Flexure Design Sequence

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 3-1

Example 120'-0" Centerline bridge 1'-0" 18'-0" 18'-0" 1'-0" 9" AASHTO Type VI Beam 4'-9" 9'-6" 4'-9" 4'-9" 9'-6" 4'-9" TYPICAL SECTION July 2005 3-2

Simple spans: 120 feet Fully prestressed beams Bonded tendons Skew angle: 0 degrees Stress limit for tension in beam concrete (corrosion conditions): severe Deck concrete ' f = 5 ksi c Beam concrete ' f c = 8 ksi 15. ' 15. E = 33000w f = ( 33000)( 0145. ) 5 = 4074 ksi c E = 5154 ksi ' f = 7 ksi ci c E = 4821 ksi ci c Prestressing steel Strand type: 270 ksi low relaxation Strand diameter: 0.6 inch Cross-sectional area per strand: 0.217 in 2 July 2005 3-3

Prestressing steel f py = 0. 90 f = 243 ksi E = 28, 500 ksi ps pu Non-prestressed reinforcement f y = 60 ksi E = 29 000 ksi s, - 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 - 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 July 2005 3-4

- Interior Beam c) The average spacing of adjacent beams: 9.5 ft (9.5) (12) = 114 in Effective flange width = 114 in - 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 - 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 July 2005 3-5

- 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 - Exterior Beam (114)(0.7906) = 90.12" 72" 9" 76.50" 22.96" 17.16" C. G. Composite Section C. G. Beam C. G. Slab 53.54" 36.38" Non-composite Section Property A (in 2 ) I (in 4 ) y b (in) y t (in) S b (in 3 ) S t (in 3 ) w (k/ft) AASHTO Type VI Beams 1085 733320 36.38 35.62 20157 20587 1.130 Property I comp (in 4 ) y bc (in) y tc (in) y slab top (in) S bc (in 3 ) S tc (in 3 ) S slab top (in 3 ) Composite Section Interior Beam 1485884 53.54 18.46 27.46 27751 80503 68443 Exterior Beam 1485884 53.54 18.46 27.46 27751 80503 68443 July 2005 3-6

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 3-7

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) Beam Slab Rail Wearing Surface Interior Beam 2034 2137 250 405 Exterior Beam 2034 2053 250 405 Live Load (Including Dynamic Allowance): 3680 k-ft (Truck + Lane) 3080 k-ft (Tandem + Lane) Distribution of Live Load Interior Girder Exterior Girder Lever Rule Equations ( m included) Special Analysis Lever Rule Equations Special Analysis Bending Moment One Lane -- X -- X -- X Multi- Lane -- X -- -- X X One Lane -- X -- X -- X Shear Multi- Lane -- X -- -- X X July 2005 3-8

INTERIOR BEAM one lane loaded two lanes loaded one lane loaded - fatigue MOMENT 0.506 0.741 0.422 SHEAR 0.740 0.918 0.617 EXTERIOR BEAM one lane loaded two lanes loaded one lane loaded - fatigue Additional Investigation one lane loaded two lanes loaded one lane loaded - fatigue MOMENT 1.042 0.848 0.868 0.793 0.942 0.661 SHEAR 1.042 0.864 0.868 0.793 0.942 0.661 Load Combinations and Load Factors Load Strength I DC DW LL IM Max-1.25 Min-0.90 Max-1.50 Min-0.65 1.75 1.75 Strength II Max-1.25 Min-0.90 Max-1.50 Min-0.65 1.35 1.35 Limit State Service I 1.00 1.00 1.00 1.00 Service III 1.00 1.00 0.80 0.80 Fatigue ---- ---- 0.75 0.75 July 2005 3-9

Beam Stresses Due to dead load and live load Service III limit state (Crack Control) Mbeam + Mslab M + M + 08. M fbottom = S S Service I limit state f top M = beam b + M S t slab rail ws LL bc M + M + M + S rail ws LL tc Beam Stresses - Interior Beam Stresses at midspan due to dead load and live load. Extreme bottom of beam fiber (Service III Limit State): ( 2034 + 2137)( 12) [ 250 + 405 + ( 0. 8)( 2728) ]( 12) f bottom = = 3710. ksi (t) 20157 27751 Extreme top of beam fiber (Service I Limit State): ( 2034 + 2137)( 12) [ 250 + 405+ 2728]( 12) f top = + = 2. 936 ksi (c) 20587 80503 July 2005 3-10

