Downloaded 475 times
![34
Resistance Factor
1 + COV Q
2
λR (Σγ i Qi )
1 + COV R
2
φ=
{ [(
Q exp βT ln 1 + COVR 1 + COVQ
2
)( 2
)]}
Dead Load Factors
γD = 1.25
λQD = 1.05
COVQD = 0.1
QD
Live Load Factors γD +γL
γL = 1.75 QL 1.4167
λQL = 1.15
FS = ≅
QD φ
COVQL = 0.2 φ
+ 1
QL ](https://image.slidesharecdn.com/lrfdshortcoursepresentation-12526274988162-phpapp03/85/Lrfd-Short-Course-Presentation-34-320.jpg)
![80
Base Resistance for Compressive
Loading
Rock with 70 < RQD < 100 (FHWA, Eqn 11.6)
qmax ( MPa) = 4.83[qu ( MPa)]
0.51
qmax = 44.9 Mpa (469 tsf)
[ ]q
Jointed Rock (FHWA, Eqn, 11.7)
0.5
(
qmax = s + ms + s0.5
)
0.5
u
qmax = 49.5 Mpa (517 tsf)
Detroit Experience
qmax = qall (FS ) = 120(2.5) qmax = 28.7Mpa (300 tsf)](https://image.slidesharecdn.com/lrfdshortcoursepresentation-12526274988162-phpapp03/85/Lrfd-Short-Course-Presentation-80-320.jpg)
More Related Content
![[프레스세미나 1주제] 16년도 프레스업종 세미나 자료](https://cdn.slidesharecdn.com/ss_thumbnails/116-160518004446-thumbnail.jpg?width=640&height=640&fit=bounds)
Lrfd Short Course Presentation
- 1. 1 Load and ResistanceFactor Design (LRFD)- Deep Foundations Donald C. Wotring, Ph.D., P.E. February 2009
- 2. 2 Presentation • This presentationis intended as a detailed internal short-course with design examples. It will also be used as a brown-bag lunch presentation, but with less detail covered.
- 3. 3 Presentation Goals 1. BasicDifferences between ASD and LRFD 2. Fundamentals of LRFD 3. Application of LRFD to Deep Foundations
- 4.
- 5. 5 Load = Resistance?????
- 6.
- 7. 7 Examples of Uncertainty • Material dimensions and location • Material strength • Failure mode and prediction method • Long-term material performance • Material weights • Prediction of potential transient loads • Load analysis and distribution methods • General uncertainty with structure function
- 8. 8 Allowable Stress Design (ΣDL + ΣLL ) ≤ Rn / FS ADVANTAGES DISADVANTAGES Simplistic Inadequate account of variability Accustomed to use Stress not a good measure of resistance Factor of Safety is subjective No risk assessment
- 9. 9 Definition - LimitState • A Limit State is a condition beyond which a structural component ceases to satisfy the provisions for which it is designed.
- 10. 10 Definition - Resistance •Resistance is a quantifiable value that defines the point beyond which the particular limit state under investigation, for a particular component, will be exceeded.
- 11. 11 Resistance Can BeDefined in Terms of • Load/Force • Stress (normal, shear, torsional) • Number of cycles • Temperature • Strain • etc.
- 12. 12 AASHTO LRFD BridgeDesign Specifications • 4th Edition, 2007 • 2008 Interim Revisions
- 13. 13 Load and ResistanceFactor Design Σηiγ i Qi ≤ φRn = Rr ADVANTAGES DISADVANTAGES Load factor applied to each load More complex than ASD combination Old habits Types of loads have different levels of Requires availability of statistical data uncertainty Resistance factors vary Accounts for variability Uniform levels of safety Risk assessment
- 14. 14 LRFD Equation Σηiγ i Qi ≤ φRn = Rr ηi Load modifier: factor relating to ductility, redundancy, and operational importance γi Load factor: statistically based multiplier applied to force effects Qi Force effect φ Resistance factor: statistically based multiplier applied to nominal resistance Rn Nominal resistance Rr Factored Resistance
- 15. 15 Limit States • Strength– strength and stability sufficient to resist the specified statistically significant load combination during design life I – Normal vehicular use without wind II – Owner-specified design vehicle without wind III – Bridge exposed to wind velocity exceeding 55 mph (WS) IV – Very high dead load to live load ratio (when DL/LL > 7, construction) V – Normal vehicular use with 55 mph wind (WL)
- 16.
- 17. 17 Limit States • Service– Restrictions on stress, deformation, and crack width under regular service conditions I– Normal operational use with 55 mph wind. Also related to deflection control in tunnels, slopes, etc. II – Yielding of steel structures and slip of slip- critical connections due to vehicular live load III – Longitudinal analysis relating to tension in prestressed concrete IV – Relating to crack control from tension in concrete columns
- 18.
