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NO PERMIT Structural Calcs78080 Calle Amigo, Suite 102 Telephone: (760) 771-9930 La Quinta, CA 92253 Cell: (760) 771-9998 Fax: (760) 808-9146 Structural Calculation For Lot 19A At 52-741 Ross Avenue The Madisoti Club La Quinta, CA. 'I'ype Of Pro'eet: Residential NSFR E� CITY OF LA QUINTA BUILDING & SAFETY DEFT. APPROVED FOR CONSTRUCTIOM 13 Ila Date: December 18, 2012 Design by: R.A. JN: 1211109 Ps G CD NO, C 67613 53 EXP. o OF Cq�-_ -20-13 RSA S}Tl �UCJ� 1 MZYG r + E�ER�IG TABLE OF CONTENTS 1. Design Criteria....................................................................................1 2. Beam Design.......................................................................................2-104 3. Lateral Design Loads (Seismic And Wind)...... 4. Shear Wall Design.............................................................................115-128 5. Foundation Design..................................................................................129-144 O F C 21 2012 DESIGN CRITERIA 2009 INTERNATIONAL BUILDING CODE 2010 CALIFORNIA BUILDING CODE SOILS BEARING PRESSURE = 1500 PSF (PER SOILS REPORT) EXTERIOR WALL = 15.00 PSF EXTERIOR WALL = 25.00 PSF with VENEER SIDING INTERIOR WALL = 10.00 PSF ROOF DEAD LOAD TILE ROOFING = 15.00 PSF SHEATHING = 2.50 PSF FRAA41NG = 2.50 PSF DRYWALL = 2.50 PSF INSULATION = 1.50 PSF MISCELLENUOUS = 6.00 PSF TOTAL DEAD LOAD = 30.00 PSF TOTAL LIVE LOAD = 20.00 PSF TOTAL LOAD = 50.00 PSF FLOOR DEAD LOAD CARPET = 2.00 PSF LIGHT WT CONC =13.00 PSF SHEATHING = 2.50 PSF FRAtIING = 3.30 PSF - DRYWALL = 2.80 PSF INSULATION = 1.40 PSF TOTAL DEAD LOAD = 25.00 PSF TOTAL LIVE LOAD = 40.00 PSF TOTAL LOAD = 65.00 PSF BALCONY DEAD LOAD DECK FINISH = 2.00 PSF SHEATHING = 2.50 PSF FRAMING = 3.30 PSF DRYWALL = 2.80 PSF LIGHT WT CONC =13.00 PSF INSULATION = 1.40 PSF TOTAL DEAD LOAD = 25.00 PSF TOTAL LIVE LOAD = 60.00 PSF TOTAL LOAD = 85.00 PSF Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM #1 HDR FRONT AND REAR OF M. EWW DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Desian Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM fix 10 No. 1, Douglas Fir -Larch L = 8.25 ft wD = 510 Ibs / ft wL = 300 Ibs / ft PD1 = 3000 Ibs L1 = 1.25 ft P" = 0 lbs L2 = 0 ft L, L, PD I PD2 wD d L = L / 3450 Camber => 0.19 inch d Kcr D+ L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cr, Condition Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1,25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wself wt = 11 Ibs / ft RLee = 5.93 kips RRigM = 3.84 kips VMax = 5.28 kips, at 9.5 inch from left end MM. = 8.99 ft-kips, at 3.55 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = N/A d = 9.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb' = N/A A = 52.3 in2 I = 393 in4 F = 170 psi Fe = 1,350 psi Sx = 82.7 in3 RB = N/A E' = 1,600 ksi F = 170 psi 1E= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1304 psi < Fb = f,; = 1.5 VMax / A = 152 psi < CHECK DEFLECTIONS d (L' Max) = 0.05 in, at 4.125 ft from left end d (KID+L, Max) = 0.18 in, at 4.125 ft from left end Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.19 in, at 3.838 ft from left end Cv Cc Cr 1.00 1.00 1.00 1350 psi F, [Satisfactory] [Satisfactory] d L = L / 360 [Satisfactory] d Kcr D + L = L / 240 [Satisfactory] CZ) ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fr' = Fc CD CP CF = 1358 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.e Fc* = Fc CD CF = 1480 psi Le = Ke L = 1.0 L = 99 in d = 9.5 in SF = slenderness ratio = 10.4 < FcE = 0.822 E'min / SF = 4390 psi E'min = 580 ksi F = FcE / Fc* = 2.966 c = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 19 psi < F� THE ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] F F y 0.918 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1361 psi < Fti [Satisfactory] GHF.CK COMBINED STRESS [NDS 2005 Sec. 3.92] (fc / F.' )2 + fb / [Fti (1 - fc / FcE)] = 0.633 < 1 [Satisfactory] 0 lA Structural _ I Beam Design Base on PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY CLIENT: BEAM *2 HDR REAR OF M. HDRM PAGE JOB NO.: DESIGN BY: R.A DATE: i2111/2012 )$ 2pOS REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM fix 6 No. 1, Douglas Fir -Larch L = 5,26 ft Wn = 510 Ibs / ft WL = 300 Ibs / ft Pni = 0 Ibs Lt = 0 It PD2 = 0 Ibs L2 = d =1I n ft L, L, PD � j PD2 „L 1 WD L 360 Camber => 0.11 inch "KWO + L = L 1 240 Does member have continuous lateral support by top diaphragm ? 0= yes, 0= no) 1 Yes Code Duration Factor C Condition 1 2 0.90 Dead Load 3 1.00 Occupancy Live Load 4 1.15 Snow Load 5 1.25 Construction Load 6 1-60 Wmd/Eart✓Zquake Load Choice 2.00 Impact Load 2 Occupancy Live Load ILYSIS -ERMINE REACTIONS, MOMENT, SHEAR wsetr = = 7 lbs / ft RLen = 2.14 kips VMax = 1.77 kips, at 5.5 inch from left end MM. = MINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 5.50 in FbE = N/A A = 30.3 in2 I = 76 in4 Sx = 27.7 in' RB = N/A IE = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1217 psi < Fb = fv = 1.5 VMS / A = 88 psi < CHECK DEFLECTIONS A (L, Max) = 0.04 in, at 2.625 ft from left end, d (Kcr D+ L, Max) = 0.11 in, at 2.625 ft from left end Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.11 in, at 2.625 ft from left end THE BEAM DESIGN IS ADEQUATE. Code Desi nation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRight = 2.14 kips 2.81 ft-kips, at 2.63 ft from left end E = Ex = 1600 ksi Fb = N/A Fb = 1,350 psi F = FbE / Fb` = N/A F = 170 psi Fe = 1,350 psi E' = 1,600 ksi Fv' = 170 psi Cv Cc Cr 1.00 1.00 1.00 1350 psi [Satisfactory] F,, [Satisfactory] < AL=L/360 < AKcr D+L=L/240 [Satisfactory] [Satisfactory] n (CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1327 psi Where F� = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c10.5 = Fc' = Fc CD CF = 1480 psi Le = KB L = 1.0 L = 63 in d = 5.5 in SF = slenderness ratio = 11.5 < Fc, = 0.822 E'mIn / SF2 = 3634 psi E'min = 580 ksi F = FCE / FC* = 2.455 c = 0.8 1 HE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 33 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] 0.897 50 1 1 y L 1 � [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1317 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )2 + % / [Fe (1 - f,./ Fes] = 0.616 < 1 [Satisfactory] 0 Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM #3, HDR RIGHT OF M. BDRM DESIGN BY R.A Structural JOB NO.: DATE: 17111/2012 REVIEW BY Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 10 No, 1, Douglas Fir -Larch L = 12.25 ft wD = 240 Ibs / ft wL = 120 Ibs / ft PDt = 0 Ibs Lt= 0 it PD2 = 0 Ibs L2= 0 ft L Foe NL • r 1 .� "'o 1 d L = L / 350 Camber => 0.30 inch dKUD+L=L/240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirwt = 11 Ibs / ft RLeR = 2.27 kips VMex = 1.98 kips, at 9.5 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 9.50 in FbE = N/A A = 52.3 in 1 = 393 in Sx = 82.7 in RB = N/A 1E = N/A Co CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1010 psi < Fb f = 1.5 VMS / A = 57 psi < CHECK DEFLECTIONS A (L, Max) = 0.10 in, at 6.125 ft from left end, d (Kcr D+L, Max) = 0.30 in, at 6.125 ft from left end Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (t 5D, Max) = 0.30 in, at 6.125 ft from left end THE BEAM DESIGN IS ADEQUATE. Code Desinnation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRight = 2.27 kips MMax = 6.97 ft-kips, at 6.13 ft from left end E = Ex = 1600 ksi Fb+= N/A Fb = 1,350 psi F = FbE / Fb* = N/A Fv = 170 psi Fe = 1,350 psi E' = 1,600 ksi Fv' = 170 psi Cv C': Cr 1.00 1.00 1.00 1350 psi F,' [Satisfactory] [Satisfactory] d L = L / 360 [Satisfactory] AKuD+L=L/240 [Satisfactory] l�J CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fti = Fc CD CP CF = 1158 psi Where F. = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s - Fc* = F� CD CF = 1480 psi LB = K. L = 1.0 L = 147 in d = 9.5 in SF = slenderness ratio = 15.5 < FCE = 0.822 E'm,n / SFZ = 1991 psi E'min = 580 ksi F = FEE / Fc* = 1.345 c = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 19 psi < F.' I IdE ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 0.782 50 1 ► FiE. F ] i [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1068 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / Fti )Z + fb / [Fti (1 - fc / F.E)] = 0.499 < 1 [Satisfactory] l� IW PROJECT : LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT BEAM #4, HIGH HDR RIGHT OF GREAT ROOT DESIGN BY: R.A Structural JOB NO.. DATE: 12I1V2012 REVIEW BY: R.A_ Wood Beam Design Base on NDS 2005 INPUT DATA 8r DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 8 No. 1, Douglas Fir -Larch L= 4 ft WD = 576 Ibs / ft WL = 340 Ibs / ft PDt = 3750 Ibs Lt = 1.5 ft PD2 = 0: Ibs L2 = 0 ft L, FoFo 1 1� 1 F112 WD 7 7 1 I d L = L / 36Q Camber => 0.05 inch dKcrD+L= L/ 240 Does member have continuous lateral support by top diaphragm ? THE BEAM DESIGN IS ADEQUATE. (1= yes, 0= no) 1 Yes Code Duration Factor C Condition Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirwt = 9 Ibs / ft RLett = 4.18 kips RR;9nt = 3.24 kips VMax = 3.61 kips, at 7.5 inch from left end MMax = 5.24 ft-kips, at 1,50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'R,in = N/A E = E,, = 1600 ksi Fb. = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb` = N/A A = 41.3 in 1 = 193 in° Fv = 170 _ psi Fe - 1,350 psi S,t = 51.6 s in RB= N/A E' = 1,600 ksi F,; = 170 psi 1E = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 1219 psi < Fb = f,' = 1.5 VMax / A = 131 psi < CHECK DEFLECTIONS d (L, M.) = 0.01 in, at 2.000 ft from left end, d (Kcr D+ L, Max) = 0.04 in, at 1.900 ft from left end Where Kr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.05 in, at 1.900 ft from left end CV Cc Cr 1.00 1.00 1.00 1350 psi [Satisfactory] F, [Satisfactory] < dL=L/360 < 4crD+L=L/240 [Satisfactory] [Satisfactory] ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1439 psi Where FI = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o 5 FI* = F. CD CF = 1480 psi Le = KQ L = 1.0 L = 48 in d = 7.5 in SF = slenderness ratio = 6.4 < FIE = 0.822 E'm,n / SF2 = 11640 psi E'min = 580 ksi F = FIE / Fc* = 7.865 c = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 psi < F.' ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 0.973 50 1 1 .f l '�' WAY ", a. F' y' F '; 6 I •t F [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1292 psi < Fe [Satisfactory] CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F.' )2 + % / [Fb' 0 - fI / FI)] = 0.600 < 1 [Satisfactory] (J fU PROJECT : LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM #5, HDR RIGHT N RIGHT OF OFFICE DESIGN BY: R.A Structural JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 13 1/2 Glulam 24F-1.8E L = 13.5 It wD = 570 Ibs / ft wL = 340 Ibs / ft PDt = 0 Ibs Lt = 0' ft PD2 = 0;: Ibs L2 = 0 ft dL=L/360 A Ka D+L= L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C� Condition 1 0.90 Dead Load 2 1,00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load L L, PD1 ; I PD2 NOD 1 Camber=> 0.35 inch THE BEAM DESIGN IS ADEQUATE. ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirwt = 16 Ibs / ft RLeft = 6.25 kips RRight = 6.25 kips VM. = 5.21 kips, at 13.5 inch from left end MM. = 21.11 ft-kips, at 6.75 ft from left end ANE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = E,, = 1800 ksi Fb = N/A d = 13.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 69.2 in 1 = 1,051 in Fv = 265 psi Fe = 2,400 psi S,t = 155.7 1n3 RB = N/A E' = 1,800 ksi Fv' = 265 psi LE= N/A CD CM Ct Ci 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 1627 psi < f,=1.5VMx/A= 113 ps CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 Fb = 2400 psi < Fv1 [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.13 in, at 6.750 ft from left end, A (Ka D+ L. Mm) = 0.37 in, at 6.750 It from left end Where Kr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.60, Max) = 0.35 in, at 6.750 ft from left end [Satisfactory] A L = L / 360 [Satisfactory] 4 Ka D + L = L / 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS FI' = F, CD CP CF = 2369 psi Where FI = 1600 psi Cp = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s - FI* = FI CD CF = 2560 psi LB = Ke L = 1.0 L = 162 in d = 13.5 in SF = slenderness ratio = 12.0 < FIE = 0.822 E'min / SF = 5309 psi E'min = 930 ksi F = FIE / F.* = 2.074 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 14 psi < Fc' - ALLOWABLE FLEXURAL STRESS IS Fti = 3840 psi, [ for Co = 1.6 ] 0.925 50 1 1 F y f- [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1670 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / Fti )Z + fb / [Fti (1 - fI / Fes] = 0.436 < 1 [Satisfactory] RA STRUCTURAL ENGINEERING 78080 CALLE CAMINO, SUITE 102 LA QUINTA, CA. 92253 (760) 771-9993 Title Block Line 6 6 -^ _ IvoQ-d' Beam ; Title : Engineer: Project Desc.: Job # ,,: v 1s BEC 20!2 12 13A;a tyERCALC, INC. 19B'.r2Q12; &nid:6:'12.9.43,11er,E 12.1:"a3 Y=lot uk EvIE Description : LOT 128, MADISON BLUB, COLUMBUS WAY, BM # 6 OVER STAIRS t:DDE.R�CES`::: Calculations per NDS 2005, ASCE 7-05 Load Combination Set: ASCE 7-05 Material Properties Analysis Method: Allowable Stress Design Fb - Tension 2400 psi E: Modulus of Elasticity Load Combination ASCE 7-05 Fb - Compr 1850 psi Ebend- xx 1800 ksi Fc - Pril 1650 psi Eminbend - xx 930 ksi Wood Species : DF/DF Fc - Perp 650 psi Ebend- yy 1600 ksi Wood Grade : 24F - V4 Fv 265 psi Eminbend - yy 830 ksi Ft 1100 psi Density 32.21 pcf Beam Bracing :Completely Unbraced D(O.3) LQ D3}) Lr(0-3) D(2) Lr(1.15) - f D 0-15 i D(018) L r{0.32} ' 5.125x15 Span = 14.250 ft 1p Service loads entered. Load Factors will be applied for calculations. A iiiXd Lads __ _ _ Point Load : D = 0.30, Lr = 0.30 k @ 4.0 ft Point Load : D = 0.30, Lr = 0.30 k @ 5.0 ft Point Load : D = 2.0, Lr =1.150 k @ 8.0 ft Uniform Load: D = 0.030, Lr = 0.020 ksf, Extent = 8.0 --» 14.250 ft, Tributary Width =16.0 ft Uniform Load: D = 0.0250 ksf, Tributary Width = 6.0 ft DESIGN SUMMARY_ Maximum Bending Stress Ratio = O.6812 1 Maximum Shear Stress Ratio Section used for this span 5.125x15 Section used for this span fo:Actual - 1,617.40psi fv:Actual FB : Allowable = 2,377.39psi Fv: Allowable Load Combination +D+Lr+H Load Combination Location of maximum on span = 7.980ft Location of maximum on span Span # where maximum occurs - Span # 1 Span # where maximum occurs Maximum Deflection Max Downward L+Lr+S Deflection 0,112 in Ratio = 1527 Max Upward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Downward Total Deflection 0.335 in Ratio = 510 Max Upward Total Deflection 0.000 in Ratio = 0 <240 11+Faxrttim FibhC' 8:5.taesse5;tr Laadaribinafions. _ Load Combination Max Stress Ratios _ Moment Values ]c-sinn n 0.439 : 1 5.125x15 116.47 psi 265.00 psi +D+Lr+H 13.039 ft Span # 1 Shear Values Segment Length Span # M V C d C FN C i Cr Cm C t C L M tb Fb V iv Fv +p 0.00 0.00 0.00 0. 00 Length=14,250 ft 1 0.451 0.292 1.00 1.00 1.00 1.00 1.00 1.00 0.99 17.19 1,073.36 2377.39 396 77.27 265.00 +D+Lr+H RA STRUCTURAL ENGINEERING Title : Job # 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project Desc.: (760) 771-9993 Title Block Line 6 Pointed: 16DEC:: ,2.12:13AM o yam `',`;::: ';•; _ 777 ;- _ .:. ,. .. ,- ... - - ;, - ,.. ::. ENERCIiLC; ]1dC.1993-2L]12; Build:5;12,5-#3; Verfi.12i;31.`: Description : LOT 12B, MADISON BLUB, COLUMBUS WAY, BM # 6 OVER STAIRS Load Combination Max Stress Ratios Moment Values Shear Values Segment Length Span # M V C d C FN C i Cr C m C t C L M ib Fb V fu Fv +D+0.750Lr+0.750L+0.5250E+H 1.00 1.00 1.00 100 1.00 0.99 0.00 0.00 0.00 0.00 Length=14.250 ft 1 0.623 0.403 1.00 1.00 1.00 1.00 1.00 1.00 0.99 23.73 1,481.39 2377.39 5.47 106.67 265.00 OvefdI(MaxjAjji �-Vkn ctored Loads: . Load Combination Span Max. "-' Defl Location in Span Load Combination Max. '+" Defl Location in Span D+Lr 1 0.3349 1.410 0 0000 0.000 yetiical ROaet 4h �.. 0nfaCt0re. - :. ;.: ::. Support notation : Far left is #1 Values in KIPS Load Combination Support 1 Support 2 Overall MAXimum 4.368 7 120 D Only 3.014 4.723 Lr Only 1.354 2.396 D+Lr 4.368 7.120 0 PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLJENT : BEAM #7, HDR FROM" OF M SUITE 2 DESIGN BY: R.A JOB NO.: DATE: 12111=12 REVIEW BY: RA 'Wood Beam` Qesiah: Base on' NDS' 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM L 6 x 8 No. 1, Douglas Fir -Larch L, L = 625 ft wu = 510 Ibs / PD, + PDz ft wL = 300 Ibs / ft Pat = 0 Ibs "D Lt= 0 ft - PD2 = 0 Ibs } L2 = 0 ft et L = L 1 360 Camber => 0.09 inch AKcrD+L=L1240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C� CDndiiinn Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseirvw = 9 Ibs / ft RLeft = 2.56 kips RRight = 2.56 kips VMax = 2.05 kips, at 7.5 inch from left end MMax = 4.00 ft-kips, at 3.13 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb* = N/A A = 41.3 in 1 = 193 in F = 170 psi Fe = 1,350 psi S,t = 51.6 in3 RB = NIA E' = 1,600 ksi F,; = 170 psi 1E= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MM. / Sx = 931 psi < Fb fv = 1.5 VMax / A = 74 psi < CHECK DEFLECTIONS A (L, Max) = 0.03 in, at 3.125 ft from left end, d (KcrD+L, Max) = 0.09 in, at 3.125 ft from left end Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.09 in, at 3.125 ft from left end J Cv Cc Cr 1.00 1.00 1.00 1350 psi Fv' [Satisfactory] [Satisfactory] A L = L / 360 [Satisfactory] AKerD+L=L1240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = F� CD CP CF = 1370 psi Where F� = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 = Fc* = Fr CD CF = 1480 psi Le = Ke L = 1.0 L = 75 in d = 7.5 in SF = slenderness ratio = 10.0 < F.E = 0.822 E'min / SFZ = 4768 psi E'min = 580 ksi F = FEE / Fj = 3.221 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 psi < F,' THE ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for Cp = 1.6 ] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1003 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F.' )Z + fb / [Fti (1 - fc / Fes] = 0.467 RN 0.925 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] Fe [Satisfactory] < 1 [Satisfactory] a fUA PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT : BEAM #8, HDR RIGHT OF GUEST SUITE 1 DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY L MEMBER SIZE 6 x 8 No. 1, Douglas Fir -Larch MEMBER SPAN L = 6.25 ft 1 PD1 + PD2 UNIFORMLY DISTRIBUTED DEAD LOAD wD = 450 Ibs / ft + UNIFORMLY DISTRIBUTED LIVE LOAD wu = 260 Ibs / ft FTTT i I I d CONCENTRATED DEAD LOADS PD1 = 2206 Ibs "D (0 for no concentrated load) L, = 1.75 ft PD2 = 0 Ibs L2 = 0 ft DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM A L = L / 360 Camber => 0.15 inch AKcro+L= L/ 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load THE BEAM DESIGN IS ADEQUATE. Code Desiffnation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 _YSIS ERMINE REACTIONS, MOMENT, SHEAR wself wt = 9 IbS / ft RLee = 3.83 kips RRlght = 2.86 kips VMax = 3.38 kips, at 7.5 inch from left end MM. = 5.70 ft-kips, at 2.30 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1600 ksi Fb N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 41.3 in 1 = 193 in Fv = 170 psi Fe = 1,350 psi S,t = 51.6 in RB = N/A E' = 1,600 ksi Fv' = 170 psi 1E= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1326 psi < Fb = f, = 1.5 VMax / A = 123 psi < CHECK DEFLECTIONS A �_, Max) = 0.03 in, at 3.125 ft from left end, A (KID I L, Max) = 0.13 in, at 2.988 ft from left end Where K, = 1.00 , (NDS 3.52) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.15 in, at 2.988 ft from left end Cv Cr Cr 1.00 1.00 1.00 1350 psi F [Satisfactory] I [Satisfactory] A L = L / 360 [Satisfactory] A KID - L_ L / 240 [Satisfactory] 0 (CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F� = F, Cp CP CF = 1370 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 = Fc` = F� CD CF = 1480 psi Le = KB L = 1.0 L = 75 in d = 7.5 in SF = slenderness ratio = 10.0 < FcE = 0.822 E'min / SF = 4768 psi E'min = 580 ksi F = FCE / F�* = 3.221 c = 0.8 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 24 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] 1 1 F y f- }y yy I I 0.925 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1399 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )2 + fb / [Fe (1 - f. / FEE)] = 0.651 < 1 [Satisfactory] IA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM #9, TYP E3M AT BRIDGE DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 4 x 10 No. 2, Douglas Fir -Larch L = 6.5 ft wD= 210.' Ibs/ft wL= 100 Ibs/ft PD1 = 0 Ibs Lt= 0 ft PD2 = 0 lbs L2= 0 ft L Poi I Poe NL Wp .I liL. it } d4_ = L / 360 Camber => 0.04 inch d Kcr D + L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Factor. Cp Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1,25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseirwr = 7 Ibs / ft RLeft = 1.03 kips VMax = 0.79 kips, at 9.25 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.50 in E'min = N/A d = 9.25 in FbE = N/A A = 32.4 in 1 = 231 in S,r = 49.9 in RB = N/A 1E= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.20 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 403 psi < Fb f = 1.5 VMax / A = 36 psi < CHECK DEFLECTIONS d (L. Max) = 0.01 in, at 3.250 ft from left end, d (Kcr D+ L, Max) = 0.03 in, at 3.250 ft from left end Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.04 in, at 3.250 ft from left end THE BEAM DESIGN IS ADEQUATE. Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 3 RRiyht = 1.03 kips MMax = 1.67 ft-kips, at 3.25 ft from left end E = Ex = 1600 ksi Fb = N/A Fb = 900 psi F = FbE / Fb. = N/A F = 180 psi Fti = 1,080 psi E' = 1,600 ksi F, = 180 psi Cv, Cc Cr 1.00 1.00 1.00 1080 psi F [Satisfactory] [Satisfactory] AL = L / 360 [Satisfactory] d Kcr D + L = L / 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F.' = F, CD CP CF = 2341 psi Where FI = 1350 psi CD = 1.60 CF = 1.20 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s - Fc* = FI CD CF = 2592 psi LB = Ke L = 1.0 L = 78 in d = 9.25 in SF = slenderness ratio = 8.4 < FIE = 0.822 E'min / SF2 = 6705 psi E'min = 580 ksi F = FCE / Fc* = 2.587 c = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 31 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fe = 1728 psi, [ for Co = 1.6 ] 1 rTFM 1 0.903 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 495 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )2 + fb / [Fti (1 - f. / F.E)] = 0.288 < 1 [Satisfactory] fLA PROJECT : LOT 12B, MADISON CLUB,,COLUMBUS WAY PAGE : Structural CLIENT: BEAM #10, HDR LEFT OF GUEST SUITE 1 DESIGN BY: RA JOB NO.: DATE: 12/1112-OU REVIEW BY: R.A. Wood Beam Desion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) 6 x 8 No. 1, Douglas Fir -Larch L = 5.25 ft wD = 240 Ibs / ft wL = 120 Ibs / ft PD1 = 2700 Ibs L1 = 4 It P❑2 0 Ibs L2 = 0 ft Ll L- PD1 ; �' PD2 Y'D DEFLECTION LIMIT OF LIVE LOAD AL = L 1 360 Camber => 0.07 inch DEFLECTION LIMIT OF LONG-TERM sKcrD+L = L 1 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor. Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load 4LYSIS FERMINE REACTIONS, MOMENT, SHEAR wseir wt = 9 Ibs / ft RLea = 1.61 kips VM. = 2.80 kips, at 7.5 inch from right end Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRi9ht = 3.03 kips MM. = 3.49 ft-kips, at 4.