- Exterior Beam Stresses at midspan due to dead load and live load. Extreme bottom of beam fiber (Service III Limit State): ( 2034 + 2053)( 12) [ 250 + 405 + ( 0. 8)( 3837) ]( 12) f bottom = = 4. 044 ksi (t) 20157 27751 Extreme top of beam fiber (Service I Limit State): ( 2034 + 2053)( 12) [ 250 + 405 + 3837]( 12) f top = + = 3052. ksi (c) 20587 80503 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 July 2005 3-11

Preliminary Strand Arrangement f f ten ten P Pe = + f A S b bottom P Pe 1 e fbottom = = P A S A S ften f P = 1 e A S bottom b b b Preliminary Strand Arrangement f pe = f pj estimated losses P e = (f pe )(A ps ) P No. Strands = P e Example Preliminary strand arrangement July 2005 3-12

Preliminary Strand Arrangement BEAM STRESSES Bottom fiber, maximum tension stress, all loads applied, Service III Limit State: Interior beam, f bottom = -3.710 ksi Exterior beam, f bottom = -4.044 ksi CONCRETE STRESS LIMIT Limit for tension, all loads applied, after all losses: f ten ' = 0. 0948 f = 0. 0948 8 = 0. 268 ksi c ESTIMATE NUMBER OF STRANDS f pt = total loss in the prestressing steel stress = 60 ksi (estimated) f pj = stress in the prestressing steel at jacking = (0.75)(270) = 202.5 ksi f pe = effective stress in the prestressing steel after losses = 202.5-60 = 142.5 ksi A ps = area of prestressing steel (per strand) = 0.217 in 2 July 2005 3-13

ESTIMATE NUMBER OF STRANDS P e = effective prestressing force in one strand = (f pe )(A ps ) = (142.5)(0.217) = 30.9 k A= area of non-composite beam = 1085 in 2 S b = section modulus, non-composite section, extreme bottom beam fiber = 20157 in 3 e = eccentricity of prestress force at midspan = -32 in (estimated) - Exterior Beam (114)(0.7906) = 90.12" 72" 9" 76.50" 22.96" 17.16" C. G. Composite Section C. G. Beam C. G. Slab 53.54" 36.38" - Interior Beam Estimated total prestress force required: ( ) ( ) P f f ten bottom = = 0. 268 2. 710 e ( ) = 13718. k 1 1 32 A Sb 1085 20157 Estimated number of strands required: P 13718. No. Strands = = = 44. 4 30. 9 P e July 2005 3-14

- Exterior Beam Estimated total prestress force required: ( 0. 268) ( 4. 044) 1 ( 32) P f f ten bottom = = 1 e A Sb 1085 20157 Estimated number of strands required: P 1504. 9 No. Strands = = = 48. 7 30. 9 P e = 1504. 9 k STRAND ARRANGEMENT - 50 strands At midspan of beam y = ( 13)( 2) + ( 13)( 4) + ( 13)( 6) + ( 11)( 8) 50 e = 4.88-36.38 = -31.50 in = 488. in At the ends of beam (harp 12 strands) ( 10)( 2) + ( 10)( 4) + ( 10)( 6) + ( 8)( 8) + ( 3)( 64) + ( 3)( 66) + ( 3)( 68) + ( 3)( 70) y = 50 = 19. 76 in e = 19.76-36.38 = -16.62 in July 2005 3-15

4 spa @ 2" 4 spa @ 2" 4 spa @ 2" At beam ends At beam centerline e = -16.62 in e = -16.77 in e = -20.46 in e = -24.14 in e = -27.82 in e = -31.50 in 8 strands 10 strands 3 strands 3 strands 3 strands 10 strands 3 strands 10 strands e = -31.50 in C. G. beam 6" 48'-0" centerline bearing 12'-0" Prestress Loss Article 5.9.5 Losses due to Elastic shortening Shrinkage Concrete creep Relaxation of steel July 2005 3-16

