PCI 6 th Edition

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PCI 6 th Edition. Flexural Component Design. Presentation Outline. What’s new to ACI 318 Gravity Loads Load Effects Concrete Stress Distribution Nominal Flexural Strength Flexural Strength Reduction Factors Shear Strength Torsion Serviceability Requirements. New to ACI 318 – 02.

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### PCI 6th Edition

Flexural Component Design

Presentation Outline
• What’s new to ACI 318
• Concrete Stress Distribution
• Nominal Flexural Strength
• Flexural Strength Reduction Factors
• Shear Strength
• Torsion
• Serviceability Requirements
New to ACI 318 – 02
• Stress limits
• Member Classification
• Strength Reduction factor is a function of reinforcement strain
• Minimum shear reinforcement requirements
• Torsion Design Method
• U = 1.4 (D + F)
• U = 1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lr or S or R)
• U = 1.2D + 1.6 (Lr or S or R) + (1.0L or 0.8W)
• U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)
• U = 1.2D + 1.0E + 1.0L + 0.2S
• U= 0.9D + 1.6W + 1.6H
• U= 0.9D + 1.0E + 1.6H
• U=1.2D + 1.6 L 2002
• U= 1.4D + 1.7L 1999

If L=.75D

i.e. a 10% reduction in required strength

Classifications
• No Bottom Tensile Stress Limits
• Classify Members Strength Reduction Factor
• Tension-Controlled
• Transition
• Compression Controlled
• Three Tensile Stress Classifications
• Class U – Un-cracked
• Class T – Transition
• Class C – Cracked
Class C Members
• Stress Analysis Based on Cracked Section Properties
• No Compression Stress limit
• No Tension Stress limit
• Increase awareness on serviceability
• Crack Control
• Displacements
• Side Skin Reinforcement
• Beams
• Columns
• Floor Member – Double Tees, Hollow Core
• Spandrels
• Tributary Area
• Floor members, actual top area
• Beams and spandrels
• Floor members  spandrels or beams  Columns
• Live Loads can be reduced based on:

Where:

KLL = 1

Lo = Unreduced live load and

At = tributary area

• Or the alternative floor reduction shall not exceed

or

Where:

R = % reduction ≤ 40%

r = .08

Member Shear and Moment
• Shear and moments on members can be found using statics methods and beam tables from Chapter 11
Strength Design
• Strength design is based using the rectangular stress block
• The stress in the prestressing steel at nominal strength, fps, can be determined by strain compatibility or by an approximate empirical equation
• For elements with compression reinforcement, the nominal strength can be calculated by assuming that the compression reinforcement yields. Then verified.
• The designer will normally choose a section and reinforcement and then determine if it meets the basic design strength requirement:
Concrete Stress Distribution
• Parabolic distribution
• Equivalent rectangular distribution
Stress-Strain relationship

is not constant

Stress Block Theory

f’c=6,000 psi

f’c=3,000 psi

Stress Block Theory
• Stress-Strain relationship
• Stress-strain can be modeled by:

Where :strain at max. stress

and :max stress

Stress Block Theory
• The Whitney stress block is a simplified stress distribution that shares the same centroid and total force as the real stress distribution

=

Equivalent Stress Block – b1 Definition

b1 = 0.85

when f’c < 3,000 psi

b1 = 0.65

when f’c > 8,000 psi

Design Strength
• Mild Reinforcement – Non - Prestressed
• Prestress Reinforcement
Strength Design Flowchart
• Figure 4.2.1.2 page 4-9
• Non-Prestressed Path
• Prestressed Path
Non-Prestressed Members
• Find depth of compression block
Depth of Compression Block

Where:

As is the area of tension steel

A’s is the area of compression steel

fy is the mild steel yield strength

Assumes compression steel yields

Flanged Sections
• Checked to verify that the compression block is truly rectangular
Compression Block Area
• If compression block is rectangular, the flanged section can be designed as a rectangular beam

=

=

Compression Block Area
• If the compression block is not rectangular (a> hf),

=

To find “a”

Determine Neutral Axis
• From statics and strain compatibility
Check Compression Steel
• Verify that compression steel has reached yield using strain compatibility
• By strain compatibility, compression steel yields if:
• If compression steel has not yielded, calculation for “a” must be revised by substituting actual stress for yield stress
• Non prestressed members should always be tension controlled, therefore c / dt < 0.375
• Add compression reinforcement to create tesnion controlled secions
Moment Capacity
• 2 equations
• rectangular stress block in the flange section
• rectangular stress block in flange and stem section

Strength Design Flowchart

Figure 4.2.1.2page 4-9

Non- Prestressed Path

Prestressed Path

This portion of the flowchart is dedicated to determining the stress in the prestress reinforcement