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb* = N/A A = 41.3 in 1 = 193 in Fv = 170 psi Fe = 1,350 psi S,t = 51.6 in RE, = N/A E' = 1,600 ksi Fv' = 170 psi IE= N/A CD CM C( Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 813 psi < Fb = f = 1.5 VMax / A = 102 psi < CHECK DEFLECTIONS d (L. Max) = 0.01 in, at 2.625 ft from left end, d (Ka D+L, Max) = 0.05 in, at 2.763 It from left end Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) - 0.07 in, at 2.900 ft from left end Cv C, Cr 1.00 1.00 1.00 1350 psi F,; [Satisfactory] [Satisfactory] d L = L / 360 [Satisfactory] d Kcr o+ L = L / 240 [Satisfactory] I 'CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F,' = FI CD CP CF = 1406 psi Where FI = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o e = Fc* = Fc CD CF = 1480 psi Le = Ke L = 1.0 L = 63 in d = 7.5 in SF = slenderness ratio = 8.4 < FIE = 0.822 E'min / SFZ = 6757 psi E'min = 580 ksi F = FIE / F,* = 4.565 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fI = F / A = 24 psi < F,' THE ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] F 1 1 r — �r f 1 1 0.950 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 886 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / Fc' )Z + fb / [Fti 0 - fc / FCE)] = 0.412 < 1 [Satisfactory] Iw PROJECT - LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM #11, HDR AT GUEST BDRM 2 DOOR DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A, Wood Beam Desiqn Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 6 x 6 No. 1, Douglas Fir -Larch L= 4 ft wD= 210 Ibs/ft wL = 100 Ibs / ft PD1 = 2150 Ibs L1 = 1 ft PD2 = 0 Ibs L2 = 0 ft L, Poi 1 1 L, 1 Foe 1 1NL Wp 1 AL = L / 360 Camber => 0.06 inch d Kcr D + L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Co Condition Code Designation 1 0A0 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseir wt = 7 Ibs / ft RLeft = 2.25 kips RRigm = 1.17 kips VMax = 2.10 kips, at 5.5 inch from left end MM. = 2.09 ft-kips, at 1.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = NIA E = EX = 1600 ksi Fb = N/A d = 5.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 30.3 in 1 = 76 in Fv = 170 psi Fe = 1,350 psi Sx = 27.7 in RB = N/A E' = 1,600 ksi F = 170 psi IE = N/A CD CM Ct G, CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 903 psi < Fb = 1350 psi [Satisfactory] f = 1.5 VMax / A = 104 psi < Fv' [Satisfactory] CHECK DEFLECTIONS d (L, Max) = 0.00 in, at 2.000 ft from left end, < d L = L / 360 [Satisfactory] d (Kcr D + L, Max) = 0.04 in, at 1.800 ft from left end < d Kcr D + L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.06 in, at 1.800 ft from left end ECK THE BEAM CAPACITY WITH AXIAL LOAD 4L LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = Fe, Co CP CF = 1399 psi Where F. = 925 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.e - Fc* = Fc CD CF = 1480 psi Le = K. L = 1.0 L = 48 in d = 5.5 in SF = slenderness ratio = 8.7 < F,E = 0.822 E'min / SF = 6260 psi Emin = 580 ksi F = FEE / FC* = 4.229 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 33 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] 1 � F T 0.946 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1002 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / Fc' )Z + fb / [Fe 0 - fc / FEE)] = 0.467 < 1 [Satisfactory] f.4 PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM #12, BM AT FRONT OF M. BATH 1 DESIGN BY: R.A Structural JOB NO-: DATE: 12/1112012 REVIEW BY: R.A. Wood Beam Desion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x I0 No. 1, Douglas Fir -Larch L= 10 ft wo = 360 Ibs / ft wL = 200 Ibs / ft PD, = 2000 Ibs L1 = 1.5 ft PD2 = 0 Ibs LZ = :. 6: ft L, yII L, PD2I ¢�. �. �. WD ]..Ili s 1• A 1 A L = L / 360 Camber => 0.27 inch A1111L=L/240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Dural -€on Factor Cn Condit)on 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load THE BEAM DESIGN IS ADEQUATE. Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseir wi = 11 Ibs / ft RLefl = 4.56 kips RRigM = 3.16 kips VMex = 4.10 kips, at 9.5 inch from left end MM. = 8.71 ft-kips, at 4.30 ft from left end o)INE SECTION PROPERTIES&ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = N/A d = 9.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 52.3 in2 I = 393 in° Fv = 170 psi Fe = 1,350 psi Sx = 82.7 in3 RB = N/A E' = 1,600 ksi Fv1 = 170 psi 1E = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1264 psi < Fb = f, = 1.5 VMax / A = 118 psi < CHECK DEFLECTIONS A (L. Max) = 0.07 in, at 5.000 ft from left end, A (Kcr D + L, Max) = 0.25 in, at 5.000 ft from left end Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.27 in, at 4.650 ft from left end Cv CD Cr 1.00 1.00 1.00 1350 psi F, [Satisfactory] I [Satisfactory] A L = L / 360 [Satisfactory] d Kcr D + L = L / 240 [Satisfactory] 1 0 jCHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAI. LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = F,, Cp Cp CF = 1286 psi Where Fc = 925 psi Cp = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 ; F�* = F� Cp CF = 1480 psi Le = Ke L = 1.0 L = 120 in d = 9.5 in SF = slenderness ratio = 12.6 < FoE = 0.822 E'min / SF2 = 2988 psi E'min = 580 ksi F = FCE / Fc' = 2.019 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fe = F / A = 19 psi < Fc' I HE ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for Co = 1.6 ] 1 1 F �l ii f T T 0.869 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1321 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + % / [Fe (1 - f. / F.E)] = 0.616 < 1 [Satisfactory] dD Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT- BEAM #13, HDR AT RIGHT OF M. BATH 1 DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Desiqn Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) L 6 x 6 No. 1, Douglas Fir -Larch L L= 4 ft Po 1I 1 Po: wD = 660 Ibs / ft 1 1 I WL = 400 Ibs / ft PD1 = 0 Ibs Li = 0 ft PD2 = 0 Ibs ✓< L2 = 0 ft DEFLECTION LIMIT OF LIVE LOAD d L = L / 360 Camber => 0.05 inch DEFLECTION LIMIT OF LONG-TERM AKcr0+L L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 _YSIS ERMINE REACTIONS, MOMENT, SHEAR Wself Ana = 7 Ibs / ft RLee = 2,13 kips RRight = 2.13 kips VMax = 1.64 kips, at 5.5 inch from left end MMex = 2.13 ft-kips, at 2.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1600 ksi Fb. - WA d = 5.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb* = N/A A = 30.3 in 1 = 76 in4 Fv = 170 psi Fe = 1,350 psi SX = 27.7 in RB = N/A E' = 1,600 ksi Fv = 170 psi lE = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 923 psi < Fb = fv = 1.5 VMx / A = 82 psi < CHECK DEFLECTIONS d (L, Max) = 0.02 in, at 2.000 ft from left end, d (Kcr D+ L, Max) = 0.05 in, at 2.000 ft from left end Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.05 in, at 2.000 ft from left end Cv C, Cr 1.00 1.00 1.00 1350 psi F, [Satisfactory] < < [Satisfactory] d L = L / 360 [Satisfactory] d Ker D . L = L / 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fo' = F, CD CP CF = 1399 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s _ FG* = F� CD CF = 1480 psi Le = K. L = 1.0 L = 48 in d = 5.5 in SF = slenderness ratio = 8.7 < Fc = 0.822 E'mm / SF = 6260 psi E'min = 580 ksi F = FcE / Fc* = 4.229 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f� = F / A = 33 psi < F� THE ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1022 psi < Fti CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / Fc' )Z + fb / [Fe (1 - f,,/ Fes] = 0.476 5 Y Y 1 0.946 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] [Satisfactory] < 1 [Satisfactory] Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT = SEAM #14, HDR AT RIGHT OF M. SUITE 1 DESIGN BY: R.A Structural JOB NO, DATE: 1211112012 REVIEW BY: R.A. I Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1./8 x 13 1/2 Glulam 24F-1.8E L = 12.25 ft wD = 78Q, Ibs / ft wL = 48Q Ibs / ft PD1 = 0' Ibs ft PD2 = 0 .. Ibs L2 = '.10 ft AL=L/,360: AKc D+L=L/240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code _Duration Factor. C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseirwt = 16 Ibs / ft . RLell = 7.82 kips VMax = 6.38 kips, at 13.5 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 13.50 in FbE = N/A A = 69.2 in 1 = 1,051 in Sx = 155.7 in3 RB = N/A 1E = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1846 psi < Fb f = 1.5 VM., / A = 138 psi < ECK DEFLECTIONS A (L, Max) = 0.13 in, at 6.125 ft from left end, A(KcrD+L, M.) = 0.34 in, at 6.125 ft from left end Where Kar = 1.00 , (NDS 3.5.2) PERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.32 in, at 6.125 ft from left end LF", L 1 Poe "'L1 V Y wD Camber => 0.32 inch THE BEAM DESIGN IS ADEQUATE. RRight = 7.82 kips MMax = 23.94 ft-kips, at 6.13 ft from left end E = Ex = 1800 ksi Fb+= N/A Fb = 2,400 psi F = FbE / Fb. = N/A Fv = 265 psi Fe = 2,400 psi E' = 1,800 ksi F, = 265 psi Cv Cc Cr 1.00 1.00 1.00 2400 psi F [Satisfactory] [Satisfactory] A L = L / 360 [Satisfactory] A Kcr D+ L = L / 240 [Satisfactory] l ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = Fc Co CP CF = 2415 psi Where F. = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s = F.* = F. CD CF = 2560 psi L.e = KB L = 1.0 L = 147 in d = 13.5 in SF = slenderness ratio = 10.9 < FcE = 0.822 E'min / S172 = 6447 psi E'min = 930 ksi F = FVE / Fc' = 2.519 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS f� = F / A = 14 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fb' = 3840 psi, [ for Co = 1.6 ] 0.943 50 1 1 t4 V F A F 1 i [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1889 psi < Fe [Satisfactory] :-CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / Fc' )2 + fb / [Fti 0 - f. / F.E)] - 0.493 < 1 [Satisfactory] C2- IA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE CLIENT., BEAM #15, RIGHT OF M. SUITE 1 COVERED F DESIGN BY: RA Structural JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A f Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLIB 5 1/8 x 13 1/2 Glulam 24F-1.8E L = 17.25. It WD = 300. Ibs / ft wL = 160 Ibs / ft PD1 = 0 Ibs L1 =' : 0' ft PD2 = 0 Ibs L2 = 0 It AL=L/360 A Kcr D + L = L ! 240 L� Poi 1 1 Po? 1 � "'b 1 It. Camber => 0.50 inch Does member have continuous lateral support by top diaphragm ? THE BEAM DESIGN IS ADEQUATE. (1= yes, 0= no) 1 Yes Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif vw = 16 Ibs / ft RLefl = 4.11 kips VMax = 3.57 kips, at 13.5 inch from left end MINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 13.50 in FbE = N/A A = 69.2 in 1 = 1,051 in Sx = 155.7 in RB = N/A 1E= N/A CD CM Ct CI CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1366 psi < Fb = f,; = 1.5 VMax / A = 77 psi < CHECK DEFLECTIONS A (L, Max) = 0.17 in, at 8.625 ft from left end, d (KcrD+L , Max) = 0.50 in, at 8.625 ft from left end Where Kc, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.50 in, at 8.625 ft from left end RRight = 4.11 kips MMax = 17.72 ft-kips, at 8.63 ft from left end E = Ex = 1800 ksi Fb. = N/A Fb = 2,400 psi F = FbE / Fb. = N/A Fv = 265 psi Fb' = 2,400 psi E' = 1,800 ksi Fv' = 265 psi Cv C, Cr 1.00 1.00 1.00 2400 psi Fv' [Satisfactory] [Satisfactory] A L = L / 360 [Satisfactory] A Kcr D. L = L / 240 [Satisfactory] !CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips T HE ALLOWABLE COMPRESSIVE STRESS IS F, = F� CD CP CF = 2145 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s _ Fc' = Fc CD CF = 2560 psi Le = Ke L = 1.0 L = 207 in d = 13.5 in SF = slenderness ratio = 15.3 < FCE = 0.822 E'min / SF2 = 3251 psi E'min = 930 ksi F = FCE / F�* = 1.270 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 14 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fe' = 3840 psi, [ for Co = 1.6 ] 0.838 50 i r �. F y ■i I F yi �- 7 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1410 psi < Fe [Satisfactory] =CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )Z + fb / [Fti (1 - f. / F.E)] = 0.369 < 1 [Satisfactory] j RA PROJECT: L7T 2^.i' f�ROfso CLUP.A. SCLIENT: ,5tif,fi6,BM;'ATR€GHTOF'GI STRUCTURAL JOB NO.: DATE Steel Beam Design mth Gravity Loading Based on AISC Mgnual 131h E INPUT DATA & DESIGN SUMMARY BEAM SECTION (WF, Tube or WT) _ > W15X67 SLOPED DEAD LOADS wDL,1 = 0-43 kips /ft wDL,z = 0 .'' kips / ft PROJECTED LIVE LOADS wLL,1 = ' .01 .: kips / ft wLL,2 = kips / ft CONCENTRATED LOADS PDL= 0 kips PLL = ::, 0 .. kips BEAM SPAN LENGTH L i = 38 ft CANTILEVER LENGTH L 2 = p ft, (0 For no cantilever) BEAM SLOPE 6 : 12 (9 = 0.00 D ) DEFLECTION LIMIT OF LIVE LOAD d LL = L 1 240:.. BEAM YIELD STRESS Fv=: �G .. ;:: ksi THE BEAM DESIGN IS ADEQUATE. (ANALYSIS DETERMINE REACTIONS, MOMENTS & SHEARS u�, 11 r w Lz L1+Lz Rz=0.5 +w�.�JLI+I +w�.=J�Li+O.SLz�—+P cash COSFI Li Li 11.97 kips R, = K"_ 11 +wu., Li+ +WLL.a)Lz+P—R2 cose (w,,�cose = 11.97 kips X1 = 19.00 ft X2= 19.00 ft X3 = 0.00 ft Mm,=0.5I CoSO+waz)Lz+PL2= 0.0 ft-kips __(_11n )� i�1Y2�� wM1 +w�, 8 113.7 ft-kips Vmax = 11.97 kips, at R1 right. ry . PAGE: co�©iF€� Fgrio DESIGN BY.- RA REVIEW BY: R4' ..:.' WF Ix Sx J bf tr tw 970 119 2.62 10.20 0.67 0.40 P w�i 1 MMin BENDING CAPACITY ABOUT MAJOR AXIS (AISC 360-05 Chapter F) l=Max(L2 , X3) = 0.00 ft, unbraced length wp , W." Y T R_ Aw Aw Re wired Conditions Double S metric a ter F Sections for WF Tube WT F7 x F9 Compact Web gF2 RT Noncom act Web x Slender Web Compact Flanges x x Noncom act Flanges Slender Flanges Applicable Section ok CHECK MMax BENDING CAPACITY ABOUT MAJOR AXIS (AISC 360-05 Chapter F) Mallowable = Mn 7 Db = 237.1 ft-kips, top flange fully supported > Mmax [Satisfactory] Mallowable =Mn / 0b = 237.1 ft-kips > MA,fla [Satisfactory] where Db = 1.67 , (AISC 360-05 F1) 9 SHEAR CAPACITY ABOUT MAJOR AXIS (AISC 360-05 Chapter G2 or G5) (confd) V alrowabre = V„ / .f]„ = 83.3 kips > Vm.a [Satisfactory] where Qv = 1.67 , (AISC 360-05 G1) CAMBER AT DEAD LOAD CONDITION L = L i/cos B = 38.00 ft, beam sloped span a = L 2/cos B = 0.00 ft, beam sloped cantilever length P = PDL cos 0 = 0.00 kips, perpendicular to beam w1 = woo COS B = 0.43 klf, perpendicular to beam w2 = wDL,2 COS 0 = 0.00 klf, perpendicular to beam Pa' (L+a)_ w,Pa+ wsa3(4L+3a) DEnd = 3EI 24EI — 24EI — 0.00 in, downward perpendicular to beam. USE C = 414" A i CANTILEVER PaL2 5wiL4 wzL2rr2 = — + _ Mid 16M 384Ef __ 32E� 0.72 in, downward perpendicular at middle of beam. USE C = 314" A T MID BEAM, DEFLECTION AT LIVE LOAD CONDITION P = PLL coa B = 0.00 kips, perpendicular to beam wr = wL1,I cost 0 = 0.20 klf, perpendicularto beam W2 = wLL,,2 COS 0 = 0.00 klf, perpendicular to beam ANH d = L Pat (L + a) — w,L3a + 3EI w,c 3 (4L + 3a) l 24EI 24EI J �s B = 0.00 in, downward to vertical direction. 2L2 / 240 = 0.00 in [Satisfactory] O.tira + PaLz 5w,L4 _ w_L2a21 = — 16ET 384EI 32PI cosB = 0.33 in, downward to vertical direction. < L y / 240 = 1.90 in [Satisfactory] D Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM #16A, RIGHT OF GREAT COVERED PA DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. l Beam )]esign Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLIB 5 1/8 x 24 Glulam 24F-1.8E L = 38 ft wD = 150 Ibs / ft wL = 100 Ibs / ft PDt = 0 Ibs Lt = 6. ft PD2 = 0 Ibs L2 = 0 ft AL=L/36b AKcrD+L = L / 240 L, LJJIF��P�-T PD2 NtT Y, wD I � 1 Camber => 1.19 inch THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor Ca Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseirwt = 29 Ibs / ft RLeft = 5.31 kips RRight = 5.31 kips VMax = 4.75 kips, at 24 inch from left end MMax = 50.42 ft-kips, at 19.00 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = Ex = 1800 ksi Ft,*= N/A d = 24.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 123.0 in 2 1 = 5,904 ina F„ = 265 psi Fe = 2,110 psi S,t = 492.0 in3 RB = N/A E' = 1,800 ksi F,; = 265 psi lE= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.88 1.00 1.00 :HECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1230 psi < F b = 2110 psi [Satisfactory] f,' = 1.5 VMax / A = 58 psi < Fv' [Satisfactory] :HECK DEFLECTIONS A (L, Max) = 0.44 in, at 19.000 ft from left end, < A L = L / 360 [Satisfactory] d (Kcr D . L, Max) = 1.23 in, at 19.000 ft from left end < d Ker D « L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 1.19 in, at 19.000 ft from left end 8 CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1744 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0-5 = 0.681 Fc' = F. CD CF = 2560 psi L. = Ke L = 1.0 L = 456 in d = 24 in SF = slenderness ratio = 19.0 < 50 FcE = 0.822 E'min / SF2 = 2118 psi E'min = 930 ksi F = FcE / Fa* = 0.827 c = 0.9 HE ACTUAL COMPRESSIVE STRESS IS f = F/A = [Satisfies NDS 2005 Sec. 3.7.1.41 c 8 psi < Fc' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fe = 3377 psi, [ for Co = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1254 psi < Fe CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / Fc' )2 + fb / [Fe 0 - fc / FcE)] = 0.373 [Satisfactory] < 1 [Satisfactory] 35 fA PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT BEAM # 17, RIGHT OF GREAT ROOM DESIGN BY : Structural JOB NO.: DATE: 12A 112012 REVIEW BY: R.A. Wood Beam Desiqn Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) iDEFLECTION LIMIT OF LIVE LOAD (DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 27 , Glulam 24F-1.8E L = 29.5 ft wD = 570 Ibs / ft wL = 180 Ibs / ft P❑t = 0 Ibs Lt = 0 ft PD2 = 0 Ibs L2 = 0 ft dL=L/360 AKcrD+L= L/ 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes L, jII� PD1 I 1 L, 1 Po2 1 NL r v'o G _ Camber => 1.02 inch THE BEAM DESIGN IS ADEQUATE. Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirwt = 33 Ibs / ft RLea = 11.55 kips RRIght = 11.55 kips VMax = 9.79 kips, at 27 inch from left end MMax = 85.17 ft-kips, at 14.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = EX = 1800 ksi Fb = N/A d = 27.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 138.4 in2 I = 8,406 in4 Fv = 265 psi Fb = 2,139 psi S, = 622.7 in3 RB = N/A E' = 1,800 ksi F = 265 psi lE= N/A CD CM Ct Ci CL CP Cv C, Cr �- 1.00 1.00 1.00 1.00 1.00 1.00 0.89 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 1641 psi < Fb = 2139 psi [Satisfactory] fv' = 1.5 VMax / A = 106 psi < Fv' [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.20 in, at 14.750 ft from left end, < d (Kcr D + L, Max) = 0.88 in, at 14.750 ft from left end < Where K,, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.50, Max) = 1.02 in, at 14.750 ft from left end d L = L / 360 [Satisfactory] d Kar D + L = L / 240 [Satisfactory] 0 ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips - ALLOWABLE COMPRESSIVE STRESS IS F,' = F, Co CP CF = 2310 psi Where F� = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s = F,* = F� CD CF = 2560 psi Le = KB L = 1.0 L = 354 in d = 27 in SF = slenderness ratio = 13.1 < F,E = 0.822 E'min / SF2 = 4447 psi E'min = 930 ksi F = FCE / F�* = 1.737 c = 0.9 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 7 psi < Fc' TiiF ALLOWABLE FLEXURAL STRESS IS Fti = 3423 psi, I for CD = 1.6 ] 1 1 TTTT-T--1 II 1 1 0.902 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1663 psi < Fti [Satisfactory] =CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )2, fb / [Fe (1 - f. / F.E)] = 0.487 < 1 [Satisfactory] 1.4 PROJECT. LOT 12B, MADISON CLUB, COLUM13US WAY PAGE: Structural CLIENT BEAM # 18; RIGHT OF GREAT ROOM DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 112 x 27 Glulam 24F-1.8E MEMBER SPAN L = 29.5 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 660 lbs/it UNIFORMLY DISTRIBUTED LIVE LOAD wL = 340 Ibs / ft CONCENTRATED DEAD LOADS PDl = 0 Ibs (0 for no concentrated load) L, = 0 ft PD2 = 0 Ibs L2 = 0 ft DEFLECTION LIMIT OF LIVE LOAD - R = L 1 360 DEFLECTION LIMIT OF LONG-TERM AKcr D, L = L 1 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseifwt = 35 Ibs / ft RLea = 15.27 kips VMeu = 12.94 kips, at 27 inch from left end L, PD 1 L� 1 PD2 1 �L wD I � � 4 Camber => 1.09 inch THE BEAM DESIGN IS ADEQUATE. RRIght = 15.27 kips MMex = 112.63 ft-kips, at 14.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1800 ksi Fb = N/A d = 27.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 148.5 M2 I = 9,021 in F = 265 psi Fe = 2,124 psi Sx = 668.3 in3 RB = N/A E' = 1,800 ksi F = 265 psi lE = N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.89 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 2023 psi < Fb = 2124 psi f, = 1.5 VMax / A = 131 psi < Fv' [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.36 in, at 14.750 ft from left end, < d (KcrD+ L, Max) = 1.09 in, at 14.750 ft from left end < Where KW = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 1.09 in, at 14.750 it from left end [Satisfactory] A L = L / 360 [Satisfactory] d K,rD+L=L/240 [Satisfactory] CSS :K THE BEAM CAPACITY WITH AXIAL LOAD LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = Fc CD CP CF = 2310 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 = FC* = F� CD CF = 2560 psi Le = KB L = 1.0 L = 354 in d = 27 in SF = slenderness ratio = 13.1 < Fe, = 0.822 E'min / SF = 4447 psi E'min = 930 ksi F = FEE / F�* = 1.737 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS f� = F / A = 7 psi < F,' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3398 psi, [ for CD = 1.6 ] y f- 1 0.902 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 2043 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + % / [Fti (1 - fc / F.E)] = 0.602 < 1 [Satisfactory] RA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT BEAM # 19,TYP FRONT OF GREAT ROOM DESIGN BY: RA JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Desiqn Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 12 No. 1, Douglas Fir -Larch L = 7.5 ft wD = 3.30 Ibs / ft wL = 120 Ibs / ft PDt = 0 Ibs Lt = 0 ft PD2 = 0 Ibs L2= 0 ft PD, ; 1 FD2 1 'N� µ�D 1 d L = L / 360 Camber => 0.03 inch AKc D+L= L/ 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor. C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseifwt = 14 Ibs / ft RLeft = 1.74 kips Vm. = 1.29 kips, at 11.5 inch from left end THE BEAM DESIGN IS ADEQUATE. Code Desi_qnatinn 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No, 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRight = 1.74 kips MM. = 3.26 ft-kips, at 3.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb. = N/A d = 11.50 in FbE = NIA Fb = 1,350 psi F = FbE / Fb* = N/A A = 63.3 in 1 = 697 in Fv = 170 psi Fe = 1,350 psi Sx = 121.2 in RB= N/A E' = 1,600 ksi F = 170 psi 1E= N/A CD CM Ct C•, CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 323 psi < Fb = fv = 1.5 VMax / A = 31 psi < CHECK DEFLECTIONS d (L, Max) = 0.01 in, at 3.750 ft from left end, A (Kcr D+ L , Max) = 0.03 in, at 3.750 ft from left end Where K. = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (t SD, Max) = 0.03 in, at 3.750 ft from left end Cv C, Cr 1.00 1.00 1.00 1350 psi Fv' [Satisfactory] [Satisfactory] d L = L / 360 [Satisfactory] d Kor D+ L = L / 240 [Satisfactory] 9 ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips -i'FIE ALLOWABLE COMPRESSIVE STRESS IS F� = F, Co CP CF = 1417 psi Where F. = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 = Fc' = F, CD CF = 1480 psi I, = Ke L = 1.0 L = 90 in d = 11.5 in SF = slenderness ratio = 7.8 < F�E = 0.822 E'min / SFZ = 7784 psi E'min = 580 ksi F = FoE / F�* = 5.260 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 16 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ] 1 � a v 0.957 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 370 psi < Fe [Satisfactory] --CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + fb / [Fe (1 - f. / F.E)] = 0.172 < 1 [Satisfactory] 01 Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 20, AT POWDER ROOM . DESIGN BY: R.A JOB NO.: DATE: 12/1112012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE 6 x 12 No. 1, Douglas Fir -Larch tom` L, ..