The Entice System is intended for exterior installations requiring the use of insulating glass.

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Edward C. Robison, PE C.R. Laurence Co., Inc.

INSULATING GLASS ENTRY WALLS The Entice System uses stainless steel clad

doors and entries walls with sidelights and transoms.

indicated in this report. Seismic loads up to SDS = 3.

finish floor or 200 lb concentrated load at 42 above the finish floor

limitations when properly designed and detailed by a qualified professional

1 Edward C. Robison, PE C.R. Laurence Co., Inc East Vernon Los Angeles, CA March 2015 SUBJ: ENTICE SYSTEM - INSULATING GLASS ENTRY WALLS The Entice System uses stainless steel clad aluminum extrusions to support 1 insulating glass units to construct glass doors and entries walls with sidelights and transoms. The Entice System is intended for exterior installations requiring the use of insulating glass. The system is designed for the following loading conditions: Wind Loads as indicated in this report. Seismic loads up to SDS = lb live load at any location or 50 plf live load at 42 above the finish floor or 200 lb concentrated load at 42 above the finish floor Installation shall be as detailed in the Entice System Installation Instructions from CRL- US Aluminum. The allowable door widths and heights may be modified from the system limitations when properly designed and detailed by a qualified professional engineer. All glass in the doors and sidelights shall be fully tempered. The Entice System when constructed in accordance with the recommendations herein comply with all applicable requirements of the 2009, 2012 and 2015 International Building Codes. Edward Robison, P.E.

2 Entice System Insulating Glass Entries Page 2 of 26 LIVE LOAD: 200# live load perpendicular at 42 above finish floor or 50 plf live load perpendicular at 42 above finish floor or 50 lbs on 1 sf at any location. DEAD LOAD D = 6.0 psf for 1 insulating glass (1/4 glass, 1/2 air space, 1/4 glass) SEISMIC LOADING: ap = 1.0; SDS= 3.0; IP = 1.0; RP = 2.5 V = 0.4aPSDSIPW(1+2z/h)/RP = (0.4/2.5)*1.0*3.0(1+2h/h)W = 1.44W D = 6.0 psf for 1 insulating glass (1/4 glass, 1/2 air space, 1/4 glass) E = 1.44*6.0 = 8.6 psf For connections Rp = 1.0 Ec = 2.5/1.0*8.6 = 21.6 psf Equivalent wind load pressure: SDS Wind Load psf Weq = 7.2*SDS : Equation may be used in lieu of the table. For earthquake loads convert the SDS to the equivalent wind load in psf that the system will need to resist. From the wind loading tables verify that system strength is adequate for this equivalent pressure. WIND LOADING ON GLASS The wind load for a specific installation shall be determined in accordance with ASCE/SEI 7-05 Fig Components and Cladding, ASCE/SEI 7-10 Chapter 30 or other approved method by a qualified professional engineer. All wind loads shown in this report are Allowable Stress Design (ASD) level wind loads, Components and Cladding. For Strength Level wind loads calculated per ASCE/SEI 7-10 Chapter 30 the wind load shall be adjusted to the ASD value by multiplying by 0.6, ASCE/SEI 7-10 Section 2.4.

3 Entice System Insulating Glass Entries Page 3 of 26 System layout, nomenclature and dimensions.

Entice System Insulating Glass Entries Page 4 of 26 SYSTEM COMPONENTS Door Stiles and Jamb Mullions Primary structural element is the aluminum extrusion core. Ix = 0.810148 Iy = 0.088711 A = 1.

4 Entice System Insulating Glass Entries Page 4 of 26 SYSTEM COMPONENTS Door Stiles and Jamb Mullions Primary structural element is the aluminum extrusion core. Ix = Iy = A = in 2 Sx = in 3 r = in Other components used to complete the stile/jamb do not add significantly to the structural strength and stiffness and aren t considered in the determination of the allowable bending strength and stiffness. Allowable aluminum stress based on ADM Table 2-24 Fta = 15 ksi (line 2) For compression: Extrusion is continuously braced by glass Fca = 15 ksi For flat element supported on one edge: b/t = /0.323 = Fca = 15 ksi Allowable bending moment based on aluminum stress: Ma =0.693in 3 *15ksi = 10,395 # = # Maximum allowable load on mullion W = *8/L 2 = xx plf for L in feet W = 10,395 #*8/L 2 = xx pli for L in inches For deflection L/175 and 0.75 Check based on deflection limit of L/175: L 5*0.7*W*L EI Solving for W: W EI/(1.595*L 3 ) substituting for EI W (10.0x10 6 * )/(1.595*L 3 ) = 5,079,298/L 3 W in pli and L in inches Determine length at which deflection will govern allowable load by setting the two equations equal: 10,395*8/(L 2 ) = (10.0x10 6 * )/(1.595*L 3 )

