LRFD Theory for Geotechnical Design Session 3 LRFD Theory for Geotechnical Design Topic 3 – Part A Deep Foundations LRFD for Highway Bridge Substructures and Earth Retaining Structures Course No. 130082A

Learning Outcomes State the performance limits that should be evaluated when designing a deep foundation Be able to select a deep foundation type Be able to select the appropriate resistance factor for each performance limit evaluated

Deep Foundation Performance Limits

Deep Foundation Design Process Detailed Flow Chart – RM page 3.3.6 Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Check the service limit state

Strength Limit State Checks Driven Piles Drilled Shafts Structural resistance Axial geotechnical resistance Driven resistance Structural resistance Axial geotechnical resistance

Structural Axial Failure

Structural Flexure Failure

Structural Shear Failure

Axial Geotechnical Resistance

Driven Resistance Pile damage

Driven Performance Limit

Driven Performance Limit

Service Limit State Checks Driven Piles Drilled Shafts Global Stability Vertical Displacement Horizontal Displacement Global Stability Vertical Displacement Horizontal Displacement

Global Stability

Displacement Dx Dz

LRFD Differences from ASD Same Determining Resistance Determining Deflection Different Comparison of load and resistance Specific separation of resistance and deflection

Deep foundation type selection B Method of support Bearing material depth Load type, direction and magnitude Constructability Cost

Deep Foundation Material Deep Foundation Types Deep Foundation Material Driven X Drilled or Bored -- Jacked / Special Driven Piles Prestressed Concrete Post-tension Cocnrete Pre-cast Concrete Cast-in-place Concrete Steel Wood Specialty / Composites Drilled Shafts

Method of Support End Bearing Side Friction Combined

Driven Low Displacement Piles

Driven High Displacement Piles

Drilled Shafts

Depth to Bearing/ Scour

Load Type and Direction Permanent/ Transient/ Cyclic Horizontal or Vertical

Load Type and Direction Wood is better for transient resistance than permanent Steel pile better cyclic resistance High horizontal loads better resisted by stiffer piles or shafts

Typical range of nominal (ultimate) resistance (kips) Load Magnitude Deep foundation type Typical range of nominal (ultimate) resistance (kips) Typical length (feet) Timber pile 75 – 200 20 – 40 Concrete pile 200 – 2,000 20 – 150 Steel H-pile 200 – 1,000 20 – 160 Pipe pile 175 – 2,500 20 – 100 Drilled shaft 750 – 10,000

Constructability Obstructions/ Rock Use low displacement steel piles -or- Drilled shafts

Equipment access Low headroom requires pile splicing Equipment size a function of pile/shaft size

Wrap Up Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Check the service limit state

Selection of Resistance factors Strength limit state Structural Resistance Geotechnical Resistance Driven Resistance (piles only) Service limit state Resistance factor = 1.0 (except global stability)

Methods for determining structural resistance Axial compression Combined axial and flexure Shear Concrete – Section 5 LRFD Specifications Steel – Section 6 Wood – Section 8

Structural resistance factors Concrete (5.5.4.2.1) Axial Comp. = 0.75 Flexure = 0.9 Shear = 0.9 Steel (6.5.4.2) Axial = 0.5-0.6 Combined Axial= 0.7-0.8 Flexure = 1.0 Shear = 1.0 Resistance factors dependent upon: Type of material Type of stress Placement conditions (confidence) LRFD Specifications Timber (8.5.2.2) Compression = 0.9 Tension = 0.8 Flexure = 0.85 Shear = 0.75

Determining Geotechnical Resistance of Piles Field methods Static load test Dynamic load test (PDA) Driving Formulae Static analysis methods

Determining Geotechnical Resistance of Piles

Static Load Test Load Elastic pile compression Settlement Pile top settlement

Dynamic Load Test (PDA)

Driving Formulas

Geotechnical Resistance Factors for Piles Method Site Variability  Static Load Test Low 0.8 – 0.9 Medium 0.7 – 0.9 High 0.55 – 0.8 Site Variability Defined in NCHRP Report 507 Range of Values of Resistance Factors Depends on Number of Static Load Tests AASHTO Table 10.5.5.2.2-2

Geotechnical Resistance Factors for Piles Method  Dynamic Test w/Signal Matching (e.g., PDA + CAPWAP) 0.65 Test 1% to 50% of Production Piles, Depending on Site Variability and Number of Piles Driven Site Variability Defined in NCHRP Report 507 AASHTO Table 10.5.5.2.2-1 & 3

Geotechnical Resistance Factors for Piles Method  Wave Equation only 0.4 FHWA-Modified Gates ENR 0.1 AASHTO Table 10.5.5.2.2-1

Geotechnical Safety Factors for Piles Basis for Design and Type Increasing Design/Construction Control of Construction Control Subsurface Exploration 3.50 X X 2.75 X 2.25 X 2.00 X 1.90 Static Calculation Dynamic Formula Wave Equation CAPWAP Analysis Static Load Test Factor of Safety (FS)

