Flexure Design Sequence - Ohio Department of … 2005 3-1 Prestressed Concrete Beam Design Workshop Load and Resistance Factor Design Flexure Design Flexure Design Sequence • Determine

July 2005 3-1

Prestressed Concrete Beam Design Workshop

Load and Resistance Factor Design

Flexure Design

Flexure Design Sequence

• Determine Effective flange width• Determine maximum tensile beam stresses

(without prestress)• Estimate eccentricity and number of strands

at midspan• Calculate prestress loss• Determine number of strands and develop

strand arrangement

Flexure Design Sequence

• Determine eccentricities• Check service stresses• Check fatigue• Calculate nominal flexural resistance• Check reinforcement limits• Determine pretensioned anchorage zone

reinforcement

July 2005 3-2

Example

Example 1

120'-0"

Example 1

Centerline bridge1'-0" 18'-0" 18'-0" 1'-0"

AASHTO Type VI Beam

4'-9" 9'-6" 4'-9" 4'-9" 9'-6" 4'-9"

9"

TYPICAL SECTION

July 2005 3-3

Example 1

• Simple spans: 120 feet• Fully prestressed beams• Bonded tendons• Skew angle: 0 degrees• Stress limit for tension in beam concrete

(corrosion conditions): severe

Example 1Deck concrete

Beam concrete

( )( )

f

w f

c

c

'

. ' ..

=

= = =

5

33000 33000 0145 5 40741 5 1 5

ksi

E ksic

fc' ==

==

8515474821

ksiE ksif ksiE ksi

c

ci'

ci

Example 1

• Prestressing steel– Strand type: 270 ksi low relaxation– Strand diameter: 0.6 inch– Cross-sectional area per strand: 0.217 in2

July 2005 3-4

Example 1Prestressing steel

Non-prestressed reinforcementfy =

=

6029 000

ksiE ksis ,

fpy = =

=

0 90 243

28 500

.

,

f ksi

E ksipu

ps

Example 1 - Interior BeamEffective flange width may be taken as the least of:

a) One-quarter of the effective span length: 120 ft

(0.25) (120) (12) = 360 in

Example 1 - Interior Beamb) 12.0 times the average thickness of the slab, 9 in, plus the

greater of: Web thickness: 8 in One-half of the top flange of the girder: (0.5) (42) = 21 in

(12) (9) + 21 = 129 in

July 2005 3-5

Example 1 - Interior Beamc) The average spacing of adjacent beams: 9.5 ft

(9.5) (12) = 114 in

Effective flange width = 114 in

Example 1 - Exterior BeamEffective flange width may be taken as one-half the effective width of the adjacent interior beam, 114 in, plus the least of:

a) One-eight of the effective span length: 120 ft

(0.125) (120) (12) = 180 in

Example 1 - Exterior Beamb) 6.0 times the average thickness of the slab, 9 in, plus the

greater of: One-half the web thickness: (0.5) (8) = 4 in One-quarter of the top flange of the girder:

(0.25) (42) = 10.5 in

The greater of these two values: 10.5 in(6) (9) + 10.5 = 64.5 in

July 2005 3-6

Example 1 - Exterior Beamc) The width of the overhang: 4.75 ft

(4.75) (12) = 57 in

Effective flange width = (0.5) (114) + 57 = 114 in

Example 1 - Exterior Beam(114)(0.7906) = 90.12"

C. G. Beam

C. G. Composite Section

53.5

4"

36.3

8"

C. G. Slab

72"

9"

76.5

0"

17.1

6"

22.9

6"

Example 1

68443 68443 Sslab top (in3)1.130w (k/ft)80503 80503 Stc (in3)20587St (in3)27751 27751 Sbc (in3)20157Sb (in3)27.46 27.46 yslab top (in)35.62yt (in)18.46 18.46 ytc (in)36.38yb (in)53.5453.54ybc (in)733320I (in4)

14858841485884Icomp (in4)1085A (in2)

Exterior Beam

Interior BeamProperty

AASHTO Type VI Beams

Property

Composite SectionNon-composite Section

July 2005 3-7

Analysis

• Loads• Distribution of live load• Load factors

Non-composite Dead Loads

• Beam (DC)• Deck slab (DC)• Diaphragms (DC)

