Edward C. Robison, PE - CRL-ARCH stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21: F t = 19 ksi and F c = 23.9-0.238[(2L ... 5 23.5 17.6 15.6 14.1 12.8 11.7

11 May 2012C.R. Laurence Co., Inc.2503 East VernonLos Angeles, CA 90058

SUBJ: ALUMINUM WINDSCREEN SYSTEM! ALUMINUM FRAMED GLASS WIND WALLS AND FENCES

!The AWS is an engineered system designed for the following criteria:! The design loading conditions are:! Concentrated load = 200 lbs (1 sf area) @ 42” above grade or,! Distributed load = 50 plf @ 42” above finish grade or,! Concentrated load = 50 lbs on 1 sf area or,! Uniform load = 10 psf or,! Seismic loads will not affect design because of the small dead loads.! Wind load as calculated based on ASCE/SEI 7-05 and as limited for the specific configuration as shown in tables 2 to 11 as applicable.

For these conditions the system will meet or exceed all requirements of the 2006 and 2009 International Building Codes and International Residential Codes along with state codes adopting the IBC and IRC and 2005 Aluminum Design Manual. The system will meet all requirements for a swimming pool enclosure when installed as recommended and in compliance with IBC Section 3109. When fall protection is required a top rail is required or a grab rail must be installed between 36” and 42” above the walking surface. Refer to the appropriate tables herein to determine allowable post spacing, heights and wind loads. The supporting structure shall be designed by others and be adequate to support the AWS with all imposed loads.

Calculation! ! ! ! PageSignature/Seals! ! ! 2Post Loading!! ! ! 3! ! Glass Strength! ! ! 12Wind load! ! ! ! 4! ! Glass Allowable Load Tables! 13 - 142” X 2–5/8” Post! ! ! 5 – 6! ! Post/Stanchion Embedment! 153” Round Post 7 – 8 Embedment in CMU! ! 162-5/8” Barrier Post 9 Base Plate Option! ! ! 17Stanchion 10 Guage!! ! 10! ! Base Plate Anchorages! ! 18 – 19Stanchion 1/4”! ! ! 11! ! Design Steps/Example! ! 20

Edward Robison, P.E.

Edward C. Robison, PE

10012 Creviston DR NW 253-858-0855Gig Harbor, WA 98329 FAX 253-858-0856

Signed 11 May 2012

Aluminum Wind Screen System - 11 May 2012 Page 2 of 20

Edward C. Robison, PE10012 Creviston DR NW 253-858-0855Gig Harbor, WA 98329 FAX 253-858-0856

Loading to Posts: Live load = 200# @ 42” heightAny location along wall(42” above finish floor)Or:50 lb on one square foot at any location on glass.Or:Wind load on solid areaOr:10 psf live load on entire area including voids.

M200 = 200#x42” = 8,400”#

M50plf = 50*42”*SMaximum spacing when fall protection is required:S = Ma/2,100”#/ft

M50= 50lb x H*12”/ft(will not govern post design)

MLL = 10psf*(S)*(H2/2)*12”/ft

MWL= W*(S)*(H2*0.55) (‘#)Wind loading typically controls post design.

Determine the maximum post heightsMa = Allowable post momentfor Wind load:MWL= W*(S)*(H2*0.55) = Ma

Solving for SS = Ma/(0.55*W*H2)

Solving for HH = [Ma/(0.55*W*S)]1/2

Allowable wind load: W = Ma/(0.55*S*H2)

S

H

1SF

50#

WIND LOAD = ?ORLL = 10 PSF

1SF

200#

42"



WIND LOADING ON WIND SCREENS AND FENCESCalculated in accordance with SEI/ASCE 7-05 Section 6.5.14 Design Wind Loads on Solid Freestanding Walls and Solid Signs. This section is applicable for free standing guardrails, wind walls, and fences :

F = qhGCfAs As = solid surface area perpendicular to the wind directionFor wind walls/fences the coefficients have the following values: G = 0.85 from section 6.5.8.1 for a rigid structure (ω=√[h*m/(⅓*0.855)] ≥ 1.0). Cf From Figure 6-20 - varies depending on height and length qh = 0.00256KzKztKdV2I Where: I = 1.0 or 0.87 from Table 6-1 Kz from Table 6-3 at the height z of the railing centroid and exposure. Kd = 0.85 from Table 6-4. Kzt From Figure 6-4 for the site topography, typically 1.0. V = Wind speed (mph) 3 second gust, Figure 6-1 or per local authority.Simplifying - Assuming 1.3 ≤ Cf ≤ 2.6 (Typical limits for fence with returns.)