Computation of Static Geotechnical Resistance RR = fRn fRn = fqpRp + fqsRs RP = AP qP RS = AS qs RS RP AASHTO 10.7.3.7.5-2

Static Analysis Methods Driven Piles Drilled Shafts a method b method l method Nordlund -Thurman method SPT-method CPT-method Side friction in Rock Tip Resistance in Rock Note that the static analysis resistance factors are much less than the field tested resistance factors. Ask Participants why (answer less uncertainty from fielded tested resistance)

Pile Group Resistance Static Geotechnical Resistance Take lesser of Note that the static analysis resistance factors are much less than the field tested resistance factors. Ask Participants why (answer less uncertainty from fielded tested resistance)

Geotechnical Resistance Factors Pile Static Analysis Methods Comp Ten  - Method 0.35 0.25  - Method 0.20  - Method 0.40 0.30 Nordlund-Thurman 0.45 SPT CPT 0.50 Group 0.60 AASHTO Table 10.5.5.2.2-1

Driven Pile Time Dependant Effects Setup Relaxation RS RS RS RS RP RP RP RP

Drilled Shaft Resistance Total Resistance A Side Resistance B Resistance D C Tip Resistance RS Displacement RP RR = fRn = fqpRp + fqsRs

Drilled Shaft Group Resistance For cohesive soils use equivalent pier approach Rn group = h x Rn single where: h = 0. 65 at c-c spacing of 2.5 diameters h = 1.0 at c-c spacing of 6 diameters For cohesionless soils, use group efficiency factor approach

Geotechnical Resistance Factors Drilled Shafts Method Comp Ten Shafts in Clay  - Method (side) 0.45 0.35 Total stress (tip) 0.40 -- Shafts in Sand b - Method (side) 0.55 O’Neill & Reese (tip) 0.50 Group (sand or clay) The resistance factors are provided for each method in AASHTO section 10.5 AASHTO Table 10.5.5.2.3-1

Geotechnical Resistance Factors Drilled Shafts Method Comp Ten Shafts in Interm. Geomat’ls (IGMs) O’Neill & Reese (side) 0.60 O’Neill & Reese (tip) 0.55 -- Shafts in Rock Side (H&K, O&R) 0.40 Side (C&K) 0.50 Tip (CGS, PMT, O&R) Load Test (all mat’ls) <=0.7 AASHTO Table 10.5.5.2.3-1

Guided Walk Through Axial Geotechnical Resistance of a Drilled Shaft in Clay Reference Manual 3.3.7.5 Example 9 Stiff Clay Su = 1500 psf E = 200 ksf  = 125 pcf e50 = 0.007 Drilled Shaft f’c = 4 ksi Ec = 3600 ksi 50’ Slide Control None Reference Reference Manual 3.3.7.5 Example 9 Drilled shaft design manual Speaking Points Have participants open their RM to example 9 and follow along Introduce problem (problem is from an example in the drilled shaft design manual for lateral analysis) Adult learning 2.5’

Guided Walk Through Determine Unit Side Resistance qs =  Su To find , check Su/pa = 1.5 / 2.12 Su/pa = 0.7 < 1.5 So  = 0.55 qs = 0.55 x 1500 psf qs = 0.825 ksf Stiff Clay Su = 1500 psf E = 200 ksf  = 125 pcf e50 = 0.007 Drilled Shaft f’c = 4 ksi Ec = 3600 ksi 50’ Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5.1b-2 Speaking Points If Su had been greater than 3200 psf (Su/pa > 1.5) alpha would be determined by = 0.55 – 0.1(Su/pa – 1.5) Which would be less than 0.55 Adult learning 2.5’

Guided Walk Through Determine Exclusion Zones Per AASHTO 10.8.3.5.1b Top 5' non contributing Bottom 1 diameter (2.5') non contributing Ls = 50’ – 5’ - 2.5’ = 42.5’ 5’ 42.5’ 50’ Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5.1 Speaking Points These exclusion zones are applicable to shafts in clay only Exclusion of top 5’ is due to disturbance during construction, cyclic lateral loads, lack of confinement, and seasonal moisture changes. Some circumstances may require more than 5’ Exclusion of bottom 1D is due to tensile cracks resulting from tip mobilization. Adult learning Ask why top 5’ might be excluded. Answers above 2.5’ 2.5’

Guided Walk Through As =  D Ls As =  (2.5’)(42.5’) As = 334 ft2 Rs = qs As Rs = (0.825 ksf)(334 ft2) Rs = 275 kips 50’ Rs = 275 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5-3 Speaking Points Total resistance is the surface area times the unit resistance Adult learning 2.5’

Guided Walk Through Point Resistance qp = Nc Su Nc = 6(1 + 0.2 (Z/D)) < 9 Nc = 6(1 + 0.2 (50/2.5)) Nc = 30 not less than 9 thus Nc = 9 qp = 9 (1.5 ksf) qp = 13.5 ksf 50’ Rs = 275 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5.1c Speaking Points This is similar to the cohesion term from the standard bearing resistance equation. Note that Nc increases with depth which accounts for overburden pressure According to AASHTO Nc does not increase without bound but some studies have indicated that the concept of limiting tip resistance may be conservative. Adult learning 2.5’