Elastic Shortening f pes = E E p ci f cgp (5.9.5.2.3a-1) Elastic Shortening For components of usual design, may calculate f cgp assuming stress in the prestressing steel 0.70f pu for low relaxation strands 0.65f pu for stress relieved strands Shrinkage Pretensioned members f psr = 17. 0 0150. H (5.9.5.4.2-1) H = average annual ambient relative humidity July 2005 3-17

Creep f = 12. 0f 7. 0 f 0 pcr cgp cdp (5.9.5.4.3-1) f cdp = 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 f pu July 2005 3-18

Relaxation Stress relieved strand f pr1 log ( 24. 0 t) fpj = 055. 10. 0 fpy f pj (5.9.5.4.4b-1) Relaxation Low relaxation strand f pr1 ( 24 0 t) log. fpj = 055. f 40. 0 fpy (5.9.5.4.4b-2) pj Relaxation Pretensioned members After transfer Stress relieved strand f 2 = 20. 0 0. 4 f 0. 2 f + f ( ) pr pes psr pcr (5.9.5.4.4c-1) July 2005 3-19

Relaxation Pretensioned members After transfer Low relaxation strand 30% of stress relieved strand f = 03200.. 04. f 0. 2 f + f 2 = 60. 012. f 006. f + f [ ( )] pes ( psr pcr ) pr pes psr pcr Prestress Loss Final total losses f pt = f pes + f psr + f pcr + f pr2 (5.9.5.1-1) f pt = f pes + f psr + f pcr + f pr1 + f pr2 Prestress loss Exterior beam Example July 2005 3-20

Prestress Loss Exterior Beam - Exterior Beam f pt = total loss in the prestressing steel stress (ksi) f pes = loss due to elastic shortening (ksi) f psr = loss due to shrinkage (ksi) f pcr = loss due to creep of concrete (ksi) f pr1 = loss due to relaxation of steel at transfer (ksi) f pr2 = 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 in 2 I = moment of inertia of non-composite section = 733320 in 4 I comp = moment of inertia of composite section = 1485884 in 4 - Exterior Beam ELASTIC SHORTENING (5.9.5.2.3a) f cgp = 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) M beam = moment due to weight of member = (2034)(12) = 24408 k-in P t = prestress force at transfer = (0.70)(270)(0.217)(50) = 2050.7 k E p = E ci = modulus of elasticity of prestressing steel = 28500 ksi modulus of elasticity of concrete at transfer = 4821 ksi July 2005 3-21

- Exterior Beam For components of the usual design, may calculate f cgp using a stress in the prestressing steel of 0.70 f pu. P ( P e) y M t t y beam fcgp = + + A I I 2050. 7 [( 2050. 7)( 3150. )]( 3150. ) ( 24408)( 3150. ) = + + 1085 733320 733320 = 36164. ksi f pes Ep = ( ) E f = cgp 28500 36164. = 2138. ksi 4821 ci - Exterior Beam SHRINKAGE (5.9.5.4.2) H = average annual ambient relative humidity = 70% ( ) [ ( )( )] f psr = 17. 0 0150. H = 17.0-0.150 70 = 650. ksi - Exterior Beam CREEP (5.9.5.4.3) f cdp = 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 f cdp should be calculated at the same section or at sections for which f cgp is calculated (ksi) M slab = moment due to slab and diaphragms = (2053)(12) = 24636 k-in M rail = moment due to curb and rail = (250)(12) = 3000 k-in M ws = moment due to wearing surface and FWS = (405)(12) = 4860 k-in July 2005 3-22

