b'solidsquareshaftproductsonPage6. This d Largest= Largest Plate Dia. (12 in = 1 ft)anchorconfigurationwillprovideultimate L 0= 7 ft + (10 x 1 ft) = 17 feetcapacity required for tension support of the wall L 0 = 17 feet Min. Horiz. Embedmentwhen spaced along the wall at 5 feet center to 7.Calculate the Critical DepthD cr :center. Use 6 x d Largest plate. (Discussed Page 32)Table 10. Projected Areas* of Helical Torque Anchor Plates (See Results for Tieback Design, next page.)Helical 6 8 10 12 14 16 D cr = 6x1 (ft) = 6 feet.Plate Dia. Dia. Dia. Dia. Dia. Dia. 8.SelectInstallationAngleandDetermineShaft Projected Area ft 2 ProductLength: Positiontheanchorsto1-1/2 Sq. 0.181 0.333 0.530 0.770 1.053 1.381 penetratethewallattwofeetbelowthesoil1-3/4 Sq. 0.175 0.328 0.524 0.764 1.048 1.375 surface. (Note:This location isthreefeet2 Sq. 0.168 0.321 0.518 0.758 1.041 1.396 down from top of basement wall.) In Step 72-7/8 Dia 0.151 0.304 0.500 0.740 1.024 1.351 above, itwasdeterminedthattherequired3-1/2 Dia 0.130 0.282 0.478 0.719 1.002 1.329 Critical Depth, (D cr ), is 6 feet, which means that4-1/2 Dia 0.086 0.239 0.435 0.675 0.959 1.286 the12diameterplatemustterminateat an5. Installation Torque: Use Equation 2 - elevation least4feetlowerthanwheretheChapter 1, or use Graph 6 from Chapter 1 shown anchorshaftpenetratedthewall.Selectanin the example above to calculate the installation installationangleof15 0 anddeterminethetorque requirement for this anchor. minimuminstalledproductlengththatwillT = T u/ k, Where, provide the needed extra soil depth requirementT u= 22,050 lb of 4 feet above the 12 plate that will insure thek = 10 (Table 12, below from Chapter 1) needed critical depth. his can be determined asT = 22,050 lb / 10 ft -1 follows:T = 2,200 ft-lb L 15 deg= 4 ft / sine 15 0(Table 13, Chapter 1)The torque must be developed for a distance that L 15 deg= 4 ft / 0.259 = 15-1/2 ftThe minimum distance from the wall to the 12is long enough to insure that the helical plates are plate when installed at a 15 0downward angle isproperlyembedded anddeveloptherequired 15-1/2 feet to insure meeting the D cr= 6 feet.tension capacity. The torque requirement must Comparing the minimum horizontal embedmentbeaveragedoveradistanceofatleastthree length of 17 feet from Step 6 to the 15-1/2 foottimesthediameterofthelargestplate.The lengthrequiredforobtainingCriticalDepthat2,200 ft-lbs must be measured continuously for a 15 0installation angle; it is clear that 17 feet ofminimumdistanceof3feetbefore terminating horizontal length of embedment from the wallTable 12. Soil Efficiency Factor k is the controlling distance. Typically Suggested TherewillbeanadditionallengthofshaftTorque Anchor Encountered Average Value, required to get to the 12 inch diameter plate toType Range k k the required distance of 17 feet due to the 15 01-1/2 Sq. Bar 9 - 11 10 downward installation angle of the shaft needed1-3/4 Sq. Bar 9 - 11 10 to achieve Critical Depth, (D cr ).Here is how to2 Sq. Bar 8.5 ( Compression) 8.5 calculate the additional shaft length needed due10 (Tension) 10 to the 15 0 downward installation angle.2-7/8 Diameter 8 - 9 9 Use the equation shown on Chapter 1 - Table 133-1/2 Diameter 7 - 8 8 for a 15 0downward angle to determine the shaft4-1/2 Diameter 6 - 7 7 length to the 12 inch diameter plate.the installation.(12 diameter plate x 3) L 15 deg= [H + (10 d largest )] x 1.0356.MinimumHorizontalEmbedment: L 15 deg= [7 ft + (10 x 1 ft] x 1.035 = 17.6 feetDeterminethe MinimumEmbedmentLength Total Shaft Length Needed: L Total= L 15+ L TipfromEquation9inChapter1orFigure3in Where, L Tip= 3D 10 dia .Chapter 1. L Total= 17.6 ft + (3 x 10)/12L 0= H + (10 x d Largest ) Where, L Total= 17.6 ft + 2.5 ft = 20.1 ftH = Height of Soil - (7 ft) Use: L Total= 20 ftat= 15 0 (Minimum)ECP Technical Design ManualTorque Anchor Design Examples2021 Earth Contact Products, L.L.C.2021-09 Chapter 5Page 90 All rights reserved'