5 Entice System Insulating Glass Entries Page 5 of 26 Solving for L: L 3 /L 2 = (10.0x10 6 * )/(1.595*10,395*8) = = 5-1 Deflection limit will control for all heights over 5-1 : W =12*10,395 #*8/L 2 = xx plf Up to 61 Allowable wind load will be controlled by L/175 up to 0.75*175 = = Determine maximum wind load at height: W 5,079,298/( ) = pli = 27 plf DOOR ALLOWABLE WIND LOAD (PSF) FOR DOOR/SIDELIGHT WIDTH HEIGHT INCHES PLF

6 Entice System Insulating Glass Entries Page 6 of 26 TRANSOM HEADER: wide aluminum extrusion that is clad and matched with closure piece that secures glass in place. Ix = in 4 Iy = in 4 A = in 2 Sx = in 3 Sy = in 3 rx = 0.520in ry = in Allowable stress: 6063-T6 aluminum For bending; Fc = Ft = 15 ksi Max = 15ksi*0.520 in 3 = 7,800 # = 650 # Vertical loads May = 15ksi*0.997 in 3 = 14,955 # = 1, # Horizontal loads Maximum span will occur with a double door without a buttress fin at center - supported only at jambs. Can conservatively assume that transom lights are supported by vertical mullions so horizontal load share to transom is concentrated at center along with load from door. Vertical moment: assume that dead load is evenly split between 4 points - 1/8L, 3/8L, 5/8L and 7/8L: M = D/2*(1/8L + 3/8L) = DL/4 D = 6.0 PSF*L/2HT dead load for each transom light L = overall width between jamb mullions (ft) HT = height of transom (ft) For horizontal loads: Pw = wl*ho/4 HO = overall height - door + transom (ft) Mw = PwL/4 = PwL 3 *12 3 /(48EI) = PwL 3 *12 3 /(48*10x10 6 *1.346) = PwL 3 /373, L/175 Pw *12/(175L 2 ) = /(L 2 ) Bending stress also must be limited so the fbh + fbv 15,000 psi Pw 4*1,246.25/L >> /(L 2 ) for L 5.14 Deflection will control for widths over 5.14 Pw = wlho/ /(L 2 ) solving for w: w 4*25638/[LHO*L 2 ] = /(HO*L 3 )

7 Entice System Insulating Glass Entries Page 7 of 26 Check Transom allowable size based on dead load on header: DL/4 = 6.0HT*L 0.5*650 # HT = 54.17/L: in feet Width Height Allowable wind load: For stress controlled spans 5.14 w [4*1,246.25/L - (6.0HT*L/4)/650]/HO = [4,985/L - HTL/433.33]/HO Allowable wind load for overall door width 4.0 ft Transom Ht Door Ht

8 Entice System Insulating Glass Entries Page 8 of 26 Allowable wind load for overall door width 4.5 ft Transom Ht Door Ht Allowable wind load for overall door width 5 ft Transom Ht Door Ht

9 Entice System Insulating Glass Entries Page 9 of 26 Allowable wind load for overall door width 5.5 ft Transom Ht Door Ht Allowable wind load for overall door width 6.0 ft Transom Ht Door Ht

10 Entice System Insulating Glass Entries Page 10 of 26 Allowable wind load for overall door width 6.5 ft Transom Ht Door Ht Allowable wind load for overall door width 7.0 ft Transom Ht Door Ht

11 Entice System Insulating Glass Entries Page 11 of 26 Allowable wind load for overall door width 7.5 ft Transom Ht Door Ht Allowable wind load for overall door width 8.0 ft Transom Ht Door Ht

