Flexure Design Sequence Determine Effective flange width - - PDF document

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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

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SLIDE 1

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

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SLIDE 2

July 2005 3-2

Example Example 1

120'-0"

Example 1

Centerline bridge 1'-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

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SLIDE 3

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 1

Deck concrete Beam concrete ( )( ) f w f

c c ' . ' .

.

= = = =

5 33000 33000 0145 5 4074

15 15

ksi E ksi

c

fc

' =

= = = 8 5154 7 4821 ksi E ksi f ksi E 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

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SLIDE 4

July 2005 3-4

Example 1

Prestressing steel Non-prestressed reinforcement fy = = 60 29 000 ksi E ksi

s

, fpy =

= =

090 243 28 500 . , f ksi E ksi

pu ps

Example 1 - Interior Beam

Effective 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 Beam

b) 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

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SLIDE 5

July 2005 3-5

Example 1 - Interior Beam

c) The average spacing of adjacent beams: 9.5 ft (9.5) (12) = 114 in Effective flange width = 114 in

Example 1 - Exterior Beam

Effective 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 Beam

b) 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

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SLIDE 6

July 2005 3-6

Example 1 - Exterior Beam

c) 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

53.54" 36.38"

  • C. G. Slab

72" 9" 76.50" 17.16" 22.96"

Example 1

68443 68443 Sslab top (in3) 1.130 w (k/ft) 80503 80503 Stc (in3) 20587 St (in3) 27751 27751 Sbc (in3) 20157 Sb (in3) 27.46 27.46 yslab top (in) 35.62 yt (in) 18.46 18.46 ytc (in) 36.38 yb (in) 53.54 53.54 ybc (in) 733320 I (in4) 1485884 1485884 Icomp (in4) 1085 A (in2) Exterior Beam Interior Beam Property AASHTO Type VI Beams Property Composite Section Non-composite Section

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SLIDE 7

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)
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SLIDE 8

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 2137 Slab 405 405 Wearing Surface 250 250 Rail 2034 2034 Beam Exterior Beam Interior Beam Live Load (Including Dynamic Allowance): 3680 k-ft (Truck + Lane) 3080 k-ft (Tandem + Lane)

Distribution of Live Load

Shear Bending Moment Multi- Lane One Lane Multi- Lane One Lane X X X X Special Analysis X

  • X
  • Equations
  • X
  • X

Lever Rule Exterior Girder

  • Special

Analysis X X X X Equations (“m” included)

  • Lever Rule

Interior Girder

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SLIDE 9

July 2005 3-9

Example 1

0.617 0.422

  • ne lane loaded - fatigue

0.918 0.741 two lanes loaded 0.740 0.506

  • ne lane loaded

INTERIOR BEAM SHEAR MOMENT

Example 1

1.042 1.042

  • ne lane loaded

0.661 0.661

  • ne lane loaded - fatigue

0.942 0.942 two lanes loaded 0.793 0.793

  • ne lane loaded

Additional Investigation 0.868 0.868

  • ne lane loaded - fatigue

0.864 0.848 two lanes loaded EXTERIOR BEAM SHEAR MOMENT

Load Combinations and Load Factors

0.75 0.80 1.00 1.35 1.75 IM 0.75 0.80 1.00 1.35 1.75 LL

  • 1.00

1.00 Max-1.50 Min-0.65 Max-1.50 Min-0.65 DW

  • 1.00

1.00 Max-1.25 Min-0.90 Max-1.25 Min-0.90 DC Fatigue Service III Service I Strength II Strength I Limit State Load

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SLIDE 10

July 2005 3-10

Beam Stresses

  • Due to dead load and live load

Service III limit state (Crack Control) Service I limit state f M M S M M M S

bottom beam slab b rail ws LL bc

= − + − + + 08

.

f M M S M M M S

top beam slab t rail ws LL tc

= + + + +

Beam Stresses

Example 1

Example 1 - Interior Beam

Stresses at midspan due to dead load and live load. Extreme bottom of beam fiber (Service III Limit State): ( )( ) ( )( )

[ ](

)

fbottom = − + − + + = − 2034 2137 12 20157 250 405 08 2728 12 27751 3710 . . ksi (t)

Extreme top of beam fiber (Service I Limit State): ( )( )