Composite Dead Loads

• Curb (DC)• Bridge rail (DC)• Overlays (DW)• Future wearing surface (DW)• Utilities (DW)

July 2005 3-8

Design Vehicular Live Load

• Article 3.6.1.2• HL-93

– Combination of • Design truck or design tandem• Design lane load

Unfactored Moments (k-ft)

2053 2137Slab

405405Wearing Surface250250Rail

20342034BeamExterior BeamInterior Beam

Live Load (Including Dynamic Allowance):

3680 k-ft (Truck + Lane)

3080 k-ft (Tandem + Lane)

Distribution of Live LoadShearBending Moment

Multi-Lane

One Lane

Multi-Lane

One Lane

XXXXSpecial Analysis

X--X--Equations--X--XLever Rule

Exterior Girder

--------Special Analysis

XXXXEquations (“m”included)

--------Lever Rule

Interior Girder

July 2005 3-9

Example 1

0.6170.422one lane loaded - fatigue0.9180.741two lanes loaded 0.7400.506one lane loaded

INTERIOR BEAMSHEARMOMENT

Example 1

1.0421.042one lane loaded

0.6610.661one lane loaded - fatigue0.9420.942two lanes loaded 0.7930.793one lane loaded

Additional Investigation0.8680.868one lane loaded - fatigue0.8640.848two lanes loaded

EXTERIOR BEAMSHEARMOMENT

Load Combinations and Load Factors

0.750.801.001.351.75IM0.750.801.001.351.75LL

----1.001.00Max-1.50Min-0.65

Max-1.50Min-0.65

DW

----1.001.00Max-1.25Min-0.90

Max-1.25Min-0.90

DC

FatigueService IIIService IStrength IIStrength I

Limit StateLoad

July 2005 3-10

Beam Stresses

• Due to dead load and live load Service III limit state (Crack Control)

Service I limit state

fM M

SM M M

Sbottombeam slab

b

rail ws LL

bc= −

+−

+ + 08.

fM M

SM M M

Stopbeam slab

t

rail ws LL

tc=

++

+ +

Beam Stresses

Example 1

Example 1 - Interior BeamStresses at midspan due to dead load and live load.Extreme bottom of beam fiber (Service III Limit State):

( )( ) ( )( )[ ]( )fbottom = −+

−+ +

= −2034 2137 12

20157250 405 08 2728 12

277513710

.. ksi (t)

Extreme top of beam fiber (Service I Limit State):( )( ) [ ]( )ftop =

++

+ +=

2034 2137 1220587

250 405 2728 1280503

2 936. ksi (c)

July 2005 3-11

Example 1 - Exterior BeamStresses at midspan due to dead load and live load.Extreme bottom of beam fiber (Service III Limit State):

( )( ) ( )( )[ ]( )fbottom = −+

−+ +

= −2034 2053 12

20157250 405 08 3837 12

277514 044

.. ksi (t)

Extreme top of beam fiber (Service I Limit State):( )( ) [ ]( )ftop =

++

+ +=

2034 2053 1220587

250 405 3837 1280503

3052. ksi (c)

Preliminary Strand Arrangement

• Calculate maximum tensile stress– Service III limit state

• Determine stress limit for tension• Set maximum tensile stress equal to stress

limit for concrete tension• Estimate eccentricity at midspan• Solve for total prestress force required


• Estimate total prestress loss• Estimate effective prestress force per strand• Estimate number of prestressing strands

July 2005 3-12


fPA

PeS

ftenb

bottom= − +

f fPA

PeS

PA

eSten bottom

b b− = − = −

⎛⎝⎜

⎞⎠⎟

1

Pf f

Ae

S

ten bottom

b

=−

−⎛⎝⎜

⎞⎠⎟

1


fpe = fpj – estimated lossesPe = (fpe)(Aps)

NoPPe

. Strands =

Example

• Preliminary strand arrangement

July 2005 3-13


Example 1

Example 1BEAM STRESSESBottom fiber, maximum tension stress, all loads applied, Service III Limit State:

