simplified design procedures for precast hybrid frames · department of civil engineering and...

UNIVERSITY OF WISCONSIN MILWAUKEE

Department of Civil Engineering and Mechanics

Simplified Design Proceduresfor Precast Hybrid Frames

Research Team:

Dr. Adeeb Rahman, Assistant Professor (PI)

Dr. Habib Tabatabai, Associate Professor (Co-PI)

Mr. Rami Hawileh, PhD Candidate

Submitted to:

PCI: Precast / Prestressed Concrete Institute

209 West Jackson Boulevard

Chicago, IL 60606-6938

Final Report

September, 04, 2004

Conclusions or recommendations in this report are the opinions of the authors. PCI assumes no responsibility for the interpretation or application of the information contained herein.

ACKNOWLEDGEMENTS

The research team at the University of Wisconsin-Milwaukee would like to acknowledge

the Precast/Prestressed Concrete Institute for providing the funds to conduct this research.

This work has been funded through the PCI-Daniel P. Jenny Research Fellowship for

2003-2004. The research team would like to thank the Industry Advisory Panel: S. K.

Ghosh, Ned Cleland, Roger Becker, and Sri Sritharan for providing technical support at

various level of this proj ect. Their recommendations, suggestions and critique continue to

be valuable in advancing the knowledge in this filed. We also would like to acknowledge

Fattah Shaikh for providing valuable technical insight and for his continued support. Our

thanks also go to Paul Johal, Research Director, who facilitated the administration and

the technical oversight of this project at the PCI office. Our thanks also is extended to A1

Ghorbanpoor, the Director of the structures laboratory at UWM and Rahim Reshadi,

Instrumentation Specialist, for providing facilitative support for parts of this project.


Table of Contents

List of Figures

List of Tables

Abstract

Chapter 1 Introduction

1.1 Background

1.2 Objectives

1.3 Hybrid Frame - Concept Description

1.4 Report Organization

Chapter 2 Fracture Criteria for the Unbonded Mild Steel Bars in Hybrid Frames

2.1 Introduction

2.2 Bending calculations for bars

2.3 Axial strain calculations for bars

2.4 Inelastically deformed geometry of bars

2.4.1 The FE model

2.4.2 Types of elements

2.4.3 Material properties

2.4.4 Boundary conditions and loads

2.4.5 Results

2.5 Low-cycle fatigue for bar fracture

2.5.1 Introduction

2.5.2 Low-cycle fatigue relationships for reinforcing steel bars

2.6 Effects of nonzero mean strain on low-cycle fatigue behavior

iv

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viii

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35.

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2.7 Effect of grout in the duct

Chapter 3 Simplified Design Procedures for Precast Hybrid Frames

3.1 Hybrid frame - PRESSS design procedure

3.2 Overview of Other Design Factors

3.3 Development of non-dimensional procedure

3.4 Summary of Proposed non-dimensional steps

3.8 Parametric studies

3.9 Results from the parametric studies

Chapter 4 Design Examples

4.1 Example # 1

4.1.1 Basic solution using PRESSS procedure

4.1.2 Solution using modified non-dimensional design procedure based on PRESSS

4.1.3 Solution using proposed design charts

4.1.4 Solution using design equations

4.2 Example # 2





4.3 Example # 3




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4.4 Example # 4





4.5 Example # 5





Chapter 5 Summary and conclusions

Appendix A EXCEL spreadsheet example

References

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iii


List of Figures

Figure 1.1 Hybrid flame

Figure 1.2 Drift vs relative strength of resisting elements

Figure 2.1 Deformed segment of the mild steel bar

Figure 2.2 Path of point D of the unbonded segment of the mild steel bar

Figure 2.3 Location of the center of rotation at Odes

Figure 2.4 Isoparametric view of the entire model

Figure 2.5 Isoparametric view of the concrete blocks

Figure 2.6 Isoparametric view of the mild steel bar

Figure 2.7 Isoparametric view of the duet grout

Figure 2.8 BEAM 188 3-D linear finite strain beam

Figure 2.9 Mild steel bar cross section

Figure 2.10 Duct grout cross section

Figure 2.11 Concrete cross section

Figure 2.12 Stress-strain curve of grade 60 mild steel bar

Figure 2.13 Stress-strain model for concrete

Figure 2.14 Deflected shape of the model

Figure 2.15 Horizontal deformation (in) of the model

Figure 2.16 Horizontal deformation (in) of the mild steel bar

Figure 2.17 Vertical deformation (in) of the model

Figure 2.18 Vertical deflection (in) of the mild steel bar

Figure 2.19 Mild steel bar axial total strain distribution

Figure 2.20 Axial total strain distribution Of the unbonded mild steel bar

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Figure 2.21 Mild steel bar axial stress distribution

Figure 2.22 Axial stress distribution of unbonded mild steel bar

Figure 2.23 Axial total strain distribution for concrete block

Figure 2.24 Axial total strain distribution for grout

Figure 2.25 Vertical deformation of the unbonded segment of the bar

Figure 2.26 Vertical deformation of the unbonded segment of the bar

Figure 2.27 Slope (rotation) of the unbonded segment of the bar

Figure 2.28 2nd derivative of the unbonded segment of the bar

Figure 2.29 Curvature of the unbonded segment of the bar

Figure 2.30 Bending strain of the unbonded segment of the bar

Figure 2.31 Plastic strain-life relationship

Figure 3.1 Location of forces at design drift

Figure 3.2 Number of cycles to failure versus the period of the structure

Figure 3.3 Sample view from EXCEL spreadsheet

Figure 3.4 [Mdes/Aphbfp,des] versus qb for 8s,max = 4%

Figure 3.5 [Mp,acs/Mdes] versus do for ~s,n~x = 4%

Figure 3.6 [Maes/As)~s,deshbfsy] versus ~ for es,max = 4%

Figure 3.7 [Mdes/Aphbfp,des] versus do for ~s,n~x = 2%

Figure 3.8 [Mp,ae~/Ma~] versus d~ for ~,max = 2%

Figure 3.9 [Maes/A~)~,ac~hbfsy] versus dO for ~,max = 2%

Figure 3.10 Ap/bh versus qb for f ’e = 5 ksi and fpi = 0.65 fpu

Figure 3.11 Ap/bh versus ~ for 0d¢s = 2% and fpi = 0.65 fpu

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72

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Figure 3.12 Ap/bh versus ~b for Odes = 2% and f ’e

Figure 3.13 rides versus 91

Figure 3,14 fp,des/fpy versus ~b for fpi = 0.65 fpu

Figure 3.15 fsy/fp,des versus ( for fpi = 0.65 fpu

Figure 3.16 fp,des/fpy versus d~ for 0d~ = 2%

Figure 3.17 f~y/fp,d~s versus d~ for Odes = 2%

= 5 ksi 79

79

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81

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vi


List of Tables

Table 2.1 Study cases 14

Table 2.2 Model dimensions 17

Table 2.3 Materiai’s properties 20

Table 2.4 FEA results 28

Table 2.5 Predicted results compared with FEA results 35

Table 2.6 Effective axial strain calculations 41

Table 2.7 Average strain in the bar due to grout effect 44

Table 3.1 Material properties of hybrid frame components 48

Table 3.2Bar strains and corresponding over-strength factors for ASTM A706 bars48

Table 3.3 Range of parameters used in the parametric studies 73

Table 4.1 Material properties and design information for example 1 86

Table 4.2 Results compared using the four methods 102







Table 419 Material properties and design information for example 5 157


vii


ABSTRACT

In the Precast Seismic Structural Systems (PRESSS) research program (Stanton and

Nakaki, 2002), the unbonded post-tensioned frame with damping (hybrid frame)

performed very well in the experimental evaluation. The hybrid frame system offers

an important feature that allows the designer to eliminate residual drift aider an

earthquake. The PRESSS report outlined a series of step-by-step design procedures

for hybrid frames. In this project, the objective is to develop simplified design

procedures for precast hybrid frames. A set of new dimensionless parameters and

procedures for the design of hybrid frame is developed to replace current dimensional

parameters. A mild steel fracture criterion under combined axial and bending strains

is also proposed. Parametric studies were performed using the developed non-

dimensional formulation. These involved solving optimization problems to minimize

overall drift and achieve zero residual drift. The results of these studies were used to

generate non-dimensional design charts and simplified equations. Design examples

representing low and high seismic zones are presented. The results of the three

procedures (PRESSS, proposed design charts, and proposed design equations) are

compared. Very close agreement was found in all cases.

oooVlll


Chapter I

INTRODUCTION

1.1 Background

In the Precast Seismic Structural Systems (PRESSS) report Volume 3 - 09 (Stanton and

Nakaki, 2002), a total of five different seismic structural systems made from precast

concrete elements were proposed. These systems formed various parts of the structural

framing in the PRESSS Phase III experimental building that was tested at the University

of California at San Diego.

The unbonded post-tensioned frame with damping (hybrid frame) performed very well in

the PRESSS evaluation. The hybrid frame system offers an important feature that allows

the designer to eliminate residual drift after an earthquake: This feature is not available in

the framing systems recognized in the 1997 Uiform Building Code (UBC). The PRESSS

report outlined a series of step-by-step design procedures for hybrid frames.

1.20b|ectives

One of the recommendations in the PRESSS report involved the development of a series

of design examples for the Hybrid frame. The objectives of this research are:

1. To develop a set of new dimensionless parameters to replace current dimensional

parameters in the calculation process.

2. To perform parametric studies based on a new non-dimensional formulation of

the PRESSS design procedures. These studies involve solving a large number of

optimization problems to minimize overall chill and achieve zero residual drift.

Design charts will be generated based on these parametric studies.


3. To develop and verify simplified equations for the design of hybrid frames. These

charts and equations will enable designers to perform the design in fewer number

of steps, while affording them the opportunity to easily see how changing one or

more parameters would affect the design.

4. To develop a mild steel fracture criterion under combined axial and bending

strains by controlling the total plastic strain in the bar below a maximum value.

5. To prepare a number of design examples that includes comparisons of PRESSS

and new procedures.

1.3 Hybrid Frame - Concept Description

The hybrid frames contain precast elements (beams and columns) that are connected by

unbonded post-tensioning steel and bonded reinforcement bars, both of which contribute

to the overall moment resistance (Figure 1). An interesting feature of the connection

between beam and column is the hybrid combination of mild steel and post-tensioning

steel where the mild steel is used to dissipate energy by yielding in tension and

compression and the post-tensioning steel is used to clamp the beam against the column.

The post-tensioning force would act as a clamping/restoring force to bring the frame back

to its original configuration after an earthquake, and would provide for shear resistance

through friction developed at the beam-column interface. For a given total moment

strength, a higher proportion of mild steel reinforcement would result in more apparent

damping, lower peak drift, and higher residual drift.

2


Mild Reinforcing (A706) --\(grouted) ~

Post-tensioningTendon (unbonded)

Reinforcing bar debondedlocally

Fiber-reinforcedGrout

Figure 1 Hybrid Frame (Stanton and Nakaki, 2002)

The general design philosophy, as presented by Stanton and Nakaki (2002), is to

minimize peak drift, maintain zero residual drift, and maximize the moment strength

provided by the mild steel reinforcement. Figure 2 shows the relationship between peak

and residual drifts, where


Drift

Peak DriftResidualDrift

Mcap,s / Mcap,tot

Figure 2 Drift vs. Relative Strength of Resisting Elements(Stanton and Nakaki, 2002)

Mcap,tot = total moment strength (Mcap,p + Mcap,s)

M~ap,p = moment strength due to the post-tensioning force

Mo~p,s = moment strength due to the yielding of the mild steel reinforcement

The process of finding the best solution is basically an optimization problem that can be

implemented in a parametric study. Such a parametric study utilizing non-dimensional

parameters is the subject of this research.

1.4 Report Organization

This report consists of five chapters as described below

Chapter 1 (Introduction) covers research objectives, hybrid frame concept, and report

organization

Chapter 2 (Fracture criteria for the unbonded mild steel bars in hybrid frames) discusses

the fracture criteria for the unbonded mild steel bars, combined bending and axial strains,

and low cycle fatigue


Chapter 3 (Simplified design procedures for precast hybrid frames) describes the

development and derivations of the simplified design procedure for precast hybrid frames

Chapter 4 (Design examples) covers five design examples for different seismic regions

that includes comparisons of PRESSS and new procedures.

Chapter 5 (Summary and conclusions) covers the summary and conclusions.

And finally the appendix A shows the formulation of the EXCEL spreadsheet for the first

design example.


CHAPTER 2

Fracture Criteria for the Unbonded Mild Steel Bars in Hybrid Frames

2.1 Introduction

From preliminary nonlinear finite element (FE) analyses of a 3-D model of a hybrid

frame system [Hawileh, 2003], it was determined that the mild steel bars in a hybrid

frame exhibit significant inelastic axial and bending strains. Once the gap at the beam-

column interface opens, relatively high levels of repetitive plastic strains develop in the

mild steel bar. This prompted a need for considering the low-cycle fatigue of the mild

steel bars including both bending and axial strains. The relatively high inelastic strains in

the FE model show the potential vulnerability of mild steel bars to low-cycle fatigue

failure. It should be noted that both the PRESSS [Stanton and Nakaki, 2002] and

National Institute of Standards and Technology (NIST) [Cheok and Stone, 1994] test

reports indicate bar fractures during cyclic testing.

A mild steel fracture criterion is therefore needed in the design procedure for hybrid

frames by controlling the total plastic strains in the mild steel bar below a maximum

value. This criterion should be based on the low-cycle fatigue behavior of reinforcing

steel.

Mander and Panthaki (1994) studied the behavior of reinforcing steel bars under low-

cycle fatigue subjected to axial-strain reversals with strain amplitudes ranging from yield

to 6%. Liu (2001) studied the low-cycle fatigue behavior of steel bars subjected to

bending strain reversals with variable amplitudes. In this study, a mild steel fracture

6


criterion under combined axial and bending strains is proposed based on the works of

Mander, Panthaki, and Liu.

2.2 Bending Strain Calculations for Bars

Liu (2001) presented the following procedure for calculating strain-displacement

relationships for bars subjected to bending when the material is stressed beyond the

elastic range. Consider the deformed bar segment shown in Figure 2.1 below.

R

Ydy

At

dx

Figure 2.1 Deformed Segment of the Mild Steel Bar

Assuming that the cross section of bar is symmetric (i.e. elastic and plastic neutral axes

are at the center of the bar), the bending strain and the radius of curvature can be

calculated as follows:

A’B’-ds (R +~)dO-R dO,st, = = --~--~ (2-1)

ds R dO 2R


1 2~

Where

R = radius of curvature

dO = angle of rotation

ab = maximum bending strain

db = bar diameter

However,

dsz ~ dx2 + dy2 =dx2 (l +( dy)21=dx2 (l +

ds = dx(1 + y,~)l/~

But

y, dy tan0

Differentiate both sides of the above equation

y. dZy dO=-aT; = Nsec 0

=~ dO = 1 + tan~ 0 =

Knowing that

(2-2)

(2-3)

(2-4)

(2-5)

(2-6)

(2-7)

(2-8)

(2-9)


ds = Rd0 (2-10)

dO 1

ds R

By inserting dO and ds values from equations 2-9 and 2-5,

Y" dx1 1 + y,2

R - dx(l + y,2 )’’2

(2-11)

(2-12)

(2-13)

2.3 Axial Strain Calculations for Bars

In Figure 2.2 shown below, the solid line CD is the initial unbonded length of the

reinforcing bar in a hybrid frame. The dashed line DE shows the path that the end D of

the unbonded segment of the mild steel bar would take as it moves from D to E.

9


D

Figure 2.2 Path of Point D of the Unbonded Segment of the Mild Steel Bar

Assuming that the center of rotation "O" for the opening of the joint at the beam-column

interface is at the neutral axis of the beam when the hybrid frame is subjected to a design

10


interface rotation of Odes, the following can be written using the same assumptions as in

PRESSS program (Figure 2.2 and 2.3).

Figure 2.3 Location of the Center of Rotation at Odes (Stanton and Nakaki, 2002)

R = (1-~-rldes)hg (2-14)

rides = distance from the compression face of the beam to the neutral axis at 0 aes divided

by the height of the beam

distance from the center of the mild steel bar to the nearest face divided by the height

of the beam

hg = beam height

0 des = maximum (design) interface rotation

Lsu = unbonded length of the mild steelbar at each interface

Let:

As (2-15)

11


Where As is the horizontal gap opening at the location of the tension bar at the beam-

column interface (Figure 2.3) and ea, a is the axial bar strain as calculated in the PRESSS

program assuming that there is no vertical movement at the end of the bar as a result of

rotation.

As= RO~ = hg (1-~ - rlaes)Oe~s (2-16)

(2-17)Lsu -

~a,a ~’a,a

And

tan a" = Ls---~" =R

However,

DE = OD 0~

Ba,a

And

AY= DE sin(-~+a")

(2-18)

(2-19)

(2-20)

(2-21)

(2-22)

(2-23)

12


(2-24)

AYtan a" - (2-25)

02des -}- °~ a,a

"-b o° a,a COS "+- Ofn

(2-26)

The axial strain in the reinforcing bar (Eaxial) can be written as:

4(L,u + AX)2 + AY2 - L~u~’axial -- (2-27)

1

a’,=,.~, :1+ 2(Oza~, + Oe2a.a )’~ COS (2-28)

2.4 Inelastically Deformed Geometry of Bars

As shown in Section 2.2 of this report, the deformed geometry of the bar is needed to

allow calculation of bending strains. Therefore, it is important to determine the

inelastically deformed shape of the reinforcing bar at Odes. A FE model can be used to

accurately predict the inelastically deformed shape of the bar. Once AN equation for the

deflected shape of the bar is derived, the bending and axial strains of the bar can be

calculated based on the equations of sections 2.2 and 2.3.

A parametric study involving multiple nonlinear FE models was performed to understand

the combined axial and bending behavior of the mild steel bar for six different cases

listed in Table 2.1 below.

13


Soo

2

3

4

5

6

Table 2.1 Study Cases

L$11

(1-~ldes-~)h 0 des (inches)

20 0.02 10

20 0.04 20

30 0.02 15

30 0.04 30

20 0.01

20 0.02 20

(rad.)

0.4636

0.7854

0.4636

0.7854

0.245

0.7854

(inches)0.398

0.784

0.597

1.176

0.200

0.396

AY(inches)

0.204

0.816

0.306

1.224

0.051

0.404

0.0398

0.0392

0.0398

0.0392

0.04

0.0198

2.4.1 The FE Model

The undeformed geometry of the mild steel bar at the beam-column interface is modeled

as shown in Figures 2.4 through 2.7 and Table 2.2 below. The partially unbonded bar and

two concrete blocks representing parts of the beam and column are modeled. The grout

around the bar in the bonded portion is also modeled. The length of the concrete block is

assumed to be 20 inches on each side of the unbonded length of the bar to provide

sufficient bar bonded length for all study cases.

14


Figure 2.4 Isoparametric View of the entire Model

~ELEMENT~ A~

2004

Figure 2.5 Isoparametric View of the Concrete Blocks

15


JA~ 2 200410:18:21

Figure 2.6 Isoparametric View of the Mild Steel Bar

JAN 2 2004i0:20:22

Figure 2.7 Isoparametric View of the Duct Grout

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Table 2.2 Model Dimensions

Material Width (in) Height (in) Diameter (in) Area (in2)

Mild Steel Bar 1 . 0.785

Grout 3 6.281

Concrete Block 7 7 41.932

2.4.2 Types of Elements

The ANSYS finite element program was utilized [ANSYS user manual, version 7.1,

2000]. The geometry, node locations, and the coordinate system for the ANSYS element

BEAM 188 are shown in Figure 2.8 below. BEAM188 is defined by nodes I and J in the

global coordinate system. Node K is always required to define the orientation of the

element. The element has six degrees of freedom at each node. The degrees of freedom at

each node include translations in x, y, and z directions, and rotations about the x, y, and z

directions. BEAM188 has linear, large rotation, and/or large strain nonlinear capabilities.