- MEMBER SPAN L = 12 ft Poi + Poz UNIFORMLY DISTRIBUTED DEAD LOAD wD = 330 Ibs / ft + UNIFORMLY DISTRIBUTED LIVE LOAD wL = 120 Ibs / ft r x CONCENTRATED DEAD LOADS PD1 = 0 Ibs °'D (0 for no concentrated load) L1 = 0 ft PD2 = 0 Ibs d L2= -0 ft DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM d L = L ! 360 Camber => 0.22 inch AKcrD+L=L/240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load THE BEAM DESIGN IS ADEQUATE. Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseir wt = 14 Ibs / ft RLea = 2.78 kips RRight = 2.78 kips VMax = 2.34 kips, at 11.5 inch from left end MM. = 8.35 ft-kips, at 6.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = N/A d = 11.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 63.3 in 1 = 697 in Fv = 170 psi Fe = 1,350 psi Sx = 121.2 in RB = N/A E' = 1,600 ksi F = 170 psi 1E= N/A CD CM Ct Ci CL CF Cv, Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 826 psi < Fb = 1350 psi f,; = 1.5 VMax / A = 55 psi < F, [Satisfactory] CHECK DEFLECTIONS d (L. Max) = 0.05 in, at 6.000 ft from left end, < d (Ku D * L. Max) = 0.19 in, at 6.000 ft from left end < Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.22 in, at 6.000 ft from left end [Satisfactory] d L = L / 360 [Satisfactory] d Kcr D I L= L/ 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = F� CD CP CF = 1290 psi Where F. = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 - Fc* = Fr CD CF = 1480 psi Le = Ke L = 1.0 L = 144 in d = 11.5 in SF = slenderness ratio = 12.5 < Foe = 0.822 E'min / SF2 = 3041 psi E'min = 580 ks1 F = F,E / Fo* = 2.055 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fo = F / A = 16 psi < F,' THE ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 1 � F F 0.872 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 874 psi < Fti [Satisfactory] -CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fo / F.' )Z + fb / [Fe (1 - fo / Fes] = 0.407 < 1 [Satisfactory] Iw PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 21, FLR BM AT STAIRS DESIGN BY: R.A JOB NO.: DATE: 12111/2012 REVIEW BY: R.A. Wood Beam 'Desian Base on NOS 2006 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN 6 x 10 L = 8.5 No. 1, Douglas Fir -Larch ft L, 'If'-� Fo ; Poz UNIFORMLY DISTRIBUTED DEAD LOAD wD = 210 Ibs / ft + UNIFORMLY DISTRIBUTED LIVE LOAD wL = 450 Ibs / ft W` CONCENTRATED DEAD LOADS PD1 = 0 Ibs (0 for no concentrated load) L, = 0 ft PD2 = 0 Ibs L2= 0 ft DEFLECTION LIMIT OF LIVE LOAD d L = L / 360 Camber => 0.06 inch DEFLECTION LIMIT OF LONG-TERM dKaD+L= L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor Cr,Condition Code ❑esignatton 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural,'Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wsetrwt = 11 Ibs / ft RLeft = 2.85 kips RRight = 2.85 kips VMax = 2.32 kips, at 9.5 inch from left end MMax = 6.06 ft-kips, at 4.25 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = NIA d = 9.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb* = N/A A = 52.3 in 1 = 393 in4 IF, = 170 psi Fe = 1,350 psi Sx = 82.7 in3 RB = N/A E' = 1,600 ksi F = 170 psi IE = NIA CD CM C( Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 879 psi < Fb = 1350 psi [Satisfactory] f = 1.5 VMax I A = 67 psi < F,; [Satisfactory] CHECK DEFLECTIONS d (L, Max) = 0.08 in, at 4.250 ft from left end, < d L = L / 360 [Satisfactory] d (Kcr D + L, maxi = 0.13 in, at 4.250 ft from left end < d Kcr D + L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.06 in, at 4.250 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 11 kips E ALLOWABLE COMPRESSIVE STRESS IS F� = F� Co CF CF = 1349 psi Where F. = 925 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)Z - F / c]o.s = Fc" = Fc CD CF = 1480 psi Le = Ke L = 1.0 L = 102 in d = 9.5 in SF = slenderness ratio = 10.7 < FEE = 0.822 E'mi„ / SF2 = 4136 psi E'min = 580 ksi F = FCE / Fc` = 2.794 c = 0.8 E ACTUAL COMPRESSIVE STRESS IS fc = F / A = 19 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 7 i f W Y T 0.912 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 937 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / Fc' )2 + fb / [Fe 0 - f. / F.E)] = 0.436 < 1 [Satisfactory] 9 Iw PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE Structural CLIENT: BEAM # 22, FLR BM AT STAIRS DESIGN BY : R.A JOB NO.: DATE: 1211112012 REVIEW BY: R.A. Wood Beam Desiqn Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 3 118 x 24 Glulam 24F-3.8E MEMBER SPAN L= 12 ft PDI Po_ UNIFORMLY DISTRIBUTED DEAD LOAD wD = 200 Ibs 1 ft i UNIFORMLY DISTRIBUTED LIVE LOAD wL = 100 Ibs 1 ft ` t { { { v { CONCENTRATED DEAD LOADS PD1 = 7120 Ibs "D ] ., (0 for no concentrated load) L1 = 3 ft PD2 = 2850 Ibs ! L2 = 7 ft DEFLECTION LIMIT OF LIVE LOAD dL = L / 360 Camber => 0.13 inch DEFLECTION LIMIT OF LONG-TERM d111D+L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, G, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 18 Ibs / ft RLeft = 8.43 kips VMax = 7.80 kips, at 24 inch from left end VIINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.13 in E'min = N/A d = 24.00 in FbE = N/A A = 75.0 in 1 = 3,600 in S,t = 300.0 in RB = N/A 1E= N/A THE BEAM DESIGN IS ADEQUATE. RRigM = 5.35 kips MMax = 24.08 ft-kips, at 4.20 ft from left end E = Ex = 1800 ksi Fb. = NIA Fb = 2,400 psi F = FbE / Fb' = N/A Fv = 265 psi Fe = 2,400 psi E' = 1,800 ksi Fv' = 265 psi CD CM Ct Ci CL CF Cv, Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 963 psi < Fb = 2400 psi [Satisfactory] fv' = 1.5 VMax / A = 156 psi < Fv' [Satisfactory] CHECK DEFLECTIONS d (L. Max) = 0.01 in, at 6.000 ft from left end, < d L = L / 360 [Satisfactory] d(KcrD+L, Mex) = 0.10 in, at 5.700 ft from left end < AKcr D+L = L / 240 [Satisfactory] Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.13 in, at 5.700 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F1, CD CP CF = 2526 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s - Fc* = Fc Co CF = 2560 psi LQ = Ke L = 1.0 L = 144 in d = 24 in SF = slenderness ratio = 6.0 < FEE = 0.822 E'min / SF2 = 21235 psi E'min = 930 ksi F = FCE / Fc* = 8.295 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS f� = F / A = 13 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] 1 1 l F — F f 1 0.987 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1003 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 & / Fc' )2 + fb / [Fe (1 - fc / F.E)] = 0.261 < 1 [Satisfactory] IA PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT! BEAM # 23, FLR BM AT STAIRS DESIGN BY: R.A Structural JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 3 1/8 x 24 L Glulam 24F-1.8E L L= 12 ft Psi PD- wD = 225 Ibs / ft T wL= 100 Ibs/ft PDt = 2850 Ibs Wo Lt = 7 ft PD2 = 0 Ibs L2= 0 ft AL = L / 360 Camber => 0.07 inch d Kcr D+L= L/ 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor_C. Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif vvn = 18 Ibs / ft RLeft = 3.24 kips RRight = 3.72 kips VM. = 3.03 kips, at 24 inch from right end MM. = 14.31 ft-kips, at 7.00 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.13 in E'min = N/A E = Ex = 1800 ksi Fb* = N/A d = 24.00 in FbE = NIA Fb = 2,400 psi F = FbE / Fb* = N/A A = 75.0 in2 1 = 3,600 in4 Fv = 265 psi Fe - - 2,400 psi Sx = 300.0 in' RB = N/A E' = 1,800 ksi F,; = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 573 psi < Fb = 2400 psi [Satisfactory] f,' = 1.5 VMax / A = 61 psi < F,; [Satisfactory] CHECK DEFLECTIONS d (L, Max) = 0.01 in, at 6.000 ft from left end, < AL = L / 360 [Satisfactory] A(Ka D+L, Max) = 0.05 in, at 6.200 ft from left end < AKaD+L = L / 240 [Satisfactory] Where Ku = 1.00 , (NDS 3.5.2) rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.07 in, at 6.200 ft from left end ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fe = F� CD Cp CF = 2526 psi Where F� = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CF = (1+F) / 2c - [(1+F)/ 2c)2 - F / c]o.s = Fc* = Fc CD CF = 2560 psi L. = K. L = 1.0 L = 144 in d = 24 in SF = slenderness ratio = 6.0 < Fc, = 0.822 E'nn / SF2 = 21235 psi E'min = 930 ksi F = FCE / Fc* = 8.295 c = 0.9 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 13 psi < F,' ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = F F y 1 1I 0.987 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 613 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F� )2, fb / [Fe (1 - fc / FcE)] � 0.160 < 1 [Satisfactory] 6 RA STRUCTURAL ENGINEERING Title: Job # 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project Desc.: (760) 771-9993 - Title Block Line 6 - ....., a,-�,c:a- -, DEC tort, 2AWil .a x,.. - BERCALC, INC 19�2012, Buiki.6.12.9. €3, Ver-fi.17:1.31`r II Description : LOT 12B, MADISON BLUB, COLUMBUS WAY, BM # 24 FL BM RIGHT OF FOYER Calculations per NDS 2005, ASCE 7-05 Load Combination Set : ASCE 7-05 Material Properties Analysis Method.- Allowable Stress Design Fb - Tension 2400 psi E:Modulus ofElasficity Load Combination ASCE 7-05 Fb - Compr 1850 psi Ebend- xx 1800 ksi Fc - Prll 1650 psi Eminbend - xx 930 ksi Wood Species DF7DF Fc - Perp 650 psi Ebend- yy 1600 ksi Wood Grade 24F - V4 Fv 265 psi Eminbend - yy 830 ksi Beam Bracing :Completely Unbraced Ft 1100 psi Density 32.21 pcf D(O.15) i ❑(0.175) L(0.28) D(2) L 1.72 0.1) L(0.16 3 L■(2.35) ■ 8.75x24 Span = 36.0 ft AO. )kd Loads.. : T ' Service loads entered. Load Factors will be applied for calculations. Load for Span Number 1 Uniform Load : D = 0.0150 ksf, Extent = 0.0 -->> 4.0 ft, Tributary Width = 6.0 ft Point Load: D = 2-0, L = 1.720 k @ 4.0 ft Uniform Load: D = 0.0250, L = 0.040 ksf, Extent = 4.0 -->> 12.0 ft, Tributary Width = 4.0 ft Point Load: D=3.0. L = 2 350 k @ 12.0 ft Uniform Load: D = 0-0250, L = 0.040 ksf, Extent = 12.0 » 36.0 ft, Tributary Width = 7-0 ft Uniform Load : D = 0.0150 ksf, Tributary Width =10.0 ft DESIGNMMARY Maximum Bending Stress Ratio = 0.941: 1 Maximum Shear Stress Ratio Section used for this span 8.75x24 Section used for this span fb : Actual = 1, 891.93 psi fv : Actual FB : Allowable = 2, 011.26 psi Fv : Allowable Load Combination +D+L+H Load Combination Location of maximum on span = 15-120ft Location of maximum on span Span # where maximum occurs = Span # 1 Span # where maximum occurs Maximum Deflection Max Downward L+Lr+S Deflection 0.756 in Ratio = 571 Max Upward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Downward Total Deflection 1-699 in Ratio = 254 Max Upward Total Deflection 0.000 in Ratio = 0 <240 Marc morn Forces & 5tres es.:fpr:Laad:C.ornbiriatiorrs Load Combination Max Stress Ratios Segment Length Span # M V C d C FN C i Cr Cm Ct CL M Moment Values T )Length = 36.0 ft 1 0.523 0.225 1.00 0-84 1.00 100 100 100 099 73.67 J+L+H 0.84 1-00 1-00 1.00 1.00 099 Length = 36.0 ft 1 0.941 0.396 1.00 084 1.00 1-00 100 1.00 0.99 13244 +D+0.750Lr+0.750L+H 0.84 1.00 1.00 1-00 100 0.99 ]esian C, 0.396 : 1 8.75x24 104.97 psi 265.00 psi +D+L+H = 0,000 ft Span # 1 tb Pb 0-00 1,052-44 2011.26 0.00 1,89193 2011.26 0-00 _ Shear Values V fv F v 0.00 000 000 836 59.75 26500 0.00 000 000 14-70 10497 265.00 000 0.00 000 RA STRUCTURAL ENGINEERING Title 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project Desc.: (760) 771-9993 - Title Block Line 6 J11 od 8eam - - - EI TUN ,S Description: LOT 12B, MADISON BLUB, COLUMBUS WAY, BM # 24 FL BM RIGHT OF FOYER Job # Prinle:16DECffiT2, 2:48AM Load Combination Max Stress Ratios Moment Values Shear Values Segment Length Span # M V C d C FN C i Cr C m C I C L M fb F'b V fv F v Length = 36 0 ft 1 0.836 0.353 1.00 0.84 1.00 1.00 1.00 1.00 0.99 117.74 1,681,98 2011.26 13-11 93.67 265.00 +D+0.750L+0.750S+H 0.84 1.00 1.00 1-00 1.00 0.99 0.00 0.00 0.00 0.00 Length = 36.0 ft 1 0.836 0.353 1.00 0.84 1.00 1.00 1.00 1.00 0.99 117.74 1,681.98 2011.26 13.11 93.67 265.00 +D+0.750Lr+0.750L+0.750W+H 0-84 1.00 1.00 1.00 1.00 0.99 0.00 0.00 0.00 0-00 Length = 36 0 ft 1 0.836 0.353 1.00 0.84 1.00 1.00 1.00 1.00 0.99 117.74 1,681.98 2011.26 13.11 93.67 265.00 +D+0.750L+0.750S+0.750W+H 0.84 1-00 1.00 1.00 1.00 099 0.00 0.00 0.00 0.00 Length = 36.0 ft 1 0.836 0-353 1.00 0.84 1.00 100 1-00 1.00 0-99 117.74 1,681.98 2011.26 1311 93.67 265.00 +D+0.750Lr+0.750L+0.5250E+H 0.84 1.00 1.00 1.00 1.00 0.99 0.00 0.00 0.00 0.00 Length = 36 0 ft 1 0-836 0.353 100 0.84 1.00 1.00 1.00 1.00 0.99 117.74 1,681.98 2011.26 1311 93.67 265.00 +D+0-750L+0.750S+0.5250E+H 0.84 1.00 1.00 1.00 1,00 0.99 0-00 0.00 0.00 0-00 Length = 36 0 ft 1 0.836 0.353 1.00 0.84 1.00 1.00 1.00 1.00 0.99 117.74 1,681.98 2011.26 1311 93.67 265.00 0veral[Maxirnum Defections -:t�.nfa*red. Loads Load Combination Span Max. "-' Defl Location _ in Span Load Combination Max- "+' Defl Location in Span D+L 1 1.6993 17.640 0.0000 0.000 ~.. Vertical Reactions .m Util`actoreU :.:.:.. J Support notation : Far left is #1 Values in KIPS Load Combination Support 1 Support 2 Overall MAXimum 15.171 12.659 D Only 8.840 6 920 L Only 6 331 5.739 D+L 15.171 12.659 RA STRUCTURAL ENGINEERING Title : Job # 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project Desc.: (760) 771-9993 Title Block Line o APO Beam Description : LOT 128, MADISON CLUB, COLUMBUS WAY, BEAM # 24 FL BM RIGHT OF FOYER 0bE=REli7ilifCL�.:;:. Calculations per AISC 360-05, ASCE 7-05 Load Combination Set: ASCE 7-05 ............. aterial Properties ...:. .. . Analysis Method: Allowable Strength Design Beam Bracing: Completely Unbraced Bending Axis: Major Axis Bending Load Combination ASCE 7-05 0(0.15 PtiNed 16DEC i 12, 2:5WA aF_R&A ,INC_ ' Q63-2012- D,, ki G. t2:9;i3 ila 6.32.'i_31 Fy : Steel Yield: 36.0 ksi E: Modulus: 29, 000.0 ksi D(0.175) L(0.28) r D(0.1) L(0.1 fi) D(0. (2) i ■ ■ i 'Y�2.35) ■ r r Span = 36A rt - W 18x76 - Apptied. toads. Service loads entered. Load Factors will be applied for calculations. Beam self weight calculated and added to loads Load for Span Number 1 Uniform Load : D = 0.0150 ksf, Extent = 0.0 » 4.0 ft, Tributary Width = 6.0 ft Point Load: D = 2.0, L =1.720 k @ 4.0 ft Uniform Load : D = 0,0250, L = 0.040 ksf, Extent = 4.0 -->> 12.0 ft, Tributary Width = 4.0 ft Point Load: D = 3.0, L = 2.350 k @ 12.0 ft Uniform Load: D = 0.0250, L = 0-040 ksf, Extent =12.0 -» 36.0 ft, Tributary Width = 7.0 ft Uniform Load: D = 0.0150 ksf, Tributary Width = 10.0 ft DESIGN SUNATARY Maximum Bending Stress Ratio = 0.741 : 1 Maximum Shear Stress Ratio = 0.148 : 1 Section used for this span W18x76 Section used for this span W18x76 Ma: Applied 144.443 k-ft Va : Applied 16.537 k Mn / Omega: Allowable 194,927 k-ft Vn/Omega : Allowable 111.384 k Load Combination +D+L+H Load Combination +D+L+H Location of maximum on span 15.480ft Location of maximum on span 0.000 ft Span # where maximum occurs Span # 1 Span # where maximum occurs Span # 1 Maximum Deflection Max Downward L+Lr+S Deflection 0,355 in Ratio = 1,215 Max Upward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Downward Total Deflection 0.874 In Ratio = 494 Max Upward Total Deflection 0.000 in Ratio = 0 <240 Maxim Forces i Stresses for Load. Combinations Load Combination Max Stress Ratios Summary of Moment Values Summary of Shear Values Segment Length Span # M V Mmax+ Mmax - Ma - Max Mnx Mnx/Omega Cb Rm Va Max Vnx Vnx/Omega +D Dsgn. L = 36.00 it 1 0.440 0.092 85.63 85-63 325.25 194.76 1.15 1-00 10.21 167-08 11138 +D+L+H Dsgn. L = 36 00 ft 1 0.741 0148 144.44 144-44 325.53 194.93 1.15 1.00 16.54 167.08 111.38 +D+0.750Lr+0.750L+H Dsgn. L = 36.00 ft 1 0.666 0.134 129.74 129.74 325.53 194.93 1.15 1.00 14.95 167.08 11138 +D+0.750L+0.750S+H Dsgn. L = 36.00 ft 1 0.666 0.134 129.74 129.74 325.53 194.93 1.15 1-00 14.95 167.08 111.38 +D+0.750Lr+0.750L+0.750W+H Dsgn. L = 36-00 ft 1 0.666 0.134 ,W7501-+0.750S+11750W+H 129.74 129.74 325.53 194.93 1.15 1.00 14.95 167.08 111.38 Dsgn. L = 36.00 ft 1 0 666 0.134 129.74 129.74 325.53 194,93 1-15 1.00 14.95 167.08 111.38 +D+0-750Lr+0.750L+0.5250E+H Dsgn. L = 36.00 ft 1 0.666 0.134 129-74 129.74 325-53 194-93 1.15 1.00 14.95 167.08 111.38 +D+0.750L+0.750S+0.5250E+H Dsgn. L = 36.00 ft 1 0.666 0.134 129.74 129.74 325.53 194.93 1.15 1.00 14.95 167.08 11138 KA 51 KUG I UKAL ENGINEEKING 78080 CALLE CAMINO, SUITE 102 LA QUINTA, CA. 92253 (760) 771-9993 Title : Engineer: Project Desc.: Job # Title Block Line 6 Re0j.-Beam Description : LOT 12B: MADISON CLUB, COLUMBUS WAY, BEAM # 24 FL BM RIGHT OF FOYER Ovef Maxiri u mi iJefieciions`y:bmf z6" Loads P,inl, �012, 2:55AM 777 Load Combination Span Max. "-"Defl Location in Span Load Combination Max. "+" Defl Location in Span D+•L 1 0.8743 17-640 a-0000 0 000 . -..arts'%i" i aad:Comhinatios':.-:11" .airared:taads: Load Combination Span Max. Downward Defl Location in Span Sport Max. Upward Defl Location in Span D Only 1 0,5188 17.640 0-0000 0-000 L Only 1 0.3555 17.640 0-0000 0.000 D+L 1 0.8743 17 640 00000 0.000 VediC&i Reactions Unfactol'ed : Support notation : Far left is #1 Values in KIPS Load Combination Support 1 _ Support 2 Overall MAXimum 16.537 14-025 D Only 10.206 8.286 L Only 6.331 5.739 D+L 16.537 14.025 a Iw PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 25, MDR AT ENTRY DOOR DESIGN BY: R.A Structural JOB NO.: DATE: 12J1112012 REVIEW BY: RA. Woof! Beam Desian Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/2 x 18. Glulam 24F-1.8E L= '17 ft wD= 810 Ibs/ft wL = 580 Ibs / ft PD1 = 0 _ Ibs LY= 0:_ ft PD2 = 0 Ibs L2 = 0 ft dL=L/360 AKuD+L=L1240. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes LF�'D1� L, II PD2 NL 4 Y Y yD I l Camber => 0.49 inch THE BEAM DESIGN IS ADEQUATE. Code Duration Factor, Factor, Cr, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 24 Ibs / ft RLefl = 12.02 kips RR;ym = 12.02 kips VMex = 9.90 kips, at 18 inch from left end MM. = 51.07 ft-kips, at 8.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1800 ksi Fb = N/A d = 18.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 99.0 in 1 = 2,673 in F = 265 psi Fe = 2,337 psi S, = 297.0 in RB = N/A E' = 1,800 ksi F,; = 265 psi 1E = N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.97 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 2063 psi < Fb = 2337 psi [Satisfactory] f,' = 1.5 VMax / A = 150 psi < F,; [Satisfactory] CHECK DEFLECTIONS d (L, Max) = 0.23 in, at 8.500 ft from left end, < d L = L / 360 [Satisfactory] d (KerD I L, Max) = 0.55 in, at 8.500 ft from left end < d Kr D+L = L / 240 [Satisfactory] Where Kr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (15D, Max) = 0.49 in, at 8.500 ft from left end ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fc' = Fo CD CP CF = 2398 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 - Fo` = F� CD CF = 2560 psi LB = Ke L = 1.0 L = 204 in d = 18 in SF = slenderness ratio = 11.3 < Foe = 0.822 E�min / SFZ = 5952 psi E'min = 930 ksi F = FEE / F�* = 2.325 c = 0.9 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 10 psi < Fe' E ALLOWABLE FLEXURAL STRESS IS Fe = 3740 psi, [ for CD = 1.6 ] 1 1 F F �i 1 0.937 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 2094 psi < Fo [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fo / F.' )2 + fb / [Fti 0 - fc / Fce)] = 0.561 < 1 [Satisfactory] e Iw PROJECT : LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 25A, HDR AT ENTRY DOOR DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 1/2 x 15 Glulam 24F-1.8E L L. MEMBER SPAN L = 17 It UNIFORMLY DISTRIBUTED DEAD LOAD wD = 375 Ibs / ft Pp1 1 P° UNIFORMLY DISTRIBUTED LIVE LOAD wL= 300 Ibs /ft CONCENTRATED DEAD LOADS PDf = 0 Ibs "D (0 for no concentrated load) L1 = 0 ft 14 PD2 = 0 Ibs 1 } L2 = 0 ft DEFLECTION LIMIT OF LIVE LOAD AL = L / 360 Camber => 0.40 inch DEFLECTION LIMIT OF LONG-TERM d KcrD+L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cr, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wself wt = 20 Ibs / ft RLeft = 5.90 kips RRIyM = 5.90 kips VMax = 5.04 kips, at 15 inch from left end MMax = 25.09 ft-kips, at 8.50 ft from left end MINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1800 ksi Fb = N/A d = 15.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb,r = N/A A = 82.5 in 1 = 1,547 in Fv = 265 psi Fe = 2,380 psi Sx = 206.3 in RB = N/A E' = 1,800 ksi F = 265 psi l E = N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.99 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1460 psi < Fb = 2380 psi [Satisfactory] fv' = 1.5 VMax / A = 92 psi < F, [Satisfactory] CHECK DEFLECTIONS d tL, Meal = 0.20 in, at 8.500 ft from left end, < d L = L / 360 [Satisfactory] d(KcrD+L, Max) = 0.47 in, at 8.500 ft from left end < AKcrD+L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.40 in, at 8.500 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = t kips 1E ALLOWABLE COMPRESSIVE STRESS IS Fe = Fc CD CP CF = 2280 psi Where Ft = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 - Ft* = Fc CD CF = 2560 psi Le = Ke L = 1.0 L = 204 in d = 15 in SF = slenderness ratio = 13.6 < FCE = 0.822 E'min / SF2 = 4133 psi E'min = 930 ksi F = FtE / Ft* = 1.614 c = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F/A = 12 psi < Fo' IE ALLOWABLE FLEXURAL STRESS IS Ft = 3808 psi, [ for CD = 1.6 ] 0.890 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1496 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (ft / F.' )2 + fb / [Fe (1 - ft / FtE)] = 0.394 < 1 [Satisfactory] 0 Reza PROJECT: BM#26, ENTRY PORCH PAGE: As har OUr CLIENT: LOT 12B, MADISON CLUB DESIGN BY: R.A. 9 P JOB NO.: DATE: 2f7/2013 REVIEW BY: R.A. Wood Beam Desion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 18 Glulam 24F-1 8E L1 L= 21 ft P. = 265 Ibs / ft WL = 420 Ibs / ft WL V y V V T V i t T If T 1 T PD1 = 0 Ibs L, = 0 ft PD2 = 0 Ibs 4 L2 = 0 ft d L = L / 360 Camber => 0.42 inch dK«D+L=L/240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C„ Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1,25 Construction Load 5 1-60 Wind/Earthquake Load 6 2,00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wselfwl = 22 Ibs / ft RLeft = 7.42 kips RRiyht = 7.42 kips VMax = 6.36 kips, at 18 inch from left end MMax = 38.97 ft-kips, at 10.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = Ex = 1800 ksi Fb = N/A d = 18.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb' = N/A A = 92.3 in 1 = 2,491 in° F, = 265 psi Fe = 2,305 psi Sx = 276.8 in RB = N/A E' = 1,800 ksi F, = 265 psi / E = N/A CD CM Cf Ci CL CF Cv C, Cf 1.00 1.00 1.00 1.00 1.00 1.00 0.96 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1690 psi < Fb = 2305 psi f, = 1.5 VMax / A = 103 psi < F, (Satisfactory] CHECK DEFLECTIONS d (L. Max) = 0.41 in, at 10.500 ft from left end, < d(KaD+L, Max) = 0.69 in, at 10.500 ft from left end < Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1 5D, Max) = 0.42 in, at 10.500 ft from left end [Satisfactory] d L = L / 360 [Satisfactory] d Ka D + L = L / 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = FI Co CP CF = 2252 psi Where FI = 1600 psi Cp = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o 5 = FI` = FI Cp CF = 2660 psi L, = K, L = 1.0 L = 252 in d = 18 in SF = slenderness ratio = 14.0 < FIE = 0.822 E'mi, / SF = 3900 psi E'min = 930 ksi F = FIE / Fc" = 1.524 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fI = F / A = 11 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3687 psi, [ for Co = 1.6 ] +I t y+4.b.y,Vv6y.