Interior beam, fbottom = -3.710 ksiExterior beam, fbottom = -4.044 ksi

CONCRETE STRESS LIMITLimit for tension, all loads applied, after all losses:

f ften c= = =0 0948 0 0948 8 0 268. . .' ksi

Example 1ESTIMATE NUMBER OF STRANDS

fpT = total loss in the prestressing steel stress = 60 ksi (estimated)

fpj = stress in the prestressing steel at jacking = (0.75)(270) = 202.5 ksi

fpe = effective stress in the prestressing steel after losses = 202.5 - 60 = 142.5 ksi

Aps = area of prestressing steel (per strand) = 0.217 in2

July 2005 3-14

Example 1ESTIMATE NUMBER OF STRANDSPe = effective prestressing force in one strand = (fpe)(Aps) =

(142.5)(0.217) = 30.9 kA= area of non-composite beam = 1085 in2

Sb = section modulus, non-composite section, extreme bottom beam fiber = 20157 in3

e = eccentricity of prestress force at midspan = -32 in (estimated)

Example 1 - Exterior Beam(114)(0.7906) = 90.12"

C. G. Beam

C. G. Composite Section

53.5

4"

36.3

8"

C. G. Slab

72"

9"

76.5

0"

17.1

6"

22.9

6"

Example 1 - Interior BeamEstimated total prestress force required:

( ) ( )( )P f f

Ae

S

ten bottom

b

=−

−⎛⎝⎜

⎞⎠⎟

=− − −

−−⎡

⎣⎢⎤⎦⎥

=1

0 268 2 7101

108532

20157

13718. .

. k

Estimated number of strands required:

No PPe

. ..

. Strands = = =1371830 9

44 4

July 2005 3-15

Example 1 - Exterior BeamEstimated total prestress force required:

( ) ( )( )P f f

Ae

S

ten bottom

b

=−

−⎛⎝⎜

⎞⎠⎟

=− − −

−−⎡

⎣⎢⎤⎦⎥

=1

0 268 4 0441

108532

20157

1504 9. .

. k

Estimated number of strands required:

No PPe

. ..

. Strands = = =1504 9

30 948 7

Example 1STRAND ARRANGEMENT - 50 strands

At midspan of beam

( )( ) ( )( ) ( )( ) ( )( )y =+ + +

=13 2 13 4 13 6 11 8

504 88. in

e = 4.88 - 36.38 = -31.50 in

Example 1At the ends of beam (harp 12 strands)

( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )y =+ + + + + + +10 2 10 4 10 6 8 8 3 64 3 66 3 68 3 70

5019 76 = in.

e = 19.76 - 36.38 = -16.62 in

July 2005 3-16

Example 1

At beam ends

4 sp

a @

2"

At beam centerline4

spa

@ 2

"

4 sp

a @

2"

Example 1

C. G. beam

12'-0"

10 strands 3 strands

3 strands3 strands

8 strands10 strands

10 strands

3 strands

48'-0"6"

e =

-16.

6 2 in

e =

-16 .

7 7 i n

e =

-20.

46 in

e =

-24.

1 4 in

e =

-27 .

8 2 i n

e =

- 31 .

50 in

e =

-31.

5 0 in

centerline bearing

Prestress Loss

• Article 5.9.5• Losses due to

– Elastic shortening– Shrinkage– Concrete creep– Relaxation of steel

July 2005 3-17

Elastic Shortening

∆fEE

fpESp

cicgp=

(5.9.5.2.3a-1)

Elastic Shortening

• For components of usual design, may calculate fcgp assuming stress in the prestressing steel – 0.70fpu for low relaxation strands– 0.65fpu for stress relieved strands

Shrinkage

• Pretensioned members

∆fpSR = −17 0 0150. . H

(5.9.5.4.2-1) H = average annual ambient relative humidity

July 2005 3-18

Creep

(5.9.5.4.3-1)

∆ ∆f f fpCR cgp cdp= − ≥12 0 7 0 0. .

∆fcdp = change in concrete stress at center of gravity ofprestressing steel due to permanent loads, with the exception of the load acting at the time theprestressing force is applied.

Relaxation

• Total relaxation– Relaxation at transfer– Relaxation after transfer

Relaxation

• At transfer• Initially stressed in excess of 0.50 fpu

July 2005 3-19

Relaxation

• Stress relieved strand

(5.9.5.4.4b-1)

( )∆ f

ffpR

pj

pypj1

24 010 0

055= −⎡

⎣⎢⎢

⎤

⎦⎥⎥

log ..