[ ](

)

ftop = + + + + = 2034 2137 12 20587 250 405 2728 12 80503 2 936 . ksi (c)

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SLIDE 11

July 2005 3-11

Example 1 - Exterior Beam

Stresses at midspan due to dead load and live load. Extreme bottom of beam fiber (Service III Limit State): ( )( ) ( )( )

[ ](

)

fbottom = − + − + + = − 2034 2053 12 20157 250 405 08 3837 12 27751 4 044 . . ksi (t)

Extreme top of beam fiber (Service I Limit State): ( )( )

[ ](

)

ftop = + + + + = 2034 2053 12 20587 250 405 3837 12 80503 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

Preliminary Strand Arrangement

  • Estimate total prestress loss
  • Estimate effective prestress force per strand
  • Estimate number of prestressing strands
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SLIDE 12

July 2005 3-12

Preliminary Strand Arrangement

f P A Pe S f

ten b bottom

= − +

f f P A Pe S P A e S

ten bottom b b

− = − = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

1 P f f A e S

ten bottom b

= − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

1

Preliminary Strand Arrangement

fpe = fpj – estimated losses Pe = (fpe)(Aps) No P Pe . Strands =

Example

  • Preliminary strand

arrangement

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SLIDE 13

July 2005 3-13

Preliminary Strand Arrangement

Example 1

Example 1

BEAM STRESSES Bottom fiber, maximum tension stress, all loads applied, Service III Limit State: Interior beam, fbottom = -3.710 ksi Exterior beam, fbottom = -4.044 ksi CONCRETE STRESS LIMIT Limit for tension, all loads applied, after all losses:

f f

ten c

= = = 0 0948 0 0948 8 0 268 . . .

'

ksi

Example 1

ESTIMATE 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

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SLIDE 14

July 2005 3-14

Example 1

ESTIMATE NUMBER OF STRANDS Pe = effective prestressing force in one strand = (fpe)(Aps) = (142.5)(0.217) = 30.9 k A= 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

53.54" 36.38"

  • C. G. Slab

72" 9" 76.50" 17.16" 22.96"

Example 1 - Interior Beam

Estimated total prestress force required:

( ) ( ) ( )

P f f A e S

ten bottom b

= − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 0 268 2 710 1 1085 32 20157 13718 . . . k Estimated number of strands required: No P P

e

. . . . Strands = = = 13718 30 9 44 4

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SLIDE 15

July 2005 3-15

Example 1 - Exterior Beam

Estimated total prestress force required:

( ) ( ) ( )

P f f A e S

ten bottom b

= − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 0 268 4 044 1 1085 32 20157 1504 9 . . . k Estimated number of strands required: No P P

e

. . . . Strands = = = 1504 9 30 9 48 7

Example 1

STRAND ARRANGEMENT - 50 strands At midspan of beam

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

y = + + + = 13 2 13 4 13 6 11 8 50 4 88 . in e = 4.88 - 36.38 = -31.50 in

Example 1

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

y = + + + + + + + 10 2 10 4 10 6 8 8 3 64 3 66 3 68 3 70 50 19 76 = in .

e = 19.76 - 36.38 = -16.62 in

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SLIDE 16

July 2005 3-16

Example 1

At beam ends 4 spa @ 2" At beam centerline 4 spa @ 2" 4 spa @ 2"

Example 1

  • C. G. beam

12'-0" 10 strands 3 strands 3 strands 3 strands 8 strands 10 strands 10 strands 3 strands 48'-0" 6" e =

  • 1

6 . 6 2 i n e =

  • 1

6 . 7 7 i n e =

  • 2

. 4 6 i n e =

  • 2

4 . 1 4 i n e =

  • 2

7 . 8 2 i n e =

  • 3

1 . 5 i n e =

  • 3

1 . 5 i n centerline bearing

Prestress Loss

  • Article 5.9.5
  • Losses due to

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

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SLIDE 17

July 2005 3-17

Elastic Shortening

∆f E E f

pES p ci cgp

=

(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

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SLIDE 18

July 2005 3-18

Creep

(5.9.5.4.3-1)

∆ ∆ f f f

pCR cgp cdp

= − ≥ 12 0 7 0 . .

∆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.

