Lrfd Short Course Presentation
Transcript of Lrfd Short Course Presentation
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Load and Resistance Factor
Design (LRFD)- Deep Foundations
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Design (LRFD)- Deep FoundationsDonald C. Wotring, Ph.D., P.E.
February 2009
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Presentation
• This presentation is intended as a detailed internal short-course with design examples. It will also be used as a brown-bag lunch
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will also be used as a brown-bag lunch presentation, but with less detail covered.
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Presentation Goals
1. Basic Differences between ASD and LRFD
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2. Fundamentals of LRFD
3. Application of LRFD to Deep Foundations
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Load < Resistance
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Load = Resistance ?????
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Load > Resistance
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Examples of Uncertainty
• Material dimensions and location
• Material strength
• Failure mode and prediction method
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• 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
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Allowable Stress Design
( ) FSRLLDL n /≤Σ+Σ
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ADVANTAGES
Simplistic
Accustomed to use
DISADVANTAGES
Inadequate account of variability
Stress not a good measure of resistance
Factor of Safety is subjective
No risk assessment
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Definition - Limit State
• A Limit State is a condition beyond which a structural component ceases to satisfy the provisions for which it is
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to satisfy the provisions for which it is designed.
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Definition - Resistance
•Resistance is a quantifiable value that defines the point beyond which the particular limit state under
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the particular limit state under investigation, for a particular component, will be exceeded.
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Resistance Can Be Defined in Terms of
• Load/Force
• Stress (normal, shear, torsional)
• Number of cycles
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• Number of cycles
• Temperature
• Strain
• etc.
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AASHTO LRFD Bridge Design
Specifications
• 4th Edition, 2007
• 2008 Interim Revisions
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• 2008 Interim Revisions
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Load and Resistance Factor Design
rniii RRQ =≤Σ φγη
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ADVANTAGES
Load factor applied to each load combination
Types of loads have different levels of uncertainty
Accounts for variability
Uniform levels of safety
Risk assessment
DISADVANTAGES
More complex than ASD
Old habits
Requires availability of statistical data
Resistance factors vary
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LRFD Equationrniii RRQ =≤Σ φγη
ηi Load modifier: factor relating to ductility, redundancy, and operational importance
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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
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Limit States
• Strength – strength and stability sufficient to resist the specified statistically significant load combination during design life
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I – Normal vehicular use without windII – Owner-specified design vehicle without windIII – 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)
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Limit States
• Service – Restrictions on stress, deformation, and crack width under regular service conditions
I – Normal operational use with 55 mph wind.
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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
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Limit States
• Extreme – Structural survival during a major event (earthquake, flood, vessel impact, ice, etc.)
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I – Earthquake
II – Other events
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Limit States
• Fatigue – Limit crack growth under repetitive loads to prevent fracture during design life
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Load Modifier
When maximum value of γi is appropriate95.0≥= IRDi ηηηη
rniii RRQ =≤Σ φγη
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0.11 ≤=
IRDi ηηη
η When minimum value of γi is appropriate
ηD Ductility load modifier
ηR Redundancy load modifier
ηD Operational importance load modifier
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Load Modifier - Ductility
Strength Limit StateηD > 1.05 Non-ductile components and connections
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= 1.00 Conventional designs according to AASHTO specs
< 0.95 Ductility enhancing measures specified beyond AASHTO specs
All other Limit StatesηD = 1.00
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Load Modifier - Redundancy
Strength Limit StateηR > 1.05 Non-redundant members
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= 1.00 Conventional redundancy
< 0.95 Exceptional redundancy
All other Limit StatesηR = 1.00
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Load Modifier – Operational
Importance
Strength Limit StateηI > 1.05 Important bridges
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= 1.00 Typical bridges
< 0.95 Relatively less important bridges
All other Limit StatesηI = 1.00
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Loads rniii RRQ =≤Σ φγη
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Load Combinations and Load Factors
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Load Factors for Permanent Loads
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Destabilizing Stabilizing
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Loading Summary
rniii RRQ =≤Σ φγη
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Load modifier – Usually = 1.0
Load factor
Load
Develop a governing load combination for each of:- Strength- Service- Extreme- Fatigue
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Probability ReviewNormal Distribution
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Probability of Failure Reliability Index, ββββNo. of standard deviations that the mean value is above 0
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Probability of Failure – Reliability
Index
Structure Pile Redundancy
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ββββ Pf
2.33 1.0%
3.00 0.13%
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Resistance Factor
( )
( )( )[ ]{ }22
2
2
11lnexp
1
1
QRT
R
Q
iiR
COVCOVQ
COV
COVQ
++++Σ
=β
γλφ
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Dead Load FactorsγD = 1.25λQD = 1.05COVQD = 0.1
Live Load FactorsγL = 1.75λQL = 1.15COVQL = 0.2
φφ
γγ4167.1
1
≅
+
+=
L
D
LL
DD
Q
Q
Q
Q
FS
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Does a low resistance value =
Inefficient Design method?
