flexural design and material properties for reinforced concrete and prestressed concrete

7/27/2019 Flexural Design and Material Properties for Reinforced Concrete and Prestressed Concrete

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Flexural Design and Material Properties



Outline

Material Properties

Reinforced concrete beams

Prestressed concrete beams

Reinforced concrete columns

Brittle failure of prestressed members



Material properties



Material properties

Concrete calculations use cylinder strength ~0.8 × cube strengthConcrete specification uses both: C40/50 has 40MPa cylinder strength ( f

ck) and 50 MPa cube strength

Covers concrete grades up to C70/85 for bridges andC90/105 for buildingsShort term Young’s Modulus for concrete obtainedfrom:

3.0

ck

cm10

822

f E



Material properties



Design strengths

Element Characteristic strength Design strength

Reinforcementyield

f yk f yd = f yk / γ s

Prestressing yield f p0.1k f pd = f p0.1k / γ s

Concrete in directcompression

f ck f cd = α cc f ck / γ c

cc allows for long term effects, eccentricities and true shape of concrete stress-strain blockRecommended value = 1.0 for buildings and 0.85 for bridgesUK National Annex for bridges and buildings gives 0.85 for bending andaxial force, 1.0 for shear Other values are used in other situations e.g. varies in strut and tie



Reinforcement

Reinforcement specified to EN 10080Plain round bars not covered by EN 1992Bars can have ductility class A, B or C; A not permitted for

bridge design

Standard reinforcement yield strength is 500 MPaDesignation of reinforcement to EN 10080 is:B500B

Class Characteristic strain atmaximum force, εuk

Minimum value of k = (f t /f y)k

A ≥ 2.5% ≥ 1.05 B ≥ 5% ≥ 1.08 C ≥ 7.5% ≥ 1.15, <1.35

Bar Yield stress , f yk Ductility class



Stress – strain curves for reinforcement

ε ud may be found in the National Annex and is recommended tobe taken as 0.9 ε uk

Benefit of using inclined branch compared to past UK practice

f yd / E s uk

kf yk

f yd = f yk / γs

0 ud

f yk

kf yk

B

A

k = ( f t / f y)k

A – IdealisedB – Design

Reinforcement



Compressive stress blocks for bending andaxial force

Possible blocks are:rectangular bilinear parabola-rectangle

All have same

maximum stress sorectangular givesgreatest bendingresistanceNo formulae for bending

resistance provided – see exampleStrain limits differ for thedifferent blocks….

f cd f cd f cd

Stress (for f ck ≤ 50 MPa) Strain

x λ x(λ =0.8)



Compressive stress blocks for bending andaxial force

c

c c2 cu2

f ck

f cd

0

(a) Parabolic-rectangular distribution

c

c c3 cu3

f ck

f cd

0(b) Bilinear distribution

c

c 0.0035

f cd

0 0.00200.0007 0.00175

(c) Alternative concrete design stress blocks for f ck ≤ 50MPa



Pre-stressing steel

Pre-stressing specified to EN 10138Part 1 gives general rules, Part 2 – wire, Part 3 – strand, Part4 - bar

Pre-stressing steel can be relaxation class 1, 2 or 3

Designation of strand to EN 10138-3 is:Y1860S7-15.7

Strand diameter (mm)Tensile stress, f pk 7 wire strand

Class Prestressing steel type ρ1000 , Relaxation lossat 1000 hours at 20

°

C1 Ordinary prestressing tendons

(wire or strand)8.0%

2 Low relaxation prestressingtendons (wire or strand)

2.5%

3 Hot rolled and processed bars 4%



• ε ud may be found in the National Annex and is recommended to be takenas 0.9 ε uk

• Need to use inclined branch to get similar results to past UK practice

Design curves for prestressing

f pd / E p uk

f pk

f pd = f p0.1k / γs

0 ud

f p0.1k f pk / γs

B

A

A – IdealisedB – Design



Concrete creep and shrinkage

Treated more rigorously than previous UK practiceCreep strain calculated from:

Long term E value is thus E cm / (1+ )Creep factor, ( ,t 0) , calculated from Annex B or from simple chartsShrinkage split into:

Autogenous shrinkage - occurs on hydration and hardening without loss of moisture;complete within a few months

Drying shrinkage - occurs through loss of moisture; complete in a number of years

• Both creep and shrinkage parameters have to be calculated based onconcrete mix (also age at loading for creep)

c

ccc

E t t

),(),( 00



Beam bending resistance – clause 6.1



Beam Bending Resistance

Assumptions to be made for bending resistance according to EC2:Plane sections remain plane;Strain in bonded reinforcement, whether in tension or compression, isthe same as the strain in the concrete at the same level;

Tensile strength of the concrete is ignored;The stresses in the concrete, reinforcement and prestressing aregiven by the design stress-strain relationships shown earlier;The initial strain in prestressing is taken into account.