- 19. 19 Limit States • Extreme– Structural survival during a major event (earthquake, flood, vessel impact, ice, etc.) I – Earthquake II – Other events
- 20.
- 21. 21 Limit States • Fatigue– Limit crack growth under repetitive loads to prevent fracture during design life
- 22.
- 23. 23 Load Modifier Σηiγ i Qi ≤ φRn = Rr ηi = η Dη Rη I ≥ 0.95 When maximum value of γi is appropriate 1 ηi = ≤ 1.0 When minimum value of γi is appropriate η Dη Rη I ηD Ductility load modifier ηR Redundancy load modifier ηD Operational importance load modifier
- 24. 24 Load Modifier -Ductility Strength Limit State ηD > 1.05 Non-ductile components and connections = 1.00 Conventional designs according to AASHTO specs < 0.95 Ductility enhancing measures specified beyond AASHTO specs All other Limit States ηD = 1.00
- 25. 25 Load Modifier -Redundancy Strength Limit State ηR > 1.05 Non-redundant members = 1.00 Conventional redundancy < 0.95 Exceptional redundancy All other Limit States ηR = 1.00
- 26. 26 Load Modifier –Operational Importance Strength Limit State ηI > 1.05 Important bridges = 1.00 Typical bridges < 0.95 Relatively less important bridges All other Limit States ηI = 1.00
- 27. 27 Loads Σηiγ i Qi ≤ φRn = Rr
- 28. 28 Load Combinations andLoad Factors
- 29. 29 Load Factors forPermanent Loads Destabilizing Stabilizing
- 30. 30 Loading Summary Σηiγ i Qi ≤ φRn = Rr Load modifier – Usually = 1.0 Load factor Develop a governing load combination for each of: Load - Strength - Service - Extreme - Fatigue
- 31. 31 Probability Review Normal Distribution
- 32. 32 Probability of Failure Reliability Index, β No. of standard deviations that the mean value is above 0
- 33. 33 Probability ofFailure – Reliability Index Structure Pile Redundancy β Pf 2.33 1.0% 3.00 0.13%
- 34. 34 Resistance Factor 1 + COV Q 2 λR (Σγ i Qi ) 1 + COV R 2 φ= { [( Q exp βT ln 1 + COVR 1 + COVQ 2 )( 2 )]} Dead Load Factors γD = 1.25 λQD = 1.05 COVQD = 0.1 QD Live Load Factors γD +γL γL = 1.75 QL 1.4167 λQL = 1.15 FS = ≅ QD φ COVQL = 0.2 φ + 1 QL
- 35.
- 36.
- 37. 37 Does a lowresistance value = Inefficient Design method? COV = 0.4 λ= 1.0 φ= 0.44 φ/λ = 0.44 COV = 0.4 λ= 1.5 φ= 0.67 φ/λ = 0.44 COV = 0.58 λ= 1.5 φ= 0.44 φ/λ = 0.29 Overpredictive Underpredictive (built in FS)
- 38. 38 Efficiency of theMethod φ/λ λ FS(λ)
- 39. 39 Summary - Wheredo we stand? Σηiγ i Qi ≤ φRn = Rr Resistance factor based on probability of failure for different methods of estimating the resistance.
- 40. 40 Limit States asApplied to Deep Foundations • AASHTO, Section 10.5 • Service Limits • Strength Limits • Extreme Limits
- 41. 41 Service Limit States φ = 1.0 • Settlements – limitation to be compared with costs of designing structure to tolerate more movement or maintenance (jacking and shimming bearings) • Horizontal movements – top of foundation and abutment movements based on tolerance of structure (bridge seat, bearing width, structure type, etc.) • Overall stability – global slope stability of earth slopes • Scour at design flood – Section 2.6.4
- 42. 42 Extreme Limit States • Scour – Check flood (Section 2.6.4) • Earthquake • Liquefaction • Ice • Vehicle or Vessel Impact φ = 1.0 general φ = 0.8 uplift
- 43. 43 Strength Limit States– Driven Piles, Drilled Shaft, and Micropile • Axial compression resistance for single pile and pile group • Uplift resistance of single pile and pile group • Pile punching failure into weaker underlying stratum • Single pile and pile group lateral resistance • Constructability, including pile drivability As part of strength limit state, the effects of downdrag, soil setup/relaxation, and buoyancy should be evaluated.