Guided Walk Through Point Resistance Rp = qp Ap Ap =  D2/4 Ap =  (2.5’)2/4 Ap = 4.9 ft2 Rp = 13.5 ksf (4.9 ft2) Rp = 66 kips Rs = 275 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5-2 Speaking Points Total tip resistance is the area of the tip times the unit tip resistance Adult learning Rp = 66 kips

Guided Walk Through Combining Side and Point Resistance RR = qs Rs + qp Rp qs = 0.45 qps = 0.4 RR = 0.45 (275) + 0.4 (66) RR = 150 kips Rs = 275 kips Slide Control On 1st click “X” comes in Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.3.5-1 Speaking Points Adding the factored end and side resistances directly is not appropriate. Adult learning Ask why it is not appropriate to add the base and side resistance. After all, this equation is straight out of AASHTO Rp = 66 kips

Guided Walk Through Combining Side and Point Resistance 1.0 1.0 Rsd / Rs Rpd / Rp Slide Control None Reference Reference Manual 3.3.7.5 Example 9 AASHTO 10.8.2.2.2 Speaking Points These curves are for clay but similar methods are available for determining the load transfer in sands and in rock. Adult learning 1.0 2.0 5.0 10.0 zt / D (%) zt / D (%) zt / D (%)

Guided Walk Through Check Relative Stiffness SR = (Z/D) (Esoil/Eshaft) < 0.01 Shaft can be considered rigid SR = (50’/2.5’) (1.39 ksil/3600 ksi) Slide Control None Reference Reference Manual 3.3.7.5 Example 9 FHWA IF 99-025, Page C-9 Speaking Points Relative stiffness simply assesses whether the compression of the shaft can be ignored. For shafts in rock the compression of the shaft can not be ignored. Adult learning SR = 0.008 < 0.01 Shaft can be considered rigid

Rs = 256 kips RP = 38 kips 0.3 Slide Control None Reference Reference Manual 3.3.7.5 Example 9 Speaking Points This graph is produced by applying the normalized curves to our specific case. Refer to the example in the RM for more specifics on how this is done Note that at the maximum resistance the base resistance is not fully developed and the side resistance is developed slightly past the peak side resistance. Adult learning 0.3

RR = qs Rs + qp Rp RR = 0.45 (256) + 0.4 (38) RR = 131 kips Guided Walk Through RR = qs Rs + qp Rp RR = 0.45 (256) + 0.4 (38) RR = 131 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 9 Speaking Points The developed values of base and side resistance are used to evaluate total factored resistance Adult learning

Driven Resistance Note that the static analysis resistance factors are much less than the field tested resistance factors. Ask Participants why (answer less uncertainty from fielded tested resistance)

Wave Equation Results

Driven Resistance Factors Concrete piles, f = 1.00 AASHTO Article 5.5.4.2.1 Steel piles, f = 1.00 AASHTO Article 6.5.4.2 Timber piles, f = 1.15 AASHTO Article 8.5.2.2

Participant Exercise Participant Workbook Page 3.3A.22

Participant Exercise Fz = 3594.0 kips

Participant Exercise Method  Qn (kips) Qr (kips) # of Piles a-method 0.4 550 220 17 PDA on 5% 0.65 Gates Formula Structural Resistance 0.6 775 358 11 220 17 465 8

Participant Exercise Comparison to ASD Service Load = 2794 kips Method FS Qn (kips) Qr (kips) # of Piles a-method 3.5 550 157 18 (17) PDA on 5% 2.25 244 12 (11) Gates Formula Structural Resistance 3 (0.33fy) 775 256 11 (8)

Wrap Up Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Check the service limit state

Topic Wrap Up Participant Workbook Page 3.3A.25

Topic Wrap Up Geotechnical resistance Structural resistance Exercise 1: List the three strength limit state checks for driven piles Geotechnical resistance Structural resistance Driven resistance Exercise 2: List the three service limit state checks for drilled shafts Horizontal deflection Vertical deflection (settlement) Global stability

Topic Wrap Up Exercise 3: Match the deep foundation type to the condition. Condition Deep granular material Loose random fill overlying rock Large horizontal loads Type Steel H-Pile Closed end pipe Large diameter drilled shaft B A C

Topic Wrap Up Exercise 4: What criteria should be used to select the geotechnical resistance factor for a driven pile? Exercise 5: Where would you find the structural resistance factors for a drilled shaft? The method used to determine the ultimate resistance. AASHTO Section 5 – Concrete Design

Learning Outcomes State the performance limits that should be evaluated when designing a deep foundation Be able to select a deep foundation type Be able to select the appropriate resistance factor for each performance limit evaluated

