0.85While the detail shown in Figure 4.6.1 is the

most economical means of providing a mechani-cal connection across the longitudinal joints, al-ternate connections are available which may bedesirable in certain circumstances. Figure4.6.2(a) shows reinforcing steel placed across thelongitudinal joint and grouted into the cores. Thisdetail might be considered when the amount of re-inforcement required in the transverse joints isgreat enough to cause congestion. Figure 4.6.2(b)shows weld anchors in the slabs and a loose platewelded across the longitudinal joint. Use of thisdetail should be carefully coordinated with thehollow core slab supplier to ensure that proper an-chorage of the weld plates in the slabs can be ac-complished.

Where the diaphragm must unload shear into alateral-resisting element, boundary element or in-terior drag strut, a condition similar to the longitu-dinal joint exists. For longitudinal shear, againshear friction can be used to design reinforcementas the collector to cross potential crack planes andtransfer the shear. Figure 4.6.3 depicts an exam-ple of such a collector detail. While drag strutsand boundary elements may have a vertical stiff-ness similar to the slab deck, the lateral-resistingelements will usually have a significantly highervertical stiffness. The collectors connecting di-rectly to the lateral-resisting elements will tend tobe rigid vertically. While strength and toughnessat such collectors is certainly important, it isequally important to consider every day perfor-mance of the structure. At rigid vertical elements,it may be desirable to allow slab camber growth or

Fig. 4.6.3 Collector detail

Bar Spacingby Design

BoundaryReinforcement

cut-outsIntermittent slab

Fig. 4.6.4 Potential effects of rigid lapconnection

Potential jointcracking andgrinding

Vertically rigidconnection

deflection to occur without distress at the connec-tion. Figure 4.6.4 shows potential damage at thefirst interior longitudinal joint when a verticallyrigid connection is used. The potential for distressis dependent on the slab span and the real appliedloads. Short, lightly loaded spans may experienceno problems.

The effect of different vertical stiffnesses maybe accounted for by:

1. Determining that distress will not affect thestrength or performance of the system,

2. Locating vertically rigid connections near theslab supports where vertical movement isminimized, or

3. Providing allowance for vertical movement inthe connection detail.

4--8

4.6.2 Transverse JointsThe transverse joints serve many functions. As

described in Section 4.6.1, reinforcement in thetransverse joints may provide the shear friction re-inforcement for shear in the longitudinal joints.The transverse joint may also have to act as a dragstrut with axial tension or compression to carry di-aphragm loads to the lateral-resisting elements. Atransverse joint may also be the chord memberwhere flexural tension is resisted. Finally, an inte-rior transverse joint disrupts the web of the hori-zontal beam where horizontal shear would have tobe transferred to maintain the composite depth ofthe diaphragm.

The design of shear friction reinforcement forlongitudinal joint shear is covered in Section4.6.1. Drag strut reinforcement is calculated sim-ply as:

As = TuÔfy

(Eq. 4.6.3)

Chord tension is resisted by reinforcement to pro-vide flexural strength to the diaphragm. It is sug-gested that the effective depth of the reinforce-ment from the compression side of the diaphragmbe limited to 0.8 times the depth of the diaphragm.Hence, the chord reinforcement is calculated as:

As = MuÔ0.8hfy

(Eq. 4.6.4)

where

h = depth of the diaphragm

φ = 0.9

Because diaphragms tend to act as tied archesrather than beams, tension in the chord reinforce-ment does not go to zero at the ends of the dia-phragm. The chord reinforcement must be an-chored at the ends of the diaphragm where a stan-dard hook at the corner will suffice. Forhorizontal shear in the web of the diaphragm, ashear parallel to the transverse joint is developed.Shear friction reinforcement perpendicular to thetransverse joint and embedded in the slab key-ways can be used to reinforce for this shear. Theapplied shear can be calculated as:

Vh = VuQI

or

Vh = Mujh

(Eq. 4.6.5)

In the first case, a unit shear is calculated and shearfriction reinforcement is distributed according tothe shear diagram. In the second case, the totalshear is calculated as the tension or compressionof the internal couple. In this case, shear frictionreinforcement is uniformly distributed over thelength between zero moment and maximum mo-ment. It is suggested that the shear friction rein-forcement be distributed according to the sheardiagram in UBC zones 3 and 4 and BOCA Seis-mic Performance Categories C, D and E to mini-mize the force redistribution required with a uni-form spacing.

Because of the orientation of the joints and theloading directions considered, the reinforcementin the transverse joint discussed above is not alladditive. Typically, the chord tension and longitu-dinal joint shear will be concurrent. The drag struttension will typically occur with loads applied inthe perpendicular direction.

4.7 CollectorsCollectors function as connections to transfer

forces into diaphragms and from diaphragms toboundary elements, drag struts or lateral-resistingelements. The preceding discussion has indicatedthat reinforcing bars may be used as collectors us-ing shear friction design procedures. As shearfriction reinforcement, the steel is used in tensionto resist a shear force. In detailing the steel, acrack plane is defined and the bars must be an-chored for full strength on each side of the crackplane. For anchorage at a transverse boundaryelement, the bars may be grouted into the keywaysor into slab cores where the top of the core is cutaway. Concrete is then used to fill the cores for thelength of the bar embedment. Based on a reviewof the literature, it is not clear when anchorage ofcollector bars in keyways is sufficient and whenthe collector bars should be placed in slab cores.There is a concern that as the boundary elementand keyway crack, anchorage for a collector bar ina keyway may be lost. Deformations and revers-ible loading in a seismic event would suggest thatanchoring collector bars in slab cores would bepreferable in more intense seismic areas. In keep-ing with code philosophy, it is suggested that barsbe anchored in slab cores in UBC zones 3 and 4

4--9

Fig. 4.9.1 Example Problem

30

8 @ 25 -0 = 200 -0’ " "’

’20’

30’

80’

30’

30’

and BOCA Seismic Performance Categories Cand greater.

In non-seismic and low seismic design situa-tions, the collectors need not be reinforcing bars.Particularly for direct connections to lateral-re-sisting elements, welded and bolted connectionswill suffice for the collector connections whenthey are compatible with the slab system used.

4.8 Topped vs. Untopped DiaphragmsWhen a composite structural topping is pro-

vided, it should have a minimum thickness of 2 to2 1/2 in (50-65 mm). The topping can then be de-signed as the diaphragm without consideration ofthe hollow core slabs. When the topping providesthe strength and stiffness for the diaphragm butthe connections are made in the hollow core slabs,shear stresses will be present at the interface of thetopping and the hollow core slabs. These stresseswill generally be well distributed throughout theinterface, but may be more highly localized nearthe connections. As discussed in Chapter 2, hori-zontal shear stresses should be kept below a nomi-nal strength of 80 psi (0.55 MPa).

The primary benefits of a composite structuraltopping are to increase stiffness and to allow easi-er continuous ties in plans with irregular shapes orlarge openings. However, in seismic areas, theadditional topping weight increases the seismic

design forces. It is suggested that a topping beconsidered in high seismic zones in buildings withplan irregularities or large diaphragm span todepth ratios.

Untopped hollow core diaphragms are sug-gested when the diaphragm force system isstraightforward and the in-plane diaphragmdeflections are acceptable. An example at the endof this chapter illustrates a procedure for deter-mining diaphragm deflections. In high seismicareas, local codes may limit the use of untopped,hollow core diaphragms.

4.9 Design ExampleGiven the building plan shown in Figure 4.9.1,

design and detail the untopped hollow core dia-phragm assuming:

a. wind design per UBCb. Zone 2A seismic per UBC.

Building data6 stories14 ft floor to floor8 in hollow core floors wt = 53.5 psfpartitions & mechanical wt = 20 psfprecast framing system wt = 32 psfexterior wall system (avg.) wt = 35 psf

4--10

Solutions:

a. Seismic Zone 0; Basic wind speed 80 mph

Use Exposure C

Design wind pressure = P = CeCqqsIwwhere

Ce = 1.53

Cq = +0.8, -- 0.5

qs = 16.4

Iw = 1.0

P = 1.53(0.8)(16.4)(1.0) = 20.1= 1.53(0.5)(16.4)(1.0) = 12.5

32.6 psf

Wind to diaphragm = w = 14(0.0326) = 0.46k/ft

• Consider load applied parallel to the slabs

Total V = 200(0.46) = 92k

Assuming a rigid diaphragm, the shear dis-tribution to the walls based on their flexural stiff-ness is:

30 ft walls: V = 40k

20 ft wall: V = 12k

The diaphragm equilibrium is:

40 k k12 40 k

0.46 k/ft

V

M

40

406

6

87ft

1739 ft-k

The factored design forces are then:

Vu30 = 1.3(40) = 52k

Vu20 = 1.3(6) = 7.8k

Mu = 1.3(1739) = 2261 ft-k

• Chord Forces:

Using the perimeter beams as chords:

Tu =MuÔ0.8h

= 22610.9(0.8)(80)

= 39.3k

Connect beams through columns for this forceplus forces due to volume change and gravityloads. (Fig. 4.9.2 Det. C)

The chord must continue through the centerwall.

As =Tufy

(φ was included in Tu)

= 39.360

= 0.66 in2

Use 2 - #6 (Fig. 4.9.2 Det. F)

• Connect diaphragm web to chords

Vuh = Mujh

j ≅ 0.8

Vuh = 22610.8(80)

= 35.3k

Distribute over length from zero moment tomaximum moment

Vuh = 35.387

= 0.41k/ft

Additionally, this connection must resist thenegative wind pressure from the exterior wallsystem.

wu = 1.3(0.0125)(14)

= 0.23k/ft

Use 300 lb/ft for structural integrity

(Fig. 4.9.2 Det. A)

The same forces must be resisted at the trans-verse joints. Use shear friction for the shearwith bars placed in the keyways perpendicular

4--11

to the transverse joint. With keyways at 3 ft oncenter:

As = 3(0.3)0.9(60)

+ 3(0.41)0.85(60)(1.4)

= 0.034 in2/keyway

Use #3 at every 2nd keyway

(Fig. 4.9.2 Det. B)

• Longitudinal shear

The maximum longitudinal joint shear is at thefirst slab joint from the30 ft shear wall. Since con-nections will be made directly from the center bayto the shear wall, only the center bay joint lengthshould be considered.

Vu = 52k

φVn = φ(0.08)hnℓ= 0.85(0.08)(8 -- 2)(20 x 12)

= 97.9k

With concerns for shrinkage cracking in joints,transverse shear friction reinforcement can beprovided in the transverse joints at each end of thecenter bay.

Avf=VuÔfyμ

= 520.85(60)(1.0)

= 1.02 in2 / 2 transverse joints

= 0.51 in2 per joint

Use 1 - #7 in transverse joint

(Fig. 4.9.2 Det. B)

• Shear connection to 30 ft wall

Vu = 52k

Additionally, negative wind pressure must beresisted across this joint, but would not be con-current with shear. Structural integrity ties willcontrol for this case.

Tu = (0.3)(20)

= 6k for bay

Using shear friction reinforcement

Avf= 1.02 in2 (from above) or

As = 60.9(60)

= 0.11 in2 does not control

Use 4 - #5 located near slab ends

(Fig. 4.9.2 Det. D)

Alternatively, mechanical connections of slabto wall could be used to transfer the sameforces.

• Shear at center 20 ft wall:

With the rigid diaphragm assumption:

Vu = 7.8k on each side of wall

Avf = 7.80.85(60)(1.0)

= 0.15 in2

Use 2 - #4 located near slab ends or usemechanical connections

(Fig. 4.9.2 Det. E)

• Consider load applied perpendicular to theslabs

Total V = 80(0.46) = 36.8k

Distribution to walls is:

V = 36.8/2

= 18.4k


18.4 k 18.4 k

0.46 k/ft

V

M

18.4

18.4

368 ft-k

4--12

The factored design forces are then:

Vu = 1.3(18.4) = 23.9k

Mu = 1.3(368) = 478 ft-k

• Chord force:

Tu = MuÔ0.8h

= 4780.9(0.8)(200)

= 3.3k

As = 3.360

= 0.06 in2

The #3 bars across the transverse joints will beadequate for the chord force. (Fig. 4.9.2 Det. B)

Longitudinal shear

Vuh = Mujh

≅ 4780.8(200)

= 3.0k will not control

• Shear connection to walls

Using shear friction reinforcement

Vu = 23.9k/30 ft wall

= 0.8k/ft controls over parallel wind

With bars in keyways at 3 ft on center

Avf = 3(0.8)0.85(60)(1.4)

= 0.034 in2 per keyway

Use #3 in every 2nd keyway

(Fig. 4.9.2 Det. F)

• Shear in transverse joint

Vu = 1.3(18.4 -- 0.46 × 30)

= 6k

Avf = 60.85(60)(1.0)

= 0.12 in2

#3 at every 2nd keyway will be adequate

(Fig. 4.9.2 Det. B)

b. Seismic Zone 2A

The building weight attributable to each flooris:

wi = 80(200)(0.0535 + 0.020 + 0.032)+14(0.035)(200 + 80)(2)

= 1962k

then

W = 6(1962)

= 11772k

• Base shear

V = ZICRw

W

Z = 0.15

I = 1.0

C = 2.75

Rw = 8

V = 0.15(1.0)(2.75)8

(11772)

= 607k

• Vertical Distribution

Ft = 0.07TV

with a site coefficient of 1.2 and C = 2.75

T = 0.4 sec < 0.7 sec

Ft = 0

Fx =(V− Ft)wxhx

Σn

i=1wihi

wi hi wxhx Fx

1962 84 164808 173

1962 70 137340 145

1962 56 109872 116

1962 42 82404 87

1962 28 54936 58

1962 14 27468 29

576828

4--13

Fig. 4.9.2 Wind design summary

A

B

C

E

F

D30

3020

30

80

30

8 @ 25 -0 = 200 -0

0.3k/ft0.41k/ft

C

39.3kChord Force

E

#7 cont.

#3 @ every2nd keyway

A

B

#4 near eachend of wall

D F

#3 every2nd keyway

2-#6

Intermittent slabcut-outs

2-#5 near eachend of bay


’ " ’ "

’ ’’’

’

’

4--14

• Diaphragm load

Fpx=Ft + Σ

n

i=xFi

Σn

i=xwi

wpx

wpx Ft + ΣFi Σwi Fpx

1962 173 1962 173

1962 318 3924 159

1962 434 5886 145

1962 521 7848 130

1962 579 9810 116

1962 607 11772 102

Minimum diaphragm load

Fpx= 0.35ZIwpx

= 0.35(0.15)(1.0)(1962)

= 103k

Maximum diaphragm load

Fpx= 0.75ZIwpx

= 0.75(0.15)(1.0)(1962)

= 221k

To keep diaphragm in the elastic range, multi-ply the diaphragm loads by 2R/5. At the roof

Fpxu = 173(2)(8)/5

= 554k

The factored roof diaphragm load by code pro-visions is

Fpxu = (1.1)(1.3)(173)

= 248k

Design roof diaphragm for a factored load of554k to keep in elastic range.

• For shear parallel to slabs

Using a rigid diaphragm, the shear distributionto the walls is:

30 ft walls: V = 241k

20 ft wall: V = 72k


241k k72 241k

2.77 k/ft

V

M

241

24136

36

87ft

10483 ft-k

• Chord forces:

Using reinforcement in a perimeter boundaryelement

As = MuÔ0.8hfy

= 104830.9(0.8)(80)(60)

= 3.0 in2

Use 4 - #8

(Fig. 4.9.3 Det. A)

• Connect diaphragm web to chord

Vuh = Mujd

≅ 104830.8(80)

= 164k

Distribute over length from zero moment tomaximum moment

Vuh = 16487

= 1.89k/ft

Additionally, this connection must resist theoutward force from the exterior wall system.Conservatively, this force will be:

T = 0.75ZIww

= 0.75(0.15)(1.0)(14 × 0.035)

4--15

= 0.055k/ft

Tu = 0.055(2)(8)/5

= 0.176k/ft

As =TuÔfy

+ VuÔfyμ

= 0.1760.9(60)

+ 1.890.85(60)(1.4)

= 0.033 in2/ft

Use #3 at 3 ft on center grouted into cores

(Fig. 4.9.3 Det. A)

At the transverse joint, the same shear parallelto the transverse joint as at the chord must betransferred. However, the tension should con-sider the inertial force from the weight of theexterior bay. Conservatively

T = 0.75ZIwp

wp = 14(0.035) + 30(0.0535 + 0.020 + 0.032)

= 3.66k/ft

T = 0.75(0.15)(1.0)(3.66)

= 0.41k/ft

Tu = 0.41(2)(8)/5

= 1.31k/ft

As = 1.890.85(60)(1.4))

+ 1.310.9(60)

= 0.051 in2/ft

Use #4 at 3 ft on center in keyways

(Fig. 4.9.3 Det. B)

• Longitudinal shear

The maximum longitudinal shear is at the firstslab joint from the 30 ft wall. Provide shearfriction reinforcement in the two transversejoints and the two boundary elements for shearresistance. Conservatively consider 5% mini-mum eccentricity being resisted only in endwalls.