12 Entice System Insulating Glass Entries Page 12 of 26 Allowable wind load for overall door width 8.5 ft Transom Ht Door Ht Allowable wind load for overall door width 9 ft Transom Ht Door Ht

13 Entice System Insulating Glass Entries Page 13 of 26 Side Light Rails: 4 Rail may be used both and sill. 10 rail for sill only. Connection uses tension piece locked onto mounting shoe which is secured to rail with #10 screw. Strength of screw mounting shoe to head or sill- #10 screw to steel Ta = 885# Screw pull-thru strength - groove under screw filled with thermal break material. Pnov = Ct1Ftu1(Dws-Dh) = 1.0*0.07 *22ksi*( ) = 670# Paov = Pnov/ns = 670/3 = 223# Shear strength: Va = 573# For bearing on aluminum channel: Pas = 2Ftu1Dt1/nu = 2*22ksi*0.19*(2*0.07)/1.95 = 600# Allowable load per screw: Tension on screw from load eccentricity on rail: M = ev For 4 rail: M = 4 *V T = M/(2.075 /2) = 4*V/(2.075 /2) = 3.855V For combined load on screw: V/Va+T/Ta 1.2 V/ V/223 = 1.2 Solving for V: V = 63# For head - requires screw for each 63# of wind load tributary to top rail. Tributary area = 1/2B 1/2H (lesser of 1/2 of light width or height) For 10 bottom rail to sill- Screw to tension tie: strength and loading same as checked for 4 rail. Typical connection to concrete will use a Tapcon screw 1/4 x 2-3/4 Based on ESR-2202 Ta = 350# Va = 430#

14 Entice System Insulating Glass Entries Page 14 of 26 Check allowable load based on aluminum: For pull through: -Head on screw will extend over vertical webs in extrusion so tear through at screw head won t occur. For bearing on aluminum channel: Pas = 2Ftu1Dt1/nu = 2*22ksi*0.19*(2*0.07)/1.95 = 600# Determine allowable load based on anchorage strength- load per screw: M = ev For 10 rail: M = 10 *V T = M/(2.075 /2) = 10*V/(2.375 /2) = 8.421V For combined load on screw: V/Va+T/Ta 1.2 V/ V/350 = 1.2 Solving for V: V = 46# per screw Use same spacing for head and sill and for 4 and 10 rails: Allowable wind load (psf) based on screw spacing in inches and light width Number of screws Light Width/ Allowable wind load in psf

15 Entice System Insulating Glass Entries Page 15 of 26 Wall Jambs Jamb consists of two primary parts- Main body is attached to wall with #10 sheet metal or self drilling screw. Screw strength: Ta = 885# Screw pull-thru strength - groove under screw filled with thermal break material. Pnov = Ct1Ftu1(Dws-Dh) = 1.0*0.07 *22ksi*( ) = 670# Paov = Pnov/ns = 670/3 = 223# Shear strength: Va = 573# For bearing on aluminum channel: Pas = 2Ftu1Dt1/nu = 2*22ksi*0.19*(2*0.07)/1.95 = 600# Allowable load per screw: Tension on screw from load eccentricity on rail: M = ev For jamb depth e = 1.25 M = 1.25 *V T = M/(1.185 ) = 1.25*V/(1.185) = 1.055V For combined load on screw: V/Va+T/Ta 1.2 V/ V/223 = 1.2 Solving for V: V = 185# Screw Light Width/ Allowable wind load in psf Spacing Inches

16 Entice System Insulating Glass Entries Page 16 of 26 Connection of retainer piece to jamb: #8 self tapping screw to extrusion: Ta = 335# Tension strength of screw will be controlled by pullout from the primary piece: 6063-T6 aluminum Pnot = KsDtcFty2 = 1.2*0.164*0.118 *25 ksi = 581 Pa = Pnot/3 = 581/3 = 194# Tension load on screw is combination of direct tension and prying load: Total tension based on M about the base of the closure strip: M = 0.88 *W *R = 0 W = horizontal load on jamb R = screw tension R = (0.88/0.28)W = W determine allowable load per screw: W = 194#/3.143 = 62# per screw Allowable wind load for screw spacing: Screw Light Width/ Allowable wind load in psf Spacing Inches

2675 in 2 Sx = 1.1691 in 3 r = 0.778 in Narrow wall jamb: Full section- Capture piece Ix = 0.6042 in 4 Ix = 0.00845 in 4 Iy = 0.1259 in 4 At = 0.95 in 2 Ac = 0.379 in 2 Sx = 0.4401 in 3 ry = 0.