Strain Compatibility

Can be used to determine ultimate bending resistanceDetermined either iteratively or algebraicallyMust be used for non-uniform sections (in compression regions)Needed if rebar stress-strain curve with rising branch usedGenerally used for prestressed concrete membersIterative approach:

Estimate location of N.A. and calculate strains in reinforcementCalculate stresses based on strainsCalculate concrete stresses based on assumed N.A. depthCalculate net tensile/compressive forces in section. If not equalrepeat calculations for modified N.A. positionTake moments about common point to determine moment resistance



Strain Distribution

Where there is equal compressive strains at both faces of a section, a reducedstrain limit of c2 is used because:

Peak stresses are reached at a strain of about c2 For pure compression, peak load occurs at about c2

For pure flexure, resistance increases beyond this point and limit of cu2 usedFor intermediate cases where whole section or outstand part is wholly in

compression, intermediate limit is appropriate to correct idealised diagram - straindiagram is rotated about intermediate pivot point to reduce strain limitFor flanged beams, mean strain under full compression under concentric loadingis limited to c2 (e.g. flanges in box girders where N.A. is in webs)“Concentric” is defined as e /h < 0.1

Pivot point

Compressive strains in flange

Actual finalstrain

limiting cases for strain

h h c2/ cu2

c2 cu2 c

c c2 cu2

f cd

0

(a) Parabolic-rectangular distribution

idealised

“real”




Parabolic stress blockFailure strain ( cu2 ) is only appropriate for parabolic stress blocks ( cu3 isused for bilinear and rectangular blocks)

f av and are based on the geometry of the stress block – they aresimplest for the rectangular stress block …..

εcu2

εs

x

d

b

As

βx

z

Fc

Fs

f cd

s

s

yk

s A f

F

bx f F avc




Stress blocksParabolic rectangular:

Bilinear:

Simplified rectangular:

2

2

1

11

cu

ccd av

n f f

1

2121

222

2

2

2

2

2

n

nn

ccu

cu

ccu

3

3501cu

ccd av . f f

2

621332

3

2

3

2

3

ccu

cu

ccu

cd av f f

2 /




Rectangular stress block RC resistance can be calculated using rectangular block (mosteconomic) and reinforcement diagram with plateau (simplest)Simple formula can then be derived for bending resistance:

εcu3

εs

x

d

b

As

x

z

Fc

Fs

ηf cd

λ x For f ck ≤ 50 MPa,recommended valuesare:

λ = 0.8

η = 1.0

2 /

cd av f f

z f A M yd sRd

bd f

A f d z

cd

s yd

21

cd

yd s

f b

f A x

1

1

3cuss

yk

E f d

x

with

which is OK provided where

assuming steel yields




For design, rearrange these equations to:

Then:

Solve the equation; the lower root is the relevant oneThe ratio of x /d is checked (for rebar yield) against:

The reinforcement area is now designed from:

where:

2

1

d

x

d

x

d

x

d

xK av 0

2

avK

d

x

d

x =>

1

1

2cuss

yk

E

f d

x

z f M A

yk

ss

xd z



C35/45 concreteB500B reinforcementTake:

cc = 0.85 s = 1.15 c = 1.5

Determine bendingresistance (this is subject of

workshop)Note there is no maximumvalue for z

1500mm

7 No. 20 diameter bars

1000 mm

50 mm

EN 1992-2: R.C. beam example



Calculation with reinforcement curve withplateau at f yd gives M Rd = 1354 kNm

Calculation repeated with rectangular block but reinforcement curve with risingbranch. This leads to M Rd = 1449 kNm

0

435

466

0.045

0

435

EN 1992-2: Bending resistance



BS5400resistance

(kNm)

Eurocode 2resistance – steel plateau

(kNm)

Eurocode 2resistance – rising branch

(kNm)1309 1354 1449

EN 1992-2: Bending resistance

Summary of bending resistances for simplebeam to BS5400 Part 4 and EN 1992-2

Generally, under-reinforced sections will benefit fromthe increased calculation effort of using the stress-strain curve with rising branch



Doubly Reinforced Rectangular beams

In heavily reinforced tension zones, where tension steel doesn’tyield:

Add compression steel to reduce concrete compression zonedepth

Allows tension reinforcement to yieldNeeded when is exceeded (yield criterion)

May be necessary to analyse sections with known compressionreinforcement for ultimate flexural resistanceEC2 uses same stress strain curve relationship for reinforcementin tension and compression (unlike BS5400 Part 4)Design approach as follows ……

