pci 6th edition - flexural component design.ppt

PCI 6th EditionPCI 6th Edition

Flexural Component Design

Presentation OutlinePresentation 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 – 02New to ACI 318 – 02

• Load Combinations• Stress limits• Member Classification• Strength Reduction factor is a function of

reinforcement strain• Minimum shear reinforcement requirements• Torsion Design Method

Load CombinationsLoad Combinations

• 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

Comparison of Load CombinationsComparison of Load Combinations

• U=1.2D + 1.6 L 2002• U= 1.4D + 1.7L 1999

If L=.75D

i.e. a 10% reduction in required strength

Ratio 1.2D1.6 .75D 1.4D1.7 .75D 0.90

ClassificationsClassifications

• 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

Copied from ACI 318 2002, ACI 318-02 table R18.3.3

Class C MembersClass 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

0

50

100

150

200

0 5,000 10,000 15,000 20,000

Concrete Strength, f'c, psi

Minimum Shear ReinforcingMinimum Shear Reinforcing

1999

2002

Avfy

bws

System LoadsSystem Loads

• Gravity Load Systems– Beams– Columns– Floor Member – Double Tees, Hollow Core– Spandrels

• Tributary Area– Floor members, actual top area– Beams and spandrels

• Load distribution– Load path– Floor members spandrels or beams Columns

• Live Loads can be reduced based on:

Where:

KLL = 1

Lo = Unreduced live load and

At = tributary area

Live Load ReductionLive Load Reduction

L Lo 0.25

15

KLLA

t

Live Load ReductionLive Load Reduction

• Or the alternative floor reduction shall not exceed

or

Where: R = % reduction ≤ 40%r = .08

R r (A

t 150)

R 23.1 1D

Lo

Member Shear and MomentMember Shear and Moment

• Shear and moments on members can be found using statics methods and beam tables from Chapter 11

Strength DesignStrength 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:

M

nM

u

Concrete Stress DistributionConcrete Stress Distribution

• Parabolic distribution

• Equivalent rectangular distribution

Stress Block TheoryStress Block Theory

• Stress-Strain relationship

– is not constant E

(f 'c)

E

(f 'c)

f’c=3,000 psi

f’c=6,000 psi


• Stress-Strain relationship– Stress-strain can be modeled by:

fc

2f ''c(

)

1 ( )2

Where :strain at max. stress

1.71f 'c

Ec

and :max stress f ''

c.9f '

c


• 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 – 1 DefinitionEquivalent Stress Block – 1 Definition

1 = 0.85

when f’c < 3,000 psi

1 = 0.65

when f’c > 8,000 psi

a1c

11.05 05

f 'c

1,000psi

Design StrengthDesign Strength

• Mild Reinforcement – Non - Prestressed

• Prestress Reinforcement

Strength Design FlowchartStrength Design Flowchart

• Figure 4.2.1.2 page 4-9

• Non-Prestressed Path

• Prestressed Path

Non-Prestressed MembersNon-Prestressed Members

• Find depth of compression block

Depth of Compression BlockDepth 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

aA

sf

y A'

sf '

y

.85f 'cb

Assumes compression steel yields

Flanged SectionsFlanged Sections

• Checked to verify that the compression block is truly rectangular

Compression Block AreaCompression Block Area

• If compression block is rectangular, the flanged section can be designed as a rectangular beam

A

compab

= =

Compression Block AreaCompression Block Area

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

=

To find “a”

Af(b b

w)h

f

AwA

comp A

f

aA

w

bw

Determine Neutral AxisDetermine Neutral Axis

• From statics and strain compatibility

c a /

Check Compression SteelCheck Compression Steel

• Verify that compression steel has reached yield using strain compatibility

3 'c d

Compression CommentsCompression Comments

• 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

c 3d'

Moment CapacityMoment Capacity

• 2 equations– rectangular stress block in the flange section– rectangular stress block in flange and stem section

Strength Design FlowchartStrength Design Flowchart

Figure 4.2.1.2page 4-9Non- Prestressed PathPrestressed Path

This portion of the flowchart is dedicated

to determining the stress in the prestress

reinforcement

Stress in StrandStress in Strand

fse - stress in the strand after lossesfpu - is the ultimate strength of the strandfps - stress in the strand at nominal strength