�vir�v F Tn^ I F 0.880 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1722 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fI / F.' )Z + fb / [Fe 0 - fc / FIE)] = 0.468 < 1 [Satisfactory] (SD RA PROJECT: LOT 1213, MADISON CLUB, COLUMBUS WAY' PAGE Structural CLIENT: BEAM # 27. BEAM AT RIGHT OF PATIO COVE DESIGN BY : R.A JOB NO.: DATE: 12/1112012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 18 Glulam 24F-1.8E L= 18 ft wD = 260 Ibs / ft wL = 480:; Ibs / ft PD1 = .. 0 Ibs L1= 0:-_ft PD2 = 0 . Ibs L2 = , ..0, _ ft AL=L/360 dKcrD+L= L/ 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor. Co Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load kLYSIS FERMINE REACTIONS, MOMENT, SHEAR wseir wt = 22 Ibs / ft RLeft = 6.86 kips VMax = 5.71 kips, at 18 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES FL I L., PD, T PD, WD �. I Camber => 0.22 inch THE BEAM DESIGN IS ADEQUATE. RRt9nt = 6.86 kips MM. = 30.86 ft-kips, at 9.00 ft from left end b = 5.13 in Emir, = N/A E = Ex = 1800 ksi Fb = N/A d = 18.00 in FbE = NIA Fb = 2,400 psi F = FbE / Fb. = N/A A = 92.3 in2 I = 2,491 in4 Fv = 265 psi Fe = 2,340 psi Sx = 276.8 in3 RE; = NIA E' = 1,800 ksi Fv- = 265 psi CE= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1338 psi < Fb f, = 1.5 VM,, / A = 93 psi < CHECK DEFLECTIONS A (L. Max) = 0.25 in, at 9.000 ft from left end, d (KcrD+L, Max) _ 0.40 in, at 9.000 ft from left end Where Ka = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.22 in, at 9.000 ft from left end Cv C, Cr 0.98 1.00 1.00 2340 psi Fv' [Satisfactory] [Satisfactory] A L = L ! 360 [Satisfactory] dKcrD + L = L / 240 [Satisfactory] f CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = F� CD CP CF = 2369 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 F�* = Fc CD CF = 2560 psi L, = K,L = 1.0L = 216 in d = 18 in SF ='slenderness ratio = 12.0 < F,E = 0.822 E'min / SF' = 5309 psi E'min = 930 ksi F = FEE / Fc` = 2.074 c = 0.9 T HE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 11 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fti = 3745 psi, [ for CD = 1.6 ] 1 / FTTT f F 0.925 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1371 psi < Fti [Satisfactory] ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F.' )Z + fb / [Fe (1 - f, / FcF)] = 0.367 < 1 [Satisfactory] RA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 28, HDR AT RIGHT OF KITCHEN DESIGN BY R.A Structural JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. !_Beam Design Base an NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 51/8 x 13 1/2 Glulam 24F-1.8E L = 12.25 ft wD = 385 Ibs / ft wL = -780 Ibs / ft P01 = 0 Ibs L, = 0 ft PD2 = 0 Ibs L2 = ' 0 ft AL=L/360 A Kcr D+ L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 WindlEarthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load L, 11 PD p t PD, 1 N` �D �� H Camber=> 0.16 inch THE BEAM DESIGN IS ADEQUATE. LYSIS -RMINE REACTIONS, MOMENT, SHEAR wseirwi = 16 Ibs / ft RLea = 7.24 kips RRlght = 7.24 kips VMax = 5.91 kips, at 13.5 inch from left end MMax = 22.16 ft-kips, at 6.13 ft from left end AINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 13.50 in FbE = N/A A = 69.2 in 1 = 1,051 in Sx = 155.7 in RB = N/A [E= N/A CD CM C( Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1708 psi < Fb = f, = 1.5 VMax / A = 128 psi < ECK DEFLECTIONS A (L. Max) = 0.21 in, at 6.125 ft from left end, A (KcrD+ L, Max) = 0.32 in, at 6.125 ft from left end Where Kr = 1.00 , (NDS 3.5.2) rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D. Max) = 0.16 in, at 6.125 ft from left end E = Ex = 1800 ksi Fb = N/A Fb = 2,400 psi F = FbE / Fb. = N/A Fv = 265 psi Fe = 2,400 psi E' = 1,800 ksi F = 265 psi Cv C, Cr 1.00 1.00 1.00 2400 psi [Satisfactory] F,; [Satisfactory] < AL=L/360 < AKul) -L=L/240 [Satisfactory] [Satisfactory] HECK THE BEAM CAPACITY WITH AXIAL LOAD UAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS F� = F, CD CP CF = 2415 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o 5 = Fc' = Fc CD CF = 2560 psi L. = KB L = 1.0 L = 147 in d = 13.5 in SF = slenderness ratio = 10.9 < F, = 0.822 E'min / SF2 = 6447 psi E'min = 930 ksi F = FcE / Fc' = 2.519 c = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 14 psi < Fc- THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] 0.943 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1752 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + % / [Fti (1 - f. / F.E)] = 0.457 < 1 [Satisfactory] 62h 1U PROJECT : LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 29, FL FLR BM AT KITCHEN DESIGN BY: R.A Structural JOB NO.: DATE; 12/1112012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 1/2 x 19 112 Glulam 24F-1.8E MEMBER SPAN L = 21 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 463 Ibs / ft UNIFORMLY DISTRIBUTED LIVE LOAD wL = 390 Ibs / ft CONCENTRATED DEAD LOADS PD1 = 0 Ibs (0 for no concentrated load) Li = 0 ft PD2 = 0 Ibs L2= 0 ft DEFLECTION LIMIT OF LIVE LOAD d L = L / 360 DEFLECTION LIMIT OF LONG-TERM dKcrD+L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factoration Factor. Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load L PD I 1 1 PD+ 1 V/L `"D Camber=> 0.52 inch THE BEAM DESIGN IS ADEQUATE. ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseif wt = 26 Ibs / ft RLeft = 9.22 kips RRight = 9.22 kips VMax = 7.80 kips, at 19.5 inch from left end MMax = 48.43 ft-kips, at 10.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1800 ksi Fb = N/A d = 19.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 107.3 in2 1 = 3,398 in F = 265 psi Fe = 2,270 psi Sx = 348.6 in3 RB= N/A E' = 1,800 ksi F„ = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.95 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1667 psi < Fb = 2270 psi [Satisfactory] f = 1.5 VMax / A = 109 psi < F [Satisfactory] CHECK DEFLECTIONS d & Max) = 0.28 in, at 10.500 It from left end, < d L / 360 L = [Satisfactory] d(Kcr D+L, Max) = 0.63 in, at 10.500 ft from left end < AKI D+L = L / 240 [Satisfactory] Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (taD, Max) = 0.52 in, at 10.500 It from left end 0 ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = F. Co CP CF = 2321 psi Where F. = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cf -5 Fc' = F, Co CF = 2560 psi Le = I, L = 1.0 L = 252 in d = 19.5 in SF = slenderness ratio = 12.9 < F,, = 0.822 E'min / SF2 = 4577 psi E'min = 930 ksi F = FEE / Fe,' = 1.788 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 9 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fti = 3632 psi, [ for Co = 1.6 ] 0.907 50 {! II1 -. ITTI F F [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1695 psi < Fti [Satisfactory] ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F� )2 + fb / [Fe (1 - f. / F.E)] = 0.468 < 1 [Satisfactory] RA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 30, HDR AT REAR OF KITCHEN DESIGN BY: RA Structural JOB NO.: DATE: 1211112012 REVIEW BY: R.A. I Beam Desigrt Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 12 ' Glulam 24F-1.8E L = 6.25 ft WD = ..,150. Ibs / ft wL = 60 Ibs / ft PD1 = 9222 Ibs L1 = 075 :ft PDz = " 0 _:: Ibs Lz = p..,- ft AL=L/366 AKerD+L=L/240. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration FactorDuration Factor. Cu Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ILYSIS "ERMINE REACTIONS, MOMENT, SHEAR wself m = 15 Ibs / ft RLeft = 8.82 kips VMax = 8.59 kips, at 12 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 12.00 in FbE = N/A A = 61.5 in 1 = 738 in SX = 123.0 in RB = N/A 1E= N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 639 psi < Fb f,; = 1.5 VMax / A = 210 psi < CHECK DEFLECTIONS A([, Max) = 0.00 in, at 3.125 ft from left end, d (Kcr D+L, Max) = 0.03 in, at 2.650 ft from left end Where Ku = 1.00 , (NDS 3.5.2) 7ETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (15D, Max) = 0.04 in, at 2.650 ft from left end L, Poi L-, 1 PD2 1 1NL '"D � Camber=> 0.04 inch THE BEAM DESIGN IS ADEQUATE. RRIght = 1.81 kips MM.. = 6.55 ft-kips, at 0.75 ft from left end E = EX = 1800 ksi Fb. = N/A Fb = 2,400 psi F = FbE / Fb. = N/A Fv = 265 psi Fe = 2,400 psi E' = 1,800 ksi Fv' = 265 psi Cv C, Cr 1.00 1.00 1.00 2400 psi [Satisfactory] Fv' [Satisfactory] < AL=L/ 360 < 14KcrD+L=L1240 [Satisfactory] [Satisfactory] ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips I ALLOWABLE COMPRESSIVE STRESS IS F� = F� CD C1 CF = 2523 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s = F,` = F� CD CF = 2560 psi Le = K. L = 1.0 L = 75 in d = 12 in SF = slenderness ratio = 6.3 < F.E = 0.822 E'min / SF2 = 19570 psi E'min = 930 ksi F = F�E / F,* = 7.645 c = 0.9 E ACTUAL COMPRESSIVE STRESS IS f, = F / A = 16 psi < F,' THE ALLOWABLE FLEXURAL STRESS IS Fn' = 3840 psi, [ for CD = 1.6 ] 1 1 F F 0.985 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 688 psi < Fe [Satisfactory] -CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F., )Z + fb / [Fe (1 - f. / F.E)] = 0.179 < 1 [Satisfactory] 0 RA PROJECT LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 31, FL FLR BM AT KITCHEN DESIGN BY: R.A Structural JOB NO.: DATE: 1211112012 REVIEW BY : P.A. I Beam Design Base on NDS 2045 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 3 1/2 x 24 Glulam 24F-1.8E L= 21 ft wD = 360 Ibs / ft WL = 140 Ibs / ft PD1 = 0 Ibs Ls = 0 ft PD2 = 0 Ibs L2 = 0 ft 14L=L1360 -4uD+L=L1240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor - CD Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wself wt = 20 Ibs / ft RLefl = 5.46 kips VMax = 4.42 kips, at 24 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES , PD, PD2 wD l Camber=> 0.34 inch THE BEAM DESIGN IS ADEQUATE. RRtght = 5.46 kips MMax = 28.67 ft-kips, at 10.50 ft from left end b = 3.50 in E'min = N/A E = Ex = 1800 ksi Fb = N/A d = 24.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 84.0 in2 I = 4,032 in a Fv = 265 psi Fe = 2,326 psi Sx = 336.0 in a RB = N/A E' = 1,800 ksi F = 265 psi lE = N/A CD CM Ct Ci CL CF CV C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.97 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1024 psi < Fb = 2326 psi [Satisfactory] f, = 1.5 VMx / A = 79 psi < Fv' [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.08 in, at 10.500 ft from left end, < A L = L / 360 [Satisfactory] 4 (Ka D + L, M.) = 0.31 in, at 10.500 ft from left end < A Kcr D I L = L / 240 [Satisfactory] Where K,r = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.34 in, at 10.500 ft from left end HECK THE BEAM CAPACITY WITH AXIAL LOAD (fAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = Fc CD CP CF = 2429 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]os - Fc` = Fc CD CF = 2560 psi LB = Ke L = 1.0 L = 252 in d = 24 in SF = slenderness ratio = 10.5 < FEE = 0.822 E'min / SF = 6934 psi E'min = 930 ksi F = FCE / Fc` = 2.709 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 12 psi < Fc' ITHE ALLOWABLE FLEXURAL STRESS IS Fe = 3722 psi, [ for CD = 1.6 ] 0.949 50 1 1 FI FT F _ l T [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1060 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )Z + % / [Fe (1 - f. / F.E)] = 0.285 < 1 [Satisfactory] 0 fV1 O 1 RVV I Vr%P%L CIVVIIYCERINU i rue: 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project Desc.: (760) 771-9993 Title Block Line 6 - - IVood Beam - Description: LOT 12B. MADISON CLUB, COLUMBUS WAY, BEAM # 32, HDR FRONT OF KITCHEN CODE. REFERENCES`": Calculations per NDS 2005, ASCE 7-05 Load Combination Set: ASCE 7-05 Material Properties Job # -.:n;cd 16 OF0 2012. 31�tP £'NERGALC, ANC. 1983-201Z 8uild:6.12.9,13, Ver6-izi,34 Analysis Method: Allowame Stress uesign Fb - Tension 2400 psi E: Modulus ofElasticify Load Combination ASCE 7-05 Fb - Compr 1850 psi Ebend- xx 1800 ksi Fc - Prll 1650 psi Eminbend - xx 930 ksi Wood Species : DF/DF Fc - Perp 650 psi Ebend-yy 1600ksi Wood Grade : 24F - V4 Fv 265 psi Eminbend - yy 830 ksi Beam Bracing :Completely Unbraced Ft 1100 psi Density 32.21 pcf D(0.275) L'0-4,4(3) L(2.5) D(4) L(5.25) D(0.15 D(1.2 ) L(0.13 i I • ■ - I i I L 5.5x21 - Span = 13.50 ft Service loads entered. Load Factors will be applied for calculations. Beam self weight calculated and added to loads Load for Span Number 1 Uniform Load : D = 0.250, L = 0.020 ksf, Extent = 0.0 -->> 7.0 ft, Tributary Width = 5.0 ft Uniform Load : D = 0.0150 ksf, Tributary Width =10.0 ft Uniform Load : D = 0.0250, L = 0.040 ksf, Extent = 0.0 ->> 8.0 ft, Tributary Width =11,0 ft Point Load: D=3.0, L=2.50k@60ft Point Load : D = 4.0, L = 5.250 k @ 8.0 ft DESIGN S[1AMOY Maximum Bending Stress Ratio = 0.931: 1 Maximum Shear Stress Ratio Section used for this span 5.5x21 Section used for this span fb : Actual = 2,193.17psi fv : Actual FB : Allowable = 2,355.21 psi Fv : Allowable Load Combination +D+L+H Load Combination Location of maximum on span = 6.008ft Location of maximum on span Span # where maximum occurs = Span # 1 Span # where maximum occurs Maximum Deflection Max Downward L+Lr+S Deflection 0.120 in Ratio = 1346 Max Upward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Downward Total Deflection 0.299 in Ratio = 541 Max Upward Total Deflection 0.000 in Ratio = 0 <240 Ailaximuhl Forces:.& Stresses for Load Cornb[nations Load Combination Max Stress Ratios Segment Length Span # M V C d C FN C i Cr Cm C t C L ✓ Length = 13.50 ft 1 IL+H Length = 13-50 ft 1 +D+0.7501-r+07501-+H 0.560 0.931 0.473 1.00 0-98 1-00 1.00 1.00 1.00 0.99 0.98 1-00 1.00 1.00 100 099 0 748 1.00 0.98 1.00 1.00 1.00 1.00 0.99 0-98 100 1.00 100 100 0.99 1 v ■ 0.748 : 1 5.5x21 198.11 psi 265.00 psi +D+L+H 0.000 ft = Span # 1 Moment Values M tb Fb 0.00 44-46 1,319-68 2355.21 000 73-88 2,193.17 235521 000 V Shear Values Iv Fb 0-00 0-00 0.00 9.64 125.23 265.00 0.00 0,00 000 15.25 198-11 265.00 0-00 0.00 0.00 Ed) RA STRUCTURAL ENGINEERING Title: Job # 78080 CALLE CAMINO, SUITE 102 Engineer: LA QUINTA, CA. 92253 Project ❑esc.: (760) 771-9993 Title Block Line 6 Nnred: 16 DEC 2012, 3:55AFA Mondam �[F=RC,41C, INC.1902(112, k1:8.12.9.13Vet-fi.i2-1,31 Description : LOT 120. MADISON CLUB, COLUMBUS WAY, BEAM # 32, HDR FRONT OF KITCHEN Load Combination Max Stress Ratios Moment Values Shear Values Segment Length Span # M V C d C FN C i Cr C m C t C L M tb F'b V fv F'v Length =13.50 ft 1 0.838 0.679 1.00 0.98 1.00 1.00 1.00 1.00 0.99 66.53 1,974.80 2355.21 13.85 179.89 265.00 +D+0.750L+0.750S+H 0.98 1.00 1.00 1.00 1.00 0.99 000 0.00 0.00 000 Length =13.50 ft 1 0.838 0.679 1.00 0.98 100 1.00 1.00 1.00 0-99 66.53 1,974.80 2355.21 13.85 179.89 265-00 +D+0.750Lr+0 750L+0.750W+H 0.98 1.00 1.00 1.00 1.00 0.99 000 0.00 0.00 000 Length =13.50 ft 1 0.838 0.679 1.00 0.98 1.00 1.00 1.00 1.00 0.99 66.53 1.974.80 2355.21 13.85 179.89 265.00 +D+0.750L+0.750S+0.750W+H 0.98 1.00 1.00 100 1.00 0.99 0-00 0.00 0.00 0-00 Length =13.50 ft 1 0.838 0.679 1.00 0.98 1.00 1.00 1.00 1.00 0.99 66.53 1,974.80 2355.21 13.85 179.89 265.00 +D+0.750Lr+0.750L+0.5250E+H 0.98 1.00 1.00 1.00 1.00 0.99 0-00 0.00 0.00 000 Length =13.50 ft 1 0.838 0.679 1.00 0.98 1.00 1.00 1.00 1.00 0.99 66.53 1,974.80 2355.21 13.85 179.89 265.00 +D+O 750L+0.750S+0.5250E+H 0.98 1.00 1.00 1.00 100 0.99 000 0.00 0.00 000 Length =13.50 ft 1 0.838 0.679 1.00 0.98 1.00 1.00 1.00 1.00 0.99 66.53 1.974.80 2355.21 13.85 179.89 265.00 Overall maximum Deflecfion_s_-_Llnfactored Loads:` Load Combination Span Max. ° ° Dell Location in Span Load Combination Max. "+" Deft Location in Span D+L t yLxP .1:Reactiar+s'. Elnfacta're :; 0-2994 6-683 0.0000 0 000 Support notation : Far left is #1 Values in KIPS Load Combination Support 1 Support 2 Overall MAXm-um 19 036 13 258 D Only 12.513 7 811 L Only 6.523 5.447 D+L 19.036 13.258 f Iw PROJECT - LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT BEAM # 33, HDR AT wine roorn DESIGN BY: R.A JOB NO.: DATE: 1211112012 REVIEW BY: R.A. Woad Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD +DEFLECTION LIMIT OF LONG-TERM 6 x 8 No. 1, Douglas Fir -Larch L = 3 ft wD = 410 Ibs / It wL = 560 Ibs / ft PD1 = 2560 Ibs L1 = 0.5 It P" = 0 Ibs L2 = 0 ft L yII PD 1 L PDz WD I r a1 AL = L 1 360 Camber => 0.01 inch Ala D+L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C„ Condition Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wself wt = 9 Ibs / ft RLeft = 3.60 kips RRight = 1.90 kips VMax = 2.99 kips, at 7.5 inch from left end MM. = 1.83 ft-kips, at 1.10 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 41.3 in2 I = 193 in° F = 170 psi Fe = 1,350 psi Sx = 51.6 in3 RB = N/A E' = 1,600 ksi Fv' = 170 psi lE = N/A CD CM Ct Ci CL CF CV C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 427 psi < Fb = 1350 psi [Satisfactory] f,' = 1.5 VMax / A = 109 psi < F, [Satisfactory] CHECK DEFLECTIONS d (L. Max) = 0.00 in, at 1.500 ft from left end, < A L = L / 360 [Satisfactory] d(Kr D+L,Max)= 0.01 in, at 1.400Itfrom left end < dKcrD+L= L/ 240 [Satisfactory] Where Kr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.01 in, at 1.400 ft from left end 0 (CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1458 psi Where F� = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.e = F�* = F� CD CF = 1480 psi LB = KB L = 1.0 L = 36 in d = 7.5 in SF = slenderness ratio = 4.8 < FEE = 0.822 E'min / SF = 20693 psi E'min = 580 ksi F = FCE / F j = 13.982 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 psi < Fc- ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 0.985 50 1 1 4 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 499 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / Fc' )2 + fb / [Fti (1 - fc / F.E)1 = 0.232 < 1 [Satisfactory] 0 Iw PROJECT : LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: CLIENT: BEAM # 34, HDR AT MECHANICAL ROOM DESIGN BY: RAStructural JOB NO- : DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 10 No. 1, Douglas Fir -Larch L = 6.25 ft wD = 935 Ibs / ft wL = 440 Ibs / ft PD7 = 2560 Ibs Lt = 0.5 ft PD2 = 0 Ibs L2 = 0 ft P" 1 PD2 r Y r r r'D 4 A L = L / 360 Camber => 0.09 inch A Kcr D + L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirva = 11 Ibs / ft RLee = 6.69 kips VMax = 5.59 kips, at 9.5 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 9.50 in FbE = N/A A = 52.3 in2 I = 393 in4 Sx = 82.7 in3 RB = N/A 1E = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1076 psi < Fb f,; = 1.5 VMax / A = 160 psi < CHECK DEFLECTIONS A (L. Max) = 0.02 in, at 3.125 It from left end, d (Kcr D + L, Max) = 0.08 in, at 3.125 ft from left end Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.09 in, at 3.125 ft from left end THE BEAM DESIGN IS ADEQUATE. Code Designation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRight = 4.54 kips MMex = 7.42 ft-kips, at 2.86 ft from left end E = EX = 1600 ksi Fb = N/A Fb = 1,350 psi F = FbE / Fb* = N/A F = 170 psi Fe = 1,350 psi E' = 1,600 ksi F, = 170 psi Cv Cc Cr 1.00 1.00 1.00 1350 psi F,; [Satisfactory] [Satisfactory] A L = L / 360 [Satisfactory] 4 Kcr D + L = L / 240 [Satisfactory] 0 HECK THE BEAM CAPACITY WITH AXIAL LOAD <JAL. LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1416 psi Where F� = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]os = Fc' = F� CD CF = 1480 psi I, = Ke L = 1.0 L = 75 in d = 9.5 in SF = slenderness ratio = 7.9 < FEE = 0.822 E'mjn / SF2 = 7649 psi E'min = 580 ksi F = FEE / F�* = 5.168 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 19 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] l F 4 � 0.957 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1133 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )Z + fb / [Fe 0 - f. / FCE)] = 0.526 < 1 [Satisfactory] 0 Iw PROJECT : LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: SEAM # 35,BM AT REAR OF sDRM 2 DESIGN BY: R.A JOB NO.: DATE: 12hV2012 REVIEW BY: RA Wood Beam Desian Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) (DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 51/8 x 12 Glulam 24F-1.8E L = 9.5 ft wD = ._ 1075__ Ibs / ft wL = 760 Ibs / ft PD1 _ " 0 _ Ibs L,= 0 =ft PD2 = . d. Ibs L2 = 0 ft dL=L/360 d KcrD+L= I-/ 240 Does member have continuous lateral support by top diaphragm ? Li 9D1 1 1 PD2 WL WD Camber => 0.23 inch THE BEAM DESIGN IS ADEQUATE. (1= yes, 0= no) 1 Yes Code Duration FactorDuration C� Condition 1 0,90 Dead Load 2 1.00 Occupancy Live Load 3 1,15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 zoo Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirwt = 15 Ibs / ft RLeR = 8.79 kips RRight = 8.79 kips Vm. = 6.94 kips, at 12 inch from left end MMex = 20.87 ft-kips, at 4.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = EX = 1800 ksi Fb = N/A d = 12.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 61.5 in2 I = 738 in4 F„ = 265 psi Fe = 2,400 psi S, = 123.0 in3 RB = N/A E' = 1,800 ksi Fv- = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 2036 psi < Fb = 2400 psi [Satisfactory] f = 1.5 VMax / A = 169 psi < F,,' [Satisfactory] CHECK DEFLECTIONS d (L, Max) = 0.10 in, at 4.750 ft from left end, < d L = L / 360 [Satisfactory] d (Ku D + L, Max) = 0.26 in, at 4.750 ft from left end < d Ker D + L = L / 240 [Satisfactory] Where K. = 1.00 , (NDS 3.52) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.23 in, at 4.750 ft from left end 1 161 jCHECK THE BEAM CAPACITY WITH AXIAL LOAD (AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = Fc CD CP CF = 2459 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c10.5 = Fc* = Fc CD CF = 2560 psi Le = KBL = 1.0L = 114 in d = 12 in SF = slenderness ratio = 9.5 < FcE = 0.822 E'min / SFz = 8470 psi E'min = 930 ksi F = FcE / Fc* = 3.309 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 16 psi < Fe' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] 0.961 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 2085 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F� )Z + fb / [Fe (1 - fc / FcE] = 0.544 < 1 [Satisfactory] R A PROJECT: LOT 12B, MADISON CLUB, COLUMBOS WAY PAGE: Structural CLIENT: BM. # 36, FL CANT BM AT BATHROOM 2 DESIGN BY: R.A. JOB NO.: DATE: 12/1/2012 REVIEW BY: R.A. Wood Beam Design Base. on NDS 2005.' INPUT DATA & DESIGN SUMMARY BEAM SECTION GLB 31/8 x 24 Glulam 24F-1.8E BEAM SPAN L t = 24 ft CANTILEVER L2 = 2 ft, (0 for no cantilever) SLOPED DEAD LOADS WDL,1 = 0.34 kips / ft `NDL,2 = 0.34 kips / ft PROJECTED LIVE LOADS WLL l = 0.16 kips / ft WLL,2 = 0.16 kips / ft CONCENTRATED LOADS PDL = 0.75 kips PLL = 0.5 kips SLOPE 0 : 12 (0 = 0.00 ° ) DEFLECTION LIMIT OF LIVE LOAD d ILL = L / 360 LONG-TERM DEFLECTION (NDS 3.5.2) AKID+L= L/ 240 Code Duration Factor, CR Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load 2 ANALYSIS DETERMINE REACTIONS, MOMENTS & SHEARS R2=0.5 "'+wu.., Li+ 1v..,+was (L.+0.5L2)Lz+PLi+L2= cosh cos9 L, L, R'= ca DD+wa, Lt+ cW- +Waa Lz+P-Rz= 5.85 kips P W«. wiz Wog., R. `JSlope 12 R,r L, X, I X, I X, I L: 0- A- THE BEAM DESIGN IS ADEQUATE. 8.40 kips Mmr, =0.5 (Zoz-o z+I'Lz3.5 ft-kips X, = 11.71 ft +w�,L X2= 11.71 ft r 1 l X3 = 0.58 ft M,u.