17 Entice System Insulating Glass Entries Page 17 of 26 INTERMEDIATE MULLION- Used between two sidelights are to split transom. Composite mullion formed by screwing the narrow jamb mullion to the narrow wall jamb. Strength will be sum of the two mullions- Narrow jamb mullion: Ix = in 4 Iy = in 4 A = in 2 Sx = in 3 r = in Narrow wall jamb: Full section- Capture piece Ix = in 4 Ix = in 4 Iy = in 4 At = 0.95 in 2 Ac = in 2 Sx = in 3 ry = in Required screw spacing between main jamb and glazing capture piece: #10 screw shear strength will be controlled by aluminum bearing in main jamb; t = Va = Dt2Ftu2/nu = 0.19 *0.112 *30ksi/1.95 = 327# each Shear flow between the sections- Centroid of section to centroid of capture y = Q = Ac*y = 0.379*0.73 = in 3 q = VQ/I = V* / = 0.46V 327# Pullout strength of screws: Pnot = KsDtcFty2 = 1.2*0.19 *0.112 *25ksi = 638# Paot = pnot/ns = 638/3 = 213# Screw tension load, T = W = V V/213 + V/(327/0.46) 1.2 Solving for V: V 197 For screws at 12 on center V = 197 plf (loading from one side of composite mullion)

18 Entice System Insulating Glass Entries Page 18 of 26 For connection between the two mullion pieces: #10 screws: Thickness to t1 = 0.07 ; t2 = 0.10 Va = D2*t1Ftu1/nu = 0.19 *2*0.07 *30ksi/1.95 = 409# each Va = Dt2Ftu2/nu = 0.19 *0.10 *30ksi/1.95 = 292# each Shear flow between the sections- Centroid of section to centroid of capture y = 0.17 Q = Ac*y = 0.95*0.17 = in 3 q = VQ/I = V*0.1615/( ) = 0.114V 292# No tension load on screws: Va = 292/0.114 = 2,557# per screw Strength of combined section: Ix = in 4 Iy = in 4 A = in 2 Sx = in 3 r = in Elements braced or stiffened- Allowable bending stress = 15 ksi Ma = in 4 *15 ksi = 21,210 # = 1,767.5 # Maximum allowable load on mullion W = 1,767.5*8/L 2 = xx plf for L in feet W = 17,536.5 #*8/L 2 = xx pli for L in inches For deflection L/175 and 0.75 Check based on deflection limit of L/175: L 5*0.7*W*L EI Solving for W: W EI/(1.595*L 3 ) substituting for EI W (10.0x10 6 *1.414)/(1.595*L 3 ) = 8,865,204/L 3 W in pli and L in inches

19 Entice System Insulating Glass Entries Page 19 of 26 Determine length at which deflection will govern allowable load by setting the two equations equal: 17,536.5*8/(L 2 ) = (10.0x10 6 *1.414)/(1.595*L 3 ) Solving for L: L 3 /L 2 = (10.0x10 6 *1.414)/(1.595*17,536.5*8) = 63.2 = 5-3 3/16 W = 12*17,536.5 #*8/L 2 = 1,683,504/L 2 For L in inches Deflection limit will control for all heights over 5-3 : Allowable wind load will be controlled by L/175 up to 0.75*175 = = For greater heights: = 0.75: W = 12*1,133,220,000/L 4 Mullion ALLOWABLE WIND LOAD (PSF) FOR DOOR/SIDELIGHT WIDTH Height Inches PLF N/A