. t

f

Relaxation

• Low relaxation strand

( )∆f

ffpR

pj

pypj1

24 040 0

055= −⎡

⎣⎢⎢

⎤

⎦⎥⎥

log ..

. t

f

(5.9.5.4.4b-2)

Relaxation

• Pretensioned members• After transfer• Stress relieved strand

( )∆ ∆ ∆ ∆f f f fpR pES pSR pCR2 20 0 0 4 0 2= − − +. . .

(5.9.5.4.4c-1)

July 2005 3-20

Relaxation

• Pretensioned members• After transfer• Low relaxation strand

– 30% of stress relieved strand

( )[ ]( )

∆ ∆ ∆ ∆

∆ ∆ ∆

f f f f

f f fpR pES pSR pCR

pES pSR pCR

2 0 3 20 0 0 4 0

6 0 012 0 06

= − − +

− − +

. . . .

. . .

2

=

Prestress Loss

• Final total losses∆fpT = ∆fpES + ∆fpSR + ∆fpCR + ∆fpR2

(5.9.5.1-1)

∆fpT = ∆fpES + ∆fpSR + ∆fpCR + ∆fpR1 + ∆fpR2

Example

• Prestress loss• Example 1• Exterior beam

July 2005 3-21

Prestress Loss

Example 1 – Exterior Beam

Example 1 - Exterior Beam∆fpT = total loss in the prestressing steel stress (ksi)∆fpES = loss due to elastic shortening (ksi)∆fpSR = loss due to shrinkage (ksi)∆fpCR = loss due to creep of concrete (ksi)∆fpR1 = loss due to relaxation of steel at transfer (ksi)∆fpR2 = loss due to relaxation of steel after transfer (ksi)e = eccentricity of prestress force at midspan = -31.50 inA = area of non-composite beam = 1085 in2

I = moment of inertia of non-composite section = 733320 in4

Icomp = moment of inertia of composite section = 1485884 in4

Example 1 - Exterior BeamELASTIC SHORTENING (5.9.5.2.3a)fcgp = sum of concrete stresses at the center of gravity of

prestressing tendons due to the prestressing force at transfer and the self weight of the member at the sections of maximum moment (ksi)

Mbeam = moment due to weight of member = (2034)(12) = 24408 k-in

Pt = prestress force at transfer = (0.70)(270)(0.217)(50) = 2050.7 k

Ep = modulus of elasticity of prestressing steel = 28500 ksi Eci = modulus of elasticity of concrete at transfer = 4821 ksi

July 2005 3-22

Example 1 - Exterior BeamFor components of the usual design, may calculate fcgp using a stress in the prestressing steel of 0.70 fpu.

( )

( )( )[ ]( ) ( )( )

f PA

P y Mcgp

t t beam= + +

= +− −

+−

=

eI

yI

ksi

2050 71085

2050 7 3150 3150733320

24408 3150733320

36164

. . . . .

.

( )∆fEE

fpESp

cicgp= = ⎛

⎝⎜⎞⎠⎟

=285004821

36164 2138. . ksi

Example 1 - Exterior BeamSHRINKAGE (5.9.5.4.2)H = average annual ambient relative humidity = 70%

( ) ( )( )[ ]∆fpSR = − =17 0 0150 70 650. . . H = 17.0 - 0.150 ksi

Example 1 - Exterior BeamCREEP (5.9.5.4.3)∆fcdp = change in concrete stress at center of gravity of

prestressing steel due to permanent loads, with the exception of the load acting at the time the prestressing force is applied. Values of fcdp should be calculated at the same section or at sections for which fcgp is calculated (ksi)

Mslab = moment due to slab and diaphragms = (2053)(12) = 24636 k-in

Mrail = moment due to curb and rail = (250)(12) = 3000 k-inMws = moment due to wearing surface and FWS = (405)(12)

= 4860 k-in

July 2005 3-23

Example 1 - Exterior Beam

( )

( )( ) ( )( )

∆f MI

M MIcdp

slab rail WS

comp

= − −+

= −−

−+ −

=

y y

ksi

comp

24636 3150733320

3000 4860 48 661485884

13156. .