The most important characteristic of BEAM 188 is that it can be used with any cross

section. Elasticity and plasticity models are supported (irrespective of cross section

subtype). This element is based on the Timoshenko beam theory. Shear deformation

effects are also included.

17


Figure 2.8 BEAM188 3-D Linear Finite Strain Beam [3]

Beam �lcmcnts arc one-dimensional line �lcmcnts in space. The cross section details arc

provided separately. The cross sectional dimensions and properties for the mild steel bar,

duct grout and concrete arc shown in Figures 2.9 through 2.11 below.

M = Centroid ~ = ShearCenterSECTION ID 3DATA SUMMARY

Section Name= bar

Area= .786146

Iyy= .049046

lyz= .228E-17

Izz= .049046

Warping Constant=0

Torsion Constant= .098091

Centroid Y= -.133E-16

Centroid Z= ,IZOE-16

Shear Center Y= .254E-16

Shear Center Z= -.133E-16

Shear Corr. MY= .856825

shear Corr. YZ= -.304E-14

Shear Corr. ZZ= .856825

Figure 2.9 Mild Steel Bar Cross Section [3]

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\\\

SECTION ID 2DATA SUMMARY

Section Name= grout

Area= 6.281

Iyy= 3.9Z4

= -.156E-15Izz

= 3.924Warping Constant

=0Torsion Constant

= 7.847Centrold Y

= .906E-16Centroid Z

= -.442E-17Shear Center Y

= .843E-16shear Center Z

= .588E-16Shear Corr, YY

= .688249Shear Corr. YZ

= .214E-15Shear Corr. ZZ

= .688249

Figure 2.10 Duct Grout Cross Section [3]SECTION ID i

X = Cen~roid D = ShearCen~er DATA SUMMARY

Section Nama= hollow

Area= 41. 932

= 196, ID6

= . 828E-05Izz

= 196,106Warping Constant

= 15.87Torsion Constant

= 3Z9.549C ~ntr old Y

= . 662E-08Centr oid Z

= -.805E-08Shear Center Y

= .291E-05Shear Center Z

= -.148E-05Shear Corr. YY

= . 634177shear Corr. YZ

= . 437E-06shear Corr. ZZ

= . 634177

Figure 2.11 Concrete Cross Section [3]

19


2.4.3 Material Properties

Various components of this model are composed of the following materials:

¯ Concrete

¯ Grout

¯ Grade 60 Reinforcing bar

The values of this constant as used in the model are shown inModulus of Elastici~_ ."

Table 2.3 below.

Table 2.3 Material Properties

Modulus ofMaterial # Material Elasticity E,Name (ksi)

1 Concrete 4900

3 Mild Steel 29,000

5 Duct Grout 3500

Stress-Strain Curve." Figures 2.12 and 2.13 below shows the stress strain diagrams for the

mild steel bar and concrete.

20


#8, Grade 60 Reinforcing Mild Steel Bar

120

100

80

60

4O

2O

0 0.02 0.04 0.06 0.08

Strain (in/in)

0.1

Figure 2.12 Stress-Strain Curve of Grade 60 Mild Steel Bar [27]

Stress-Strain Model of unconfined concrete for f "c = 7400 psi

8O00

7000

6000

5000

4000

3000

2000

1000

00.0005 0.001 0.0015 0.002 0.0025 0.003

Strain (in/in)

+Series 1 ]

Figure 2.13 Stress-Strain Model of Concrete [31]

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Densi_ty (~): The densities of steel (mild steel and post-tensioning) and concrete were

assumed to be 490 lb/ft3 and 150 lb/ft3, respectively.

Poisson’s Ratio (v): The Poisson’s Ratio (v) was assumed to be 0.3 for steel and 0.2 for

concrete.

2.4.4 BoundalW Conditions and Loads

The entire concrete block nodes on the left side of the bar’s unbonded length were

restrained (fixed) in all 6 directions (Figure 2.4). All of the concrete block nodes at the

right end of the unbonded mild steel bar were simultaneously deformed to achieve the

following concurrent displacements and rotation

¯ A horizontal displacement AX

¯ A vertical displacement AY

¯ A rotation

The grout and reinforcing bar elements were not restrained in either block. The values of

these applied displacements for the six different cases are listed in Table 2.1. The relative

magnitudes of these displacements are consistent with the deformations in hybrid frames.

2.4.5 Results

The FE model provides full fields of stress and strain throughout the model. Figure 2.14

shows the deflected shape of the entire model. Figures 2.15 through 2.18 show the

horizontal and vertical deformations of the model and the bar for the Model No. 1. The

total (elastic + plastic) axial strains and stresses for. bar, grout, and concrete are shown in

Figures 2:19 through 2.24. The results for the bar axial and bending strains are listed in

Table 2.4.

22


DISPLACEHENT

SUB =24TIME=tO0D~X =.748385

2004

NODAL SOLUTION

STEP=ISUB =24TIME=f00uxRSYS=ODMX =.7q8385SMX =.468

Figure 2.14 Deflected Shape of the Model

JAN 210:26:27

0 . 104 .208 . 312 .416.052 .156 .26 .364 .468

Figure 2~15 Horizontal Deformation (in) of the Model

23


NODAL SOLUTION

SUBTIHE=IO0

D~ =.71212~S~ =.~07Sll

JAN 2 200~110:33:27

0 .090558 .181116 .271674 ,362232.045279 .135837 .226395 .316953 ¯407511

Figure 2.16 Horizontal Deformation (in) of the Mild Steel Bar

NODAL SOLUTION

STEP=ISUB =24TIHE=IO0UY (AVG)RSYS=ODHX =.748385SHX

JAN 2 200410:27:05

0 .129778 .259556 .389333 .519111¯ 064889 .194667 .324444 .454222 .584

Figure 2.17 Vertical Deformation (in) of the Model

24


NODAL SOLUTION

STEP=~SUB =24

uY (

DHX =.712124SKX =. 584

JAN 2 200410:34:41

0 .12~778 .259556 .389333 .519111.064889 ,194667 .324444 .454222 ,584

Figure 2.18 Vertical Deflection (in) of the Mild Steel Bar

NODAL SOLUTION

STEP=ISUB =24TIHE=I00EPTOX (AVG)RSYS=0DMX =.712124

SMX =.042933

JAN 2 200410:36:09

0 .009541 .019081 .028622 .038163¯ 00477 .014311 .0Z3852 .033392 ,042933

Figure 2.19 Mild Steel Bar Axial Total Strain Distribution

25


NODAL SOLUTION

TIHE~I00EPTOX

DHX =.712124~NN ~0S~X =.042933

JAN 2 200410:38:45

¯ 03 .032889 .’035778 .038667 . 041556¯ 031444 ¯ 034333 .037222 .040111 .043

Figure 2.20

NODAL SOLUTTON

STEP=ISUB =24TI~E=IO0sx (AVG)RSYS=0DMX =.712124

S~X =85. 584

Axial Total Strain Distribution of Unbonded Mild Steel Bar

JAN 2 2004I0:35:34

0 19.01’9 38.037 57.056 76.0759.509 28.528 " 47.547 66.565 85.584

Figure 2.21 Mild Steel Bar Axial Stress Distribution

26


NODAL SOLUTION

STEP=iSUB =24TIME=f00SX (AVG)RSYS=0DMX =.712124

S~X =85.58~

JAN 2 2004i0:40:34

76 78.111 80.222 82.333 84.44477.056 79.167 81.278 83.389 85.5

Figure 2.22

NODAL SOLUTION

STEP=ISUB =24TIHE=IO0EPTOX (AVG)RS¥S=0DMX =.7q8385

S~X =.264E-O3

Axial Stress Distribution of Unbonded Mild Steel Bar

JAN 2 200410:30:30

0 .588E-0~ .118E-03 .176E’03 .235E-03.29~E-04 .881E-04 ,147E-03 .206E-03 .264E-03

Figure 2.23 Axial Total Strain Distribution for Concrete Block

27


NODAL SOLUTION

SUS =24TIldE=f00£PTOX

D~X =.723177

SHX =.264E-03

JAN 2 200410:42:52

0 .586E-04 .i17E-03 ,176E-03 .234E-03.293E-04 .879E-0~ .147E-03 .205E-03 .264E-03

Figure 2.24 Axial Total Strain Distribution for Grout

Table 2.4 FEA Results

CaseStudyNo.

AverageAxialStrain

{~avg

0.0398

MaximumBendingStrain

~b

0.0035 0.0433

/(%)

8.8

2 0.0392 0.0038 0.0430 9.7

3 0.0398 0.0025 0.0423 6.2

4 0.0392 0.0026 0.0418 6.7

5 0.0400 0.0029 0.0429 7.2

6 0.0198 0.0019 0.0217 9.5

28


The above figures show the development of plastic strains in the mild steel, bar with its

highest value at the fixed end of the bar (Figure 2.20). The vertical deformation across

the unbonded segment of the bar is plotted in Figure 2.25 for Model No.land in Figure

2.26 for all model cases.

Vertical Displacement vs x

0.200

0.150

0.100

0.050

0.000

-0.050

y = -O.O002x3 + 0.0035x2 + O.O001x - 5E-05 ~i11~~

R~= 1

2 4 6 8 10

x (in)

Case 1~Poy. (Case 1)

Figure 2.25 Vertical Deformation of the Unbonded Segment of the Bar

Vertical Displacement vs x

1.400

1.200

1.000

0.800

0.600

0.400

0.200

0.0000 5 10 15 20 25 30 35

x (in)

r~ - Case 1~ Case 2

Case 3Case 4

:::::::::::::::: Case 5-- " --_" Case 6

Figure 2.26 Vertical Deformation of the Unbonded Segment of the Bar

29


In all six model cases, the vertical deformation data for the bars were fitted into a 3rd

order polynomial. The R2 value of this regression analysis is 1.0 in all cases, which

means that the regression results provide perfect, fit for the FE results. The equations for

the vertical deflection, slope, second derivative, curvature, and bending strain take the

following form:

y = ax3 + bx2 + cx + d

y’ = 3ax2 + 2bx + c

(2-29)

(2-30)

y" = 6ax + 2b (2-31)

1 y"R

(1 + y,2)~-

(2-32)

6ax +2b3 (2-33)

(2-34)

The slope (rotation), second derivative, curvature, and bending strain of the bar are

plotted in Figures 2.27 through 2.30 for all six cases.

30


0.05

0.04

0.03

0.01

">, 0.03

10 15 20 25 30 35

x (in)

Case 1Case 2Case 3Case 4Case 5Case 6

Figure 2.27 Slope (Rotation) of the Deflected Shape of the Unbonded Segment ofthe Bar

y" vs x

0.01

0.008

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

x (in)

-- -- -- Case 1m . Case2

" " - Case 3-- Case 4-- Case 5m . Case 6

Figure 2.28 2nd Derivative of the Deflected Shape of the Unbonded Segment of theBar

31


1/R (curvature) vs x

0.01

0.008

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

x (in)

-- - -- Case 1m . Case 2

- - " Case 3--"--’- Case 4~ Case 5m . Case 6

Figure 2.29 Curvature of the Unbonded Segment of the Bar

(bending strain) vs x

0.005

0.004

0.003

0.002

,.~ 0.001

0

-0.001

-0.002

-0.003

x (in)

-- -- - Case 1m - Case2- - - Case 3~ Case 4~ Case 5~ - Case 6

Figure 2.30 Bending Strain in the Unbonded Segment of the Bar

It can be noticed from Figures 2.29 and 2.30 that the maximum curvature and bending

strain occur at the fixed end of the bar (at x = 0). A general equation for the bending

32


strain in the bar can be determined using constants a, b, c, and d based on the prescribed

boundary conditions as follows:

Boundary_ Conditions:

Atx=O

y=0 ~d=0

y’=0~c=0

At x = Lsu

y= Ay

Odes

0desLs,, -2Aya = (2-35)

Lsu 3

b = 3Ay-OdesLsuZsu 2 (2-36)

Using Ay and Lsu values from equations 2-24 and 2-17,

Odes2 2 ~- 2(0 de, + e .,.) sin( + a")

a = (2-37)

( (1-- rld= ----~- ) hg Ode" l2~,,,. J

According to the FE results the values of both the curvature and bending strain are

maximum at the fixed end of the bar (at x =0).

33


Therefore, Substituting Equations 2-30 and 2-31 in Equation 2-33 yields and considering

thatx=c =d=0:

:=> e~,.m~x 2 ~. R Jmax = bdb (2-40)

where ~b,max is the maximum bending strain along the length of the bar. Substituting for b

from equation 2-38,

3(Odes+o°a,a) sin(+~z")-O~,.~_g,~l )db

-

Or

(2-41)

z ~ dbeb.n~ = 3(02d~ +e o.,) sin( +g") -Odes (2-42)

Table 2.5 below compares predicted a, b, and ~b,max values with the FEM results. It can be

noticed that from Table 2,5 that the predicted values for ~b.ma× (Equation 2-42) is slightly

greater than those of the FE results for all model cases. This is because the grout is

modeled in the FE model which will results in reducing the bending strain in the bar. The

grout effect is not included in the predicted equations (Equations 2-35, 2-36, and 2-42).

34


Model

Case

NO.

1

2

3

4

5

6

Fig

xl0-s

-20

-9

-8

-4

-30

-5

Table 2.5

a

Calculated

xlO-s

-20.8

-10.4

-9.24

-4.62

-41.6

-5ol

Predicted Results Compared with FEA Results

b I~b,max

FE Calculated

xl0-3 xl0-3

3.5 4.12

3.8 4.12

2.5 2.75

2.6 2.75

4.3 4.12

1.9 2

FE

xl0~

3.5

3.8

2.5

2.6

4.3

1.9

Calculated

xl0~

4.12

4.12

2.75

2.75

4.12

2

Calculted/FE

(%)

18

8

10

6

4

5

2.5 Low-Cycle Fatigue for Bar Fracture

2.5.1 Introduction:

High-cycle fatigue and low-cycle fatigue fractures are two different modes of damage in

members subjected to cyclic loading. Low-cycle fatigue is associated with high strain

ranges and low relatively number of cycles would be needed to produce fatigue failure. In

low-cycle fatigue, significant plastic deformations occur during each cycle. On the other

hand, high-cycle fatigue is associated with lower stress/strain ranges (strain cycles are in

the elastic range) and larger number of cycles would be needed to produce fatigue failure.

Low-cycle fatigue lives typically range from one up to 105 cycles, and high-cycle fatigue

cycle lives are greater than 105 cycles [Collins, 1993].

35


2.5.2 Low-Cycle Fatigue Relationships for Reinforcing Steel Bars:

Manson and Coffin (1955) proposed the following empirical equation to estimate the

general fatigue life of a material:

A._~_~ = 2N~.)b + ~f (2Nf ) (2-43)2

Where:

A, = total strain range (*m~x - ~min)

c~’f = fatigue strength coefficient

Nf = number of cycles to failure

E = modulus of elasticity

b = fatigue strength exponent (ranges from -0.05 to -0.15)

~’f = fatigue ductility coefficient

c = fatigue ductility exponent (ranges from -0.5 to -0.8)

The first term in the above equation represents the elastic strain component (high-cycle

fatigue) and the second term represents the plastic strain component (low-cycle fatigue).

The mild steel bar in hybrid frames is subjected to large inelastic deformations. Mitchel

(1979) and Koh and stephens (1991) found that for most low-cycle fatigue analyses the

elastic part can be neglected, for which the above equation can be simplified.

-~- = ~fA~ ,(2Nf)C(2-44)

Mander (1994) experimentally evaluated the low-cycle fatigue behavior of reinforcing

steel bars subjected to cyclic axial strain amplitudes ranging from yield to 6%. He

evaluated the experimental results with existing low-cycle fatigue models found in the

literature..The experimental data were fit to existing fatigue equations. As a result, low-

36


cycle fatigue life relationships were developed for reinforcing steel bars. The following

fatigue life relationships are all based on the work of Mander (1994).

The relationship between plastic-strain amplitude (~ap) and low cycle fatigue life for axial

deformations of steel bars is as follows:

A~p _ 0"08(2NI)-°5(2-45)~ap- 2

Where

~.p = plastic strain amplitude -2

(2-46)

range of plastic strain = ep,m~- (2-47)

~o,max = maximum plastic strain in a cycle

o~p,m~. = minimum plastic strain in a cycle

NI = number of cycles to failure

The total axial strain amplitude of the mild steel deformed reinforcement subjected to

strain cycles ranging from zero to ~s,max can also be calculated using the above low cycle

fatigue equations as follows [Mander, 1994]:

A~_ _0.0795t2tv )-°’448" "’’.,fga2

(2-48)

Where:

gs.m~ =maximumtotal strain ina cycle

=:’ gs,m.x = 2~a (2-50)

37


Although many of the test performed by Mander (1994) were on ASTM A615

reinforcing bars, the author recommends the above equation for all steel types. Additional

tests on ASTM A706 bars are being performed by the writers of this report.

!n a separate study by Chin Liu (2001), mild steel bar specimens were subjected to

bending reversals and low-cycle fatigue life relationships were developed. The results of

constant displacement amplitude tests were plotted and a best fit relationship of the

following form was derived:

~,~, = A~b = 0.23(2NI)-°32 (2-51)2

Where

plastic bending strain amplitude = ~b.m.x- ~b,r~n2

(2-52)

Agb =. range of bending plastic strain =

~b,max = maximum bending plastic strain in a cycle

~b,.~. = minimum bending plastic strain in a cycle

Nf = number of cycles to failure

Figure 2.31 below displays the plastic strain-life relationships under both axial and

bending loadings.

(2-53)

38


Plastic Strain vs 2Nf

0.16

0.14

= 0.12

E "’& 0.1< ,,t

0.08"~ ,~ 0.06

0.04

0.02

05 10 15 20 25

2Nf

Figure 2.31 Plastic Strain-Life Relationship

It can be noticed from Figure 2.31 that for a given number of cycles to failure, low cycle

bending fatigue can achieve higher maximum strain amplitude than low cycle axial

fatigue. In other words, for the same maximum strain amplitude in the bar, specimens

subjected to cyclic bending strains would have longer life than pure axial strains. This

difference is expected to occur because under bending, only the extreme fibers of the

cross section develop maximum strain

The low-cycle fatigue equations under bending action can be converted to equivalent

fatigue equations under axial loading [Chin Liu, 2001]. If the number of cycles to failure

is assumed to be the same for both the axial and bending equations, the effective strain

amplitude under axial loading can be calculated as follows:

A,~_ , (2-54)

39


t t c2

2 °°a = °°fa 2gf (2-55)

1

c2

Where:

eb -- strain amplitude under cyclic bending

effective strain amplitude under cyclic axial loading

cl = fatigue ductility exponent under bending

c2 = fatigue ductility exponent under axial loading

(2-56)

(2-57)

Based on the above, the effective strain amplitude under bending for the mild steel bar is:

-g~ = 0.63(gb)~4 (2-58)2

(2-59)

and

The effective axial strain amplitude and the corresponding maximum effective axial

strain for the six FE model cases are listed in Table 2.6 below:

40


Table 2.6 Effective Axial Strain Calculations

CaseStudy Ax/Ls. ~b ~a’

NO.1 0.0398 0.0035 0.00046

2 0.0392 0.0038 0.00052

3 0.0398 0.0025 0.00028

4 0.0392 0.0026 0.00030

5 0.0399 0.0029 0.00035

6 0.0198 0.0019 0.00019

It can be noticed from Table 2.6 that the value of the effective axial strain amplitude (~a’)

is much smaller than the axial strain in the bar. However, the bars in the study by Liu

[Liu, 2001] were subjected to bending reversals only without subjecting the bar to.plastic

axial strain throughout the section. The above equation (2-59) is not directly applicable to

bars in hybrid frames because the additional axial strain due to bending in hybrid frames

is superimposed on a fully plastic section from the axial loading effects.