Relaxation

  • Total relaxation

– Relaxation at transfer – Relaxation after transfer

Relaxation

  • At transfer
  • Initially stressed in excess of 0.50 fpu
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SLIDE 19

July 2005 3-19

Relaxation

  • Stress relieved strand

(5.9.5.4.4b-1) ( )

∆f

f f

pR pj py pj 1

24 0 10 0 055

= − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥

log . . . t f

Relaxation

  • Low relaxation strand

( )

∆f f f

pR pj py pj 1

24 0 40 0 055 = − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ log . . . t f (5.9.5.4.4b-2)

Relaxation

  • Pretensioned members
  • After transfer
  • Stress relieved strand

( )

∆ ∆ ∆ ∆

f f f f

pR pES pSR pCR 2

20 0 04 02

= − − +

. . . (5.9.5.4.4c-1)

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SLIDE 20

July 2005 3-20

Relaxation

  • Pretensioned members
  • After transfer
  • Low relaxation strand

– 30% of stress relieved strand

( )

[ ]

( )

∆ ∆ ∆ ∆ ∆ ∆ ∆ f f f f f f f

pR pES pSR pCR pES pSR pCR 2

0 3 20 0 0 4 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
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SLIDE 21

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 in A = 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 Beam

ELASTIC 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

  • f 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

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SLIDE 22

July 2005 3-22

Example 1 - Exterior Beam

For components of the usual design, may calculate fcgp using a stress in the prestressing steel of 0.70 fpu.

( ) ( )( )

[ ](

) ( )( )

f P A P y M

cgp t t beam

= + + = + − − + − = e I y I ksi 20507 1085 20507 3150 3150 733320 24408 3150 733320 36164 . . . . . .

( )

∆f E E f

pES p ci cgp

= = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 28500 4821 36164 2138 . . ksi

Example 1 - Exterior Beam

SHRINKAGE (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 Beam

CREEP (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-in Mws = moment due to wearing surface and FWS = (405)(12) = 4860 k-in

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SLIDE 23

July 2005 3-23

Example 1 - Exterior Beam

( ) ( )( ) ( )( )

∆f M I M M I

cdp slab rail WS comp

= − − + = − − − + − = y y ksi

comp

24636 3150 733320 3000 4860 48 66 1485884 13156 . . .

( )( ) ( )( )

∆ ∆ f f f

pCR 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 y

comp bc

= − = − = 4 88 5354 4866 . . . in

Example 1 - Exterior Beam

RELAXATION AT TRANSFER (5.9.5.4.4b) t = time estimated in days from stressing to transfer = 2 days fpj = initial stress in the tendon at the end of stressing = 202.50 ksi fpy = specified yield strength of prestressing steel = 243 ksi

( ) ( )( )

[ ]

( )

∆f f f

pR pj py pj 1

24 0 40 0 055 24 0 2 40 0 202 50 243 055 202 50 2 41 = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = log . . . log . . . . . . t f ksi

Example 1 - Exterior Beam

RELAXATION AFTER TRANSFER (5.9.5.4.4c)

( )

[ ]

( )( ) ( )( )

[ ]

∆ ∆ ∆ ∆ f f f f

pR pES pSR pCR 2

03 200 04 02 2138 02 650 3419 099 = − − + − + = . . . . . . . . . = 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

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SLIDE 24

July 2005 3-24

Example 1 - Exterior Beam

EFFECTIVE STRESSES fpe = fpj - ∆fpT fpj = 202.50 ksi fpe = 202.50 - 65.47 = 137.03 ksi fpe = fpt - ∆fpT fpj = 202.50 + 2.41 = 204.91 ksi fpt = 202.50 ksi fpe = 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 losses Pe = (fpe)(Aps) P f f A e S

ten bottom b

= − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

1 No P Pe . Strands =

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SLIDE 25

July 2005 3-25

Example

  • Strand arrangement

Strand Arrangement

Example 1

Example 1

fbottom = extreme bottom beam fiber tensile stress from applied loads interior beam = -3.710 ksi exterior 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 ksi exterior beam = 202.50 – 61.82 = 140.68 ksi

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SLIDE 26

July 2005 3-26

Example 1

A = 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

Example 1 - Exterior Beam

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

No P P

e

. . . . strands = = = 1519 9 305 49 8

Total force required in strands:

( ) ( ) ( )

P f f A e S

ten bottom b

= − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 0 268 4 044 1 1085 2146 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
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SLIDE 27

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 1

Temporary stresses before losses: Tension 0 0948 0 0948 7 0 251 0 2 . . . .