Stress in Strand

fse - stress in the strand after losses

fpu - is the ultimate strength of the strand

fps - stress in the strand at nominal strength

Stress in Strand
• Typically the jacking force is 65% or greater
• The short term losses at midspan are about 10% or less
• The long term losses at midspan are about 20% or less
Stress in Strand
• Nearly all prestressed concrete is bonded
Stress in Strand
• Prestressed Bonded reinforcement

gp = factor for type of prestressing strand, see ACI 18.0

= .55 for fpy/fpu not less than .80

= .45 for fpy/fpu not less than .85

= .28 for fpy/fpu not less than .90 (Low Relaxation Strand)

rp = prestressing reinforcement ratio

Compression Block Height

Assumes compression steel yields

Prestress component

Where

Aps - area of prestressing steel

fps - prestressing steel strength

Compression Steel Check
• Verify that compression steel has reached yield using strain compatibility
Moment Capacity
• 2 Equations
• rectangular stress block in flange section
• rectangular stress block in flange and stem section
Flexural Strength Reduction Factor
• Based on primary reinforcement strain
• Strain is an indication of failure mechanism
• Three Regions
Member Classification
• On figure 4.2.1.2
e < 0.002 at extreme steel tension fiber or

c/dt > 0.600

= 0.70 with spiral ties

= 0.65 with stirrups

Compression Controlled
e > 0.005 at extreme steel tension fiber, or

c/dt < 0.375

f= 0.90 with spiral ties or stirrups

Tension Controlled
0.002 < e < 0.005 at extreme steel tension fiber, or

0.375 < c/dt < 0.6

f = 0.57 + 67(e) or

f = 0.48 + 83(e) with spiral ties

f = 0.37 + 0.20/(c/dt) or

f = 0.23 + 0.25/(c/dt) with stirrups

Transition Zone
Strand Slip Regions
• ACI Section 9.3.2.7

‘where the strand embedment length is less than the development length’

f =0.75

Limits of Reinforcement
• To prevent failure immediately upon cracking, Minimum As is determined by:
• As,min is allowed to be waived if tensile reinforcement is 1/3 greater than required by analysis
Limits of Reinforcement
• The flexural member must also have adequate reinforcement to resist the cracking moment
• Where

Correction for initial stresses on non-composite, prior to topping placement

Section after composite has been applied, including prestress forces

Horizontal Shear
• ACI requires that the interface between the composite and non-composite, be intentionally roughened, clean and free of laitance
• Experience and tests have shown that normal methods used for finishing precast components qualifies as “intentionally roughened”
Horizontal Shear, Fh Positive Moment Region
• Based on the force transferred in topping (page 4-53)
Horizontal Shear, Fh Negative Moment Region
• Based on the force transferred in topping (page 4-53)
Unreinforced Horizontal Shear

Where

f – 0.75

bv – width of shear area

lvh - length of the member subject to shear, 1/2 the span for simply supported members

Reinforced Horizontal Shear

Where

f – 0.75

rv - shear reinforcement ratio

Acs - Area of shear reinforcement

me - Effective shear friction coefficient

Prestress Concrete Shear Capacity

Where:

• ACI Eq 11-9
• Effective prestress must be 0.4fpu
• Accounts for shear combined with moment
• May be used unless more detail is required
Prestress Concrete Shear Capacity
• Concrete shear strength is minimum is
• Maximum allowed shear resistance from concrete is:
Shear Capacity, Prestressed
• Resistance by concrete when diagonal cracking is a result of combined shear and moment

Where:

Vi and Mmax - factored externally appliedloads e.g. no self weight

Shear Capacity, Prestressed
• Resistance by concrete when diagonal cracking is a result of principal tensile stress in the web is in excess of cracking stress.

Where:

Vp = the vertical component of effective prestress force (harped or draped strand only)

Vcmax
• Shear capacity is the minimum of Vc, or if a detailed analysis is used the minimum of Vci or Vcw
Shear Steel

If:

Then:

Shear Steel Minimum Requirements
• Non-prestressed members
• Prestressed members

Remember

both legs of a stirrup count for Av

Torsion
• Current ACI
• Based on compact sections
• Greater degree of fixity than PC can provide
• Provision for alternate solution
• Zia, Paul and Hsu, T.C., “Design for Torsion and Shear in Prestressed Concrete,” Preprint 3424, American Society of Civil Engineers, October, 1978. Reprinted in revised form in PCI JOURNAL, V. 49, No. 3, May-June 2004.
Torsion