Vu = 241 + (0.05 x 200)(554)/200

= 269k

Avf = 2690.85(60)(1.0)

= 5.27 in2 / 4 joints

= 1.32 in2 per joint

In boundary elements, add chord requirement

At first jointMu = 241(3) -- 32(2.77)/2

= 711 ft-k

As = 1.32+ 7110.9(0.8)(80)(60)

= 1.53 in2

4 - #8 ok

(Fig. 4.9.3 Det. A)

In transverse jointsAs = 1.32 in2

Use 2 - #8

(Fig. 4.9.3 Det. B)

• Shear connection to 30 ft wall:

Transfer shear to wall and drag strut

Vu = 25380

= 3.16k/ft

Avf = 3.160.85(60)(1.0)

= 0.062 in2/ft

Use #4 hairpins at 3 ft on center

(Fig. 4.9.3 Det. D)

Drag strut reinforcement

Tu = (80− 30)2

(3.16)

= 79k

As = 790.9(60)

= 1.46 in2

Use 2 - #8

(Fig. 4.9.3 Det. C)

4--16

• Shear connection at 20 ft wall

Vu = 36k

over building width

Vu = 3680

= 0.45k/ft

Avf= 0.450.85(60)(1.0)

= 0.009 in2/ft

Use #4 dowels at 8 ft on center

(Fig. 4.9.3 Det. F)


Tu = (80− 20)2

(0.45)(2)

= 27k

As = 270.9(60)

= 0.5 in2

Use 2 - #5

(Fig. 4.9.3 Det. E)

• In-plane deflection of diaphragm

Idealize the diaphragm section as

8"

8"

x

956"

nA4.3" s

with 4000 psi concrete in chordEc = 3835

with 5000 psi concrete in slabEc = 4300

normalize on slab concretenchord = 0.89

ATchord = 0.89(64)

= 57 in2

nsteel = 6.74

nAs = 6.74(3.16)

= 21.3 in2

57(x -- 4) + 4.3(x -- 8)2/2 = 21.3(956 -- x)

find x = 87.9 in

Icr = 57(87.9 -- 4)2 + 4.3(87.9 -- 8)3/3+ 21.3(956 -- 87.9)2

= 17,184,000 in4

= 829 ft4

As a rigid diaphragm, the factored load deflec-tion between end shear walls is:

∆ = 5384

(2.77)(200)4

(4300)(829)(12)− 72(200)3

48(4300)(829)(12)

= 1.07 in (ignoring shear deflections)

As a flexible diaphragm with rigid supports thedeflection will be substantially smaller. Thediaphragm deflection plus the deflection of thelateral-resisting system is used to evaluate thegravity load support members for integritywhen deformed.

• Consider load applied perpendicular to theslabs

Total Vu = 554k

Distribution to walls is

Vu = 554/2

= 277k

4--17


277k 277k

6.93 k/ft

V

M

277

277

5544 ft-k

u

u

• Chord force

Tu = MuÔ0.8h

= 55440.9(0.8)(200)

= 38.5k

As = 38.560

= 0.64 in2

The #4 bars across the transverse joints at 3 fton center will be adequate. (Fig. 4.9.3 Det. B)

Longitudinal shear

Vuh = Mujd

≅ 55440.80(200)

= 34.7k will not control

• Shear connection to walls

With 5% eccentricity

Vu = 1.1(277) = 304.7k

Transfer shear to wall and drag strut

Vu = 304.7/200 ft

= 1.52k/ft

Loading parallel to slabs controls


Tu = (200− 30)2

(1.52)

= 129.2k

As = 129.20.9(60)

= 2.39 in2

Chord reinforcement from load parallel toslabs controls.

• Shear in transverse joint

In center bay

wp = 20(200)(0.0535 + 0.020 + 0.032)+ 20(14)(0.035)(2)

= 442k

Conservatively use

V = 0.75ZIw

= 0.75(0.15)(1.0)(442)

= 49.7k

Vu = 49.7(0.55)(2)(8)/5

= 87.5k per joint including5% eccentricity

Load parallel to slabs will control

See Figure 4.9.3 for summary

4--18

Fig. 4.9.3 Seismic design summary

A

B

F

G

D30

3020

30

80

30

8 @ 25 -0 = 200 -0

E

A

#4 @ 8ft. o.c.

C E G

#3 @3ft o.c.in cores

4 - #8 cont.

2 - #8 cont. #4 @ 3ft. o.c.in keyways

D

#4 @ 3ft. o.c.

F

#4 @8ft. o.c.

2 - #5 cont.

2 - #8 cont.

B

C



’

’’

’

’

’

’ " "’

#4 @ 3ft. o.c.

2-#8 cont. 2-#5 cont.#3 @ 3ft. o.c.

grout in cores

4 - #8cont.

CHAPTER 5

5--1

CONNECTIONS IN HOLLOW CORE SLABS

5.1 GeneralConnections will be required in hollow core

slab systems for a wide variety of reasons. Chap-ter 4 described the connection requirements for ahollow core diaphragm as an element for lateralstability. Most connection requirements will befor localized forces ranging from bracing a parti-tion or beam to hanging a ceiling.

Connections are an expense to a project and, ifused improperly, may have detrimental effects bynot accommodating volume change movementsthat occur in a precast structure. Connections maydevelop forces as they restrain these movements.In specifying connection requirements, the actualforces in the connection must be addressed. If noforce can be shown to exist, the connection shouldnot be used. Again, cost is reduced and undesir-able restraining forces will not be developed.When a connection is determined to be necessary,the force in the connection should be specified es-pecially when at an interface between a hollowcore slab and another material. The extent of de-tailing to be left to the hollow core slab suppliershould be those items that will be supplied withthe product.

5.2 DetailsCommon details are shown in Sections 5.3, 5.4,

5.5, and 5.6 to cover a number of conditions whereforces will probably exist that need to be trans-mitted into or through a hollow core slab. Theconditions cover common detailing situationswhen hollow core slabs are used and are intendedto give the specifier an idea of the possibilities thatexist. The commentary provided with each detailis intended to give a better understanding of themerits of each detail. The emphasis is that theseprovide a guide which can be used as a basis forbetter discussions with local producers. The de-tails are only conceptual and would require de-tailed information to be used on a project.

Differences between wet cast and dry cast hol-low core slabs will be evident in the embedded an-chors that can be provided. Without forms to se-cure anchors to, dry cast systems may be limitedto shallow anchors that can be tied directly tostrands or to inserts that can be placed after cast-ing. Wet cast systems can accommodate a widervariety of anchors placed directly in the form priorto casting. Therefore, anchor details in the hollowcore slabs are not shown. Connection possibilitiesneed to be explored with the local producers.

5--2

5.3 Typical Details with Concrete Beams

Other connection details perform functions similar to those shown. Consult the local PCI producer for information on relative economy anddesign capabilities.

Fig. 5.3.1

Fig. 5.3.2

Design Considerations:

• Can transfer internal diaphragm forces• Can be designed as structural integrity tie

Fabrication Considerations:

• Advantageous to have no hardware in slab• Beam embedments must line up with slab

joints• Accommodates variations in slab length

Erection Considerations:

• Advantageous to have connection completedby follow-up crew

• Difficult for welder to hold loose plate inposition




• May increase beam reinforcement forshallower beam

• Layout must have opposing slab joints lined up


• Clean and simple

Bearing strip

P.C. or C.I.P.concretebeam

Topping ifrequiredGrout

Reinforcementgrouted in slabkeyway

Bearing strip


Withreturns

with headedstud anchors

Plate with deformedbar grouted in

slab keyway Topping ifrequired

5--3

5.3 (Continued)

Fig. 5.3.3

Fig. 5.3.4


• With large factors of safety, friction maytransfer nominal forces

• Additional structural integrity ties may berequired






• Can transfer internal diaphragm forces• Can be designed as structural integrity tie• Consider concrete cover on reinforcement

over beam


• Slab layout must have opposing joints lined up




Bearing strip


Topping ifrequired

Bearing strip


Topping ifrequired

Grout

Reinforcement drapedover beam and groutedin slab keyway

Grout

5--4

5.3 (Continued)

Fig. 5.3.5

Fig. 5.3.6


• Can transfer internal diaphragm forces• Will develop volume change restraint forces

that must be considered in design ofconnections


• Slab manufacturing system must allow bottomweld anchors

• Beam inserts must align with slab insertsallowing fabrication tolerances


• Connections can be completed by follow-upcrew

• Access for welding may require ladders orscaffold

• Spacer may be required to make weld


• Can transfer internal diaphragm forces• Can be designed as structural integrity tie• Horizontal shear from beam cap must be

transferred• Opposing slab joints must line up


• Clean and simple for slabs


• Beam may have to be shored until cap iscured

• Horizontal shear reinforcement may presentsafety hazard for erector

• Core dams must be placed


Bearing strip


Topping ifrequired

Damcores

Reinforcement perdesignConcrete

Reinforcementgrouted inslab keyway

Bearing strip


Topping ifrequired

withheadedstud anchors

Weld Plate(alt. ends)

5--5

5.3 (Continued)

Fig. 5.3.7

Fig. 5.3.8


• Can transfer internal diaphragm forces• Can be designed as structural integrity tie• Horizontal shear in composite beam must be

transferred• Opposing slab joints must line up




• Beam may have to be shored until topping iscured

• Horizontal shear reinforcement may presentsafety hazard for erector

• Core dams must be placed


• Can transfer diaphragm shear• Can provide lateral brace for beam• Potential for negative moment in slabs


• Slab insert difficult to install. Because oftolerance on sawcut ends, the insert should beinstalled after slabs are cut to length

• Beam and slab inserts must align


• If required for lateral beam stability, weldingmay have to be completed as slabs are set


Bearing strip



Plate as requiredby design

Topping ifrequired

or deformed bar

stud anchors with headed

Bearing strip


Topping

Damcores

Reinforcement perdesign


5--6

5.3 (Continued)

Fig. 5.3.9

Fig. 5.3.10


• Can transfer diaphragm shear• Can provide lateral brace for beam• Potential to develop negative moment in slabs


• Connection can be completed with a follow-upcrew

• Lateral bracing for beam will not be provideduntil keyway grout cures


• Plates in beam must align with slab jointsallowing tolerance






• Clean and simple• Keyway dimensions may limit the

reinforcement diameter


Plate with headedstud anchors

P.C. or C.I.P.concrete

beam

Bearingstrip

Topping ifrequired

Plate with deformed baranchor grouted inslab keyway


beam

Bearingstrip

Topping ifrequired

Reinforcement groutedin slab keyway

5--7

5.3 (Continued)

Fig. 5.3.11

Fig. 5.3.12


• Can transfer diaphragm shear• Can be designed as structural integrity tie


• Clean and simple for both beam and slabs


• Reinforcement must be tied in place• Concrete must be cast around reinforcement• Edge form is required for cast-in-place

concrete• Dowels from beam may present safety hazard


• Can transfer internal diaphragm forces• Will develop volume change restraint forces

that must be considered in design ofconnection


• Slab manufacturing system must allow bottomweld inserts

• Beam and slab inserts must align withallowance for tolerance







beam

Bearingstrip

Topping ifrequired

Reinforcement groutedin slab keyway

Longitudinalbar as req’d.


beam

Bearingstrip

Topping ifrequired

Weld Plate (alt. ends)

with headedstud anchor

5--8

5.3 (Continued)

Fig. 5.3.13

Fig. 5.3.14


• Can transfer diaphragm shear• Torsional and lateral beam restraint can be

provided• Will develop volume change restraint forces



• Slab manufacturing system must allow bottomweld inserts

• Beam and slab weld anchors must align withallowances for tolerance






• This detail is not recommended because ofinstallation difficulties which may result in anunreliable connection


• Great difficulty aligning bars with keyways


• Potential difficulties in bending bars• Possible fracture of bent bars• Second rebar bend may be required to align

with slab joints• Cast-in-place concrete required around

reinforcement• Edge forming required



Topping ifrequired

Weld plate


Bearingstrip


beam

Bearingstrip

Longitudinalbar as req’d.

Toppingif req’d.

Field bendinto slab

keyway andgrout

DO NOT USE

5--9

5.4 Typical Details with Walls

Fig. 5.4.1

Fig. 5.4.2


• Can transfer diaphragm shear• Can be designed as structural integrity tie• Can provide lateral brace for wall• Consider axial force path through slab ends• Opposing slab joints must line up


• Clean and simple for slabs• Small tolerance for placement of bars in walls• Tolerance on length of slabs to accommodate

bars in joint


• With longitudinal bar, have potential congestion• Slab erection must consider tight tolerance on

butt joint gap• With precast walls, consider method of

installing vertical dowel


• Can transfer diaphragm shear• Can be designed as structural integrity tie• Can provide lateral brace for wall• Opposing slab joints must line up




• Clean and simple• Wall is not braced until grout is placed and

cured


P.C. or C.I.P.concretewall

Bearingstrip

Dowels asrequired

Topping ifrequired


Grout

Wall

Longitudinalbar as req’d

P.C. or C.I.P.concretewall

Bearingstrip

Dowels asrequired

Topping ifrequired


Grout

Longitudinalbar as req’d

5--10

5.4 (Continued)

Fig. 5.4.3

Fig. 5.4.4


• Can transfer diaphragm shear• Can provide lateral brace for wall with proper

bar detailing• Consideration should be given to forces

developed as slab ends rotate


• Simple for slab erection• The mason can set bars independent of the

slab joints• Some block cutting may be required for bars

from keyways




• Can transfer diaphragm shear• Can provide lateral brace for wall with proper

detailing• Consideration should be given to forces

developed as slab ends rotate


• Simple for slab erection• The mason can set bars independent of the

slab joints• Grout at slab end may be difficult to place




Mortar bedto levelblock course

Topping ifrequired


Bearingstrip

Bearingwall

Groutedbearingcourse

Mortar bedto levelblock course

Topping ifrequired

Hooked bargrouted inslab keywayBearingstrip

Bearingwall

Solidbearingcourse

Hooked barin wall

5--11

5.4 (Continued)

Fig. 5.4.5

Fig. 5.4.6


• This detail is not recommended because ofinstallation difficulties which may result inan unreliable connection



• Mason will have great difficulty locating bars atslab joints

• Potential difficulties to field bend bars includingfracture

• Second bend may be required to align barswith joints


• Wall will not be braced at this level




• Small tolerance in slab layout


Bearingwall

Bearingstrip

Bar to be fieldbent into slabkeyway or fielddrilled into wall

Toppingif req’d

DO NOT USE

Toppingif req’d

Non-bearingwall

Clearance

5--12

5.4 (Continued)

Fig. 5.4.7

Fig. 5.4.8


• Walls may not be laterally braced• Consideration should be given to forces

developed from deflections or camber growth• Drypack may be required under slab for axial

load transfer




• Allowance must be made for slab camber• Wall will not be laterally braced at this level• Small tolerance in slab layout


• Can transfer diaphragm shear• Can provide lateral brace for wall• Consideration should be given to forces

developed from deflection or camber growth• Consider axial load path


• If not done in field, slots and holes must be cutfor steel

• In stack casting system slots and holes mightnot be practically cut in plant


• Allowance must be made for slab camber• If not done in plant, holes and slots must be

cut for steel• Wall is not braced until steel is grouted


Wall

Clearance

Allo

wfo

r ca

mbe

r

Toppingif req’d

Wall Grout

GroutToppingif req’d

Deformed baror steel strap

5--13

5.4 (Continued)

Fig. 5.4.9

Fig. 5.4.10


• Wall thrust from earth pressure can be resisted• Can transfer diaphragm shear only with special

detailing of keyway and reinforcement• For long spans consider effects of restraint of

vertical movement




• Edge joint must be grouted which may notbe standard practice


• Can transfer diaphragm shear• Can provide lateral brace for wall• Consideration should be given to forces

developed from deflections or camber growth


• If not done in field, edge core must be cutopen

• In stack casting operation, holes might not bepractically cut in plant


• If not done in plant, holes must be field cut intoedge core

• Mason may have to cut block to installreinforcement



wall

Allo

wfo

r ca

mbe

r

Toppingif req’d

Reinforcementgrouted intobroken core

Grout atbars

Wall

5--14

5.4 (Continued)

Fig. 5.4.11

Fig. 5.4.12


• Can transfer diaphragm shear• Can provide lateral brace for wall• Connection capacity must be verified by test




• Minimum edge distances must be maintained• No interfacing tolerances


• Can transfer diaphragm shear• Can provide lateral brace for wall• Consider effects of vertical restraint• Connection capacity must be verified by test




• Minimum edge distances must be maintained• No interfacing tolerances


BearingStripBond

Beam

Field Drill

Reinforcing BarDriven In Hole

Dry Pack

BondBeam

Field Drill

Reinforcing BarDriven In Hole

5--15

5.5 Typical Details with Steel Beams

Many connection details shown perform similar functions. Consult the local PCI producer for information on relative economy and designcapabilities.