• 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

fse0.5f

pu


• Nearly all prestressed concrete is bonded

Stress in Strand Stress in Strand

• Prestressed Bonded reinforcement

p = 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)

p = prestressing reinforcement ratio

fpsf

pu 1

p

1

pfpu

f 'c

d

dp

'

Determine Compression BlockDetermine Compression Block

Compression Block HeightCompression Block Height

Where

Aps - area of prestressing steel

fps - prestressing steel strength

aA

psf

ps A

sf

y A'

sf '

y

.85f 'cb

Assumes compression steel yields

Prestress component

Flange Sections CheckFlange Sections Check

Compression Steel CheckCompression Steel Check

• Verify that compression steel has reached yield using strain compatibility

3 'c d

Moment CapacityMoment Capacity

• 2 Equations– rectangular stress block in flange section– rectangular stress block in flange and stem section

Flexural Strength Reduction FactorFlexural Strength Reduction Factor

• Based on primary reinforcement strain

• Strain is an indication of failure mechanism

• Three Regions

Member ClassificationMember Classification

• On figure 4.2.1.2

Compression ControlledCompression Controlled

< 0.002 at extreme steel tension fiber or

• c/dt > 0.600

= 0.70 with spiral ties = 0.65 with stirrups

Tension ControlledTension Controlled

> 0.005 at extreme steel tension fiber, or

• c/dt < 0.375

= 0.90 with spiral ties or stirrups

Transition ZoneTransition Zone

• 0.002 < < 0.005 at extreme steel tension fiber, or

• 0.375 < c/dt < 0.6

= 0.57 + 67() or = 0.48 + 83() with spiral

ties

= 0.37 + 0.20/(c/dt) or

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

Strand Slip RegionsStrand Slip Regions

• ACI Section 9.3.2.7‘where the strand embedment length is less than the development length’

=0.75

Limits of ReinforcementLimits 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

As,min

3 f '

c

fy

bwd

200bwd

fy

Limits of ReinforcementLimits of Reinforcement

• The flexural member must also have adequate reinforcement to resist the cracking moment– Where

M

n1.2M

cr

McrS

bc

P

A

Pe

Sb

fr

M

nc

Sbc

Sb

1

Section after composite has been

applied, including prestress forces

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

Critical SectionsCritical Sections

Horizontal ShearHorizontal 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 RegionHorizontal Shear, Fh Positive Moment Region

• Based on the force transferred in topping (page 4-53)

Horizontal Shear, Fh Negative Moment RegionHorizontal Shear, Fh Negative Moment Region

• Based on the force transferred in topping (page 4-53)

Unreinforced Horizontal ShearUnreinforced Horizontal Shear

Fh 80b

vl

vh

Where – 0.75bv – width of shear arealvh - length of the member subject to shear, 1/2 the span for simply supported members

Reinforced Horizontal ShearReinforced Horizontal Shear

Where – 0.75

v - shear reinforcement ratio

Acs - Area of shear reinforcement

e - Effective shear friction coefficient

Fh (260 0.6

vf

y) b

vl

vh

Acs

Fh

ef

y

Shear Friction CoefficientShear Friction Coefficient

e

1000 Acr

Vu

Fh

Shear Resistance by Non-Prestressed ConcreteShear Resistance by Non-Prestressed Concrete

• Shear strength for non-prestressed sections

V

c2 f '

cb

wd

Prestress Concrete Shear CapacityPrestress 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

Vc 0.6 f '

c 700

Vud

Mu

bw

d

Vud

Mu

1

Prestress Concrete Shear CapacityPrestress Concrete Shear Capacity

• Concrete shear strength is minimum is

• Maximum allowed shear resistance from concrete is:

V

c2 f '

cb

wd

V

c5 f '

cb

wd

Shear Capacity, PrestressedShear Capacity, Prestressed

• Resistance by concrete when diagonal cracking is a result of combined shear and moment

Vci0.6 f '

cb

wd V

d

ViM

cr

Mmax

Where:

Vi and Mmax - factored externally applied loads e.g. no self weightVd - is un-factored dead load shear

Shear Capacity, PrestressedShear Capacity, Prestressed

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

V

cw 3.5 f '

c 0.3f

pc bwd V

p

Where:Vp = the vertical component of effective prestress force (harpedor draped strand only)

VcmaxVcmax

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

Shear SteelShear Steel

If:

Then:

V

u V

c

v

sV

n V

c or v

s

Vu

V

c

Shear Steel Minimum RequirementsShear Steel Minimum Requirements

• Non-prestressed members

• Prestressed members

Av0.75 f '

cb

ws

fy

50bw

s

fy

Av

Apsf

pus

80fyd

d

bw

Remember

both legs of a stirrup count for Av

TorsionTorsion

• 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.