f cdp - Exterior Beam y= y y b = 488. 3638. = 3150. in y = y y = 488. 5354. = 4866. in comp ( M + M ) Mslab y y rail WS = I I = comp ( 24636)( 3150. ) ( 3000 + 4860)( 48. 66) 733320 f = 12. 0f 7. 0 f bc pcr cgp cdp comp ( )( ) ( )( ) 1485884 = 12. 0 36164. 7. 0 13156. = 3419. ksi = 13156. ksi - Exterior Beam RELAXATION AT TRANSFER (5.9.5.4.4b) t = time estimated in days from stressing to transfer = 2 days f pj = initial stress in the tendon at the end of stressing = 202.50 ksi f py = specified yield strength of prestressing steel = 243 ksi f pr1 ( 24 0 t) log. f pj = 055. fpj 40. 0 fpy log [( 24. 0)( 2) ] 202 50 =. 055. ( 202. 50) = 2. 41 ksi 40. 0 243 - Exterior Beam RELAXATION AFTER TRANSFER (5.9.5.4.4c) fpr2 = 03200. [. 04. fpes 02. ( fpsr + fpcr) ] = 0.3 20.0-0.4 2138. 02. 650. + 3419. = 099. ksi [ ( )( ) ( )( )] TOTAL LOSSES f pt = f pes + f psr + f pcr + f pr1 + f pr2 = 21.38 + 6.50 + 34.19 + 2.41 + 0.99 = 65.47 ksi (32.3%) f pt = f pes + f psr + f pcr + f pr2 = 21.38 + 6.50 + 34.19 + 0.99 = 63.06 ksi July 2005 3-23

- Exterior Beam EFFECTIVE STRESSES f pe = f pj - f pt f pj = 202.50 ksi f pe = 202.50-65.47 = 137.03 ksi f pe = f pt - f pt f pj = 202.50 + 2.41 = 204.91 ksi f pt = 202.50 ksi f pe = 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 f pe = f pj total losses P e = (f pe )(A ps ) ften fbottom P = 1 e A S P No. Strands = b P e July 2005 3-24

Example Strand arrangement Strand Arrangement f bottom = extreme bottom beam fiber tensile stress from applied loads interior beam = -3.710 ksi exterior beam = -4.044 ksi f ten = allowable tensile stress in concrete after losses = 0.268 ksi f pj = stress in prestressing steel at jacking = (0.75)(270) = 202.50 ksi f pe = 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 July 2005 3-25

A = area of non-composite beam = 1085 in 2 S b = section modulus of non-composite section, extreme bottom beam fiber = 20157 in 3 A ps = area of prestressing steel (per strand) = 0.217 in 2 e = eccentricity of prestress force at midspan = -31.50 in - Exterior Beam Total force required in strands: ( ) ( ) P f f ten bottom = = 0. 268 4. 044 e ( ) = 1519. 9 k 1 1 2146. A Sb 1085 20157 Number of strands required: Effective prestressing force in one strand after all losses = P e = (f pe )(A ps ) = (140.96)(0.217) = 30.5 k P 1519. 9 No. strands = = = 49. 8 305. P e 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 3-26

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 Temporary stresses before losses: Tension ' 0. 0948 f ci = 0. 0948 7 = 0. 251 ksi > 0. 2 ksi Compression ' 060. f = ci 060. 7 = 42. ksi ( )( ) Stresses at service limit state after losses: Tension ' 0. 0948 f c = 0. 0948 8 = 0. 268 ksi Stresses at service limit state after losses: Compression Due to effective prestress and permanent loads ' 045. f = 045. 8 = 36. ksi c ( )( ) Due to live load and one-half the sum of effective prestress and permanent loads ' 040. f c = ( 040. )( 8) = 32. ksi July 2005 3-27

Stresses at service limit state after losses: Compression Due to effective prestress, permanent loads, and transient loads ' 06. φw f c = ( 06. )( φw)( 8) flange width 114 = = 12. 7 < 15 flange depth 9 φw = 10. ' 06. φ w f = 06. 10. 8 = 48. ksi c ( )( )( ) Stress Summary EXAMPLE 1 - BEAM STRESS SUMMARY Interior and Exterior Beams Maximum Limit Release Tensile stress 0.030 ksi 0.200 ksi Compressive stress 3.670 ksi 4.2 ksi Effective prestress and dead loads Tensile stress none 0.268 ksi Compressive stress 2.519 ksi 4.8 ksi July 2005 3-28

EXAMPLE 1 - BEAM STRESS SUMMARY Interior and Exterior Beams Effective prestress, dead loads, and live load Tensile stress Compressive stress Live load and half the sum effective prestress and dead loads Compressive stress Maximum 0.252 ksi 2.123 ksi 1.348 ksi Limit 0.268 ksi 4.8 ksi 3.2 ksi 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 M r = φ M n (5.7.3.2.1-1) φ = 1.00 M n = Nominal flexural resistance July 2005 3-29