This issue requires an experimental evaluation program to address the combined axial-

bending strain effects on low-cycle fatigue of bars. In lieu of this experimental

evaluation, the proposed conservative approach is to find the maximum total strain

variation in the mild steel bar by adding the maximum bending strain with the axial strain

as follows [Collins, 1993]:

~tot,~ = %~,~ + ~b (2-60)

The ~tot,1 can be conservatively used to estimate low-cycle fatigue life using Mander’s

recommended equation (Eq. 2-48).

41


2.6 Effects of Nonzero Mean Strain On Low-Cycle Fatigue Behavior

The mild steel bars in the hybrid frame are subj ected to nonzero mean plastic strains. The

minimum strain emin in the bars is approximately equal to approximately zero. That is, the

strain ranges from zero up to some maximum strain value (say 4%) and then back to zero.

Therefore test results should be adjusted to account for the nonzero mean strain effects if

needed.

A few investigators [12] have studied the effects of nonzero mean strain under low-cycle

fatigue. The experimental results indicate that the effect of compressive mean strain is the

same as the effect of tensile mean strain on low-cycle fatigue if their magnitudes are the

same. The effects of nonzero mean stress are of primary importance only in high-cycle

fatigue where the sign of the mean stress being compressive or tensile may increase or

decrease the fatigue life (Koh and Stephens, 1991).

The mean strain effects were found to be negligible for the range of strain amplitudes

(1% to 6%) used in an experimental study on low-cycle behavior of reinforcing steel bars

[Mander and Panthaki, 1994]. Thus the mean stress and mean strain effects on the low-

cycle behavior of bars can be neglected in design. The total strain range is therefore

considered.

2.7 Effect of Grout in the Duct

Raynor and Lehman [2002] studied the bond characteristics of bars grouted in light-

gauge metal ducts. Growth in the unbonded length of the bar was noticed under high

cyclic strain. This leads to a non-uniform strain distribution at each end of the unbonded

length of the bar as shown in Figures 2.19 through 21. Raynor and Lehman (2002)

42


represented the bar elongation due to strain penetration by an equivalent additional

unbonded length Lua as follows:

Where

(2-61)

f~u = steel bar ultimate strength (90 to 110 ksi for ASTM A706 bars)

f~y = steel bar yield strength (ksi)

fg = grout strength (ksi)

Thus the average axial strain would be equal to:

AL(2-62)~"’~g - Ls.. +

Where

~ = ~/(~ + ~X)~ +~r~ -~,~

Table 2.7 below compares the above equation that calculates the average strain in the bar

due to strain penetration with the FE results for the six study cases. The assumed grout

strength and the yield strength of the bar are as follows:

f’g = 8 ksi

fsy = 60.9 ksi

fsu = 85 ksi

db=l in.L,a- 0.81(85-60.9) =0.86 in.

db (8)1’5

43


Table 2.7 Average Strain in the Bar Due to Grout Effect

ModelCase Lsu AX AY AL ~avg =

No. (in.) (in.) (in.) (in.) AL/(Ls.+2L.a)

10 0.398 0.204 0.4 0.034

2 20 0.784 0.816 0.8 0.037

3 15 0.597 0.306 0.6 0.036

4 30 1.176 1.224 1.2 0.0378

5 0.2 0.051 0.2 0.03

20 0.396 0.404 0.4 0.018

I~avg

(FE)

0.036

0.037

0.037

0.0375

0.035

0.019

(FE/Eq.)%

5.8

16

5

It can be noticed from Table 2.7 that the FE results and the empirical equation that

calculates the additional equivalent unbonded length in the bar are veryclose (within 15

%). Therefore the equation proposed by Raynor and Lehman (2002) is considered valid

and can be used in both analysis and design.

44


Chapter 3

Simplified Design Procedures for Precast Hybrid Frames

In this chapter, the PRESSS design procedures are first described followed

presentation of the proposed non-dimensional procedures and simplified equations.

by a

3.1 Hybrid Frame - PRESSS Design Procedure

The design equations proposed in the PRESSS report are based on the following

assumptions (Stanton and Nakaki, 2002):

1. The design forces (Mdes) and drifts (Odes) are known.

2. The overall dimensions of the frame members are known.

3. At each interface, the centroid of the Post-tensioning tendon is located at the mid-

depth of the beam section.

4. The Post-tensioning tendon is at incipient yield at the design drift.

5. the Post-tensioning tendon is unbonded for the entire length of the frame.

6. Equal top and bottom mild reinforcement is used.

7. Properties of the materials are known.

The PRESSS design equations use deformation compatibility and equilibrium of forces to

calculate the forces and the resulting moment capacity at the interface between precast

beams and columns.

The forces acting on a joint between precast beam and column in a hybrid frame

subjected to a design interface rotation of 0~les are shown in Figure 3.1.

45


a) Dimensions and Displacements b) Forces

Figure 3.1 Location of Forces at Design Drift (Stanton and Nakaki, 2002)

The parameters shown in Figure 3.1 are defined as follows: -

Odes = interface rotation at the design limit state

Ap = axial deformation of post-tensioning steel

As = tensile axial deformation of mild steel reinforcement

A’s = compressive axial deformation of mild steel reinforcement

~ = distance from the center of the mild steel bar to the nearest face divided by the height

of the beam

hg = depth of grout pad at beam-column interface (equal to the height of the beam h)

rides = distance from the neutral axis to the beam compression face divided by the beam

height

46


~des = distance from the center of the equivalent compression block to the beam

compression face divided by the beam height (~d~s = 0.5[~’qd~)

Fp,~es = post-tensioning force in the post-tensioning tendon

Fc,~tes = concrete compressive force at the beam-column interface

Fs,~s = force in tension mild steel reinforcement at the design drit~

F~,,~s = force in compression mild steel reinforcement at the design drift

[~1 = factor defined in ACI 318-99 (section 10.2.7.3). I~1 shall be taken as 0.85 for f’c of

4000 psi and less. For f’c greater than 4000 psi, 131 shall be reduced continuously at a

rate of 0.05 for each 1000 psi of strength in excess of 4000 psi [1~1 = 0.85 - 0.05(t’c

- 4)], but 131 shall not be taken less than 0.65.

According to the PRESSS procedures, a total of 17 steps are required to design the Post-

tensioning tendon and the mild steel reinforcement at the interface. Step 16 calculates the

required unbonded length and elongation of the mild steel reinforcement.

The PRESSS design process is based on an iterative approach and utilizes a number of

dimensional design parameters. The PRESSS design procedure steps (Stanton and

Nakaki, 2002) are summarized as follows:

47


Step 1: Establish Material Properties

Table 3.1 Material Properties of Hybrid Frame Components

Material Property

Concrete and interface grout compressive strengths

Equivalent depth of Whitney compressive stress block

Post - tensioning tendon initial strain, initial stress, modulus

of elasticity, yield stress, and yield strain

Mild steel reinforcement yield stress, modulus of elasticity,

minimum tensile strength, and maximum design axial strain

Mild steel reinforcement over-strength factors (tension and

compression)

Symbol

f ’e, f ’g

ades

8pi, fpi, Ep, fpy, ~py

fsy, Es, fsu, *s,m~x

~s,des and Ls’ ,des

The yield strength of the reinforcement is multiplied by the material, over-strength factors

to calculate the stress in the reinforcement at a certain strain specified by the designer.

Table 3.2 below lists the mean over-strength factors and their corresponding bar strains as

proposed in the PRESSS report (Stanton and Nakaki, 2002).

Table 3.2 Bar Strains and Corresponding Over-Strength Factors for ASTM A706 Bars(Stanton and Nakaki, 2002)

System State Strain ~ ks,

First Yield 0.002 1.0 1.0

Design 0.04 1.35 1.0

Max. Credible 0.08 1.5 1.0

48


Step 2: Obtain Design Loads (M~es) and Drifts (0aes, in percent)

Design loads can be obtained by using either Forced-Based Design (FBD) or

Displacement-Based Design (DBD).

Step 3: Estimate Frame Beam Dimensions

hg _< In/3 (3-1)

bg _> 0.3 hg (3-2)

Where:

bg = beam width

In = clear span of beam between column faces

Step 4: Establish Constants

A term Afp~ is defined below, which is subsequently used in step 11 to calculate stress

change in the tendon.

Af,~ =0.5EpO,~eshg (3-3)

Where:

lpu = unbonded length of pre-stressing tendon tributary to one interface

lpu = ln/2 + he/2

he = column width

lpu = half of the average span length if the span lengths along the tendon are not equal.

Step 5: Determine the Proportion of the PT and Deformed Reinforcement Moment

Strengths.

In the first iteration, an initial estimate can be made as follows:

Mp,des ~ 0.55 Mdes (3-4)

49


Ms,,~es = Maes -

Where:

Mdes = total resisting moment at design limit state

Mp,des = resisting moment provided by PT tendon

(3-5)

Ms,de~ = resisting moment provided by tension mild steel reinforcement

Step 6: Estimate the required area of PT Tendon (Ap)

Initially (1st iteration) assume that the compression force in the beam is located 0.05hg

fi:om the compression face or the depth of Whitney stress block (a) is equal to 0.1hg. This

assumption leads to:

Ap = MP":les (3"6)(0.45hg)f,y

Step 7: Estimate the Required Area of Deformed Reinforcement (As)

Using the same assumption as in Step 6 such that:

A, = Ms’d~ (3-7)( 0. 9 5-~ ) h g/~s,ae, f sy

Step 8: Estimate the Neutral Axis Parameter at the Design Drift (~d~)

The relative location of the neutral axis can be found using the assumption that was made

in Step 6 such that:

0.1 (3-8)

This initial estimate of the neutral axis location will be adjusted during the iterative

design procedure.

Step 9: Calculate the Stress in the Tension Mild Steel Reinforcement

f~,d~ = 2~,d~ f,~ (3-9)

50


The strain in the tension reinforcement corresponds to es,max

Step 10: Calculate the Stress in the Compression Mild Steel Reinforcement

f,’,aes =

Step 11: Calculate the Stress and Elongation in the Prestressing Tendon at 0aes

Ap = Oaeshg(O.5- r]aes)

lpu

Substituting the term Afpoo from equation 3-3 in equation 3-12 yields the following:

Af,o = Afp=, (1 - 2r/,~ )

£o= fpy-Af

if f po > f pi then f po = f pi

fp,d :fpo÷Af,

else f p,~ : f py

Step 12: Calculate the Forces at 0~

The forces in the PT steel,

follows:

(3-10)

(3-11)

(3-12)

(3-13)

(3-14)

(3-15)

(3-16)

(3-17)

tension, and compression deformed reinforcement are as

(3-18)

(3-19)

(3-20)

(3-21)

51


The compression force (Fc,des) between the grout and concrete is calculated based on the

equilibrium of forces at the section.

Step 13: Calculate the Locations of the Compression Force and the Neutral Axis

The depth of the Whitney equivalent stress block (a~es), the relative position of the

compressive force as a fraction of overall height ((Xdes), and the relative position of neutral

axis as a fraction of overall height (Tide~) can be calculated as follows:

Fc,desa~ 0.85f~’ b (3 -22)

= aa~ (3-23)~ des 2h

The above-calculated value of T~des is then compared with the previous value (Step 8 or

previous Step 13), and Steps 11-13 are repeated until the computed and the old and new

values of Tides converge.

Step 14: Calculate the Moment Strength of the Section at 0ae~

Taking moments of the forces in the PT and deformed reinforcement about the centroid

of the compression force yields:

M~,,ae, = F~,,a~h(0"5 - aa~,) (3-25)

M,,a,, = ~,a,,h(l_ff _aa~) (3-26)

M,,,a~ = F,,,,~h(ff -aa~) (3-27)

~ the tot~ moment strength

M~p,o~z = M p,aes + Ms,a~ + M~,.a~ (3-28)

52


The total moment strength should be greater or equal to the design moment

~ M~,p,be, m > Ma~ (3-29)

If Mcap,beam < Mdes, increase Ap and As and repeat Steps 11-14 until convergence is

achieved.

Step 15: Evaluate the Restoring Properties of the Beam

At zero drift, the restoring moment provided by the PT tendon should be greater than the

corresponding moment due to the mild reinforcement, to ensure the re-centering of the

frame after an earthquake. Both sets of reinforcement (top and bottom) would be in

compression. Thus the forces, depth of Whitney equivalent stress block, location of

compression force and neutral axis, and the moments at zero drift can be calculated as

follow:

Fso = As/Zs,,,~f~y (3-30)

F~,o = A~A,,,desf, y (3-31)

Fpo = Apf pi (3-32)

Fco = Fpo - F,o - F,,o (3-33)

Fc0ao - (3-34)

0.85f~’ b

aoao = -z-s. (3-35)

a0r/0 = ~ (3-36)

and

53


M ~,o = Fpoh(0.5 -ao) (3-37)

M,o = Fsoh(1-~-ao) (3-38)

Ms,o = F¢oh(ff -ao) (3-39)

To ensure self-centering

Moo > Ms0 + Ms’0 (3-40)

Where:

Fs0 = force in tension mild steel reinforcement at zero drift

Fs,0 = force in compression mild steel reinforcement at zero drift

Fp0 = force in PT tendon at zero drift

Feo = force between beam concrete and grout at zero drift

ao = depth of Whitney compression block at zero drift

oto = distance from the center of the compression block to the beam compression face

divided by the beam height at zero drift

rio = distance from the neutral axis to the beam compression face divided by the beam

height at zero drift

Moo = resisting moment provided by PT tendon at zero drift

M~0 = resisting moment provided by tension mild steel reinforcement at zero drift

Ms,o = resisting moment provided by compression mild steel reinforcement at zero drift

If self-centering criterion in equation 3-40 is not met, then select a higher ratio of

Ml~,deJMs,des and repeat steps 5 - 15 until the criterion is met.

54


Step 16: Calculate the Elongation and Required Unbonded Length of the Mild Steel BarAs = 0d~h(1-~’-r/a,,)

(3-41)

The strain in the mild steel deformed bar should be equal or less to the maximum usable

strain (~,max). The unbonded length of the bar must therefore satisfy the following

condition:

lsu > As (3-42)

Step 17: Confine Compression Region

The common assumption that plane sections remain plane would not be valid at the

beam-column interface because the PT tendon is unbonded and pre-stressed. Also the

beam end deformation is concentrated in a single crack [Stanton and Nakaki, 2002].

Therefore, the concrete strains cannot be calculated from the curvature and a plastic hinge

length.

Stanton and Nakaki [2002] recommended a plastic hinge length, taken as a function of

the compression zone depth such that

lph = kph Tides h (3-43)

Where:

kph = 1 (Without experimental validation)

~ - Od~(rla~h) - O~e~ (3-44)lph kph

Where:

ec = concrete compression strain at beam compression face

The compression region, should be confined if the calculated concrete strain is greater

than the ultimate concrete strain (_= 0.003).

55


3.2 Overview of Other Design Factors

A more simplified design procedure including user-friendly design charts and equations

would be helpful in gaining more widespread acceptance and utilization of" this

significant technology for precast seismic - resistant construction. This can be achieved

by developing dimensionless parameters that would be utilized in non-dimensional

equations applicable to most cases. The derivations of the proposed non-dimensional

design parameters and equations are described in the following.

The proposed material over-strength factors suggested by the PRESSS report are based

on the authors’ observations (Xs,aes = 1.35 and Ls’,0es = 1). However, according to section

3.3.3 of the ACI T1.2-03 the tensile strength of the mild steel reinforcement must be

taken as the specified minimum tensile strength fu and the compressive strength must be

taken as 1.25fy (in absence of test data on the stress-strain properties of the mild steel

reinforcement). A more rational procedure based on inelastic cyclic behavior of

reinforcing steel bars is needed for calculating the reinforcing steel’s over-strength

factors in tension and compression. Although, discussions on finding Ls,0es and ~s’,des

based on experimental results are given later in this section, the values recommended by

ACI T1.2-03 were used in the examples provided in chapter 4.

A potential failure mode of the hybrid frame is the fracture of the mild steel bars. Tests of

hybrid frame connections by NIST (Cheok and Stone, 1994) indicated fracture of bars

during cyclic loading. Thus the prediction of steel fracture is a very important aspect and

should be considered in the design procedure based on low-cycle fatigue behavior of

bars. The PRESSS procedures do not currently address this issue specifically.

Section 3.3.2 of the ACI T1.2-03 recommends that ASTM A 706 reinforcing bars be used

56


in hybrid frames and specifies a maximum strain (Ss,max) that is 2% less than the strain at

the minimum elongation indicated in ASTM A 706 for a given bar size. For a #8 bar, this

limit would be 10%. This limit could be as high as 12% for other bar sizes.

Priestly and Seible (1996) recommend that the maximum allowable strain in the mild

steel reinforcement under cyclic loading should be limited to 75% of ultimate strain and

6% in absence of better information. Fatigue tests of ASTM A706 bars (with or without

non-zero mean strain) are currently not available. The low-cycle fatigue behavior of

A706 deformed-steel reinforcing bars will be evaluated experimentally at the University

of Wisconsin Milwaukee. The deformed bars will be subjected to cyclic axial strain

amplitudes ranging from 2% to 10% with non-zero mean strain.

As discussed in chapter 2, Mander (1994) experimentally evaluated the low-cycle fatigue.

behavior of ASTM A615 grade 40 ordinary deformed reinforcing steel bars subjected to

cyclic axial strain amplitudes ranging from yield to 6%. The plastic-strain amplitude

relationship with fatigue life for reinforcing steel bars is given by Mander (1994) as:

!~’~’p _ 0.08(2Ni)-o., (3-45)%- 2

Where:

g.~, = plastic strain amplitude - ~m~x- ~n2

Aep = plastic strain range = ~m~x- ~rnin

%~x ----- maximum plastic strain in a cycle

gm~ = minimum plastic strain in a cycle

(3-46)

number of cycles to failure

57


0.0032~ Nf - (gap)2 (3-47)

The hysteretic behavior of bars subjected to cyclic deformations should be considered

when determining the stress level in the bar at maximum design strain. The maximum

tensile stress fmax,a" (in ksi) corresponding to ~ap can be calculated using the following

equation proposed by Mander (1994) for reinforcing steel bars:

fmax,rO%p = 51.6(2N:)-°"541 (3-48)

Substituting for Nffrom Equation 3-47 yields:

fm~x,r = 145(~.p)°~8 (3-49)

fmax,T2s,des =~ (3-50)

f,

Using the above equation, the )~s,aes values for ~ap of 0.04 and fy of 60 ksi would be 1.354,

which is close to the 1.35 values suggested by Stanton and Nakaki.