TorsionTorsion

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

1.0D 1.2D+1.6L

TorsionTorsion

• In order to neglect Torsion

Where:

Tu(min) – minimum torsional strength provided by concrete

T

uT

u(min)

Minimum Torsional StrengthMinimum Torsional Strength

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

is the prestress factor

T

u(min) 0.5 f '

c x2 y

Prestress Factor, Prestress Factor,

• For Prestressed Members

Where:

fpc – level of prestress after losses

110fpc

fc̀

Maximum Torsional StrengthMaximum Torsional Strength

• Avoid compression failures due to over reinforcing

Where:

Tn(max)

1

3K

t f

c̀x2y

1K

tV

t

30CtT

u

2

Tn(max)

T

u

Kt 12 10

fpc

fc̀

Ct

bwd

x2y

Maximum Shear StrengthMaximum Shear Strength

• Avoid compression failures due to over reinforcing

Vn(max)

10 f

c̀b

wd

130C

tT

u

KtV

t

2

Vn(max)

V

u

Torsion/Shear RelationshipTorsion/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

Tc

T 'c

1T '

cT

u

V'c

Vu

2

Torsion/Shear RelationshipTorsion/Shear Relationship

• Determine the shear carried by the concrete

Vc

V'c

1V'

cV

u

T 'c

Tu

2

Torsion Steel DesignTorsion Steel Design

• Provide stirrups for torsion moment - in addition to shear

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

At

Tu

T

c

s

tx

1y

1f

y

t0.66 0.33y

1x

11.5

Torsion Steel DesignTorsion Steel Design

• Minimum area of closed stirrups is limited by

Av2A

t min50

bws

fy

( )2 200b

ws

fy

Longitudinal Torsion SteelLongitudinal Torsion Steel

• Provide longitudinal steel for torsion based on equation

or

• Whichever greater

A

l

2At(x

1 y

1)

s

Al

400xfy

T

u

Tu

Vu

3Ct

2A

t

s

x1 y

1

Longitudinal Steel limitsLongitudinal Steel limits

Al

400xfy

T

u

Tu

Vu

3Ct

2A

t

s

x1 y

1

The factor in

the second equation need not exceed

2At

s

50bw

fy

112f

pc

fc̀

50bw

fy

Detailing Requirements, StirrupsDetailing 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 SteelDetailing 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 RequirementsServiceability Requirements

• Three classifications for prestressed components– Class U: Uncracked– Class T: Transition– Class C: Cracked

t7.5 f '

c 7.5 f '

c

t12 f '

c

t12 f '

c

Stress

Uncracked SectionUncracked 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 SectionTransition 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 SectionCracked 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 CalculationCracked 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

Cracked Transformed Section Property Calculation Steps

Step 1 – Determine if section is crackedStep 2 – Estimate Decompression Force in StrandStep 3 – Estimate Decompression Force in mild

reinforcement (if any)Step 4 – Create an equivalent force in topping if presentStep 5 – Calculate transformed section of all elements and

modular ratiosStep 6 – Iterate the location of the neutral axis until the

normal stress at this level is zeroStep 7 – Check Results with a a moment and force

equilibrium set of equations

Steel StressSteel 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.

Simple ExampleSimple Example

Page 4-31

Deflection Calculation – Bilinear Cracked Section

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 InertiaEffective Moment of Inertia

• Alternative method

Ie

Mcr

Ma

3

Ig 1

Mcr

Ma

3

Icr

or based on stress

Mcr

Ma

1 ftl f

r

fl

Where: ftl = final stress fl = stress due to live load fr = modulus of rupture

Prestress LossesPrestress 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 ShorteningElastic Shortening

• Caused by the prestressed force in the precast member

Where:Kes = 1.0 for pre-tensioned membersEps = modulus of elasticity of prestressing tendons (about 28,500 ksi)Eci = modulus of elasticity of concrete at time prestress is appliedfcir = net compressive stress in concrete at center of gravity of prestressing force immediately after the prestress has been applied to the concrete