=1 �Se+wU,Jg 2�T = 34.3 ft-kips Vmax = 7.40 kips, at R2 left. DETERMINE SECTION PROPERTIES AND DESIGN FACTORS L u = M-(X3 , L2) = 2.0 ft, (NDS 2005 Table 3.3.3) GLB 3 1/8 x 24 Properties b = 3.13 in Fb = 2,400 psi lE = 4.1 ft, (Tab 3.3.3 footnote 1) d = 24.00 In F„ = 265 psi, (NDS 97 CH included) RB = 11.0 < 50 A = 75.0 in E' = 1,800 ksi E'y= 1,600 ksi D (cQnt'd) S, = 300.0 in Fe = 2,303 psi FbE = 9185 psi I = 3,600 in F, = 265 psi Fb = 2400 psi E = E, = 1800 ksi E'min = 930 ksi F = FbE / Fb* = 3.83 CD CM Ct Ci CL Cp Cv Cc Cr 1.00 1.00 1.00 1.00 0.98 1.00 0.96 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES Cantilever: fb' = MM„ /SX = 140 psi < Fb = 2303 psi [Satisfactory] Middle Span: fb = MMax /SX = 1371 psi < Fb = 2343 psi [Satisfactory] Shear: fv' = 1.5 V Max /A = 148 psi < Fv' [Satisfactory] (neglected d offset conservatively) CHECK DEFLECTION AT LIVE LOAD CONDITION L = L 1/cos 0 = 24.00 ft, beam sloped span a = L 2/COS 0 = 2.00 ft, beam sloped cantilever length P = Pu COS 0 = 0.50 kips, perpendicular to beam W1 =WLl COS2 0 = 0.16 klf, perpendicular to beam W 2 = W LL,2 COS2 B = 0.16 klf, perpendicular to beam w,L3a w,a3{4L+3a}1 DEnd = CPa2(L+a) 3EI 24EI + 24EI J cos B = -0.04 in, uplift to vertical direction. < 2 1-2 / 360 = 0.13 in [Satisfactory] PaI,2 Sw,L4 wLxa2 L' Amid = — I fiEI + 38 — 32EI COS B = 0.18 in, downward to vertical direction. < L, / 360 = 0.80 in [Satisfactory] CHECK DEFLECTION AT LONG-TERM LOAD, Kr DL + LL, CONDITION P = PKcrDL+LL COS B = 1.25 kips, perpendicular to beam Ku = 1.00 , (NDS 3.5.2) W+ = Kcr WDL,1 COS 0 +WLl COS2 B = 0.50 klf, perpendicular to beam W2 = Kcr WDL,2 COS 0 +Wu,2 COS2 B = 0.50 klf, perpendicular to beam Pa2(L+a) w�L3a w,a3(4L+3a)1 End = 3EI — 24EI + 24EI J cos 6 = -0.14 in, uplift to vertical direction. < 2 L2 / 240 = 0.20 in [Satisfactory] PaL`2 5wrL4 1�Lyr72 OMid = + — 1 fiEl 384EI 32EI ]COS B = 0.56 in, downward to vertical direction. < L, / 240 = 1.20 in [Satisfactory] 0 J RA PROJECT - LOT 12B, MADISON CLUB, COLUMBOS WAY PAGE: BM. # 37, FLAT BEDROOM 2 DESIGN BY: R.A. Structural JOB NO.: CANT BM DATE: 12/11 0 2 REVIEW BY . R.A. Wood`Beam Design Base`on NDS 2005> INPUT DATA & DESIGN SUMMARY BEAM SECTION GLB 31/8 x 24 Glulam 24F-1.8E BEAM SPAN L j = 24 ft CANTILEVER L2 = 2 ft, (0 for no cantilever) SLOPED DEAD LOADS WDL,1 = 0.06 kips / ft wDL,2 = 0.06 kips / ft PROJECTED LIVE LOADS wLL,l = 0.03 kips / ft wLL,Z = 0.03 kips / ft CONCENTRATED LOADS PDT = 3 kips PLL = 3 kips SLOPE 0:12(0= 0.00 °) R, DEFLECTION LIMIT OF LIVE LOAD d LL = L / 360 LONG-TERM DEFLECTION (NDS 3.5.2) dKcrD+L= L/ 240 Code Duration Factor. C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ra 1P WLLI WLL2 WO2 Wo. R. I Slope 12 L, A- THE BEAM DESIGN IS ADEQUATE. ANALYSIS DETERMINE REACTIONS, MOMENTS & SHEARS ll ll Lz L]+Lz = R2=0.5( _1s4+wu,JLi+ cosB+wa,J(Li+O.SLz)L' +P Li 7.77 kips Ri-(c I +W." ILi+I � B+w..=JLz+P-Rz= 0.57 kips 11 MM,� = 0.5 -W+w�,)Lz+PL2 (c_o_si = 12.2 ft-kips X, = 6.36 ft X2 = 6.36 ft (Xz+Xs�� X3= 11.28 ft -I MAf.- a +wu.,) - - 1.8 ft-kips $ Vmax = 7.59 kips, at R2 left. DETERMINE SECTION PROPERTIES AND DESIGN FACTORS L u = M-(X3 , LZ ) = 11.3 ft, (NDS 2005 Table 3.3.3) GLB 3 1/8 x 24 Properties b = 3.13 in Fb = 2,400 psi 1E = 23.2 ft, (Tab 3.3.3 footnote 1) d = 24.00 in Fv = 265 psi, (NDS 97 CH included) RB = 26.2 < 50 A = 75.0 inZ E' = 1,800 ksi E'yz 1,600 ksi 13, S. = 300.0 in Fb' = 1,503 psi I = 3,600 in F, = 265 psi E = EX = 1800 ksi E'min = 930 ksi CD CM Ct C; CL CF 1.00 1.00 1.00 1.00 0.63 1.00 CHECK BENDING AND SHEAR CAPACITIES Cantilever: fb' = MMin /SX = 487 psi < Middle Span: fb = MMax /SX = 73 psi < FbE = 1629 psi Fb = 2400 psi F = FbE / Fb* = 0.68 Cv Cc Cr 0.96 1.00 1.00 Fb = 1503 psi [Satisfactory] Fb = 2400 psi [Satisfactory] Shear: f„' = 1.5 V Max /A = 152 psi < F„' [Satisfactory] (neglected d offset conservatively) ECK DEFLECTION AT LIVE LOAD CONDITION L = L 1/cos 0 = 24.00 ft, beam sloped span a = L2/COS 0 = 2.00 ft, beam sloped cantilever length P = Pu cos 0 = 3.00 kips, perpendicular to beam W1 = WL,I cos2 0 = 0.03 klf, perpendicular to beam W2 =WLL,2 COS 0 = 0.03 klf, perpendicular to beam CPaz(L+a) _tiy,L3a N1za3(4L +3a) E'I 24 DEnd 3.EI + 24EI J cos B = 0.02 in, downward to vertical direction. < 2 L2 / 360 = 0.13 in [Satisfactory] PaL2 5w,L4 w�L2aZ A,t�d — [ 16EI 384E1 ' 32EI I cos B = -0.02 in, uplift to vertical direction. < Li / 360 = 0.80 in rSatisfactorvl ;HECK DEFLECTION AT LONG-TERM LOAD, Kcr DL + LL, CONDITION P = PKcr DL+LL. COS 0 = 6.00 kips, perpendicular to beam Ku = 1.00 , (NDS 3.5.2) Wt = Kcr Wog t cos 0 +WLc,l cos2 0 = 0.09 klf, perpendicular to beam W2 =Kcr WoL,2 COS 0 +Wtt,2 cos2 0 = 0.09 klf, perpendicular to beam Pr2a(L+a� _ yy,L�a wza3(4L+3a�l End = 3FI 24EI + 24EI J cos B = 0.03 in, downward to vertical direction. < 2 L2 / 240 = 0.20 in [Satisfactory] _ PaLZ 5w,L4 _ wzLZazl A,ya 16E1 + 384E1 32EI J cos B = -0.01 in, uplift to vertical direction. 240 = 1.20 in [Satisfactory] (cont'd) go RA PROJECT LOT 12B, MADISON CLUB, COLUMBOS WAY PAGE: CLIENT : BM. # 38, FL CANT BM AT BEDROOM 3 DESIGN BY: R.A. Structural JOB NO.: DATE: 12/1/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY BEAM SECTION GLB 3 1/8 x 24 Glulam 24F-1.8E BEAM SPAN L = 21.5 ft CANTILEVER = 2 ft, (0 for no canuiever) SLOPED DEAD LOADS wDL,1 = 0.505 kips / ft wDL,2 = 0.505 kips / ft PROJECTED LIVE LOADS wLL,i = 0.16 kips / ft wLL,2 = 0.16 kips / ft CONCENTRATED LOADS PDL = 0.75 kips PLL = 0.5 kips SLOPE 0:12(0= 0.00 °) DEFLECTION LIMIT OF LIVE LOAD d LL = L / 360 LONG-TERM DEFLECTION (NDS 3.5.2) dKWD+L=L/240 Code Duration Factor,C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load 2 ANALYSIS DETERM INE REACTIONS, MOMEENTS & SHEARS R2=0.Ss9casB+wu,J(L,+0.5LZ)Lz+PLC+L2= 9.91 11 11 L, L, Ri=(Z9_ +w.,)Li+( IL9+wu,,)L2+P—R2= 6.97 kips I w1, wLL2 wou w�, l R. I I._ X. � x; I X, J A- THE BEAM DESIGN IS ADEQUATE. kips Mmn = 0.5 w°`_ (coso +wu, Ls+PL2 = 3.8 ft-kips X, = 10.48 It Xz = 10.48 ft l r X3 = 0.54 ft (-W I = 36.5 ft-kips B+w., Vma = 8.58 kips, at R2 left. MINE SECTION PROPERTIES AND DESIGN FACTORS L u = Max(X3 , L2) = 2.0 ft, (NDS 2005 Table 3.3.3) GLB 3 1/8 x 24 Properties b = 3.13 in Fb = 2,400 psi ZE = 4.1 ft, (Tab 3.3.3 footnote 1) d = 24.00 in Fv = 265 psi, (NDS 97 CH included) RB = 11.0 < 50 A = 75.0 in E' = 1,800 ksi E'y= 1,600 ksi (confd) I S. = 300.0 in Fe = 2,327 psi I = 3,600 in F, = 265 psi E = Ex = 1800 ksi E'min = 930 ksi CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 0.98 1.00 --CK BENDING AND SHEAR CAPACITIES Cantilever: fb' =MMin 1S, = 153 psi < Middle Span: fb = MMax /SX = 1461 psi < FbE = 9185 psi Fb = 2400 psi F = FbE / Fb: = 3.83 Cv Cc Cr 0.97 1.00 1.00 Fb = 2327 psi [Satisfactory] Fb = 2367 psi [Satisfactory] Shear: f„' = 1.5 VMax /A = 172 psi < F„' [Satisfactory] (neglected d offset conservatively) CHECK DEFLECTION AT LIVE LOAD CONDITION L = L 1/cos 0 = 21.50 ft, beam sloped span a = L2/cos 0 = 2.00 ft, beam sloped cantilever length P = PLL cos 0 = 0.50 kips, perpendicular to beam w1 = wL l cosZ 0 = 0.16 klf, perpendicular to beam w2 =WLL,2 COS 0 = 0.16 klf, perpendicular to beam w,L3a+WV q3{4L+3,2i DEnaCP02(L+a) _ = 3EI cos9= 24EI 24EI -0.03 in, uplift to vertical direction. < 2 LZ / 360 = 0.13 in [Satisfactory] PaL2 + 5w,L4 OM`d — [— _ w.L2a2 cos B = 0.11 in, downward to vertical direction. 16EI 384EI 32EI < L, / 360 - 0.72 in [Satisfactory] CHECK DEFLECTION AT LONG-TERM LOAD, Ka DL + LL, CONDITION P = PKcrDL+U Cos B = 1.25 kips, perpendicular to beam Kcr = 1.00 , (NDS 3.5.2) wt = Kcr wOL,1 cos B + wLL1, COS B = 0.67 klf, perpendicular to beam w2 = Kcr wDL,2 cos B + wLL,2 COS B = 0.67 klf, perpendicular to beam wza3(4L+3a)l DEna—rAa2(L+a�_x,,L3a L 3E1 + 24EI 24EI JI cos B = -0.13 in, uplift to vertical direction. < 2 L2 / 240 = 0.20 in [Satisfactory] PaL2 Sw,L'r w:L22 _ + OMrd = [ _ cos B = 0.48 in, downward to vertical direction. 16EI 384EI 32EI < L, / 240 = 1.08 in [Satisfactory] RA PROJECT, LOT 126, MADISON CLUB, COLUMBUS WAY PAGE Structural CLIENT: BEAM * 39 ,HDR FRONT OF HALLWAY DESIGN BY. R.A JOB NO.: DATE: 12/1112012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) (DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 12 Glulam 24F-1.8E L = • 10.5 it wD = 935 lbs / ft wL = 54D lbs / ft PD1 = 0, Ibs 0' : ft PD2 = 0 _ Ibs L2= dL=L/360: dKuD+L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cry Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load L 9D1 1 11 1 PD2 11 IL I r r 11 WD �. A 1 Camber => 0.29 inch THE BEAM DESIGN IS ADEQUATE. ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseif wt = 15 Ibs / ft RLaft = 7.82 kips RRIght = 7.82 kips VM. = 6.33 kips, at 12 inch from left end MM. = 20.53 ft-kips, at 5.25 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'mir, = N/A E = Ex = 1800 ksi Fb = N/A d = 12.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 61.5 in2 I = 738 in4 F = 265 psi Fe = 2,400 psi S,r = 123.0 in3 RB = N/A E' = 1,800 ksi F = 265 psi lE= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMa,r / S,r = 2003 psi < Fb = 2400 psi [Satisfactory] f,; = 1.5 VMa, / A = 154 psi < F,; [Satisfactory] CHECK DEFLECTIONS A (L. Max) = 0.11 in, at 5.250 ft from left end, < d L = L / 360 [Satisfactory] d (Kcr D+ L, Max) = 0.31 in, at 5.250 ft from left end < d Ku D + L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.29 in, at 5.250 ft from left end IA-962 CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = Fc CD CP CF = 2429 psi Where Fc = 1600 psi Cp = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]"' = F�* = Fe Cp CF = 2560 psi LB = Ke L = 1.0 L = 126 in d = 12 in SF = slenderness ratio = 10.5 < F,, = 0.822 E'min / SFZ = 6934 psi Emin = 930 ksi F = FoE / Fc* = 2.709 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS % = F / A = 16 psi < Fc' i HE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for Co = 1.6 THE ACTUAL FLEXURAL STRESS IS fe = (M + Fe) / S = 2052 psi < CHECK COMBINED STRESS [NDS 2006 Sec. 3.9.2; (f. / F.' )2 + % / [Fb' (1 - f� / FEE)] = 0.536 1 1 T 0.949 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] Fe [Satisfactory] < 1 [Satisfactory] PROJECT: LOT.126, MADISON CLUB, COLUMBUS WAY RA CLIENT: SEAM 40 .HDR FRONT OF p0 2J820�12 Structural JOB NO. = J Beam Desi n Base on NDS 2005 NPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/8 x 12 Glulam 24F-1.8E L = (5.25 ft wD = 935 Ibs / ft wL = 540 Ibs / ft PD1 = 7000 Ibs Li= 1 ft PD2 = p Ibs L2 = :,=: 0_ ft AL=L1360 PAGE: DESIGN BY: R.A REVIEW BJ LL. Poi I + PD2 'NL �vD T camber => 0.07 inch A Kcro+L=L/ 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Conte Code Duration Factor C Oead Load 1 0.90 occupancy Live Load 2 1.00 Snow Load 3 1.15 Construction Load 4 1.25 wrid[Farthquake Load 5 1.60 Impact Load 6 2.00 2 Occupancy Live Load Choice => OWALYSIS DETERMINE REACTIONS, MOMENT. SHEAR = 10.54 kips Wse1f Wt 15 Ibs I ft VMS _ 9.05 kips, at 12 inch from left end DETERMINE SECTION PROPERTIES& ALLOIWAABLE STRESSES b = 5.13 in E�min = N/A d = 12.00 in FbE — a in 2 I = 738 in A = 61.5 s N/A Sx = 123.0 in RB - /E= N/A RRiaet = 5.78 kips MMax = 11.19 ft-kips, at 2.28 ft from left end E = EX = 1800 ksi Fb = N/A N/A Fb 2,400 Psi F = FbE / Fb- = ' = 2,400 psi Fb, Fv = 265 Psi Psi Fv - 265 P I-'- � = 1,800 ksi CF Cv Cc Co CM 1. Ct Ci CL CL00 1.00 1.00 1.00 1.00 1.00 1.00 Cr 1.00 CHECK BENDING AND SHEAR CAPACITIES Fb = 2400 psi [Satisfactory] fb = MMax is x = 1091 Psi F ' [Satisfactory] f,=1.5VMax/A= 221 Psi v � ALL 1360 d (L, Max) — =CK DEFLECTIONS - - 0.01 in, at 3.125 ft from left end, L 1240 d KcrD+L — A(KcrD+L,Max) =0.06 in, at 2.913 ft from left end Where 1.00 , (NDS 3.5.2) 'Cr TERMiNE CAMBER AOT075 (DEADa{ 2 913 WEIGHT) end d (1.5D, Max) [Satisfactory] [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F,' = F� CD C1 CF = 2523 psi Where F� = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]" = F�* = F� CD CF = 2560 psi Lb = K, L = 1.0 L = 75 in d = 12 in SF = slenderness ratio = 6.3 < F�, = 0.822 E'mj, / SF = 19570 psi E'min = 930 ksi F = FcE / Fr* = 7.645 C = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 16 psi < Fc' 1 0.985 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 1140 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F� )Z + % / [Fe (1 - f. / FA] = 0.297 < 1 [Satisfactory] RA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 41 ,HDR REAR OF BDRM 3 DESIGN BY : R.A JOB NO.: I I DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Desiun Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 1/8 x 13 1/2 Glulam 24F-1.8E L L MEMBER SPAN L = 625 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 935 . Ibs / ft PD1 +PD2 UNIFORMLY DISTRIBUTED LIVE LOAD wL = , 5.40. Ibs / ft CONCENTRATED DEAD LOADS PD1 = 16006. Ibs (0 for no concentrated load) L1 = 1 ft [ PD2 = 0,:.': Ibs L2 = G. ift DEFLECTION LIMIT OF LIVE LOAD AL = L / 360 Camber => 0.06 inch DEFLECTION LIMIT OF LONG-TERM AKcrD+L= L/ 240' . THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 16 Ibs / ft RLea = 13.06 kips RRight = 6.26 kips Vm. = 11.38 kips, at 13.5 inch from left end Mm. = 13.14 ft-kips, at 2.06 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = EX = 1800 ksi Fb = N/A d = 13.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 69.2 in 1 = 1,051 in Fv = 265 psi Fe = 2,400 psi Sx = 155.7 in Re = N/A E' = 1,800 ksi Fv' = 265 psi 1E= N/A CD CM Ct C, CL CIF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / SX = 1013 psi < Fb = 2400 psi [Satisfactory] f, = 1.5 VMax / A = 247 psi < Fv' [Satisfactory] CHECK DEFLECTIONS A (L' Max) = 0.01 in, at 3.125 ft from left end, < d L = L / 360 [Satisfactory] d (Ka D + L, Max) = 0.05 in, at 2.913 ft from left end < d Kcr D+ L = L 1240 [Satisfactory] Where Ka = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.06 in, at 2.913 ft from left end 9 F CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fr- = F, Cp CP CF = 2531 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 F,.' = IF, CD CF = 2560 psi I, = Ke L = 1.0 L = 75 in d = 13.5 in SF = slenderness ratio = 5.6 < F,E = 0.822 E'min / SF = 24769 psi E'min = 930 ksi F = FEE / F�* = 9.675 c = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 14 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for Co = 1.6 ] 1 Y 4 � 0.989 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1056 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + fb / [Fe 0 - f. / F.E)] = 0.275 < 1 [Satisfactory] (0 fLA PROJECT, LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 42 , BM REAR OF BDRM 3 DESIGN BY: R,A JOB NO.: DATE: 12-111/2012 REVIEW BY : R.A. Wood Beam Desion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 12 No. 1, Douglas Fir -Larch L = 9.25 ft wD = 300 Ibs / ft wL = 200 Ibs / ft PD1 = 0 Ibs L,= 0 ft PD2 = 0 Ibs L2 = 0 ft Li PDI 1 Foe �Il I A L = L / 360 Camber => 0.07 inch AKID+L= L/ 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration FactorDuration Factor�Co Condition Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseir M = 14 Ibs / ft RLef, = 2.38 kips RRigM = 2.38 kips y VMax = 1.88 kips, at 11.5 inch from left end MM. = 5.49 ft-kips, at 4.63 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1600 ksi Fb. = N/A d = 11.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 63.3 in 1 = 697 in F = 170 psi Fe = 1,350 psi S,r = 121.2 in RB = N/A E' = 1,600 ksi Fv' = 170 psi 1E= N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 544 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 VMax / A = 45 psi < F, [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.03 in, at 4.625 ft from left end, < A L = L / 360 [Satisfactory] d Xcr D + L. Max) = 0.08 in, at 4.625 ft from left end < A Kcr D + L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.07 in, at 4.625 ft from left end ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1. kips C ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1378 psi Where F. = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 _ F,,* = F� CD CF = 1480 psi L8 = Ke L = 1.0 L = 111 in d = 11.5 in SF = slenderness ratio = 9.7 < FEE = 0.822 E'min / SF2 = 5117 psi E'min = 580 ksi F = FCE / Fc* = 3.458 c = 0.8 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 16 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 0.931 50 FTTT1 1 I Y i [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 591 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )2 + fb / [Fe (1 - fc / FEE)] = 0.275 < 1 [Satisfactory] . ` PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 43 , HDR AT GARAGE DOOR DESIGN BY: R:A JOB NO.: DATE: 12/1112012 REVIEW BY: R.A 1 Beam Desn Base an NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) L)LFi ECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 10 No. 1, Douglas Fir -Larch L = 9.25 ft WD= 110 Ibs/ft WL = 80 Ibs / ft PDf = 0 Ibs L, = 0 ft PD2 = 0 Ibs L2= 0 ft L L, PD 1 I PD2 wL r j W r y 1 A L = L / 360 Camber => 0.05 inch dKcrD+L=L1240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code DuratiDn Factor C Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wself Wt = 11 Ibs / ft RLeft = 0.93 kips VMax = 0.77 kips, at 9.5 inch from left end MMax = DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES THE BEAM DESIGN IS ADEQUATE. Code Desl nation 1 Select Structural, Douglas Fir -Larch 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 RRi9nt = 0.93 kips 2.15 ft-kips, at 4.63 ft from left end b = 5.50 in E'min = N/A E = Ex = 1600 ksi Fb'= N/A d = 9.50 in 2 FbE = N/A a Fb = 1,350 psi F = FbE / Fb* = N/A A = 52.3 in in' I = 393 in Fv = 170 psi Fe = 1,350 psi S,t = 82.7 RB = N/A E' = 1,600 ksi Fv- = 170 Psi 1E = N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 312 psi < Fb = 1350 psi [Satisfactory] f, = 1.5 VMax / A = 22 psi < F, [Satisfactory] -HECK DEFLECTIONS A (L, Max) = 0.02 in, at 4.625 ft from left end, < A L = L / 360 [Satisfactory] 4(Kcr D+L. Mu) =0.05 in, at 4.625 ft from left end < dKcrD+L=L1240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.52) )EfERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.05 in, at 4.625 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F,' = FC CD CP CF = 1320 psi Where FC = 925 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s FC" = FC CD CF = 1480 psi Le = Ke L = 1.0 L = 111 in d = 9.5 in SF = slenderness ratio = 11.7 < F,, = 0.822 E'mni/ SFZ = 3492 psi E'min = 580 ksi F = FcE / FC* = 2.360 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 19 psi < F.' - ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 1 1 F F 0.892 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 370 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )2 + fb / [Fti 0 - f. / FcF)] = 0.172 < 1 [Satisfactory] 0 PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE Structural CLIENT: BEAM # 44 , FL FLR BM AT GARAGE DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam`Design Base on NDS 2005' INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) (DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM L GLB 3 1/8.x 19 1/2 Glulam 24F-1.8E L, L= �'Dl P62 w❑ = 930;;;:. Ibs / ft + wL = T40 Ibs / ft W` y �• P❑1 = 0 Ibs WD I J L1 = 0 ft P❑2 = 0 Ibs 1 Lz = 0 ft AL = L 1 360 Camber => 0.26 inch AK,D.L= L 1 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor CE, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirvin = 15 Ibs / ft RLeft = 10.95 kips RRigM = 10.95 kips VM. = 8.21 kips, at 19.5 inch from left end MMa� = 35.59 ft-kips, at 6.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.13 in E'min = N/A E = Ex = 1800 ksi Fb. = N/A d = 19.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 60.9 in 1 = 1,931 in Fv = 265 psi Fe = 2,400 psi Sx = 198.0 in RB = NIA E' = 1,800 ksi Fv' = 265 psi IE= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 2156 psi < Fb = 2400 psi [Satisfactory] f = 1.5 VMax / A = 202 psi < Fv' [Satisfactory] (CHECK DEFLECTIONS A (L, Max) = 0.14 in, at 6.500 ft from left end, < d (KcrD+ L. Max) = 0.31 in, at 6.500 ft from left end < Where K, = 1.00 , (NDS 3.52) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.26 in, at 6.500 ft from left end A L = L 1360 [Satisfactory] A Ker D , L = L / 240 [Satisfactory] IN (CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = Fr, Cp C, CF = 2494 psi Where F� = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s - Fc" = Fc CD CF = 2560 psi Le = Ke L = 1.0 L = 156 in d = 19.5 in SF = slenderness ratio = 8.0 < FCE = 0.822 E'min / SF2 = 11945 psi E'min = 930 ksi F = FcE / Fj = 4.666 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 16 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fti = 3840 psi, [ for CD = ` 1 � V F F y 0.974 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] I -HE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 2205 psi < Fti [Satisfactory] (CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / Fc' )Z + fb / [Fe (1 - f. / F.E)] = 0.575 < 1 [Satisfactory] 1A PROJECT: LOT 126, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 45 , FL FLR SM AT GARAGE DESIGN BY: R.A JOB NO.: DATE: 12111/2012 REVIEW BY: RA Wood Beam Desion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 3 118 x 24 Glulam 24F-1.BE L = 17.5 ft wD = 100 Ibs / ft wL = 60 Ibs / ft PD, = 11000 Ibs L, = 1.5 ft PD2 = 0 Ibs L2 = 0 ft AL=L/360 dKuD+L=L/240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C. Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1 A 5 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseitwt = 18 Ibs / ft RLea = 11.61 kips VMax = 11.26 kips, at 24 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.13 in E'min = N/A d = 24.00 in FbE = N/A A = 75.0 in2 I = 3,600 in4 S,t = 300.0 in3 Ra = N/A lE = N/A CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 702 psi < Fb fv' = 1.5 VMax / A = 225 psi < CHECK DEFLECTIONS A (L, Max) = 0.02 in, at 8.750 ft from left end, A (Ku D I L, Maxi = 0.14 in, at 8.025 ft from left end Where K�, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) 4 (1.5D, Max) = 0.18 in, at 8.025 ft from left end PD11 1PD2 wL 1 1 'VD I I I I r 9 1 1 1� 1 Camber=> 0.18 inch THE BEAM DESIGN IS ADEQUATE. RRi9ht = 2.50 kips MM. = 17.55 ft-kips, at 3.68 ft from left end E = EX = 1800 ksi Fb. = N/A Fb = 2,400 psi F = FbE / Fb. = NIA F, = 265 psi Fe = 2,396 psi E' = 1,800 ksi Fv- = 265 psi Cv Cc Cr 1.00 1.00 1.00 2396 psi [Satisfactory] F,; [Satisfactory] < AL=L/360 dKuD+L=L/240 [Satisfactory] [Satisfactory] 8 ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips 'rHE ALLOWABLE COMPRESSIVE STRESS IS F� = Fr CD CP CF = 2478 psi Where F,, = 1600 psi CD = 1.60 CF = 1.00 (Lumberonly) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.e _ Fc* = Fe, CD CF = 2560 psi I, = KB L = 1.0 L = 210 in d = 24 in SF = slenderness ratio = 8.8 < FoE = 0.822 E'mm / SF2 = 9985 psi E'min = 930 ksi F = FEE / F, = 3.900 c = 0.9 i'HE ACTUAL COMPRESSIVE STRESS IS f� = F / A = 13 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3834 psi, [ for CD = 1.6 ] �� 1 f 1 4 b i F F 0.968 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 742 psi < Fe [Satisfactory] --GK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )Z + fb / [Fe (1 - f. / F.E)] = 0.194 < 1 [Satisfactory] JIA n ,.J/.w IUA PROJECT: LOT 12B, MADISO CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM AT GARAGE DESIGN BY: R.A JOB NO.: `I0DATE: 12/1112012 REVIEW BY: R.A. Wood Beam Design Base on NOS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 1/8 x,12 Glulam 24F-1.8E MEMBER SPAN L = 2.5 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 4.50 Ibs / ft UNIFORMLY DISTRIBUTED LIVE LOAD wL = 100 Ibs / ft CONCENTRATED DEAD LOADS PD1 = 110.00 Ibs (0 for no concentrated load) L1 = 0.5 ft PD2 = 0 Ibs L2 = 0 ft DEFLECTION LIMIT OF LIVE LOAD A L = L / 360 DEFLECTION LIMIT OF LONG-TERM A KUD I L = L / 240 Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes L, II PD1 1 1 L, + PD2 11f ��TI T N` T 1 11 i } 1 1 N'D 1 Camber=> 0.00 inch THE BEAM DESIGN IS ADEQUATE. Code Duration Factor, Cn Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1AS Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 15 Ibs / ft RLeft = 9.51 kips RRight = 2.91 kips VMax = 8.94 kips, at 12 inch from left end Mm. = 4.68 ft-kips, at 0.50 ft from left end AINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = Ex = 1800 ksi Fb = N/A d = 12.