According to Mander (1994), the equivalent number of equi-amplitude cycles of building

motion in an earthquake (Nc) can be conservatively estimated using the following

equation:

-1

Nc = 7T5- (3-51)

Where

T = fundamental period of the structure in seconds

Chang and Mander [ 1994] report that the number of equi-amplitude cycles (Ne) depends

on the earthquake and the period of the structure as shownin Figure 3.2 below.

58


I0

0.10,01 0.1 t 10

Period (see)

Figure 3.2 Number of Equi-Amplitude Cycles versus the Period of the Structure(Chang and Mander, 1994)

Therefore, substituting Nc from equation 3-51 as Nf in equation 3-45 yields the following:

Sap = 0.021T1/6 (3-52)

The mild steel bar in the hybrid frame is subjected to strains ranging froln zero to

maximum tension strain. The minimum strain would be approximately zero,

’~min ~ 0 (3-53)

Equation 3-52 can therefore be used to estimate the maximum design strain in the bar for

a given period of the structure.

59


3.3 Development of Non-Dimensional Procedure

The step-by-step PRESSS procedures can be restated form as

described below.

Step 1: Establish Material properties (Same as PRESSS Design Step 1)

Step 2: Obtain Design Loads (Mdcs) and Drifts (0des~ °/_~


The PRESSS procedures specify the following two limits for beam dimensions

hg < in/3 (3-54)

bg > 0.3 h (3-55)

However, knowing that

lpu = ½ Ls for spans of equal length along the entire frame

= (average length of spans)/2 for spans of unequal length

Ls = span length (center-to-center of columns)

The following two non-dimensional parameters can be defined.

in a non-dimensional

~b- lpu (3-56)

and

b (3-57)hg

Therefore, based on the original PRESSS limits

~ _> 2 (3-58)

and

q/ > 0.3 (3-59)

60


Step 4: Determine the Proportion of the PT

Strength

Assume a value for ~o:

M p,des

Therefore,

Mp,des = ~oMaes

and

M~,de~ = (1 - (0)Mdes

and Deformed Reinforcement Moment

- ratio of PT moment to total moment strength (3-60)

(3-61)

(3-62)

Step 5: Estimate the Neutral Axis Parameter at the Design Drift (rl~e_~s

0.1r/aes = ~ Or any other initial estimate value (3-63)

The initial estimate of the neutral axis location (0.1/131) will be adjusted during the

iterative design procedure.

Steo 6." Calculate the Strain in the Prestressing Tendon at Odes

The following proposed procedure is slightly different from the PRESSS procedure in

that the strain (and not stress) in the post-tensioning tendon is checked to ensure that the

tendon would not yield.

zx, =

The corresponding change in strain is:

(3-64)

(3 -65)

The PT steel must not yield, and therefore its strain should be kept at or below yield as

follows:

61


~p = epi + A~p _< a’gpy (3-66)

Where:

c~’ is a factor to allow the designer to control and limit strain in the PT tendon at any

desired level below yield. It should be noted that the ~x’ is not included in the original

PRESSS procedures.

a’_< 1 (3-67)

If this condition (Eq. 3-66) is not satisfied, then increase the unbonded length of the PT

steel (qb = lpu/h).

The PT steel stress, fp,aes can then be obtained from the o-g curve for Grade 270

prestressing strands (PCI design handbook) as follows:

For ~py--~ 0.0086 ~ fp,des = Ep ~p (3-68)

Define the following non-dimensional parameters

X- f p,de, (3-69)

~- fP"~ (3-70)

N- fp,d~ (3-71)

Y- f~Y (3-72)f;

R- fsy (3-73)f p,de~

62


Ste~ 7: Calculate the Relative Locations of the Compression Force and the Neutral Axis

The depth of the Whitney equivalent stress block (a~es), the relative position of the

compressive force as a fi:action of overall height (c~es), and the relative position of neutral

axis as a fraction of overall height (rides) can be calculated as follows:

Fcd~ (3-74)aaes --

0.85f~’ bg

- aaes (3-75)cram - 2h-~-

Where:

Fc.d~ = F~,.d~ + Fs.d~ - Fs,,a~ (3-76)

Fp.,~ = Apf p,,~ (3-77)

F~,a~ = A~2s.~f~y (3-78)

F,,,~e, = As2,..a~f,y (3-79)

Define the following non-dimensional parameters

Ap (3-80)b~hg

Apr -A~

Q_ L’ (3-82)

According to Section 18.4.2 of the ACI 318-02 provisions

~, < 0.45 Q (3-83)

Substituting for Fc,des from equations 3-76 to 3-79 in equation 3.75 yields:

63


_ ades

Therefore,

(3-84)

(>85)

(3-86)

The above calculated value of rides is then compared with the previous value in Step 5,

and Steps 6 and 7 are repeated until the rlaes values converge. The non-dimensional

design relationships involving area of PT tendon (Ap) and area of mild steel

reinforcement (As) for the new rides values are:

Md~ - 0"5(1-fllr/ae*) (3-87)Aphgfp,aes ~

M,~,=~ _ (1-g" - O.5fl, r&s)- 0-88)AshgZ s,desf sy 1-co

and

ip RA,,a~ (1-~-O.S flfla~,)

As 0.5(1- fllr/d~, )(1-- 1)�o

(3-89)

Ste_~__8: Calculate the Moment Strength of the Sction at 0de__~

Taking moments of the PT and deformed reinforcement forces about the centroid of the

compression force yields:

M~,,a= = Fp,q~h(0.5 - aa~) (3-90)

M,,.s = F,,a~,h(1-~-aa=) (3-91)

(3-92)

64


The total moment strength

M~p,be,m = Mp,des + M,,des +

The total moment strength should be greater or equal to the design moment

M cap,beam >-- Males

Therefore, the following non-dimensional relationship can be written

0.5(1- fl~r/d~) + rR---[As,a~(1-~’-0.Sfllr/a~)+ As,,a~(~’-0.5fllr/a~)t

The above condition would be the objective function if a

performed. The objective function should be achieved by changing rlaes, ~0, and

Steo 9: Evaluate the Restoring Properties of the Beam (Constraint Equations)

= - - s0

a0--

a0~o =~

ao

M,o = F~hg(O.5-ao)

M~o = F~oh~ (1- £’- ao)

M~,o = F~’ohg (~ -eto)

(3 -93)

(3 -94)

_> 0’5(1-fllr/a~) (2-95)

design optimization is

(3 -96)

(3-97)

(3-98)

(3-99)

(3-100)

(3-101)

(3-102)

(3-103)

(3-104)

(3-105)

65



Mpo > Mso + Ms,o (3-106)

The above requirement for self-centering can be written in non-dimensional form:

r > 22s,,dosK (3-107)

where,

K- Zy (3-108)

Step 10: Calculate the Elongation and Required Unbonded Length of the Mild Steel Bar

In this step, the PRESSS procedures are modified to include the effect of bending strain

(~b,max). the contribution of additional effective unbonded length of bars dueto straining

in the bonded regions is also considered. The bar’s horizontal deformation (AX), axial

strain (ea), and bending strain (~b) should be calculated using the equations presented in

Section 2.3 as follows:

Let

As O ae, h g (1- rl ae, - £ )~’a,a ----- ~ ----- (3-109)

where,

Lsu = unbonded length of mild steel reinforcement at each interface

From the relationship developed in section 2.3 of this report,

1

~,= l+2(02ae, +e2.,.)icos +a" +(02ae. +~ .,a -1 (3-!10)

66


(3-111)

Where,

~z" = tan-1/O~ / (3-112)

Growth in the unbonded length of the bar is expected to occur under high cyclic strain.

Raynor and Lehman [2002] studied the bond characteristics of bars grouted in light-

gauge metal ducts. A non-uniform strain distribution occurs at each end of the unbonded

region of the bar that penetrates into the bonded region of the bar. According to their

study, the elongation due to strain penetration is equivalent to the peak strain acting

uniformly over an equivalent additional unbonded length (Lua) on each end. The

additional equivalent debonded length can be calculated using the following equation

[Raynor, 2002]:

L"---a-~ = 0"81(f~u-f~Y) (3-113)do

Thus, the total unbonded bar length (Lot) is:

Lu,=L,u +2L.. (3-114)

The maximum total strain that the bar is subjected to is the sum of axial and bending

strains (~a×ial + ~b,m~x). As discussed earlier, the conservative approach proposed is to

ensure that the total strain is less than ~s,ma×, the maximum strain that the bar can be

subjected to basedon low-cycle fatigue tests.

’~,=;,,~ + ’~t,,~x < ’~s,~x (3-115)

67


We can then calculate eU,m~x (equation 2-111) and subtract it from es,m~x to obtain the

maximum allowable axial strain (earl).

Gall : Gs,max -- Gb,nmx (3-116)

To ensure satisfactory performance, the following condition must be met

Gox~l _< GatI (3-117)

However, since the actual effective unbonded length is L~t and not Lsu, the calculated

axial strain can be reduced by the ratio (Lsu/Lut).

G,,,, >G,,xi,~, ~ Lut)- (3-1t8)

Lut >_ Ls" ~°t - (X-rld~" -£)O~ l G’~i"’I (3-119)hg hg GalI Ga,a k. Gall J

For typical values of ~a,~ and 0~,s, the ratio of ~x~___L is very close to 1. Therefore,

>_ (1 - V e, -- (3-120)

and

L,. (Required) = Lu, - 2L,u, (3-121)


Gc -- . lph kph

kph = 1 (Without experimental validation)

If ec > 0.003 provide confinementreinforcement in the compression region.

68


3.4 Summary of Proposed Non-Dimensional Steps

1. Establish material properties

2. Obtain the applied moment and design drift (Maes and 0aes)

3. Specify d? = Lpu/h

4. determine the portion of moment strength (03)

5. Estimate the neutral axis parameter (~lae~)

6. Calculate the strain in the prestressing tendonAp _ 0~os (0.5-qa~)

¯A~r - lpu

~

7. Calculate the relative locations of the compression force and the neutral axis

Ap7-

bghg

N - fp,ae~f;

f ~d.es

G~,des -- 1.7

Mao~ _ 0.5(1-fl~r/a~)Aphgf p,des o)

69


M~¢, = (1-g" - 0.5fl~r/~,)

Ap R&, ,~ (1- g" - 0.Sfl~r/~,)

As 0.5(1- tilt/des)(1-1)

8. Calculate the moment strength of the section (objective

optimization problem)

0.5(1--j~lTldes) -’b -~I,~s,des(1--~--O.5~lTldes)-b ~s,,des(~--O.51~llldes)l

9. Evaluate the restoring properties of the beam (constrained

optimization problem)

function in an

> 0.5(1- fl, r/,~)

equation in an

10. Calculate the required unbonded length of the deformed reinforcement

0 des + °~ a,a COS +

2 ~ ~3(0 ~ + g o.) sin( + a") - Od~ db

L~._~_~ = 0.8 I(f,, - f**)db

L., =Ls. +2L..

%~.~ + eb,~x < es,m.x

70


3.5 Parametric Studies

The above dimensionless parameters were varied in an extensive parametric study

(approximately 1200 runs) involving an optimization process. The objective of the

optimization program was to ensure that the moment capacity of the section is greater

than Mdes and satisfying additional constraints discussed below.

M cap,bea~n = M p,aes + M s,ae, + M s,,aes > M ,u~

R A, -:::> 0"5(1--)~l/’]des) + 7[ "’a~(1-~-O’Sfl’rlae’)+ ~s"des(~--O’5t~’~des)~ > 0"5(1-fl’~/a~’)co.’. The objective function is equal to:

0.5(1-fl, r/a~) + RE,~,.a~(1-~’-0.Sfllqa~)+ I~s,,des(~--O.5~l~des)~-O’5(1-~lTldes) =0

The Microsoft EXCEL’s Solver feature was used to find the optimum solution by

changing HOes, m, and y subjected to following constraints:

(to achieve zero residual drift)

(to ensure that PT steel would not yield)

(Q =f{:/fro) (to meet ACI 318-02 section 18.4.2 requirements)

Figure 3.3 below shows some features of the Excel spreadsheet. The formulation of the

whole spreadsheet is given in Appendix A. Table 3.3 below lists the range of parameters

used in the parametric studies.

71


ksi

0.03419

M~ ~M,o +M,,o

CMcu~e ti~ re]a6ve.loc ations of the compression £o~ce md the neutral

St~l-. 10 CldCULlte ~he moment ,trengffi of~ section = ~

x[~ =o-�- ~)+~,.,(¢-~)] ~ (o.5- ~. )(o~-~+ ’ ~ ¯ ~

_take OA~ to the le~ side ~the equation

Figure 3.3 Sample view from EXCEL Spreadsheet

72


Table 3.3 Range of Parameters Used in the Parametric Studies

Parameter

4-L~

(%)

(%)

f~’ (ksi)

Minimum Maximum

2 20

0.5 4

1 8

0.5 0.65

4 8

3.6 Results from the Parametric StudiesBased on the results of the parametric studies, a complete set of charts

dimensional parameters were gene{ated as shown in Figures 3.4 through

simplified design equations were recommended.

The design charts that follow are based on the following basic properties

En (ksi) 28500

f~u (ksi) 270

f~y (ksi) = 0.9 fp,, 243

Es (ksi) 29000

f,~, (ksi) 60

~ 0.055

with non-

3.19 and

A designer can calculate the required areas of post-tensioning tendon (Ap) and mild steel

reinforcement (As) by utilizing Figures 3.4, 3.6, 3.7, and 3.9and using the following

procedures for a specified M~es, Odes, and Vs,m~x :

1. Specify d~ = Lpu/hg

73


2. Go to either one of Figures 3.14 through 17, determine ~2, and calculate fp,des

3. Go to Figure 3.4 or Figure 3.7, determineMaes , and calculate Ap

M des4. Go to Figure 3.6 or Figure 3.9, determine As2c,,ae~h~,f,y, and calculate As

It can be noticed from Figure 3.4 that the required area of PT tendon Ap increases as Odes

increases for a given do, Mdes, and Ss,ma~. On the other hand, Av decreases as

increases for a given do, Mae~, and Odes (Figure 3.7). The reverse is true for the area of the

mild steel reinforcement As.

The stress in the PT tendon fo,des at a specified design drift Odes increases as dO increases. It

also increases when the initial stress in the PT tendon fpi increases as shown in Figures

3.14 through 3.17.

The designer can also calculate the ratio of the PT moment to the total moment strength

co from Figure 3.5 and Figure 3.8. The relative area of the PT tendon to the beam’s cross

section 3’ can be calculated from Figures 3.10through 3.12. The relative location of the

neutral axis qdes can be calculated from Figure 3.13. It can be noticed from Figure 3.13

that rides increases as the compressive strength of the concrete f’c increases.

74


0.96

0.94

~ 0.92~.~ 0.9

~ 0.88~ 0.86~ 0.84

0,82

0.8

Mdo,lAphgfp,do, Vs ~[S,,max = 4%]

10 15 20 25

¢ = LpJh~

values ofe.oo (%)+0.5--=--1~ 1.5--x~2

+3--+--3.5~4

Figure 3.4 [Mdes/Aphgfp,des] versus � for ~s, max ----" 4%

0.55

0.54

0.53

0.52

0.51

0.5

0.49

0.48

0,47

0.46

~0 = Mp,des/Mdes VS �[E;s,max " 4%]

0 5 10 15 20

� = Lpu/h~

25

values ofOdes (%)

---~-’- 0.5--~--- 1---b--- 1.5---X--" 2~2.5"--~3--+--3.5~4

Figure 3.5 [Mp,des/Mdes] versus � for Ss,max = 4%

75


1.95

1.90

1.85

1.8o

1.75

1.70

1.65

[ Mdesl(As~,s,deshgfsy)] V$ �

[Ss,max = 4%]

5 10 15 20

� = Lpu/hg

Figure 3.6 [Mdes/As~,s, deshgfsy] versus � for 8s,max = 4%

values ofede.~ (%

%2

"-~’- a-’-+-- &5~4

25

0.96

0.940.92

0.9

~ 0.880.86

0.84

0.820.8

Mdes/Aphgfp,des VS ~p[Odes ---- 2%]

/

0 5 10 15 20

� = Lpu/hg

25

values of~s,max (%)

---~- 4~5~6

Figure 3.7 [Mdes/Aphgfp,des] versus � for Odes = 2%

76


0) = Mp,des / Mdes Vs ¢[Odes = 2%]

0.57

0.55

~0.53

0.51

0.49

0.47

0.455 10 15 20

� = Lpu/hg

25

values of~s,max (%)"-~-’- 1

"--~-- 2---)*--2.51~̄’~ 3

~5~6~7~8

Figure 3.8 [Mp,des/Mdes] versus � for Odes = 2%

2.052.001.951.901.851.801.751.701.65

[Mdes/As hgZs,desfsy)]Vs ~[Odes = 2%]

20

� = Lpu/h

25

values of

~s,max (%)+1---I-- 1.5~r-- 2

+3+3.5---t--- 4~5

---~-- 7+8

Figure 3.9 [Mdes/As~,s,deshgfsy] versus � for Odes = 2%

7?