LRFD Theory for Geotechnical Design Session 3 LRFD Theory for Geotechnical Design Topic 3 – Part B Deep Foundations LRFD for Highway Bridge Substructures and Earth Retaining Structures Course No. 130082A

Learning Outcome Apply the rigid cap method to evaluate the strength limit state checks

Where We Are Going … Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Compute load effects in piles using rigid cap method Compare load effects to factored resistances for piles Check the service limit state

Rigid Cap Model Centroid of Pile group X Y Z

Distribution of Axial Loads My Fz Pi Mx X -xi yi Y Z

Distribution of Horizontal Loads X Fx Hi Y Z

Horizontal Response Qt P Ht Mt y y Properties A, E, I y

P-y Curve development Typical required soil parameters Su f  k 50 Slide Control None Reference COM624P User’s Manual FHWA-SA-91-048 Speaking Points P-y curves are developed using familiar soil parameters for strength and unit weight. The deformation of the soil is defined by the soil parameters k and e50. k is the coefficient of variation of subgrade reaction. It is not the same as the modulus of subgrade reaction used for pavement design even though it has the same units of force/ length cubed. It represents the rate of increase of soil response with respect to depth. e50 is the strain at 50% of the ultimate strength as measured in a triaxil test. It is representative of the modulus of elasticity of the soil. Equations for developing P-y curves are available from different sources for a variety of soils and rocks. Many of them are preprogrammed into the various analysis software that is available. k – coefficient of variation of subgrade reaction 50 - strain at 50% of ultimate strength

P-y Results for Single Element

Variation of Stiffness (EI) Reinforced Concrete Shaft

Pile Head Fixity Strength Limit State Service Limit State Dx Dx Moment

Group Effects Fx H2 H1

P-y Interaction Effects Original curve Modified curve Pm * P y

Output for multiple loads Applied Horizontal Load Resulting Deflection Maximum Moment Slide Control 1st click highlights deflection 2nd click highlights applied load 3rd click highlights maximum moment Reference Speaking Points The P-y analysis should be run for multiple loading conditions The result is a table of maximum deflections, slopes, and moments for each condition Adult learning

Horizontal Load (kips) Deflection (in) Maximum Moment (in-kips) Slide Control None Reference Speaking Points It is convenient to plot the results of a P-y analysis as shown The results are for one pile using one P multiplier The analysis can be extended to different P multipliers and plotted on the same plot We will use a plot like this to evaluate the strength and service limit states of a deep foundation later in this topic. Adult learning

Computer P-y Modeling

Horizontal Loads, Pile Moment Dx Dx Fx H2 H1 M2 M1

Where We Are Going … Guided Walk Through… Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Compute load effects in piles using rigid cap method Compare load effects to factored resistances for piles Check the service limit state

Guided Walk Through Participant Workbook Page 3.3B.7

Guided Walk Through 5’-0” 46’-6” 5’-0” 6’-0” 15’-6” 15’-0” 15’-6” 4’-6” 3’-6” 23’-0” 12’-0”

Guided Walk Through HP 12x53 Centroid 18” 36” 36” 36” 18” 18” 60” 60”

Guided Walk Through Applied Loads Strength V load case Fx = 38.4 kips Fy = 109.1 kips Fz = 3594.0 kips Mx = 3196.5 k-ft My = -8331.9 k-ft Fz Fy -My Mx Fx

Guided Walk Through Example calculation, pile 9: Fz = 3594.0 kips Mx = 3196.5 k-ft n = 20 piles yi = 18 in (1.5 ft)  xi2 = 1000 ft2 My = -8331.9 k-ft  yi2 = 225 ft2 xi = 60 in (5 ft) P9 = 243 kips

Guided Walk Through

Guided Walk Through Fy Dy Dy assumed to be 0.15”

Guided Walk Through Pm = 0.7 0.5 0.35 Load (kips) Deflection (in) 10 Pm = 0.7 0.5 8 7.2 kips 0.35 5.9 kips 6 Load (kips) 4.5 kips 4 2 Deflection (in) 0.1 0.2 0.15 in -200 -340 k-in Max. Moment (k-in) -390 k-in -400 0.35 -450 k-in 0.5 -600 0.7

Guided Walk Through Row Pm Hy Mmax 1 0.35 4.5 kips -340 k-in 2 3 0.5 0.7 7.2 kips -450 k-in Sum of Hy forces times piles per column = (22.1 kips/column) (5 columns) = 110.5 kips 110.5 kips close to 109.1 kips

Guided Walk Through Dx Fx Dx assumed to be 0.05”

Guided Walk Through 0.85 0.70 Load (kips) Pm = 1.0 Deflection (in) 2.5 2.2 kips 0.70 2.0 kips 2.0 1.8 kips 1.5 Load (kips) Pm = 1.0 1.0 0.5 Deflection (in) 0.025 0.075 0.05 in -33 Max. Moment (k-in) -66 -75 k-in 0.70 -80 k-in -90 k-in 0.85 -100 1.0