20 Entice System Insulating Glass Entries Page 20 of 26 GLASS BUTTRESS FINS: Cantilevered fins: Allowable load will be controlled by strength of glass connection to ceiling bracket. From center of reaction to centerline of top connection bolts: M = RH R = Horizontal Load at patch fitting (bottom) D = Fin dead Load = A*9.5psf = H ft*1 *9.5psf For inward wind force: Allowable load on bolts: Check shear strength of sex-bolt in mate through glass to brackets. Bolt shear strength (double shear) Av = 0.22 in 2 Vn = 40.5ksi*(0.22 in 2 )= 8,910# Vs = 8,910/2 = 4,455# Check glass shear at bolts: Holes are from bottom V = 3,000psi*( *0.719 *2) = 7,819# (Allowable shear) Check glass bearing at bolts: 7/8 diameter holes B = 6,000psi*(0.875 *0.75 ) = 3,938# controls Allowable moment on buttress fin based on glass bearing stress: Taking elastic rotation about edge of fin: Rb = 3,938*( )/9.5 = 8,860.5# Abg = 8,860.5/(0.719*6,000 psi) = b1 = /2 = 8.473; b2 = /2 = 6.098; b3 = /2 = Ma = 3,938*( / /8.473) = 57,092 # = 4, #

21 Entice System Insulating Glass Entries Page 21 of 26 Determine reduction in allowable moment for dead load: V = D/3 = 9.5H/3 = 3.167H (H in feet) Revised allowable moment: Ma = (3, H)* (moment in ft-lbs) Determine allowable load on fin based on height: RH = (3, H)* = 4, H R = 4,757.89/H Ma (ft-lb) 3938 For glass attachment bolts Height ft R: Allowable Load lbs To determine allowable wind load for buttress fin: w = 4R HO*(L+B) Allowable wind load for 4 tall fin, 12 overall height, 6 wide door and 5 wide sidelight: R = 1185 lbs (lesser value from chart above or from chart for specific anchorage method used) HO = 12 L = Overall door width (two 3 foot doors) B = 5 w = 4*1185 = 35.9 psf 12*(6+5) Allowable wind load on the store front will be the lesser wind load for the sidelight, door, transom or fin.

22 Entice System Insulating Glass Entries Page 22 of 26 Attachment to ceiling: Determine load distribution on anchors For outward force rotation of bracket about edge of ceiling bearing - Assume first fastener carries no tension loadcompression zone is centered on this fastener. From M about first fastener: M = Mfin - d = Mfin -( / / / )*Tfastener Solving for Tfastener: Tfastener = Mfin/ For maximum fin moment = 57,092 # Required anchor strength = 57,092 #/ = 2,706# Strength of 1/4 stainless steel countersunk cap screws- ASTM F879 or F835 Condition AF Tn = 2,703# Ta = 2,703/2 = 1,352# Ma = Ta* d = 1,352#* = 28,524 # = 2,377 # 5 anchor layout will only develop 1/2 of the allowable moment on the fin. Determine allowable load on fin based on height: PH = (Ma-3.167H)* = 2, # H P = 2,871.89/H Ma (ft-lb) 2377 (5) 1/4 SS Cap screws to steel framing Height ft R Allowable Load lbs Strength of 1/4-28 fine thread countersunk screw NAS or equivalent strength Fu 160 ksi At = Tn = *160ksi = 5,824# Ta = 5,824/2 = 2,912# Ma = Ta* d = 1,352#* = 28,524 # = 2,377 # 5 anchor layout will develop full allowable moment on glass fin when using the 1/4-28 fine thread countersunk screw NAS or equivalent strength

23 Entice System Insulating Glass Entries Page 23 of 26 Anchorage to wood: Required embedment length into solid wood: From NDS Table 11.2A for #16 wood screws into wood with G 0.50 (DFL, SP or SCL beams) D = Dr = W = 172#/in. Adjustment for load duration: CD = 1.6 for wind CD = 1.6 for live loads CD = 0.9 for dead loads Tension strength of #16 screw: Tn = 110ksi* in 2 = 3,774# Ta = 3,774/2 = 1,887# Allowable moment: Ma = Ta* d = 1,887#* = 39,811 # = 3, # Required embedment controlled by live or wind loads: W = 172*1.6 = 275 pli e = 1887#/275#/in = 6.86 Required length of screws: l = = 7.92 Use 8 length Ma (ft-lb) (5) #16 x 8 Countersunk wood screws Height ft R Allowable Load lbs For 6 screw: e = = Ta = *275 #= 1,358# Ma = Ta* d = 1,358#* = 28,650 # = 2, # Ma (ft-lb) (5) #16 x 6 Countersunk wood screws Height ft R Allowable Load lbs

Entice System Insulating Glass Entries Page 24 of 26 DOOR FRAME: Top Rails: Uses aluminum extrusion 4 high by 2-3/8 wide clad with stainless steel or other metal sheet. Ix = 2.29 in 4 Iy = 2.