0.25%

0.24%

0.23%

0.22%

0.21%

0.20%

0.19%

0.18%

y = A~/bhg Vs �If "g = 5 Ksi, fpi = 0.65 fpu]

5 10 15 20

~ = L~/ho

25

values of6des (%)

Figure 3.10 Ap/bhg versus ~ for f’c = 5 ksi and fpi = 0.65

7 = Ap/bhg Vs �[f "c = 5 Ksi,edes = 2%]

0.35%

0.30%

0.25%

0.20%

0.15%5 10 15 20 25

( = Ipu/hg

+ fpi = 0.5 fpu+ fpi = 0.55 fpu--&--fpi = 0.6 fpu--x--fpi = 0.65 fpu

Figure 3.11 Ap/bh versus ~ for edes = 2% and fpl = 0.65 fpu

78


7 = Ap/bh Vs �[Odes = 2%, fpi--’-- 0.65 fpu]

0.40%

0.35%

0.30%

0.25%

0.20%

0.15%10 15 20 25

~ = Lp,,/h

~ f "c = 5ksi~f’c 6ksi~f’c 7 ksi

!--x--f "c 8ksi

Figure 3.12 Ap/bh versus ~? for Odes ----- 2% and f’e = 5 ksi

~s Vs I~

0.2

0.19

0.18

0.17

0.16

0.15

0.14

0.13

0.12

0.11

0.10.6 0.65 0.7 0.75 0.8 0.85 0.9

Figure 3.13 Tides versus

79


~ =fp,deslfpy VS �

[fpi -- 0.65 fpu]

1.00

0,95

0.90

0.85

0.80

0.75

0.705 10 15 20

� = Lpu/hg

25

~1.5---X---- 2.--)k~-- 2.5

---e--3

---~-" 3.5~4

Figure 3.14 fp,des/fpy versus ( for fpi = 0.65 fpu

0.350.340.330.320.31

~.~,~0.300.290.280.27O.260.25

R = fsylfp,desVS (~

[fpi = 0.65 fpu]

0 5 10 15 20

~ = LpJh25

(%)

-’~’- 0.5"-~-- 1

1.5"-’X’-- 2~2.5"-e--3~3.5

~4

Figure 3.15 fsy/fp,des versus � for fpl = 0.65 fpu

80


E~ = fp,des/fpy VS � = Ipu/hg[O = 2%]

1.00

0.95

0.90

~~0.85

" 0.80

n,’ 0.75

0.70

0.65

0.600 5 10 15 20

~ = Ip,,/hg

25

---,-fpi = 0.5 fpu--~-- fpi = 0.55 fpu~ fpi = 0.6 fpu--x--fpi = 0.65 fpu

Figure 3.16 fp,des/fpy versus ( for 0d~s = 2%

R = fsy/fp,des VS � = Ipu/hg[0 = 2 %]

0.45

0.40

0.35

0.30

0.25

0.205 10 15 20 25

� = Ipu/hg

+ fpi = 0.5 fpu--~-- fpi = 0.55 fpu~ fpi = 0.6 fpu---X---fpi = 0.65 fpu

Figure 3.17 fsy/fp,des versus ~ for Odes ---- 2%

81


Since the curves in the above figures show smooth variations, simple design equations

that are functions of Odes, I~s,max, �, and 131 can be derived from the design charts. The

proposed design equations are as follows:

Mp des’ -- 0.8 90)0.04~s,max"0"0371 (3-123)des

Maes 0.0371= 1.125rPo.o4~s,m~x (3-124)Aphgfp,des

Mde~ = 0.882Wo.o4~s,m~x*’°34 (3-125)A~Zs,ae~ h g f p,a~

Ap7 - - ~’~v~v~ (in percent) (3-126)

r/aes = -0.2fll +0.3 (3-127)

Where:

030.04 = the ratio of the PT moment to the total moment strength corresponding as,n~× = 4%

Mdes~Po.o4 - at es,n~ = 4% (3-128)

Aphgfp,des

Mdesaro.o4 = at e,,n~ = 4% (3-129)Asd,,,a~sh g f p,a~

1. For d~ = Lpu/hg -< 8

Mp des

COo.o4 = ____a__, = (5.160a~, + 0.49)#(z7°~+°’°3~Mdes

(3-130)

Mdes 0 9)¢(2’70m +0’03)q~o.o4 = = (-6"7~a~, + ¯Aphg fp,,t~s

(3-131)

M,~ = (22.230ae~ + 1.7)¢.(3.464om+o.o2,6)mo.o4 = A~2,,a~h g f l,,a~ (3-132)

82


0aes = interface design drift expressed in fraction

2. For d~ = Lpu/hg > 8

Mp,des 0.47) #-(1.9oa~ +o.oo42)O90.04 = = (3.79~a= +Mdes

(3-133)

Maes 0.94) 0(1"90aes+0.0042)CPo.o4 - _ (-5.810a~ +Aphgfp,d~

(3-134)

(3-135)

Ap fOrfc~= 5 ksi ~’ldfpi = 0.65fpu (3-136)~"~ - b~,h~,

~’r~ = (-20aes + 0"215)0(38~0~=+°°25) (in percent) (3-137)

v~ = is a factor that accounts for different values off~’

= 0.208f ’-0.0355 (3-138)

Vp~ = is a factor that accounts for different values of fp~fou

Vp~ = -1.82fpi + 2.18 (3-139)

The results of the three procedures (PRESSS, proposed design charts, and proposed

design equations) are compared for each of the five examples in the following chapter.

The following steps must be followed when utilizing the new design procedures:

A. Procedures when using Design Charts:

1. Obtain the applied moment and design drift (Mae~ and 0a~).

2. From the estimated fundamental period of the structure T, calculate Nc using

equation 3-51 and Ss,max using equation 3-45 or equation 3-52.

83


3. Specify d~ =/-~u/h

4. Calculate fp,des by using Figures 3.14-3.17.

5. Calculate Ap by using Figures 3.4 and 3.7.

6. Calculate As using Figures 3.6 and 3.9.

7. Calculate ,/using Figures 3.10 through 3.12..

8. Calculate Lsu

B. Procedures when using Design Equations:

1. Obtain the applied moment and design drift (Mdes and Odes).

2. From the estimated fundamental period of the structure T,

equation 3-51 and ~s,max using equation 3-45 or equation 3-52.

3. Specify d~ = Lpu/h

4. Calculate r/d~ =--0.2fl1+0.3

5. Calculate Asp using equation 3-65.

6. Check if 8p _< a’ %y using equation 3-66.

7. Calculate fp,dcs using equation 3-67.

8. Calculate Ap using equation 3-124.

9. Calculate As using equation 3-125.

10. Calculate ), using equation 3-126.

11. Calculate Lsu

calculate Nf using

84


Chapter 4

Design Examples

In order to validate, verify, and illustrate the effectiveness of the developed design charts

and simplified equations, a set of five design examples are considered. Each example

includes four separate calculation processes: the original PRESSS procedures, the

modified non-dimensional procedures, the developed non-dimensional charts, and

proposed simplified equations. A comparison and discussion of various solutions is then

given. It should be noted that )~s,des and )~’s,de~ values recommended in ACI T1.2-03 were

used in these exampes

4.1 Example # 1:

Reference: S.K. Ghosh Associates Inc. (PCI Seismic Seminar, Chicago, March 2002)

The design parameters in this example are identical to the PRESSS example at the PCI

convention. However, the calculation steps are revised and corrected to achieve an

optimum solution.

Seismic Data:

Location: Charleston, SC

Site Class D

5-story Office Buildings

3 equal bays (span length = 45 if)

hg = 42 in., bg = 22.5 in

Moment-Resisting Frame System (Hybrid Frame)

85


This example is solved using the original PRESSS calculation, the modified non-

dimensional procedures, the developed non-dimensional charts, and the proposed

simplified equations. Table 4.1 below shows the material properties and information that

are used in the four different design procedures. The results are compared in Table 4.2.

Table 4.1: Material Properties and Design Information for Example 1

Parameter

f ’c (ksi)

f ’g (ksi)

fpy (ksi)

fpu (ksi)

(ksi)Es (ksi)fsy (ksi)

Ls

~,, (It)

Ep (ksi)

Value

5

5.5

243

270

175.5

29000

¯ 60

0.055

45

22.5

28500

0,8[31

~3s,max 0.046

)~s 1.4

Ls’ 1.25

Sy 0.00207

Spy 0.0085

~pi 0.00616

Males (£t-kips) 1235

Odes (%) 1.96

Comments

Compressive strength of concrete

Compressive strength of grout

Yield strength of Gr. 270 strand

Minimum strength of Gr. 270 strand

Initial stress in strand

Modulus of elasticity of mild steel

Yield strength of Gr. 60 steel bars

Cover factor (See Figure 3.2)

Tendon length c/c of columns

(All spans are of equal length)

Ls/2

Modulus of elasticity of Gr. 270 strand

131 = 0.85 - 0.05 (f ’c - 4)

Maximum sustainable strain in the mild reinforcing steel

(See section 4.1.2)

Tension over-strength factor (ACI T1.2-03)

Compression over-strength factor (ACI T 1.2-03)

fsy

fpy fpi/E.

Obtained using Displacement Based Design (DBD)

Calculated Drift

86


4.1.1 Basic Solution Using PRESSS Procedure

The following is an iterative process that follows steps 1 through 15 based on initial

assumptions. For the sake of brevity, only the initial and final values of parameters are

given below.

~ Establish Material Properties and Design Information

See Table 4.1

Step 2: Obtain Design Loads (Mdes) and Drifts (Odes %)

See Table 4.1


hg </n = 42.67x12 =170.7 in. OK3 3

bg >0.3hg =0.3x42=12.6in. OK


= 0.5x 28,500× 0.0196x42

43.4 ksi22.5x12


Strength

In the PRESSS procedure, an initial assumption of Mp,des/Maes is made, which in this case

was 0.65. However after several iterations of the steps shown here, the optimum solution

converged on a Mp,desiMdes ratio of 0.51.

Mp,des =- 0.51 Mdes = 630. ft-kips (Optimum solution after several iterations)

Ms,des = Mdes - Mp,des = 605 it-kips

87


Step 6: Estimate the Required Area of PT Tendon (Ap)

Assume initially that the compression force in the beam is located at 0.05hg (a = 0. lhg)

from the compression face. This assumption leads to:

Ap = MP’des

(0.45hg)f,y

630x12= 1.65 in2

0.45 × 42 × 243



(0.95-g’)hg~,d~,fsy

605x12= = 2.3 in2

(0.95-0.055) x 42 x 1.39 x 60

Step 8: Estimate the Neutral Axis Parameter at the Design Drift (rides)

The location of the neutral axis can be found using an initial assumption of a!hg -= 0.1 that

was made in Step 6. This assumption results in an initial value of Tides = 0.1/~1. However

after several iterations of the steps shown here, the optimum solution converged on a rides

value of 0.14.

a/h~,r/d~ =~ = 0.14


= 1.4 × 60 = 84 ksi

88


Please note that the PRESSS procedure recommends a LS,des value of 1.35, but allows for

a more rational determination of this parameter. In this report, a ~s,des value of 1.4 is used

according to the provisions of section 3.3 of ACI T1.2-03.


= 1.25 x 60 = 75 ksi

Please note that the PRESSS procedure recommends a Ls’,des value of 1, but allows for a

more rational determination of this parameter. In this report, a )Vs,,des value of 1.25 is used


Step 11: Calculate the Stress and Elongation in the Pre-Stressing Tendon at 0aes

A, = Od h,(O..5- Vd )

= 0.0196x 42x (0.5- 0.14) -- 0.296 in

Ap 0.296Afp =--gp =

lpu 22.5 x 12

f,o= f,y-Af,

-- 243 - 31.2 =211.8 ksi

iffpo > f,, = 0.65x270 = 175.5 ksi

then fpo = fpi = 175.5 ksi

f p,ae, = f po + Af p

= 175.5 + 31.2 = 206.7 ksi

x28,500= 31.2ksi

else f p,de, = f py

89


0.5(1- flfld. )h ~f p,d~

630x120.5(1- 0.8 x 0.14) x 42 x 206.7

= 1.96 in~

And

Ms,des

( l -- ~ -- 0.5 fllrides )hg,¢Cs,desf sy

605 x 12(1- 0.055-0.5 x 0.Sx 0.14)x 42x 1.4x 60

= 2.31 in2

Step 12: Calculate the Forces at Od~s

= 1.96x 207 = 405.71 kips

Fs,d~, = A,~,,de~fsy

= 2.31xl.4x60 = 194 kips

F,,cles = AsFI,s,,desfo,

= 2.31xl.25x 60 = 173.25 kips

Fc,ae, = Fp,ae, + F,,des-Fs,,aes

= 405.7 +194 - 173.25 = 426.5 kips


426.5= 4.5 in.

0.85 f_’b_ 0.85x5x22.5

4.5 = 0.054

_ ad~ = 4.5 =0.134 (Optimum value is 0.14, OK.)rides fllh g 0.Sx 4~

90


1~des = O. 14

O~des = 0.5X fll ×lTdes

= 0.5 x 0.8 x 0.14 = 0.056

Step 14: Calculate the Moment Strength of the section at Odes

Taking moments of both the PT and deformed reinforcement forces about the centroid of

the compression force yields:

= 405.7 x 42(0.5-0.056)/12 = 631.1 ft-kips

Ms.ae, = F~,a~hg(1-~-aae~)

= 194 x 42(1 - 0.055 - 0.056)/12 = 604.1 ft-kips

= 173.25 x 42(0.055 - 0.056)/12 = - 0.6 g&ips

Therefore the total moment stren~h c~ be dete~ned as follows:

M ~.p,b~ = M ~,a~ + M ~,a~ + M ,’,a~

= 631.1 + 604.1 - 0.6 = 1234.6 fl-kips

g cap,b~.~ ~ g aes

1234.6 a 1235 fi-kips, OK (Optimum Solution)


= 2.31xl.25x60 = 173.3 kips

F,,o = As~s,,aesf sy

= 2.31xl.25x60 = 173.3 kips

91


=2×175.5 =351 kips

Fco =Fpo-Fso-F~’o

= 351- 173.3 - 173.3 = 4.4 kips

~Feo

0.85£4.4

0.85×5x22.5= 0.046 in

_ 0.046 = 0.000552x 42

ao

0.046

0.8x 42--= 0.00137

Mpo = Fpohg (0.5 - Of 0)

= 351x 42(0.5 - 0.00055)/12 = 613.6 ft-kips

M,o = F~ohg(1-~-ao)

= 173~3 x 42(1- 0.055 - 0.00055)/12 = 572.9 if-kips

M,,o = F ’ohg( -ao)= 173.3 x 42(0.055 - 0.00055)/12 = 33 ~-~ps


Mpo > Mso + Ms’o

613.6 ~ 572.9+ 33 = 606 fl-kips, OK

92



As = 0,~hg(1-g"-r/a~s)

= 0.0196 x 42(1 - 0.055 - 0.14) = 0.66 in

0.66- - 14.4 in0.046

Provide 14.5 in. of unbonded length

Please note that the PRESSS procedure recommends a es,m~x value of 0.04, but allows for

a more rational determination of this parameter. In this report, a new methodology for

calculating ~s,max is proposed (See sections 3.5 and 4.1.2 for details), which results in a

es,m~x value of 0.046.


lph = kph 1Ides hg

= lx0.14x42 = 5.88 in.

~c- tgdes(r/’~eshg) - odes =0.0196lph kph

0.003 Provide proper confinement to prevent spalling

4.1.2 Solution Using Modified Non-Dimensional Design Procedure based on PRESSS

This modified procedure utilizes non-dimensional parameters and does not involve

iterative calculations as in PRESSS. However, an optimization procedure is performed

within Microsoft Excel. The following procedures also utilize the proposed low-cycle

fatigue criterion for the.mild reinforcing steel. See section 3.5 for details.

Let,

93


Y = fsy/f "e = 12

Z = Isy/I py = 0.247

K = fsy/f po = 0.34

O = f’c/fpi = 0.028

Please note that the following calculations for ~s,max are not included in the original

PRESSS procedure.

Calculate the estimated fundamental period of the structure as follows:

T = 0.03 (hn)3/4 = 0.03 (65.5)3/4 = 0.691 sec.

Where hn is the overall height of the building

The number of cycles to failure can then be calculated as

Nc = 7 T -1/3 = 7 (0.691)-i/3 = 7.92 cycles

Enter the value of Nc for Nf in the following equation

~,~, = 0.08(2N~.)-°’45 = 0.023

and

~s,max = 2~av = 0.046

fs,des=~,s,des fsy = 84 ksi

fs’,des=Xs’,des fsy = 75 ksi

Steo 1." Establish Material Properties and Design Information

See Table 4.1.

Steo 2.’ Obtain Design Drift

See Table 4.1

94


~ Estimate Frame Beam Dimensions

~ = Lpu/hg = 6.43

Ste~ 4: Determine the Proportions of Moment Strength

~o = 0.5 (Optimum solution calculated using the EXCEL Solver)

~ Estimate The Neutral Axis Parameter

rides = 0.14 (Optimum solution calculated using the EXCEL Solver)

~des = 0.5 ~11]Ses = 0.055

Ste~ 6: Calculate the Strain in the Pre-Stressing tendon at Odes

This step is a modified form of the PRESSS procedure to ensure that the post-tensioning

steel would not deform beyond yieldAp _ 0d~ (0"5--r/d~) = 0.0011

~p = ~pi + A~ = 0.0073

Check if

Let ~z’ =t

0.0073 < 0.0085 OK

fp,des = Ep 8p = 207 ksi

Let,

R- Lyf p,des

= 0.29

=.0.85

95



Let,

Apy - = 0.21% (Optimum Solution)

bghg

1.96- - 22.2 in(bg)°ptimum0.0021x42

bg = 22.5 in. is OK ~

Check if

0.45 Q = 1.41%, OK (ACI 318-02 Provisions)

and

_ 2" N + --()t~,de~ = 0.14~L~0.85fll r

Mdes

Aphgfp.d~

~Ap =

and

_ 0.5(1- fllr/a~) = 0.87

1235x12 I= 1.96 in21

0.87 x 42 x 207

M~ = (1--g"--0"5fllr/d~)=1.83Ashg,%s.aes f~y 1 -c0

1235x12A~ = =12.3 in~l

1.83×42xl.4×60

96


Ap RZ~ ,~= (1 - ~" - 0.5~6~r/,~)v = -- = = 0.85

A~ 0.5(1 _ tilt/des)(!_ 1)

Steo 8: Calculate the moment strength of the section at Odes

M ~,,~,,, = M ~,,a~ + M s,,~ + M ,,,ae~ >Mae.,

R ~ 0.5(1-/~lV~,)

.’. The objective Nnction is equal to:

O.50- ~ ~ + ~ [~s,~ (I- £ - 0.S~) + &,~ (~- 0.~~- 0.S(1- ~l~ =0

The EXCEL Solver is used to find the optimum solmion by ch~ging ~a~s, m, an6 ~ Subjected to

The following constraints:

1) z- > 2A~,,a~K

2) a’. <. atoC’py

3) V < 0.45 Q

4) (Tides)calculated ---- (]]des)assumed

(]]des)optimum - 0.14

(O))optimum = 0.51

(~)optimum -- 0.21%

Steo 9." Evaluate the Restoring Properties of the Beam (Constrained Equations)


Mp0 -> M~0 + M~,0

r > 2&,,a=K=0.85

97


Step 10: Calculate the required unbonded length of the deformed reinforcement

Let,

~’.,o = Q~x = 0.046

ct~=tan-ltOa~ )=0.403rad.

1

~-~-+ +(02aes + e ~,~ -1=0.0458and

2 2 ~

L~ = 0.8 l(f~ - f,),) _ 1.51

Thus, the total strain is:

£axial q" ~b,max ~ ~s,max

e~. = es,m,x - e~,r~

~,u = 0.046- 0.00283 = 0.043

98


L,__c, > (1- r/a~ -(’)Oae~ = 0.365h O°atl

For h = 42 in.

L,t > 42× 0.365 = 15.33 in

Lsu = Lut - 2L,,,

= 15.33 - 2(1.51)= 12.31 in.

Provide Lsu = ~


lph = kph Tlaes hg

= lxO.14x42 = 5.88 in.


4.1.3 Solution Using Proposed Design Charts

1. Go to Figure 3.14 and Determine:

fp,des / fpy = 0.87

~ fp,des = 0.87 x 243 = 211.41 ksi


M des

Aphgf p,aes- 0.885

1235x12 = .~~ Ap = 0.885 x 42 x 211.41


99


1.79

As = 1235x12 =[2.35 in2J1.79x42xl.4x60

4.1.4 Solve Using Design Equations

1. Calculate Tides

2. Calculate ASp

3. Check if ep < ~x’ ~py

4. Calculate fp,aes

5. For d~ = Lpu/hg ~< 8

r/d~, = -0.2fll +0.3

=-0.2(0.8)+0.3 = 0.14

Ap _ ~s (0.5_qaes) = 0.0011Agp- lpu ~

ep = gpi + A~p = 0.0073

Check if

~p ~

Let a’ = 1

0.0073 < 0.0085 OK

~p,des ---- Ep % --- 207 ksi

M~o~ = 1.125~00.04~s,max0’0371Aphgfp,des

rPo.o~ = (--6.70des + 0.9)~(z70~+°°3)

= (--6.7 X 0.0196 + 0.9) X 6.43(2’v×°’°~96÷°’°3) = 0.897

100


M des= El. 125 x 0.046°’°3711 x 0.897 = 0.897

Aphg fp,~t~s

1235x12=> Ap - 0.897 x 42 x 207

6. For d~ = Lpuih< 8

Mdes -0.034= 0.882nro.o4e~,~x

aro.o4 = (22.230a~ + 1.7)~-(3"4640~+0’0216)

= (22.23 × 0.0196 + 1.7) x 6.43-(3’464×0’0196+0"0216) = 1.8

Mdes = E0.882 x 0.046-°’°341 x 1:8 = 1.76A~2~,a~h~ fp,a~

~As =1235x12 = 2.38 in2

1.76 x 42 x 1.4 x 60

The results of the four methods are compared in Table 4.2 below

101


Table 4.2: Results Compared Using the Four Methods

Method

PRESSS Procedure

Non-DimensionalProcedure

Charts

Design Equations

Ap (in2)

1.96

1.96

1.9

1.9

As (in2)

2.31

2.3

2.36

2.38

L,u (in)

14.5

12.5

12.5

12.5

Remarks

Bending strainand grout

effects are notincluded

Bending strainand grouteffects areincluded



Use 13 - ½ in. diameter PT strands (Ap,provided = 2 in2), along with 3 #8 bars

(As, provided = 2.4 in2).