Guided Walk Through Column Pm Hx Mmax 1 0.7 1.8 kips -75 k-in 2 3 4 0.85 2.0 kips -80 k-in 5 1.0 2.2 kips -90 k-in Sum of Hx forces times piles per row = (9.6 kips/row) (4 rows) = 38.4 kips 38.4 kips = 38.4 kips

Guided Walk Through Shear (kips) -2 2 4 6 8 100 Depth (in) 200 300

Guided Walk Through For load case Strength V: Max. axial load (Pile 5) = 326 kips Min. axial load (Pile 16) = 32 kips Maximum combined loading (Pile 5) Axial load = 326 kips Moment (x-direction) = -37.5 kip-ft Moment (y-direction) = -7.5 kip-ft (no uplift) Max. shear = 7.2 kips in y-direction (Piles 1, 2, 3, 4, 5 at the top of pile)

Where We Are Going … Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Compute load effects in piles using rigid cap method Compare load effects to factored resistances for piles Check the service limit state

Guided Walk Through Driven HP 12 x 53 4’ Loose Silty sand f = 31o sat = 110 pcf 35’ Hard Clay Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Same subsurface conditions as in other examples but rock is replaced by a hard overconsolidated clay. Go through parameters Adult learning Remind participants to follow along in their RM. Su = 8000 psf sat = 125 pcf OCR = 10 >100’

Guided Walk Through Pn Structural Resistance – Axial compression As = 15.5 in2 (after corrosion loss) Fy = 50 ksi l = 0 in Pn Pn = 0.66lFyAs = 0.660(50)(15.5) Pn = 775 kips AASHTO Articles 6.9.4.1-1, 10.7.3.12.1

Guided Walk Through Mnx Mny Structural Resistance – Flexure Resistance zx = 74 in3 zy = 32.2 in3 Fy = 50 ksi Mnx x Mny Mnx = (50 ksi)(74 in3) = Mny = (50 ksi)(32.2 in3) = 3700 k-in 1610 k-in

Guided Walk Through Vny Structural Resistance – Shear Resistance D = 11.78 in tw = 0.435 in Fy = 50 ksi C = 1.0 y x Vny Vp = (0.58)(50 ksi)(11.78 in)(0.435 in) VpC = 149(1.0) = 149 kips AASHTO Articles 6.10.7.2-1,6.10.7.2-2,6.10.7.3.3a

Guided Walk Through Combined Compression and Flexure f = 0.7 for Pr, 1.0 for Mr Shear f = 1.0 for Vr Axial Compression f = 0.6 for Pr

Guided Walk Through qs =  'v and qp = Nt 'v Geotechnical Resistance – Axial compression Use the beta method fro axial resistance in sand and clay. qs =  'v and qp = Nt 'v Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Ref: AASHTO 10.7.3.8.6c-1 Speaking Points The beta method is an effective stress method Adult learning

Guided Walk Through For Sand 0.28 Slide Control None Reference Reference Manual 3.3.7.5 Example 3 FHWA NHI-05-042 page 9-52 Speaking Points Determining Beta for sand layer Adult learning 0.28

For Sand 28 Slide Control None Reference Reference Manual 3.3.7.5 Example 3 FHWA NHI-05-042 page 9-53 Speaking Points Determining bearing capacity coefficient for sand layer Adult learning

For Clay 1.5 Slide Control None Reference Reference Manual 3.3.7.5 Example 3 AASHTO 10.7.3.8.6c Speaking Points Determining Beta for Clay layer Adult learning 1.5

Cum. side friction (kips) Total Resistance (kips) Guided Walk Through Tip resistance in clay qp = 9 Su Depth (ft) Average s'v (ksf) Cum. side friction (kips) Qp = qp Ap (kips) Total Resistance (kips) 5 0.12 0.67 6.6 7.3 Slide Control None Reference Reference Manual 3.3.7.5 Example 3 AASHTO 10.7.3.8.6e Speaking Points The chart is an example and would be extended to greater depths using the beta and point bearing equations appropriate for the layer. This was done in the RM and is plotted on the next slide Adult learning

Axial Geotechnical Resistance vs. Depth 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft) Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Point out that this is a useful tool to analyze a number of complications and differing resistances as the design progresses. It can serve as a way for the geotechnical engineer to communicate the geotechnical resistance to the structural engineer Adult learning Side Friction Point Resistance Total Resistance

Estimate Required Length 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft) Assume   Q =  Pn  Pn = 0.6 (775 kips)  Pn = 465 kips   Q = stat Rnstat stat = 0.25 Rnstat = 465 kips/0.25 Rnstat = 1860 kips 1860 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Use of the static analysis resistance factor for the estimate of length is intended to account for method bias. However, it also includes additional uncertainty associated with use of a static analysis as the only means of resistance verification. Thus it may result in excessively long estimated lengths if more sophisticated methods are used to verify resistance during construction (ie. Dynamic monitoring or static load test). AASHTO allows some flexibility in estimation of contract length based on experience Adult learning Side Friction Point Resistance Total Resistance Dest = 108’