Nominal Flexural Resistance Without compression and non-prestressed tension reinforcement a Mn = Apsfps dp 2 f ps = Average stress in prestressing steel Strength Limit State For practical design, use rectangular compressive stress distribution Depth of compressive stress block a = 1 c Stress in Prestressing Steel For rectangular section behavior ' ' Apsfpu+ Af s y Af s y c = b k A f ' 085. f β + c 1 ps d (5.7.3.1.1-4) pu p July 2005 3-30

Stress in Prestressing Steel For components with bonded tendons f f k c ps = pu 1 d (5.7.3.1.1-1) p f k = 2104. f py pu (5.7.3.1.1-2) = 028. for low relaxation strand Example Nominal flexural resistance Exterior beam Nominal Flexural Resistance - Exterior Beam - Midspan July 2005 3-31

- Exterior Beam A ps = area of prestressing steel = 10.850 in 2 f pu = specified tensile strength of prestressing steel = 270 ksi f py = specified yield strength of prestressing steel = 243 ksi A s = area non-prestressed tension reinforcement = 0 in 2 ' A s = area of compression reinforcement = 0 in 2 f y = yield strength of tension reinforcement = 60 ksi ' f = yield strength of compression reinforcement = 60 ksi y ' f c = compressive strength of concrete = 5 ksi b = width of compression flange = 114 in b w = width of web = 8 in d p = distance from extreme compression fiber to the centroid of the prestressing tendons (in) 1 = stress block factor = 0.80 - Exterior Beam 1. Factored moments, M u M u = (1.25)(2034 + 2053 + 250) + (1.5)(405) + (1.75)(3837) = 12744 k-ft - Exterior Beam 2. Depth of compression block k = 0.28 (low relaxation strands) d p = e + y t + t slab = 31.50 + 35.62 + 9 = 76.12 in For rectangular section behavior: ' ' A f + A f A f ps pu s y s y c = k A f ' pu 085. f β b+ c 1 ps d p ( 10850. )( 270) + 0 0 = = 735. in (. )( )(. )( ) + 270 085 5 080 114 ( 028. )( 10850. ) 7612. ( )( ) a= β 1 c= 080. 735. = 588. in July 2005 3-32

- Exterior Beam 3. Stress in prestressing steel at nominal flexural resistance, components with bonded tendons f = f k c ps pu ( ) ( ) 1 = d p 270 1 0 28 735.. 7612 = 262. 70 ksi. - Exterior Beam 4. Factored flexural resistance M = A f ps a d p 2 n ps = ( )( ) 588. 1 10850. 262. 70 7612. = 17382 k - ft 2 12 M r = (1.0)(17382) = 17382 k-ft > 12744 k-ft o.k. Reinforcement Limits Amount of prestressed and non-prestressed reinforcement Maximum Minimum July 2005 3-33

Reinforcement Limits Maximum amount of prestressed and nonprestressed reinforcement should satisfy c 042. (5.7.3.3.1-1) d d e e A f d = A f + A f d + A f ps ps p s y s ps ps s y (5.7.3.3.1-2) Effective Depth b de d s dp Aps A s Reinforcement Limits Article 5.7.3.3.2 Minimum amount of prestressed and nonprestressed tensile reinforcement Adequate to develop a factored flexural resistance at least equal to the lesser of 1.2 M cr 1.33 factored moments July 2005 3-34

Cracking Moment ( ) Sc Mcr = Sc fr + fcpe Mdnc 1 Sf S 5.7.3.3.2-1 nc c r - Exterior Beam The minimum amount of prestressed and non-prestressed reinforcement M dnc = total unfactored dead load moment acting on the monolithic or non-composite section S nc = section modulus for the extreme fiber of the monolithic or non-composite section where tensile stress is caused by externally applied loads S c = section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads f r = modulus of rupture f cpe = compressive stress in concrete due to effective prestress forces only Reinforcement limits Exterior beam Example July 2005 3-35