COV = 0.4λ= 1.0φ= 0.44φ/λ = 0.44
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Underpredictive (built in FS)Overpredictive
φ/λ = 0.44
COV = 0.4λ= 1.5φ= 0.67φ/λ = 0.44
COV = 0.58λ= 1.5φ= 0.44φ/λ = 0.29
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Efficiency of the Method
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φ/λφ/λφ/λφ/λ
FS(λλλλ)
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Summary - Where do we stand?
rniii RRQ =≤Σ φγη
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Resistance factor based on probability of failure for different methods of estimating the resistance.
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Limit States as Applied to Deep
Foundations
• AASHTO, Section 10.5
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• Service Limits
• Strength Limits
• Extreme Limits
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Service Limit States
• Settlements – limitation to be compared with costs of designing structure to tolerate more movement or maintenance (jacking and shimming bearings)
φ = 1.0
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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
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Extreme Limit States
• Scour – Check flood (Section 2.6.4)
• Earthquake
• Liquefaction
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• Liquefaction
• Ice
• Vehicle or Vessel Impact
φ = 1.0 generalφ = 0.8 uplift
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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
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• 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.
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Strength Limit Resistance Factors
• Presented as a function of soil type (sand, clay). Sand = drained shear strength and Clay = undrained shear strength!!!!!
• β = 3.5 (P of 1 in 5,000)
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• 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.
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Driven Piles
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Driven Piles
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What is a difficulty (driven piles)?
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Pile Length Estimate for 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
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Field testing (e.g., PDA w/ CAPWAP) is used for driving criteria.
nstatstatndrdyn RR φφ =
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MDOT Bridge Design Manual (7.03.09)
Pile Type Rndr (k)
12” O.D., 0.25” 350
Cast-in-place Concrete Piles
Pile Type Rndr (k)
HP10x42 300
Steel H-Piles
Nominal Driving Resistance Values, Rndr
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12” O.D., 0.25” 350
14” O.D., 0.312” 400
14” O.D., 0.438” 500
Pile Type Rndr (k)
Timber 150
Timber Piles
HP10x42 300
HP10x57 450
HP12x53 400
HP12x74 600
HP12x84 650
HP14x73 600
HP14x89 700
HP14x102 800
HP14x117 900
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Structural Compressive Resistance -
Steel
AFP ynλ66.0= λ < 2.25
AF88.0If fully embedded,
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λ > 2.25
λAF
P yn
88.0=
E
F
r
kL y2
=π
λ Euler Equation
If fully embedded, λ = 0
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Structural Pile Resistance Values