Fig. 5.5.1

Fig. 5.5.2


• Top beam flange should be consideredunbraced


• Clean and simple for slabs• Beam flange width must be sufficient for slab

bearing length


• Unsymmetrical loading may cause beaminstability


• Can transfer internal diaphragm forces• Provides lateral brace for steel beam


• Slab layout must align slab joints• Stabilizer bars might be field or shop installed

depending on local regulations or agreements• Beam flange width must be sufficient for

minimum slab bearing


• Grouting of slabs must include the butt joint• Steel erection may require that stabilizer bars

be field installed• Steel beam will not be laterally braced until

grout cures• Unsymmetrical loading may cause beam

instability

Steel beam withstabilizer barsto brace top flange

2" Joint

Toppingif req’d

GroutReinforcementgrouted inslab keyway

2" (Min.) toppingif required

Steel beamNote: Top flange isunbraced

1" Min. joint

5" Min. flange6" Recommendedflange

5--16

5.5 (Continued)

Fig. 5.5.3

Fig. 5.5.4


• Can transfer internal diaphragm forces• Provides lateral brace for steel beam• Will develop volume change restraint forces



• Slab manufacturing system must allow forinstallation of bottom weld anchors


• Welding of slabs to beam should be done aserection proceeds to laterally brace beams


• Can transfer diaphragm shear• Provides lateral brace for steel beam• Potential torsion on steel beam should be

considered• Will develop volume change restraint forces



• Slab manufacturing system must allow forinstallation of bottom weld anchors


• Welding of slabs to beam should be done aserection proceeds to brace beam



Steel beam

Toppingif req’d

Grout Weld plate(weld at alternateends of slabs)

Steel beam

1" (Min.) past centerlineof beam

Weld plate (weld at alternate ends)

Toppingif req’d

5--17

5.5 (Continued)

Fig. 5.5.5

Fig. 5.5.6


• Can transfer diaphragm shear• Provides lateral brace for steel beam




• Welding of bars must be coordinated with slaberection for alignment

• Depending on forces to be transferredconcrete may have to be cast along edge

• Beam will not be braced until keyway groutcures


• Internal diaphragm forces can betransferred only through topping

• Provides lateral brace for steel beam• Consider potential torsion on beam during

slab erection


• Beam flange width must be sufficient forminimum slab bearing

• Slab notching will require a hand operation infield or, preferably, in plant


• Slab erection will be very difficult with thisdetail on both slab ends. Slabs must be slidinto beams possibly through access holes inflanges

• Beams will not be braced during slab erection


Steel beam

Deformed bargrouted inslab keyway

Steel Beam6"8"

flangeRecommendedMin. flange

Notch slab(as required)

1/2"

Clr

.

Note:Difficult erection ifthis detail occurs atboth ends of slab

1"Clr. Clr.

1"Additional

reinforcementover steel

beam


5--18

5.5 (Continued)

Fig. 5.5.7

Fig. 5.5.8


• Internal diaphragm forces can be transferredonly through topping

• Provides lateral brace for steel beam• Consider potential torsion in beam during slab

erection


• Angle legs must be sufficient for minimum slabbearing

• Beam depth must be sufficient for clearanceunder top flange


• Slab erection will be very difficult if this detailoccurs at both slab ends. Slabs will have tobe slid into beams possibly through accessholes in flanges

• Beams will not be braced during slab erection


• Torsion design must consider erectiontolerance

• Lintel must be securely anchored at span ends• Connection to slab may be required to brace

lintel




• Watch for stability of lintel prior to slab erection


Steel Beam3"4" Recommended leg

Min. Leg

1/2"

Clr

.

Note:Difficult erection ifthis detail occurs atboth ends of slab

1"Clr. Clr.

1"Additionalreinforcement

over steelbeam


Optionalreinforcement

grouted inslab keyway

Continuous ’swelded to steelbeam

Wall

Toppingif req’d

Channel andplate lintel

5--19

5.5 (Continued)

Fig. 5.5.9

Fig. 5.5.10


• Butt joint must be grouted to brace verticalangle legs

• Lintel must be securely anchored at span ends




• Lintel must be securely anchored prior tosetting slabs


• Clearance must be allowed for slab camber• Beam will not be braced until topping is cast


• Camber must be monitored to stay withinclearance


• Erection may be very difficult if slab supportbeams are also raised


Topping ifreq’d

Double angle (min. 4" leg)or WF (min. 8" flange)lintel

Additionalreinforcement

ToppingA

llow

for

cam

ber

5--20

5.6 Typical Cantilever Details

Fig. 5.6.1

Fig. 5.6.2


• Wall bracing or transmitting diaphragm shearwould only be accomplished by questionablefriction

• Additional structural integrity ties may berequired


• None other than top reinforcement required forcantilever




• Can transfer diaphragm shear• Provides lateral brace for wall


• If not field drilled, slots in keyways andaligning holes in masonry are required

• If not field drilled, alignment will be difficult


• If not preformed, holes must be drilled throughslabs into masonry

• Wall may not be braced until grout cures• Grout placement may be difficult


Cores filled withinsulation overexterior wall

Bearingstrip

Wall

Wall

Bearingstrip

Insulation Dowel grouted inslabs at keyway

5--21

5.6 (Continued)

Fig. 5.6.3

Fig. 5.6.4


• This detail is not recommended because ofinstallation difficulties which may result inan unreliable connection



•Mason will have great difficulty aligningdowels with slab joints

• Most keyway configurations will requirenotches for dowels

• Field bending of dowels into keyways will bevery difficult


• Wall will not be braced by slabs• Depending on end support conditions wall may

have to support edge slab• No thermal break provided between interior

and exterior


• Depending on bearing conditions the overhangdimension may be limited by the producer’sability to install transverse reinforcement


• None


Bearingstrip

Wall

Dowels field bentand grouted inslab keyways

DO NOT USE

Keyway filledwith grout

Verify max. dimensionwith slab supplier

Bearing wallbeyond

Non-bearingwall

Allo

w f

or c

ambe

r

Preferred end ofbearing wall

5--22

5.6 (Continued)

Fig. 5.6.5


• Wall will not be braced by slabs• Depending on end support conditions wall may

have to support edge slab• No thermal break provided between interior

and exterior


• When transverse reinforcement cannot beinstalled, steel strap must serve as externalreinforcement

• Anchorage of a steel strap to the slabs willdepend on the producer’s ability to install topweld anchors


• Depending on end support conditionstemporary shoring may be required until steelstrap is installed and keyways are grouted


Verify max. dimensionwith slab supplier

Bearing wallbeyond

Non-bearingwall

Allo

w f

or c

ambe

r

Preferred end ofbearing wall

Steel strap (weld totop anchors or expansionbolt to slabs)

5--23

5.7 Miscellaneous Details

Fig. 5.7.1


"A"

"A"

Bearing

Hollow coreslab

Feather edgewith latex, concealin wall, orrecess whenno topping

PLAN

SECTION "A-A"

HEADER DETAIL

Header angles

5--24

5.7 (Continued)

Fig. 5.7.2


"A"

"A"

BearingHollow core

slab

PLAN

SECTION "A-A"

HEADER DETAIL

Header angles

"B" "B"

SECTION "B-B"

Feather edge with latex,conceal in wall, orrecess whenno topping

5--25

5.7 (Continued)

Fig. 5.7.3


Light straps;for ceiling andduct work

Hanger thru bolt;for heavy loads

only with sufficientExpansion bolt;

bottom thickness loadsonly for verticalToggle bolt;

CHAPTER 6

6--1

FIRE RESISTANCE OF ASSEMBLIESMADE WITH HOLLOW CORE SLABS

6.1 IntroductionOne of the attributes of hollow core slab

construction is excellent fire resistance. Morethan 30 standard fire tests (ASTM E119) havebeen conducted on hollow core floor assemblies.The January, 1994 issue of Underwriters Labora-tories, Inc. “Fire Resistance Directory” includesmore than 50 design numbers for hollow coreslabs which qualify for ratings of 1, 2, 3, or 4hours. Constructions which conform to these de-signs are assigned ratings by most U.S. buildingcodes.

As an alternative to UL ratings, model codesnow include prescriptive requirements which canbe used to establish fire endurance ratings. Foreach fire endurance rating, strand cover andequivalent thickness provisions are given. Use ofsuch provisions eliminates the need for fire testsor UL ratings.

Most U.S. building codes will also assign rat-ings to hollow core assemblies which do not con-form with the UL designs if it can be shown by cal-culations made in accordance with proceduresgiven in the PCI manual, “Design for Fire Resis-tance of Precast, Prestressed Concrete” (PCIMNL 124-89)38 that they qualify for the requiredfire endurance. Readers can obtain more detailedinformation from that manual on fire resistance ofhollow core slab assemblies as well as informa-tion on fire resistance of concrete beams, walls,and protection of connections.

In Canada, The National Building Code ofCanada requires that fire resistance ratings be de-termined either on the basis of results of tests con-ducted in accordance with CAN/ULC-S101-M,“Standard Methods of Fire Endurance Tests ofBuilding Construction and Materials”, or on thebasis of Appendix D, “Fire Performance Rat-ings”. While the general principles set forth in thisManual are fully valid in that they are based onmaterials properties and structural engineeringprocedures, users of the Manual are cautioned thatin Canada, fire resistance ratings should be deter-mined strictly in accordance with applicable

building code requirements.

6.2 Heat Transmission Through Floors orRoofs

The standard fire test method, ASTM E119,limits the average temperature rise of the unex-posed surface, i.e., the surface of floor or roof notexposed to fire, to 250 degrees F (120 degrees C)during a fire test. This criterion is often called theheat transmission end point.

For solid concrete slabs, the temperature rise ofthe unexposed surfaces depends mainly on theslab thickness and aggregate type. Figure 6.2shows the relationship between slab thickness andfire endurance as determined by the heat transmis-sion end point criterion.

6.2.1 Equivalent ThicknessThe information in Figure 6.2 is applicable to

hollow core slabs by entering the graph with the“equivalent thickness” of the unit instead of thethickness. Equivalent thickness can be calculatedby dividing the net area of the cross section of ahollow core unit by the width of the unit.

Fig. 6.2 Fire endurance (heat transmission) ofhollow core units

2 3 4 5 6 71-1/20

1

2

3

4

5

Fire

End

uran

ce, H

r

Equivalent Thickness, in.,

of Hollow Core Unit

Ligh

twei

ght (

100

pcf)

Sand

-Lig

htw

eigh

t (11

5 pc

f)

Carbo

nate

Aggre

gate

Silice

ous A

ggreg

ate

6--2

In Figure 6.2, concrete aggregates are desig-nated as lightweight, sand-lightweight, carbon-ate, or siliceous. Lightweight aggregates includeexpanded clay, shale, slate, and slag which pro-duce concretes having unit weights between about95 and 105 pcf (1520 - 1680 kg/m3) without sandreplacement. Lightweight concretes in whichsand is used as part or all of the fine aggregate andweigh less than about 120 pcf (1920 kg/m3) aredesignated as sand-lightweight. For normalweight concrete, the type of coarse aggregate in-fluences the fire endurance; the type of fine aggre-gate has only a minor effect. Carbonate aggre-gates include limestone, dolomite, and limerock,i.e., those consisting mainly of calcium or magne-sium carbonate. Siliceous aggregates includequartzite, granite, basalt, and most hard rocks oth-er than limestone or dolomite.

6.2.2 Toppings, Undercoatings, or RoofInsulation

All 8 in (200 mm) deep hollow core units whichare currently manufactured in North Americaqualify for at least a one-hour fire endurance asdetermined by heat transmission and some qualify

for two hours or more. The addition of toppings,undercoatings, fire resistive ceilings, roof insula-tion, or filling the cores with dry aggregates willincrease the heat transmission fire endurance.Figure 6.2.2.1 shows graphically the thickness ofspray applied undercoating required for heattransmission fire endurances of 2, 3 and 4 hours.Figure 6.2.2.2 shows the thickness of sand-light-weight concrete, insulating concrete and highstrength gypsum concrete overlays required for 2,3 and 4 hours. Figure 6.2.2.3 shows data for 2 and3 hr. roofs with mineral board or glass fiber boardinsulation with 3-ply built-up roofing. Datashown in Figures 6.2.2.1, 6.2.2.2 and 6.2.2.3 ap-ply directly to hollow core slabs made with sili-ceous aggregates and are conservative for slabsmade with carbonate aggregates or with light-weight aggregates.

Example 6.2.1 Equivalent ThicknessDetermine the thickness of topping required toprovide a 3 hr. fire endurance (heat transmission)for the generic hollow core slab shown in Figure1.7.1. Both the slab and the topping are made withcarbonate aggregate concrete.

Fig. 6.2.2.1 Hollow core units undercoated with spray applied materials(Heat transmission fire endurance)

3.5 4.0 5.04.5 5.50

0.2

0.4

0.6

0.8

1.0

Thi

ckne

ss o

f SM

For

VM

C, i

n.

Equivalent Thickness, in.

of Hollow Core Unit

4 hr

3 hr

2 hr

Hollow core slab made withsiliceous aggregate concrete

Sprayed mineral fiber orvermiculite cementitous material

6--3

Fig. 6.2.2.2 Floors with overlays of sand-lightweight concrete (120 pcf maximum), insulating concrete(35 pcf maximum), and high strength gypsum concrete

3.5 4.0 5.04.5 5.50

0.5

1.0

1.5

2.0

2.5

Ove

rlay

Thi

ckne

ss, i

n.

Equivalent Thickness, in., of Hollow Core Units

4 hr

3 hr

2 hr

Overlay

Hollow Core slab made withsiliceous aggregate concrete

Sand-Lightweight ConcreteOverlay Overlay

Insulating Concrete

5.54.53.5 4.0 5.0

High Strength Gypsum Concrete

Overlay

4.03.5 4.5 5.0 5.5

2 hr

3 hr

4 hr

0

1

2

3

4

1 in. G.C. Overlay

1/2 in. G.C. Overlay

Fire

End

uran

ce, H

r.

Equivalent Thickness, in., ofHollow Core Units

6--4

Fig. 6.2.2.3 Roofs with insulation board and 3-ply built-up roofing(Heat transmission fire endurance)

Mineral board or glass

Hollow core slab made withsiliceous aggregate concrete

4.03.5 4.5 5.0 5.5

or G

lass

Fib

er B

oard

Equivalent Thickness, in., ofHollow Core Unit

3-Ply built-up roofing

fiber board insulation

Thi

ckne

ss, i

n., o

f M

iner

al B

oard

0

0.25

0.50

0.75

3 hr

2 hr

Solution:Equivalent thicknessteq = Area/width

= 154/36

= 4.28 inFrom Figure 6.2, the thickness of carbonate ag-

gregate concrete required for 3 hr. is 5.75 in. Thus,the thickness of topping needed is:

5.75 -- 4.28 = 1.47 in

Example 6.2.2Determine if a hollow core slab roof will quali-

fy for a 2 hr. fire endurance (heat transmission) ifthe slabs are made with carbonate aggregate con-crete, have an equivalent thickness of 4.28 in, andthe roof insulation consists of a layer of 3/4 in thickmineral board. The roofing is a standard 3-plybuilt-up roof.

Solution:From Figure 6.2.2.3 it can be seen that with an

equivalent thickness of 4.28 in, a layer of mineralboard 0.16 in thick with 3-ply roofing qualifies for2 hours even if the slabs are made with siliceousaggregates. With carbonate aggregate concrete,

the required thickness would be even less. Thus,the roof qualifies for a fire endurance significantlylonger than 2 hours.

6.2.3 CeilingsGypsum wallboard used as ceilings increases

the fire endurance of the assemblies. Very few firetests have been conducted utilizing concretefloors with gypsum wallboard ceilings, and nosuch tests have been conducted utilizing hollowcore units. To be effective, gypsum wallboardmust remain in place throughout most of the fireendurance period. Because most hollow coreunits by themselves have heat transmission fireendurances of one hour to two hours and longer,the wallboard must remain in place during fire ex-posure for long periods of time. For a fire endur-ance of 3 hours, a layer of 5/8 in (16 mm) Type Xgypsum wallboard can be used. The wallboardshould be installed as shown in Figure 6.2.3.

6.3 Structural Fire Endurance of Floor or RoofAssemblies

During standard fire tests, specimens must sup-port the anticipated superimposed loads through-

6--5

Fig. 6.2.3 Details of 3 hr. assembly consisting of hollow core slabs with a gypsum wall boardceiling

1. Precast concrete hollow core slabs - Minimum equivalent thickness = 2.75 in

2. Grout - (Not Shown) - Sand-cement grout along full length of joint.

3. Hanger Wire - No. 18 SWG galvanized steel wire. Hanger wire used to attached wallboard furring channels to precast concrete units.Wire to be located at each intersection of furring channels and joints between hollow core slabs, but not to exceed 4 ft o.c.

4. Wallboard Furring Channels - No. 26 ga. galvanized steel, 7/8 in high, 2 3/4 in base width, 1 3/8 in face width and 12 ft long. Channelsto be installed perpendicular to hollow core slabs and spaced 24 in o.c., except at wallboard butt joints where they are spaced 6 1/2 ino.c. Channels secured to concrete units with double strand of hanger wire looped through fasteners. At furring channel splices, chan-nels to be overlapped 6 in and tied together with hanger wire at each end of splice.

5. Wallboard - 5/8 in thick, 4 ft wide, Type X, installed with long dimension perpendicular to furring channels. Over butt joints, a 3 in widepiece of wallboard to be inserted with ends extending a minimum 6 in beyond board width.

6. Wallboard Fasteners - 1 in long, Type S, bugle head screws. Fasteners spaced 12 in on center along each furring channel except atbutt joints where fasteners spaced 8 in on center. At butt joints, fasteners located 3 1/4 in from board edge. Along side joints, fasten-ers located 3/4 in from board edge.

7. Joint System - (Not Shown) - Paper tape embedded in cementitious compound over joints, and covered with two layers of cementi-tious compound with edges feathered out. Wallboard fastener heads covered with two layers of cementitious compound.

Restrained Unrestrained

1

4

5

3

3

56 4

1 6

5

Side JointEnd Joint

5/8"

7/8"

Max. 4

1

3/4"

’

out the fire endurance period. Failure to supportthe loads is called the structural end point.

The most important factor affecting the struc-tural fire endurance of a floor or roof assembly isthe method of support, i.e., whether the assemblyis simply supported and free to expand (“unre-strained”) or if the assembly is continuous or ther-mal expansion is restricted (“restrained”).

6.3.1 Simply Supported SlabsFigure 6.3.1.1 illustrates the behavior of a sim-

ply supported slab exposed to fire from beneath.Because strands are parallel to the axis of the slab,

the ultimate moment capacity is constant through-out the length:φMn = φApsfps(dp -- a/2) (Eq. 6.3.1)See Chapter 2 for evaluating fps.