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 61.5 in2 I = 738 in° F = 265 psi Fe = 2,400 psi S,t = 123.0 in3 Ra = N/A E' = 1,800 ksi Fv' = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 457 psi < Fb = 2400 psi [Satisfactory] f,; = 1.5 VMax / A = 218 psi < F,; [Satisfactory] CHECK DEFLECTIONS A (L. Max) = 0.00 in, at 1.250 ft from left end, < A L = L / 360 [Satisfactory] d (Kr D+ L, Max) = 0.00 in, at 1.100 ft from left end < A KcrD+L = L 1240 [Satisfactory] Where Kr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.00 in, at 1.100 ft from left end HECK THE BEAM CAPACITY WITH AXIAL LOAD (IAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS F� = F, Co CP CF = 2555 psi Where F. = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)Z - F / c]o.s F�* = F� Co CF = 2560 psi Le = Ke L = 1.0 L = 30 in d = 12 in SF = slenderness ratio = 2.5 < F,E = 0.822 E'min / SF2 = 122314 psi E'min = 930 ksi F = FEE / Fc* = 47.779 C = 0.9 ACTUAL COMPRESSIVE STRESS IS f� = F / A = 16 psi < Fr' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] 0.998 50 1 1 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 506 psi < Fti [Satisfactory] ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )Z + fb / [Fti (1 - f, / Fes] = 0.132 < 1 [Satisfactory] 1A PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE CLIENT: BEAM # 47 , HDR AT GYM DESIGN BY: R.A Structural JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 5 1/8 x 12 Glulam 24F-1.8E MEMBER SPAN L = 9 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 880 Ibs / ft UNIFORMLY DISTRIBUTED LIVE LOAD wL = 82. Ibs / ft CONCENTRATED DEAD LOADS PD1 = 0 Ibs (0 for no concentrated load) Lt = 0 ft PD2 = 0 Ibs L2 = s 0 ft DEFLECTION LIMIT OF LIVE LOAD A L = L / 350 DEFLECTION LIMIT OF LONG-TERM AKID +L = L / 240 Does member have continuous lateral support by top diaphragm ? L, lII PD I L, 1 PD2 r wD 4 Camber a 0.15 inch THE BEAM DESIGN IS ADEQUATE. (1= yes, 0= no) 1 Yes Code _Duration Factor. Ca Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 15 Ibs / ft RLOt = 7.72 kips RRIght = 7.72 kips VMax = 6.00 kips, at 12 inch from left end MMax = 17.36 ft-kips, at 4.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'R,ID = N/A E = Ex = 1800 ksi Fb = N/A d = 12.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb. = N/A A = 61.5 in2 I = 738 in4 F = 265 psi Fe = 2,400 psi S,t = 123.0 in3 RB= N/A E' = 1,800 ksi Fv1 = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1694 psi < Fb = 2400 psi [Satisfactory] f = 1.5 VMm / A = 146 psi < Fv' [Satisfactory] CHECK DEFLECTIONS A (L. Max) = 0.09 in, at 4.500 ft from left end, < A L = L / 360 [Satisfactory] A (KcrD+L, Max) = 0.19 in, at 4.500 ft from left end < A Kcr D + L = L / 240 [Satisfactory] Where Ka = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.15 in, at 4.500 ft from left end ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS F,' = F� CD CP CF = 2472 psi Where F. = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s = Fc` = Fe Co CF = 2560 psi Le = KB L = 1.0 L = 108 in d = 12 in SF = slenderness ratio = 9.0 < FcE = 0.822 E'min / SF2 = 9438 psi E'min = 930 ksi F = FcE / F j = 3.687 c = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 16 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for CD = 1.6 ] 1 1 F F y f- 0.966 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] - ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1743 psi < Fti [Satisfactory] =CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F.' )2 + fb / [Fe 0 - f. / F.E)] = 0.455 < 1 [Satisfactory] 01 RA PROJECT: LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT BEAM #F 48 , BM AT BRIDGE DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Desian Base on NDS 2005 INPUT DATA 8r DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 8 No. 1, Douglas Fir -Larch L = 6.5 ft wD= 145 Ibs/ft wL = 240 Ibs / ft PDt = .. 0 Ibs Lt = 0 ft PD2 = 0 Ibs L2 = 0 ft L, P" t 1 PD2 r v , I"• 1 1 1:1 L = L 1360 Camber => 0.03 inch A Kcr D+ L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration_ Factor, Cn Condition Code Desianalion 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wseif wt = 9 Ibs / ft RLeft = 1.28 kips RRight = 1.28 kips VMax = 1.03 kips, at 7.5 inch from left end MM. = 2.08 ft-kips, at 3.25 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E = EX = 1600 ksl Fb. = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb* = N/A A = 41.3 in 1 = 193 in4 Fv = 170 psi Fe = 1,350 psi Sx = 51.6 in Ra = N/A E' = 1,600 ksi F, = 170 psi lE = N/A Co CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 484 psi < Fb = 1350 psi [Satisfactory] f, = 1.5 VMax / A = 38 psi < Fv' [Satisfactory] CHECK DEFLECTIONS d (L, Mex) = 0.03 in, at 3.250 ft from left end, < d L = L / 360 [Satisfactory] d (Kcr D+L, Max) - 0.05 in, at 3.250 ft from left end < d Kcr D+ L = L / 240 [Satisfactory] Where Kcr = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (t.sD, Max) = 0.03 in, at 3.250 ft from left end ICHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1359 psi Where F. = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 = Fj = F,, CD CF = 1480 psi 1 = Ke L = 1.0 L = 78 in d = 7.5 in SF = slenderness ratio = 10.4 < F,E = 0.822 E'min / SF = 4408 psi E'mh = 580 ksi F = FEE / Fc* = 2.978 c = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 psi < Fc' - ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 ] 1 � F __. F 0.918 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] 7 HE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 557 psi < Fti [Satisfactory] (CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / Fc' )Z + fb / [Fe 0 - f. / FEE)] = 0.260 < 1 [Satisfactory] fLA PROJECT: LOT 12B; MADISON CLUB, COLUMBUS WAY PAGE: Structural CLIENT: BEAM # 49 , TYP. BM AT BRIDGE DESIGN BY: R.A JOB NO.: DATE: 12/11/2012 REVIEW BY: R.A. Wood Beam Design Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE MEMBER SPAN UNIFORMLY DISTRIBUTED DEAD LOAD UNIFORMLY DISTRIBUTED LIVE LOAD CONCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM 6 x 12 No. 1, Douglas Fir -Larch L= 13 ft wD = 145 , Ibs / It wL = 240 ' Ibs / ft PD1 = 0 Ibs L, = 0 - ft PD2 = 0 Ibs L2 = 0 It L� II PDi1 11 L lPD2 1 w` WD 1 A L = L / 360 Camber => 0.14 inch d Kcr D + L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, Cn Condition Code Designation 1 0.90 Dead Load 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir -Larch 3 1.15 Snow Load 3 No. 2, Douglas Fir -Larch 4 1.25 Construction Load 4 Select Structural, Southern Pine 5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR Wseirvw = 14 Ibs ! ft RLea = 2.59 kips RR;ght = 2.59 kips VMax = 2.21 kips, at 11.5 inch from left end MM. = 8.42 ft-kips, at 6.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'ml„ = N/A E = Ex = 1600 ksi Fb. = N/A d = 11.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fb. = N/A A = 63.3 in 1 = B97 in Fv = 170 psi Fb' = 1,350 psi Sx = 121.2 in RB= NIA E' = 1,600 ksi Fv' = 170 psi 1E= N/A CD CM C( Ci CL CF Cv C, Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 834 psi < Fb = 1350 psi [Satisfactory] f,' = 1.5 VMax / A = 52 psi < F,; [Satisfactory] CHECK DEFLECTIONS d (L, Mex) = 0.14 in, at 6.500 ft from left end, < d (Kcr D+L, Max) = 0.23 in, at 6.500 ft from left end < Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D, Max) = 0.14 in, at 6.500 ft from left end d L = L ! 360 [Satisfactory] d KcrD+L = L / 240 [Satisfactory] 9 CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F,' = F, CD CP CF = 1248 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o 5 - F�* = F� CD CF = 1480 psi Le = Ke L = 1.0 L = 156 in d = 11.5 in SF = slenderness ratio = 13.6 < Foe = 0.822 E'min / SF2 = 2591 psi E'min = 580 ksi F = FE / Fj = 1.751 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 16 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fti = 2160 psi, [ for CD = 1.6 THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 881 psi < CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / Fc )Z + fb / [Fy' (1 - f. / Fes] = 0.411 0.843 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] Fo [Satisfactory] < 1 [Satisfactory] fA PROJECT- LOT 12B, MADISON CLUB, COLUMBUS WAY PAGE. Structural CLIENT: BEAM # 50 , FL FLR BM AT GARAGE DESIGN BY R.A JOB NO.: DATE: 12I1112012 REVIEW BY: RA - Wood Hearn lesion Base on NDS 2005 INPUT DATA & DESIGN SUMMARY MEMBER SIZE GLB 3 1/8 x 24 Glulam 24F-1.8E L ` L, MEMBER SPAN L = 15 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 725 Ibs / ft PDi PD2 UNIFORMLY DISTRIBUTED LIVE LOAD wL = W Ibs / ft CONCENTRATED DEAD LOADS PDT = 0 Ibs °D (0 for no concentrated load) L, = 0 ft Poe = 0 Ibs 1 } L3= 0 ft DEFLECTION LIMIT OF LIVE LOAD AL = L 1360 Camber => 0.20 inch DEFLECTION LIMIT OF LONG-TERM 6KwD+L- L1 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes Code Duration Factor, C.-, Condition 1 0.90 Dead Load 2 1.00 Occupancy Live Load 3 1.15 Snow Load 4 1.25 Construction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR v'/setrwt = 18 Ibs / ft RLea = 9.62 kips RRreht = 9.62 kips VMax = 7.06 kips, at 24 inch from left end MM. = 36.08 ft-kips, at 7.50 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 3.13 in E'min = N/A E = EX = 1800 ksi Fb. = N/A d = 24.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 75.0 in 1 = 3,600 in Fv = 265 psi Fe = 2,400 psi Sx = 300.0 in 3 RB = N/A E' = 1,800 ksi Fv' = 265 psi 1E= N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1443 psi < Fb = 2400 psi [Satisfactory] f,' = 1.5 VMax / A = 141 psi < F,; [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.09 in, at 7.500 ft from left end, < A L = L / 360 [Satisfactory] d (Kr D. L, Max) = 0.23 in, at 7.500 ft from left end < d K, D+ L = L / 240 [Satisfactory] Where Kc, = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D, Max) = 0.20 in, at 7.500 ft from left end D�Dlb( CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = F� CD CP CF = 2503 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2C)2 - F / c]° s 0,978 F�* = Fc CD CF = 2560 psi I_ = Ke L = 1.0 L = 180 in d = 24 in SF = slenderness ratio = 7.5 < 50 F,, = 0.822 E'min / SF = 13590 psi E'm;n = 930 ksi F = FCE / F�* = 5.309 c = 0.9 THE ACTUAL COMPRESSIVE STRESS IS t v [Satisfies NDS 2005 Sec. 3.7.1.41 f,, = F / A = 13 psi < Fc' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fti = 3840 psi, [ for CD = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1483 psi < Fti [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f� / Fc' )Z + % / [Fe (1 - fc / FcE)] = 0.387 < 1 [Satisfactory] rV-% JODUIUIdi C119111CCIIIn9 78080 Calle Amigo, Suite #102 La Quinta, CA. 92253 (760)771-9993 TltleBlock Lines Wood Beam 1"11 Description : 13M#51 Material Properties Analysis Method: Allowable Stress Design Load Combination 2006 IBC & ASCE 7-05 Wood Species ; Douglas Fir - Larch Wood Grade : No.1 Beam Bracing ; Completely Unbraced Applied Loads Beam self weight calculated and added to loads Uniform Load : D = 0.210 , Tributary Width =1.0 ft, (Tile weiqht) rDESIGN SUMMARY I Wid Dsgnr: Project Desc.: Project Notes JOD IF Printed: 7 FEB 2013, 1:13PM F-RCALC, INC. 1983-2011. Bidd>6.11.4.5, Verb,11.4,1 Calculations per NDS 2005, ASCE 7-05 Fb - Tension 1350 psi E : Modulus of Elasticity Fb - Compr 1350 psi Ebend- xx 1600 ksi Fc - Prll 925 psi Eminbend - xx 580ksi Fc - Perp 625 psi Fv 170 psi Ft 675 psi Density 32.21 pcf .21 12x12 Span =21.0ft Service loads entered. Load Factors will be applied for calculations. Maximum Bending Stress Ratio = 0.4631 Maximum Shear Stress Ratio Section used for this span 12x12 Section used for this span fb : Actual 625.23psi fv : Actual FB : Allowable = 1,350.00psi Fv : Allowable Load Combination +D Load Combination Location of maximum on span = 10.500ft Location of maximum on span Span # where maximum occurs = Span # 1 Span # where maximum occurs Maximum Deflection Max Downward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Upward L+Lr+S Deflection 0.000 In Ratio = 0 <360 Max Downward Total Deflection 0.453 in Ratio = 556 Max Upward Total Deflection 0.000 in Ratio = 0 <240 0.153 : 1 12x12 25.96 psi 170.00 psi +D 0.000 ft Span # 1 Maximum Forces & Stresses for Load Combinations_ Load Combination Max Stress Ratios Summary of Moment Values Summary of Shear Values Segment Length Span # M V C d C FN C r C m C t Mactual fb-desian Fb-allow Wartllal fv�lpcinn Fv-allnu *D 1.000 1.000 1.000 1.000 Length = 21.0 ft 1 0.463 0.153 1.000 1.000 1.000 1.000 1.000 13.21 Overall Maximum Deflections - Unfactored Loads Load Combination Span Max. " " Defl Location in Span Load Combination D Only 1 0.4532 10.605 Maximum Deflections for Load Combinations - Unfactored Loads 625.23 1,350.00 2.29 25.96 170.00 Max. "+" Defl Location in Span 0.0000 0.000 W Load Combination Span Max. Downward Defl Location in Span Max. Upward Defl Location in Span D Only 1 0.4532 10.605 0.0000 0.000 Vertical Reactions - Unfactored Support notation : Far left is #1 Values in KIPS Load Combination Support 1 Support 2 Overall MAxmum 2.516 2.516 D Only 2,516 2.516 .v-.-u..(., p�i.ymw,i-y 78080 Calle Amigo, Suite #102 Dsgnr: La Quinta, CA. 92253 Project Desc.: (760)771-9993 Project Notes Title Block Line 6 Pnnied. 7 FEB 2013. 1:13PM Wood Beam >r r- INC.19B3.2o11. e1,aa:s.11.4.5. Ver.6A1.4.1 Description : Bk#51 13 10 7 3 I 8EN4•••�. 2.00 4.10 6.20 830 10.40 12.50 14.60 16.70 18.80 20.90 Distance (ft) • rD 2.6 E O 2.00 4.10 6.20 830 10.40 12.50 14.60 16.70 18.90 20.90 Distance (ft) ■ tD 2.00 4.10 6.20 8.30 10.40 12.50 14.60 16.70 18.80 20.90 DistaKe (ft) ■ DO•ly 1v 1 1 UI- "I I V I I I—n IV - itic JUU $ 78080 Calle Amigo, Suite #102 Dsgnr: La Quinta, CA. 92253 Project Desc.: (760)771-9993 Project Notes Title Block Line 6 _ Primed: 7FEB 2013, 1:11PM Wood Beam ENERCALC, INC- IW3-2011, Build: IIII .5, VW6-11.4.1 Description : BM#52 Material Properties Analysis Method: Allowable Stress Design Load Combination 2006 IBC & ASCE 7-05 Wood Species Douglas Fir - Larch Wood Grade No.1 Beam Bracing Completely Unbraced Fb - Tension 1350 psi Fb . Compr 1350 psi Fc • Pril 925 psi Fc - Perp 625 psi Fv 170 psi Ft 675 psi ■ Span = 38.0 ft Applied Loads Beam self weight calculated and added to loads Uniform Load : D = 0.20 , Tributary Width =1.0 ft, (Tile weiqht) (DESIGN St1h NARY Maximum Bending Stress Ratio = 0.585 1 Section used for this span 12x20 fb : Actual 743.47psi FB : Allowable = 1,270.30psi Load Combination +D Location of maximum on span = 19,000ft Span # where maximum occurs = Span # 1 Maximum Deflection Calculations per NDS 2005, ASCE 7-05 E : Modulus of Elasticity Ebend-xx 1600ksi Eminbend - xx 580ksi Density 32.21 pcf I Service loads entered. Load Factors will be applied for calculations. Maximum Shear Stress Ratio Section used for this span fv : Actual Fv : Allowable Load Combination Location of maximum on span Span # where maximum occurs 0.172 : 1 12x20 = 29.25 psi 170.00 psi +D 0.000 ft Span # 1 Max Downward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Upward L+Lr+S Deflection 0.000 in Ratio = 0 <360 Max Downward Total Deflection 1.041 in Ratio = 438 Max Upward Total Deflection 0.000 in Ratio = 0 <240 Maximum Forces & Stresses for Load Combinations Load Combination Max Stress Ratios 4 Summary of Moment Values Summary of Shear Values Segment Length Span # M V C d C FN C r C m C t Mactual fbdesign Fb-allow Vactual fv-design Fv-allow +D 0.947 1.000 1.000 1.000 Length = 38.0 ft 1 0.585 0.172 1.000 0.947 1.000 1.000 1.000 45.15 743.47 1,270.30 4.37 29.25 170.00 Overall Maximum Deflections - Unfactored Loads Load Combination Span Max. " " Defl Location in Span Load Combination Max. "+ Defl Location in Span D Only 1 1.0405 19.190 W 0.0000 0.000 Maximum Deflections for Load Combinations - Unfactored Loads Load Combination Span Max. Downward Defl Location in Span Max. Upward Defl Location in Span 0 Only 1 1.0405 19.190 0.0000. 0,000 Vertical Reactions - Unfactored Support notation : Far left is #1 Values in KIPS Load Combination Support 1 Support 2 T ' _ Overall MAXimum 4.753 4.753 D Only 4.753 4.753 - "' 78080 Calle Amigo, Suite #102 I itie : Job # Dsgnr: La Quinta, CA. 92253 Project Desc. (760)771-9993 Project Notes �TitfeMock Line 6 Wood Beam - - -- - -- - - - -- -- -- - Printed: 7FEB Zi3, 1:11PM . t ► ERCALC, INC. 1983-2011, Build:6.11.4.5, Ver.6.11.4.1 Description : BM#52 46 35 r 23 12 BEAM--->. BEAM--• > -03 6 -0.5 .0.0 -1.1 ��•-• �o..a 30.Z1 34.01 37A1 Distance (ft) ■ +D ""1 nAl 30.21 34.01 37,01 Distance (ft) i +D -'^• �caa Zb.41 30.21 34.01 37.81 Distance (ft) ■ DO.Iy RIH PROJECT: LOT 1 2l3, MAD}SDN (--LuB... PAGE CLIENT : DESIGN BY R.A. Structural JOB NO.: 1 11109 DATE' 12I1W011 REVIEW BY R.A_ Tables for Wood Post Design on NDS 2005 DURATION FACTOR (1.0, 1.15, 1.25, 1.6) CD = 1,00 (NDS 2.3.2) COMMERCIAL GRADE (# 1 or# 2) # 2 Post Axial Canacit►► for ❑ouolas Fir -Larch it 2. (kind Height Section Size (ft) 4x4 4x6 4x8 4x 10 4x 12 6x6 6x8 6x 10 6x 12 8x8 8x10 6 10.89 16.85 21.64 27.34 33.25 19.59 26.72 33.85 40.97 37.92 48.03 7 8.68 13.51 17.62 22.21 27.01 18.89 25.76 32.63 39.49 37.33 47.28 8 6.96 10.87 14.22 18.00 21.89 17.99 24.54 31.08 37.63 36.59 46.34 9 5.66 8.85 11.60 14.72 17.90 16.91 23.06 29.21 35.35 35.69 45.21 10 4.67 7.31 9.59 12.19 14.82 15.66 21.35 27.05 32.74 34.61 43.84 11 3.90 6.12 E. 10.23 12.44 14.32 19.52 24.73 29.93 33.34 42.23 12 3.31 5.19 6.83 8.69 10.57 12.96 17.67 22.39 27.10 31.87 40.37 13 2.84 4.46 5.86 7.46 9.08 11.67 15.91 20.15 24.39 30.23 38.30 14 2.46 3.86 5.08 6.47 7.87 10.47 14.28 18.09 21.90 28.46 36.05 15 2.15 3.38 4.45 5.67 6.89 9.41 1 12.83 16.25 19.67 26.62 33.72 16 1.90 2.98 3.92 5.00 6.08 8.46 11.54 14.62 17.70 24.76 31.37 17 1.69 2.65 3.49 4.44 5.40 7.63 10.41 13.19 15.96 22.96 29.08 18 1.51 2.37 3.12 3.97 4.83 6.91 9.42 11.93 14.44 21.23 26.89 19 1.36 2.13 2.80 3.57 4.35 6.27 8.55 10.83 13.12 19.62 24.85 20 1.23 1.92 2.53 3.23 3.93 5.72 7.79 9.87 11.95 18.13 22.96 21 1.11 1.75 2.30 2.94 3.57 5.22 7.12 9.02 10.92 16.76 21.23 22 1.02 1.59 2.10 2.68 3.26 4.79 6.53 8.28 10.02 15.52 19.66 23 0.93 1.46 1.92 2.45 2.98 4.41 6.01 7.62 9.22 14.39 18.23 24 0.85 1.34 1.77 2.26 2.74 4.07 5.55 1 7.03 8.51 13.36 16.93 25 0.79 1.24 1.63 2.08 2.53 3.77 5.13 6.50 7.87 12.43 15.75 26 0.73 1.15 1.51 1.92 2.34 3.49 4.76 6.03 7.30 11.59 14.68 27 0.68 1.06 1.40 1 1.79 1 2.17 3.25 4.43 5.61 1 6.80 10.82 13.71 28 0.63 0.99 1.30 1.66 2.02 3.03 4.13 5.23 6.34 10.13 12.83 29 0.59 0.92 1.22 1 1.55 1.89 1 2.83 3.86 4.89 5.92 1 9.49 12.02 30 0.55 0.86 1.14 1.45 1 1.76 1 2.65 3.62 4.58 5.54 1 8.91 11.29 Post Axial Caoacit►r for Southern Pine # 2. 1kinel Height Section Size (ft) 4x4 4x6 4x8 4x 10 4x 12 6 x 6- 6x8 6x 10 6x 12 8x8 8x 10 6 11.10 17, 27 22.54 28.44 34,17 14.96 20.41 25.85 31.29 28.68 36.32 7 8.79 13.73 1T98 22.77 27.49 14.57 19.87 25.16 30.46 28.33 35.89 8 7.02 10.98 14.41 18.30 22.15 14.07 19.18 24.30 29.42 27.91 35.36 9 5.69 8.91 11.71 14.89 18.05 13.45 18.35 2324 28.13 27.41 34.72 10 4.69 7.35 9.66 12.30 14.91 12.72 17.35 21.98 26.61 26.80 33.95 11 3.92 6.15 8.09 10.30 12M 11.90 16.22 20.55 24.87 26.09 33.05 12 3.32 5.21 6.86 8.74 10.61 11.00 15.00 19.00 23.00 25.26 32.00 13 2.85 4.47 5.88 7.50 9.10 10.09 1176 17.42 21.09 24.32 30.81 14 2.47 3.87 5.10 6.50 7.89 9.20 12.54 15.88 19.23 23.27 29.47 15 2.16 3.39 4.46 5.69 6.91 8.36 11.40 14.44 17.48 22.12 28.02 16 1.90 2.99 3.93 5.01 6.09 7.59 10.35 13.11 15.87 20.91 26.48 17 1.69 2.65 3.49 4.45 5.41 6.89 9.40 11.91 14.42 19.66 24.91 18 1.51 2.37 3.12 3.98 4.84 6.27 8.56 10.84 13.12 18.42 1 23.33 19 1.36 2.13 2.81 3.58 4.35 5.72 7.80 9.88 11.96 17.21 1 21.80 20 1.23 1.93 1 2.54 3.24 3.93 5.23 7.13 1 9.04 10.94 16.05 20.33 21 1.11 1.75 2 30 2.94 3.57 4.80 6.54 8.28 10.03 i4.95 18.94 22 1.02 1.60 2.10 2.68 3.26 4.41 6.01 7.62 9.22 13.93 17.65 23 0.93 1.46 1.93 2.46 2.99 4.06 5.54 7.02 8.50 12.98 16.45 24 0.86 1.34 1.77 2.26 2.74 3.76 5.12 6.49 7.86 12.11 15.34 25 0.79 1.24 1.63 2.08 1 2.53 3.48 4.75 6.01 7.28 11.31 1T33 26 0.73 1.15 1.51 1.93 2.34 3.23 4.41 5.59 6.76 10.58 13.40 27 0.68 1.06 1.40 1.79 2.17 3.01 4.11 1 5.20 6.30 9.90 12.54 28 0.63 0.99 1.30 1.66 2.02 2.81 3.83 4.85 5.88 9.28 11.76 29 0.59 0.92 1.22 1.55 1.89 2.63 3.58 4.54 5.50 8.72 11.04 30 0.55 0.86 1.14 1.45 1.76 2.46 3.36 4.25 5.15 820 10.39 Note: 1. The bold values require steel bearing plate based on, Fc1, 625 psi. 2. The table values are from Wood Column software at www.Engineering-intemational.com . `o� a L' AM I [I�av ;�klz-RIllwm rza F .0 .k 1 :F11 w. MIQtMIT7rI E PD) LOCATION: L& QURiTA9, C.A (ZUNDMF--. 927M) ss = Mas CL6" - &we = Lolo .5m &6w SHMECDERMCATEGGRYD .16135-6) SITE Cr AM D (ASSUMM) 1 3.5-2) OCCUPAWCTCATMORY-IM ALA SEIMC RV941RIANM FACTOR I= IL90- RESPOWMMODIECATk (ASCE IE 12,21) ,V-ffAFDT,P7,4A& PAAgEr- R 6.5 (HF CATALOG "'g) TABLE 122-1) AT CANTMEM MMLC0LZMM.