Resistance during pile driving φ = 1.0
Axial resistance for compression subject Combined axial and flexural
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Axial resistance for compression subject to damage where pile type is required
H-piles φ = 0.5Pipe piles φ = 0.6
Axial resistance for compression not subject to damage
H-piles φ = 0.6Pipe piles φ = 0.7
Combined axial and flexural resistance for undamaged pile
Axial H-piles φ = 0.6Axial pipe piles φ = 0.7Axial pipe piles φ = 1.0
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Drilled Shafts
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Micropiles
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Summary
• LRFD – statistically based method to account for the probability of failure
▫ Compared with ASD
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▫ Compared with ASD
▫ Limit states and resistance
▫ Load factors and combinations
▫ Resistance factors
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Example 1 - Estimate Pile Length
Fill
Clay
γ= 130 pcf
γ= 125 pcf
10’• Pier Factored Load = γQ = 1.25(3640) = 4550 kips
• Assume PDA w/ CAPWAP � φdyn = 0.65
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Clay γ= 125 pcfsu = 2.5 ksf
120’
• Assume PDA w/ CAPWAP � φdyn = 0.65
•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
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Pile Caps
12-inch Pipe Piles 14-inch Pipe Piles
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42”
42”
180”
138”
49”
49”
259”
112”
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Example 1 - Estimate Pile Length
Fill
Clay
γ= 130 pcf
γ= 125 pcf
10’Assume PDA w/ CAPWAP � φdyn = 0.65Assume λ-method � φstat = 0.40
φ
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Clay γ= 125 pcfsu = 2.5 ksf
120’
Pile Type Rndr (k) Rstat (k)
12” O.D., 0.25” 350 570
14” O.D., 0.312” 400 650
ndrndrstat
dynnstat RRR 625.1==
φφ
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Depth Summary
40
50
0 200 400 600 800 1000
Qs = Rnstat (kips)
12 OD
59
60
70
80
90
100
110
120
130
140
De
pth
(ft
)
12 OD
14 OD
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Settlement
2D/3 = 73’
• Unfactored Pier Load 3640 kips• Service Factored Load = γQ = 1.0(3640) = 3640 kips
ksfftk 1.215.172/3640 2 ==σStress at top of Piles
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2D/3 = 73’
I Stress at 2D/3
ksfftk 5.15.2422/3640 2 ==σ
0.36
0.66
0.96
Clay AssumptionsOCR = 1.2 Cc = 0.2 Cr/Cc = 0.1eo = 0.5 Cr = 0.02
17’
20’
20’
AB
C
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Settlement
Point σσσσ’vo(ksf)
I ∆σ∆σ∆σ∆σ(ksf)
σσσσ’vf(ksf)
σσσσ’p(ksf)
Ho/1+eo(in)
Sp(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
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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
+=
vo
vfr
o
op C
e
HS
'
'log
1 σσ
If σσσσ’vf < σσσσ’p If σσσσ’vf > σσσσ’p
+
+=
p
vf
vo
p
c
rc
o
op C
CC
e
HS
'
'log
'
'log
1 σσ
σσ
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Example 2
Strength V Limit – Rigid Cap ModelApplied Factored Loads
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Applied Factored Loads
Fx = 38.4 kipsFy = 109.1 kipsFz = 3,594.0 kips
Mx = 3,196.5 k-ftMy = -8,331.9 k-ft Loose Sand
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Calculate Pile Axial Loads
∑∑==
++= n
ii
iy
n
ii
ixzi
x
xM
y
yM
n
FP
1
2
1
2243