102


4.2 Example # 2:

Reference: S.K. Ghosh and David A. Fanella, Seismic and Wind Design of Concrete

Buildings based on 2000 IBC, ASCE 7-98, and ACI 318-99 code, 2003

Seismic Data:

¯ Location: Atlanta, GA (zip code 30350)

¯ Seismic Design Category (SDC): B

¯ SiteClassC

¯ 3-Story School Building

¯ 5 equal bays (span length = 30 ft)

¯ hg = 18.5 in.,.bg = 54 in.

¯ Moment-Resisting Frame System (Hybrid Frame)





103



Parameter

f ’¢ (ksi)

f ’~ (ksi)

fpy (ksi)

f, (ksi)f,i (ksi)~ (ksi)f~y (ksi)

L~ (if)

~s,max

~,,s’

~y

Spy

Epi

M,~¢s (ft-kips)

O~os (%)

Value

4

7

243

270

175.5

29000

60

0.055

30

15

28500

0.85

0.043

1.4

1.25

0.00207

0.0085

0.00616

525

Comments









Tendon length c/c of columns


Ls/2Modulus of elasticity of Gr. 270 strand

131 = 0.85 - 0.05 (f ’e - 4)


(See section 4.2.2)

Tension over-strength factor (ACI T1.2-03)

Compression over-strength factor (ACI T1.2-03)

fsy/Es

fpy

Obtained using Force Based Design (FBD)

Calculated Drift

104





given below.

Step 1: Establish Material properties and Design Information

See Table 4.3

Step 2:. Obtain Design Loads (Mdes) and Drifts (Odes %)

See Table 4.3

Step Estimate Frame Beam Dimensions3."

hg <--=l" 27.67x12 -110.7 in. OK3 3

bg >0.3hg =0.3x18.5=5.6 in. OK


= o.5E O

18.5- 14.5 ksi= 0.5 x 28,500x 0.01x

15x12


Strength

Mp,des a 0.48 Mdes = 252 ft-kips (Optimum solution after several iterations)

Ms,des = Mdes - Mp,des = 273 ft-kips



from the Compression face. This assumption leads to:

105


(0.45hg)fpy

252 x 12 - 1.5 in20.45 x 18.5 x 243

Step 7: Estimate the Required Area of Deformed.Reinforcement (As)


Ms,des

(0.95-£)hg~s,a~f~y

273x12(0.95-0.055) x 18.5 x 1.4 x 60

- 2.34 in2

Step 8: Estimate the Neutral Axis Parameter at the Design Drift (Tides)

The location of the neutral axis can be found using an initial assumption of a/hg = 0.1 that

was made in Step 6. This assumption yields to an initial value Ofrldes = 0.1/~1. However

after several iterations of the steps shown here, the optimum solution converged on a rldes

value of 0.13.

a/hg~]des = -- = O. 13


= 1.4 x 60 = 84 ksi

Please note that the PRESSS procedure recommends a )~s,des value of 1.35, but allows for

a more rational determination of this parameter. In this report, a )~s,aes value of 1.4 is used


106



fs’.ae, =

= 1.25 x 60 = 75 ksi

Please note that the PRESSS procedure recommends a ~s’,des value of 1, but allows for a

more rational determination of this parameter. In this report, a £s’,des value of 1.25 is used


Step 11: Calculate the Stress and Elongation in the Pre-Stressing Tendon at Odes

= 0de hg(0.5-

= 0.01x 18.5x (0.5 -0.13) = 0.068 in

As 0.068 x 28Afp = --Ep = 500 = 10.84 ksi

lpu 15×12 ’

fpo= 243 - 10.84 = 232.16 ksi

if fpo >- fpi =0.65×270=175.5 ksi

thenfpo = f~,i = 175.5 ksi

f p.a~, = f ,o + Af p

= 175.5 + 10.84 = 186 ksi

elSe f p,des = f py

Mp,desAp = 0.5(1- f

252x120.5(1-0.85 x 0.13) x 18.5 x 186

And

= 2.04 in~

107


273x12

(1-0.055 - 0.5 x 0.85 x O.13)x 18.5 x 1.4x 60= 2.37 in2

Step 12: Calculate the Forces at Odes

= 2.04x 186 = 379.4 kips

F~,gm = As2,,amf~y

= 2.37×1.4×60 = 199 kips

F,,ae~ = As 2s,,ge, f ~,,

= 2.33× 1.25 x 60 = 177.8 kips

= 379.4 +199 - 177.8 = 400.6 kips

Step 13: Calculate the Locations of the Compression Force ~d the Neutral ~is

400.6= 2.18 in.

a des 0.85/’79 0.85 x 4 x 54

agm 2.18ages - - 0.059

2hg 2x18.5

_ ages 2.18 = 0.138 (Optimum value is 0.13, OK.)r/am - fl~g = 0.85 x 18.5

r/am = 0.13

ades ---- 0.5 X/~1 X ?]des

= 0.5X 0.85x 0.13 = 0.055



the compression force yield:

108


= 379.4 x 18.5(0.5-01055)/12 = :260.3 ft-kips

= 199 x 18.5(1-0.055-0.055)/12 = 273 ft-kips

Ms,,des= 177.8x18.5(0.055-0.055)/12 = 0

Therefore the total moment strength can be determined as follows:

M cap,~,ea,n = M p,aes + M.~,a~ + M s,,a,~s

= 260.3 + 273 + 0 = 533.3 R-kips

g cap,beam >-- Maw

533.3 > 525 ft-kips, OK (Optimum Solution)


= 2.37x1.25x 60 = 177.75 kips

= 2.37xl.25× 60 = 177.75 kips

Fpo:Apfpo

= 2,04x 175.5 = 358.5 kips

Feo = Fpo - F~o - Fso

= 358.5 - 177.75 - 177.75 = 3 kips

Feoa° = 0.85f~bg

3- = 0.016 in0.85x4x54

109


0.016= -- = 0.000442×18.5

a0

0.0160.85x18.5

- 0.001017

Mpo = F.ohg(0.5-ao)= 358.5 × 18.5(0.5 - 0.00044)/12 = 276.1 it-kips

Mso = F, ohg(1-~’-ao)

= 177.75 x 18.5(1 - 0.055 - 0.00044)/12 = 258.5 ft-kips

M ,,o = F,’oh(( - ao)

= 177.75 x 18.5(0.055-0.00044)/12 = 14.9 ft-kips


Mpo > Mso + Ms’o

276.1 > 258.8 + 14.9 = 273.7 ft-kips, OK


As = Odeshg(1-~-rlae,)

= 0.01x18.5(1- 0.055- 0.13) -- 0.15 in

0.15- -3.5 in0.043

Provide 3.5 in. of unbonded length

Please note that the PRESSS procedure recommends a es,max value of 0.04, but allows for


110


calculating [is,m,x is proposed (See sections 3.5 and 4.2.2 for details), which results in a

~s,max value of 0.043.


-- lxO.13x18.5 =2.4in.

O’~es(rla~shb) 0°es 0.01lph kph

[ic > 0.003 Provide proper confinement to prevent spalling





fatigue criterion for the mild reinforcing steel. See section 3.5 for details.

Let,

Y = fsy/f ’o = 15

Z = fsy I f py = 0.247

K = fsylf po = 0.34

Q = f’¢/fpi = 0.0228

Please note that the following calculations for [is,max, ~.s,des, and Ls’,des are not included

in the original PRESSS procedure.


T = 0.03 (hn)3/4 = 0.03 (39)3/4 = 0.47 sec.

111




Nc = 7 T -1/3 = 7 (0.47)-1/3 = 9 cycles

Enter the value of No for Nf in the following equation

~’ap = 0.08(2Nf )-o.4s = 0.0215

and

es,m.x = 2e~ = 0.043

fs,des=~,s,des, fsy = 84 ksi

fs’,~es=Xs’,aes, fsy = 75 ksi


See Table 4.3

~ Obtain Design Drift

See Table 4.3

Ste~ 3: Estimate Frame Beam Dimensions

~ = Lpu/hg = 9.73

~ Determine the Proportions of Moment Strength

co = 0.48 (Optimum solution calculated using the EXCEL Solver)

Ste~ 5: Estimate The Neutral Axis Parameter

~laes = 0.13 (Optimum solution calculated using the EXCEL Solver)

ades -- 0.5 ~]Tl~es = 0.055

Ste~ 6: Calculate the Strain in the Pre-Stressing tendon at 0aes

This step is a modified form of the PRESSS procedure to ensure that the post,tensioning

112


steel would not deform beyond yield.Ap _ 0aes (0.5 -r/a~) = 0.00038

A~p- lpu~

~p = epi + A~p = 0.0065

Check if

°°p~ ~py

Let a’ = 1

0.0065 < 0.0085 OK

fp,des = Ep Ep = 186 ksi

Let,

R- f~Y -0.32f p,des


Let,

Ap7 - - 0.19% (Optimum Solution)

bghg

2.04 ¯(b)0ptimum : = 58 in

0.0019x18.5

Check if

113


0.45 Q = 1.79%, OK (ACI 318-02 Provisions)

andqa~ - O.~5,B1 N + -~(A~,a=-~’,a=) =0.13

Mdes - 0"5(1-fllr/ae*) - 0.91Aphgfp,aes co

525x12 12.01 in21"0.93 x18.5 x186

and

Md~ = (1-~’--0"5fl:/des) = 1.73Ashg ~,s,aesf sy 1-co

As =525x12

-12.34 in:]1.73x18.5xl.4x60

Ap R/~, a~ (1- £ - O.5 flfla~ ) = O.85

As 0.5(1 _ fl, r/a~s 1( 1-- - 1)

Ste~ 8: Calculate the moment strength of the section at Odes

Mcap,beam = Mp,aes + M,,~, + M,,,ae~ > Mdes

The objective ~nction is equal to:

> 0.5(1-

0.5(1 - fl~ r/a~, ) _- 0

The EXCEL Solver is used to find the optimum solution by changing ~ldes, 0), and ~, Subjected to

114


the following constraint:

1) r >_ 2~s,,desK

2) O°p ~

3) 3’ -< 0.45 Q

4) (~des)calculated = (~des)assumed

(flOes)optimum = 0.13

(0))optimum ---- 0.48

(3’)optimum = 0.19 %

Ste~ 9." Evaluate Restoring Properties of the Beam (Constrained Equations)


Mp0 > Mso + Ms,0

22.s,,desK = 0.85


Let,

~.,. = ~s,n~x = 0.043

O:" = tan-~ = 0.228 rad.

115


and

3 2 + 2 )~sin(((9 ~, ~ o,~+~z’) -Odes (h~)=O.O059

L~ _ 0.81(Z~-~) =1

db (f~)l.5


¯ ~at~ = ~,~ax --~a,max

~.¢~ = 0.043 -- 0:0059 = 0.0371

1.,,,1>_ (1-- rldes --¢)Od~ =0.22

Forhg = 18.5 in.

Lut >_ 18.5x 0.22 = 4.1 in

Lsu = L~ - 2L~

=4.1- 2(1)=2.1 in.

Provide L,~ = ~


lph = kph lqdes hg

116


= 1×0.13×18.5 =2.4in.




fp,des / fpy = 0.77

~ fp,des = 0.77 x 243 = 187.11 ksi


525x12= 11.96 in2]~Ap = 0.93x18.5x187.11


Mdes= 1.71

Ashg’~s des L)

525x12 IAs = - 2.37 in21

1.71x18.5xl.4x 60

4.2.4 Solve Using Design Equations

1. Calculate Tides

r/d~ = --0.2fl~ +0.3= --0.2(0.88)+0.3 = 0.13.

2. Calculate Asp

117


Ap _ Odes

A~’p- 1~ ~ (0"5-r/a~) -- 0"00038

3. Check if ep < c~’ ~py

~p = epi + A~p = 0.0065

Check if

~p ~ a ~’py

Let g’ = 1

0.0065 < 0.0085 OK

4. Calculate fp,des

fp,des = Ep. I~p = 186 ksi

5. For d~ = L~uih -> 8

Mdes = 1.1 25~Oo.o4gs,r~x°’°371Aphgfp,aes

M~es 0.94)~(L90~ +0.0042)~00.04 -- = (-5.8 10an +Aphg fp,cles

(/90.04 ~ (-5.81 × 0.01 + 0.94) x 9.73(t"9×°"°1+°"°°42) = 0.93

M,a~s _ [1.125 x 0.043°’°37’ I x 0.93 = 0.93~ Aphg fp,aes

525x1211,97 in2]~Ap - 0.93x18.5x186 =

6. For d~ = Lpu/h > 8

= O. 8 82~70.04~.s,max_O.034

118


nro.o4 = Md~ = (13.650ae, + 1.65)~-(L970"~÷°’°°13)

~ro.o4 = (13.65 x 0.01 + 1.65) x 9.73"0’97×°’°1÷°’°°a3) = 1.7

= I0.882 x 0.043-°’°341 x 1.7 = 1.67

~ As_ 525x12 _ 12.42 in~l1.67x18.5xl.4x60



Method

PRESSS Procedure


Ap (inz)

2.04

2.01

As (inz)

2.37

2.34

(in.)

3.5

2.5

Charts

Design Equations

1.96

1.97

2.37

2.42

2.5

2.5

Remarks






Use 14 - ½ in. diameter PT strands (Ap,provided = 2.14 in2), along with 4 #8 bars

(As,provided = 3.12 in2).

119


4.3 Example # 3:


Buildings based on 2000 IBC, ASCE 7-98, and ACI 318-99 code, 2003.

Seismic Data:

. Location: San Francisco, CA (zip code 94105)

¯ Seismic Design Category (SDC): D

¯ Site Class D

¯ 12-Story School Buildings

¯ 7 equal bays (span length = 26 r)

¯ hg = 26 in., bg = 28 in.






120



Parameter Value Comments

f ’c (ksi) 6 Compressive strength of concrete

f ’g (ksi) 8 Compressive strength of grout

fpy (ksi) 243 Yield strength of Gr. 270 strand

fpu (ksi) 270 Minimum strength of Gr. 270 strand

fpi (ksi) 175.5 Initial stress .in strand

Es (ksi) 29000 Modulus of elasticity of mild steel

fsy (ksi) 60 Yield strength of Gr. 60 steel bars

~ 0.055 Cover factor (See Figure 3.2)

Tendon length c/c of columnsL, (~) 26


Lpu ( ft) 13 L~/2

Ep (ksi) 28500 Modulus of elasticity of Gr. 270 strand

~1 0.75 ~1 = 0.85 - 0.05 (f ’c - 4)

Maximum sustainable strain in the mild reinforcing steelI~s,max

0.05(See section 4.3.2)

L, 1.4 Tension over-strength factor (ACI T1.2-03)

L~, 1.25 Compression over-strength factor (ACI T1.2-03)

ey 0.00207 fsy iEs

Spy 0.0085 fpy/Ep

~pi 0.00616 fpi/Ep

Maes (ft-kips) 737 Obtained using Force Based Design (FBD)

Odes (%) 1.65 Calculated Drift

121





given below.


See Table 4.5

Step 2:. Obtain Design Loads (Mdes) and Drifts (Odes %)

See Table 4.5


hg <ln= 23.67X12_95 in. OK3 3bg > 0.3hg = 0.3 x 26 = 7.8 in. OK

Step 4: Establisti Constants

h

= 0.5x 28,500x 0.0165 x26 - 39.19 ksi13x12


Strength

Mp,des = 0.51 Mdes = 375.9 R-kips (Optimum solution alter several iterations).

Ms,des = Mdes -- Mp,des = 361.1 It-kips

122





A _ Mp,des

P (0.45hg)fpy

375.0×12 = 1.59 in20.45× 26x 243



Ms,des

(0.95361.1x12 = 2.22 in2

(0.95-0.055) x 26 x 1.4 x 60

Step 8: Estimate the Neutral Axis Parameter at the Design Drift (rlaes)

The location of the neutral axis can be found using an initial assumption of a/h = 0.1 that

was made in Step 6. This assumption yields to an initial value of ~ldes = 0.1/~1. However

after several iterations of the steps shown here, the optimum solution converged on a

value of 0.15.

a/hg _ 0.15


= 1.4 × 60 = 84 ksi

123


Please note that the PRESSS procedure recommends a Ls,des value of 1.35, but allows for

a more rational determination of this parameter. In this report, a )~s,de~ value of 1.4 is used



= 1.25 x 60 = 75 ksi

Please note that the PRESSS procedure recommends a Ls’,aes value of 1, but allows for a

more rational determination of this parameter. In this report, a )~s’,des value of 1.25 is used


Step 11: Calculate the Stress and Elongation in the Pre-Stressing Tendon at Odes

= 0.0165 x 26x (0.5-0.15) = 0.15 in

Afp =--Ep = 0.15 x28,500= 27.4ksilp, 13x12

fpo = f,y-/,f,= 243 - 27.4 = 215.6 ksi

iffpo >- fpi -- 0.65x270 =175.5 ksi

then fpo = fpi = 175,5 ksi

fp,a~ = fpo + Af,

= 175.5 + 27.4 = 202.9 ksi

else f p,aes = f ~y

124


375.9x120.5(1 - 0.75 x 0.15) x 26 x 202.9

= 1.93 in2

And

As = Ms’aes

361.1x12(1-0.055-0.5 x 0.75 x 0.15) x 26 x 1.4 x 60

= 2.23 in2


= 1.93x 202.9 = 391.6 kips

= 2.23x 1.4x 60 = 187.32 kips

Fs,des = A,2,,,a=f~y

= 2.23x 1.25 x 60 = 167.25 kips

= 391.6 +187.32 - 167.25 = 412 kips

Steo 13: Calculate the Locations of the Compression Force and the Neutral Axis

Fc,d~ 412= = 2.89 in.0.85f~’bg 0.85 x 6 x 28

ades _ 2.89ad~~ = -- --.-- 0.056

2hg 2x26

_ a~= = 2.89 = 0.148 (Optimum value is 0.15, OK.)rl,~es - flah----7 0.75 x 26

125


r/~ =0.15

ade~ = 0.5x fll × rl,~s

= 0.5 x 0.75x 0.15 = 0.056

Step 14: Calculate the Moment Strength of the section at 0d,s



= 391.6 × 26(0.5- 0.056)/12 = 376.7 ft-kips

Ms,ae, = F~,a~hg (1-~" - aden)

= 187.32 x 25(1-0.055-0.055)/12 = 350.8 fi-~ps

M ¢,d~~ = F¢,a~h , ( ~ - a a~ )

= 157.25 x 26(0.055-0.056)/12 = - 0.36 ~-~ps

Therefore the total moment stren~h can be dete~ed ~ follows:

M~,,~ = M~,ae~ + M,,a~ + M~,,a~

= 376.7 + 360.8 - 0.36 = 737.1 r-kips

M cap,bea,n >- M des

537.1 = 737 r-kips, OK (Optimum Solution)


= 2.23× 1.25 × 60 = 167.25 kips

F ,o == 2.23 × 1.25 × 60 = 167.25 kips

126


Fpo = A~,f ~,o

= 1.93x175.5 = 338.7 kips

Fco=Fpo-Fso-Fs’o

= 338.7-167.25-167.25 = 4.22 kips

Fcoa° = 0.85fbg

4.22- = 0.03 in0.85x6x28

0.03- = 0.000572x26

ao

0.030.75 x 26

- 0.0015

Mpo = Fpohg (0.5 - ao)

= 338.7 x 26(0.5-0.00057)/12 = 366.5 ft-kips

M,o = Fsohg (1 - ~" - ~zo)

= 167.25 x 26(1-0.055-0.00057)/12 = 342.2 ft-kips

Ms,o = F,’ohg(~-ao)

= 167.25 × 26(0.055-0.00057)/12 = 19.72 fl-kips


Mpo > Mso + Ms,o

366.5 > 342.2 + 19.72 = 362 ft-kips, OK

127



As = Odeshg(1--~--r]des)

= 0.0165 x 26(1-0.055-0.15) = 0.34 in

AS

0.34- - 6.8 in

0.05Provide 7 in. ofunbonded length

Please note that the PRESSS procedure recommends a ~s,max value of 0.04, but allows for


calculating ~s,max is proposed (See sections 3.5 and 4.3.2 for details), which results in a

. es,max value of 0.05.


lob = kph rides hg

= lx0.15×26 =3.9 in.