For Cf = 1.3: F = qh*0.85*1.3 = 1.11 qhFor Cf = 2.6: F = qh*0.85*2.6 = 2.21qh

Wind Load will vary along length of fence in accordance with SEI/ASCE 7-05 Figure 6-20.Typical exposure factors for Kz with height 0 to 15’:Exposure B C DKz = 0.70 0.85 1.03

Centroid of wind load acts at 0.55h on the fence.Typical wind load for I = 1.0 and Kzt = 1.0Table 1: Wind load in psf Cf = 1.3 Wind load in psf Cf = 2.60 Wind Speed B C D B C DV 0.00169V2 0.00205V2 0.00249V2 0.00337V2 0.00409V2 0.00495V2

85 12.2 14.8 17.9 24.3 29.5 35.890 13.7 16.6 20.2 27.3 33.1 40.1100 16.9 20.5 24.9 33.7 36.9 49.5110 20.5 24.8 30.1 40.7 49.5 59.9120 24.3 29.6 35.8 48.5 58.9 71.3130 28.6 34.7 42.0 56.9 69.1 83.7140 33.1 40.2 48.8 66.0 80.1 97.1Where fence ends without a return the wind forces may be as much as 1.667 times Cf=2.6 value.When I = 0.87 is applicable (occupancy category I) multiply above loads by 0.87.MINIMUM WIND LOAD TO BE USED IS 10 PSF.SPECIFIER SHALL VERIFY WIND LOADS FOR THE SPECIFIC INSTALLATION IN ACCORDANCE WITH SEI/ASCE 7-05 Section 6.5.14 AND FIGURE 6-20.



POST OPTIONS: 2” X 2–5/8” PostArea: 1.135 sq inIxx: 0.855 in4 Iyy: 0.611 in4

rxx: 0.868 in ryy: 0.734 in and J = 1.292 in4

Cxx: 1.3125 in Cyy: 1.00 inSxx: 0.651 in3 Syy: 0.611 in3

Allowable stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21:Ft = 19 ksi and Fc = 23.9-0.238[(2LbS)/(IJ)1/2]1/2 Fc = 23.9-0.238[(2*72*0.611)/(0.611*1.292)1/2]1/2 = 21.5 ksi but ≤ 21 ksiMa = Sxx*Fb = 0.611 in3*19 ksi = 11,609”# = 967.4’#

Post Variations:90˚ Corner PostArea: 1.200 sq inIxx: 1.002 in4 Iyy: 1.002 in4



Allowable stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21:Ft = 19 ksi and Fc = 23.9-0.238[(2LbS)/(IJ)1/2]1/2 Fc = 23.9-0.238[(2*72*0.735)/(0.735*1.669)1/2]1/2 = 21.1 ksi but ≤ 21 ksiMa = Sxx*Fb = 0.735 in3*19 ksi = 13,965”# = 1,163.7’#

135˚ Corner PostArea: 1.503 sq inIxx: 1.582 in4 Iyy: 1.002 in4



Allowable stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21:Ft = 19 ksi and Fc = 23.9-0.238[(2LbS)/(IJ)1/2]1/2 Fc = 23.9-0.238[(2*72*0.707)/(0.707*1.868)1/2]1/2 = 21.7 ksi but ≤ 21 ksiMa = Sxx*Fb = 0.707 in3*19 ksi = 13,433”# = 1,119.4’#Note: Loading from both sides will contribute to bending.



The standard straight post will typically govern the wind screen design. Use the equations derived in page 2 determine the allowable wind loads based on the post strength (post directly core mounted in grout or other method that will develop the fill post strength.)

