Short course for UNIVERSIDAD NACIONAL DE INGENIERIA January 26-29, 2016
Planning and Design for Rehabilitation of Rivers Using Large Wood Metodología para Reforestar Ríos Degradados por Actividades Humanas usando Técnicas de Bioingeniería
9.0 Design calculations and tools
Course overview Day I (Jan 26)—Foundational topics • Review of information resources (design handbooks and spreadsheets) for large wood • Is wood appropriate for your site?—criteria for screening (Planning) • Three design approaches • Key issues for large wood design
Day 2 (Jan 27)—Designing large wood structures • Case study I—Little Topashaw Creek, Mississippi • Design life for wood structures/selection of design event or condition
Day 3 (Jan 28)—Risk, uncertainty and construction • Sensitivity and Monte Carlo analyses
• Constructability assessment • Case study II—Trinity River, California • Monitoring
Day 4 (Jan 29)--Field trip
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• Introductions
• Types of wood structures • Findings of recent research on drag and lift coefficients • “Road testing” selected design spreadsheets
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Force balance • • • •
Buoyancy Drag Lift Differential hydrostatic force
• Stabilizing (resisting) forces • • • • •
Ballast Friction Embedment Pilings Anchors
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• Destabilizing (driving) forces
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• Buoyancy
• Drag • Lift
Fb dVd wVw CD A wU Fd 2g 2 CL A wU o FL 2g
2 o
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Destabilizing (driving) forces
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This formula assumes the entire structure is submerged.
Buoyant force Fb dVd wVw
• Fb is net buoyant force • Subscripts d and w are for wood and water • and V are specific weight and volume • Vd = Vw = volume of displaced water = volume of submerged wood
Ratio of Net Buoyant Force to Drag Force 1000
Assuming d/w = 0.45
100
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• Easy(?) to compute, but don’t overlook!
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1
0.1
0
1
2
3
4
Stream velocity, m/s
5
5
• Treat logs as cylinders • Add up all volumes of all logs whether they are buried or not…EXCEPT for vertical piles • Treat submerged logs and those above the water line separately • Equations 2 and 3, p. 48, Knutson and Fealko (2014) To be conservative, use
wood = 0.45 water
in both equations.
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Buoyant force example
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CL A wU FL 2g
2 o
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Drag and lift forces
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CL A wU FL 2g
2 o
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What should you do about floating debris loading?
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What should you do about debris loading?
• The accumulation dimensions may be estimated as • Depth = flow depth (if less than 4 to 10 ft) • Width, W = “sturdy log length” • So, Area for drag and lift equations will be this W x depth
W
w
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• Assume floating logs and branches will collect and increase the effective size of the large wood structure
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Estimating sturdy log length, W
• Upstream channel width • Maximum length of sturdy logs • In USA, 30 ft + channel width/4
W Shields Engineering, LLC
• W = “sturdy log length” • Sturdy log length = longest abundant log…AND • Sturdy log length = length of logs that are strong enough to resist breakage • The smallest of….
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Estimating “sturdy log length�
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Work drag and lift example • CD= 0.9 for fully submerged conditions and • CD= 1.5 for conditions where the water surface is within one (typical) log diameter of the top of the structure. • Let’s use CD= 1.1 for this example
• CL
• Knutson and Fealko (2014) note lift forces typically ignored in LW design • In the absence of better information, CL may be assumed = 1.0 for complex large wood structures that are submerged. Lift may be assumed = 0 for large wood that is not fully submerged. • Let’s use CL = 1.0 for this example
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• CD--In the absence of better information
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Work drag and lift example • Channel width = 50 m • Mean depth = 3 m, depth upstream of structure = 3.5 m • Approach velocity = 3 m/s
• Find Drag Coefficient • • • • •
Sturdy log length = 10 + 50/4 = 22.5 m, say 15 m Wood Area = 15 x 3.5 = 53 m2 Blockage, B = Wood Area/XS area of flow = 53/(3 x 50) = 0.35 Blockage factor by Gippel = 1/(1-B)2= 2.37 Effective CD = 1.1 x 2.37 = 2.6
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• Hydraulics
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Work drag and lift example [2.6 x 53 x 9810 x 32]/(2x 9.81) =
CL A wU FL 2g
2 o
[1.0 x 53 x 9810 x 32]/(2x 9.81) =
7.13E4 N Equations 4 and 19, Knutson and Fealko (2014)
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6.15E5 N
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• Ballast • Friction • Embedment • Pilings • Anchors
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Stabilizing (resisting) forces
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Vertical resisting force--Ballast • Boulders and soil handled separately • Boulders—treat as spheres to compute volume Shields Engineering, LLC
• Force = Fsub + Fdry (Eq 5-7, p. 49, Knutson and Fealko)
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Vertical resisting force--Ballast
the density of water
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• Soil ballast called “embedment” by Knutson and Fealko (2014), Equation 8, p. 51
We will return to this equation after we define each term….