'

fci = = ksi > ksi Compression

( )( )

0 60 0 60 7 4 2 . . .

'

f ksi

ci =

= Stresses at service limit state after losses: Tension 0 0948 00948 8 0 268 . . .

'

fc = = ksi

Example 1

Stresses at service limit state after losses: Compression Due to effective prestress and permanent loads

( )( )

0 45 0 45 8 36 . . .

'

f ksi

c =

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

( )( )

0 40 0 40 8 32 . . .

'

f ksi

c =

=

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SLIDE 28

July 2005 3-28

Example 1

Stresses at service limit state after losses: Compression Due to effective prestress, permanent loads, and transient loads

( )( )( )

0 6 0 6 10 8 4 8 . . . . f ksi

c '

φw = = flange wid

w

th flange depth < 15 = = = 114 9 12 7 10 . . φ

( )( )( )

0 6 0 6 8 . . fc

'

φ φ

w w

=

Stress Summary

Example 1

4.8 ksi 2.519 ksi Compressive stress 0.268 ksi none Tensile stress Effective prestress and dead loads 4.2 ksi 3.670 ksi Compressive stress 0.200 ksi 0.030 ksi Tensile stress Release Limit Maximum Interior and Exterior Beams EXAMPLE 1 - BEAM STRESS SUMMARY

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SLIDE 29

July 2005 3-29

4.8 ksi 2.123 ksi Compressive stress 3.2 ksi 1.348 ksi Compressive stress Live load and half the sum effective prestress and dead loads 0.268 ksi 0.252 ksi Tensile stress Effective prestress, dead loads, and live load Limit Maximum Interior and Exterior Beams EXAMPLE 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)

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SLIDE 30

July 2005 3-30

Nominal Flexural Resistance

  • Without compression and non-prestressed

tension reinforcement

M A f d a

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

c A f A f A f b k A f d

ps pu s y s y c ps pu p

= + − +

' ' '

. 085

1

f β (5.7.3.1.1-4)

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SLIDE 31

July 2005 3-31

Stress in Prestressing Steel

  • For components with bonded tendons

f f k c d k f f

ps pu p py pu

= − ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = − ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ =

1 2 104 028 . . 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

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SLIDE 32

July 2005 3-32

Example 1 - Exterior Beam

Aps = area of prestressing steel = 10.850 in2 fpu = specified tensile strength of prestressing steel = 270 ksi fpy = specified yield strength of prestressing steel = 243 ksi As = 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 in bw = width of web = 8 in dp = 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 Beam

  • 1. Factored moments, Mu

Mu = (1.25)(2034 + 2053 + 250) + (1.5)(405) + (1.75)(3837) = 12744 k-ft

Example 1 - Exterior Beam

  • 2. Depth of compression block

k = 0.28 (low relaxation strands) dp = e + yt + tslab = 31.50 + 35.62 + 9 = 76.12 in For rectangular section behavior:

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

c A f A f A f k A f d

ps pu s y s y c ps pu p

= + − + = + − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =

' ' '

. . . . . . . . 085 10850 270 085 5 080 114 0 28 10850 270 7612 7 35

1

f b in β

( )( )

a c = = = β1 080 7 35 588 . . . in

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July 2005 3-33

Example 1 - Exterior Beam

  • 3. Stress in prestressing steel at nominal flexural resistance,

components with bonded tendons

( ) ( )

f f k c d

ps pu p

= − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 270 1 0 28 7 35 7612 262 70 . . . . ksi

Example 1 - Exterior Beam

  • 4. Factored flexural resistance

( )( )

M A f d a

n ps ps p

= − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 10850 262 70 7612 588 2 1 12 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

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July 2005 3-34

Reinforcement Limits

  • Maximum amount of prestressed and non-

prestressed reinforcement should satisfy

c d d A f d A f d A f A f

e e ps 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

Reinforcement Limits

  • 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

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July 2005 3-35

Cracking Moment

( )