Solving for SS = 967.4’#/(0.55*W*H2) = 1,759’#/(W*H2) Example determine required post spacing for 20 psf wind load and 4’-0” screen height:S = 1,759’#/(20*42) = 5’-6”

Solving for HH = [1,759’#/(W*S)]1/2

Example determine maximum screen height for 20 psf wind load and 6’-0” post spacing:H = [1,759’#/(20*6)]1/2 = 3’ - 10”

Allowable wind load: W = 1,759’#/(S*H2)Example determine maximum wind load for 4’ screen height and 6’-0” post spacing:W = 1,759’#/(6*42) = 18.3 psfTable 2: 2" Post Post strength (ft-#)= Post strength (ft-#)= 967.4

Wind load Post SpacingPost SpacingScreen Height 3 4 4.5 5 5.5 6

3 65.1 48.9 43.4 39.1 35.5 32.63.5 47.9 35.9 31.9 28.7 26.1 23.94 36.6 27.5 24.4 22.0 20.0 18.3

4.5 29.0 21.7 19.3 17.4 15.8 14.55 23.5 17.6 15.6 14.1 12.8 11.7

5.5 19.4 14.5 12.9 11.6 10.6 NA6 16.3 12.2 10.9 NA NA NA

Based on post strength, assumes anchorage method will develop the full post strength.NA = Not Allowed.Maximum spacing when fall protection is required:S = Ma/2,100”#/ftS = 11,609”#/2,100”#/ft = 5.528’ = 5’6 1/3”



3” Round PostArea: 1.145 sq inIxx: 0.829 in4 Iyy: 0.964 in4




90˚ Corner PostArea: 1.218 sq inIxx: 1.070 in4 Iyy: 1.070 in4




135˚ Corner PostArea: 1.313 sq inIxx: 1.143 in4 Iyy: 1.275 in4 Ixy: 1.309 in4


Cxx: 1.556 in Cyy: 1.556 in Cxy: 1.610 inSxx: 0.688 in3 Syy: 0.688 in3 Sxy: 0.813 in3

Allowable stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21:Ft = 19 ksi and Fc = 23.9-0.238[(2LbS)/(IJ)1/2]1/2 Fc = 21 ksiMa = Sxy*Fb = 0.813 in3*19 ksi = 15,447# = 1,287.2’#Note: Loading from both sides will contribute to bending.



The standard straight post will typically govern the wind screen design. Use the equations derived in page 2 determine the allowable wind loads based on the post strength (post directly core mounted in grout or other method that will develop the fill post strength.)

Solving for SS = 1,017.6’#/(0.55*W*H2) = 1,850.2’#/(W*H2) Example determine required post spacing for 20 psf wind load and 4’-0” screen height:S = 1,850.2’#/(20*42) = 5’-9”

Solving for HH = [1,850.2#/(W*S)]1/2

Example determine maximum screen height for 20 psf wind load and 6’-0” post spacing:H = [1,850.2’#/(20*6)]1/2 = 3’-11”

Allowable wind load: W = 1,850.2’#/(S*H2)Example determine maximum wind load for 4’ screen height and 6’-0” post spacing:W = 1,850.2’#/(6*42) = 19.3 psf

Table 3: Allowable wind loads (psf) on 3” Round Post3" Post Post strength (ft-#)= Post strength (ft-#)= 1017.6


3 68.5 51.4 45.7 41.1 37.4 34.33.5 50.3 37.8 33.6 30.2 27.5 25.24 38.5 28.9 25.7 23.1 21.0 19.3

4.5 30.5 22.8 20.3 18.3 16.6 15.25 24.7 18.5 16.4 14.8 13.5 12.3

5.5 20.4 15.3 13.6 12.2 11.1 10.26 17.1 12.8 11.4 10.3 NA NA

Based on post strength, assumes anchorage method will develop the full post strength.NA = Not Allowed.

Maximum spacing when fall protection is required:S = Ma/2,100”#/ftS = 12,211”#/2,100”#/ft = 5.815’ = 5’ -9 3/4”



2-3/4” X 2–5/8” PostBarrier System PostArea: 1.74 sq inIxx: 1.634 in4 Iyy: 1.336 in4



Allowable stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21:Ft = 19 ksi and Fc = 23.9-0.238[(2LbS)/(IJ)1/2]1/2 Fc = 23.9-0.238[(2*72*1.188)/(1.634*2.051)1/2]1/2 = 21.6 ksi but ≤ 21 ksiMa = Sxx*Fb = 1.188 in3*19 ksi = 22,572”# = 1,881’#

Use the equations derived in page 2 determine the allowable wind loads based on the post strength (post directly core mounted in grout or other method that will develop the fill post strength.)