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Vsoili is just the volume of soil directly over log i
soil is the bulk density of the soil soil’ is the difference between the saturated soil bulk density and
Work embedment example • • • • • •
No. of embedded logs = 8 L, Log length = 10 m Lem, Embedment depth = 5 m Log diameter = 0.4 m D, Burial depth = 2 m Dw, Depth to groundwater = 1 m
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• Logs
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Work embedment example • • • •
Bulk density of soil = 1.88E4 N/m3 Void ratio = 0.38 Saturated bulk density of soil = 2.15E4 N/m3 Submerged bulk density of soil = 2.15E4 – 0.981E4 = 1.17E4 N/m3
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• Soil
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Work embedment example
Vsoil sub = 5 x 0.4 x 1 = 2.0 Vsoil dry = 5 x 0.4 x 1 = 2.0 Fsoil (1 log) = 2.0 x 1.17E4 + 2.0 x 1.88E4 = 6.10E4 For 8 logs = 4.88E5 N
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Vertical force due to soil
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Horizontal resisting force--Ballast
Dsub = D – Dw Ddry = Dw
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Horizontal resisting force due to soil
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Work embedment example Dsub = D – Dw= 1 Ddry = Dw = 1
sv = (1 x 1.17E4) + (1 x 1.88E4) = 3.05E4 N/m2 Kp = (1+ sin 30)/(1 – sin 30) = 3.0 Fpassive = -0.5 x 3 x 3.05E4 x 5 x 0.4 = -9.15E4 N for one log or -7.32E5 N for 8 logs
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Horizontal force due to soil
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Vertical resisting force--pilings
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• • • •
• • • •
Number of pilings = 9 wood = 4.41E3 N/m3 Lpiles, burial depth = 3 m hload, height above potential scour depth that load is applied Diameter of pilings = 0.4 m Soil friction angle, f = 30 deg Ks, coefficient of lateral earth pressure = 1.28 Ks = (Ka + Kp +Ko)/3
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Work pilings example
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s’ = 3 x 1.17E4 = 3.51E4 N/m2 Fpiles-v = 9 x p x 0.4 x 3 x [1.28 x tan(2/3 x 30) *3.51E4 + (0.4/4) (9.81-4.41)E3] = 33.9 [1.62E4 - 0.54E3] = 5.31E5 N
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Work pilings example
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Horizontal resisting force--pilings
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Fpiles-h = -9 x (33 x (1/2) x 0.899E4 x 0.4 x 3.0)/(2 + 3)
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Work pilings example
= -2.63E5 N 27
Two approaches for analyzing forces and moments
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• Treat the entire LW structure as a single mass (Wright, Brooks, TS14J, TS14E, LWNM, Knutson and Fealko) • Perform a force/moment balance for each member (Rafferty)
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Flow force computations flow logs
cables
Earth anchors
stream bed
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• Using wood, CD, and CL analyze forces and moments using a free body diagram • This approach lends itself to spreadsheet applications
Fb
FL
Cables and earth anchors are used for restraint here, but other methods may be better!
3
Fc L1
q1
Fd
hs
Fc q2
L2
1
Fa
29 2
Ls
Fa
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If your structure is like this‌.
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Then free body diagram‌.
Note that we are treating the entire LW structure as a single body. What are the assumptions we are making when we do this?
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Force balance
Driving forces Resisting forces
SFx = 0 = Fdrag - Fcable (sinq1- sinq2) SFy = 0 = Fbuoyant + FLift- Fcable(cosq1 + cosq2) q1 = arctan(hs/L1)
q2= arctan(hs/L2)
Ignores contribution of bed friction and ballast to resisting forces Ignores geotechnical forces on buried members Doug Shields, Jr.
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Summation of forces at point 3
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www.friendofrivers.com
Simplifications to force/moment analysis • Don’t assume cables • Simply compute driving forces as
• Simply compute resisting forces as • Horizontal = friction • Vertical = ballast
• Ignore moment analysis • Ignore geotechnical forces on buried members • This approach used by Australians for ELJ-type structures with ballast fill (Brooks 2006)
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• Horizontal = Drag • Vertical = Lift + Net Buoyancy
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Comparison of design tools 2006 Brooks 2003 Wright
2007 TS14J
2007 TS14E
2011 Abbe 2014 Knuston Brooks Fealko 2014 Rafferty
Buoyancy
X
X
X
X
X
X
X
Drag
X
X
X
X
X
X
X
Lift
X
X
Hydrostatic Force
X
Friction
X
Ballast
X
X X
X
X
Piles Embedment
*
X
Anchors
Scour
X
Spacing Sheet
X X
mine
X On web, but locked
x
X
X
X
Vertical only
X
X
X
X
X
X
X X mine
*equations provided, but not in spreadsheet
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Quantity
? mine
On web
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Brooks 2006 spreadsheet definitions Soil fill and piles not Design spreadsheet adapted from Table 9.2 in Brooks 2006 included in Log Log spreadsheet
Type No. of logs/layer Key footer 1 Key logs 4 Cross spanners 2 Longitudinal logs (row 1) 4 Longitudinal logs (row 2) 3 Cross spanners 4 Longitudinal logs (row 1) 4 Longitudinal logs (row 2) 3 Diagonal logs 2 Totals 27
Structure dimensions
Log length 9 10 9 10 10 9 10 10 10
diameter 0.35 0.55 0.35 0.45 0.35 0.35 0.45 0.35 0.35
radius 0.18 0.28 0.18 0.23 0.18 0.18 0.23 0.18 0.18
Log volumes 0.87 9.50 1.73 6.36 2.89 3.46 6.36 2.89 1.92 35.98
Width
Height
Length
XS area
Rootwad radius 0.44 0.69 0.44 0.56 0.44 0.44 0.56 0.44 0.44
Cumulati Rootwad Rootwad Cumulative e width, thickness volumes height, m m 0.70 0.14 0.35 1.10 2.18 0.90 6.60 0.70 0.28 1.25 0.90 1.19 1.70 0.70 0.42 2.05 0.70 0.56 2.40 0.90 1.19 2.85 0.70 0.42 3.20 0.70 0.28 3.55 6.67
Structure Volume Wood volume
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xxx
Why is it not a bad idea to neglect burial of base?