1.2D+1.6L

1.0D

For members loaded two sides, such as inverted tee beams, find the worst case condition with full load on one side, and dead load on the other

Torsion
• In order to neglect Torsion

Where:

Tu(min) – minimum torsional strength provided by concrete

Where:

x and y - are short and long side, respectively of a component rectangle

g - is the prestress factor

Minimum Torsional Strength
Prestress Factor, g
• For Prestressed Members

Where:

fpc – level of prestress after losses

Maximum Torsional Strength
• Avoid compression failures due to over reinforcing

Where:

Maximum Shear Strength
• Avoid compression failures due to over reinforcing
Torsion/Shear Relationship
• Determine the torsion carried by the concrete

Where:

T’c and V’c - concrete resistance under pure torsion and shear respectively

Tc and Vc - portions of the concrete resistance of torsion and shear

Torsion/Shear Relationship
• Determine the shear carried by the concrete
Torsion Steel Design
• Provide stirrups for torsion moment - in addition to shear

Where

x and y - short and long dimensions of the closed stirrup

Torsion Steel Design
• Minimum area of closed stirrups is limited by
Longitudinal Torsion Steel
• Provide longitudinal steel for torsion based on equation

or

• Whichever greater
Longitudinal Steel limits

The factor in

the second equation need not exceed

Detailing Requirements, Stirrups
• 135 degree hooks are required unless sufficient cover is supplied
• The 135 degree stirrup hooks are to be anchored around a longitudinal bar
• Torsion steel is in addition to shear steel
Detailing Requirements, Longitudinal Steel
• Placement of the bars should be around the perimeter
• Spacing should spaced at no more than 12 inches
• Longitudinal torsion steel must be in addition to required flexural steel (note at ends flexural demand reduces)
• Prestressing strand is permitted (@ 60ksi)
• The critical section is at the end of simply supported members, therefore U-bars may be required to meet bar development requirements
Serviceability Requirements
• Three classifications for prestressed components
• Class U: Uncracked
• Class T: Transition
• Class C: Cracked

Stress

Uncracked Section
• Table 4.2.2.1 (Page 4.24)
• Easiest computation
• Use traditional mechanics of materials methods to determine stresses, gross section and deflection.
• No crack control or side skin reinforcement requirements
Transition Section
• Table 4.2.2.1 (Page 4.24)
• Use traditional mechanics of materials methods to determine stresses only.
• Use bilinear cracked section to determine deflection
• No crack control or side skin reinforcement requirements
Cracked Section
• Table 4.2.2.1 (Page 4.24)
• Iterative process
• Use bilinear cracked section to determine deflection and to determine member stresses
• Must use crack control steel per ACI 10.6.4 modified by ACI 18.4.4.1 and ACI 10.6.7
Cracked Section Stress Calculation
• Class C member require stress to be check using a Cracked Transformed Section
• The reinforcement spacing requirements must be adhered to
Cracked Transformed Section Property Calculation Steps

Step 1 – Determine if section is cracked

Step 2 – Estimate Decompression Force in Strand

Step 3 – Estimate Decompression Force in mild reinforcement (if any)

Step 4 – Create an equivalent force in topping if present

Step 5 – Calculate transformed section of all elements and modular ratios

Step 6 – Iterate the location of the neutral axis until the normal stress at this level is zero

Step 7 – Check Results with a a moment and force equilibrium set of equations

Steel Stress
• fdc – decompression stress

stress in the strand when the surrounding concrete stress is zero – Conservative to use, fse (stress after losses) when no additional mild steel is present.

Deflection Calculation – Bilinear Cracked Section
• Deflection before the member has cracked is calculated using the gross (uncracked) moment of inertia, Ig
• Additional deflection after cracking is calculated using the moment of inertia of the cracked section Icr
Effective Moment of Inertia
• Alternative method

Where:

ftl = final stress

fl = stress due to live load

fr = modulus of rupture

Prestress Losses
• Prestressing losses
• Sources of total prestress loss (TL)

TL = ES + CR + SH + RE

• Elastic Shortening (SH)
• Creep (CR)
• Shrinkage (SH)
• Relaxation of tendons (RE)
Elastic Shortening
• Caused by the prestressed force in the precast member

Where:

Kes = 1.0 for pre-tensioned members

Eps = modulus of elasticity of prestressing tendons (about 28,500 ksi)

Eci = modulus of elasticity of concrete at time prestress is applied

fcir = net compressive stress in concrete at center of gravity of prestressing force immediately after the prestress has been applied to the concrete

fcir

Where:

Pi = initial prestress force (after anchorage seating loss)

e = eccentricity of center of gravity of tendons with respect to center of gravity of concrete at the cross section considered