Solving for SS = 1,881’#/(0.55*W*H2) = 3,420’#/(W*H2) Example determine required post spacing for 30 psf wind load and 5’-0” screen height:S = 3,420’#/(30*52) = 4’ 7”

Solving for HH = [3,420#/(W*S)]1/2

Example determine maximum screen height for 30 psf wind load and 5’-0” post spacing:H = [3,420’#/(30*5)]1/2 = 4’-9”

Allowable wind load: W = 3,420’#/(S*H2)Example determine maximum wind load for 4’ screen height and 6’-0” post spacing:W = 3,420’#/(5*52) = 27.4 psf

Table 4: Allowable wind loads (psf) on 2-5/8” Barrier Post2-5/8" post Post strength (ft-#)= Post strength (ft-#)= 1881


3 126.7 95.0 84.4 76.0 69.1 63.33.5 93.1 69.8 62.0 55.8 50.8 46.54 71.3 53.4 47.5 42.8 38.9 35.6

4.5 56.3 42.2 37.5 33.8 30.7 28.15 45.6 34.2 30.4 27.4 24.9 22.8

5.5 37.7 28.3 25.1 22.6 20.6 18.86 31.7 23.8 21.1 19.0 17.3 15.8

Based on post strength, assumes anchorage method will develop the full post strength.

2 5/8"

2 3/

4"

1/8"



10 Gauge StanchionStanchions are break formed from HR Steel A1011 or A572 Grade 50 alloy steel (Fy ≥ 50 ksi) powder coated, or 304 Stainless steel, ASTM A666 1/8 hard (Fy ≥ 50 ksi).t = 0.135”A = 0.546 in2

Ixx = 0.124 in4

Iyy = 0.197 in4

Cxx = 0.926”Cyy = 0.750”Zxx = 0.231 in3

Zyy = 0.310 in3

The stanchions are installed so that primary bending axis is YY with essentially no bending in the XX direction.Stanchion strength: t/b = 1.25/0.135 = 9.26 < 20Compression buckling of the flange is prevented because of confinement in grout and in the post above the grout therefore stanchion will develop the full plastic section at yield:øMnyy = øFyZyyø = 0.9Determine the service moment on the stanchions based on a typical load factor of 1.6 (live or wind loads).Ms = øMn/1.6

øMnyy = 0.9*50 ksi*0.310 = 12,555”# = 1,046.25’#Ms = 12.555”#/1.6 = 7,847”# = 653.9’#S = 653.9’#/(0.55*W*H2)

Table 5: Allowable wind loads (psf) on 10 gauge Stanchion (Any post)PST4 stanchion Post strength (ft-#)= Post strength (ft-#)= 653.9


3 44.0 33.0 29.4 26.4 24.0 22.03.5 32.4 24.3 21.6 19.4 17.6 16.24 24.8 18.6 16.5 14.9 13.5 12.4

4.5 19.6 14.7 13.0 11.7 10.7 NA5 15.9 11.9 10.6 NA NA NA

5.5 13.1 NA NA NA NA NA6 11.0 NA NA NA NA NA

Based on post strength, assumes anchorage method will develop the full post strength.NA = Not Allowed.Maximum spacing when fall protection is required:S = Ma/2,100”#/ftS = 7,847”#/2,100”#/ft = 3.737’ = 3’ -8 7/8”



1/4” StanchionStanchion is made from A1011 steel sheet or 304 stainless steel (1/8 hard) with Fy ≥ 50 ksi. Ixx = 0.189 in4 Iyy = 0.278 in4

Sxx = 0.221 in3 Syy = 0.371 in3

Zxx = 0.291 in3 Zyy = 0.489 in3

r = 0.550 in J = 0.070 in4

t/b = 1.25/.25 = 5 < 20 therefore local buckling won’t control and full yield strength will be developed in stanchion.

