stone design material in this section is drawn primarily from neh 654 ts 14c and 14k jon fripp...
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Stone DesignStone Design
Material in this section is drawn primarily from NEH 654 TS 14C and 14KMaterial in this section is drawn primarily from NEH 654 TS 14C and 14K
Jon FrippNDCSMC
Ft. Worth, TX
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Rock Design
• Uses of rock
• Sizing methods
• Examples
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Why use rock?
High stress areasHigh riskEmergency situations
•Quick response•Ecological Implications?•Geomorphic Implications?
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Why use rock?
Result: Static channel boundary.Is this what is needed?
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Photo from Meg Jonas
Photo from Jim Ludlam
May be part of Streambank Soil Bioengineering
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USACE WES Rip Rap test facility
How Big?What Gradation?What Shape?What Density?What Quality?
What do we need to know?
If we are going to use rock – we need to do it right
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Size
Answer: The particle size for which 50% of the sample is finer.
What is D50?
How big does the rock need to be?
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Stone Sizing
FF = Force of flowing water
FD = Drag force
FL = Lift force
FW = Submerged weight
FC = Contact or interlock force
Empirical methods were developed for specific applications.
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Stone Sizing
Bottom Line: Match the rock sizing method with the intended use.
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High vs. Low Energy
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Mild slope - Low Energy
Steep Slope - High EnergyNote: there are exceptions
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Parallel flow – low energy
Impinging flow – high energyNote: there are exceptions
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What variables do the equations account for?
•Usually stone size is the dependent variable•Is it part of a defined gradation?
•Independent variables can include:•Velocity•Depth•Energy slope•Bed slope•Side slope•Rock shape•Rock density•Others?
•Are those important?
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USGS
44.201.050
Vd Arizona field data where riprap performed without damage.
d50 is median stone size (inches)V is channel velocity (fps)
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Report 108 – (1970)
4/50 eRSd Developed for roadside drainage channels.
d50 is median stone size (inches)Se is energy slope (ft/ft)R is hydraulic radius (ft)
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LANE’S Far West States Method
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USACOE - Maynord52
1
5
30
.
DgK
V.
WS
WDTCvCsCSFd
dm = Stone size (ft); m percent finer by weight
D water depth in feetCs Stability Coefficient Z=2 or flatter C=0.30Cs = Stability coefficient (0.3 for angular rock, 0.375 for rounded rock)Cv = Velocity distribution coefficient (1.0 for straight channels or inside of bends, calculate for outside of bends)CT = Thickness coefficient (use 1.0 for 1 D100 or 1.5 D50, whichever is greater))
USACE WES Rip Rap test facility
For slopes < 2%
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USACOE
R= center-line bend radius W = water surface width
)log(2.0283.1 WRCv
2
2
1 sin
sin1 K
rockangular for degrees) 40 (typically repose of angle
horizontalrock with of angle
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USACOE – Boulder Design
)1(
)(18
SG
Sdepthd f
d = Minimum stone size (ft) depth = channel depthSf = channel friction slope SG = specific gravity of the stone
EMRRP-SR-11
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Abt Rock Design
23.535.1 43.056.050 SqD
D50 is median stone size (inches)S channel slope in (ft/ft)q is unit discharge (ft3/ft)
Slopes: 2% to 20%
Rock design for spillways or loose rock grade controlSteven R. Abt, and Terry L. Johnson (1991)
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529.05.150 923.112 qSD
0.02 < S < 0.1 For Slopes between 2% and 10%
ARS Rock Chutes
0.10<S<0.40
529.058.050 233.012 qSD
D50 is median stone size (inches)S channel slope ft/ftq is unit discharge (ft3/ft)
For Slopes between 10% and 40%
K. M. Robinson, C. E. Rice, and K. C. Kadavy (1998)
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Rock Chute Design Spreadsheet
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Given:GS = 2.65, Unit wt of stone 165.36 lb/ft3
Bottom Width = 40 ftn = 0.045Slope = 0.06 ft/ftDepth = 3.5 ft
Find:Appropriate rock size
Velocity = 16.7 ft/sQ = 2,340 ft3/sCritical Depth = 4.7 ft
Maynord D50 = 1.9 ftLane’s FWS D50 = 3.2 ftAbt and Johnson D50 = 1.3 ftARS rock chute D50 = 1.1 ft
Not appropriate for givens
Example Problem
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Given:GS = 2.6Bend Radius = 350 ftChannel width = 50 ft Side slope = 2:1Slope = 0.01 ft/ftDepth = 5 ft
Find:Appropriate rock size using Lane’s FWS technique
Example Problem
Use graph in back of handout
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Stone Sizing – Final Thoughts•Use a rock sizing method appropriate for your application
•Use several methods and look for convergence•But do not expect exact convergence
•Use a factor of safety appropriate for your situation
•Assess significant threats to life and property
•Size may need to be larger than what the equations indicate as sufficient to resist flows
•To resist ice and debris•For habitat enhancement•For aesthetic purposes•To reduce vandalism and theft
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Questions?