M S f f M S S S f

cr c r cpe dnc c nc c r

= + − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤

1

5.7.3.3.2-1

Example 1 - Exterior Beam

The minimum amount of prestressed and non-prestressed reinforcement Mdnc = total unfactored dead load moment acting on the monolithic or non-composite section Snc = section modulus for the extreme fiber of the monolithic

  • r 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 rupture fcpe = compressive stress in concrete due to effective prestress forces only

Example

  • Reinforcement limits
  • Example 1
  • Exterior beam
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July 2005 3-36

Reinforcement Limits

Example 1 - Exterior Beam - Midspan

Example 1 - Exterior Beam

Aps = area of prestressing steel = 10.850 in2 fpu = specified tensile strength of prestressing steel = 270 ksi fpy = specified yield strength of prestressing steel = 243 ksi As = area of non-prestressed tension reinforcement = 0 in2 fy = yield strength of tension reinforcement = 60 ksi dp = distance from extreme compression fiber to the centroid

  • f the prestressing tendons = 76.12 in

c = distance from the extreme compression fiber to the neutral axis = 7.35 in de = corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement (in)

Example 1 - Exterior Beam

The maximum amount of prestressed and non-prestressed reinforcement

( )( )( ) ( )( )

d A f d A f d A f A f c d

e ps ps p s y s ps ps s y e

= + + = + + = = = 10850 262 70 7612 10850 262 70 7612 7 35 7612 010 . . . . . . . . . in

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July 2005 3-37

Example 1 - Exterior Beam

The minimum amount of prestressed and non-prestressed reinforcement Mdnc = 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

  • r 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 Beam

fcpe = 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 P A P S

cpe e e nc

= − = − − = e ksi 1526 4 1085 1526 4 3150 20157 37922 . . . . f f

r c

= = = 0 24 0 24 8 0 6788 . . .

'

ksi

Example 1 - Exterior Beam

( ) ( )( ) ( )

M S f f M S S

cr c r ce dnc c nc

= + − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 27751 0 6788 37922 49044 27751 20157 1 1 12 8800 k - ft . . Scfr = (27751)(0.6788) = 18837 k-in = 1570 k-ft 1.2Mcr = (1.2)(1570) = 1884 k-ft 1.33 (factored moment) = (1.33)(12744) = 16950 k-ft Lesser = 1884 k-ft Actual flexural resistance = 17382 k-ft > 1884 k-ft o.k.

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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 A

r s s

= (5.10.10.1-1)

Pretensioned Anchorage Zone

  • 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

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July 2005 3-39

Pretensioned Anchorage Zone

Example 1

Example 1

Factored 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 =

  • verall depth of precast member = 72 in

Pt = prestressing force at transfer = 1953.2 k Pr = (Pt)(0.04) = (1953.2)(0.04) = 78.13 k

h 4 72 4 18 0 = = . in

Example 1

Bursting resistance provided by vertical reinforcement in the ends of pretensioned beams at service limit state: P f A

r s s

= A P f P

s r s r

= = = = 20 7813 20 391 . . in

2

Using pairs of No. 4 bars, As = 0.40 in2, the number of pairs of bars required: 391 0 40 9 8 . . . =

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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

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July 2005 3-41

Development Length

f f ld

ps pe

Distance from free end of strand S t e e l s t r e s s transfer length

Transfer Length

  • 60 strand diameters – 5.11.4.1

Development Length l d

ps pe b

k f f

≥ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

2 3 d

(5.11.4.2-1) k = 1.6 for precast, prestressed beams

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July 2005 3-42

Example

  • Development length

Development and Transfer Length

Example 1 - Exterior Beam

Example 1 - Exterior Beam

TRANSFER LENGTH (5.11.4.1) db = nominal strand diameter = 0.6 in 60 db = (60)(0.6) = 36 in

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July 2005 3-43

Example 1 - Exterior Beam

BONDED STRAND (5.11.4.2) db = nominal strand diameter = 0.6 in fps = 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 f

ps pe b

− ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 2 3 16 262 70 2 3 14068 0 6 162 2 d = in . . . . .

Thank You

George Choubah, P.E. FHWA- Federal Lands Highway Bridge Office (703) 404-6244 George.choubah@fhwa.dot.gov