Stanchion bending strength for bending about YY (typical post installation) Mn = 0.489 x 50ksi = 24,450”#Determine allowable service load for typical LF = 1.6 for wind or live loads.Ms = øMn/1.6Ms = 0.9*24,450”#/1.6 = 13,753”# = 1,146.1’#S = 1,146.1’#/(0.55*W*H2)

Table 6: Allowable wind loads (psf) on 1/4” Stanchion (with any post)PST8 stanchion Post strength (ft-#)= Post strength (ft-#)= 1146.1


3 77.2 57.9 51.5 46.3 42.1 38.63.5 56.7 42.5 37.8 34.0 30.9 28.44 43.4 32.6 28.9 26.0 23.7 21.7

4.5 34.3 25.7 22.9 20.6 18.7 17.25 27.8 20.8 18.5 16.7 15.2 13.9

5.5 23.0 17.2 15.3 13.8 12.5 11.56 19.3 14.5 12.9 11.6 10.5 NA

Based on post strength, assumes anchorage method will develop the full post strength.NA = Not allowedMaximum spacing when fall protection is required:S = Ma/2,100”#/ftS = 13,753”#/2,100”#/ft = 6.549’ = 6’ -6 9/16”

ALTERNATIVE STANCHIONS:Custom stanchions may be produced using higher strength steels and alternative configurations to provide greater strength. For a stanchion made using steel with a different yield strength the allowable loads shall be adjusted by multiplying the tabulated value by

Fyhs/50 whereFyhs = Yield strength of steel used based on mill certification or testing.



GLASS STRENGTH

All glass is fully tempered glass conforming to the specifications of ANSI Z97.1, ASTM C 1048-97b and CPSC 16 CFR 1201. The minimum Modulus of Rupture for the glass Fr is 24,000 psi. Safety Factor of 4.0 is applicable to the glass when subject to human impact.

Allowable glass bending stress: 24,000/4 = 6,000 psi. – Tension stress calculated.

Bending strength of glass for the given thickness: I = 12”* (t)3 = (tave)3 in3/ft 12For deflection (I) use tave. S = 12”* (t)2 = 2* (tmin)2 in3/ft 6For bending (S) use tmin.

For 1/4” glass, tmin = 0.219”, tave = 0.2315”S = 2*(0.219)2 = 0.0959 in3/ft

Mallowable = 6,000psi*0.0959 in3/ft = 575.5”#/ft = 47.96’#/ft I = 0.23153 = 0.0124 in4


Mallowable = 6,000psi*0.1705 in3/ft = 1,023.2”#/ft = 85.26’#/ft I = 0.3123 = 0.0304 in4


Mallowable = 6,000psi*0.252 in3/ft = 1,512.3”#/ft = 126.0’#/ft I = 0.38053 = 0.0551 in4


Mallowable = 6,000psi*0.44 in3/ft = 2,640”#/ft = 220’#/ft I = 0.53 = 0.125 in4



GLASS IN SIMPLE SPANSFor panels simply supported on two opposite sides the moment and deflection are calculated from basic beam theory (applicable when glass is installed without structural top and bottom rails and is supported in posts only): Mw = W*L2/8 for uniform load W and span L or Mp = P*L/4 for concentrated load P and span L, highest moment P @ center

Δmax= (5/384)*wl4/(EI) = (5/374)*(w/12)l4/(10,400,000t3) = (wl4)/(9.34x109*t3)l = glass span in inches

When glass is designed for a safety factor of 4.0 or greater the deflection will not govern the allowable loading

Table 7: Allowable wind load (psf) for post spacing based on glass strengthPost spacing, feetPost spacing, feet

Glass thickness 3 3.5 4 4.5 5 5.5 61/4" 42.6 31.3 24.0 18.9 15.3 12.7 10.75/16" 75.8 55.7 42.6 33.7 27.3 22.5 18.93/8" 112.0 82.3 63.0 49.8 40.3 33.3 28.01/2" 195.6 143.7 110.0 86.9 70.4 58.2 48.9

Table 8: Check maximum glass span (post spacing, feet) for 200# concentrated load:Glass height (feet)Glass height (feet)Glass height (feet)

Glass thickness 3 3.5 4 4.5 5 5.5 61/4" 2.9 3.4 3.8 4.3 4.8 5.3 5.85/16" 5.1 6.0 6.8 7.7 8.5 9.4 10.23/8" 7.6 8.8 10.1 11.3 12.6 13.9 15.11/2" 13.2 15.4 17.6 19.8 22.0 24.2 26.4