1

1

2cuss

yk

E

f d

x




For equilibrium assuming all reinforcement yields:

Used to determine required compression reinforcement so tensionreinforcement yields. x is first set in above so the tension rebar yields:

Moment determined from:

d

b

As

d′

A′s

βx Fc

Fs

f cd

F′s

εcu2

εs

x ε′s

s

s

yk

s

s

yk

avssc A f

A f

bx f F F F yk

savss

f

bx f A A

=>

1

1

2cuss

yk

E

f d

x

d F xF d F M scs Rd




Also need to check compression reinforcement yields for formula to bevalid

If reinforcement does not yield, strain compatibility method is used.

d

b

As

d′

A′s

βx Fc

Fs

f cd

F′s

εcu2

εs

x ε′s

21

1

cuss

yk

E

f d

x



Flanged Beams

If N.A. in compression flange at ULS, previous equations can be usedWhen rectangular stress block used, equations still OK for N.A. depth 1/(~1.25) × flange thickness where f ck<50 MPa)

Also OK if web in compression, flange in tension – different b usedIf flange and web in compression, section should be analysed usingstrain compatibility methodStrictly, variable strain limits should be applied to flanged sections

h

d c

b b

εcu3

εs

x

d

b

As

0.5x

z

Fc

Fs

ηf cd

λ x

z f A M yd s

bd f

A f d z

cd

s yd

21

cd

yd s

f b

f A x

1

1

3cuss

yk

E

f d

x

with

provided where



Prestressed Concrete Beams

General assumptions same as for reinforced concreteInitial strain of prestressing tendons considered for ultimateresistance ( )Strain compatibility can also be used; pre-strain to be added tostrain diagram calculated at failure (see example)Unbonded tendons

Cannot be treated using same general rulesStrain in tendons does not increase at same rate as strain inconcrete at same level

xP xP t mPt d ,,

εcu3

εs

x

d

b

A p

0.5x

z

Fc

Fs

ηf cd

λ x

ε p

prestrain



Columns



Reinforced Concrete Columns

Same assumptions as used for bending designCompression failure in flexure defined by strain limit cu2

- Strain is adjusted depending on position of N.A. and whether section is in pure flexure or axial load (as beams)

cu2 = 0.0035 for class 50/60 concrete in flexure

- Modified strain c2 under combined bending and axial load is 0.0020(assumes whole section in compression)

Calculated strengths are usually relatively insensitive for variations in assumptions of ultimate concrete strain.- Caution needed for heavily reinforced sections

Minimum applied moments to be considered in design- Axial loads applied at minimum eccentricities (max [ h /30, 20mm])In slender columns, additional second order moments must beallowed for




Strain Compatibility Assume reinforcement area and estimate N.A.Set extreme fibre compressive strain to ε cu2 (or ε cu3 )Calculate strains throughout section and stresses in reinforcementIterate for strain limit if whole section is in compression

Calculate axial load and moment section can resist.

Verification through further iterationDetermine moment resistance for given axial force; verify this momentresistance exceeds coexistent moment, or

Applied moment and axial force increased pro-rata together; verify loadfactor exceeds unity.

M M max

N

N bal

N u Calculates a pointon this graph




Axial Load with Uniaxial BendingEquations for flanged beam with N/A in flange applyFor equilibrium

=>

Taking moments about Application of N:

N

M

h

b

d′

d A′s

As

x

0.0035

ε′s

εs

f′s

f cd

- f s - F s

F′s

Fc

βx

ssc F F F N

ssssav A f A f bx f N

d

h A f d

h A f x

hbx f M ssssav

222




Axial Load with Biaxial BendingThrough rigorous analysis interaction diagram can bedeveloped for N Ed , M Edz and M Edy

Shape of diagram represented by

M Rdi are the moment resistances about each axis in theabsence of axial force

Not suited for design as “ a” depends on reinforcement sohave to guess reinforcement, then check it

0.1

a

Rdy

Edy

a

Rdz

Edz

M M

M M



Brittle fracture of prestressed members



Brittle Failure of Members with prestress

Prestressed beams must not to fail in a brittle manner due to corrosionor failure of individual tendons.Potential problem if tendons corroding but the concrete remainsuncracked – can’t see signs of the damage.For prestressed beams, new requirements to safeguard against this

Protection can be achieved by one of three ways:a) Ensure remaining cables, after corrosion or failures has led tocracking, are adequate to carry frequent combination designmoment at ULSb) Minimum Reinforcement

c) Provide proven monitoring (inspection regime)

flexural design and material properties for reinforced concrete and prestressed concrete

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