.

( )( ) ( )( )∆ ∆f f fpCR cgp cdp= −

= − =

12 0 7 0

12 0 36164 7 0 13156 3419

. .

. . . . . ksi

y y yb= − = − =4 88 36 38 3150. . . in

y y ycomp bc= − = − =4 88 5354 48 66. . . in

Example 1 - Exterior BeamRELAXATION AT TRANSFER (5.9.5.4.4b)t = time estimated in days from stressing to transfer = 2 daysfpj = initial stress in the tendon at the end of stressing = 202.50 ksifpy = specified yield strength of prestressing steel = 243 ksi

( )

( )( )[ ] ( )

∆fffpR

pj

pypj1

24 040 0

055

24 0 240 0

202 50243

055 202 50 2 41

= −⎡

⎣⎢

⎤

⎦⎥

= −⎡⎣⎢

⎤⎦⎥

=

log ..

.

log ..

. . . .

t f

ksi

Example 1 - Exterior BeamRELAXATION AFTER TRANSFER (5.9.5.4.4c)

( )[ ]( )( ) ( )( )[ ]

∆ ∆ ∆ ∆f f f fpR pES pSR pCR2 0 3 20 0 0 4 0 2

2138 0 2 650 3419 0 99

= − − +

− + =

. . . .

. . . . .

= 0.3 20.0 - 0.4 ksi

TOTAL LOSSES∆fpT = ∆fpES + ∆fpSR + ∆fpCR + ∆fpR1 + ∆fpR2

= 21.38 + 6.50 + 34.19 + 2.41 + 0.99 = 65.47 ksi (32.3%)

∆fpT = ∆fpES + ∆fpSR + ∆fpCR + ∆fpR2= 21.38 + 6.50 + 34.19 + 0.99 = 63.06 ksi

July 2005 3-24

Example 1 - Exterior BeamEFFECTIVE STRESSES fpe = fpj - ∆fpTfpj = 202.50 ksifpe = 202.50 - 65.47 = 137.03 ksi

fpe = fpt - ∆fpTfpj = 202.50 + 2.41 = 204.91 ksifpt = 202.50 ksifpe = 202.50 - 63.06 = 139.44 ksi

Strand Arrangement

• Develop strand pattern• Determine actual eccentricities• Compare with estimate• Solve for total prestress force required• Calculate force in one strand after all losses• Calculate number of strands required

Strand Arrangement

fpe = fpj – total lossesPe = (fpe)(Aps)

Pf f

Ae

S

ten bottom

b

=−

−⎛⎝⎜

⎞⎠⎟

1

NoPPe

. Strands =

July 2005 3-25

Example

• Strand arrangement

Strand Arrangement

Example 1

Example 1fbottom = extreme bottom beam fiber tensile stress from

applied loads interior beam = -3.710 ksiexterior beam = -4.044 ksi

ften = allowable tensile stress in concrete after losses = 0.268 ksi

fpj = stress in prestressing steel at jacking = (0.75)(270) = 202.50 ksi

fpe = effective stress in prestressing steel after losses interior beam = 202.50 – 61.54 = 140.96 ksiexterior beam = 202.50 – 61.82 = 140.68 ksi

July 2005 3-26

Example 1A = area of non-composite beam = 1085 in2

Sb = section modulus of non-composite section, extreme bottom beam fiber = 20157 in3

Aps = area of prestressing steel (per strand) = 0.217 in2

e = eccentricity of prestress force at midspan = -31.50 in


Number of strands required:Effective prestressing force in one strand after all losses = Pe= (fpe)(Aps) = (140.96)(0.217) = 30.5 k

No PPe

. ..

. strands = = =1519 9

30549 8

Total force required in strands:( ) ( )

( )P f f

Ae

S

ten bottom

b

=−

−⎛⎝⎜

⎞⎠⎟

=− − −

−−⎡

⎣⎢⎤⎦⎥

=1

0 268 4 0441

10852146

20157

1519 9. ..