Table 9: Check maximum glass span (post spacing, feet) for 50 plf live load:Glass height (feet)Glass height (feet)Glass height (feet)

Glass thickness 3 3.5 4 4.5 5 5.5 61/4" 4.8 5.2 5.5 5.9 6.2 6.5 6.85/16" 6.4 6.9 7.4 7.8 8.3 8.7 9.03/8" 7.8 8.4 9.0 9.5 10.0 10.5 11.01/2" 10.3 11.1 11.9 12.6 13.3 13.9 14.5



GLASS GLAZED IN STRUCTURAL BOTTOM SHOE AND POSTS (3 SIDES)When glass is supported at posts and along the bottom the glass stresses are determined from flat plate theory where: M = 1/8*wb2/[1+2(b/2h)3] or solving for w: w = 8Ma[1+2(b/2h)3]/b2

Table 10: Allowable wind load (psf) for post spacing based on glass strength3'6" Height Post spacing, feetPost spacing, feet

Glass thickness 3 3.5 4 4.5 5 5.5 61/4" 49.3 39.2 32.9 29.0 26.5 25.0 24.15/16" 87.7 69.6 58.5 51.6 47.2 44.4 42.83/8" 129.6 102.9 86.5 76.2 69.7 65.6 63.31/2" 226.3 179.6 151.0 133.1 121.7 114.6 110.5

4'0" Height Post spacing, feetPost spacing, feet Glass thickness 3 3.5 4 4.5 5 5.5 6

1/4" 47.1 36.6 30.0 25.7 22.8 20.9 19.75/16" 83.8 65.0 53.3 45.7 40.6 37.2 34.93/8" 123.8 96.1 78.8 67.5 60.0 55.0 51.61/2" 216.2 167.7 137.5 117.9 104.8 96.0 90.1


1/4" 45.8 35.0 28.2 23.7 20.6 18.5 17.05/16" 81.4 62.2 50.1 42.1 36.6 32.8 30.23/8" 120.3 92.0 74.1 62.2 54.1 48.5 44.61/2" 210.0 160.6 129.3 108.6 94.5 84.7 77.9


1/4" 44.9 34.0 27.0 22.4 19.2 16.9 15.35/16" 79.9 60.5 48.1 39.8 34.1 30.1 27.13/8" 118.0 89.3 71.1 58.8 50.4 44.4 40.11/2" 206.1 156.0 124.1 102.8 88.0 77.5 70.0


1/4" 44.0 32.9 25.8 20.9 17.6 15.1 13.35/16" 78.2 58.4 45.8 37.2 31.2 26.9 23.73/8" 115.5 86.4 67.7 55.0 46.2 39.7 35.01/2" 201.7 150.8 118.1 96.1 80.6 69.4 61.1



POST/STANCHION ANCHORAGE:Direct embedment of post:Post may be directly embedded in grout or concrete. The post end (embedded portion) shall be powder coated or otherwise protected from direct contact with the grout or concrete.

Minimum post embedment shall be adequate to develop the full post bending strength. The concrete slab or footing shall be designed to support the imposed post moment equal to the full post bending strength. Bending is resisted be shear failure in concrete by the post prying out.The concrete strength is determined in accordance with ACI 318-08 Appendix D

(ACI Appendix D D.6.3.1 and Eq D-4)Vcp = kcpNcbkcp = 2.0 (hef ≥ 2.5”)Ncb = [ANc/ANco]ϕed,Nϕc,Nϕcp,NNb

ANc= (1.5*6”+4)*(1.5*6”*2+1.5”)=253.5in2

ANco= 9*62 = 324in2

ϕed,N = 0.7+0.3*4/(1.5*6) =0.833ϕc,N = 1.4 (post installed)ϕcp,N= 4/(2.5*6) = 0.267 ≥ 1.5/2.5 = 0.6Nb = 17*1.0*√3000*61.5 = 13,685Ncb = 253.53/324*0.833*1.4*0.6*13,685 = 7,493#Vcp = 2*7,493 = 14,986#øMncp = 0.65[0.85*(5/6)*6”*14,986#] = 41,399”#Ma = 41,399”#/1.6 = 25,874”# > 13,753”# will develop full stanchion/post strength.