-R= 1-5 (ASM 7-05, -RJ90U.NDA-Nr-Y-YA'MMR p = U40 - iMM7 BASH,qEEARV=p3cLSmU(1-4zlt)jxW = (ASM 7-05, EQU 17-9-1 & IU-2) BASE SEEAR V = L30 ic [1.010 x L01(1-4 x 6.5)] T OA45W - MMITT-1 ASSUNEa) MOST C0NSWVAlTVE.Fj" Al�- =20- 17.S Pff WIND E)MOSURE CAJMORY- C NVIRD lh-fiNXRXAlqCE FACTOR 11 = luvo BASIC UrRqD SPEMD 90. APR TOP043RAPEEIC FACTM JKZr = mo wiMD3FAdCT0Rpi=X-xKzrxlxpm 1 MjIMo M45- -(ASIQE 7-D5. FIGURE 6-2) (A-SM-7-0. SECT 6-5 6-3) - (� 7-0. TABLE 6-1) (AS(M 7-05. MGMM 6-1) (ASCR T-05. SECT. 15 5-7-2) - (.AS&- 7-0%13QU- 6-1) D LATERAL ANALYSIS SECTION 1 LONGITUDINAL ROOF AVERAGE HEIGHT = 25.00 FT TOP PLATE AVERAGE HEIGHT = 22.00 FT LOWER PLATE AVERAGE HEIGHT = 10.00 FT 25 FT. MAX ROOF HEIGHT WIND FORCE = 24.03 PSF WIND LOAD AT UPPER LEVEL = 24.03 x ( 25.00 - 12.00 - 10/2) = 192.00 PLF WIND LOAD AT LOWER LEVEL = 24.03 x ( 22.00 - 10.00/2 - 10.00/2) = 288.00 PLF TOTAL WIND = 480.00 PLF ROOF SEISMIC LOAD = 0.145x(30.00xl5.00+15.00x2x5+10.00x5x1 ) = 94.00 PLF FLOOR SEISMIC LOAD = 0.145x(25.00xl5.00+15.00x2x10+10.00xOx1)=76.00 PLF SEISMIC GOVERNS = 170.00 PLF REDISTRIBUTE SHEAR 94.00x22 + 76.00x10 = 2828.00 SHEAR AT ROOF = 94.00 x 22.00 x 170.00 / 2828.00 = 124.00 PLF SHEAR AT FLR = 76.00 x 10.00 x 170.00 / 2828.00 = 47.00 PLF ROOF MAX. SHEAR = 192.00 x 29.00 / 2 x 15.00 = 186.00 PLF CHORD FORCE = 192.00 x 29.00 x 29.00 / 8 x 15.00 = 1346.00 LBS USE '/z" CDX BLOCKED WITH 8D' S AT 6", 12" O/C USE (12) 16D'S PER TOP PLATE SPLICE FLOOR MAX. SHEAR = 480.00 x 29.00 / 2 x 15.00 = 464.00 PLF CHORD FORCE = 480.00 x 29.00 x 29.00 / 8 x 15.00 = 3233.00 LBS USE 5/8"CDX BLOCKED WITH 8D' S AT 3", 12" O/C USE (24) 16D' S PER TOP PLATE SPLICE Li LATERAL ANALYSIS SECTION 2 LONGITUDINAL ROOF AVERAGE HEIGHT = 25.00 FT TOP PLATE AVERAGE HEIGHT = 22.00 FT LOWER PLATE AVERAGE HEIGHT = 10.00 FT 25 FT. MAX ROOF HEIGHT WM FORCE = 24.03 PSF WIND LOAD AT UPPER LEVEL = 24.03 x ( 25.00 - 12.00 - 10/2) = 192.00 PLF WIND LOAD AT LOWER LEVEL = 24.03 x ( 22.00 - 10.00/2 - 10.00/2) = 288.00 PLF TOTAL WIND = 480.00 PLF ROOF SEISMIC LOAD = 0.145x( 30.00x25.00+15.00x2x5+lO.00x5xl ) = 138.00 PLF FLOOR SEISMIC LOAD = 0.145x(25.00x25.00+15.00x2x10+10.00xOxl)= 134.00 PLF SEISMIC GOVERNS = 272.00 PLF REDISTRIBUTE SHEAR 13 8.00x22 + 134.00x 10 = 43 76.00 SHEAR AT ROOF = 138.00 x 22.00 x 272.00 / 4376.00 = 189.00 PLF SHEAR AT FLR = 134.00 x 10.00 x 272.00 / 4376.00 = 83.00 PLF ROOF MAX. SHEAR = 192.00 x 25.00 / 2 x 25.00 = 96.00 PLF CHORD FORCE = 192.00 x 25.00 x 25.00 / 8 x 55.00 = 600.00 LBS USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE FLOOR MAX. SHEAR = 480.00 x 25.00 / 2 x 55.00 = 240.00 PLF CHORD FORCE = 480.00 x 25.00 x 25.00 / 8 x 25.00 = 1500.00 LBS USE 5/8"CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (12) 16D'S PER TOP PLATE SPLICE a SECTION 3 LONGITUDINAL ROOF AVERAGE HEIGHT = 30.00 FT TOP PLATE AVERAGE HEIGHT = 22.00 FT LOWER PLATE AVERAGE HEIGHT = 10.00 FT 30 FT. MAX ROOF HEIGHT WIND FORCE = 24.92 PSF WIND LOAD AT UPPER LEVEL = 24.92 x (30.00 - 14.00 - 10/2) = 275.00 PLF WIND LOAD AT LOWER LEVEL = 24.92 x (24.00 - 12.00/2 - 10.00/2) = 350.00 PLF TOTAL WIND = 625.00 PLF ROOF SEISMIC LOAD = 0.145x( 30.00x27.00+15.00x2x5+lO.00x5xl ) = 147.00 PLF FLOOR SEISMIC LOAD= 0.145x(25.00x27.00+15.00x2x10+10.00x2x10)= 170.00 PLF SEISMIC GOVERNS = 317.00 PLF REDISTRIBUTE SHEAR 147.00x24 + 170.00x10 = 5228.00 SHEAR AT ROOF = 147.00 x 24.00 x 317.00 / 5228.00 = 214.00 PLF SHEAR AT FLR = l70.00 x l O.00 x 317.00 / 5228.00 = 103.00 PLF ROOF MAX. SHEAR = 275.00 x 27.00 / 2 x 22.00 = 167.00 PLF CHORD FORCE = 275.00 x 27.00 x 27.00 / 8 x 22.00 = 1139.00 LBS USE 1/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (10) 16D' S PER TOP PLATE SPLICE FLOOR MAX. SHEAR = 625.00 x 27.00 / 2 x 22.00 = 383.00 PLF CHORD FORCE = 625.00 x 27.00 x 27.00 / 8 x 22.00 = 2588.00 LBS USE 5/8"CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (18) 16D'S PER TOP PLATE SPLICE r LATERAL ANALYSIS SECTION 4 LONGITUDINAL ROOF AVERAGE HEIGHT = 30.00 FT TOP PLATE AVERAGE HEIGHT = 24.00 FT 30 FT. MAX ROOF HEIGHT WIND FORCE = 24.92 PSF WIND LOAD = 24.92 x (30.00 - 24.00/2) = 450.00 PLF INTERIOR SECTION SEISMIC LOAD = 0.145 x ( 30.00x42.00 + 25.00x2xl2) = 235.00 PLF SEIMIC GOVERNS = 235.00 PLF MAX. SHEAR = 23 5. 00 x 30.00 / 2 x 42.00 = 84.00 PLF CHORD FORCE = 235.00 x 30.00 x 30.00 / 8 x 42.00 = 630.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 5 LONGITUDINAL ROOF AVERAGE HEIGHT = 15.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF VaND LOAD = 21.54 x ( 15.00 - 10.00/2 ) = 215.00 PLF SEISMIC LOAD = 0.145 x ( 25.00x84.00 + 15.00x4x5) = 348.00 PLF SEISMIC GOVERNS = 348.00 PLF MAX. SHEAR = 348.00 x 18.00 / 2 x 84.00 = 38.00 PLF CHORD FORCE = 348.00 x 18.00 x 18.00 / 8 x 84.00 = 168.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 6 LONGITUDINAL ROOF AVERAGE HEIGHT =15.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF WIlVD LOAD = 21.54 x ( 15.00 - 10.00/2 ) = 215.00 PLF SEISMIC LOAD = 0.145 x ( 30.00x22.00 + 15.00x2x5) = 117.00 PLF SEISMIC GOVERNS = 117.00 PLF MAX. SHEAR = 117.00 x 34.00 / 2 x 22.00 = 90.00 PLF CHORD FORCE = 117.00 x 34.00 x 34.00 / 8 x 22.00 = 768.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE LATERAL ANALYSIS SECTION 1 TRANSVERSE ROOF AVERAGE HEIGHT = 25.00 FT TOP PLATE AVERAGE HEIGHT = 22.00 FT LOWER PLATE AVERAGE HEIGHT = 10.00 FT 25 FT. MAX ROOF HEIGHT WIND FORCE = 24.03 PSF WIND LOAD AT UPPER LEVEL = 24.03 x ( 25.00 - 12.00 - 10/2 } = 192.00 PLF WIND LOAD AT LOWER. LEVEL = 24.03 x (22.00 - 10.00/2 - 10.00/2) = 288.00 PLF TOTAL WIND = 480.00 PLF ROOF SEISMIC LOAD = 0.145x( 30.00x30.00+l5.00x2x5+lO.00x5xl ) = 160.00 PLF FLOOR SEISMIC LOAD = 0.145x(25.00x30.00+15.00x2x10+10.0OxOxl)= 152.00 PLF SEISMIC GOVERNS = 312.00 PLF REDISTRIBUTE SHEAR 160.00x22 + 152.00x10 = 5040.00 SHEAR AT ROOF = 160.00 x 22.00 x 312.00 / 5040.00 = 218.00 PLF SHEAR AT FLR = 152.00 x 10.00 x 312.00 / 5040.00 = 94.00 PLF ROOF MAX. SHEAR = 218.00 x 25.00 / 2 x 30.00 = 91.00 PLF CHORD FORCE =lfl@00 x 25.00 x 25.00 / 8 x 30.00 = 568.00 LBS USE'/z" CDX UNBLOCKED WITH 8D'S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE FLOOR MAX. SHEAR = 480.00 x 25.00 / 2 x 30.00 = 200.00 PLF CHORD FORCE = 480.00 x 25.00 x 25.00 / 8 x 30.00 = 1250.00 LBS USE 5/8"CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (10) 16D' S PER TOP PLATE SPLICE •- SECTION 2 TRANSVERSE ROOF AVERAGE HEIGHT = 30.00 FT TOP PLATE AVERAGE HEIGHT = 22.00 FT LOWER PLATE AVERAGE HEIGHT = 10.00 FT 30 FT. MAX ROOF HEIGHT WIND FORCE = 24.92 PSF WIND LOAD AT UPPER LEVEL = 24.92 x ( 30.00 - 14.00 - 10/2) = 275.00 PLF WIND LOAD AT LOWER LEVEL = 24.92 x (24.00 - 12.00/2 - 10.00/2) = 350.00 PLF TOTAL WIND = 625.00 PLF ROOF SEISMIC LOAD = 0.145x( 30.00x7O.00+l5.00x2x5+lO.00x5x4 ) = 355.00 PLF FLOOR SEISMIC LOAD= 0.145x(25.00x80.00+15.00x2x10+10.00x4x10)= 391.00 PLF SEISMIC GOVERNS = 746.00 PLF REDISTRIBUTE SHEAR 355.00x24 + 391.00x10 = 12430.00 SHEAR AT ROOF = 355.00 x 24.00 x 746.00 / 12430.00 = 511.00 PLF SHEAR AT FLR = 391.00 x 10.00 x 746.00 / 12430.00 = 235.00 PLF ROOF MAX. SHEAR = 511.00 x 26.00 / 2 x 70.00 = 95.00 PLF CHORD FORCE = 511.00 x 26.00 x 26.00 / 8 x 70.00 = 617.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE FLOOR MAX. SHEAR = 746.00 x 26.00 / 2 x 80.00 = 121.00 PLF CHORD FORCE = 746.00 x 26.00 x 26.00 / 8 x 80.00 = 788.00 LBS USE 5/8"CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE 0 LATERAL ANALYSIS SECTION 3 TRANSVERSE ROOF AVERAGE HEIGHT = 30.00 FT TOP PLATE AVERAGE HEIGHT = 24.00 FT 30 FT. MAX ROOF HEIGHT WIND FORCE = 24.92 PSF WIND LOAD = 24.92 x ( 30.00 - 24.00/2 ) = 450.00 PLF INTERIOR SECTION SEISMIC LOAD = 0.145 x ( 30.00x60.00 + 15.00x2x12) = 313.00 PLF SEMC GOVERNS = 313.00 PLF MAX. SHEAR = 313.00 x 36.00 / 2 x 60.00 = 94.00 PLF CHORD FORCE = 312.00 x 36.00 x 36.00 / 8 x 60.00 = 845.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 4 TRANSVERSE ROOF AVERAGE HEIGHT = 20.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF MrM LOAD = 22.97 x ( 20.00 - 10.00/2 ) = 345.00 PLF SEISMIC LOAD = 0.145 x ( 30.00x48.00 + 15.00x2x5 + 10.00x2x5) = 245.00 PLF WIND GOVERNS = 345.00 PLF MAX. SHEAR = 345.00 x 20.00 / 2 x 48.00 = 72.00 PLF CHORD FORCE = 345.00 x 20.00 x 20.00 / 8 x 48.00 = 359.00 LBS USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE 8 SHEAR WALL DESIGN SW #1 LEFT OF GUEST BATH 2 TOTAL LOAD WIND INWARD ONLY= 192.00 x 29.00/2 x 0.8/1.3 = 1713.00 LBS. TOTAL LOAD SEISMIC= 124.00 x 29.00/2 = 1798.00 LBS. SHEAR WALL LENGTH = 4.25 FT. SHEAR WALL = 1798.00 / 4.25 = 423.00 PLF USE SHEAR WALL TYPE 12 WITH 16D'S AT 3" O.0 SILL PLATE NAILING MAX. DRAG = 638.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3911.00 LBS USE (3) SIMPSON CS16 STRAPS EACH END SW #2 RIGHT OF GUEST BDRM 2 TOTAL LOAD WIND INWARD ONLY= 192.00 x 29.00/2 x 0.8/1.3 = 1713.00 LBS. TOTAL LOAD SEISMIC= 124.00 x 29.00/2 = 1798.00 LBS SHEAR WALL LENGTH = 4.75 + 4.75 = 9.50 FT. SHEAR WALL = 1798.00 / 9.50 = 190.00 PLF USE SHEAR WALL TYPE 11 WITH 16D'S AT 3" O.0 SILL PLATE NAILING MAX. DRAG = 478.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1546.00 LBS USE (2) SIMPSON CS16 STRAPS EACH END SW #3 RIGHT OF OF GUEST BDRM 1 TOTAL LOAD WIND INWARD ONLY= 192.00 x 31.00/2 x 0.8/1.3 = 1832.00 LBS. TOTAL LOAD SEISMIC= 189.00 x 29.00/2 = 2741.00 LBS. SHEAR WALL LENGTH = 8.00 + 12.00 = 20.00 FT. SHEAR WALL = 2741 / 20.00 = 137.00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O.0 SILL PLATE NAILING MAX. DRAG = 464.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 387.00 LBS NO SIGNIFICANT UPLIFT, NO HOLDOWNS SW #4 LEFT OF GUEST BEDROOM 1 TOTAL LOAD WIND INWARD ONLY= 192.00 x 25.00/2 x 0.8/1.3 = 1477.00 LBS. TOTAL LOAD SEISMIC= 189.00 x 25.00/2 = 2363.00 LBS. SHEAR WALL LENGTH = 4.50 + 4.50 + 4.00 + 4.00 = 17.00 FT. SHEAR WALL = 2363.00 / 17.00 = 139.00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O/C SILL PLATE NAILING MAX. DRAG = 765.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1140.00 LBS USE SIMPSON CS16 STRAP EACH END SW #5 LEFT OF GARAGE TOTAL LOAD WIND INWARD ONLY= 480.00 x 32.00/2 x 0.8/1.3 = 4726.00 LBS. TOTAL LOAD SEISMIC= 170.00 x 32.00/2 = 2720.00 LBS. USE HARDY FRAME USE (2) HFX-24x10 1 1/8" HS TOTAL CAPACITY 4330.00 x 2 = 8660.00 LBS. SEE FOUNDATION DESIGN D SHEAR WALL DESIGN SW #6 RIGHT OF POWDER/MECHANICAL TOTAL LOAD WIND INWARD ONLY= 480.00 x 32.00/2 x 0.8/1.3 = 4726.00 LBS. TOTAL LOAD SEISMIC= 170.00 x 32.00/2 = 2720.00 LBS. SHEAR WALL LENGTH = 15.50 FT. SHEAR WALL = 4726.00 / 15.50 = 305.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 728.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1883.00 + 1546.00 (ABOVE) = 3429.00 LBS USE SHVIPSON HDU5 HOLDOWN EACH END SW #7 RIGHT OF GARAGE TOTAL LOAD WIND INWARD ONLY= 480.00 x 31.00/2 x 0.8/1.3 = 4579.00 LBS. TOTAL LOAD SEISMIC= 272.00 x 31.00/2 = 4216.00 LBS. SHEAR WALL LENGTH = 4.00 + 12.00 = 16.00 FT. SHEAR WALL = 4579.00 / 16.00 = 286.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 1584.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2190.00 LBS USE SMIPSON HDU5 HOLDOWN EACH END SW #8 LEFT OF GYM TOTAL LOAD WIND INWARD ONLY= 480.00 x 31.00/2 x 0.8/1.3 = 4579.00 LBS. TOTAL LOAD SEISMIC= 272.00 x 31.00/2 = 4216.00 LBS. SHEAR WALL LENGTH = 4.50 + 4.50 = 9.00 FT. SHEAR WALL = 4579.00 / 9.00 = 509.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 2290.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3789.00 + 1140.00 (ABOVE) = 4929.00 LBS USE SEAPSON HDU5 HOLDOWN EACH END SW #9 FRONT OF 2 GUEST BDRM 2 TOTAL LOAD = 218.00 x 15.00/2 = 1635.00 LBS. SHEAR WALL LENGTH = 11.00 FT. SHEAR WALL = 1635.00 / 11.00 = 149.00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O.0 SILL PLATE NAILING MAX. DRAG = 698.28 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 209.00 LBS NO SIGNIFICANT UPLIFT, NO HOLDOWNS SW #10 REAR OF GUST BDRM 2 TOTAL LOAD = 218.00 x 15.00/2 = 163 5. 00 LBS. SHEAR WALL LENGTH = 12.00 FT. SHEAR WALL = 1635.00 / 12.00 = 13 7. 00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O.0 SILL PLATE NAILING MAX. DRAG = 480.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 81.00 LBS NO SIGNIFICANT UPLIFT, NO HOLDOWNS cl, SHEAR WALL DESIGN SW 911 FRONT OF GUEST BDRM BATH 1 TOTAL LOAD = 218.00 x 25.00/2 = 2725.00 LBS. SHEAR WALL LENGTH = 11.00 + 11.50 = 22.50 FT. SHEAR WALL = 2725.00 / 22.50 = 121.00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O.0 SILL PLATE NAILING MAX. DRAG = 46.75 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 571.00 LBS NO SIGNIFICANT UPLIFT, NO HOLDOWNS SW #12 REAR OF GUEST BDRM 1 TOTAL LOAD = 218.00 x 25.00/2 = 2725.00 LBS. SHEAR WALL LENGTH = 7.00 FT. SHEAR WALL = 2725.00 / 7.00 = 389.00 PLF USE SHEAR WALL TYPE 11 WITH 16D'S AT 3" O.0 SILL PLATE NAILING MAX. DRAG = 907.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3192.00 LBS USE (2) SIMPSON CS16 STRAPS EACH END SW #13 FRONT OF CAR GARAGE TOTAL LOAD = 480.00 x 15.00/2 = 3600.00 LBS. SHEAR WALL LENGTH = 912.00 FT. SHEAR WALL = 3600.00 / 12.00 = 3 00. 00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1920.00 LBS. USE (14) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2399.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW # 14 REAR OF CAR GARAGE TOTAL LOAD = 480.00 x 15.00/2 = 3600.00 LBS. SHEAR WALL LENGTH = 912.00 FT. SHEAR WALL = 3600.00 / 12.00 = 300.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1920.00 LBS. USE (14) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2399.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW # 15 FRONT OF GYM TOTAL LOAD = 480.00 x 25.00/2 = 6000.00 LBS. SHEAR WALL LENGTH = 11.50 FT. SHEAR WALL = 6000.00 / 11.50 = 522.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 1920.00 LBS. USE (22) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4450.00 LBS USE SIlVIPSON HDU5 HOLDOWN EACH END SHEAR WALL DESIGN SW #16 REAR OF GARAGE TOTAL LOAD = 480.00 x 25.00/2 = 6000.00 LBS. SHEAR WALL LENGTH = 4.50 + 5.50 = 10.00 FT SHEAR WALL = 6000.00 / 10.00 = 600.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 1998.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4750.00 LBS USE SIMPSON HDUS HOLDOWN EACH END SW #17 LEFT OF M. SUITE 2 CLOSET TOTAL LOAD = 275.00 x 10.00/2 = 1375.00 LBS. SHEAR WALL LENGTH = 18.00 FT. = 16.00 AROUND WINDOW SHEAR WALL = 1375.00 / 16.00 = 85.00 PLF USE SHEAR WALL TYPE 10 WITH 16D'S AT 6" O.0 SILL PLATE NAILING MAX. DRAG = 446.0 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 271.00 LBS NO SIGNIFICANT UPLIFT, NO HOLDOWNS SW #18 LEFT OF M. UITE 2 M. BATHROOM TOTAL LOAD = 275.00 x 29.00/2 = 3988.00 LBS. SHEAR WALL LENGTH = 16.00 FT. SHEAR WALL = 3988.00 / 16.00 = 249.00 PLF USE SHEAR WALL TYPE 1 I WITH 16D'S AT 4" O.0 SILL PLATE NAILING MAX. DRAG = 979.00 LBS. USE (8) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1571.00 LBS USE (1) SIMPSON CS16 STRAPS EACH END SW #19 LEFT OF M. SUITE 2 TOTAL LOAD = 275.00 x 41.00/2 = 5638.00 LBS. SHEAR WALL LENGTH = 19.00 FT. SHEAR WALL = 5638.00 / 19.00 = 297.00 PLF USE SHEAR WALL TYPE I WITH 16D'S AT 4" O.0 SILL PLATE NAILING MAX. DRAG= 1409.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2017.00 LBS USE (2) SIMPSON CS16 STRAPS EACH END SW #20 RIGHT OF M. SUITE 2 TOTAL LOAD = 275.00 x 23.00/2 = 3163.00 LBS. SHEAR WALL LENGTH = 5.25 + 4.25 = 9.50 FT. SHEAR WALL =3163.00/9.50=333.00PLF USE SHEAR WALL TYPE 11 WITH 16D'S AT 4" O.0 SILL PLATE NAILING MAX. DRAG = 799.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3117.00 LBS USE (2) SRVIPSON CS16 STRAPS EACH END 8 SHEAR WALL DESIGN SW #21 LEFT OF OFFICE TOTAL LOAD = 235.00 x 30.00/2 = 3525.00 LBS. SHEAR WALL LENGTH = 12.00 FT. SHEAR WALL = 3525.00 / 12.00 = 294.00 PLF USE SHEAR WALL TYPE 11 WITH 16D'S AT 4" O/C SILL PLATE NAILING MAX. DRAG= 2034.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1736.00 LBS USE (2) SIMPSON CS16 STRAPS EACH END SW #22 LEFT OF BDRM 3 TOTAL LOAD = 625.00 x 20.00/2 = 6250.00 LBS. SHEAR WALL LENGTH = 22.00 FT. SHEAR WALL = 6250.00 / 22.00 = 284.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 1160.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1557.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #23 RIGHT OF BATHROOM 3 TOTAL LOAD = 625.00 x 42.00/2 = 13125.00 LBS. SHEAR WALL LENGTH = 17.50FT. SHEAR WALL = 13125.00 / 17.50 = 750.00 PLF USE SHEAR WALL TYPE 14 WITH 5/8" x 12" A.B. AT 12" O.0 W/ 3x SILL MAX. DRAG = 2920.00 LBS. USE (24) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 6658.00 LBS USE SIMPSON HDU8 HOLDOWN EACH END SW #24 RIGHT OF BDRM 2 TOTAL LOAD WIND INWARD = 625.00 x 50.00/2 x 0.8/1.3 = 9615.00 LBS. TOTAL LOAD SEISMIC= 317.00 x 50.00/2 = 7925.00 LBS. SHEAR WALL LENGTH = 16.00 FT. SHEAR WALL = 9615.00 / 16.00 = 601.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 2404.00 LBS. USE (18) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 5209.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #25 REAR OF BDRM 3 TOTAL LOAD WIND = 625.00 x 41.00/2 = 12813.00 LBS. TOTAL LOAD SEISMIC= 317.00 x 41.00/2 + 235.00 x 31.00/2 + 348.00 x 15.00/2 + 117.00 x 32.00/2 = 14623.00 LBS. SHEAR WALL LENGTH = 5.25 + 6.00 + 6.75 + 6.00 + 6.25 = 30.25 FT. SHEAR WALL = 14623.00 / 30.25 = 484.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 2924.00 LBS. USE (22) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 5209.00 LBS USE SMPSON PIDU5 HOLDOWN EACH END 0 SHEAR WALL DESIGN SW #26 LEFT OF COAT AND M SUITE 1 TOTAL LOAD WIND INWARD = 625.00 x 41.00/2 x 0.8/1.3 = 7885.00 LBS. TOTAL LOAD SEISMIC= 235.00 x 31.00/2 + 117.00 x 51.00/2 + 102.00x 22.00/2 = 8125.00 LBS, SHEAR WALL LENGTH = 6.00 + 8.00 + 8.00 = 22.00 FT. SHEAR WALL = 8125.00 / 22.00 = 370.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 1625.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3009.00 LBS USE SHVIPSON HDU5 HOLDOWN EACH END SW #27 LEFT OF SHOWER TOTAL LOAD = 215.00 x 16.00/2 = 1720.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 1720.00 / 8.00 = 215.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 688.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1550.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #28 REAR OF M SUITE 2 TOTAL LOAD = 511.00 x 24.00/2 = 6132.00 LBS. SHEAR WALL LENGTH = 5.00 + 10.00 = 15.00 FT. SHEAR WALL = 6123.00 / 15.00 = 409.00 PLF USE SHEAR WALL TYPE 12 WITH 16D'S AT 3" O.0 SILL PLATE NAILING MAX. DRAG = 1530.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 30125.00 LBS USE (2) SUVIPSON CS16 STRAPS EACH END SW #29 FRONT OF M. SUITE 2 TOTAL LOAD WIND= 275.00 x 24.00/2 + 450.00 x 39.00/2 = 12075.00 LBS. TOTAL LOAD SEISMIC = 511.00 x 24.00/2 + 313.00 x 39.00/2 = 12235.00 LBS. SHEAR WALL LENGTH = 11.00 + 9.00 + 10.00 = 30.00 FT. SHEAR WALL = 12235.00 / 30.00 = 408.00 PLF USE SHEAR WALL TYPE 12 WITH 16D'S AT 3" O.0 SILL PLATE NAILING MAX. DRAG = 2243.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2429.00 LBS USE (2) SEVIPSON CS 16 STRAPS EACH END SW #30 REAR OF M. SUITE 1 TOTAL LOAD = 450.00 x 39.00/2 + 345.00 x 20.00 = 12225.00 LBS. SHEAR WALL LENGTH = 20.00 FT. SHEAR WALL = 12225.00 / 20.00 = 611.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 2445.00 LB S. USE (18) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 5233.00 LBS USE SRVIPSON HDU5 HOLDOWN EACH END SHEAR WALL DESIGN SW #31 REAR OF FIRST FLLOR TOTAL LOAD = 746.00 x 26.00/2 = 9698.00 LBS. SHEAR WALL LENGTH = 5.50 + 10.00 + 7.00 = 22.50 FT. SHEAR WALL = 9698.00 / 22.50 = 431.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 2445.00 LBS. USE (18) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3393.00 LBS USE SIM[PSON HDU5 HOLDOWN EACH END SW #32 AT HALLWAY AND FRONT OF KITCHEN TOTAL LOAD WI D= 625.00 x 26.00/2 + 450.00 x 36.00/2 = 16225.00 LBS. TOTAL LOAD SEISMIC= 746.00 x 26.00/2 + 313.00 x 36.00/2 = 15332.00 LBS SHEAR WALL LENGTH = 5.50 + 10.00 + 7.00 + 7.00 = 29.50 FT. SHEAR WALL = 16225.00 / 29.50 = 550.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG= 1955.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4491.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #33 STEEL COLUMNS AT REAR PATIO CANTILEVER STEEL COLUMN R=1.5 SEISMIC FACTOR = 348.00 x 6.5 / 1.5 = 1508 PLF TOTAL SEIMIC LOAD = 1508 x 15.00 / 2 = 113100 LBS USE (4) STEEL COLUMNS LATERAL LOAD AT EACH COLUMN = 11310 / 4 = 2800.00 LBS = 2.8 K SEE STEEL COLUMN AND FLAG POLE FOOTING DESIGN W #34 TYP STEEL COLUMN AT REAR ENTRY PATIO AND GREAT ROOM CANTILEVER STEEL COLUMN R=1.5 TOTAL LOAd = 25.00 x 26.00 x 0.145 =94.25 SEISMIC FACTOR = 94.25 x 6.5 / 1.5 = 408.00 TOTAL SEMC LOAD = 408.00 x 12.00 / 2 = 2450.00 LBS USE (2) STEEL COLUMNS LATERAL LOAD AT EACH COLUMN = 2450 / 2 = 1226.00 LBS = 122K SEE STEEL COLUMN AND FLAG POLE FOOTING DESIGN SW #35 FRONT OFM. SUITE 1 TOTAL LOAD = 345.00 x 20.00 = 3450.00 LBS. SHEAR WALL LENGTH = 7.00 + 8.00 = 15.00 FT. SHEAR WALL = 3450.00 / 15.00 = 230.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 1526.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1940.00 LBS USE SIlVIPSON HDU5 HOLDOWN EACH END (B SHEAR WALL DESIGN SW #36 FRONT OF MECHANICAL ROOM TOTAL LOAD = 215.00 x 10.00/2 = 1075.00 LBS. SHEAR WALL LENGTH = 3.25 + 3.25 = 6.50 FT. SHEAR WALL = 1075.00 / 6.50 = 166.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 OK WITH REDUCTION FOR HEIGHT TO WIDTH RATIO MAX. DRAG = 625.00 LBS. USE (6) 16D' S PER TOP PLATE SPLICE MAX. UPLIFT = 1491.00 LBS USE SIM[PSON HDU5 HOLDOWN EACH END SW #37 FRONT OF BATH TOTAL LOAD = 215.00 x 16.00/2 = 1720.00 LBS. USE HARDY FRAME USE (1) HFX-24x10 1 1/8" HS TOTAL CAPACITY 2450.00 LBS. SEE FOUNDATION DESIGN SW #38 REAR OF BATH TOTAL LOAD = 215.00 x 16.00/2 = 1720.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 1720.00 / 8.00 = 215.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 850.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1350.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END D Reza PROJECT : 'LOT 128, N1An(5/7N CLUE PAGE: . _; ;( .:' .<_;: •. Asgharpau CLIENT: TSHEAR-WALL 17, LEFT..OFCLOSET. DESIGN BY: r JOB NO.: 12(]16' ,; DATE . sisr2uTz"; REVIEW BY: Wood Shear Wali with an Opening Based on IBC 06 f COO 071 NDS 05 V diD INPUT DATA — — LATERAL FORCE ON DIAPHRAGM: VlkWIND = ,, : 77 • .., plf,forwind M (SERVICE LOADS) Vdie� SFIWIc plf,for selsmic _ DIMENSIONS: L, = 7 .:. .. ft, Lz = SCRAP 2 . R Ls = ..::9 :.. it Ht = 3. .. ft, H2 = A. tl , H3 = 3 .:. ft ^� w KING STUD SECTION 1. pcs, b = ..:2..... I:: In , h = ..::: in _ SPECIES (1 = DFL, 2= SP) 1. DOUGLAS FIR -LARCH GRADE (1, 2.3.4, 5, or 6 ) .4 ... No. 2 _ EDGE STUD SECTION 2 I:rs, i2 = 2 in • h = G :.:: ' in = SPECIES (1 = DFL, 2 = SP) 1 DOUGLAS FIR -LARCH GRADE (1, 2, 3, 4, S. or 6 ] c : No. 2 PANEL GRADE (0 or 1) = ::: 1 ;' .. •:= Sheathirla and Single -Floor MINIMUM NOMINAL PANEL THICKNESS = :r : in COMMON NAIL SIZE (0=6d, 1=8d, 2=10d) 11 THE SHEAR WALL DESIGN IS ADEQUATE. SPECIFIC GRAVITY OF FRAMING MEMBERS 0.52 STORY OPTION (1=ground level, 2=upper leveq 2 ... upper level shear wall SUMMARY BLOCKED 3/8 SHEATHING WITH 8d COMMON NAILS 0 6 in O.C. BOUNDARY & ALL EDGES / 12 in O.C. FIELD, SILL PLATE ATTACHMENT#14 WOOD SCREWSx 5" LONG AT 4" O.C. HOLD-DOWN FORCES: TL = 0.77 k , TR = 0.77 k (USE CS16 SIMPSON HOLD-DOWN) MAX STRAP FORCE: F = 0.13 k (USE SIMPSON CS22 OVER WALL SHEATHING WrFH FLAT BLOCKING) KING STUD: 1 - 2" x 6' DOUGLAS FIR -LARCH No. 2, CONTINUOUS FULL HEIGHT. EDGE STUD: 2 - 2" x 6' DOUGLAS FIR -LARCH No. 2, CONTINUOUS FULL HEIGHT. SHEAR WALL DEFLECTION: A = 0.15 in 1 } V + V/2 V/2 �1 z L1 2 L TL TR ASSUME 1NFi•QCTUa}j_POINT• AT MID(7f„� []F WINDe7W L1 L2/2 L2/2 L3 F1 _ I F2 _ F2 F31 •� F4 I 2 F4 F4 I T FS F6 F7 FS F9 F10 F11 F12 F5 _ n F8 N F5 F8 F13 F14 F5 F-B L 7 8 F5 FS F15 F16 F17 F18 F5 _ F19 F8 _ F20 a �I s 9 � F21 F21 � 1 1 � F21 I 12 F22 + F23 F23 11 F24 TL FR1= — 9 DDY I Wr2 IYI 0 LIAL PAN V _i R cont'd ANALYSIS CHECK MAX SHEAR WALL DIMENSION RATIO h /w = 0.6 < .5 :':..;. [Satisfactory] DETERMINE FORCES & SHEAR STRESS OF FREE -BODY INDIVIDUAL PANELS OF WALL INDMDUAL PANEL W (ft) H (ft) MAX SHEAR STRESS (pit) NO. FORCE (M NO. FORCE (Ibf) 1 7.00 3.00 70 F1 488 F13 385 2 1.00 3A0 128 F2 128 F14 385 3 1.00 3.00 128 F3 642 F15 561 4 9,00 3.00 71 F4 385 F16 176 5 7.00 2.00 88 F5 616 F17 171 6 9.00 2.00 86 F6 128 F18 556 7 7.00 2.00 88 F7 128 F19 128 8 9,00 2.00 86 F8 770 F20 128 9 7.00 3.00 70 F9 209 F21 385 10 1.00 3.00 128 F10 176 F22 486 11 1.00 3.00 128 F11 171 F23 128 12 9.00 3.00 71 F12 214 F24 642 DETERMINE REQUIRED CAPACITY va = 128 plf, ( 1 Side Panel Required, the Max Nail Spacing = THE SHEAR CAPACnIFS PFR IRC Tahla 7)gnR d 1 11 IRr`T.hle 9'i11_I_1 Panel Grade Common Nail Min. Pcnotmti (m) Min. 'lluckness Cm) Blocked Nail Spacing Boundary @All Edges 6 4 1 3 2 Sheathing and Single -Floor 8d 1 1/2 1 318 220 1 320 1 410 530 Ivurc. 