. k

Service Limit State

• Concrete stresses in beam– Release

• Initial prestress, beam dead load• Service I limit state

– Construction• Effective prestress, dead loads (Beam, Slab, Rail

and Wearing Surface)• Service I limit state

July 2005 3-27

Service Limit State

• Concrete stresses in beam– Service

• Effective prestress, dead load, live load• Service I limit state for compressive stresses• Service III limit state for tensile stresses

– Service• Live load, half sum of effective prestress and

permanent loads• Service I limit state for compressive stresses

Example 1Temporary stresses before losses:

Tension0 0948 0 0948 7 0 251 0 2. . . .'fci = = ksi > ksi

Compression( )( )0 60 0 60 7 4 2. . .' f ksici = =

Stresses at service limit state after losses:Tension

0 0948 0 0948 8 0 268. . .'fc = = ksi

Example 1Stresses at service limit state after losses:

CompressionDue to effective prestress and permanent loads

( )( )0 45 0 45 8 36. . .' f ksic = =

Due to live load and one-half the sum of effective prestress and permanent loads

( )( )0 40 0 40 8 32. . .' f ksic = =

July 2005 3-28

Example 1Stresses at service limit state after losses:

CompressionDue to effective prestress, permanent loads, and transient loads

( )( )( )0 6 0 6 10 8 4 8. . . . f ksic'φw = =

flange wid

w

thflange depth

< 15= =

=

1149

12 7

10

.

.φ

( )( )( )0 6 0 6 8. . fc'φ φw w=

Stress Summary

Example 1

4.8 ksi2.519 ksiCompressive stress0.268 ksinone Tensile stress

Effective prestress and dead loads4.2 ksi3.670 ksiCompressive stress

0.200 ksi0.030 ksiTensile stressRelease

LimitMaximumInterior and Exterior BeamsEXAMPLE 1 - BEAM STRESS SUMMARY

July 2005 3-29

4.8 ksi2.123 ksiCompressive stress

3.2 ksi1.348 ksiCompressive stress

Live load and half the sum effective prestress and dead loads

0.268 ksi0.252 ksiTensile stress

Effective prestress, dead loads, and live load

LimitMaximumInterior and Exterior BeamsEXAMPLE 1 - BEAM STRESS SUMMARY

Fatigue Limit State

• Fully prestressed concrete components• No need to check fatigue when tensile stress

in extreme fiber at service III limit state after all losses meets tensile stress limits

Strength Limit State

• Factored flexural resistance• Mr = φ Mn

– φ = 1.00– Mn = Nominal flexural resistance

(5.7.3.2.1-1)

July 2005 3-30

Nominal Flexural Resistance

• Without compression and non-prestressed tension reinforcement

M A f da

n ps ps p= −⎛⎝⎜

⎞⎠⎟2

fps = Average stress in prestressing steel

Strength Limit State

• For practical design, use rectangular compressive stress distribution

• Depth of compressive stress block – a = 1c

Stress in Prestressing Steel

• For rectangular section behavior

cA f A f A f

b k Afd

ps pu s y s y

c pspu

p

=+ −

+

' '

'.085 1 f β

(5.7.3.1.1-4)

July 2005 3-31

Stress in Prestressing Steel

• For components with bonded tendons

f f kc

d

kff

ps pup

py

pu

= −⎛

⎝⎜⎜

⎞

⎠⎟⎟

= −⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

1

2 104

0 28

.

. for low relaxation strand

(5.7.3.1.1-1)

(5.7.3.1.1-2)

Example

• Nominal flexural resistance

• Example 1• Exterior beam

Nominal Flexural Resistance

Example 1 - Exterior Beam - Midspan

July 2005 3-32

Example 1 - Exterior BeamAps = area of prestressing steel = 10.850 in2

fpu = specified tensile strength of prestressing steel = 270 ksifpy = specified yield strength of prestressing steel = 243 ksiAs = area non-prestressed tension reinforcement = 0 in2

= area of compression reinforcement = 0 in2

fy = yield strength of tension reinforcement = 60 ksi= yield strength of compression reinforcement = 60 ksi= compressive strength of concrete = 5 ksi

b = width of compression flange = 114 inbw = width of web = 8 indp = distance from extreme compression fiber to the

centroid of the prestressing tendons (in)1 = stress block factor = 0.80

As'

fy'

fc'

Example 1 - Exterior Beam1. Factored moments, MuMu = (1.25)(2034 + 2053 + 250) + (1.5)(405) + (1.75)(3837) = 12744 k-ft

Example 1 - Exterior Beam2. Depth of compression blockk = 0.28 (low relaxation strands)dp = e + yt + tslab = 31.50 + 35.62 + 9 = 76.12 inFor rectangular section behavior:

( )( )( )( )( )( ) ( )( )

cA f A f A f

k Afd

ps pu s y s y

c pspu

p

=+ −

+

=+ −

+ ⎛⎝⎜

⎞⎠⎟=

' '

'.