Check 4” embedment with 5.5” minimum edge distance:ANc= (1.5*4+5.5)*(1.5*4”*2+1.5”)=155.25 in2

ANco= 9*42 = 144in2

ϕed,N = 0.7+0.3*5.5/(1.5*4) =0.975ϕcp,N= 5.5/(2.5*4) =0.55 ≥ 0.6Nb = 17*1.0*√3000*41.5 = 7,449Ncb = 148.5/144*0.975*0.6*1.4*1*7,449 = 6,577#Vcp = 2*6,577 = 13,154#øMncp = 0.65[0.85*(5/6)*4”*13,154#] = 24,225”#Ma = 24,225”#/1.6 = 15,141”# > 13,753”# will develop full stanchion/post strength.

Check for direct embedment of 2-5/8” barrier post – increase minimum edge distance to 5”:ANc= (1.5*6”*2)*(1.5*5”*2+2.625”)=317.25in2

ϕed,N = 0.7+0.3*5/(1.5*6) =0.867Ncb = 317.25/324*0.867*1.4*0.375*13,685 = 6,099#Vcp = 2*6,099 = 12,199#øMncp = 0.65[0.85*(5/6)*6”*12,199#] = 33,700”#



CONCRETE MASONRY UNIT CONSTRUCTION (CMU) When stanchions or posts are embedded into the grouted cells of CMU:The CMU wall shall be designed for the imposed moments from the posts.The stanchion shall be embedded a minimum of 15” into the CMU.The minimum wall thickness shall be 8” nominal.A bond beam with (2) #3 bar or larger shall be constructed along the top course or as other engineering requires to accommodate the AWS loading. The reinforcement bar shall pass between the stanchion/post and each face of the wall.

Additional reinforcement may be required depending on project requirements and specific AWS configuration.Maximum allowable moment for this detail is 9,600”# per post.

Other CMU wall configurations shall be engineered to support the imposed loads from the AWS posts.

Surface mounted base plate applications shall be engineered for the specific application.



BASPLATE MOUNTED STANCHIONStanchion is welded to base plate be inserting stanchion through a water-jet cut hole in the base plate and then fully welded from the bottom and ground flat.

Check base plate strength: 6”x6”x7/16” A36 steel or stainless steelZ = 6”*0.4375”2/4 = 0.287in3

Maximum allowable bolt tension loadøMn = 0.9*Z*Fy = 0.9*0.287*36 = 9,302#”øTn = øMn/(2*a)a = 1.5”øTn = 9,302/(2*1.5) = 3,100#Ts = 3,100/1.6 = 1,938#

Maximum allowable moment on base plate:Ma = 2Ts*(6”-0.75”)Ma = 2*1,938#*5.25” = 20,349”#Base plate strength is adequate to develop the full stanchion plastic moment.

Base Plate Anchorage to Concrete- Powers Fasteners Wedge-Bolt:Base plate mounted to concrete slab with 3/8” screw-in concrete anchors with 3” minimum embedment. Anchor strength from Powers Fasteners catalog for the specified anchors in accordance with ESR-2526.



T = (1,935#)/√4,000*√3,000 = 1,676# For 3,000 psi concreteAdjustment for anchor spacing = 4.5”

Cs = 1.00 edge distance is adequate > 3” no adjustment required for edge distance.T’ = 1.0*1,676# = 1,676#

Allowable moment on base plate based on anchor strength:Determine compression area on base plate based on concrete bearing:a = 2*1,676/(0.85*2*3,000 psi*5.75”) = 0.114”Ma = 2*1,676*(5.25-0.114/2) = 17,406”#

Base plate anchorage is adequate.

Powers Fasteners Wedge Bolt 3/8” x 4” screw-in concrete anchors with 3” minimum embedment in accordance with ESR-2526.