Odes (rlaeshb ) -- Odes -- 0.0165ooc --

lph kph

~o > 0.003 Provide proper confinement to prevent spalling.






Let,

128


Y = fsy/f’e = 10

Z = fsy / f py = 0.247

K = fsy/f po = 0.34

O = f’c/fpi = 0.034

Please note that the following calculations for es,max are not included in the original

PRESSS procedure.


T = 0.03 (h,03/4 = 0.03 (147.5)3/4 = 1.27 sec.



Nc= 7 T -1/3 = 7 (1.27)-a/3 = 6.46 cycles


O°ap = 0.08(2Nf)-°45 = 0.025

and

es,ma~, = 2Cap = 0.05


fs’,~es=Xs’,Oes fs~, = 75 ksi

Step 1: Establish Material Properties and Design Information

See Table 4.5

Ste~ 2: Obtain Design Drift

See Table 4.5

Ste~ 3." Estimate Frame Beam Dimensions

129


= / hg = 5.57

Ste~ 4." Determine the Proportions of Moment Strength

~0 = 0.51 (Optimum solution calculated using the EXCEL Solver)

Ste~ 5." Estimate The Neutral Axis Parameter

floes = 0.15 (Optimum solution calculated using the EXCEL Solver)

~des ---- 0.5 ~l.Tl~es --- 0.055

Steo 6: Calculate the Strain in the Pre-Stressing tendon at Odes


steel would not deform beyond yield.

Ap _ Odes (0.5-r/de~) = 0.00105A~:p- lpu ~

ep = O~pi + Aep = 0.0072

Check if

Let a’ =1

0.0072 < 0.0085 OK

fp,des ---- Ep ~p = 205 ksi

Let,

130


R - f~y - 0.29f p,des

Steo 7: Calculate the Relative Locations of the Compression Force and the Neutral Axis

Let,

Ap~" - - 0.26% (Optimum Solution)

bghg

1.91- 28.3 in(bg)optimum 0.0026 X 26

Check if

0.45 Q = 2.05 %, OK (ACI 318-02 Provisions)

= 0.055

and

- ~’ IN+ Y~d~ 0.85Pl/ --(~s’de’r -~,’,~e~) =0.15

Md~ ~- 0"5(1-fllqa~)=0.87Aphgfp,aes (0

737x12 = [1.91 in2[~Ap = 0.87 x 26 x 205

and

131


M,~o, = (1-4" - 0.5fl~r/,~e,) = 1.82A~hg2.~.d~, f~y 1-co

A~ = 737x12 = L2.22 in2l1.82x 26xl.4x60

Ap R2s,de~ (1--~--O.5 fl,~Tde,) = O.85

As 0.5(1 - flf/ae,)(~l - 1)

Ste~ 8: Calculate the moment strength of the section at Odes

M eap,beam = M p,des + M ,,de~ + M ~,,d~ >- M d~

===> 0.5(1-flf/a~) + _~EA~,d~(l_~_O.5]3~rld~)+ )L,,.d~(~_O.Sfl~r/d,,)~ > 0.5(1-flf/a~s)co

" The objective function is equal to:

The EXCEL Solver is used to find the optimum solution by changing qdes, ¢9, and 3’ Subjected to


1) r > 22s,,desK

2) ep < ~’~’py

3) 3,< 0.45 Q

4) (Tides)calculated = (rides)assumed

(~ldes)0ptimum = 0.15

(O))optimum -- 0.51

(It)optimum = 0.26 %

132


Ste~ 9: Evaluate the Restoring Properties of the Beam (Constrained Equations)


Mp0 > Ms0 + Ms,0

r > 2)t.s,,,~sK = 0.85


Let,

~.,. = e~,max = 0.05

2 23(0 des+~" ~,a) sin(+~") -0~ db =0.005

O°a ,a

L,,,, _ 0.8 l(f~,, - f~y) = 0.86db (f,~)1.5


133


’~s,~x - Zb,n~,x

0.05 -- 0.005 = 0.045

Lu---t-’ > (1-r/din-~)Oaes =0.29

For hg = 26 in.

L,, > 26 x 0.29 = 7.54 in

Ls, = L,,t - 2L~

= 7.54 - 2(0.86) = 5.82 in.

Provide L~ = ~


lph = kph qdes hg

= lx0.15×26 = 3.9 in.

O’~es (rldeshb ) -- Odes -- 0.0165,zq’c --

lph kph

0.003 Provide proper confinement to prevent spalling.



fp,~es / fpy = 0.84

~ fp,de~ = 0.84 x 243 = 204.12 ksi


134


737x12 = ~=~Ap 0.88x26x204.12


As =737x12

=[2.24 in2l1.Sx. 26xl.4x60

4.3.4 Solution Using Proposed Design Equations

1. Calculate

2. Calculate ASp

3. Check if

4. Calculate fp,des

r/aN = -0.2fll +0.3=-0.2(0.75)+0.3 = 0.15

A° - Odes (0.5- r/aes) = 0.001!Xgp- lpu ~

ep = ~pi + Aep = 0.0072

Check if

~p ~ I~ ~py

Let a’ = 1

0.0072 < 0.0085 OK

135


5. For @ = L~u/h < 8

Males _ 1. ] 25~0o.ogg~,max°’°37~

Aphgfp,~es

(]%.04 = (--6’70d~ + 0"9)¢= (--6.7 X 0.0i65 + 0.9) x 5.57(2’7×0"0165+°"°3) = 0.897

Md= _ [1.125 x 0.05°’°371~ x 0.897 = 0.9Aphgfp,d~

737 x12 = ~~Ap 0.9x 26x 204.12

6. For qb = I_~uih _< 8

Ma~ = 0.8 82g0.0.04 ~s,max_0.o34As2s,de, h g f v.de,

nro.o,, = (22.230a~ + 1,7)lfi-(3’4640°~s+0’0216)

= (22.23 X 0.0165 + 1.7) X 5.57-{3"464×°’°16s+°’°216) = 1.8

Mdes = I0.882 x 0.05-°’°3~ 1 x 1.8 = t.76

737x12in2l~ A~ -- --

1.76x 26x 1.4x 60


136


Table 4.6 Results Compared Using the Four Methods

Method

PRESSS Procedure


Charts

Design Equations

Ap (in2)

1.93

1.91

1.89

1.85

As (inz)

2.23

2.22

2.24

2.3

L~u (in.)

7

6

Remarks






Use 13 - 1/2 in. diameter PT strands (Ap,provided = 2 in2), along with 3 #8 bars

(As,provlded = 2;4 in2).

137


4.4 Example # 4:


Buildings based on 2000 IBC, ASCE 7-98, and ACI 318-99 code~ 2003.

Seismic Data:

¯ Location: Berkeley, CA (zip code 94705)

¯ Seismic Design Category (SDC): E

¯ Site Class D



¯ hg = 32 in., bg = 28 in.

¯

This example is solved

dimensional procedures,

Moment-Resisting Frame System (Hybrid Frame)

using the original PRESSS calculation, the modified non-

the developed non-dimensional charts, and the proposed



138


Table 4.7 Material Properties and Design Information for Example 4

Parameter

f ’~ (ksi)

f ’g (ksi)

fpy (ksi)

fpu (ksi)

f,i (ksi)Es (ksi)

fsy (ksi)

Value

6

9

243

270

175.5

29000

60

0.055

Ls(fl) 26

Lp. (ft) 13

Ep (ksi) 28500

1~1 0.75

Ss,max 0.05

)~s 1.4

Xs, 1.25

Sy 0.00207

Spy 0.0085

Spi 0.00616

Mdes (ft-kips) 1170

Odes (%) 1.9.

Comments









Tendon length c/c of colunms


Ls/2

Modulus of elasticity of Gr. 270 strand

131 = 0.85 - 0.05 (f ’c - 4)


(See section 4.4.2)

Tension over-strength factor (ACI T 1.2-03)

Compression over-strength factor (ACI T1.2-03)

fsy/Es

Obtained using Force Based Design (FBD)

Calculated Drift

139



The following is an iterative process that follows steps 1 through 15 based-on initial


given below.


See Table 4.7


See Table 4.7


hg <l,,= 23.67x12_94.5 in. OK3 3

bg > 0.3hg = 0.3 x 32 = 9.6 in. OK


hgA fsJ~ = 0"5EpOaes lp-~

= 0.5x28,500x 0.019x32 - 55.54 ksi

13x12


Strength

Mp,des = 0.52 Mdes = 608.4 ft-kips (Optimum solution after several iterations)

Ms,des = Mdes - Mp,des = 561.6 fl-kips

140





Mp,des

(0.45hg)f,y

608.4 × 12= - 2.09 in2

0.45 × 32 × 243

Step 7: Estimate the Required Area of Deformed Reinforcement (A~)


(0.95 - ~)h g~s,aesf sy

573.3x12(0.95 - 0.055) x 32 x 1.4 x 60

= 2.8 in2

Step 8: Estimate the Neutral Axis Parameter at the Design Drift

The location of the neutral axis can be found using an initial assumption of a!hg = 0.1 that

was made in Step 6. This assumption yields to an initial value of rlde~ = 0.1/[~1. However

after several iterations of the steps shown here, the optimum solution converged on a ]]des

value o f 0.15.

r/a~ - = 0.15


= 1.4 x 60 = 84 ksi

141


Please note that the PRESSS procedure recommends a Ls,des value of 1.35, but allows for

a more rational determination of this parameter. In this report, a )~s,des value of 1.4 is used



= 1.25 x 60 = 75 ksi

Please note that the PRESSS procedure recommends a ~s’,des value of 1, but allows for a

more rational determination of this parameter. In this report, a Ls’,aes value of 1.25 is used


Step 11: Calculate the Stress and Elongation in the Pre-Stressing Tendonat 0d~s

A, = 0~ho(0.5- ~)

= 0.019x 32 x (0.5 - 0.15) = 0.213 in

0.213Afp = --Ep = ~x 28,500 = 38.9 ksi

lp, 13x12

fpo = fpy-Afp

= 243 - 38.9 = 204.1 ksi

iffpo > fpi : 0.65 x 270 : 175.5 ksi

then fpo = f~i = 175.5 ksi

f p,ae, = f po + Af p

= 175.5 + 38.9 = 214.4 ksi

else f p,d~ = f py

142


0.5(1-- flfld~ )h g f p,d~,

608.4 x 120.5(1- 0.75 x 0.15) x 32 x 214.4

= 2.4 in2

And

Ms,desAs-

(1- ¢ - 0 . S fl~ rl des ) h g 2 ~ ,des f ~y

561.6x12(1 - 0.055 - 0.5 x 0.75 x O. 15) x 32 x 1.4 x 60

= 2.82 in2


= 2.4× 214.4 = 514.56 kips

Fs,a~ = As2,.desf , y

= 2.82x 1.4x 60 = 237.88 kips

= 2.82x 1.25x 60 = 211.5 kips

Fc,aes = gp,aes + F,,a~ -

= 514.56 +237.88 - 211.5 = 540.94~ps


Fc,d~ _ 540.94 =3.79in.0.85 f~’bg 0.85x6x28

ades _ 3.79ad~, = -- 0.059

2hg 2x32

= ade..~_~ = 3.79 = 0.157 (Optimum value is 0.15, OK.)rld~ fllhg 0.75x32

143


r/an = 0.15

~z~ =0.5x/~1 xr/~s

= 0.5x 0.75x 0.15 = 0.056

Step 14: Calculate the Moment Strength of the section at 0aes



Mv,ae, == 514.56 × 32(0.5 - 0.056)/12 = 609.2 ft-kips

= 237.88 x 32(1- 0.055 - 0.056)/12 = 563.9 ~-kips

= 237.88 x 32(0.055 - 0.056)/12 = - 0.64 ~-kips

Therefore the total moment strength can be dete~ined ~ follows

M cap,O~~ = M p,a~~ + M ~,a~ + M ~’,a~

= 609.2 + 563.9 - 0.64 = 1172.5 fi-kips

M cap,O~ ~ M aes

1172.5 ~ 1170 ~-kips, OK (~timum Solution)


F,o =

= 2.82×1.25×60 = 210 kips

Fs,o =

= 2.82 × 1.25 × 60 = 210 kips

144


F~,o = A~,f~,o

= 2.4x175.5 = 421 kips

Fco = F~,o - Fso - F~’o

=421-210-210= 1 kips

Fco

o.85f&1

- = 0.007 in0.85 x 6 x 28

ao

2hg

0.007= -- = 0.00011

2x32

ao

0.0070.75 x 32

- 0.00029

M o=F, oh (0.5- o)= 421x 32(0.5 - 0.00011)/12 = 553.5 ft-kips

M,o = F, ohg(1-£-ao)

= 210x 32(1 - 0.055 - 0.00011)/12 = 525.3 R-kips

M ,,o = F~’oh(~ - ao)

= 210x 32(0.055 - 0.00011)/12 = 26.9 ft-kips


Mvo- > Mso + Ms,o

553.5 > 525.2 + 26.9 = 552.2 it-kips, OK

145



As : Odeshg (1-~" - ~Tdes)

= 0.019×32(1-0.055-0.15) = 0.48 in

AS

0.48-9.7 in

0.05Provide 10 in. of unbonded length

Please note that the PRESSS procedure recommends a [3s,max value of 0.04, but allows for


calculating gs,m,x is proposed (See sections 3.5 and 4.4.2 for details), which results in a

gs,max value of 0.05.


lph = kph rides hg

= lxO.15x32 = 4.8 in.

~c- Od~(r/a=h~) - Oaes -0.019lob

gc > 0.003 Provide proper confinement to prevent spalling.




within Microsoft Excel. The following procedures also utifize the proposed low-cycle


Let,

146


Y = fsy/f’c = 10

Z = fsy I f py = 0.247

K -- fsylf po = 0.34

0 " f’c/fpi = 0.034

Please note that the following calculations for ~s,m~x is not included in the original

PRESSS procedure.


T = 0.03 (hn)3/4 = 0.03 (147.5)3/4 = 1.27 see.



N~ = 7 T -1/3 = 7 (1.27)-a/3 = 6.46 cycles

Enter the value of N~ for Nf in the following equation

~,,p = 0.08(2Ny )-0.45 = 0.025

and

8s,max = 213ap ---- 0.045

fs,des=Xs,des fsy = 84 ksi

fs’,des=~Ls’,des fsy "- 75 ksi


See Table 4.7.

Ste~ 2." Obtain Design Drift

See Table 4.7.

147


Ste~ 3." Estimate Frame Beam Dimensions

~ = Lpu/hg= 4.88


m = 0.52 (Optimum solution calculated using the EXCEL Solver)


rides = 0.i 5 (Optimum solution calculated using the EXCEL Solver)

ades "-0.5 ~lqg~s = 0.055

Steo 6." Calculate the Strain in the Pre-Stressing tendon at 0aesAp _ Ode‘ (0.5-qde,) = 0.001376

A~,- lp~

o~p ~ O°pi q- A~p = 0.00753

Check if

Let a’ = 1

0.00753 < 0.0085 OK

fp,des = Ep Vp = 215 ksi

f~’ = 0.28f p,des

148


Calculate the Relative Locations of the Compression Force and the Neutral Axis

Let,

Apy - - 0.25% (Optimum Solution)

bghg

2.4(bg)°ptimum

0.0025 x 28 - 34.3 in

Check if


Ey

- N +a aes 1.7

and

= 0.O55

= 0..15

Mdes _ 0.5(1 --/~l/Taes) = 0.85

Ap h g f p ,des

1170x12 = ~=:>Ap

0.85x32x215

and

Mdes = (1--~’--0"5fl~r/des) = 1.86Ashg~,s,ae, f~y 1-09

As =1170x12 = 2.81 inz

1.86x32x1.4x60

- 149


Ap R)~sdes(1--~’--0.Sfllr/de~) 0.85

A~ 0.5(1_ fl, r/~s)(l_ 1)

Ste~ 8." Calculate the momem strength of the section at Odes

M cap,beam = M p,aes + M ~,ae~ + M s,,des > M a~

~ 0.5(1-flflaes) +0.5(1- flf/~)

.’. The objective function is equal to:

0-5(1 - flf/a~,) + ~_E,~.~.a~ (1 _ ~. _ 0.5fl~r/a~)+ A¢ ae~ (~. _ 0.5,Bf/a~)~-’ 0.5(1- fl, r/a~,)co =0

The EXCEL Solver is used to find the optimum solution by changing floes, m, and ~, Subjected to


1) v > 2~<desK

3) r <- 0.45 Q

4) (ndes)c. cu ate = (qde )a su ed

0Ides)optimum - 0.15

(O)optimum - 0.52

(r)opt~mum = 0.25 %

~ Evaluate the Restoring Properties of the Beam (Constrained Equations)


Mpo >- Ms0 + Ms,0

150


r > 25Ls,,a~sK = 0.85


Let,

°~a,a = ~s,max = 0.05

~i.l 1 + 2 (02ae~ 2 - 1 = 0.04982= +~ .,~)Tcos +a" +(02.a~

and

3(02des + ¢2aa)-~ sin( + a") - Odesdb = 0.004

L,,~ 0.81(f~,-f,y) 0.72do (f~)"s


O~axial "1-O~b,max <~ OC’s,max

~at! = ’F’s,max -- ~b,max

z~o = 0.05 - 0.004 = 0.046

151


Lu,> (1-r/aes -£)O~e, =0.33

For hg = 32 in.

L.,t > 32x0.33 =10.56 in

Lsu = Lut - 2L~,

= 10.56-2(0.72) = 9.06 in.

Provide L~u = @


lph = kph ’q~es hg

= lx0.15x32 = 4.8 in.