Mg = bending moment due to dead weight of prestressed member and any other permanent loads in place at time of prestressing

Kcir = 0.9 for pretensioned members

Creep
• Creep (CR)
• Caused by stress in the concrete

Where:

Kcr = 2.0 normal weight concrete

= 1.6 sand-lightweight concrete

fcds = stress in concrete at center of gravity of prestressing force due to all uperimposed permanent dead loads that are applied to the member after it has been prestressed

fcds

Where:

Msd = moment due to all superimposed permanent dead and sustained loads applied after prestressing

Shrinkage
• Volume change determined by section and environment
• Where:

Ksh = 1.0 for pretensioned members

V/S = volume-to-surface ratio

R.H. = average ambient relative humidity from map

Relative Humidity

Page 3-114 Figure 3.10.12

Relaxation
• Relaxation of prestressing tendons is based on the strand properties

Where:

Kre and J - Tabulated in the PCI handbook

C - Tabulated or by empirical equations in the PCI handbook

Relaxation Table
• Values for Kre and J for given strand
• Table 4.7.3.1 page 4-85
Relaxation Table Values for C
• fpi = initial stress in prestress strand
• fpu = ultimate stress for prestress strand
• Table 4.7.3.2 (Page 4-86)
Prestress Transfer Length
• Transfer length – Length when the stress in the strand is applied to the concrete
• Transfer length is not used to calculate capacity
Prestress Development Length
• Development length - length required to develop ultimate strand capacity
• Development length is not used to calculate stresses in the member
Beam Ledge Design
• For Concentrated loads where s > bt + hl, find the lesser of:
Beam Ledge Design
• For Concentrated loads where s < bt + hl, find the lesser of:
Beam Ledge Reinforcement
• Ledge reinforcement should be provided by 3 checks
• As, cantilevered bending of ledge
• Al, longitudinal bending of ledge
• Ash, shear of ledge
Transverse (cantilever) bending reinforcement, As

Uniformly spaced over width of 6hl on either side of the bearing

Not to exceed half the distance to the next load

Bar spacing should not exceed the ledge depth, hl, or 18 in

Beam Ledge Reinforcement
Longitudinal Ledge Reinforcement
• Placed in both the top and bottom of the ledge portion of the beam:

Where:

dl - is the depth of steel

U-bars or hooked bars may

be required to develop

reinforcement at the end

of the ledge

Required for attachment of the ledge to the web

Distribution and spacing of Ash reinforcement should follow the same guidelines as for As

Hanger Reinforcement
Ash is not additive to shear and torsion reinforcement

“m” is a modification factor which can be derived, and is dependent on beam section geometry. PCI 6th edition has design aids on table 4.5.4.1

Hanger (Shear) Ledge Reinforcement
Dap Design

(1) Flexure (cantilever bending) and axial tension in the extended end. Provide flexural reinforcement, Af, plus axial tension reinforcement, An.

Dap Design

(2) Direct shear at the junction of the dap and the main body of the member. Provide shear friction steel, composed of Avf + Ah, plus axial tension reinforcement, An

Dap Design

(3) Diagonal tension emanating from the re-entrant corner. Provide shear reinforcement, Ash

Dap Design

(4) Diagonal tension in the extended end. Provide shear reinforcement composed of Ah and Av

Dap Design

(5) Diagonal tension in the undapped portion. This is resisted by providing a full development length for As beyond the potential crack.

5 Main Areas of Steel

Tension - As

Shear steel - Ah

Diagonal cracking – Ash, A’sh

Dap Shear Steel - Av

Dap Reinforcement
Tension Steel – As
• The horizontal reinforcement is determined in a manner similar to that for column corbels:
Shear Steel – Ah
• The potential vertical crack (2) is resisted by a combination of As and Ah
ShearSteel – Ah
• Note the development ld of Ah beyond the assumed crack plane. Ah is usually a U-bar such that the bar is developed in the dap
Diagonal Cracking Steel – Ash
• The reinforcement required to resist diagonal tension cracking starting from the re-entrant corner (3) can be calculated from:
Dap Shear Steel – Av
• Additional reinforcement for Crack (4) is required in the extended end, such that:
Dap Shear Steel – Av
• At least one-half of the reinforcement required in this area should be placed vertically. Thus:
Dap Limitations and Considerations
• Design Condition as a dap if any of the following apply
• The depth of the recess exceeds 0.2H or 8 in.
• The width of the recess (lp) exceeds 12 in.
• For members less than 8 in. wide, less than one-half of the main flexural reinforcement extends to the end of the member above the dap
• For members 8 in. or more wide, less than one-third of the main flexural reinforcement extends to the end of the member above the dap