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2013-07-19_REVISION - M1977311
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2013-07-19_REVISION - M1977311
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Last modified
6/15/2021 2:28:38 PM
Creation date
7/25/2013 3:49:09 PM
Metadata
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Template:
DRMS Permit Index
Permit No
M1977311
IBM Index Class Name
REVISION
Doc Date
7/19/2013
Doc Name
AM1 AR6 RESPONSE
From
COTTER
To
DRMS
Type & Sequence
AM1
Email Name
TC1
Media Type
D
Archive
No
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Attachment #2 <br />Riprap Sizing Calculations (1 page) <br />Federal Highway Administration Method (4 pages) <br />Cotter SR4 3A <br />Riprap was sized using the Federal Highway Administration HEC -11 methodology (Chapter 4: Design <br />Guidelines for Rock Riprap). Copies of the relevant portions of this document are included in this <br />Attachment #2 following these calculations. In general, this method provides a required D50 size for <br />riprap in a straight channel with known side - slopes, flow- depth, and discharge velocity. The equation is <br />metric and is as follows: D50 = 0.00594 (V)3 / do-5 W-5 where V = channel velocity (meters/sec), d = <br />normal depth (meters), and K is a variable dependent on the riprap side -slope and angle of repose. <br />K = (1 —sin 26 /sin2�)05, where 0 = riprap side -slope angle from horizontal (@ 3:1) , and = riprap angle of <br />repose (from horizontal, conservatively estimated at 1:1). <br />Therefore: K = 0.7746 (2:1 sides), K = 0.8944 (3:1 sides), K = 0.9900 (10:1 sides) <br />Proposed Road/Runoff Diversion Ditch, Steep Slope (Section F -F), 3:1 side slopes <br />Note: This section analyzed at steepest slope (12 %) and lowest n (0.035) to give highest velocity <br />V = 5.92 fps = 1.8044 mps (meters per sec) (from worksheets in attachment #1) <br />d = 0.54ft = 0.1646m (meter) <br />D. = 0.00594 (V)s / do.s K'.s <br />D50 = 0.00594 (1.8044)'/ (0.1646)0.5 (0.8944)' S = 0.1017 meter = 4.00 inches <br />Therefore: 6" D50 specified is adequate <br />Pond Discharge Chute (Section D -D), 10% slope, 3:1 side slopes <br />Note: This section analyzed at lowest n (0.035) to give highest velocity <br />V = 5.17 fps = 1.5758 mps (meters per sec) (from worksheets in attachment #1) <br />d = 0.29 ft = 0.0884 m (meter) <br />D50 = 0.00594 (V)3 / do .5 K1 -5 <br />D50 = 0.00594 (1.5758)3 / (0.0884)0-5(0.8944)1 *5 = 0.0924 meter = 3.64 inches <br />Therefore: 6" D50 specified is adequate <br />
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