- 44. 44 Strength Limit ResistanceFactors • Presented as a function of soil type (sand, clay). Sand = drained shear strength and Clay = undrained shear strength!!!!! • β = 3.5 (Pf of 1 in 5,000) • Wave equations are for EOD only, if used for BOR, the resistance values need to be lowered. “In general, dynamic testing (signal matching) should be conducted to verify the nominal pile resistance at BOR in lieu of driving formulas.” • Don’t reduce skin friction for uplift calcs. The resistance factor accounts for this. • A load factor of 1.0 should be used for pile drivability analysis. • The ENR news formula has had the FS=6 removed.
- 45.
- 46.
- 47.
- 48. 48 What is adifficulty (driven piles)?
- 49. 49 Pile Length Estimatefor Contract Documents • Static analysis is only usually used to establish the pile length estimate for contract documents. Field testing (e.g., PDA w/ CAPWAP) is used for driving criteria. φdyn Rndr = φstat Rnstat
- 50. 50 MDOT BridgeDesign Manual (7.03.09) Nominal Driving Resistance Values, Rndr Steel H-Piles Cast-in-place Concrete Piles Pile Type Rndr (k) Pile Type Rndr (k) HP10x42 300 12” O.D., 0.25” 350 HP10x57 450 14” O.D., 0.312” 400 HP12x53 400 14” O.D., 0.438” 500 HP12x74 600 HP12x84 650 Timber Piles HP14x73 600 Pile Type Rndr (k) HP14x89 700 Timber 150 HP14x102 800 HP14x117 900
- 51. 51 Structural Compressive Resistance- Steel Pn = 0.66λ Fy A λ < 2.25 If fully embedded, 0.88 Fy A λ=0 Pn = λ > 2.25 λ 2 kL Fy λ = Euler Equation rπ E
- 52. 52 Structural PileResistance Values Resistance during pile driving φ = 1.0 Axial resistance for compression subject Combined axial and flexural to damage where pile type is required resistance for undamaged pile H-piles φ = 0.5 Axial H-piles φ = 0.6 Pipe piles φ = 0.6 Axial pipe piles φ = 0.7 Axial resistance for compression not Axial pipe piles φ = 1.0 subject to damage H-piles φ = 0.6 Pipe piles φ = 0.7
- 53.
- 54.
- 55. 55 Summary • LRFD –statistically based method to account for the probability of failure ▫ Compared with ASD ▫ Limit states and resistance ▫ Load factors and combinations ▫ Resistance factors
- 56. 56 Example 1 -Estimate Pile Length • Pier Factored Load = γQ = 1.25(3640) = 4550 kips Fill γ= 130 pcf 10’ Clay γ= 125 pcf • Assume PDA w/ CAPWAP φdyn = 0.65 su = 2.5 ksf 120’ •Driven: Rr = φdynRndr Pile Type Rndr (k) Rr (k) #Piles 12” O.D., 0.25” 350 227.5 20 14” O.D., 0.312” 400 260 18
- 57. 57 Pile Caps 12-inch Pipe Piles 14-inch Pipe Piles 49” 112” 42” 138” 49” 42” 259” 180”
- 58. 58 Example 1 -Estimate Pile Length Assume PDA w/ CAPWAP φdyn = 0.65 Fill γ= 130 pcf 10’ Assume λ-method φstat = 0.40 Clay γ= 125 pcf φdyn su = 2.5 ksf 120’ Rnstat = Rndr = 1.625Rndr φstat Pile Type Rndr (k) Rstat (k) 12” O.D., 0.25” 350 570 14” O.D., 0.312” 400 650