Reinforcement Limits - Exterior Beam - Midspan - Exterior Beam A ps = area of prestressing steel = 10.850 in 2 f pu = specified tensile strength of prestressing steel = 270 ksi f py = specified yield strength of prestressing steel = 243 ksi A s = area of non-prestressed tension reinforcement = 0 in 2 f y = yield strength of tension reinforcement = 60 ksi d p = distance from extreme compression fiber to the centroid of the prestressing tendons = 76.12 in c = distance from the extreme compression fiber to the neutral axis = 7.35 in d e = corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement (in) e - Exterior Beam The maximum amount of prestressed and non-prestressed reinforcement A f d + A f d ps ps p s y s ( 10850. )( 262. 70)( 7612. ) + 0 d = = = 7612. in e A f + A f ps ps s y ( 10850. )( 262. 70) + 0 c 735. = = 010. d 7612. July 2005 3-36

- Exterior Beam The minimum amount of prestressed and non-prestressed reinforcement M dnc = total unfactored dead load moment acting on the monolithic or non-composite section = 2034 + 2053 = 4087 k-ft = 49044 k-in S nc = section modulus for the extreme fiber of the monolithic or non-composite section where tensile stress is caused by externally applied loads = 20157 in 3 S c = section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads = 27751 in 3 f r = modulus of rupture (ksi) - Exterior Beam f cpe = P e = 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) effective prestress force = 1526.4 k ' f = 0. 24 f = 0. 24 8 = 0. 6788 ksi f r cpe c ( 1526 4)( 3150) Pe Pe e 1526. 4.. = = A S 1085 20157 nc = 37922. ksi - Exterior Beam M S ( f f ) M S c = + cr c r ce dnc 1 Snc = ( )( + ) ( ) 27751 0 6788 37922 49044 27751 20157 1 1.. 12 = 8800 k -ft S c f r = (27751)(0.6788) = 18837 k-in = 1570 k-ft 1.2M cr = (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. July 2005 3-37

Pretensioned Anchorage Zone Bursting resistance provided by vertical reinforcement in ends of pretensioned beams at service limit state Pr = fsas (5.10.10.1-1) Not less than 4% prestress force at transfer Total vertical reinforcement located within a distance h/4 from end of beam 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 Pretensioned anchorage zone Example July 2005 3-38

Pretensioned Anchorage Zone Factored Bursting Resistance (5.10.10.1) f s = stress in steel not exceeding 20 ksi A s = total area of vertical reinforcement located within the distance h/4 from the end of the beam (in 2 ) h = overall depth of precast member = 72 in P t = prestressing force at transfer = 1953.2 k P r = (P t )(0.04) = (1953.2)(0.04) = 78.13 k h 72 = = 18. 0 in 4 4 Bursting resistance provided by vertical reinforcement in the ends of pretensioned beams at service limit state: P = f A r s s A P P 7813. r r s = = = = 391. in 2 f 20 20 s Using pairs of No. 4 bars, A s = 0.40 in 2, the number of pairs of bars required: 391. = 98. 040. July 2005 3-39

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 3-40

Steel stress Development Length l d transfer length f ps f pe Distance from free end of strand Transfer Length 60 strand diameters 5.11.4.1 Development Length 2 3 l d k f ps f pe d b (5.11.4.2-1) k = 1.6 for precast, prestressed beams July 2005 3-41

Example Development length Development and Transfer Length - Exterior Beam - Exterior Beam TRANSFER LENGTH (5.11.4.1) d b = nominal strand diameter = 0.6 in 60 d b = (60)(0.6) = 36 in July 2005 3-42

- Exterior Beam BONDED STRAND (5.11.4.2) d b = nominal strand diameter = 0.6 in f ps = average stress in the prestressing steel at the time for which the nominal resistance of the member is required = 262.70 ksi f pe = effective stress in the prestressing steel after losses = 140.68 ksi k = 1.6 k f 2 f ( ) ( ) ( ) 2 16 262 70 = 3 d =.. 140 68 0 6 162 2 3... in ps pe b Thank You George Choubah, P.E. FHWA- Federal Lands Highway Bridge Office (703) 404-6244 George.choubah@fhwa.dot.gov July 2005 3-43