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 Example 1 120'-0" Example 1 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
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 ' = f 5 ksi c ( )( ) 15 . = 15 . ' = = E 33000 w f 33000 0145 . 5 4074 ksi c c Beam concrete ' = f c 8 ksi = E 5154 ksi c = f ' 7 ksi ci E = 4821 ksi ci Example 1 • 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
Example 1 Prestressing steel f py = = 090 . f 243 ksi pu E = 28 500 , ksi ps Non-prestressed reinforcement f y = 60 ksi = E 29 000 , ksi s 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 July 2005 3-4
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 July 2005 3-5
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. Slab 9" 22.96" C. G. Composite Section 17.16" 76.50" 72" C. G. Beam 53.54" 36.38" Example 1 Non-composite Section Composite Section AASHTO Interior Exterior Property Property Type VI Beams Beam Beam A (in 2 ) 1085 I comp (in 4 ) 1485884 1485884 I (in 4 ) 733320 y bc (in) 53.54 53.54 y b (in) 36.38 y tc (in) 18.46 18.46 y t (in) 35.62 y slab top (in) 27.46 27.46 S b (in 3 ) 20157 S bc (in 3 ) 27751 27751 S t (in 3 ) 20587 S tc (in 3 ) 80503 80503 w (k/ft) 1.130 S slab top (in 3 ) 68443 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) Interior Beam Exterior Beam Beam 2034 2034 Slab 2137 2053 Rail 250 250 Wearing Surface 405 405 Live Load (Including Dynamic Allowance): 3680 k-ft (Truck + Lane) 3080 k-ft (Tandem + Lane) Distribution of Live Load Bending Moment Shear One Multi- One Multi- Lane Lane Lane Lane Lever Rule -- -- -- -- Equations (“m” Interior X X X X included) Girder Special -- -- -- -- Analysis Lever Rule X -- X -- Equations Exterior -- X -- X Girder Special X X X X Analysis July 2005 3-8
Example 1 MOMENT SHEAR INTERIOR BEAM one lane loaded 0.506 0.740 two lanes loaded 0.741 0.918 one lane loaded - fatigue 0.422 0.617 Example 1 MOMENT SHEAR EXTERIOR BEAM one lane loaded 1.042 1.042 two lanes loaded 0.848 0.864 one lane loaded - fatigue 0.868 0.868 Additional Investigation one lane loaded 0.793 0.793 two lanes loaded 0.942 0.942 one lane loaded - fatigue 0.661 0.661 Load Combinations and Load Factors Limit State Load Strength I Strength II Service I Service III Fatigue Max-1.25 Max-1.25 DC 1.00 1.00 ---- Min-0.90 Min-0.90 Max-1.50 Max-1.50 DW 1.00 1.00 ---- Min-0.65 Min-0.65 LL 1.75 1.35 1.00 0.80 0.75 IM 1.75 1.35 1.00 0.80 0.75 July 2005 3-9
Beam Stresses • Due to dead load and live load Service III limit state (Crack Control) + + + 08 . M M M M M = − − beam slab rail ws LL f bottom S S b bc Service I limit state + + + M M M M M = beam slab + rail ws LL f top S S t 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): [ ] ( ( )( ) ( )( ) ) + + 2034 + 2137 12 250 405 08 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
Example 1 - Exterior Beam Stresses at midspan due to dead load and live load. Extreme bottom of beam fiber (Service III Limit State): [ ] ( ( )( ) ) ( )( ) + 250 + 405 + 08 3837 . 12 2034 2053 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 P Pe = − + f f ten bottom A S b ⎛ ⎞ P Pe 1 e − = − = ⎜ − ⎟ f f P A ten bottom ⎝ ⎠ A S S b b − f f ten bottom = P ⎛ ⎞ 1 e ⎜ − ⎟ ⎝ ⎠ A S b Preliminary Strand Arrangement f pe = f pj – estimated losses P e = (f pe )(A ps ) P . Strands = No P e Example • Preliminary strand arrangement July 2005 3-12
Preliminary Strand Arrangement Example 1 Example 1 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 0 0948 . f ' 0 0948 8 . 0 268 . ksi ten c Example 1 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
Example 1 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) Example 1 - Exterior Beam (114)(0.7906) = 90.12" C. G. Slab 9" 22.96" C. G. Composite Section 17.16" 76.50" 72" C. G. Beam 53.54" 36.38" Example 1 - Interior Beam Estimated total prestress force required: ( ) ( ) − = − 0 268 . − − 2 710 . f f P = = 13718 . k ten bottom ( ) ⎛ ⎞ ⎡ − − ⎤ 1 e 1 32 − ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ 1085 20157 A S ⎣ ⎦ b Estimated number of strands required: P 13718 . Strands = = = No . 44 4 . P 30 9 . e July 2005 3-14
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