.

. . . ..

.

085

10850 270 0 0

085 5 080 114 0 28 10850 2707612

7 35

1 f b

in

β

( )( )a c= = =β1 0 80 7 35 588. . . in

July 2005 3-33


3. Stress in prestressing steel at nominal flexural resistance, components with bonded tendons

( ) ( )f f k cdps pu

p

= −⎛

⎝⎜

⎞

⎠⎟ = − ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥=1 270 1 0 28 7 35

7612262 70. .

.. ksi


4. Factored flexural resistance

( )( )

M A f d an ps ps p= −⎛

⎝⎜⎞⎠⎟

= − ⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥⎛⎝⎜

⎞⎠⎟=

2

10850 262 70 7612 5882

112

17382 k - ft. . . .

Mr = (1.0)(17382) = 17382 k-ft > 12744 k-ft o.k.

Reinforcement Limits

• Amount of prestressed and non-prestressed reinforcement– Maximum– Minimum

July 2005 3-34


• Maximum amount of prestressed and non-prestressed reinforcement should satisfycd

dA f d A f d

A f A f

e

eps ps p s y s

ps ps s y

≤

=+

+

0 42. (5.7.3.3.1-1)

(5.7.3.3.1-2)

Effective Depth

b

As

Aps

ds

de

dp


• Article 5.7.3.3.2• Minimum amount of prestressed and non-

prestressed tensile reinforcement• Adequate to develop a factored flexural

resistance at least equal to the lesser of– 1.2 Mcr

– 1.33 factored moments

July 2005 3-35

Cracking Moment

( )M S f f MSS

S fcr c r cpe dncc

ncc r= + − −

⎛⎝⎜

⎞⎠⎟ ≤1

5.7.3.3.2-1

Example 1 - Exterior BeamThe minimum amount of prestressed and non-prestressed reinforcementMdnc = total unfactored dead load moment acting on the

monolithic or non-composite section Snc = section modulus for the extreme fiber of the monolithic

or non-composite section where tensile stress is caused by externally applied loads

Sc = section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads

fr = modulus of rupturefcpe = compressive stress in concrete due to effective

prestress forces only

Example

• Reinforcement limits• Example 1• Exterior beam

July 2005 3-36


Example 1 - Exterior Beam - Midspan

Example 1 - Exterior BeamAps = area of prestressing steel = 10.850 in2

fpu = specified tensile strength of prestressing steel = 270 ksifpy = specified yield strength of prestressing steel = 243 ksiAs = area of non-prestressed tension reinforcement = 0 in2

fy = yield strength of tension reinforcement = 60 ksidp = distance from extreme compression fiber to the centroid

of the prestressing tendons = 76.12 inc = distance from the extreme compression fiber to the

neutral axis = 7.35 inde = corresponding effective depth from the extreme

compression fiber to the centroid of the tensile force in the tensile reinforcement (in)

Example 1 - Exterior BeamThe maximum amount of prestressed and non-prestressed reinforcement

( )( )( )( )( )

dA f d A f d

A f A fcd

eps ps p s y s

ps ps s y

e

=++

=+

+=

= =

10 850 262 70 7612 010850 262 70 0

7612

7 357612

010

. . .. .

.

..