If the slab is uniformly loaded, the moment dia-gram will be parabolic with a maximum value atmidspan of:

M = wℓ2

8(Eq. 6.3.2)

6--6

Fig. 6.3.1.1 Moment diagrams for simply sup-ported beam or slab before andduring fire exposure

Fire

@ 0 Hr

M = applied moment

M = moment capacityn

M = reduced moment capacity

M =

n

@ 2 Hr

appliedmoment

θ

Wherew = dead plus live load per unit of length,

k/in

ℓ = span length, inAs the material strengths diminish with ele-

vated temperatures, the retained moment capacitybecomes:

Mnθ = Apsfpsθ(dp -- aθ /2) (Eq. 6.3.3)in which θ signifies the effects of high tempera-tures. Note that Aps and dp are not affected, but fpsis reduced. Similarly, a is reduced, but the con-crete strength at the top of the slab, f′c, is generallynot reduced significantly because of its lowertemperature.

Fig. 6.3.1.2 Temperature-strength relationships for hot-rolled and cold-drawn steels

700

Temperature, F

Perc

ent o

f St

reng

th A

t 70

F

High strengthalloy steel bars(tensile strength)

Hot-rolled steel(yield strength)

Cold-drawnprestressing steel250 or 270 ksi(tensile strength)

200 400 600 800 1000 1200 1400

20

40

60

80

100

o

o

6--7

Fig. 6.3.1.3 Temperatures withincarbonate aggregate concreteslabs during fire tests

Fig. 6.3.1.4 Temperatures withinsiliceous aggregate concreteslabs during fire tests

1/2 1 1-1/2 2 3 4500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1/4 in.

1/2 in.

u = 3/4 in. F

rom Exposed Surfa

ce

1 in.

1-1/2 in.

2 in.

3 in.

4 in.

Carbonate AggregateConcrete (Normal Weight)

Fire Test Time, Hr

Tem

pera

ture

,

Fo

1/2 1 1-1/2 2 3 4500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1/4 i

n.1/2

in.

u = 3/

4 in.

From

Exp

osed

Sur

face

1 in.

1-1/2

in.

2 in.

3 in

.

4 in.

Siliceous AggregateConcrete

Fire Test Time, Hr

(Normal Weight)

oT

empe

ratu

re,

F

Flexural failure can be assumed to occur whenMnθ is reduced to M. From this expression, it canbe seen that the fire endurance depends on the ap-plied loading and on the strength-temperaturecharacteristics of the steel. In turn, the duration ofthe fire before the “critical” steel temperature isreached depends upon the protection afforded tothe reinforcement.

Test results have shown that the theory dis-cussed above is valid, not only for hollow corefloors, but also for roofs with insulation on top ofthe slabs.

Figure 6.3.1.2 shows the relationship betweentemperature and strength of various types of steel.Figure 6.3.1.3, 6.3.1.4 and 6.3.1.5 show tempera-tures within concrete slabs during standard firetests. The data in those figures are applicable tohollow core slabs. By using the equations givenabove and the data in Figure 6.3.1.2 through6.3.1.5, the moment capacity of slabs can be cal-culated for various fire endurance periods, as il-lustrated in the following example:

Fig. 6.3.1.5 Temperatures within sand-lightweight concrete slabsduring fire tests

1/2 1 1-1/2 2 3 4500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600Sand-LightweightConcrete

Fire Test Time, Hr

1/4 in

.

1/2 in

.

u = 3/

4 in.

From E

xpos

ed S

urfac

e

1 in.

1-1/

2 in.

2 in.

3 in.

oT

empe

ratu

re,

F

6--8

Example 6.3.1Determine the maximum safe superimposed loadthat can be supported by an 8 in deep hollow coreslab with a simply supported unrestrained span of25 ft and a fire endurance of 3 hr.Given:h = 8 in; u = 1.75 in; six 1/2 in 270 ksi strands;Aps = 6(0.153) = 0.918 in2; b = 36 in; dp = 8 -- 1.75= 6.25 in; wD = 54 psf; carbonate aggregate con-crete; ℓ = 25 ft

Solution:

(a) Estimate strand temperature at 3 hr. from Fig-ure 6.3.1.3, θs at 3 hr. at 1.75 in above fire-ex-posed surface = 925 degrees F.

(b) Determine fpuθ from Figure 6.3.1.2. Forcold-drawn steel at 925 degrees F:

fpuθ = 33% fpu = 89.1 ksi

(c) Determine Mnθ and w

fpsθ = 89.11− 0.280.80 0.918

366.2589.1

5

= 86.8 ksi

aθ =0.91886.80.85536

= 0.52in

Mnθ= 0.918(86.8)(6.25 -- 0.52/2)/12

= 39.8 ft-kips

w =839.81000

(25)23= 170 psf

wL = w -- wD = 170 -- 54 = 116 psf

(d) Calculate maximum allowable wL at roomtemperature

fps = 270 1− 0.280.80 0.918

366.252705

= 249 ksi

a =0.9182490.85536

= 1.49 in

φMn = 0.9(0.918)(249)(6.25 -- 0.75)/12

= 94.3 ft-kips

wu =894.31000

(25)23= 402 psf

With load factors of 1.4 (dead load)+ 1.7 (live load):

wL =402− 1.454

1.7= 192 psf

Conclusion: wL = 116 < 192; 116 psf governs

Note: Fire endurance for heat transmissionshould also be checked

Table 6.3.1 shows values of u for simply sup-ported unrestrained hollow core slabs for variousmoment ratios and fire endurance of 1, 2, and 3hours. The values shown are based onApsfpu/bdpf′c = 0.05 and can be reduced by 1/16 infor Apsfpu/bdpf′c = 0.10.

Table 6.3.1 “u” inches, for simply supported unrestrained hollow core slabs*

1 0.50 1 1/4 (32) 1 1/16 (27) 1 1/16 (27)1 0.40 1 1/16 (27) 15/

16 (24) 15/16 (24)1 0.30 15/16 (24) 13/16 (21) 13/16 (21)2 0.50 1 15/16 (49) 1 13/16 (46) 1 13/16 (46)2 0.40 1 3/4 (44) 1 9/16 (40) 1 9/16 (40)2 0.30 1 9

/16 (40) 1 5/16 (33) 1 5/16 (33)3 0.50 2 1/2 (64) 2 5/16 (59) 2 1/8 (54)3 0.40 2 3/16 (56) 2 (51) 1 15/16 (49)3 0.30 1 15/16 (49) 1 11/16 (43) 1 11/16 (43)

*“u” is distance between center of strands and bottom of slab with all strands having same “u”. Basedon Apsfpu/bdpf′c = 0.05; conservative for values greater than 0.05.

FireEndurance

(hr)M/Mn

Aggregate TypeSiliceous Carbonate Sand-Lightweight

(in) (mm) (in) (mm) (in) (mm)

6--9

Fig. 6.3.2 Equivalent concrete cover thickness for spray-applied coatings

0 1/4 1/2 3/4 1 1-1/4 1-1/20

1

2

3

4

5

Equ

ival

ent C

oncr

ete

Cov

er T

hick

ness

, in.

Thickness of Spray-Applied Insulating Material, in.

Vermicu

lite Cem

entio

us Mate

rial (V

CM) (slab

s)

Sprayed

Minera

l Fiber

(SMF) (slab

s)

6.3.2 Effect of Spray-Applied CoatingsThe fire endurance of hollow core slabs can be

increased by the addition of a spray-applied coat-ing of vermiculite cementitious material orsprayed mineral fiber. Figure 6.3.2 shows therelationship between thickness of spray-appliedcoatings and equivalent concrete cover. Thus, ifstrands are centered 3/4 in (19 mm) above the bot-tom of a hollow core slab and if 1/4 in (6 mm) ofsprayed mineral fiber is applied, the u distance tobe used in Figures 6.3.1.3, 6.3.1.4 or 6.3.1.5 is 3/4in (19 mm) plus the equivalent cover of 0.9 in (23mm) obtained from Figure 6.3.2.

6.3.3 Structurally Continuous SlabsContinuous members undergo changes in

stresses when subjected to fire, resulting fromtemperature gradients within the structural mem-bers, or changes in strength of the materials at hightemperatures, or both.

Figure 6.3.3.1 shows a continuous beam whoseunderside is exposed to fire. The bottom of thebeam becomes hotter than the top and tends to ex-pand more than the top. This differential tempera-ture causes the ends of the beam to tend to lift fromtheir supports thereby increasing the reaction at

Fig. 6.3.3.1 Moment diagrams for continuous 2-span beam before and during fireexposure

Fire Fire

M

At 0 Hr

M

M

n

n

M

-

+

n

M

+

-

MAt 3 Hr

x1θ

M nθ

x0

-

-

+

+

6--10

Fig. 6.3.3.3 Symmetrical uniformly loaded mem-ber continuous at both supports

Fig. 6.3.3.2 Uniformly loaded membercontinuous at one support

x

wM nθ

x1x2 0x

θnM

Mn

θ

+

-

-

w

M nθM θn

w2

8M

θn

x0 x2 x0

Mn

θ

- -

-

+

the interior support. This action results in a redis-tribution of moments, i.e., the negative moment atthe interior support increases while the positivemoments decrease.

During the course of a fire, the negative mo-ment reinforcement (Figure 6.3.3.1) remainscooler than the positive moment reinforcementbecause it is better protected from the fire. Thus,the increase in negative moment can be accom-modated. Generally, the redistribution that occursis sufficient to cause yielding of the negative mo-ment reinforcement. The resulting decrease inpositive moment means that the positive momentreinforcement can be heated to a higher tempera-ture before a failure will occur. Therefore, the fireendurance of a continuous concrete beam is gen-erally significantly longer than that of a simplysupported beam having the same cover and loadedto the same moment intensity.

It is possible to design the reinforcement in acontinuous beam or slab for a particular fire en-durance period. From Figure 6.3.3.1, the beamcan be expected to collapse when the positive mo-ment capacity, M+

nθ, is reduced to the value indi-cated by the dashed horizontal line, i.e., when theredistributed moment at point x1, from the outersupport, Mx1

= M+nθ.

Figure 6.3.3.2 shows a uniformly loaded beamor slab continuous (or fixed) at one support and

simply supported at the other. Also shown is theredistributed applied moment diagram at failure.

Values for M+nθ can be calculated by the proce-

dures given for “Simply Supported Slabs”.Values for M−

nθ and xo can be calculated:

M−nθ = wℓ2

2 wℓ2 2M+

n

wℓ2 (Eq. 6.3.4)

xo = 2M−

nθ

wℓ (Eq. 6.3.5)

In most cases, redistribution of moments oc-curs early during the course of a fire before thenegative moment capacity has been reduced bythe effects of fire. In such cases, the length of xo isincreased, i.e., the inflection point moves towardthe simple support. For such cases,

xo = 2M−n

wℓ (Eq. 6.3.6)

Figure 6.3.3.3 shows a symmetrical beam orslab in which the end moments are equal. In thatcase:

M−nθ = wℓ2∕8−M+

nθ (Eq. 6.3.7)

andwx2

28

= M+nθ (Eq. 6.3.8)

In negative moment regions, the compressivezone is directly exposed to fire, so calculations fordθ and aθ must be modified by (a) using f′cθ fromFigure 6.3.3.4 and (b) neglecting concrete hotterthan 1400 degrees F (760 degrees C).

6--11

Fig. 6.3.3.4 Compressive strength of concrete at high temperatures

70 200 400 600 800 1000 1200 1400

70 200 400 600 800 1000 1200 1400

20

40

60

80

100

120

0 0

20

40

60

80

100

120

Perc

ent o

f O

rigi

nal C

ompr

essi

ve S

tren

gth

Sand-Lightweight

Siliceous

Carbonate

Original Strength = f’Average f’ = 3900 psiStressed to 0.4f’ during heating

c

c

c

Temperature, Fo

6.3.4 Detailing PrecautionsIt should be noted that the amount of moment

redistribution that can occur is dependent uponthe amount of negative reinforcement. Tests haveclearly demonstrated that the negative moment re-inforcement will yield, so the negative momentcapacity is reached early during a fire test, regard-less of the applied loading. The designer must ex-ercise care to ensure that a secondary type of fail-ure will not occur. To avoid a compression failurein the negative moment region, the amount of neg-ative moment reinforcement should be smallenough so thatωθ, i.e., Asfyθ /bθdθ f′cθ, is less than0.30, before and after reductions in fy, b, d and f′care taken into account. Furthermore, the negativemoment bars or mesh must be long enough to ac-commodate the complete redistributed momentand change in the inflection points. It should benoted that the worst condition occurs when the ap-plied loading is smallest, such as dead load pluspartial or no live load. It is recommended that atleast 20% of the maximum negative moment rein-forcement be extended throughout the span.

Example 6.3.2Determine the amount of negative moment re-

inforcement needed to provide a 3 hr. fire endur-ance for sand-lightweight hollow core slabs, 8 indeep, 5 ksi concrete, 48 in wide, with six 7/16 in 270ksi strands and 2 in (4 ksi) composite topping.Slabs span 25 ft of an exterior bay (no restraint tothermal expansion). Dead load = 65 psf, live load= 100 psf. Strands are centered 1 3/4 in above bot-tom of slab. The value for M+

nθ can be calculated(by using the procedure discussed for simply sup-ported slabs) to be 39.0 ft-kips. From Eq. 6.3.4(for use in Eq. 6.3.4):wℓ2 = 4(65 + 100)(25)2/1000

= 412.5 ft-kips

M−nθ = 412.5

2− 412.5 2

(39.0)412.5

= 26.9 ft-kips

Determine As neglecting concrete above 1400 de-grees F in negative moment region. From Figure6.3.1.5 neglect 3/4 in above bottom, and assumesteel centered in topping.d = 10 -- 3/4 -- 1 = 8.25 in

6--12

Assume f′cθ in compressive zone = 0.8f′c = 4 ksiAssume d -- aθ/2 = 8.1 in

As =26.912608.1

= 0.66 in2

check aθ =0.66600.85448

= 0.24 in

d -- aθ /2 = 8.25 -- 0.12 = 8.13 in ≅ 8.1 OKUse 6 x 6 - W2.1 x W2.1 WWF throughout plus #4Grade 60 at 16 in in negative moment region.

As = 80.021+ 4816

0.20 = 0.768 in2

Calculate xo for dead load plus one-half live load.

M−nθ = 0.768

0.66(26.9) = 31.3 ft-kips

loading = 4(0.065 + 0.050) = 0.46 k/ft;Mn = 34.0 ft-kips (calculated for room tem-peratures)

From Eq. 6.3.6

xo =2M−

n

wℓ = 2(34.0)0.46(25) = 5.91 ft

Half of #4 bars should extend 7 ft each side of inte-rior support and half 5 ft.Use #4 Grade 60, 12 ft long at 16 in and stagger.

6.4 Restraint to Thermal ExpansionIf a fire occurs beneath a portion of a large floor

or roof, such as beneath a concrete floor slab inone interior bay of a multi-bay building, theheated portion will expand and push against thesurrounding unheated portion. In turn the un-heated portion exerts compressive forces on theheated portion. The compressive force, or thrust,acts near the bottom of the slab when the firestarts, but as the fire progresses, the line of thrustrises and the thermal gradient diminishes and theheated concrete undergoes a reduction in elasticmodulus. If the surrounding slab is thick andheavily reinforced, the thrust forces can be quitelarge, but they will be considerably less than thosecalculated by use of elastic properties of concreteand steel, together with appropriate coefficients ofexpansion. At high temperatures, creep and stressrelaxation play an important role. Nevertheless,the thrust is generally great enough to increase thefire endurance significantly, in some instances bymore than 2 hours. In most fire tests of restrainedassemblies, the fire endurance is determined by

Fig. 6.4.1 Moment diagrams for axiallyrestrained beam during fire exposure.Note that at 2 hr. Mnθ is less than Mand effects of axial restraint permitbeam to continue to support load.

Fire

T TTd

MM n@ 0 Hr, T=0

M nθM =T(d - - )a_θ

2T ∆

@ 2 Hr.

T

(curved due to beam deflection)

temperature rise of the unexposed surface ratherthan by structural considerations, even though thesteel temperatures often exceed 1200 degrees F(650 degrees C).

The effects of restraint to thermal expansioncan be characterized as shown in Figure 6.4.1.The thermal thrust, T, acts in a manner similar toan external prestressing force, which tends to in-crease the positive moment capacity.

Methods for calculating fire endurance of “re-strained” floors or roofs are given in PCI MNL124-89. It is seldom necessary to make such cal-culations, as noted below. The beneficial effectsof restraint are recognized in ASTM E119. Thestandard presents a guide for determining condi-tions of restraint. The guide includes Figure 6.4.2.In most cases, the interior bays of multi-bay floorsand roofs can be considered to be restrained andthe magnitude and location of the thrust are gener-ally of academic interest only. It should be notedthat Figure 6.4.2 indicates that adequate restraintcan occur in interior bays and exterior bays offramed buildings when:

“The space between the ends of precast unitsand the vertical faces of supports, or between theends of solid or hollow core slab units does not ex-ceed 0.25 percent of the length for normal weightconcrete members or 0.1 percent of the length forstructural lightweight concrete members”.

Sketches illustrating typical conditions de-scribed above are shown in Figure 6.4.3.