~c > 0.003 Provide proper confinement to prevent spatling



fp,des / fpy = 0.89

~ fp,aes = 0.89 x 243 = 216.27 ksi


= 0.86

1170×12~Ap 0.86x32x216.27 2.35 in2]


152


Mdes = 1.82AshgA~,ae~ f~y

As = 1170x12 -[2.86 in211.82x32x1.4x60


1. Calculate rlaes

2. Calculate A~p

.3. Check ifsp < c~’ ~py

4. Calculate fp,des

5. For.~ = Lpuih < 8

r/des = --0.2fll +0.3

= --0.2(0.75)+0.3 = 0.15

mp _ OdesAgp - lpu

~b (0"5--r/d=) = 0.001 38

0.00753 < 0.0085 OK

fp,des = Ep Sp = 215 ksi

Md~ = 1.125~o.o48s,max0"°371Aphgfp,aes

~Oo.o4 = (--6.70d~, + 0.9)~b(z7°.~+°.°3)

= (-6.7 x 0.019 + 0.9) x 4.88(2’v×°’°19+°’°3~ = 0.878

153


M desAphgfp,des

l. 125 x 0.050’0371J × 0.878 = 0.88

1170xi2 = [2.32in2]~ Ap = 0.88x32x215

Mdes= O. 8 821~70.04g’s,max"0’034

~J’0.04 = (22.230d~ + 1.7)¢"(3’46406+°"°216)

= (22.23 x 0.019 + 1.7) x 4.88-(3’464x0’019+0"0216) = 1.85

M des= I0.882X 0.05-0"0341 X 1.85 = 1.81

A~d’~,~hg fp,ae~

=:> A~ = 1170x12 =!2.88 in~]1.81x32xl.4x60


154


Table 4.8 Results Compared Using the Four Methods

Method

PRESSS Procedure


Ap (inz)

2.4

2.4

As (inZ)

2.82

2.81

Lsu (in.)

10

9.5

Charts

Design Equations

2.35

2.32

2.86

2.88

9.5

9.5

Remarks




Bending strainand grouteffects areincludedBending strainand grouteffects areincluded

Use 16 - ½ in. diameter PT strands (Ap,provided = 2.45 in2), along with 4 #8 bars


155


4.5 Example # 5:


Buildings based on 2000 IBC, ASCE 7-98, and ACI 318-99 code, 2003.

Seismic Data:

¯ Location: New York, NY (zip code 10013)

¯ Seismic Design Category (SDC): C

¯ Site ClassD



¯ hg = 22 in., bg = 22 in.






156



Parameter Value Comments

f ’c (ksi) 4 Compressive strength of concrete

f ’g (ksi) 6 Compressive strength of grout

fpy (ksi) 243 Yield strength of Gr. 270 strand

fpu (ksi) 270 Minimum strength of Gr. 270 strand

fpi (ksi) 175.5 Initial stress in strand

Es (ksi) 29000 Modulus of elasticity of mild steel

fsy (ksi) 60 Yield strength of Gr. 60 steel bars

~ 0.055 Cover factor (See Figure 3.2)

Tendon length c/c of columnsLs (it) 26


Lpu (It) 13 Ls/2

F_~ (ksi) 28500 Modulus of elasticity of Gr. 270 strand

131 0.85 1~1 = 0.85 - 0.05 (f ’c - 4)

Maximum sustainable strain in the mild reinforcing steelt;s,max 0.05

(See section 4.5.2)

Tension over-strength factorX~ 1.4

(ACI T02-03)

Compression over-strength factor1.25

(AC[ T02-03)

t;y 0.00207 fsy iEs

Spy 0.0085 fpy/F__,p

Spi 0.00616 fpi/Ep

M~es (it-kips) 288 Obtained using Force Based Design (FBD)

Odes (%) 1.8 Calculated Drift

157





given below.

-Step 1: Establish Material properties and Design Information

See Table 4.9


See Table 4.9

3: Estimate Frame Beam DimensionsStep

hg < l, _ 23.67×12- 3 ~: = 94.5 in.

bg > 0.3hg = 0.3 x 22 = 6.6 in.


hgA fp<~ = 0.SEp(Ta~ l p--~

= 0.5 x 28,500× 0.018x

OK

OK

22 - 36.17 ksi13x12


Strength

Mp,des ~ 0.5 Maes = 144 ft-kips (Optimum solution after several iterations)

Ms,des = Mdes -- Mp,des = 144 ft-kips

158





(0.45hg)fpy

144x12 = 0.72 in2= 0.45 x 22 x 243



Ms,dos(0.95

144x12 = 1.04 in2- (0.95 - 0.055) x 22 x 1.4 x 60

Step 8: Estimate the Neutral Axis Parameter at the Design Drift (Tides)

The location of the neutral axis can be found using an initial assumption of a!lag = 0.1 that

was made in Step 6. This assumption yields to an initial value of rides = 0.1/131. However

after several iterations of the steps shown here, the optimum solution converged on a Tides

value of 0.13.

rla~ = a/hg = 0.13


= 1.4 x 60 = 84 ksi

159


Please note that the PRESSS procedure recommends a Ls,aes value of 1.35, but allows for

a.more rational determination of this parameter. In this report, a ~.s,des value of 1.4 is used



= 1.25 × 60 = 75 ksi

Please note that the PRESSS procedure recommends a ~,s’,des value of 1, but allows for a

more rational determination of this parameter. In thisreport, a ~,s’,des value of 1.25 is used


Step 11: Calculate the Stress and Elongation in the Pre-Stressing Tendon at

zx, = O hg(O.5-, des)

= 0.018 x 22x (0.5- 0.13) = 0.146 in

AfpAp 0.146

=--Ep .- x28,500 = 26.8 ksi13x12

f,o== 243 - 26.8 = 216.2 ksi

iffpo > foi = 0.65x270 =175.5 ksi

thenfpo = f~; = 175.5 ksi

= 175.5 + 26.8 = 202 ksi

else f ~,,a~ = for

160


Mp,des

0.5(1- fl, rlaes )h g f p,ae,

144×120.5(1 - 0.85 x O. 13) x 22 x 202

= 0.89 in2

144x12(1 - 0.055 - 0.5 x 0.85 x 0.13) x 22 x 1.4 x 60

- 1.04 in2


Fp,des = Apf p,des

= 0.89x 202 = 180 kips

= 1.04× 1.4× 60 = 87.4 kips

Fs,aes = As2s,,aesfsy

= 1.04x 1.25 x 60 = 78 kips

Fc,aes = Fp,aes + Fs,aes -F,,,,~,s

= 180 + 87.4 - 78 = 189.4 kips


Fc,~ _ 189.4 = 2.53 in.aae~

0.85fc’bg 0.85 x 4 x 22

_ aae~ _ 2.53 = 0.135 (Optimum value is 0.13, OK.)r/a~ -/3,~-g 0.85 x 22

161


0.13

0.5x/31 ×

0.5x 0.75x0.15 = 0.055


Taking moment of both the PT and deformed reinforcement forces about the centroid of

the compression force yield:

= 180x22(0.5-0.055)/12 = 146.85 ft-kips

Ms,,~ = F~,,~hg(1- ~ - a,~es)

= 87.4× 22(1-0.055-0.055)/12 = 142.6 ft-kips

M s ,,aes = Fs,,aesh g ( ~ - Ot ctes)

= 78× 22(0.055- 0.055)/12 -- 0

Therefore the total moment strength can be determined as follows:

Mcap,bearn = Mp,des + Ms,d~, + Ms’,des

= 146.85 + 142.6 = 289.5 fl-kips

M cap,bea~n >- Mac,

289.5 > 288 ft-kips, OK (Optimum Solution)

Step 15: Evaluate the Restoring P~operties of the Beam

= 1.04xl.25x 60 = 78 kips

= 1.04 × 1.25 × 60 = 78 kips

162


Fpo = Apf po

= 0.89x 175.5 = 156 kips

Fco =Fpo-Fso-Fs’o

= 156 -78-78 =Okips

Fco

o.s5f&0

=Oin0.85x4x22

aoa° 2hg

02x 22

ao

00.85×22

Mpo = Fpohg (0.5 --O~0)

= 156 × 22(0.5 - 0)/12 = 143 it-kips

M,o = F, ohg(1-~-ao)

= 78 x 22(1 - 0.055 - 0)/12 = 135 ft-kips

M,,o= F,’ohg(~-cto)

= 78x 22(0.055-0)/12 = 7.9 ft-kips


Mpo > Mso + Ms,o

143 > 135 + 7.7,9 = 142.9 R-kips, OK

163



A., = O,a~hg(1-~-r/,~.~)

= O.O18x 22(1-0.055-7 0.13) = 0.32 in

0.32- - 6.4 in Provide 6.5 in.

0.05

Please note that the PRESSS procedure recommends a gs,max value of 0.04, but allows for


calculating ~s,n~x is proposed (See sections 3.5 and 4.5.2 for details), which results in a

13s,max value of 0.05.


= lx0.13x22 =2.86 in.

~c > 0.003 Provide proper confinement to prevent spalling.






Let,

164


Y = fsy/f’c = 15

Z = fsy/f py= 0.247

K = fsy/f po = 0.34

Q = f "e/f~,~ = 0.023

Please note that the following calculations for es,m~x is not included in the original

PRESSS procedure.


T = 0.03 (hn)3/4 = 0.03 (147.5)3/4 = 1.27 sec.

Where h, is the overall height of the building


Nc = 7 T -1/3 = 7 (1.27) -l/3 = 6.46 cycles


~o~ = 0.08(2~Vs)-°45 = O.025

and

gs,max = 2ga = 0.05


fs’,des’-~,s’,des fsy = 75 ksi


See Table 4.9.

Ste~ 2." Obtain Design Drit~

See Table 4.9.

165


Ste~ 3: Estimate Frame Beam Dimensions

g0 = Lpu / hg = 7.09


~0 = 0.5 (Optimum solution calculated using the EXCEL Solver)


rl~es = 0.13 (Optimum solution calculated using the EXCEL Solver)

~des ----0.5 l~ll"]Ses = 0.055

Ste~ 6." Calculate the Strain in the Pre-Stressing tendon at Odes


steel would not deform beyond yield

Ap _ 0d~ (0.5-r/~s) = 0.00094Asp- lpu’ ~

6"p = 6"pi + ASp = 0.0071

Check if

Let a’ =1

0.0071 < 0.0085 OK

fp,des = Ep I~p = 202 ksi

Let,

166


R- f~y =0.3f p,des

Ste~ 7.- Calculate the Relative Locations of the Compression Force and the Neutral Axis

Let,

Ap;r’ - = 0.18% (Optimum Solution)

bghg

0.88( = - 23.5 in,bg-optimum 0.0018 X 22

Check if


and

_ ~" IN+ Y(2~,ae,-2s’a~)t=0.13o.85/3, T ’

M,~,~ _ 0.5(1- fl~rla~) = 0.88Aphg f l~,aes 0)

288x12 = 0.88 in2l~ Ap = 0.88x 22x 202

and

167


M,~ = (1- ~ - 0.5fl~r/,~)

A~ .h ~ A ~,,~ f ~y 1-co

288×12A~ =

1.8×22×1.4×60

=1.8

Ap R~,a~, (1- ~ - O.5 ~arla~ ) _ 0.85A~ 0.5(1_ flf/.~)(l_ 1)

Stel~ 8: Calculate the moment strength of the section at

0.5(1- fir/a=)

The objective function is equal to:

0.5(1-flf/~) + -~[~,~,u~(1-~-0.5fl~rl~)+2¢,~=(~-O.5flflu~)~ 0.5(1-fl~r/~)o9 =0

The EXCEL Solver is used to find the optimum solution by changing floes, co, and y Subjected to


!) r ~ 21<desK

2) Sp < a’~py

3) ~/-< 0.45 Q

4) 0Ides)calculated -- 0Ides)assumed

(1"Ides)optimum "- 0.13

((D)optimum -- 0.5

(Y)optimum = 0.18 %

168


Ste~ 9". Evaluate the Restoring Properties of the Beam (Constrained Equations)


Mp0 > Ms0 + Ms’o

22s,,a~K = 0.85


Let

%° = ~s,m.x = 0.05

= 0.346 rad.

~,~g.~ 1+2 2 2 7 + 0 a~,+e ~,. -1=0.04984"~0 des + ~ a,a COS + O~t 2 2

and


169


~,m~ -- ~b,r~x

0.05 -- 0.00572 = 0.044

L.~ >_ (1-q~-£’)0~ -0.33h

For h = 22 in.

Lut > 22 x 0.33 = 7.26 in

Ls. = Lut - 2L.a

-- 7.26 - 2(1.35) = 4.62 in.

Provide Lsu = ~

Step 11: Confine Compression Re,on

lph = kph ~a~ hg

= 1×0.13x22 = 2.86 in.

Odes(rla~hg) -- Odes -- 0.018kph




fp,des / fpy = 0.84

~ fp,aes = 0.84 x 243 = 204.12 ksi


170


288x12 - [0.86 in2=:> Ap = 0.9x 22x 204.12


1.77x22x1.4x60


1. Calculate rld~

2. Calculate Agp

3. Check if e.p < ~’ ~py

4. Calculate fp,aes

r/d~ = --0.2fl1+0.3

= --0.2(0.85)+0.3 = 0.13

Ap _ 0,~es (0.5_r/d~s) = 0.00094AOOp- lpu ~

°~p = ~:pi + A~’p = 0.0071

Check if

OC’p __< O~’OPpy

Let a’ = 1

0.0071 < 0.0085 OK

fp,d~ = Ep e.p = 202 ksi

171


5. For ~ -- Lpuih < 8

M~ - 1.125(,Oo.o4~,~x°’°371

Aphgfp,aes

00.04 = (-6.70d~ + 0.9)~(z7°°~+°°3)

= (-6.7 x 0.018 + 0.9) x 7.09(~’7×°’°~8+°’°3~ = 0.91

M~es _ E1.125 x 0 05°.°z7~1 x 0.91 = 0.92

288x12[0.85 in~~ Ap 0.92 x 22 x 202

6. For ~ = Lpuih < 8

= 0.882~0.04~s,max"0"034

~9"0.04 = (22.230a~ + 1.7)~-(3"4640a~+0’0216)

= (22.23 x 0.018 + 1.7) x 7.09"(3’464x0"018+°’0~16) = 1.782

Mdes = I0.882x 0.05-°’°~41xl.782 =1.74

:=> As = 288x12 =[1.07 in~[1.74x 22x 1.4x 60


172



Method

PRESSS Procedure


Cha~s

Design Equations

Ap (in2)

0.89

0.88

0.86

0.85

As (in2)

1.04

1.04

1.05

1.07

(in.)

6.5

5

Remarks






Use 6 - 1/2 in. diameter PT strands (Ap,provided = 0.92 in2), along with 2 #8 bars


173


Chapter 5

Summary and Conclusions

The Unbonded Post-Tensioned frame with Damping (hybrid frame) performed very well

in the PRESSS research program. It formed one part of the structural framing in the

PRESSS Phase III experimental building that was tested at the University of California at

San Diego. The hybrid frame was one of a total of five different seismic structural

systems proposed in the PRESSS report. The hybrid frame system has an important

feature that allows the designer to eliminate residual drift after an earthquake. This

feature is not available in the framing systems recognized in the 1997 UBC. The PRESSS

report outlined a series of step-by-step design procedures for hybrid frames.-

In this report, a set of new dimensionless parameters to replace current dimensional

parameters in the PRESSS calculation process were developed. Proposals for

modifications to same PRESSS procedures were also made. Parametric studies based on

this new non-dimensional formulation of the PRESSS design procedures were performed.

A large number of optimization problems (approximately 1200) were solved to achieve

optimum solution in each case.

Based on the results of the parametric study, a complete set of charts with non-

dimensional parameters were generated and simplified design equations were

recommended. These charts and equations would enable designers to perform the design

in fewer number of steps, while affording them the opportunity to see how changing one

or more parameters would affect the design.

It was observed from a preliminary nonlinear finite element (FE) analysis of a 3-D model

of a hybrid frame system [Hawileh, 2003], that the mild steel bars in a hybrid frame

174


exhibit significant inelastic and cyclic axial and bending strains. Once the gap at the

beam-column interface opens, relatively high levels of plastic strains develop in the mild

steel bar. This prompted a need for considering the low-cycle fatigue of the mild steel

bars including both the bending and axial strains. It should be noted that both the

PRESSS [Stanton and Nakaki, 2002] and National Institute of Standards and Technology

(NIST) [Cheok and Stone, 1994] test reports indicated bar fractures during cyclic testing.

A mild steel fracture criterion is therefore needed in the design procedure for hybrid

frames by controlling the total plastic strains in the mild steel bar below a maximum

value. In this study, a mild steel fracture criterion under combined axial and bending

strains is proposed based on the works of Mander (1994) and Liu (2001).

Mander and Panthaki (1994) studied the behavior of reinforcing steel bars under low-

cycle fatigue subjected to axial-strain reversals with strain amplitudes ranging from yield

to 6%. Liu (2001) studied the low-cycle fatigue behavior of steel bars subjected to

bending strain reversals with variable amplitudes.

The deformed geometry of the bar is needed to allow calculation of bending strains.

Therefore, it is important to determine the inelastically deformed shape of the reinforcing

bar at design drift. A series of FE analysis were performed to predict the inelastically

deformed shape of the bar. Six different model cases were solved using the FE model to

study the combined axial and bending behavior of the mild steel bar.

An equation for the deflected shape of the bar was derived from the FE model, and the

bending and axial strains of the bar were calculated based on the equations of sections 2.2

and 2.3.

175


A set of five design examples were prepared coveting low to high seismic zones using

the PRESSS and newly developed simplified design procedures. The results of the three

procedures (PRESSS, proposed design charts, and proposed design equations) were in

close agreement (within 10 %).

176


Step 1

fg

p~Ep

f~ofpi = fpo

Es

Sy

Spy

~pi

~s,rnax

fpi / fpu

~ (Cover)

~s,des

~,s’,des

Step 2

Drift, Odes

Step 3

~ - 1h b

APPENDIX A

EXCEL Spreadsheet Example # 1

Establish material properties

ksi Y = fsy/f "g 12.00

ksi Z = fsy/f py 0.24691

K = fsy/f po 0.34

ksi

ksi

ksi

ksi

ksi

ksi

ksi

in/in

Q = f "c/fpi

fs,des=~,s,des¯ fsy

fs,,des=Xs,,des. fsy

0,046

Obtain Design Drift

Estimate Frame Beam Dimensions

2

0.03134

84

75

ksi

ksi

bb _> 0.3~/ hb

177


Step 7 Estimate The Neutral Axis Parameter

0.1375

C70-D70

Constraint = zero1.6E-

O9

Step 8 Calculate the elongation and stress in the prestressing tendon at eaes

A~:p= ~-~-(0,$ - qd.) =0.001104977

~p "-" I~pi -1-Al~p = 0.007262871 ksi

O°p ~ ~py

fp,des = 8p. Ep =

f p ,des

_ fp ,des

fpy

fp,des

207 ksi

Step 4 Determine the proportions of momentstrength

assume w

Mdes

fp,dosAphb

M des

fsy ’As ’L s,desh b

178


Step 5

Estimate

Ap

Estimate Required required percentage of Tendon Area,

7Ap

bbhb

~s,d~.s (1 ~-- O~des)

T < 0.45Q=1.41 %

Step 6

As _ 7bbhb z

Estimate Required Mild Reinforcement Area, As

Step 9 Evaluate Restoring Properties of the beam section

Mpo >- Mso + M,~o

K0.85470085 0.00

Step 10

Pc,desades -- 0.85f’c bb

Calculate the neutral axis location from the concrete compression force

ades

~des = ~ades2hb

in

Pc,des(~des = --1.7f’chbbb

179


Step 11 Calculate the moment strength of the section at edes

0.867573take 0.45/w to the left side of the equation

180


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184


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