- 59. 59 Depth Summary Qs = Rnstat (kips) 0 200 400 600 800 1000 40 50 12 OD 60 14 OD 70 Depth (ft) 80 90 100 110 120 130 140
- 60. 60 Settlement • Unfactored Pier Load 3640 kips • Service Factored Load = γQ = 1.0(3640) = 3640 kips Stress at top of Piles σ = 3640k / 172.5 ft 2 = 21.1ksf 2D/3 = 73’ I Stress at 2D/3 A 17’ 0.96 σ = 3640k / 2422.5 ft 2 = 1.5ksf B 20’ 0.66 C 20’ 0.36 Clay Assumptions OCR = 1.2 Cc = 0.2 Cr/Cc = 0.1 eo = 0.5 Cr = 0.02
- 61. 61 Settlement Point σ’vo I ∆σ σ’vf σ’p Ho/1+eo Sp (ksf) (ksf) (ksf) (ksf) (in) (in) A 5.8 0.96 1.44 7.24 6.96 136 0.68 B 6.9 0.66 0.99 7.89 8.28 160 0.19 C 8.2 0.36 0.54 8.74 9.84 160 0.09 Σ 0.96 If σ’vf < σ’p If σ’vf > σ’p Ho σ 'vf Ho C σ ' p σ 'vf Sp = Cr log σ ' Sp = C σ ' + log σ ' Cc log r 1 + eo vo 1 + eo c vo p
- 62. 62 Example 2 Strength VLimit – Rigid Cap Model Applied Factored Loads Fx = 38.4 kips Fy = 109.1 kips Fz = 3,594.0 kips Mx = 3,196.5 k-ft My = -8,331.9 k-ft Loose Sand
- 63. 63 Calculate Pile AxialLoads Fz M x yi M y xi 3594 3196.5(1.5) − 8331.9(−5) Pi = + n + n P = + + = 243 n 14 ∑ yi2 ∑ xi2 20 225 1000 i =1 i =1 Fz = 3,594.0 kips Mx = 3,196.5 k-ft My = -8,331.9 k-ft n = 20 piles xi = -60 in (-5 ft) yi = 18in (1.5 ft) 36 in Σxi2 = 1,000 ft2 Σyi2 = 225 ft2 60 in s2 ( ) ∑ x (row) = 12 n n 2 − 1 2 i
- 64. 64 Lateral Loading –LPILE Pile CTC P-Multiplier, Pm Spacing Row 1 Row 2 Rows 3 (loading and direction) higher 3B 0.7 0.5 0.35 5B 1.0 0.85 0.7
- 65. 65 Lateral Loading –LPILE Row Pm Hy (k) Mm (k-in) 1 0.35 4.5 -340 2 0.35 4.5 -340 3 0.5 5.9 -390 4 0.7 7.2 -450 Σ 110.5
- 66. 66 Lateral Loading –LPILE Row Pm Hz (k) Mm (k-in) 1 0.7 1.8 -75 2 0.7 1.8 -75 3 0.7 1.8 -75 4 0.85 2.0 -80 5 1.0 2.2 -90 Σ 38.4
- 67. 67 Pile Loading Summary Maximum Shear (Row 4 piles) 7.2 kips Maximum axial load in any pile (Pile 20) 327 kips Maximum combined loading (Pile 20) Pu 327 kips Mux -37.5 k-ft Muy -7.5 k-ft Alternative to rigid cap model, use FBPier
- 68. 68 Drivability Evaluation(GRL Weap) Require a PDA/CAPWAP Field Evaluation φdyn = 0.65 Nominal must be > 327/0.65=503 kips Nominal Driving Resistance • HP12x53 • Delmag D 12-32 •Rndr = 550 kips at 120bpf • 10 bpi = 5 b/0.5inch Factored Driving Resistance Rrdr = φdynRndr = 0.65(550) = 358 kips
- 69. 69 Drivability Check Driving Stresses SteelPiles in Compression or Tension σ DR = 0.9ϕ DA Fy ϕ DA = 1.0 σ DR = 0.9(1.0)50 = 45ksi > 37.5
- 70. 70 Geotechnical Resistance -Static Estimate the Depth of Penetration φdyn Rn = 358kips = φstat Rnstat Use SPT method φ = 0.3 Rnstat = 358/0.3 = 1193 kips !!! Very long piles would be required if all loose sand!! For our geology, the piles would end bear on till or bedrock and develop full capacity within a few feet of penetration (usual for H-piles). End bearing on rock (φ = 0.45). Pile tips required and driving resistance and criteria will be based on dynamic testing in the field (PDA/CAPWAP).