6--13

Fig. 6.4.2. Examples of typical restrained and unrestrained construction classifications (fromAppendix X3 of ASTM E119-88)

Fig. 6.4.3. Typical examples of restrained floors or roofs of precast construction

I. Wall Bearing:Single span and simply supported end spans of multiple baysa

(1) Open-web steel joists or steel beams, supporting concrete slab, precast units or metal decking unrestrained(2) Concrete slabs, precast units or metal decking unrestrainedInterior spans of multiple bays:(1) Open-web steel joists, steel beams or metal decking, supporting continuous concrete slab restrained(2) Open-web steel joists or steel beams, supporting precast units or metal decking unrestrained(3) Cast-in-place concrete slab systems restrained(4) Precast concrete where the potential thermal expansion is resisted by adjacent constructionb restrained

II. Steel Framing:(1) Steel beams welded, riveted, or bolted to the framing members restrained(2) All types of cast-in-place floor and roof systems (such as beam-and-slabs, flat slabs, pan joints, and

waffle slabs) where the floor or roof system is secured to the framing members restrained(3) All types of prefabricated floor or roof systems where the structural members are secured to the framing

members and the potential thermal expansion of the floor or roof system is resisted by the framingsystem or the adjoining floor or roof constructionb restrained

III.Concrete Framing:(1) Beams securely fastened to the framing members restrained(2) All types of cast-in-place floor or roof systems (such as beam-and-slabs, flat slabs, pan joists, and

waffle slabs) where the floor system is cast with framing members restrained(3) Interior and exterior spans of precast systems with cast-in-place joints resulting in restraint equivalent

to that which would exist in condition III(1) restrained(4) All types of prefabricated floor or roof systems where the structural members are secured to such

systems and the potential thermal expansion of the floor or roof system is resisted by the framingsystem or the adjoining floor or roof constructionb restrained

IV. Wood ConstructionAll Types unrestrained

aFloor and roof systems can be considered restrained when they are tied into walls with or without tie beams, the walls beingdesigned and detailed to resist thermal thrust from the floor or roof system.

bFor example, resistance to potential thermal expansion is considered to be achieved when:(1) Continuous structural concrete topping is used.(2) The space between the ends of precast units or between the ends of units and the vertical face of supports is filled with concrete

or mortar.(3) The space between the ends of precast units and the vertical faces of supports, or between the ends of solid or hollow core

slab units does not exceed 0.25 percent of the length for normal weight concrete members or 0.1 percent of the length for structurallightweight concrete members.

To be considered as restrained:c1 + c2 < 0.0025 for normal weight concretec1 + c2 < 0.0010 for lightweight concrete

Example: Determine maximum value of c1 + c2 for normal weight hollow core slabs with a clear span of 30 ftSolution: c1 + c2 = 0.0025(30 x 12) = 0.90 in

ℓ

Hollow-Core Slabs or Double Teesc1 c2

Hollow-Core or Solid Slabsc1 2c

ℓ

6--14

Example 6.4.1Hollow core floor slabs were installed in a

building several years ago when a 1 hr. fire endur-ance was required. The occupancy of the buildingwill be changed and the floors must qualify for a 3hr. fire endurance. What can be done to upgradethe fire endurance?Given:

Slabs are 4 ft wide, 8 in deep, prestressed withfive 3/8 in 270 ksi strands located 1 in above thebottom of the slab, and span 24 ft. Slabs are madewith 5000 psi siliceous aggregate concrete, havean equivalent thickness of 3.75 in, and weigh 47psf. The slabs are untopped and the superimposedload will be 50 psf.Solution:

There are a number of possible solutions. Theappropriate solution will depend on architecturalor functional requirements and economics.

For some parts of the building, the slabs mightbe made to qualify as “restrained” in accordancewith Figure 6.4.2 and Figure 6.4.3, in which casethose slabs would qualify structurally for 3 hours,but would still have to be upgraded to qualify for 3hours by heat transmission.

A gypsum wallboard ceiling installed as shownin Figure 6.2.3 would provide three hours bothstructurally and for heat transmission. Calcula-tions of the ultimate capacity and stresses shouldbe made to assure that the added weight of the ceil-ing can be adequately supported.

A spray-applied undercoating of vermiculitecementitious material or sprayed mineral fibercan also be used. For heat transmission, the re-quired thickness for three hours of undercoating is0.6 in (see Figure 6.2.2.1). From Figure 6.3.2, itcan be seen that with a thickness of 0.6 in of VCMor SMF, the equivalent thickness of concrete cov-er is more than 2 in. Thus, the equivalent “u” dis-tance is more than 2 in plus 1 in or more than 3 in.From Figure 6.3.1.4, with u more than 3 in, thestrand temperature will be less than 600 degrees Fat three hours, so the strength of the prestressingsteel will be 65% of its 70 degrees F strength (Fig-ure 6.3.1.5) or more than 0.65 x 270 = 175.5 ksi.Calculations can be made in accordance with theprocedures in the section headed “Simply Sup-ported Slabs”, but if the strand strength is reducedless than about 50% of its room temperature

strength, the assembly will generally be satisfac-tory structurally.

CHAPTER 7

7--1

ACOUSTICAL PROPERTIES OFHOLLOW CORE FLOOR SLABS

7.1 GlossaryAirborne Sound - sound that reaches the point ofinterest by propagation through air.

Background Level - the ambient sound pressurelevel existing in a space.

Decibel (dB) - a logarithmic unit of measure ofsound pressure or sound power. Zero on the deci-bel scale corresponds to a standardized referencespressure (20 µPa) or sound power (10-12 watt).

Flanking Transmission - transmission of sound byindirect paths other than through the primary bar-rier.

Frequency (Hz) - the number of complete vibra-tion cycles per second.

Impact Insulation Class (IIC) - a single figure rat-ing of the overall impact sound insulation meritsof floor-ceiling assemblies in terms of a referencecontour (ASTM E989).

Impact Noise - the sound produced by one objectstriking another.

Noise - unwanted sound.

Noise Criteria (NC) - a series of curves, used asdesign goals to specify satisfactory backgroundsound levels as they relate to particular use func-tions.

Noise Reduction (NR) - the difference in decibelsbetween the space-time average sound pressurelevels produced in two enclosed spaces by one ormore sound sources in one of them.

Noise Reduction Coefficient (NRC) - the arithme-tic average of the sound absorption coefficients at250, 500, 1000 and 2000 Hz expressed to the near-est multiple of 0.05 (ASTM C423).

Reverberation - the persistence of sound in an en-closed or partially enclosed space after the sourceof sound has stopped.

Room Criteria (RC) Curves - a revision of the NCcurves based on empirical studies of backgroundsounds.

Sabin - the unit of measure of sound absorption(ASTM C423).

Sound Absorption Coefficient (α) - the fraction ofrandomly incident sound energy absorbed orotherwise not reflected off a surface (ASTMC423).

Sound Pressure Level (SPL) - ten times the com-mon logarithm of the ratio of the square of thesound pressure to the square of the standard refer-ence pressure of 20 µPa. Commonly measuredwith a sound level meter and microphone, thisquantity is expressed in decibels.

Sound Transmission Class (STC) - the singlenumber rating system used to give a preliminaryestimate of the sound insulation properties of apartition system. This rating is derived from mea-sured values of transmission loss (ASTM E413).

Sound Transmission Loss (TL) - ten times thecommon logarithm of the ratio, expressed in deci-bels, of the airborne sound power incident on thepartition that is transmitted by the partition and ra-diated on the other side (ASTM E90).

Structureborne Sound - sound that reaches thepoint of interest over at least part of its path byvibration of a solid structure.

7.2 GeneralThe basic purpose of architectural acoustics is

to provide a satisfactory environment in which de-sired sounds are clearly heard by the intended lis-teners and unwanted sounds (noise) are isolated orabsorbed.

Under most conditions, the architect/ engineercan determine the acoustical needs of the spaceand then design the building to satisfy thoseneeds. Good acoustical design utilizes both ab-sorptive and reflective surfaces, sound barriers

7--2

and vibration isolators. Some surfaces must re-flect sound so that the loudness will be adequate inall areas where listeners are located. Other sur-faces absorb sound to avoid echoes, sound distor-tion and long reverberation times. Sound is iso-lated from rooms where it is not wanted by se-lected wall and floor-ceiling constructions.Vibration generated by mechanical equipmentmust be isolated from the structural frame of thebuilding.

Most acoustical situations can be described interms of: (1) sound source, (2) sound transmis-sion path, and (3) sound receiver. Sometimes thesource strength and path can be controlled and thereceiver made more attentive by removing dis-traction or made more tolerant of disturbance.Acoustical design must include consideration ofthese three elements.

7.3 Approaching the Design ProcessCriteria must be established before the acousti-

cal design of a building can begin. Basically a sat-isfactory acoustical environment is one in whichthe character and magnitude of all sounds arecompatible with the intended space function.

Although a reasonable objective, it is not al-ways easy to express these intentions in quantita-tive terms. In addition to the amplitude of sound,the properties such as spectral characteristics,continuity, reverberation and intelligibility mustbe specified.

People are highly adaptable to the sensations ofheat, light, odor, sound, etc. with sensitivities va-rying widely. The human ear can detect a soundintensity of rustling leaves, 10 dB, and can toler-ate, if even briefly, the powerful exhaust of a jetengine at 120 dB, 1012 times the intensity of therustling sound.

7.3.1 Dealing with Sound LevelsThe problems of sound insulation are usually

considerably more complicated than those ofsound absorption. The former involves reduc-tions of sound level, which are of the greater or-ders of magnitude than can be achieved by absorp-tion. These reductions of sound level from spaceto space can be achieved only by continuous, im-pervious barriers. If the problem also involvesstructure borne sound, it may be necessary to

introduce resilient layers or discontinuities intothe barrier.

Sound absorbing materials and sound insulat-ing materials are used for different purposes.There is not much sound absorption from an 8 in(200 mm) hollow core concrete slab; similarly,high sound insulation is not available from a po-rous lightweight material that may be applied toroom surfaces. It is important to recognize that thebasic mechanisms of sound absorption and soundinsulation are quite different.

7.4 Sound Transmission LossSound transmission loss measurements are

made at 16 frequencies at one-third octave inter-vals covering the range from 125 to 4000 Hz. Thetesting procedure is ASTM Specification E90,Laboratory Measurement of Airborne SoundTransmission Loss ofBuilding Partitions. To sim-plify specification of desired performance charac-teristics, the single number Sound TransmissionClass (STC) was developed.

Airborne sound reaching a floor or ceiling pro-duces vibration in the slab and is radiated with re-duced intensity on the other side. Airborne soundtransmission loss of a floor-ceiling assembly is afunction of its weight, stiffness and vibrationdamping characteristics.

Weight is concrete’s greatest asset when it isused as a sound insulator. For sections of similardesign, but different weights, the STC increasesapproximately 6 units for each doubling of weightas shown in Figure 7.4.1.

Fig. 7.4.1 Sound Transmission Class as afunction of weight of floor

40 50 60 70 80 90 10040

45

50

55

60

65

hollow core slabs

statistical tolerance 2.5 STC+_STC = 0.1304 W + 43.48

Weight Per Unit Area, psf

Soun

d T

rans

mis

sion

Cla

ss (

STC

)

7--3

Fig. 7.4.2 Acoustical test data of hollow core slabs (normal weight concrete)

100

125

160

200

250

315

400

500

630

800

1000

1250

1600

2000

2500

3150

4000

5000

10

20

30

40

50

60

70

16010

100

125

20

30

40

200

250

315

500

400

630

1600800

1000

1250

2000

2500

3150

70

50

60

5000

4000

Impa

ct S

ound

Pre

ssur

e L

evel

dB

Soun

d T

rans

mis

sion

Los

s, d

B

8 in. hollow core. STC-50

6 in. hollow core. STC-48

Impact Insulation

6 in. hollow core (bare).IIC-23

8 in. hollow core (bare).IIC-28

6 in. hollow corewith carpet and pad. IIC-69

with carpet and pad. IIC-738 in. hollow core

Frequency, Hz

Sound Transmission Loss

Precast concrete floors and roofs usually do notneed additional treatments in order to provide ad-equate sound insulation. If desired, greater soundinsulation can be obtained by using a resilientlyattached layer(s) of gypsum board or other build-ing material. The increased transmission loss oc-curs because the energy flow path is now in-creased to include a dissipative air column andadditional mass.

The acoustical test results of both airbornesound transmission loss and impact insulation of 6and 8 in (150 and 200 mm) hollow core slabs areshown in Figure 7.4.2. Table 7.4.1 presents theratings for various floor-ceiling assemblies.

7.5 Impact Noise ReductionFootsteps, dragged chairs, dropped objects,

slammed doors, and plumbing generate impactnoise. Even when airborne sounds are adequatelycontrolled there can be severe impact noise prob-lems.

The test method used to evaluate systems forimpact sound insulation is described in ASTMSpecification E492, Laboratory Measurement ofImpact Sound Transmission Using the TappingMachine. As with the airborne standard, mea-surements are made at 16 one-third octave inter-vals but in the range from 100 to 3150 Hz. For per-formance specification purposes, the single num-ber Impact Insulation Class (IIC) is used.

Hollow core floors in combination with resil-ient materials effectively control impact sound.One simple solution consists of good carpeting onresilient padding. Table 7.4.1 shows that a carpetand pad over a bare slab will significantly increasethe impact noise reduction. The overall efficiencyvaries according to the characteristics of the car-peting and padding such as resilience, thicknessand weight. So called resilient flooring materials,such as linoleum, rubber, asphalt vinyl, etc. arenot entirely satisfactory directly on concrete, norare parquet or strip wood floors when applied di-rectly. Impact sound also may be controlled by

7--4

1. 6 in (150 mm) hollow core slabs 48 232. Assembly 1 with carpet and pad 48 693. Assembly 1 with 1/2 in (13 mm) wood block flooring adhered directly 48 484. Assembly 1 with 1/2 in (13 mm) wood block flooring adhered to 1/2 in (13 mm)

sound-deadening board underlayment adhered to concrete 49 495. Assembly 1 with 3/4 in (19 mm) gypsum concrete 50 416. Assembly 1 with 3/4 in (19 mm) gypsum concrete on 1/2 in (13 mm) sound-

deadening board underlayment adhered to concrete 50 507. Assembly 1 with carpet and pad on 3/4 in (19 mm) gypsum concrete on

1/2 in (13 mm) sound-deadening board underlayment adhered to concrete 50 72

8. 8 in (200 mm) hollow core slabs 50 289. Assembly 8 with carpet and pad 50 73

10. Assembly 8 with 1/2 in (13 mm) wood block flooring adhered directly 51 4711. Assembly 8 with 1/2 in (13 mm) wood block flooring adhered to 1/2 in (13 mm)

sound-deadening board underlayment adhered to concrete 52 5412. Assembly 8 with 1/2 in (13 mm) wood block flooring adhered to 1/2 in (13 mm)

plywood adhered to 7/16 in (11 mm) sound-deadening board underlaymentadhered to concrete 52 55

13. Assembly 8 with 5/16 in (8 mm) wood block flooring adhered to 1/4 in (6 mm)polystyrene underlayment adhered to concrete 50 51

14. Assembly 8 with vinyl tile adhered to 1/2 in (13 mm) plywood adhered to7/16 in (11 mm) sound-deadening board underlayment adhered to concrete 50 55

15. Assembly 8 with vinyl tile adhered to 1/4 in (6 mm) inorganic felt supportedcushion underlayment adhered to concrete 50 51

16. Assembly 8 with vinyl tile adhered to 1/8 in (3 mm) polyethylene foam under-layment adhered to concrete 50 58

17. Assembly 8 with 1 1/2 in (38 mm) concrete topping with carpet and pad 50 7618. Assembly 8 with 1 1/2 in (38 mm) concrete topping with vinyl tile adhered to

concrete 50 4419. Assembly 8 with 1 1/2 in (38 mm) concrete topping with vinyl tile adhered

to 3/8 in (9 mm) plywood adhered to 1/2 in (13 mm) sound-deadening boardadhered to concrete 52 55

20. Assembly 8 with 1 1/2 in (38 mm) concrete with 1/2 in (13 mm) wood blockflooring adhered to 1/2 in (13 mm) sound-deadening board adhered to concrete 51 53

21. Assembly 8 with 1 1/2 in (38 mm) concrete with 5/16 in (8 mm) wood blockflooring adhered to foam backing adhered to concrete 51 54

22. Assembly 8 with 3/4 in (19 mm) gypsum concrete with 5/16 in (8 mm) woodblock flooring adhered to foam backing adhered to concrete 50 53

23. Assembly 11 with acoustical ceiling 59 6124. Assembly 8 with quarry tile, 1 1/4 in (32 mm) reinforced mortar bed with

0.4 in (10 mm) nylon and carbon black spinerette matting 60 5425. Assembly 24 with suspended 5/8 in (16 mm) gypsum board ceiling with

3 1/2 in (90 mm) insulation 61 62

AssemblyNo. Description STC IIC

Table 7.4.1 Airborne sound transmission and impact insulation class ratings from laboratorytests of hollow core slab floor-ceiling assemblies

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providing a discontinuity in the structure such aswould be obtained by adding a resilient-mountedplaster or drywall suspended ceiling.

7.6 Absorption of SoundA sound wave always loses part of its energy as

it is reflected by a surface. This loss of energy istermed sound absorption. It appears as a decreasein sound pressure of the reflected wave. Thesound absorption coefficient is the fraction of en-ergy incident but not reflected per unit of surfacearea. Sound absorption can be specified at indi-vidual frequencies or as an average of absorptioncoefficients (NRC).

A dense, non-porous concrete surface typicallyabsorbs 1 to 2% of incident sound and has an NCRof 0.015. In the case where additional sound ab-sorption of precast concrete is desired, a coatingof acoustical material can be spray applied, acous-tical tile can be applied with adhesive, or an acous-tical ceiling can be suspended. Most of the sprayapplied fire retardant materials used to increasethe fire resistance of precast concrete and otherfloor-ceiling systems can also be used to absorbsound. The NCR of the sprayed fiber types rangefrom 0.25 to 0.75. Most cementitious types havean NCR from 0.25 to 0.50.