For alternative anchors to concrete anchor may be designed based on ACI 318-08 Appendix D for tension withdrawal based on a calculated ultimate strength for the anchor group (2 anchors in tension):Ncbg ≥ (1.6/0.65)*(M/5.19”) = 0.4743M whereM = greater of MWL or MLL as calculated on page 2

BASE PLATE ANCHORAGE TO STEEL3/8” A307 or ASTM F593 Group 1 or 2 Condition CW stainless steel bolts into 1/4” tapped steel or with nuts.Tensile area of 3/8” threaded rod (UNC) = 0.0775 in2

Rod strength øPn= (0.75*60ksi) * 0.0775 in2 = 3,488#Check thread strength into standoff – minimum thread embed = 1/4”Internal thread stripping area = 0.828 in2 for 3/8 – 16 threadsStrength of threads øPn = 0.65*0.58*Asn*t*Ftu = 0.65*0.58*0.828*(1/4)*75ksi = 4,541# Shear strength øVn= 0.65*0.5*60 ksi*0.0775 in2 = 1,511#



ATTACHMENT TO WOOD:Check required embedment for 3/8” lag screws:From National Design Specification for Wood Construction Table 11.2AG ≥ 0.49 (pressure treated Doug-Fir or denser wood)W = 296#/”CD = 1.6 for wind loads (NDS Table 2.3.2)CM = 0.7 (NDS Table 10.3.3) where moisture content of wood may exceed 19%. W’ = 296#/”*1.6*0.7 = 332#/”Allowable tension load on 3/8” lag screw stainless steelTa = AFyt = 0.0775in2*75 ksi/3 = 1,938#

Required embedment depth e:e = 1,405/332#/” = 4.23”Required lag screw length = 4.23” + 0.5 + 7/32” = 4.95” When there is decking between the base plate and the solid limber backing the lag screw length shall be increased by the decking thickness,

Recommend using 3/8” x (5”+td) stainless steel lag screws, (4) per base plate.Where td = material thickness between the base plate and the solid lumber backing

Minimum blocking under base plate is 6x8 nominal.

Attachment of blocking to joists:Required number of screws for 1/4” x 3” screws:Z’ = 159#*1.6*0.7 = 178#From ∑M about center of block:solving for N:N = 2,810/(178) = 16 screws each end

For through bolts:Use 3/8” bolts to minimum 4x solid blocking with 2” square plate washers on the backside under the nut. Blocking shall be adequately secured.

Wood framing shall have adequate strength to carry the imposed loads from the posts.

FASCIA MOUNTED POSTSPosts may be fascia mounted to steel. concrete, CMU or wood using a minimum of two 3/8” anchors designed for the imposed loads and moments as calculated.

ALTERNATIVESAlternative anchors may be designed based on the post moments as calculated.



DESIGN STEPS:1) Determine wind load using Table 1 or ASCE SEI 7-05 for other conditions.2) Select Post Spacing using Tables 2 – 4 for directly embedded posts or Tables 5 or 6 for stanchion mounted posts.3) Select glass thickness from Tables 7 – 9 (No bottom rail) or Table 10 (with bottom rail).4) Select anchors based on substrate.

DESIGN EXAMPLE:5’ Tall AWS windscreen surface mounted to wood deck, 100 mph exposure D wind load.1) From Table 1 wind load = 24.9 psf.

2) Since posts are surfaced mounted to wood must use base plate mounted stanchion. Can use any post – select 2” post.From Table 5 Cannot use 10 gauge stanchion because allowable wind load is too low.From Table 6, 1/4” stanchion, post spacing is between 3’ and 4’ on center, interpolating :S = 3’+(30.6-24.9)/(30.6-22.9) = 3.7’ or from the equation for the post spacing:S = 2*1,146.1’#/(W*H2) = 2,292.2/(24.9*52) = 3.68’ Use 3’8” as maximum post spacing.

3) Glass will not use a structural bottom rail therefore select glass thickness from Tables 7- 9 for a maximum height of 5’ and spacing of 3’8”: From Table 7, 1/4” glass, S= 3.5’, W = 31.3 psf and S = 4’, W = 24.0 so interpolating at S = 3’8”, W > 24.9 psf : 31.3-(2”/6”)(31.3-24.0) = 28.9 psfSelect 1/4” glass.

4) Attached to wood framing use 3/8” x 6” lag screws (page 17).



Documents

Edward C. Robison, PE - CRL-ARCH stress in aluminum post in accordance with AWS Table 2-20 or Table 2-21: F t = 19 ksi and F c = 23.9-0.238[(2L ... 5 23.5 17.6 15.6 14.1 12.8 11.7