ES K

esE

psf

cirE

ci

fcirfcir

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 consideredMg = bending moment due to dead weight of prestressed member and any other permanent loads in place at time of prestressingKcir = 0.9 for pretensioned members

fcirK

cir

Pi

Ag

P

ie2

Ig

Mge

Ig

CreepCreep

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

Where:Kcr = 2.0 normal weight concrete = 1.6 sand-lightweight concretefcds = 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

CR K

crE

psE

ci fcir f

cds

fcdsfcds

Where:

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

fcds

Msde

Ig

ShrinkageShrinkage

• Volume change determined by section and environment

• Where:

Ksh = 1.0 for pretensioned membersV/S = volume-to-surface ratio R.H. = average ambient relative humidity from

map

SH 8.210 6 Ksh

Eps1 0.06V S 100 R.H.

Relative HumidityRelative Humidity

Page 3-114 Figure 3.10.12

RelaxationRelaxation

• Relaxation of prestressing tendons is based on the strand properties

Where:

Kre and J - Tabulated in the PCI handbookC - Tabulated or by empirical equations in the PCI handbook

RE K

re J SHCR ES

C

Relaxation TableRelaxation Table

• Values for Kre and J for given strand

• Table 4.7.3.1 page 4-85

Relaxation Table Values for CRelaxation 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 LengthPrestress Transfer Length

• Transfer length – Length when the stress in the strand is applied to the concrete

• Transfer length is not used to calculate capacity

t se bl f 3 d

lt f

se3 db

Prestress Development LengthPrestress Development Length

• Development length - length required to develop ultimate strand capacity

• Development length is not used to calculate stresses in the member

ldl

t f

ps f

se

ld f

se3 db

fps f

se

Beam Ledge GeometryBeam Ledge Geometry

Beam Ledge DesignBeam Ledge Design

• For Concentrated loads where s > bt + hl, find

the lesser of:

Vn3 f '

ch

l 2b

l b b

th

l

Vn f '

ch

l 2b

l b b

th

l2d

e

Beam Ledge DesignBeam Ledge Design

• For Concentrated loads where s < bt + hl, find

the lesser of:

Vn1.5 f '

ch

l 2b

l b b

th

l s

Vn f '

ch

l b

l b b

th

l

2

d

e s

Beam Ledge ReinforcementBeam Ledge Reinforcement

• For continuous loads or closely spaced concentrated loads:

• Ledge reinforcement should be provided by 3 checks– As, cantilevered bending of ledge– Al, longitudinal bending of ledge– Ash, shear of ledge

Vn24 h

l f '

c

Beam Ledge ReinforcementBeam Ledge Reinforcement

• 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

As

1

fy

Vu

a

d

N

u

hl

d

0.2 N

u

Vdl

Longitudinal Ledge ReinforcementLongitudinal 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

Al

200bl b dl

fy

Hanger ReinforcementHanger Reinforcement

• Required for attachment of the ledge to the web

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

Ash

Vu

fy

m

Hanger (Shear) Ledge ReinforcementHanger (Shear) Ledge 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

Dap DesignDap Design

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


(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


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


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


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

Dap ReinforcementDap Reinforcement

5 Main Areas of Steel• Tension - As

• Shear steel - Ah

• Diagonal cracking – Ash, A’sh

• Dap Shear Steel - Av

Tension Steel – AsTension Steel – As

• The horizontal reinforcement is determined in a manner similar to that for column corbels:

AsA

f A

n

1

fy

Vu

a

d

N

u

hl

d

and 0.2

Nu

Vdl

Shear Steel – AhShear Steel – Ah

• The potential vertical crack (2) is resisted by a combination of As and Ah

Ah

2Vu

3 fy

e

An

Shear Steel – AhShear Steel – 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 – AshDiagonal Cracking Steel – Ash

• The reinforcement required to resist diagonal tension cracking starting from the re-entrant corner (3) can be calculated from:

Ash

Vu

fy

and .75

Dap Shear Steel – AvDap Shear Steel – Av

• Additional reinforcement for Crack (4) is required in the extended end, such that:

V

n A

vf

y A

hf

y2bd f '

c

Dap Shear Steel – AvDap Shear Steel – Av

• At least one-half of the reinforcement required in this area should be placed vertically. Thus:

Av

1

2fy

V

u

2bd f '

c

Dap Limitations and ConsiderationsDap 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

Questions?Questions?

pci 6th edition - flexural component design.ppt

Documents