1000
)5(9.8331
225
)5.1(5.3196
20
359414 =−−++=P
63
36 in
60 in
== ii 11
Fz = 3,594.0 kipsMx = 3,196.5 k-ftMy = -8,331.9 k-ftn = 20 pilesxi = -60 in (-5 ft)yi = 18in (1.5 ft)Σxi
2 = 1,000 ft2
Σyi2 = 225 ft2
( )112
)( 22
2 −=∑ nns
rowxi
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Lateral Loading – LPILE
Pile CTC Spacing(loading
P-Multiplier, Pm
Row 1 Row 2 Rows 3
64
(loading direction)
Row 1 Row 2 Rows 3 and
higher
3B 0.7 0.5 0.35
5B 1.0 0.85 0.7
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Lateral Loading – LPILERow Pm Hy (k) Mm
(k-in)
1 0.35 4.5 -340
2 0.35 4.5 -340
3 0.5 5.9 -390
65
3 0.5 5.9 -390
4 0.7 7.2 -450
ΣΣΣΣ 110.5
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Lateral Loading – LPILERow Pm Hz (k) Mm
(k-in)
1 0.7 1.8 -75
2 0.7 1.8 -75
3 0.7 1.8 -75
66
3 0.7 1.8 -75
4 0.85 2.0 -80
5 1.0 2.2 -90
ΣΣΣΣ 38.4
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Pile Loading Summary
Maximum Shear (Row 4 piles) 7.2 kips
Maximum axial load in any pile (Pile 20) 327 kips
67
Maximum combined loading (Pile 20)Pu 327 kipsMux -37.5 k-ftMuy -7.5 k-ft
Alternative to rigid cap model, use FBPier
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Drivability Evaluation (GRL Weap)
Require a PDA/CAPWAP Field Evaluationφdyn = 0.65
68
Nominal Driving Resistance• HP12x53• Delmag D 12-32•Rndr = 550 kips at 120bpf• 10 bpi = 5 b/0.5inch
Nominal must be >327/0.65=503 kips
Rrdr = φdynRndr = 0.65(550) = 358 kips
Factored Driving Resistance
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Drivability
Check Driving Stresses
Steel Piles in Compression
69
yDADR Fϕσ 9.0=
Steel Piles in Compression or Tension
0.1=DAϕ
5.374550)0.1(9.0 >== ksiDRσ
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Geotechnical Resistance - Static
Estimate the Depth of Penetration nstatstatndyn RkipsR φφ == 358
70
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).
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Structural Resistance - Compression
AFP ynλ66.0=
Nominal Compressive Resistance (Section 6.9.4.1)
5.0=φCompression only, damage likely
71
AFP yn 66.0=
0=λ Fully embedded
kipsAFP yn 775)5.15(50 ===
Factored Compressive Resistance
kipsPP nr 5.542)775(7.0 === φ
Good for lower portion of pile where damage is more likely
5.0=φ
6.0=φCompression only, damage unlikely
7.0=φCombined
kipsPP nr 5.387)775(5.0 === φ
combined
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Structural Resistance - Shear
wyn DtCFV 58.0=Nominal Shear Resistance (Section 6.10.9.2)
HP12x53D = 11.78 in <=D
Check shear buckling ratio
)000,29(5Ek
72
D = 11.78 intw = 0.435 in
Factored Shear Resistance
kipsVV nr 6.148)6.148(0.1 === φ
3.601.27 <=wt
D3.60
50
)000,29(512.112.1 ==
yF
Ek
C = 1.0
kipsVn 6.148435.0)78.11(50)0.1(58.0 ==
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Structural Resistance – Flexural
Nominal Flexural Resistance (Section 6.12.2.2)
zFM yn = zx 74.0 in3
73
zFM yn = zx 74.0 inzy 32.2 in3
3700)74(50 ==nxM
1610)2.32(50 ==nyM k-in
k-in
Factored Flexural Resistancenr MM φ=
3700)3700(0.1 ==rxM
0.1=φ
1610)1610(0.1 ==ryM k-in
k-in
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Combined Loading
Nominal Combined Loading(Section 6. 9.2.2)
2.0<uP0.1≤
++ uyuxuMMP
74
2.0<r
u
P0.1
2≤
++
ry
uy
rx
ux
r
u
MM
M
P
P
2.0≥r
u
P
P0.1
9
8 ≤
++
ry
uy
rx
ux
r
u
M
M
M
M
P
P
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Combined Loading