24 Entice System Insulating Glass Entries Page 24 of 26 DOOR FRAME: Top Rails: Uses aluminum extrusion 4 high by 2-3/8 wide clad with stainless steel or other metal sheet. Ix = 2.29 in 4 Iy = in 4 A = in 2 Sx = in 3 Sy = in 3 rx = in ry = in Allowable stress: 6063-T6 aluminum b/t = 0.845/0.15 = 5.6 or 1.263/0.178 = For bending Fc = Ft = 15 ksi Max = 15ksi*1.117in 3 = 16,755 # vertical loads Max = 15ksi*1.854in 3 = 27,810 # horizontal loads Door top rail won t control wind load for any configuration. Bottom Rails: Options- Use same extrusion as top rail. Low profile - rail extrusion alone. Deep profile- 10 deep (nominal)- Adds second extrusion bolted to 4 deep rail. 1/4 cap screw into threaded hole - Thread depth = 0.3 Pa = Pnot/3 = 0.58AsntcFtu2 /3 = 0.58*0.539*0.3 *22ksi/3 Pa = 688# Tensile strength of screw = 2,700#/2 = 1,350# Allowable force on bottom extrusion based on screw strength: Ma = 688#*(2.375 /2) = 817 # e = max P = 817/5.625 = 145# per screw. Connection strength will be okay for loads on bottom panel.

Entice System Insulating Glass Entries Page 25 of 26 GLASS STRENGTH Allowable stresses on glass: Modulus of Rupture Fr 24,000 psi Safety Factor = 4: Fba = 24,000/4 = 6,000 psi for live loads and Fbal

Allowable stress for seismic loads is the same as for wind loads. Permanent loads Fbap = 0.

25 Entice System Insulating Glass Entries Page 25 of 26 GLASS STRENGTH Allowable stresses on glass: Modulus of Rupture Fr 24,000 psi Safety Factor = 4: Fba = 24,000/4 = 6,000 psi for live loads and Fbal = 0.75*6,000 = 4,500 psi for load durations over 3 minutes for wind loads (5 psf partition load) F3sec = 10,600 psi (recommend limiting to 9,600 psi because glass is only end supported). Allowable stress for seismic loads is the same as for wind loads. Permanent loads Fbap = 0.31*10,600psi = 3,286 psi 1/2 tempered glass (GL-4) Glass allowable loads and probability of breakage evaluated using ASTM E a. In accordance with ASTM E a. For monolithic tempered glass installed in 1 insulating glass Glass type factor = 3.6 for fully tempered glass both plies (1/4 ) Load share factor = 2.00 Allowable glass moment on IG glass: Ma = 2*3.6/4*10,600psi*(2* ) = 1,830 #/ft Allowable wind load on glass: LR = NFL*GTF*LS For live loads- the 50 plf line load may be modeled as 25 psf uniform load. - Adjust for live load: LL = 25psf*10.6ksi/6ksi = psf

Entice System Insulating Glass Entries Page 26 of 26 Maximum glass light size: From ASTM E1300 Fig A1.5 Determine required NFL for allowable load of 44.17 psf: NFL = 44.17/(20.9psf/kPa*3.6*2) = 0.

26 Entice System Insulating Glass Entries Page 26 of 26 Maximum glass light size: From ASTM E1300 Fig A1.5 Determine required NFL for allowable load of psf: NFL = 44.17/(20.9psf/kPa*3.6*2) = 0.29 kpa Maximum height from figure = 200 = 16-8 From flat plate theory- For wind load: M = 1/8*pb 2 /(1+2α 3 ) Try 5 x 18 maximum glass size: α = 5/18 = 0.28 M = [1/8*44.17*5 2 /(1+2* )]*12 = 1,586.7 #/ft 1,830 # Glass strength is adequate for up to 5 x18 maximum. Glass strength won t control allowable wind load on system.

Allowable Holes in VERSA-LAM Beams

Allowable Holes in VERSA-LAM Beams VERSA-LAM Products 23 An Introduction to VERSA-LAM Products When you specify VERSA-LAM laminated veneer headers/beams, you are building quality into your design. They are excel lent as floor and roof framing