If an acoustical ceiling were added to Assem-bly 11 of Table 7.4.1 (as in Assembly 23), thesound entry through a floor or roof would be re-duced 7dB. In addition, the acoustical ceilingwould absorb a portion of the sound after entryand provide a few more decibels of quieting. Useof the following expression can be made to deter-mine the intra-room noise or loudness reductiondue to the absorption of sound.

NR = 10 log Ao +AaAa

(Eq. 7.6.1)

whereNR = sound pressure level reduction, dB

Ao = original absorption, Sabins

Aa = added absorption, SabinsValues for Ao and Aa are the products of the ab-

sorption coefficients of the various room materi-als and their surface areas.

A plot of this equation is shown in Figure 7.6.1.For an absorption ratio of 5, the decibel reductionis 7dB. Note that the decibel reduction is the

same, regardless of the original sound pressurelevel and depends only on the absorption ratio.This is due to the fact that the decibel scale is itselfa scale of ratios, rather than difference in soundenergy.

While a decibel difference is an engineeringquantity which can be physically measured, it isalso important to know how the ear judges thechange in sound energy due to sound condition-ing. Apart from the subjective annoyance factorsassociated with excessive sound reflection, the earcan make accurate judgments of the relative loud-ness between sounds. An approximate relationbetween percentage loudness, reduction of re-flected sound and absorption ratio is plotted inFigure 7.6.2

The percentage loudness reduction does not de-pend on the original loudness, but only on the ab-sorption ratio. (The curve is drawn for loudnesswithin the normal range of hearing and does notapply to extremely faint sounds.) Referring againto the absorption ratio of 5, the loudness reductionis read from Figure 7.6.2 as approximately 40 per-cent.

7.7 Acceptable Noise CriteriaAs a rule, a certain amount of continuous sound

can be tolerated before it becomes noise. An “ac-ceptable” level neither disturbs room occupantsnor interferes with the communication of wantedsound.

The most widely accepted and used noise crite-ria today are expressed as the Noise Criterion(NC) curves, Figure 7.7.1a. The figures in Table7.7.1 represent general acoustical goals. They canalso be compared with anticipated noise levels inspecific rooms to assist in evaluating noise reduc-tion problems.

The main criticism of NC curves is that they aretoo permissive when the control of low or highfrequency noise is of concern. For this reason,Room Criterion (RC) Curves were developed(Figure 7.7.1b).39,40 RC curves are the result ofextensive studies based on the human response toboth sound pressure level and frequency and takeinto account the requirements for speech intelligi-bility.

A low background level obviously is necessarywhere listening and speech intelligibility is im-

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Fig. 7.6.1 Relation of decibel reduction ofreflected sound to absorption ratio

Fig. 7.6.2 Relation of percent loudness reductionof reflected sound to absorption ratio

Fig. 7.7.1a NC (Noise Criteria) Curves

0 2 4 6 8 10 12 14 16 18 200

2

4

6

8

10

12

14

Absorption Ratio, A /A2 1

SPL

Red

uctio

n, d

B

0 2 4 6 8 10 12 14 16 18 200

10

20

30

40

50

60

Absorption Ratio, A /A2 1

Perc

ent L

oudn

ess

Red

uctio

n

10

20

30

40

50

60

70

80

90

63 125 250 500 2000 8000Oct

ave

Ban

d So

und

Pres

sure

Lev

el, d

B r

e 20

mic

ropa

scal

s

Octave Band Center Frequencies, Hz

NC-65NC-60NC-55NC-50NC-45NC-40NC-35NC-30

NC-25NC-20NC-15

Approximatethreshold ofhearing forcontinuousnoise

Fig. 7.7.1b RC (Room Criteria) Curves

Region A: High probability that noise-induced vibration levels inlightweightwall/ceilingconstructionswill beclearly feelable;antic-ipate audible rattles in light fixtures, doors, windows, etc.Region B: Noise-induced vibration levels in lightweight wall/ceil-ing constructions may be moderately feelable; slight possibility ofrattles in light fixtures, doors, windows, etc.Region C: Below threshold of hearing for continuous noise.

1063 125 250 500 1000 2000 4000

Oct

ave

Ban

d So

und

Pres

sure

Lev

el, d

B r

e 20

mic

ropa

scal

s

Octave Band Center Frequencies, Hz

31.516

20

30

40

50

60

70

80

90

A

B

C

RC

50

45

4035

30

25

portant. Conversely, higher levels can persist inlarge business offices or factories where speechcommunication is limited to short distances.Often it is just as important to be interested in theminimum as in the maximum permissible levelsof Table 7.7.1. In an office or residence, it is desir-able to have a certain ambient sound level to as-sure adequate acoustical privacy between spaces,thus, minimizing the transmission loss require-ments of unwanted sound (noise).

These undesirable sounds may be from an exte-rior source such as automobiles or aircraft, or theymay be generated as speech in an adjacent class-room or music in an adjacent apartment. Theymay be direct impact-induced sound such as foot-falls on the floor above, rain impact on a light-weight roof construction or vibrating mechanicalequipment.

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Thus, the designer must always be ready to ac-cept the task of analyzing the many potentialsources of intruding sound as related to their fre-quency characteristics and the rates at which theyoccur. The level of toleration that is to be expectedby those who will occupy the space must also beestablished. Figures 7.7.2 and 7.7.3 are the spec-tral characteristics of common noise sources.

Fig. 7.7.2 Sound pressure levels - exterior noisesources

Fig. 7.7.3 Sound pressure levels - interior noisesources

63 125 250 500 1000 2000 4000 80000

20

40

60

80

100

120

Frequency, Hz.

Soun

d Pr

essu

re L

evel

, dB

automobiles - 20 ft

heavy truck - 20 ft

bus

jet aircraft takeoff500 ft

propeller aircrafttakeoff-500 ft

63 125 250 500 1000 2000 4000 80000

20

40

60

80

100

120

Frequency, Hz.

Soun

d Pr

essu

re L

evel

, dB

business machinetabulating room

teenager levelstero phonograph

riveting

typical office

bed or dining room

kitchen

NC OR RCTYPE OF SPACE CURVE

1. Private residences 25 to 302. Apartments 30 to 353. Hotels/motels

a. Individual rooms or suites 30 to 35b. Meeting/banquet rooms 30 to 35c. Halls, corridors, lobbies 35 to 40d. Service/support areas 40 to 45

4. Officesa. Executive 25 to 30b. Conference rooms 25 to 30c. Private 30 to 35d. Open-plan areas 35 to 40e. Computer/business

machine areas 40 to 45f. Public circulation 40 to 45

5. Hospitals and clinicsa. Private rooms 25 to 30b. Wards 30 to 35c. Operating rooms 25 to 30d. Laboratories 30 to 35e. Corridors 30 to 35f. Public areas 35 to 40

6. Churches 25 to 30**7. Schools

a. Lecture and classrooms 25 to 30b. Open-plan classrooms 30 to 35**

8. Libraries 30 to 359. Concert Halls **

10. Legitimate theatres **11. Recording studios **12. Movie theatres 30 to 35

* Design goals can be increased by 5 dB whendictated by budget constraints or when noiseintrusion from other sources represents a lim-iting condition.

**An acoustical expert should be consulted forguidance on these critical spaces.

Table 7.7.1 Recommended category classi-fication and suggested NoiseCriteria range for steady back-ground noise as heard in variousin-door functional activityareas*39

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Sound Pressure Level - (dB)

Frequency (Hz) 63 125 250 500 1000 2000 4000 8000Stereo Source Noise (teenager) 60 72 84 82 82 80 75 60(Figure 7.7.3)Bedroom Room Criteria 50 45 40 35 30 25 20 15

RC 30 (Figure 7.7.l)Required Insulation 10 27 44 47 52 55 55 45

Sound Pressure Level - (dB)Frequency (Hz) 125 250 500 1000 2000 4000Required Insulation 27 44 47 52 55 558 in H.C. (Figure 7.4.2) 34 39 46 53 59 64Deficiencies ---- 5 1 ---- ---- ----

With these criteria, the problem of sound isola-tion now must be solved, namely, the reductionprocess between the high unwanted noise sourceand the desired ambient level. For this solution,two related yet mutually exclusive processes mustbe incorporated, i.e., sound transmission loss andsound absorption.

7.8 Establishment of Noise Insulation objec-tives

Often acoustical control is specified as to theminimum insulation values of the dividing parti-tion system. Municipal building codes, lendinginstitutions and the Department of Housing andUrban Development (HUD) list both airborneSTC and impact IIC values for different living en-vironments. For example, the HUD minimumproperty standards41 are:

LOCATION STC IIC

Between living units 45 45Between living units and 50 50public space

Once the objectives are established, the design-er then should refer to available data, e.g., Fig.7.4.2 or Table 7.4.1 and select the system whichbest meets these requirements. In this respect,concrete systems have superior properties andcan, with minimal effort, comply with these crite-ria. When the insulation value has not been speci-fied, selection of the necessary barrier can be de-

termined analytically by (1) identifying exteriorand/or interior noise sources, and (2) by establish-ing acceptable interior noise criteria.

Example 7.8.1Assume a precast prestressed concrete apart-

ment building with hollow core floor slabs. Thefirst step is to determine the degree of acousticalinsulation required of the floor-ceiling assemblyby using Figures 7.4.1 and 7.7.3

The 500 Hz requirement, 47 dB, can be used asthe first approximation of the floor STC category.

The selected floor should meet or exceed theinsulation needs at 11 frequencies. However, toachieve the most efficient design conditions, cer-tain limited deficiencies can be tolerated. Experi-ence has shown that the maximum deficienciesare 3 dB on one frequency point.

7.9 Leaks and FlankingThe performance of a building section with an

otherwise adequate STC can be seriously reducedby a relatively small hole or any other path whichallows sound to bypass the acoustical barrier. Allnoise which reaches a space by paths other thanthrough the primary barrier is called flanking.Common flanking paths are openings arounddoors or windows, at electrical outlets, telephoneand television connections, and pipe and duct pe-netrations. Suspended ceilings in rooms wherewalls do not extend from the ceiling to the roof orfloor above allow sound to travel to adjacentrooms.

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Fig. 7.9.1 Effect of safing insulation seals

No closure 14 STCWith steel vent plate closure 28 STCWith 4 in thick safing insulation 30 STC

steel bent plate added 42 STCWith 6 in thick safing insulation 38 STC

steel bent plate added 45 STC

CombinedTransmission Loss

steel bent plate

exteriorwall

mineral woolinorganic

insulation

gap

concrete floor

Anticipation and prevention of leaks begins atthe design stage. Flanking paths (gaps) at the pe-rimeters of interior precast walls and floors aregenerally sealed during construction with grout ordrypack. In addition, all openings around pe-netrations through walls or floors should be assmall as possible and must be sealed airtight. Thehigher the STC of the barrier, the greater the effectof an unsealed opening.

Perimeter leakage more commonly occurs atthe intersection between an exterior curtain walland floor slab. It is of vital importance to seal thisgap in order to retain the acoustical integrity of thesystem as well as provide the required fire stop be-tween floors. One way to achieve this seal is toplace a 4 pcf (64 kg/m3) density mineral woodblanket between the floor slab and the exteriorwall. Figure 7.9.1 demonstrates the acousticalisolation effects of this treatment.

In exterior walls, the proper application of seal-ant and backup materials in the joints betweenunits will not allow sound to flank the wall.

If the acoustical design is balanced, the maxi-mum amount of acoustic energy reaching a spacevia flanking should not equal the energy trans-mitted through the primary barriers.

Although not easily quantified, an inverse rela-tionship exists between the performance of an ele-ment as a primary barrier and its propensity totransmit flanking sound. In other words, the prob-ability of existing flanking paths in a concrete

structure is much less than in one of steel or woodframe.

In addition to using basic structural materials,flanking paths can be minimized by:

1. Interrupting the continuous flow of energywith dissimilar material, i.e., expansion or con-trol joints or air gaps.

2. Increasing the resistance to energy flow withfloating floor systems, full height and/ordouble partitions and suspended ceilings.

7.10 Human Response to Building VibrationsModern buildings often use components with

low weight-to-strength ratios, which allow longerspans with less mass. This trend increasingly re-sults in transient vibrations which are annoying tothe occupants. Unlike equipment vibration, a per-son often causes the vibration and also senses it.These vibrations usually have very small ampli-tudes (less than 0.05 in [1 mm]) and were not not-iced in older structures with heavier framing andmore numerous and heavier partitions, which pro-vided greater damping and other beneficial dy-namic characteristics.

This problem is not well understood. Predict-ing human response to floor motion and the dy-namic response to floor motion and the dynamicresponse of a floor system to moving loads are de-veloping technologies. A number of discomfortcriteria have been published44-51, but they oftengive contradictory results.

The vibration problem is most effectivelytreated by modifying the structural system. Thenatural period (or its inverse, frequency), stiff-ness, mass, and damping are the structural param-eters related to vibration control. Stiffness is in-creased by providing greater section propertiesthan may be required for supporting loads. An in-crease in mass improves the natural frequency, butincreases deflections and stresses, so by itself isonly partially effective in controlling vibrations.For example, increasing the depth of a flexuralmember will aid greatly in vibration control, butincreasing the width will not.

Recent research has emphasized the effect thatdamping plays in the human perception of vibra-tion. In a study of 91 floor systems it was con-cluded that with damping greater than 5.5 to 6 per-cent of critical, structural systems were accept-

7--10

able; systems with less were not46. Damping isusually attributed to the existence of partitions,supported mechanical work, ceilings and similaritems, but is really not well understood. Guidesfor quantifying damping effect are scarce, andthose that are available are very approximate49-51.

7.11 Vibration Isolation for MechanicalEquipment

Vibration produced by equipment with unbal-anced operating or starting forces can usually beisolated from the structure by mounting on aheavy concrete slab placed on resilient supports.This type of slab, called an inertia block, providesa low center of gravity to compensate for thrustssuch as those generated by large fans.

For equipment with less unbalanced weight, a“housekeeping” slab is sometimes used below theresilient mounts to provide a rigid support for themounts and to keep them above the floor so theyare easier to clean and inspect. This slab may alsobe mounted on pads of precompressed glass fiberor neoprene.

The natural frequency of the total load on resil-ient mounts must be well below the frequencygenerated by the equipment. The required weightof an inertia block depends on the total weight ofthe machine and the unbalanced force. For a longstroke compressor, five to seven times its weightmight be needed. For high pressure fans, one tofive times the fan weight is usually sufficient.

A floor supporting resiliently mounted equip-ment must be much stiffer than the isolation sys-tem. If the static deflection of the floor ap-proaches the static deflection of the mounts, thefloor becomes a part of the vibrating system, andlittle vibration isolation is achieved. In general,the floor deflection should be limited to about 15percent of the deflection of the mounts.

Simplified theory shows that for 90% vibrationisolation, a single resilient supported mass (isola-tor) should have a natural frequency of about 1/3the driving frequency of the equipment. The natu-ral frequency of this mass can be calculated by:52

fn = 188 1∕∆i (Eq. 7.11.1)

where:fn = natural frequency of the isolator, CPM

∆i = static deflection of the isolator, in

From the above, the required static deflectionof an isolator can be determined as follows:

fn = fd/3 = 188 1∕∆i or

∆i = (564/fd)2 (Eq. 7.11.2)

and:

∆f ≤ 0.15 ∆i (Eq. 7.11.3)

where:

fd = driving frequency of the equipment

∆f = static deflection of the floor system causedby the weight of the equipment, includinginertia block, at the location of the equip-ment.

Example 7.11.1 - Vibration IsolationGiven:

A piece of mechanical equipment has a drivingfrequency of 800 CPM.Problem:

Determine the approximate minimum deflec-tion of the isolator and the maximum deflection ofthe floor system that should be allowed.Solution:

Isolator, ∆i = (564/800)2 = 0.50 in.Floor, ∆f = 0.15(0.50) = 0.07 in.

Additional BibliographyL.L. Beranek; Noise Reduction, McGraw-HillBook Co., New York, 1960.

Ceramic Tile Institute of America and AmericanEnka Company unpublished floor/ceiling tests.

C.M. Harris, Handbook of Noise Control,McGraw-Hill Book Co., New York, 1967.

C.M. Harris, C.E. Crede; Shock & VibrationHandbook - 2nd ed., McGraw-Hill Book Co.,New York, 1976.

A. Litvin, H.W. Belliston; “Sound TransmissionLoss Through Concrete and Concrete MasonryWalls”, ACI Journal, December, 1978.

CHAPTER 8

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GUIDE SPECIFICATION FOR PRECAST,PRESTRESSED HOLLOW CORE SLABS

This Guide Specification is intended to be used as a basis for the development of an officemaster specification or in the preparation of specifications for a particular project. In eithercase, this Guide Specification must be edited to fit the conditions of use.

Particular attention should be given to the deletion of inapplicable provisions. Necessaryitems related to a particular project should be included. Also, appropriate requirementsshould be added where blank spaces have been provided

The Guide Specifications are on the left. Notes to Specifiers are on the right.

GUIDE SPECIFICATIONS NOTES TO SPECIFIERS

1. GENERAL

1.01 Description

A. Work Included:

1. These specifications cover manufacture,transportation, and erection of precast,prestressed concrete hollow core slabs in-cluding grouting of joints between adja-cent slab units.

1.01.A This Section is to be in Division 3 ofConstruction Specifications Institute format.

B. Related Work Specified Elsewhere:1. Cast-in-Place Concrete: Section

_________.

2. Architectural Precast Concrete: Section_________.

1.01.B.1 Includes structural or non-structuraltopping. See Section 2.5 for discussion of com-posite, structural topping.