.

in

July 2005 3-37

Example 1 - Exterior BeamThe minimum amount of prestressed and non-prestressed reinforcementMdnc = total unfactored dead load moment acting on the

monolithic or non-composite section = 2034 + 2053 = 4087 k-ft = 49044 k-in

Snc = section modulus for the extreme fiber of the monolithic or non-composite section where tensile stress is caused by externally applied loads = 20157 in3

Sc = section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads = 27751 in3

fr = modulus of rupture (ksi)

Example 1 - Exterior Beamfcpe = compressive stress in concrete due to effective

prestress forces only (after allowance for all prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads (ksi)

Pe = effective prestress force = 1526.4 k

( )( )f PA

PScpe

e e

nc

= − = −−

= e ksi1526 4

10851526 4 3150

2015737922. . ..

f fr c= = =0 24 0 24 8 0 6788. . .' ksi

Example 1 - Exterior Beam( )

( )( ) ( )

M S f f M SScr c r ce dnc

c

nc

= + − −⎛⎝⎜

⎞⎠⎟

= + − −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥⎛⎝⎜

⎞⎠⎟

=

1

27751 0 6788 37922 49044 2775120157

1 112

8800

k - ft

. .

Scfr = (27751)(0.6788) = 18837 k-in = 1570 k-ft1.2Mcr = (1.2)(1570) = 1884 k-ft1.33 (factored moment) = (1.33)(12744) = 16950 k-ftLesser = 1884 k-ftActual flexural resistance = 17382 k-ft > 1884 k-ft o.k.

July 2005 3-38

Pretensioned Anchorage Zone

• Bursting resistance provided by vertical reinforcement in ends of pretensioned beams at service limit state

• Not less than 4% prestress force at transfer• Total vertical reinforcement located within

a distance h/4 from end of beam

P f Ar s s=

(5.10.10.1-1)


• Confinement reinforcement - 5.10.10.2– In bottom flange– Shaped to enclose strands– Distance 1.5d from beam ends– Minimum bar size

• No. 3 deformed

– Maximum spacing• 6 inches

Example

• Pretensioned anchorage zone

July 2005 3-39


Example 1

Example 1Factored Bursting Resistance (5.10.10.1)fs = stress in steel not exceeding 20 ksi

As = total area of vertical reinforcement located within the distance h/4 from the end of the beam (in2)

h = overall depth of precast member = 72 inPt = prestressing force at transfer = 1953.2 k

Pr = (Pt)(0.04) = (1953.2)(0.04) = 78.13 k

h4

724

18 0= = . in

Example 1Bursting resistance provided by vertical reinforcement in the ends of pretensioned beams at service limit state:P f Ar s s=

A Pf

Ps

r

s

r= = = =20

781320

391. . in2

Using pairs of No. 4 bars, As = 0.40 in2, the number of pairs of bars required:3 910 40

9 8..

.=

July 2005 3-40

Development Length

• Gradual build up of strand force • Transfer and flexural bond lengths • Determine resistance in the end zone

Development Length

• Prestress force initially varies linearly – Zero at the point where bonding starts – Maximum at the transfer length

Development Length

• Prestress force increases in a parabolic manner between the transfer and development lengths

• Reaches the tensile strength at the development length

July 2005 3-41

Development Length

f

f

ld

ps

pe

Distance from free end of strand

Stee

l stre

ss

transferlength

Transfer Length

• 60 strand diameters – 5.11.4.1

Development Length

l d ps pe bk f f≥ −⎛⎝⎜

⎞⎠⎟

23

d

(5.11.4.2-1)

k = 1.6 for precast, prestressed beams

July 2005 3-42

Example

• Development length

Development and Transfer Length


Example 1 - Exterior BeamTRANSFER LENGTH (5.11.4.1)db = nominal strand diameter = 0.6 in

60 db = (60)(0.6) = 36 in

July 2005 3-43

Example 1 - Exterior BeamBONDED STRAND (5.11.4.2)db = nominal strand diameter = 0.6 infps = average stress in the prestressing steel at the time for

which the nominal resistance of the member is required = 262.70 ksi

fpe = effective stress in the prestressing steel after losses = 140.68 ksi

k = 1.6

( ) ( ) ( )k f fps pe b−⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=23

16 262 70 23

140 68 0 6 162 2 d = in. . . . .

Thank You

George Choubah, P.E.FHWA- Federal Lands Highway Bridge Office(703) [email protected]

Documents

Flexure Design Sequence - Ohio Department of … 2005 3-1 Prestressed Concrete Beam Design Workshop Load and Resistance Factor Design Flexure Design Flexure Design Sequence • Determine