Guided Walk Through Steps to perform drivability analysis: Estimate total soil resistance and distribution Select hammer Model driving system and soil resistance Run wave equation analysis Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Indicate we will go through each of the steps Adult learning

Estimate Resistance Distribution 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft)   Q = dyn Rn stat = 0.65 Rn = 465 kips/0.65 Rn = 715 kips 715 kips 20% 40% Dest = 70’ 60% Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Point out Calculation of Rn based on dynamic monitoring Estimation of length based on dynamic analysis Estimation of percent end bearing Estimation of skin friction distribution Adult learning 80% 100% Side Friction Point Resistance Total Resistance EB = 10%

Guided Walk Through Select dynamic properties of soil Skin quake = 0.1 default per WEAP manual Skin damping = 0.2 From WEAP manual Toe quake = 1/120 of pile width Toe damping = 0.15 per FHWA NHI-05-042 page 17-68 Slide Control None Reference Reference Manual 3.3.7.5 Example 3 References for parameters are on the slide Speaking Points Experience with similar soils is the best guide to estimating the dynamic response parameters These parameters are what is determined during a signal matching analysis. Adult learning

Guided Walk Through Identify pile properties (HP12x53) As = 15.5 in2 Es = 300000 ksi g s = 490 pcf Identify hammer properties (Delmag 30-23) Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Select a hammer that is commonly used in the area Adult learning Helmet weight = 2.15 kip Cushion Area = 283.5 in2 Cushion E = 280 ksi Cushion Thickness = 2 in

715 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Note that this hammer never produces Rn = 715 kips Thus need a bigger hammer Also note that driving stress is allready over 45 ksi thus a bigger hammer may not help Adult learning

Bigger hammer (Delmag 46-13) 58 ksi 715 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points As expected, bigger hammer achieves the desired resistance but the driving stress is 58 ksi at this resistance Adult learning

Guided Walk Through Evaluate driving stress dr = 0.9 da fy (permissible driving stress) da = 1.0 dr = 0.9 (1.0) 50 ksi dr = 45 ksi 45 ksi < 58 ksi (driving stress exceeded) Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points Compare permissible driving stress to anticipated driving stress. No Good! Thus we can not develop the full structural resistance of this pile because we can not drive it to that resistance without damaging the pile We need to find what resistance we can drive it to Adult learning Ask what else could be done to resolve this problem Answer: increase the resistance factor by using better construction control (ie static load tests) this will reduce the Rn required. What is the maximum resistance that can be developed without exceeding the permissible driving stress?

45 ksi 550 kips 17 BPI Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points At the permissible driving stress of 45 ksi, the Rn = 550 kips Penetration rate is 17 bpi which is slightly higher than desirable but within the reasonable range Note that this leaves very little flexibility for hammer selection. Contractor must supply a hammer that produces the maximum permissible driving stress, no less and no more. Adult learnin 17 BPI

Guided Walk Through Factored resistance limited by driving stress (driven resistance) RR = dyn Rn dyn = 0.65 RR= 0.65 (550 kips) RR = 358 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 3 Speaking Points This is the axial resistance limited by drivability not the axial geotechnical resistance It will be refrw3ed to as the driven resistance. Adult learning

Guided Walk Through Axial geotechnical performance ratio = 326/465 = 0.7 Axial structural performance ratio = 326/465 = 0.7 Combined axial and flexural performance ratio = 0.78* Driven performance ratio 326 / 358 = 0.91 Shear performance ratio = 7.2 / 256 = 0.03 USE the same notes as on the original slide *AASHTO Eqn. 6.9.2.2-2

Estimate Required Length for Actual Factored load 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft) 1304 kips   Q = 326 kips   Q = stat Rnstat stat = 0.25 Rnstat = 326 kips/0.25 Rnstat = 1304 kips Dest = 91’ Side Friction Point Resistance Total Resistance

Wrap Up Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Compute load effects in piles using rigid cap method Compare load effects to factored resistances for piles Check the service limit state

Guided Walk Through Non-linear Column and Cap Beam Non-linear Soil Response Flexible Membrane Pile Cap T-z - Non-linear Pile Material P-y (& P-x) Q-z

Guided Walk Through Beam seat elevation Applied Loads Fz Fy -My Fx -My Mx Beam seat elevation Rock Loose Sand Applied Loads

Guided Walk Through Axial Results Shear Results Moment Results Pile 5 Max. Axial = 327 kips Min. Axial = 32 kips Shear Results Shear = 7.2 kips Moment = - 37.5 k-in Moment Results Pile 5 Pile 16 Rigid Cap Results

Guided Walk Through Axial geotechnical performance ratio = 327/465 = 0.7 Axial structural performance ratio = 327/465 = 0.7 Combined axial and flexural performance ratio = 0.73* Driven performance ratio 327 / 358 = 0.91 Shear performance ratio = 7.59 / 256 = 0.03 (0.7) (0.78) (0.91) (0.03) *AASHTO Eqn. 6.9.2.2-2