3. Precast Structural Concrete: Section_________.

1.01.B.3 Beams, columns, etc.Prestressed concrete may be specified in Section_________.

4. Structural Metal Framing: Section_________.

1.01.B.4 Includes support framework not sup-plied by Hollow Core Slab Manufacturer.

5. Masonry Bearing Walls: Section_________.

1.01.B.5 Include any inserts or anchoring de-vices required for slab connections.

6. Underlayments: Section_________. 1.01.B.6 Underlayment may be any of the follow-ing general types: asphaltic concrete, gypsumconcrete, latex concrete, mastic underlayment.


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7. Caulking and Sealants: Section_________.

1.01.B.7 Caulking between slab edges at exposedunderside of floor members and/or perimetercaulking may be included in this section.

8. Holes for Mechanical Equipment: Sec-tion _______.

1.01.B.8 Holes may be drilled or cut and trimmedwith a chisel. Cut outline of hole through lowerportion of slab from underside, after which the topside may be removed from above. Do not cut pre-stressing strand without permission of engineer.

9. Painting: Section _________. 1.01.B.9 Prime coat should be a latex base paint.Finish coat may be an oil base, flat wall or emulsi-fied finish

10. Carpet and Pad: Section _________. 1.01.B.10 Specify minimum 55 oz. pad when nocast-in-place topping is used

11. Roofing and Roof Insulation:Section _________.

1.01.B.11 Non-absorbent rigid board insulation1” or more in thickness should be used on roofs.Check local energy code for exact requirements.

1.02 Quality AssuranceA. The precast concrete manufacturing plant

shall be certified by the Precast/PrestressedConcrete Institute (PCI) Plant CertificationProgram. Manufacturer shall be certified atthe time of bidding in Category C2.

1.02.A Structural Precast Products must meet therequirements of PCI Manual, MNL-116.

In Canada, the manufacture, transportation anderection of precast prestressed hollow core slabsis governed by the Canadian Standards Associa-tion Standard A23.4-94, “Precast Concrete - Ma-terials and Construction”.

Assurance of plant capability to produce qualityprecast concrete products is set by the CSA Stan-dard A23.4-94. This Standard forms the basis of acertification program which sets rigid capabilitycriteria for precast manufacturers, their person-nel and operations.

B. Erector Qualifications: Regularly engagedfor at least _______ years in the erection ofprecast structural concrete similar to the re-quirements of this project.

1.02.B Usually 2 to 5 years.

C. Qualifications of Welders: In accordancewith AWS D1.1.

D. Testing: In general compliance with applica-ble provisions of Precast/Prestressed Con-crete Institute MNL-116, Manual for QualityControl for Plants and Production of PrecastPrestressed Concrete Products.

1.02.C Qualified within the past year.


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E. Requirements of Regulatory Agencies: Alllocal codes plus the following specifications,standards and codes are a part of these specifi-cations:1. ACI 318-Building Code Requirements

for Structural Concrete.2. AWS D1.1-Structural Welding Code -

Steel.3. AWS D1.4-Structural Welding Code - Re-

inforcing Steel.4. ASTM Specifications - As referred to in

Part 2 - Products, of this Specification.

1.02.E Always include the specific year or edi-tion of the specifications, codes and standardsused in the design of the project and made part ofthe specifications. Fire safety and resistance re-quirements are specified in local or model codes.When required, fire rated products shall be clear-ly identified on the design drawings.

For projects in Canada, the National BuildingCode of Canada governs design. Canadian Stan-dards Association Standards A23.3-94, “Designof Concrete Structures” and A23.4-94, “PrecastConcrete - Materials and Construction” also ap-ply. Fire resistance is specified in the NationalBuilding Code and the National Fire Code.

1.03 Submittals

A. Shop Drawings

1. Erection Drawings

a. Plans locating and defining all hollowcore slab units furnished by themanufacturer, with all openings largerthan 10 in (250 mm) shown and lo-cated.

b. Sections and details showing connec-tions, edge conditions and supportconditions of the hollow core slabunits.

c. All dead, live and other applicableloads used in the design.

1.03.A.1.a Openings shown on erection draw-ings are considered in the slab design. Verify slabadequacy for any other openings with the Engi-neer of Record.

d. Estimated cambers. 1.03.A.1.d Floor slabs receiving cast-in-placetopping. The elevation of top of floor and amountof concrete topping must allow for camber of pre-stressed concrete members.

2. Production Drawings

a. Plan view of each hollow core slabunit type.

b. Sections and details to indicate quanti-ties, location and type of reinforcingsteel and prestressing strands.

c. Lifting and erection inserts.

1.03.A.2 Production drawings are normally sub-mitted only upon request.

d. Dimensions and finishes.

e. Prestress for strand and concretestrength.


8--4

f. Estimated camber at release.

g. Method of transportation.

B. Product Design Criteria1. Loadings for design

a. Initial handling and erection stresses.b. All dead and live loads as specified on

the contract drawingsc. All other loads specified for hollow

core slab units where applicable.2. Design calculations of products not com-

pleted on the contract drawings shall beperformed by a registered engineer expe-rienced in precast prestressed concretedesign and submitted for approval uponrequest.

3. Design shall be in accordance with ACI318 or applicable codes.

C. Permissible Design Deviations1. Design deviations will be permitted only

after the Architect/Engineer’s written ap-proval of the manufacturer’s proposed de-sign supported by complete design cal-culations and drawings.

2. Design deviation shall provide an installa-tion equivalent to the basic intent withoutincurring additional cost to the owner.

1.03 B and C Contract drawings normally will beprepared using a local precast prestressed con-crete hollow core slab manufacturer’s design dataand load tables. Dimensional changes whichwould not materially affect architectural andstructural properties or details usually are per-missible.

Be sure that loads shown on the contract draw-ings are easily interpreted. For instance, on mem-bers which are to receive concrete topping, besure to state whether all superimposed dead andlive loads on precast prestressed members do ordo not include the weight of the concrete topping.It is best to list the live load, superimposed deadload, topping weight, and weight of the member,all as separate loads. Where there are two differ-ent live loads (e.g., roof level of a parking struc-ture) indicate how they are to be combined.Where additional structural support is requiredfor openings, design headers in accordance withhollow core slab manufacturer’s recommenda-tions.

D. Test Report: Reports of tests on concrete andother materials upon request.

2. PRODUCTS

2.01 Materials 2.01 Delete or add materials that may be re-quired for the particular job.

A. Portland Cement:1. ASTM C150 - Type I or III

B. Admixtures: 2.01.B Verify ability of local producer to use ad-mixtures

1. Air-Entraining Admixtures: ASTMC260.

2. Water Reducing, Retarding, Accelerat-ing, High Range Water Reducing Admix-tures: ASTM C494.

C. Aggregates:

1. ASTM C33 or C330.


8--5

D. Water:

Potable or free from foreign materials inamounts harmful to concrete and embeddedsteel.

E. Reinforcing Steel:

1. Bars:Deformed Billet Steel: ASTM A615.Deformed Rail Steel: ASTM A616.Deformed Axle Steel: ASTM A617.Deformed Low Alloy Steel: ASTMA706.

2. Wire:Cold Drawn Steel: ASTM A82.

2.01.E.1 When welding of bars is required, weld-ability must be established to conform to AWSD1.4.

F. Prestressing Strand:1. Uncoated, 7-Wire, Stress-Relieved

Strand: ASTM A416 (including supple-ment) - Grade 250K or 270K.

2. Uncoated, Weldless 2- and 3-Wire Strand:ASTM A910

3. Indented, 7-Wire, Stress-Relieved Strand:ASTM A886 (including supplement)

2.01.F Low-relaxation strand is the predominantstrand in use. References to stress-relieved strandare from the ASTM titles.

G. Welded Studs: In accordance with AWSD1.1.

H. Structural Steel Plates and Shapes: ASTMA36.

2.01.H When required for anchorage or lateralbracing to structural steel members, some meth-ods of manufacturing hollow core slabs precludethe use of anchors and inserts

I. Grout:1. Cement grout: Grout shall be a mixture of

not less than one part portland cement tothree parts fine sand, and the consistencyshall be such that joints can be completelyfilled but without seepage over adjacentsurfaces. Any grout that seeps from thejoint shall be completely removed beforeit hardens.

2.01.I Grout strengths of 2000 psi to 3000 psi(13.8 - 20.7 MPa) can generally be achieved withthe proportions noted. Rarely is higher strengthgrout required. Non-shrink grout is not requiredfor satisfactory performance of hollow core slabsystems.

J. Bearing Strips:1. Random Oriented Fiber Reinforced:

Shall support a compressive stress of 3000psi (20.7 MPa) with no cracking, splittingor delaminating in the internal portions ofthe pad. One specimen shall be tested foreach 200 pads used in the project.

2.01.J.1 Standard guide specifications are notavailable for random-oriented, fiber-reinforcedpads. Proof testing of a sample from each groupof 200 pads is suggested. Normal design workingstresses are 1500 psi (10.3 MPa), so the 3000 psi(20.7 MPa) test load provides a factor of 2 overdesign stress. The shape factor for the test speci-mens should not be less than 2.


8--6

2. Plastic: Multi-monomer plastic stripsshall be non-leaching and supportconstruction loads with no visible overallexpansion.

2.01.J.2 Plastic pads are widely used with hollowcore slabs. Compression stress in use is not nor-mally over a few hundred psi and proof testing isnot considered necessary. No standard guidespecifications are available.

3. Tempered Hardboard. 2.01.J.3 Hardboard bearing strips should not beused in areas where undesirable staining is pos-sible or where bearing strips may be continuallywet.

4. Untempered Hardboard

2.02 Concrete Mixes

A. 28-day compressive strength: Minimum of____ psi.

B. Release strength: Minimum of ____ psi.

2.02.A and B Verify with local manufacturer.5000 (35 MPa) psi for prestressed products is nor-mal practice, with release strength of 3000 psi(20.7 MPa).

C. Use of calcium chloride, chloride ions or oth-er salts is not permitted.

2.03 Manufacture

A. Manufacturing procedures shall be in com-pliance with PCI MNL-116.

B. Manufacturing Tolerances: Manufacturingtolerances shall comply with PCI MNL-116.

C. Openings: Manufacturer shall provide forthose openings 10 in (250 mm) round orsquare or larger as shown on the structuraldrawings. Other openings shall be locatedand field drilled or cut by the trade requiringthem after the hollow core slab units havebeen erected. Openings and/or cutting of pre-stressing strand shall be approved by Archi-tect/Engineer and manufacturer before dril-ling or cutting.

2.03.C This paragraph requires other trades tofield drill holes needed for their work, and suchtrades should be alerted to this requirementthrough proper notation in their sections of thespecifications. Some manufacturers prefer toinstall openings smaller than 10 in (250 mm)which is acceptable if their locations are properlyidentified on the contract drawings

D. Patching: Will be acceptable providing thestructural adequacy of the hollow core unit isnot impaired.

3. EXECUTION

3.01 Product Deliver, Storage, and Handling

A. Delivery and Handling:


8--7

1. Hollow core slab units shall be lifted andsupported during manufacturing, stock-piling, transporting and erection opera-tions only at the lifting or supportingpoint, or both, as shown on the shop draw-ings, and with approved lifting devices.Lifting inserts shall have a minimum safe-ty factor of 4. Exterior lifting hardwareshall have a minimum safety factor of 5.

2. Transportation, site handling, and erec-tion shall be performed with acceptableequipment and methods, and by qualifiedpersonnel.

B. Storage:

1. Store all units off ground.2. Place stored units so that identification

marks are discernible.3. Separate stacked members by battens

across full width of each slab unit.4. Stack so that lifting devices are accessible

and undamaged.5. Do not use upper member of stacked tier

as storage area for shorter member orheavy equipment.

3.02 Erection

A. Site Access: The General Contractor shall beresponsible for providing suitable access tothe building, proper drainage and firm levelbearing for the hauling and erection equip-ment to operate under their own power.

B. Preparation: The General Contractor shall beresponsible for:

1. Providing true, level bearing surfaces onall field placed bearing walls and otherfield placed supporting members.

3.02.B Construction tolerances for cast-in-placeconcrete, masonry, etc., should be specified inthose sections of the specifications.

2. All pipes, stacks, conduits and other suchitems shall be stubbed off at a level lowerthan the bearing plane of the prestressedconcrete products until after the latter areset.

3.02.B.2 Should be in Electrical, Mechanical,and Plumbing sections of project specifications.


8--8

C. Installation: Installation of hollow core slabunits shall be performed by the manufactureror a competent erector. Members shall belifted by means of suitable lifting devices atpoints provided by the manufacturer. Bearingstrips shall be set, where required. Temporaryshoring and bracing, if necessary, shall com-ply with manufacturer’s recommendations.Grout keys shall be filled.

D. At Slab Ends (where shown on Drawings):Provide suitable end cap or dam in voids as re-quired.

3.02.D If a bearing wall building, special caremust be taken. Delete when end grouting is not re-quired.

E. For areas where slab voids are to be used aselectrical raceways or mechanical ducts pro-vide a taped butt joint at end of slabs, makingsure the voids are aligned.

3.02.E Delete when voids not used for electricalor mechanical.

F. Alignment: Members shall be properlyaligned and leveled as required by the ap-proved shop drawings. Variations betweenadjacent members shall be reasonably leveledout by jacking, loading, or any other feasiblemethod as recommended by the manufacturerand acceptable to the Architect/Engineer.

3.02.F Tolerances should comply with industrytolerances published in ”Tolerances for Precastand Prestressed Concrete”, Prestressed ConcreteInstitute, JR307, 1985.3

3.03 Field Welding

A. Field welding is to be done by qualified weld-ers using equipment and materials compatiblewith the base material.

3.04 Attachments

A. Subject to approval of the Architect/Engineer,hollow core slab units may be drilled or ”shot”provided no contact is made with the pre-stressing steel. Should spalling occur, it shallbe repaired by the trade doing the drilling orthe shooting.

3.05 Inspection and Acceptance

A. Final observation of erected hollow core slabunits shall be made by Architect/Engineer forpurposes of final payment.

REFERENCES

1. PCI Design Handbook - Precast and Pre-stressed Concrete, Fifth Edition, Precast/Prestressed Concrete Institute, Chicago, IL1997.

2. ACI Committee 318, “Building Code Re-quirements for Structural Concrete (ACI318-95) and Commentary (ACI318R-95)”, American Concrete Institute,Farmington Hills, MI, 1995.

3. PCI Committee on Tolerances, “Toler-ances for Precast and Prestressed Con-crete”, PCI JOURNAL, Vol. 30, No. 1,January-February, 1985, pp. 26-112.

4. PCI Technical Activities Council, PCICommittee on Building Code, “PCI Stan-dard Design Practice,” PCI JOURNAL, V.42, No. 2, March-April 1997, pp 34-51.

5. Zia, Paul, Preston, H. Kent, Scott, NormanL, and Workman, Edwin B., “EstimatingPrestress Losses”, Concrete International,June, 1979, pp 32-38.

6. Martin, L.D., “A Rational Method for Es-timating Camber and Deflection of Pre-cast, Prestressed Concrete Members”, PCIJOURNAL, January-February, 1977.

7. ACI Committee 301, “Standard Specifica-tions for Structural Concrete (ACI301-96)”, American Concrete Institute,Farmington Hills, MI, 1996.

8. Scott, Norman L., “Performance of Pre-cast, Prestressed Hollow Core Slab withComposite Concrete Topping”, PCIJOURNAL, March-April, 1973, pp 64-77.

9. Martin, Leslie D. and Scott, Norman L.,“Development of Prestressing Strand inPretensioned Members”, ACI Journal, Au-gust 1976, pp 453-456.

10. Anderson, Arthur R., and Anderson, Rich-ard G., “An Assurance Criterion for Flex-ural Bond in Pretensioned Hollow CoreUnits”, ACI Journal, August, 1976, pp457-464.

11. Discussion and Closure, “Development ofPrestressing Strand in Pretensioned Mem-

bers”, ACI Journal, March, 1977, pp136-137.

12. Discussion and Closure, “An AssuranceCriterion for Flexural Bond in Preten-sioned Hollow Core Units”, ACI Journal,March, 1977, pp 137-140.

13. Zia, Paul and Mostafa, Talat, “Develop-ment Length of Prestressing Strands”, PCIJOURNAL, September-October, 1977, pp54-65.

14. Discussion and Closure, “DevelopmentLength of Prestressing Strands”, PCIJOURNAL, July-August, 1978, pp97-107.

15. Buckner, C. Dale, “A Review of StrandDevelopment Length for PretensionedConcrete Members”, PCI JOURNAL, V.40, No. 2, March-April, 1995, pp 84-105.

16. Discussion and Closure, “A Review ofStrand Development Length for Preten-sioned Concrete Members”, PCI JOUR-NAL, V. 41, No. 2, March-April, 1996, pp112-116.

17. Martin, Leslie D. and Korkosz, Walter J.,“Strength of Prestressed Concrete Mem-bers at Sections Where Strands are notFully Developed”, PCI JOURNAL, V. 40,No. 5, September-October, 1995, pp58-66.

18. Brooks, Mark D., Gerstle, Kurt H., andLogan, Donald R., “Effective of InitialStrand Slip on the Strength of HollowCore Slabs”, PCI JOURNAL, V. 33, No.1, January-February, 1988, pp 90-111.

19. LaGue, David J., “Load Distribution Testson Precast Prestressed Hollow Core SlabConstruction”, PCI JOURNAL, Novem-ber-December, 1971, pp 10-18.