- 71. 71 Structural Resistance -Compression Nominal Compressive Resistance (Section 6.9.4.1) Pn = 0.66λ Fy A Compression only, damage likely φ = 0.5 λ =0 Fully embedded Compression only, damage unlikely Pn = Fy A = 50(15.5) = 775kips φ = 0.6 Combined Factored Compressive Resistance φ = 0.7 Pr = φPn = 0.5(775) = 387.5kips Good for lower portion of pile where damage is more likely Pr = φPn = 0.7(775) = 542.5kips combined
- 72. 72 Structural Resistance -Shear Nominal Shear Resistance (Section 6.10.9.2) Vn = 0.58CFy Dt w HP12x53 Check shear buckling ratio D = 11.78 in D Ek 5(29,000) tw = 0.435 in = 27.1 < 60.3 1.12 = 1.12 = 60.3 tw Fy 50 C = 1.0 Vn = 0.58(1.0)50(11.78)0.435 = 148.6kips Factored Shear Resistance Vr = φVn = 1.0(148.6) = 148.6kips
- 73. 73 Structural Resistance –Flexural Nominal Flexural Resistance (Section 6.12.2.2) M n = Fy z zx 74.0 in3 zy 32.2 in3 M nx = 50(74) = 3700 k-in M ny = 50(32.2) = 1610 k-in Factored Flexural Resistance M r = φM n φ = 1.0 M rx = 1.0(3700) = 3700 k-in M ry = 1.0(1610) = 1610 k-in
- 74. 74 Combined Loading Nominal CombinedLoading(Section 6. 9.2.2) Pu Pu M ux M uy < 0.2 + + ≤ 1.0 Pr 2 Pr M rx M ry Pu Pu 8 M ux M uy ≥ 0.2 + + ≤ 1.0 Pr Pr 9 M rx M ry
- 75. 75 Combined Loading Pu 327 kips Pr 542.5 kips Pu/Pr 0.6 > 0.2 Mux -37.5 k-ft Mrx 308.3 k-ft Mrx 0.12 Muy -7.5 k-ft Mry 134.2 k-ft Mry 0.06 Pu 8 M ux M uy = 0.6 + 8 (0.12 + 0.06) = 0.76 < 1.0 + + Pr 9 rx M ry M 9
- 76. 76 Summary – StrengthV Limit State Structural Performance Ratios Driven 327/358 = 0.91 Geotechnical N/A Axial Compression only 327/387.5 = 0.84 Combined Axial and Flexural 0.76 Shear 7.2/148.6 = 0.05
- 77. 77 Example 3 -Drilled Shafts
- 78. 78 Example 3 -Drilled Shafts
- 79. 79 Example 3 -Drilled Shafts Limestone Parameters qu = 11,500 psi (79.3 Mpa) RQD = 80% ~ 1 fracture per foot Tight clean joints Drilled Shaft D = 3.3 m (10.8 ft) L = 5 ft, 10 ft, and 15 ft L Limestone Reference FHWA-IF-99-025
- 80. 80 Base Resistance forCompressive Loading Rock with 70 < RQD < 100 (FHWA, Eqn 11.6) qmax ( MPa) = 4.83[qu ( MPa)] 0.51 qmax = 44.9 Mpa (469 tsf) [ ]q Jointed Rock (FHWA, Eqn, 11.7) 0.5 ( qmax = s + ms + s0.5 ) 0.5 u qmax = 49.5 Mpa (517 tsf) Detroit Experience qmax = qall (FS ) = 120(2.5) qmax = 28.7Mpa (300 tsf)
- 81. 81 Base Resistance Say s =4(10)-2 m = 0.7
- 82. 82 Side Resistance forCompression Loading Smooth Rock (FHWA, Eqn 11.24) 0.5 0.5 qu f 'c f max = 0.65 pa ≤ 0.65 pa fmax = 1.02 Mpa (10.6 tsf) pa pa
- 83. 83 Nominal Resistance Values BaseResistance qmax = 300 tsf φ = 0.5 RBN = 27,605 tons Shaft Resistance fmax = 10.6 tsf φ = 0.55 RSN = 1,800 tons (L = 5 ft) 3,600 tons (L = 10 ft) 5,400 tons (L = 15 ft)
- 84.
- 85. 85 Strain Incompatibility QT1 QT QT1 = Total load on head at point where δT1 socket side shear failure develops. Some base resistance has developed. δT1+∆δ QT = Ultimate resistance of drilled See Appendix C of FHWA-IF-99-025 plunging
- 86. 86 Drilled Shaft – Nominal Results 30000 1.0 1.0 0.9 0.9 Skin Resistance/Ultimate Skin Resistance 25000 End Bearing/Ultimate End Bearing 0.8 0.8 Nominal Resistance (tons) 0.7 0.7 20000 0.6 0.6 15000 0.5 0.5 5 ft 0.4 0.4 10000 10 ft 0.3 0.3 15 ft 0.2 0.2 5000 0.1 0.1 0 0.0 0.0 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 Settlement/Shaft Diameter (%) Settlement/Shaft Diameter (%) Settlement/Shaft Diameter (%) For L = 10 ft δ/D = 1% Rn = 13,700 tons RSN(mob)/RSN = 0.7 δ = 0.1*10.8*12 = 1.3 inch RBN(mob)/RBN = 0.4
- 87. 87 Drilled Shaft –Nominal and Factored Values For L = 10 ft δ/D = 1% Rn = 13,700 tons Shaft Resistance RSN(mob)/RSN = 0.7 RSN = 3,600 tons RSN(mob) = 2,520 tons Base Resistance RBN(mob)/RBN = 0.4 RBN = 27,605 tons RBN(mob) = 11,1180 tons Factored Resistance Rr = φRn = 0.5(11,180)+0.55(2,520) = 6,976 tons