Accounting for Scour 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft) Scoured   Q = 358 kips   Q = stat Rnstat stat = 0.25 Rnstat = 326 kips/0.25 Rnstat = 1432 kips RS scour = 20 kips 20 kips 1432 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 5 Speaking Points The depth vs resistance chart developed in example 3 can be used to account for several different subsurface situations Should account for bias in method used to estimate scour resistance lost which is not shown in this example. Adult learning Ask participants to open their reference manual to the examples 5 – 8 and follow along Dest = 96’ Side Friction Point Resistance Total Resistance

Accounting for Scour Required driven resistance during construction   Q = 358 kips   Q = dyn Rndr – RS scour Rndr =   Q / dyn+ RS scour dyn = 0.65 Rndr = 326 kips/0.65 + 20 kips Rndr = 571 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 5 Speaking Points Note there is no resistance factor applied to resistance in the scour zone Adult learning

Accounting for Downdrag 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft)   Q = 358 kips + DD DD DD = 1.8 RS scour = 20 kips DD = 20 kips   Q = 394 kips Rnstat = 394 kips/0.25 Rnstat = 1576 kips Settling 20 kips 1576 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 6 Speaking Points Same as scour except now the skin friction reverses and becomes a factored load. Load factor is estimated based on reliability of the beta method. 1.8 is not in the 2006 interims. Adult learning Dest = 100’ Side Friction Point Resistance Total Resistance

Accounting for Downdrag Required driven resistance during construction   Q = 358 kips + DD DD DD = 1.0 Since resistance in downdrag zone determined by signal matching   Q = 358 kips + 1.0 (20 kips) = 378 kips   Q = dyn Rndr – RS downdrag Rndr =   Q / dyn+ RS downdrag dyn = 0.65 Rndr = 378 kips/0.65 + 20 kips Rndr = 602 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 6 Speaking Points Note if static analysis is the sole means of construction control then the load factor applied to downdrag would not be 1.0 Adult learning

Accounting for Set up 20 40 60 80 100 120 140 500 1000 1500 2000 Resistance (kips) Depth (ft)   Q = 358 kips Rnstat = 358 kips/0.25 Rnstat = 1432 kips 1432 kips Set up Slide Control None Reference Reference Manual 3.3.7.5 Example 7 Speaking Points This is the same as for the example with no set up since the curve is developed for the final conditions after set up Adult learning Dest = 95’ Side Friction Point Resistance Total Resistance

Accounting for Set Up Required driven resistance during construction   Q = 358 kips Rndr =   Q / ( S2) - R1dr S1 / S2 + R1dr S1 = 1.0 (no strength change expected in layer 1) S2 = 1.5 (50% strength gain in layer 2)  = 0.25 (static analysis only) R1dr = 25.6 kips (resistance in layer 1) Rndr = 358 kips/(0.25 x1.5) – 25.6 kips (1.0)/1.5 + 25.6 kips Rndr = 963 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 7 Speaking Points The required driving resistance is different than for the normal case. For development of the equation for required driving resistance see the RM Adult learning

Accounting for Set up Resistance (kips) 500 1000 1500 2000 963 kips 20 500 1000 1500 2000 963 kips 20 R1dr = 25.6 kips Rndr = 963 kips 25.6 kips 40 60 Depth (ft) 80 Set up Slide Control 1st click line for dynamic analysis at EOD appears 2nd click line for dynamic analysis at BOR appears Reference Reference Manual 3.3.7.5 Example 7 Speaking Points A curve representing the resistance during driving can be added to the plot Note that it results in the same estimated depth. Adult learning Ask how thy would use this curve if it was determined based on signal matching of end of drive blows and restrike blows after set up has occurred. Answere : Rndr = 358/(0.65x 1.5) – 25.6 (1)/1.5 + 25.6 = 376kips Dest = 61’ ft Rn = 550 kips (550 x 0.65 = 358 kips) 100 Dest = 95’ Side Friction Point Resistance Total Resistance 120 140

End Bearing on Hard Rock Assume structural resistance is much less than geotechnical resistance. Assume potential damage to pile RR =  Pn Pn = 775 kips  = 0.5 (due to potential for damage) RR = 0.5 (775 kips) = 388 kips Slide Control None Reference Reference Manual 3.3.7.5 Example 8 Speaking Points On hard rock the geotechnical resistance is assumed to be present without having to develop it by driving so the task is to get the pile to rock and stop Adult learning Ask how you would prevent damage during driving Answers: limt # of blows, watch for damage, use smaller hammer, dynamic monitoring to check maximum driving stress. Estimate length based on depth to rock Control driving to prevent damage

Topic Wrap Up Participant Workbook Page 3.3B.29

Given a load case with loading directions as depicted in the adjacent figure: Fy My Fz a. Which pile will have the highest axial load? b. Which pile will have the lowest axial load? c. Which pile will be subject to the highest horizontal load? d. Which pile will be subject to the highest bending moments? X 1 Fx Mx 4 Y Z Many times at this point the foundation size will not be known and only a single factored vertical load can be estimated 2 3 4 1 2 2 5D c-c