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Brooks 2006—prepared for this course xxx
Layer 1 2 3 4a 4b 5 6a 6b 7
Design spreadsheet adapted from Table 9.2 in Brooks 2006
Type No. of logs/layer Key footer 1 Key logs 4 Cross spanners 2 Longitudinal logs (row 1) 4 Longitudinal logs (row 2) 3 Cross spanners 4 Longitudinal logs (row 1) 4 Longitudinal logs (row 2) 3 Diagonal logs 2 Totals 27 Structure dimensions
Log length 9 10 9 10 10 9 10 10 10
Width 6.60
Vertical forces
Log diameter 0.35 0.55 0.35 0.45 0.35 0.35 0.45 0.35 0.35
d Buoyant force, Fb Velocity Lift coefficient, CL Lift force, FL Specific weight of ballast, bl Volume of ballast Weight of ballast, Wbl Submerged weight of ballast, Wbl(sub) Factor of safety, Fsv
9,810
Friction angle for bed sediments, f
40 Equation nos. in notes Coefficient of refer to those in friction between structure and bed 4,415 Brooks 2006 0.84 230,135 Force normal, Fn 1,301,496 6.51 Force of friction, Ff 1,092,085 Drag 1.0 coefficient, CD 1 37,957 Drag force, Fd 372,362 18,000 N/m3 191.65 m3 3,449,644 N
Rootwad radius 0.44 0.69 0.44 0.56 0.44 0.44 0.56 0.44 0.44
Cumulativ Rootwad Rootwad Cumulative e width, thickness volumes height, m m 0.70 0.14 0.35 1.10 2.18 0.90 6.60 0.70 0.28 1.25 0.90 1.19 1.70 0.70 0.42 2.05 0.70 0.56 2.40 0.90 1.19 2.85 0.70 0.42 3.20 0.70 0.28 3.55 6.67
Structure Volume Wood volume 234.30 42.65
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1,569,588 N 5.85
Log volumes 0.87 9.50 1.73 6.36 2.89 3.46 6.36 2.89 1.92 35.98
Height Length XS area 3.55 10 17.57
Horizontal forces w
Log radius 0.18 0.28 0.18 0.23 0.18 0.18 0.23 0.18 0.18
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Structure No.
Factor of safety, Fsh
2.93
Knutson and Fealko Shields Engineering, LLC
• Another spreadsheet prepared for this course
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Wright, River Design Group • Soil • Boulders
• • • •
Sliding Embedment Spacing Scour
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• Buoyancy • Ballast
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Wright, River Design Group
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• Embedment depth. • Empirical formula does not seem to match any of the others.
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Wright, River Design Group
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No spreadsheet Equations provided, some errors.
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Abbe and Brooks 2011
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Factors of safety Moment analysis is often ignored for LW design Shields Engineering, LLC
• FOS = S Fstab/ S Fdestab or….. • FOS = S Mstab/ S Mdestab or….. • FOSx = S Fxstab/ S Fxdestab [“sliding”], and • FOSy = S Fystab/ S Fydestab [“floating”],and • FOSx = S Mxstab/ S Mxdestab [“overturning”] • FOSy = S Mystab/ S Mydestab [“rotating”] • Start with FOSy…..because….
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Quick and dirty FOS
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Typical FOS for design (p. 6-40, LWNM)
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• Shields and Alonso (2007) Monte Carlo…Fs = 2.3 to 2.5 • TS 14J…Fs for anchoring > 2 • Rafferty…1 < Fs < 3 depending on the level of risk posed by structure failure
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Typical FOS for design
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Design calculations summary • Can be complex or simple • individual members or composite structure • forces or forces and moments • • • •
Buoyancy Drag Lift Impacts from floating wood
• Restraining forces • • • •
Ballast Friction Embedment Pilings
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• Driving forces==
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