20. Van Acker, A., “Transversal Distributionof Linear Loadings in Prestressed HollowCore Floors”, BMA/MKT 84/006, Sep-tember, 1983.

21. Johnson, Ted and Ghadiali, Zohair, “LoadDistribution Test on Precast Hollow Core

Slabs with Openings”, PCI JOURNAL,September-October, 1972, pp 9-19.

22. Pfeifer, Donald W. and Nelson, TheodoreA., “Tests to Determine the Lateral Dis-tribution of Vertical Loads in a Long-SpanHollow Core Floor Assembly”, PCIJOURNAL, Vol. 28, No. 6, November-De-cember, 1983, pp. 42-57.

23. Aswad, Alex and Jacques, Francis J., “Be-havior of Hollow Core Slabs Subject toEdge Loads”, PCI JOURNAL, V. 37, No.2, March-April, 1992, pp 72-83.

24. Stanton, John F., “Response of HollowCore Slab Floors to Concentrated Loads”,PCI JOURNAL, V. 37, No. 4, July-Au-gust, 1992, pp 98-113.

25. Stanton, John F., “Proposed Design Rulesfor Load Distribution in Precast ConcreteDecks”, ACI Structural Journal, V. 84, No.5, September-October, 1987, pp 371-382.

26. Rosenthal, I., “Full Scale Test of Continu-ous Prestressed Hollow Core Slab”, PCIJOURNAL, Vol. 23, No. 3, May-June,1978, pp. 74-81.

27. Harris, Harry G., and Iyengar, Srikanth,“Full Scale Tests on Horizontal Joints ofLarge Panel Precast Concrete Buildings”,PCI JOURNAL, Vol 25, No. 2, March-April, 1980, pp. 72-92.

28. Johal, L.S. and Hanson, N.W., “Design forVertical Load on Horizontal Connectionsin Large Panel Structures”, PCI JOUR-NAL, Vol. 27, No. 1, January-February,1982, pp 62-79.

29. PCI Committee on Precast Bearing WallBuildings, “Considerations for the Designof Precast Concrete Bearing Wall Build-ings to Withstand Abnormal Loads”, PCIJOURNAL, Vol. 21, No. 2., March-April,1976, pp. 18-51.

30. Fintel, Mark and Schultz, Donald M., “APhilosophy for Structural Integrity ofLarge Panel Buildings”, PCI JOURNAL,Vol. 21, No. 3, May-June, 1976, pp. 46-69.

31. PCI Manual for Structural Design of Ar-chitectural Precast Concrete, PCIMNL-121-77, Prestressed Concrete Insti-tute, Chicago, 1977.

32. Uniform Building Code, “Structural Engi-neering Design Provisions”, V. 2, Interna-tional Conference of Building Officials,Whittier, CA, 1994.

33. The BOCA National Building Code,Thirteenth Edition, Building Officials &Code Administrators International, Inc.,Country Club Hills, IL, 1996.

34. Cosper, Steven J., Anderson, Arthur R.,Jobse, Harold J., “Shear Diaphragm Ca-pacity of Untopped Hollow Core FloorSystems”, Concrete TechnologyAssociates, Technical Bulletin 80B3,1981.

35. Clough, D.P., “Design of Connections forPrecast Prestressed Concrete Buildings forthe Effects of Earthquake”, National Sci-ence Foundation, 1985.

36. Moustafa, Saad E., “Effectiveness ofShear-Friction Reinforcement in Shear Di-aphragm Capacity of Hollow Core Slabs”,PCI JOURNAL, Vol. 26, No. 1, January-February, 1981, pp 118-132.

37. Design and Detailing of Untopped HollowCore Slab Systems for Diaphragm Shears,Structural Engineer’s Association of Ari-zona, 1981/82.

38. PCI Fire Committee, “Design for Fire Re-sistance of Precast Prestressed Concreter-Second Edition”, Precast/Prestressed Con-crete Institute, Chicago, IL, 1989.

39. ASHRAE: ASHRAE Systems Handbookfor 1984. American Society of Heating,Refrigerating & Air Conditioning Engi-neers, Inc., New York, 1984.

40. Blazier, W.E., “Revised Noise Criteria forDesign and Rating of HVAC Systems”,paper presented at ASHRAE SemiannualMeeting, Chicago, IL, January 26, 1981.

41. Berendt, R.D., Winzer, G.E., Burroughs,C.B.; “A Guide to Airborne, Impact and

Structureborne Noise Control in Multi-family Dwellings”, prepared for FederalHousing Administration, U.S. Govern-ment Printing Office, Washington, D.C.,1975.

42. Sabine, H.J., Lacher, M.B., Flynn, D.R.,Quindry, T.L.; “Acoustical and ThermalPerformance of Exterior ResidentialWalls, Doors & Windows”, National Bu-reau of Standards, U.S. Government Print-ing Office, Washington D.C., 1975.

43. IITRI; “Compendium of Materials forNoise Control”, U.S. Department ofHealth, Education & Welfare, U.S. Gov-ernment Printing Office, Washington,D.C., 1980.

44. “Vibrations of Concrete Structures”, Pub-lication SP-60, American Concrete Insti-tute, Detroit, MI.

45. Galambos, T.V., Gould, P.C., Ravindra,M.R., Surgoutomo, H., and Crist, R.A.,“Structural Deflections - A Literature andState-of-the-Art Survey”, Building Sci-ence Series, Oct., 1973, National Bureauof Standards, Washington, D.C.

46. Murray, T.M., “Acceptability Criterion forOccupant-Induced Floor Vibration”,Sound and Vibration, November, 1979.

47. “Design and Evaluation of OperationBreakthrough Housing Systems”, NBSReport 10200, Amendment 4, September,1970, U.S. Department of Housing andUrban Development, Washington, D.C.

48. Wiss, J.F. and Parmelee, R.H., “HumanPerception of Transient Vibrations”, Jour-nal of the Structural Division, ASCE, Vol.100, No. ST4, April, 1974.

49. “Guide to Floor Vibrations”, Steel Struc-tures for Buildings-Limit States DesignCSA S16.1-1974, Appendix G. CanadianStandards Association, Rexdale, Ontario.

50. “Guide for the Evaluation of Human Ex-posure to Whole-Body Vibration”, In-ternational Standard 2631, InternationalOrganization for Standardization, 1974.

51. Murray, T.M., “Design to Prevent FloorVibration”, Engineering Journal, AISC,Third Quarter, 1975.

52. Harris, C.M. and Crede, C.E., Shock andVibration Handbook, 2nd Edition,McGraw-Hill, New York, NY, 1976.

INDEXA

Acoustical properties 1---3, 7---1. . . . . . . . . . . . . . . .

Admixtures 1---1, 1---2, 8---4. . . . . . . . . . . . . . . . . . .

Aggregates 1---2, 6---1, 6---2, 6---4, 8---4. . . . . . . . . .

Air entrainment 1---1. . . . . . . . . . . . . . . . . . . . . . . . .

Allowable live load 1---5. . . . . . . . . . . . . . . . . . . . . . .

BBearing strips 3---12, 3---13, 8---5, 8---6, 8---8. . . . . .

Bonding agents 2---16. . . . . . . . . . . . . . . . . . . . . . . . .

Boundary element 4---4 to 4---8, 4---15. . . . . . . . . . .

CCamber 1---4, 1---14, 2---3, 2---11 to 2---13, 2---14,. .

2---15, 2---16, 8---3, 8---4

Cantilevers 3---10 to 3---12. . . . . . . . . . . . . . . . . . . .

Caulking 8---2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Certification 8---2. . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chord 4---4, 4---5, 4---8. . . . . . . . . . . . . . . . . . . . . . . .

Collector 4---4, 4---5, 4---7, 4---8, 4---9. . . . . . . . . . . .

Composite ties 2---16. . . . . . . . . . . . . . . . . . . . . . . . .

Composite topping 1---3, 2---13, 2---15, 2---16, 3---10,4---9

Connections 1---3, 1---5, 3---10, 4---1, 4---2, 4---4, 4---7to 4---9, 5---1

Continuity 3---10, 6---5, 6---9, 6---10. . . . . . . . . . . . . .

Contract documents 1---3, 1---5, 4---1, 8---4, 8---6. . .

Control joints 2---16. . . . . . . . . . . . . . . . . . . . . . . . . .

Cracking moment 2---19. . . . . . . . . . . . . . . . . . . . . . .

Creep 2---11 to 2---16, 6---12. . . . . . . . . . . . . . . . . . . .

Creep losses 2---1, 2---3, 2---5. . . . . . . . . . . . . . . . . .

Curling 2---16. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

D

Damping 7---2, 7---9, 7---10. . . . . . . . . . . . . . . . . . . .

Debonded strands 2---2, 2---20, 3---10, 3---11. . . . . .

Decibel 7---1, 7---5, 7---6. . . . . . . . . . . . . . . . . . . . . .

Deflection 1---3, 1---4, 3---1, 3---10, 3---12. . . . . . . . .

Deflections 2---1, 2---5, 2---11 to 2---16, 4---3, 4---7,.4---9

Design responsibility 1---5, 4---1. . . . . . . . . . . . . . . .

Design strength 2---2, 2---19, 2---20. . . . . . . . . . . . .

Details - Cantilevers 5---20. . . . . . . . . . . . . . . . . . . .

Details - Concrete beam 5---2. . . . . . . . . . . . . . . . . .

Details - Miscellaneous 5---23. . . . . . . . . . . . . . . . . .

Details - Steel beams 5---15. . . . . . . . . . . . . . . . . . . .

Details - Walls 5---9. . . . . . . . . . . . . . . . . . . . . . . . . . .

Development length 2---19 to 2---21, 3---9, 3---12. . .

Diaphragm flexibility 4---2 to 4---4. . . . . . . . . . . . . .

Diaphragms 1---3, 2---15, 2---16, 3---10, 3---14, 4---1,.to 4---9, 5---1

Differential shrinkage 2---15, 2---16. . . . . . . . . . . . .

Drag strut 4---4, 4---5, 4---7, 4---8. . . . . . . . . . . . . . . .

Dy-Core 1---1, 1---8. . . . . . . . . . . . . . . . . . . . . . . . . . .

Dynaspan 1---1, 1---8. . . . . . . . . . . . . . . . . . . . . . . . .

EEffective resisting section 3---2, 3---3, 3---6 to 3---8.

Elematic 1---1, 1---9. . . . . . . . . . . . . . . . . . . . . . . . . .

End restraint 1---3, 6---5, 6---8, 6---12, 6---13. . . . . . .

End slip 2---20, 2---21. . . . . . . . . . . . . . . . . . . . . . . . .

Equivalent live load 1---5. . . . . . . . . . . . . . . . . . . . . .

Equivalent thickness 1---2, 1---3, 6---1, 6---2, 6---4,.6---14

Equivalent uniform load 1---5, 1---6, 3---7. . . . . . . .

Erection drawings 1---5, 8---3. . . . . . . . . . . . . . . . . .

Extruder 1---1, 2---9. . . . . . . . . . . . . . . . . . . . . . . . . .

FFinishes 1---4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Fire endurance 3---10, 6---1, 6---2, 6---4, 6---5, 6---7 to6---14

Fire rating 1---2, 1---3, 1---5, 1---6, 8---3. . . . . . . . . . .

Fixed form 1---1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Flanking 7---1, 7---8, 7---9. . . . . . . . . . . . . . . . . . . . .

Flexicore 1---1, 1---9. . . . . . . . . . . . . . . . . . . . . . . . . .

Flexural bond length 2---19, 2---20. . . . . . . . . . . . . .

Flexural design 2---1. . . . . . . . . . . . . . . . . . . . . . . . . .

Flexural strength 1---6, 2---1, 2---6, 2---9, 2---19 to. .2---21, 4---8

G

Grout 1---2, 2---16, 3---12, 4---6, 8---5. . . . . . . . . . . . .

Grout column 3---12, 3---13. . . . . . . . . . . . . . . . . . . .

H

Headers 3---8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Heat transmission1---3, 6---1, 6---2, 6---4, 6---8, 6---14

Horizontal joints 3---12 to 3---15. . . . . . . . . . . . . . .

Horizontal shear 2---16, 4---8, 4---9. . . . . . . . . . . . . .

I

Impact insulation 7---3. . . . . . . . . . . . . . . . . . . . . . . .

Impact Insulation Class 1---3, 7---1, 7---3, 7---4. . . . .

Impact noise 7---1, 7---3. . . . . . . . . . . . . . . . . . . . . . .

Inclined shear 2---10, 3---4, 3---6, 3---7. . . . . . . . . . . .

K

Keyways 1---2, 1---3, 3---1, 3---10, 4---5, 4---6, 4---8. . .

L

Lateral loads 4---1, 4---4 to 4---6. . . . . . . . . . . . . . . . .

Lateral-resisting elements 4---1 to 4---9. . . . . . . . . .

Load concentrations 2---20, 3---1, 3---3, 3---8. . . . . .

Load distribution 3---1 to 3---3. . . . . . . . . . . . . . . . .

Load tables 1---3, 1---5 to 1---7, 2---1, 2---16. . . . . . .

Longitudinal shear 4---6, 4---7. . . . . . . . . . . . . . . . . .

Loss of Prestress 2---1, 2---3 to 2---5, 2---11, 2---13. .

MManufacturing 1---1, 1---5, 1---8, 2---2, 2---3, 2---9,. .

3---1, 3---10, 8---1, 8---2, 8---5, 8---6

NNon-shrink grout 1---2, 8---5. . . . . . . . . . . . . . . . . . .

Normal end slip 2---21. . . . . . . . . . . . . . . . . . . . . . . .

OOpenings 1---3, 1---5, 3---1, 3---8, 3---9, 3---15, 4---9,.

8---3, 8---4, 8---6

PPartially developed strand 2---19 to 2---21. . . . . . . .

PCI Standard Design Practice 2---1. . . . . . . . . . . . .

Production drawings 8---3. . . . . . . . . . . . . . . . . . . . .

QQuality assurance 2---15, 8---2. . . . . . . . . . . . . . . . . .

RRelease strength 2---2, 2---3. . . . . . . . . . . . . . . . . . . .

SSandwich panels 1---4. . . . . . . . . . . . . . . . . . . . . . . . .

Seating losses 2---1. . . . . . . . . . . . . . . . . . . . . . . . . . .

Seismic base shear 4---2. . . . . . . . . . . . . . . . . . . . . . .

Service load stresses 2---1, 2---3, 2---5. . . . . . . . . . . .

Shear friction 4---5 to 4---8. . . . . . . . . . . . . . . . . . . .

Shear reinforcement 2---9, 2---11, 3---7. . . . . . . . . .

Shrinkage 1---2, 2---15, 2---16. . . . . . . . . . . . . . . . . .

Shrinkage cracks 1---2, 3---1, 4---11. . . . . . . . . . . . . .

Shrinkage losses 2---1, 2---3. . . . . . . . . . . . . . . . . . . .

Slab thickness 1---3, 1---5, 1---6, 3---11. . . . . . . . . . . .

Slip form 1---1, 1---2. . . . . . . . . . . . . . . . . . . . . . . . . .

Slump 1---1, 1---2, 2---20. . . . . . . . . . . . . . . . . . . . . . .

Sound absorption 7---1, 7---2, 7---5, 7---8. . . . . . . . . .

Sound insulation 7---1 to 7---3. . . . . . . . . . . . . . . . . .

Sound transmission 1---3. . . . . . . . . . . . . . . . . . . . . .

Sound Transmission Class 1---3, 7---1, 7---2, 7---4. . .

Sound transmission loss 7---1 to 7---3, 7---8. . . . . . . .

Span-depth ratio 3---3. . . . . . . . . . . . . . . . . . . . . . . .

Spancrete 1---1, 1---10. . . . . . . . . . . . . . . . . . . . . . . .

Spray-applied coatings 6---9, 6---14. . . . . . . . . . . . . .

Steel relaxation 2---1, 2---4. . . . . . . . . . . . . . . . . . . . .

Strain compatibility 2---6, 2---7, 2---9, 2---20, 2---21,3---11

Stress-strain diagram 2---7, 2---8. . . . . . . . . . . . . . . .

Structural end point 6---5. . . . . . . . . . . . . . . . . . . . . .

Structural integrity 3---10, 3---14, 4---1, 4---4, 4---5. .

TTolerances 1---4, 1---6, 1---14, 1---15, 8---6 to 8---8. . .

Topping 1---3, 1---4, 2---13, 2---15, 2---16, 3---10, 4---54---9, 6---2, 6---4, 6---11, 8---1 to 8---4

Torsion 3---1, 3---2, 3---8. . . . . . . . . . . . . . . . . . . . . . .

Transfer length 2---10, 2---19 to 2---21, 3---12. . . . . .

Transfer stresses 2---1, 2---2. . . . . . . . . . . . . . . . . . . .

Transverse bending 3---1, 3---3. . . . . . . . . . . . . . . . .

Transverse reinforcement 3---1. . . . . . . . . . . . . . . . .

UUltra-Span 1---1, 1---11. . . . . . . . . . . . . . . . . . . . . . .

Unit weight 1---2, 1---14. . . . . . . . . . . . . . . . . . . . . . .

VVibration isolators 7---2, 7---10. . . . . . . . . . . . . . . . .

WWall panels 1---1, 1---4. . . . . . . . . . . . . . . . . . . . . . . .

Water-cement ratio 2---20. . . . . . . . . . . . . . . . . . . . .

Web shear 2---18, 3---4, 3---7. . . . . . . . . . . . . . . . . . .

Weep holes 1---4. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Weights 1---2, 1---6. . . . . . . . . . . . . . . . . . . . . . . . . . .

Documents

hollow core