Learning Outcome Apply the rigid cap method to evaluate the strength limit state checks

LRFD Theory for Geotechnical Design Session 3 LRFD Theory for Geotechnical Design Topic 3 – Part C Deep Foundations LRFD for Highway Bridge Substructures and Earth Retaining Structures Course No. 130082A

Learning Outcome Be able to perform a rigid cap analysis of a driven pile group at Service Limit State

Where We Are Going … Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Check the service limit state

Axial Response of a Single Element (Approximate method) Qtop Dztop Point bearing only Dztop = Dzp + Qtop L/ (A E) Constant side friction only Dztop = Dzp + Qtop L/ (2 A E) L Linear increasing friction only Dztop = Dzp + Qtop L/ (3 A E) Dzp Pile Properties A, E

Axial Response of a Group

Perform Rigid Cap Analysis, Driven Pile E Guided Walk Through

Guided Walk Through HP 12x53 Centroid 18” 36” 36” 36” 18” 18” 60” 60”

Guided Walk Through Applied Loads Fx = 31.8 kips Fy = 86.1 kips Fz = 2794 kips Mx = 2547.7 k-ft My = -6306.9 k-ft Fz Fy -My Mx Fx

Guided Walk Through

Guided Walk Through Mx Fz Fy Average load, PB = 88.8 kips PC =190.6 kips PB PC

Guided Walk Through Fy = 86.1 kips / 5 rows Fy = 17.2 kips/row Assume deflection = 0.11” 1 2 3 4 Fy = H1 + H2 + H3 + H4 Fy = 3.7+3.7+4.6+5.5 Fy = 17.5 kips 0.35 0.35 0.5 0.7 Pm

HP 12x53 in loose sand, fixed x-x axis Guided Walk Through HP 12x53 in loose sand, fixed x-x axis 10 Pm = 0.7 0.5 8 0.35 5.5 kips 6 Load (kips) 4.6 kips 4 3.7 kips 2 Deflection (in) 0.11 in 0.1 0.2

Guided Walk Through Qtop Dztop Estimate Dzp=0.03 in @ Qp=500 k Assume point bearing: L = 384 in Dztop = 0.46 in = 0.00092(Qtop) Dzp QP

Guided Walk Through Pile head displacements Dztop, Pile B = 0.00092 (88.8 kips) = 0.082 in. Dztop, Pile C = 0.00092 (190.6 kips) = 0.175 in. Dy for both piles is 0.11 in.

Guided Walk Through Given coordinates: A = (72 , -333) B = (18.11 , Dztop, Pile B) C = (126.11 , Dztop, Pile C) D = (72.11 , zD) B D C Find zD by similar triangles Find a of line BC Use trigonometry to find: DyA, DzA +y +z

Guided Walk Through Initial coordinates, A (72, -333) B D C Final coordinates, A (72.40, -332.87) Displacement of A DyA = 0.40 in DzA = 0.13 in +y +z

Guided Walk Through FB Pier Analysis DyA = 0.50 in DzA = 0.13 in Rigid Cap DyA = 0.40 in DzA = 0.13 in

Wrap Up Decide deep foundation type Select resistance factor Compute resistances Layout foundation group and analyze at the strength limit state Check the service limit state

Participant Exercise Participant Workbook Page 3.3C.10

Participant Exercise 1 2 3 4 Pm 0.7 0.7 0.7 0.85 1.0 5

HP 12x53 in loose sand, fixed y-y axis Participant Exercise HP 12x53 in loose sand, fixed y-y axis 2.0 1.6 1.2 1.0 0.85 0.7 Load (kips) 0.8 0.4 0.0 0.01 0.03 0.05 Deflection (in.)

Participant Exercise Average loads in XZ plane PB = (26+60+94+128)/4 = 77 kips PC = (152+186+220+254)/4 = 203 kips Horizontal Reactions Displacement assumed to be 0.04 in Fx = 31.8 kips / 4 rows = 8 kips/row H1+H2+H3+H4+H5 = 1.5+1.5+1.5+1.7+1.8 = 8 kips, OK Settlement as a Function of Qtop Dztop = 0.00092Qtop

Participant Exercise Pile Head Displacements Pile B: Dztop = 0.071 in, Dx = 0.04 in Pile C: Dztop = 0.187 in, Dx = 0.04 in Displaced Geometry zD = 3 (0.129) a = 0.02769o Final coordinates, A = (138.20, -332.87) Displacement DxA = 0.20 in, DzA = 0.13 in

Results Participant Exercise FB Pier Analysis DxA = 0.23 in DzA = 0.13 in Rigid Cap DxA = 0.20 in DzA = 0.13 in Results

Learning Outcome Be able to perform a rigid cap analysis of a driven pile group at Service Limit State

Read More About It