Pu 327 kipsMux -37.5 k-ftMuy -7.5 k-ft
Pr 542.5 kipsMrx 308.3 k-ftMry 134.2 k-ft
Pu/Pr 0.6 > 0.2Mrx 0.12Mry 0.06
75
( ) 0.176.006.012.09
86.0
9
8 <=++=
++
ry
uy
rx
ux
r
u
M
M
M
M
P
P
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Summary – Strength V Limit State
Structural Performance Ratios
Driven 327/358 = 0.91
76
Driven 327/358 = 0.91Geotechnical N/AAxial Compression only 327/387.5 = 0.84Combined Axial and Flexural 0.76Shear 7.2/148.6 = 0.05
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Example 3 - Drilled Shafts
77
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Example 3 - Drilled Shafts
78
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Example 3 - Drilled Shafts
79
Limestone Parametersqu = 11,500 psi (79.3 Mpa)RQD = 80%~ 1 fracture per foot
L
~ 1 fracture per footTight clean joints
Drilled ShaftD = 3.3 m (10.8 ft)L = 5 ft, 10 ft, and 15 ft
ReferenceFHWA-IF-99-025
Limestone
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Base Resistance for Compressive
Loading
80
[ ] 51.0max )(83.4)( MPaqMPaq u=Rock with 70 < RQD < 100 (FHWA, Eqn 11.6)
qmax = 44.9 Mpa (469 tsf)[ ]max )(83.4)( MPaqMPaq u=
( )[ ] uqsmssq5.05.05.0
max ++=
( ) )5.2(120max == FSqq all
Jointed Rock (FHWA, Eqn, 11.7)
Detroit Experience
qmax = 44.9 Mpa (469 tsf)
qmax = 49.5 Mpa (517 tsf)
qmax = 28.7Mpa (300 tsf)
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Base Resistance
81
Says = 4(10)-2
m = 0.7
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Side Resistance for Compression
Loading
82
5.05.0' fq
Smooth Rock (FHWA, Eqn 11.24)
5.05.0
max
'65.065.0
≤
=
a
ca
a
ua p
fp
p
qpf fmax = 1.02 Mpa (10.6 tsf)
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Nominal Resistance Values
83
Base Resistanceqmax = 300 tsf φ = 0.5RBN = 27,605 tonsRBN = 27,605 tons
Shaft Resistancefmax = 10.6 tsf φ = 0.55RSN = 1,800 tons (L = 5 ft)
3,600 tons (L = 10 ft)5,400 tons (L = 15 ft)
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Strain Incompatibility
84
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Strain Incompatibility
85
QT1 QT
δT1
QT1 = Total load on head at point where socket side shear failure develops. Some base resistance has developed.
δT1+∆δ
developed.
QT = Ultimate resistance of drilled
See Appendix C of FHWA-IF-99-025
plunging
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Drilled Shaft – Nominal Results
20000
25000
30000
No
min
al
Re
sist
an
ce (
ton
s)
0.6
0.7
0.8
0.9
1.0
En
d B
ea
rin
g/
Ult
ima
te E
nd
Be
ari
ng
0.6
0.7
0.8
0.9
1.0
Sk
in R
esi
sta
nce
/U
ltim
ate
Sk
in R
esi
sta
nce
86
0
5000
10000
15000
0 1 2 3 4
No
min
al
Re
sist
an
ce (
ton
s)
Settlement/Shaft Diameter (%)
5 ft
10 ft
15 ft
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4
En
d B
ea
rin
g/
Ult
ima
te E
nd
Be
ari
ng
Settlement/Shaft Diameter (%)
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4
Sk
in R
esi
sta
nce
/U
ltim
ate
Sk
in R
esi
sta
nce
Settlement/Shaft Diameter (%)
For L = 10 ftδ/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
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Drilled Shaft – Nominal and Factored
Values
87
For L = 10 ftδ/D = 1%Rn = 13,700 tonsRn = 13,700 tons
Shaft ResistanceRSN(mob)/RSN = 0.7 RSN = 3,600 tons RSN(mob) = 2,520 tons
Base ResistanceRBN(mob)/RBN = 0.4 RBN = 27,605 tons RBN(mob) = 11,1180 tons
Factored ResistanceRr = φRn = 0.5(11,180)+0.55(2,520) = 6,976 tons