1 Ile 11 d.111.: - a1 i- 11u11- s nave Ieuureu oy spiacn[c gTavity Tartar per ltm iu nrae a, VE FLOOR SILL PLATE ATTACHMENT (NDS 2005, Table 11Q & Table 11L) SILL PLATE ATTACHMENT #14 WOOD SCREWS x S' LONG AT N' O.C. THE HOLD-DOWN FORCES- 6 in ) Va. (plf) Wall Seismic at mid -story (Ibs) Overturning Moments (ft-lbs) Resisting Moments (ft-lbs) Safety Factors Net Uplift (Ibs) SEISMIC 50 288 10440 Left 0 0.9 Tr = 580 Right 0 0.9 Tp = 580:� WIND 77 13880 Left 0 213 TL = 770. Right 0 2G TR = 770 \l Holddown SIMPSON (I L is I R Values snould include upper level UPLIFT forces if applicable) �CK MAXIMUM SHEAR WALL DEFLECTION: ( IBC Section 2305.3.2) 3 A=AB.dig+Ashes+ANau Ly+Ac&.i m fy,=!y p +vah+0.75he„+haa = 0.151 in, Aso < EALw Gt L.&x ,allowable, Aso = 0 429 in Where: v. = 128 plf, , ASD L„, = 18 ft E = 1.7E+06 psi [Satisfactory] (ASCE 7-0512.8.8) A = . 16.50.: in` h = 10 ft G = 9.0E+04 psi Ca = :: 4. t = 0721 in en = 0.000 in d. = 0:1.5 r in (ASCE 7-05 Tab 12.2-1 & Tab l l .5-1) A. = ::0:02 '. h� (ASCE 7-05 Tab 12.12-1) KING STUD CAPACITY P. = A*' kips Fr = 1350 psi Cc = 1.60 Cp = 0.38 A = 8.25 in260 E = 10 ksi CF = 1.10 F, = 894 psi > f� = 21 psi [Satisfactory] EDGE STUD CAPACFFY PRE= ::0.77 : kips, (this value should include upper level DOWNWARD loads if applicable) F� = 1350 psi Co = 1.60 Cp = 0.38 A = 16.50 in2 E = 1600 ksi CF = 1.10 F�= 894 psi > f� = 47 psi [Satisfactory] 63 RA PROJECT: LOT.92i3,'hAADISOhi CLUE PAGE CLIENT: SHEA WALL 6M RIGHT OF COVERED FA110 AT REAR OF }iCUSE DESIGN BY RA Structural JOB NO.: DATE: REVIEW BY: RA Cantliever Column & Footing Design Based on AISC 360, AGI 318, and 160 1805.7 INPUT DATA & DESIGN SUMMARY COLUMN SECTION (Tube, Pipe, or WF) ..............:.....:....:... .. .. W Shape COLUMN YIELD STRESS Fy = :. 36 ksi CANTILEVER HEIGHT H = 10-. It COLUMN TOP LATERAL LOAD F = =.` .28...s: F kips, ASD (Strong Axis Bending only) COLUMN TOP GRAVITY LOAD P = ':. 5- .` kips, ASD DIAMETER OF POLE FOOTING b = 3 ft ALLOW SOIL PRESSURE Qe = 1.5 ksf LATERAL SOIL CAPACITY Pp = 0-1 ksf / ft RESTRAINED @ GRADE ?(1=yes,0=no) 0 No Use 3 ft dia x 6.65 ft deep footing unrestrained @ ground level b" THE DESIGN IS ADEQUATE. YSIS K COMBINED COMPRESSION AND BENDING CAPACITY OF COLUMN (AISC 360-05, H1) Pr+8�M" M,x+ M 'y for Pr>_0.2 8 1 PC 9 May) PC = 0.97 < 1.0 [Satisfactory] Pr + M1+Mry for Pr <0.2 2Pc Pc Where Pr = 5.00 kips MrX = 28.00 ft-kips M,y = 0:: . ft-kips KL y = 20 . ft, weak axis unbraced axial length Pc = Pn SJc = 43 11.67 = 25.49 kips, (AISC 360-05 Chapter E) > Pr [Satisfactory] Mcx = Mn 7 Ob = 53.34 / 1.67 = 31.94 ft-kips, (AISC 360-05 Chapter F) > M„r [Satisfactory] Mcy = Mn / ,Qb = 17.07 11.67 = 10.22 ft-kips, (AISC 360-05 Chapter F) > Mry [Satisfactory] DESIGN POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8) By trials, use pole depth, d = 5.655 ft Lateral bearing @ bottom, S3 = 2 Pp Min(d , 12') = 3.39 ksf Lateral bearing @ d/3, S, = 2 Pp Mind13, 12') = 1.13 ksf Require Depth is given by 2�1+ 1� 4.361, for nonconstrained d = J 5.655 ft [Satisfactory] 4.25Ph for constrained bS3 Where P = F = 2.80 kips A = 2.34P/(b Sj) = 1.93 h = Mmax /F = 10.00 ft CHECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2) 9s0il = P/(rr b214) = 0.71 kA (net weight of pole footing included.) < Qe [Satisfactory] CHECK STRONG AXIS LATERAL DEFLECTION 0=FH3= - 3EI 0.74 in < 240 - 1.00 in [Satisfactory] RA PROJECT: I o7... _2e, raaD� r.clug PAGE: CLIENT: SI Eti uvA[e T ENTRY.PORGH. - DESIGN BY: RA Structural JOB NO.: DATE: REVIEW BY: RA Cantilever Column & Footing Design Based on AISC 360i ACI 318; and IBC 1805,7 INPUT DATA & DESIGN SUMMARY COLUMN SECTION (Tube, Pipe, or WF) Fi556K6X318 •;. COLUMN YIELD STRESS Fy = .:.36 .:: ksi CANTILEVER HEIGHT H = 9 :ft COLUMN TOP LATERAL LOAD F = .1.267 kips, ASD (Strong Axis Bending only) COLUMN TOP GRAVITY LOAD P = 3 kips, ASD DIAMETER OF POLE FOOTING b = 3 ft ALLOW SOIL PRESSURE Qe = 7. .. ksf LATERAL SOIL CAPACITY PP = 0,3 ksf / ft RESTRAINED @ GRADE ?(1=yes,0=no) 0. .. No Use 3 ft dia x 4.10 ft deep footing unrestrained @ ground level THE DESIGN IS ADEQUATE. Tube m U�r4 ANALYSIS CHECK COMBINED COMPRESSION AND BENDING CAPACITY OF COLUMN (AISC 360-05, H1) Pr + g +Al,y for P' >_ 0.2 PC 9�Lj M" J PC = 0.42 < 1.0 [Satisfactory] P' + (M, Af-+AlI'y for P'<0.2 2Pc Mcy) PC Where Pr = 3.00 kips Mrx = 11.40 ft-kips My = 0 ft-kips iCL y = 18 It, weak axis unbraced axial length PG = P„ 1 S7c = 170 / 1.67 = 101.87 kips, (AISC 360-05 Chapter E) > Pr [Satisfactory] Mcx = Mn 7 Db = 47.40 / 1.67 = 28.38 ft-kips, (AISC 360-05 Chapter F) > Mrx [Satisfactory] Mcy = Mn / Ob = 47.40 / 1.67 = 28.38 ft-kips, (AISC 360-05 Chapter F) > M y [Satisfactory] DESIGN POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8) By trials, use pole depth, d = 4.095 ft Lateral bearing @ bottom, S3 = 2Pp Mind, 12') = 2.46 ksf Lateral bearing @ d/3, S, = 2 Pp Min(d/3 , 12') = 0.82 ksf Require Derrpth is given by 2 L1+ 1 + 4.33 h J for nonconstrained d - J - 4.095 ft [Satisfactory] 4.25PIi for constrained Where P = F = 1.27 kips A= 2.34P/(bS1)= 1.21 h = Mmax /F = 9.00 ft :HECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2) qsa! = P7('r b214) = 0.42 ksf, (net weight of pole footing included.) < Qa [Satisfactory] CHECK STRONG AXIS LATERAL DEFLECTION A=FH3= - 3EI 0.47 in < 2 H/ 240: - 0.90 in [Satisfactory] T1 lZr ..... .......... RA PROJECT: 5hear:Walf #S.FRpNT O.F GARAGE ....: PAGE CLIENT: LOT I.26,:: MADISON CLi3B DESIGN BY: .R:A;`:: Structural JOB NO.: DATE: REVIEW BY' Footinq Des lan of Smear Wail Baser on ACI 31B-05 PUT DATA kLL LENGTH kLL HEIGHT ALL THICKNESS OTING LENGTH FOOTING WIDTH FOOTING THICKNESS FOOTING EMBEDMENT DEPTH ALLOWABLE SOIL PRESSURE DEAD LOAD AT TOP WALL LIVE LOAD AT TOP WALL TOP LOAD LOCATION nALL SELF WEIGHT LATERAL LOAD TYPE (0=wind,1=seismic) HIND LOADS AT WALL TOP P, LW --a� L = . .,.12 ti P w h B = 3 it T= ...1f1 .. ,n Pr D Pr,DL = 1:. kips Pr,LL ="i ::. kips L 1 L w — a = :: 4. ft L Pw = L'- 1 ,. ..: kips 0 wind F = 9-3C ` lops THE FOOTING DESIGN IS ADEQUATE. M = .: 0 ft-kips CONCRETE STRENGTH 2.5 ksi REBAR YIELD STRESS fy = .. GO ' `. ksi TOP BARS, LONGITUDINAL •1 # BOTTOM BARS, LONGITUDINAL 4 :" # 5 BOTTOM BARS, TRANSVERSE # @ 12 in o.c. ANALYSIS HECK OVERTURNING FACTOR (IBC 06 1605.2.1, 1801.2.1, & ASCE 7-05 12.13.4) F=MR/Mo= 3.37 > 1.6/09 for wind Where Pf = 7.83 kips (footing selfweight) Mo = F (h + D) + M = 16 ft-kips (overturning moment) MR = (Pr,DL) (LI + a) + Pf (0.5 L) + Pw (LI + 0.5Lw) = 53 SOIL CAPACITY (ALLOWABLE STRESS DESIGN) Ps = 5.4 kips (soil weight in footing size) P = (Pr,DL + Pr,LL) + Pw + (Pf - Ps) = 4.53 kips (total vertical net load) MR = (Pr,DL + Pr, LL) (L, + a) + Pf (0 5 L) + Pw (Lf + 0.5Lw) = 58 e = 0.5 L - (MR - MO) / P = -3.30 It (eccentricity from middle of footing) Pf 1+Lei L I _ for e < — qA,64x = BL 6 2P for e > L = -0.08 ksf — 3B(O.5L —e)' 6 Where e = -3.30 ft, < (L / 6) FOOTING CAPACITY (STRENGTH DESIGN) Mu,R = 1-2 IPr,DL (LI + a) + Pf (0.5 L) + Pw (LI + 0.5Lw)] + 0.5 Pr, LL(Lf + a) _ Mu,o = 1.6 [F(h + D) + M] = 25 ft-kips Pu = 1.2 (Pr pL + Pf + Pw) + 0.5 Pr, LL = 11 kips eu = 0-5L - (Mu,R - Mu,o) / Pu = 2.37 ft Pu 1+6e L L for eu c BL 6 = 0.69 ksf 2p , for eu > L 3B(O.5L—eu) 6 < == Not Reqiiired [Satisfactory] ft-kips (resisting moment without live load) ft-kips (resisting moment with live load) < 4/3qa [Satisfactory] 66 ft-kips 0 qu,na PU w 1M 1Z, (contd) BENDING MOMENT & SHEAR AT EACH FOOTING SECTION .� am -10 10 1 0 -10 ® V -20 1- I-- Where P- iOmin = 0.0018 .f y 0.85/3, f c Eu p„�,,y = T - 0.0129 [Satisfactory] .fY su+Et Section 0 1/10 L 2/10 L 3/10 L 4/10 L 5l10 L 6/10 L 7/10 L 8/10 L 9/10 L L Xu (ft) 0 1.20 2.40 3.60 4.80 6.00 720 8.40 9.60 10.80 12-00 Pu.w (klf) 0.0 0.0 00 0.0 0.0 0.9 0.0 0.0 0-0 0.0 0.0 Mu,w (ft-k) 0 0 0 0 0 -14 -29 -31 -33 -35 -37 Vu,w (BPS) 0 0 0 0 0 -21 -2 -2 -2 -2 -2 Pu,f {ksf? 0-3 0.3 0.3 0.3 03 0.3 0.3 0.3 0.3 0.3 0.3 Mu,f (ft-k) 0 -1 -2 -5-9 -14 -20 -28 -06 �6 S6 Vuf (kips) 0 -1 -2 -3 -4-5 -0-7 � -0-9 qu (kst] -0.7 -0.6 -0.5 -0.5 -0.4 -0.3 -0.2 -0.2 -0.1 0.0 0.0 Mu,q Pk) 0 1 5 12 20 30 42 54 67 80 94 Vu,q (kips) 0 2 4 6 8 9 10 11 11 11 11 E Mu (ft-k) 0 1 3 7 11 2 -7 � -2 -1 0 E Vu (kips) 0 1 3 3 4 -17 2 2 2 1 0 Location Mu,� d (in) PregD (�prov0 Vu,max $Vt = 2 � b d (f�)os Top Longitudinal -7 ft-k 14.69 0.0002 0.0023 17 kips 45 kips Bottom Longitudinal 11 ft-k 14.69 0.0018 0.0023 17 kips 45 kips Bottom Transverse 0 ft-k / ft 14.19 0.0000 0.0000 0 kips / ft 14 nips / Reza PROJECT bar our CLIENT _- JOB NO. Maximum Load For 2 sq. ft. Pad Footing (1500psf) DATE: PAGE: DESIGN BY: R.A. REVIEVV BY . R.A. INPUT DATA DESIGN SUMMARY COLUMN WIDTH c, = 0 in FOOTING WIDTH COLUMN DEPTH c2 = 0:.:., in FOOTING LENGTH BASE PLATE WIDTH b, = 4: in FOOTING THICKNESS BASE PLATE DEPTH b2 = 4 in LONGITUDINAL REINF. FOOTING CONCRETE STRENGTH f.' = Z5 ksi TRANSVERSE REINF RE13AR YIELD STRESS fy = 40: ksi AXIAL DEAD LOAD PDT = 2.5- k AXIAL LIVE LOAD PLL = 2.& k LATERAL LOAD (O=WIND, 1=SEISMIC) _ -_ 1.; Seismic,SD SEISMIC AXIAL LOAD PLAT = 0; k, SD SURCHARGE % = 0: ksf SOIL WEIGHT ws = 00,1 kcf FOOTING EMBEDMENT DEPTH Dr = -2; `- ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Qa = 1.5 ksf FOOTING WIDTH B = 2 ft FOOTING LENGTH L = 2 ft BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. ANALYSIS DESIGN LOADS (IBC SEC.1605.3 2 & ACI 318-05 SEC.9 2 1) CASE 1: DL + LL P = 11 kips 1.2 DL + 1.6 LL CASE 2: DL+LL+E/1.4 P = 5 kips 1 2 DL + 1.0 LL + 1.0 E CASE 3: 0.9 DL + E / 1.4 P = kips 0.9 DL + 1.0 E ( SOIL BEARING CAPACITY (ACI 318-05 SEC.15 2.2) B = 2.00 ft L = 2.00 ft T = 12 in 3 # 4 @ 9 inoc. 3 # 4 @ 9 moc. F T d Q -- P CASE 1 CASE 2 CASE 3 HL +g,t.+ (0.1 5 — ,,,$)T = 1.29 ksf, 1.29 ksf, 0.60 ksf q MAX k Q a . [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads GN FOR FLEXUR E (ACI 318-05SECA5.4.2. 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2. & 12.5) 0.85f,-I — p=Pu.n - fil j Y eu+Er LONGITUDINAL d 8.75 b 24 q u,max 1.75 Mu 1.47 P 0.000 Pmin 0.000 As 0.07 RegD 1 # 4 Max. Spacing 18 in o.c. USE 3 # 4 Q 9 in o.c. Pmax 0.019 Check Pprod ' Pmax [Satisfactory] L ; Pu = 7 kips Pu = 6 kips Pu = 2 kips P,rr,. _ ,MIN 0,0018 T 4 P d 3 ) TRANSVERSE 8.50 24 1.75 1.47 0.000 0.000 0,08 1 # 4 18 in o.c. 3 # 4 @ gin o.c. 0.019 [Satisfactory] (2) (coned) CHECK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11 3) of,,,=2obd f�, LONGITUDINAL TRANSVERSE V 0.66 0.73 0 0.75 0.75 Wn 15.8 15.3 Check V„ <Wn [Satisfactory] [Satisfactory] PUNCHING SHEAR (ACI 318-05 SEC.15.5 2, 11.12.1.2. 11 12.6. & 13.5 3.2) of',=(2+Y)Of,14p = 54.98 kips where = 0.75 (ACI 318-05, Section 9.3.2.3) P� = ratio of long side to short side of concentrated load = 1.00 bo = c, + c2 + b1 + b2 + 4d = 42.5 in AP = bo d = 366.6 in y = MIN(2 , 4 / p� 40 d I bo) = 2.0 ]— rh'+L'+dh:+c'+d 5.63 kips < d V n [Satisfactory] Vu=Pu,Max Bj 2 Y 2 )]= 0\�u Reza PROJECT : Max. Load For 2.5 sq. it. Pad Footing (15.00psf PAGE As har OUr CLIENT: DESIGN BY: R.A. g p JOB NO.: DATE: REVIEW BY .- R.A. Pad Footling Desfarr 6asat �ii ACE.'�3ft+. INPUT DATA COLUMN WIDTH c, = COLUMN DEPTH c2 = BASE PLATE WIDTH b, = BASE PLATE DEPTH bZ = FOOTING CONCRETE STRENGTH fc, = REBAR YIELD STRESS fy = AXIAL DEAD LOAD PDT _ AXIAL LIVE LOAD PLL _ LATERAL LOAD (O=WIND, 1=SEISMIC) _ SEISMIC AXIAL LOAD PLAT = SURCHARGE qs = SOIL WEIGHT ws - FOOTING EMBEDMENT DEPTH Df = FOOTING THICKNESS T = ALLOW SOIL PRESSURE Qa = FOOTING WIDTH B = FOOTING LENGTH L = BOTTOM REINFORCING # THE PAD DESIGN IS ADEQUATE. 0 in 0 in 4 in 4 -n 2,5 ksi 40 ksi 4.5 k 4.5 k 1 Seismic,SD 0 k, SD 0 ksf 0.11 kcf 2 ft 12 in 1:5 ksf 2.5 ft 2.5 ft 4 ANALYSIS DESIGN LOADS (IBC SECA605.3.2 & ACI 318-05 SEC.9.2.1) DESIGN SUMMARY FOOTING WIDTH FOOTING LENGTH FOOTING THICKNESS LONGITUDINAL REINF TRANSVERSE REINF. B = 2.50 ft L = 2.50 ft T = 12 in 3 # 4 @ 12 in o-c. 3 # 4 @ 12 moc IP T -� 1 ��7% c' j 1 n j Z' 1 b i CASE 1: DL + LL P = 9 kips 1.2 DL + 1.6 LL Pu = 13 kips CASE 2: DL+LL+E11.4 P - 9 kips 1.2DL+1.0LL+1.0E Pu = 10 kips CASE 3: 0-9 DL+ 1-11.4 P 4 kips 0.9 DL+ 1.0 E Pu = 4 kips CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2) CASE 1 CASE 2 CASE 3 q,,.,, = gL +q ti.+ (0.15 — it,,)T = 1.48 ksf, 1.48 ksf, 0.69 ksf q MAX < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads. DESIGN FOR FLEXURE (ACI 318-05 SEC.15.4 2,10.2. 10.3.5, 10.5.4. 7.12.2, 12.2. & 12.5) 0.85fc I- I-{i 0.85 Pacer= ��fC Ett -A1IN 0.0018T 4 P= r f1, Pun- ( d • 3P) .y LONGITUDINAL TRANSVERSE d 8.75 8.50 b 30 30 q u max 2.02 2.02 Mu 3.43 3.43 P 0,000 0.001 Amin 0.001 0.001 Aq 017 0.18 RegD 1 7 4 1 # 4 Max. Spgdng 18 in o.c. 18 in o.c. USE 3 # 4 @ 12 in o.c. 3 # 4 12 in o.c. Amax 0.019 0.019 Check Pprod `Pmax [Satisfactory] [Satisfactory] 0 K FLEXURE SHEAR (ACI 318-05 SEC.9 3.2.3, 15.5.2, 11.1.3 1, & 11.3) OT'n = 2obd fc LONGITUDINAL TRANSVERSE Vu 2.21 2.31 0.75 0.75 Wn 19.7 19.1 Check Vu <OVn [Satisfactory] [Satisfactory] CHECK PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2, 11.12.1.2, 11.12.6. & 13.5.3.2) WL'n=(2+Y)0 ,Ap = 54.98 kips where = 0.75 (ACI 318-05, Section 9.3.2.3) p� = ratio of long side to short side of concentrated load = 1.00 130 = ct + c2 + bi + b2 + 4d = 42.5 in AP = 130 d = 366.6 in y = MIN(2 , 4 I p� , 40 d / b0) = 2.0 Vu=Pu,max 1 BL( 2 +d II -2 -+dJJ= 11.02 kips < 0 V n [Satisfactory] j (cont'd) 0 /� Reza PROJECT: Max. Load For 30 sq. fL Pad Foaling (1500psf) PAGE : 1�;S har OUr CLIENT: DESIGN BY: R.A. g p JOB NO.: DATE : REVIEW BY: R.A. Pad FbbtJn >t pKt!I66d66_-AG14 INPUT DATA DESIGN SUMMARY COLUMN WIDTH c, = 0 in FOOTING WIDTH COLUMN DEPTH cl = 0 in FOOTING LENGTH BASE PLATE WIDTH b, = 4 in FOOTING THICKNESS BASE PLATE DEPTH bZ = 4 in LONGITUDINAL REINF. FOOTING CONCRETE STRENGTH fc' = 2.5 ksi TRANSVERSE REINF- REBAR YIELD STRESS fy = 40 ksl AXIAL DEAD LOAD Pa = 6.5 k AXIAL LIVE LOAD P« = 6.5 k LATERAL LOAD (O=WIND, 1=SEISMIC) = 1 Seismic,S❑ SEISMIC AXIAL LOAD PLAT = 0 k, SO SURCHARGE qs = 0 ksf SOIL WEIGHT ws = 0,1i.• kcf FOOTING EMBEDMENT DEPTH Of = 2 ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Q. = 1.5 ksf FOOTING WIDTH B = 3 ft FOOTING LENGTH L = 3 ft BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. B = 3.00 It L = 3.00 ft T = 12 in 3 # 4 @ 15 inoc 3 # 4 @ 15 inoc 1 % f"td Q �- �— i I f _YS1S ;N LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1) 1: DL + LL P = 13 kips 1.2 DL + 1.6 LL Pu - 18 kips 2: DL + LL + E / 1.4 P = 13 kips 1 2 DL + 1.0 LL + 1 0 E Pu = 14 kips 3� 0.9 DL + E / 1.4 P = 6 kips 0.9 DL + 1.0 E Pu - 6 kips K SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2) CASE 1 CASE 2 CASE 3 grou.r = BL + q S.+ (0.15 - I,,,5) T = 1.48 ksf, 1.48 ksf, 0.69 ksf q MAX < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads. DESIGN FOR FLEXURE (ACI 318-05 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, & 12.5) O.SS j�.I 1- I'0 38M" - 0.85i. f T 4 l E�� -M/h` 0.0018- P= f ,>r� P,iic- fy Eu+EI Pm�- ( d � 3P .y LONGITUDINAL TRANSVERSE d 875 8.50 b 36 36 q u,max 2.02 2.02 Mu 6.09 6.09 p 0.001 0 001 Pmin 0.001 0.001 As 031 0.32 Req❑ 2 # 4 2 # 4 Max. Spacing_ 18 in o.c. 18 in o.c. USE 3 # 4 @ 15 in o.c. 3 # 4 @ 15 in o.c. Pmax 0.019 0.019 Check P,,d ` Pmax [Satisfactory] [Satisfactory] (cont'd) FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3) oG'n=2obd fc LONGITUDINAL TRANSVERSE V 4.17 4.30 0 0.75 0.75 OVA 23.6 23.0 Check V < On [Satisfactory] [Satisfactory] PUNCHING SHEAR (ACI 318-05 SEC.15.5.2, 11.12.1.2. 11.12.6. & 13.5.3.2) of',=(2+y)of,AP = 54.98 kips where = 0.75 (ACI 318-05, Section 9.3.2.3 ) PC = ratio of long side to short side of concentrated load - 1.00 ba = cl+c2+bl+b2+4d = 42.5 in Ap = ba d = 366.6 in y = MIN(2 , 4 / P� , 40 d / bo) = 2.0 _ 1 b, + c, 1l/6 • + cz 1 [Satisfactory) /'u=�'u,max 1 BL( 2 +dJl -� +d1= 16.61 kips < �V„ ry] Reza PROJECT Max. Load For 3.5 sq. ft. Pad Footing (1500psf) PAGE As harn OUr CLIENT: DESIGN BY: R.A. g M JOB NO.: DATE, REVIEW BY : R.A_ Pad Footing Deli n-Based0; AC[ 318-05.' INPUT DATA DESIGN SUMMARY COLUMN WIDTH C, = 0 In FOOTING WIDTH B = 3.50 ft COLUMN DEPTH CZ = 0 in FOOTING LENGTH L = 3.50 ft BASE PLATE WIDTH b, = 4 in FOOTING THICKNESS T = 12 in BASE PLATE DEPTH b2 = 4 in LONGITUDINAL REINF. 4 # 4 @ 12 in D.C. FOOTING CONCRETE STRENGTH f� = 2.5 ksl TRANSVERSE REINF. 4 # 4 @ 12 in o c REBAR YIELD STRESS fy = 40 ksl P AXIAL DEAD LOAD PD1 = 8.5 k T AXIAL LIVE LOAD P, = 8.6 k LATERAL LOAD (O=WIND, 1=SEISMIC) = i Seismic,S❑ �- �� SEISMIC AXIAL LOAD PLAT = 0 k. SD SURCHARGE qs = 0 ksf o d n SOIL WEIGHT Ws = 0.11 kcF FOOTING EMBEDMENT DEPTH Dr = 2 ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Qa = 1.5 ksf FOOTING WIDTH B = 3.5 ft FOOTING LENGTH L = 3.5 ft O7 BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. ANALYSIS DESIGN LOADS pBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1) CASE 1- DL + LL P - 17 kips 1.2 DL+ 1.6 LL Pu = 24 kips CASE 2 DL + LL + E / 1.4 P = 17 kips 1.2 DL + 1.0 LL + 1.0 E Pu = 19 kips CASE 3: 0.9 DL + E 11.4 P = 8 kips 0.9 DL + 1.0 E Pu = 8 kips CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2) CASE 1 CASE 2 CASE 3 gniAx = L + q y.+ (0.15 - m,,S•) T = 1.43 ksf, 1.43 ksf, 0.66 ksf q MAX < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads DESIGN FOR FLEXURE (ACI 318-05 SEC.15.4.2, 10.2, 10 3.5. 10.5.4, 7.12.2, 12.2, & 12.5) l Y 0.85 jc , i - ] - 0.383hn,+,d= j�. - 0-85 j3 jr e jt l P= j PAIAV f, eu+gr Pun.=M/NIr 00018d 3PI • v LONGITUDINAL TRANSVERSE d 8.75 8.50 b 42 42 q u,max 1.94 1.94 Mu 9.44 944 P 0.001 0.001 pmu, 0.001 0.001 As 0-48 0.50 RegD 3 # 4 3 # 4 Max_ . Spacing 18 to D.C. 18 In D.C. USE 4 # 4 Q) 12 in D.C. 4 # 4 A 12 in o.c. Amex 0.019 0-019 Check p rod < pmax [Satisfactory] (Satisfactory] 0 CHECK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3. 15.5.2. 11.1.3.1. & 11.3) of', = 20bd fC LONGITUDINAL TRANSVERSE Vu 6.38 6.52 075 0.75 On 27.6 26.8 Check V. < OVn [Satisfactory] (Satisfactory] PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2, 11,12.1.2, 11.12.6. & 13.5.3 2) �vn=(2+y)0 fCap - 54.98 kips where = 075 (ACI 318-05, Section 9.3.2.3) p� = ratio of long side to short side of concentrated load - 1.00 bo = ct + c2 + b, + bZ + 4d = 42.5 in AP = bo d = 366.6 in yI B� l = MIN(2, 4//(3�, 40d/blo) = 2.0 V I— 2 J l bi+ci+d�l±+c +d)J= 22.28 ki s <:� = Pu, max p 0n [Satisfactory] (confd) 9 Reza PROJECT Max. Load For 4.0 sq. ft. Pad Footing (1500psf) PAGE: AS har ❑UC CLIENT: DESIGN BY: R.A. P JOB NO- - DATE REVIEW BY: R.A Pat! FOWL 'Nseti'txrt-AG1 31: INPUT DATA DESIGN SUMMARY COLUMN WIDTH C, = 0 in FOOTING WIDTH B = 4.00 ft COLUMN DEPTH CZ = 0 In FOOTING LENGTH L = 4.00 ft BASE PLATE WIDTH b, = 4 In FOOTING THICKNESS T = 12 in BASE PLATE DEPTH bZ = 4 in LONGITUDINAL REINF. 4 # 4 @ 14 in o c FOOTING CONCRETE STRENGTH fc' = 2.5 ksl TRANSVERSE REINF. 4 # 4 @ 14 in o.c REBAR YIELD STRESS fy = 40. ksi AXIAL DEAD LOAD PDL = 11.5 k P AXIAL LIVE LOAD PLL = 11.6 k LATERAL LOAD (O=WIND, 1=SEISMIC) - 1 5eismid,SD SEISMIC AXIAL LOAD PLAT - 0 X. SD d% SURCHARGE qs - 0 ksf o `❑ SOIL WEIGHT ws = 0.11 kof FOOTING EMBEDMENT DEPTH Df = 2 ft FOOTING THICKNESS T = 12 in ' C , ALLOW SOIL PRESSURE Qa = 1.5 ksf + FOOTING WIDTH B = 4 ft FOOTING LENGTH L = 4 ft O7 4 BOTTOM REINFORCING # 4 b+ � I THE PAD DESIGN IS ADEQUATE. -- ALYSIS iIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC 9.2.1) �E 1 DL + LL P = 23 kips 1.2 DL + 1.6 LL Pu = 32 kips ;E 2: DL + LL + E / 1.4 P = 23 kips 1.2 DL + 1.0 LL + 1.0 E Pu = 25 klps E 3: 0.9 DL + E / 1.4 P = 10 kips 0.9 DL + 1.0 E Pu = 10 kips ICK SOIL BEARING CAPACITY (ACI 318-05 SECA 5.2.2) P CASE 1 CASE 2 CASE 3 q"'"a BL +q ,.+ (0.15 - m,,t )T = 1.48 ksf, 1.48 ksf, 0.69 ksf q MAX < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads. IGN FOR FLEXURE (ACI 318-05 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7 12.2, 12.2, & 12 5) 085 1- I- Air' 0.85f3 f, c,, P= ` y fy, �rri cr '0 A41N(0.0018d 3P) • LONGITUDINAL TRANSVERSE d 8.75 8.50 G 48 48 q u.max 2.01 2.01 Mu 1479 14.79 p 0.001 0.001 Pmm 0.002 0.002 As 0.76 0.78 RegD 4 # 4 4 # 4 Max. Spacing 18 in o.c. 18 In o.a USE 4 # 4 @ 14 in o.c. 4 # 4 @ 14 in o.c. Pmax 0.019 0.019 Check pProd' Pmax [satisfactory] [Sawactory] 0 (cont'd) FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5,2, 11.1.3.1, & 11.3) of"n=2obd fc LONGITUDINAL TRANSVERSE V, 9.56 9.73 0.75 0.75 mV, 31.5 30.6 Check V. < QV„ [Satisfactory] [Satisfactory] CHECK PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2, 11.12.1.2, 11.12.6. & 13.5 3,2) 01,,,=(2+Y) cAp = 54.98 kips where = 0.75 (ACI 318-05, Section 9.3.2.3) p� = ratio of long side to short side of concentrated load — 1.00 130 = ct+c2+b1+b2+4d = 42.5 in Ap = b0 d = 366.6 in y I ( MIN(2, 41 Dc, 40 d / b0) = 2.0 Vu=Nrr.max1 BLI bi+ci+d)(b'Zc'+d)]= 30.62 kips < 0 V n [Satisfactory] 0 Reza PROJECT: Max. Load For 4.5 s ft. Pad Foofin 150O s 4. q• 9 f P. � PAGE As har our CLIENT: DESIGN BY: R.A g p JOB NO.: DATE: REVIEW BY : R.A. Pad Footiri ��esi n-�B4Sed on ACI 318-05 INPUT DATA DESIGN SUMMARY COLUMN WIDTH C, = 0 in FOOTING WIDTH B = 4.50 ft COLUMN DEPTH C2 = 0 in FOOTING LENGTH L = 4.50 ft BASE PLATE WIDTH b, = 6 In FOOTING THICKNESS T = 15 BASE PLATE DEPTH b2 = 6 in LONGITUDINAL REINF. in 4 # 5 @ 16 FOOTING CONCRETE STRENGTH fc' = 2.5 ksi TRANSVERSE REINF. in o c. 4 # REBAR YIELD STRESS fy = 40 ksi 5 @ 16 in o.c. AXIAL DEAD LOAD PoL = 14.5 k !P AXIAL LIVE LOAD PLL = 14.5 k LATERAL LOAD (O=WIND, 1=SEISMIC) 1 Seismic,SD SEISMIC AXIAL LOAD PLAT 0 k, SD %%777777 SURCHARGE qs = 0 ksf n� d SOIL WEIGHT ws = 011 kef t FI \ i N FOOTING EMBEDMENT DEPTH Df = 2 ft �' ~ FOOTING THICKNESS T = 15 in ALLOW SOIL PRESSURE Qa = 1.5 ksf C FOOTING WIDTH B = 4.5 ft FOOTING LENGTH L = 45 ft m �I BOTTOM REINFORCING # 5 / I J b THE PAD DESIGN IS ADEQUATE. y L ANALYSIS DESIGN LOADS (IBC SEC.1605 3.2 & ACI 318-05 SEC 9.2.1) CASE 1: DL + LL P = 29 kips 1.2 DL + 1.6 LL Pu = 41 kips CASE 2: DL + LL + E / 1.4 P = 29 kips 1 2 DL + 1.0 LL + 1.0E Pu = 32 kips CASE 3: 0.9 DL + E 11.4 P = 13 kips 0.9 DL + 1.0 E Pu = 13 kips CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2) CASE 1 CAS CASE 3 q,r ,.,. = BL +q �.+ (0.15 - ,�,.ti.)T = 1.48 ksf, 48 ksf, 0.69 ksf q MAx < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads. ]ESIGN FOR FLEXURE (ACI 318-05 SEC.15.4.2, 10.2, 10.3.5, 10.5.4. 7.12.2. 12 2, & 12.5) 0.85fc, 1- I- m 0.383h,O" 0.85,6 j, err P= fy J PI/.Ll= fy, ett+er P.im=rLf/NI00018d 3P) J LONGITUDINAL TRANSVERSE d 11.69 11.38 b 54 54 q u,max 2.00 2 00 Mu 20.37 20.37 P _ 0 001 0,001 Pmin 0.001 0.001 A, 0.78 0.80 RegD 3 # 5 3 # 5 Max. Spacing 16 in o.c. 18 in o.c. USE 4 # 5 @ 16 in o.c. 4 # 5 @ 1.6 in o.C. Pmax 0,019 0.019 Check Pproc < Pmax [Satisfactory) [Satisfactory] (2q (cont'd) ECK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3) 0l'n=2obd fc LONGITUDINAL TRANSVERSE Vi, 10.38 10.62 0.75 0.75 Wn 47.3 46.1 Check V < Wn [Satisfactory] [Satisfactory] CHECK PUNCHING SHEAR (ACI 318-05 SEC.15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2) 01',=(2+3')O f,Ap = 100.54 kips where m = 0.75 (ACI 318-05, Section 9.3.2.3) oc = ratio of long side to short side of concentrated load - 1.00 bo = cl + cz + b, + b2 + 4d = 58.1 in AP = be d = 670.3 in y = MIN(2, 4/(3c, 40d/bc) = 2.0 1' —P 1— 1 "+ci+d b' +d11 37.66 kips < V u — u, max BL 2 �� 11 = P n [Satisfactory] EO Reza PROJECT Max. Load For 5.0 sq. ft. Pad Footing (1500 pso PAGE: As har OUr CLIENT. DESIGN BY: R.A. g p JOB NO.: DATE : REVIEW BY: R.A. Pad Foo.#ina D"jcin. Based on ACI 318-05 INPUT DATA DESIGN SUMMARY COLUMN WIDTH c, = 0 in FOOTING WIDTH B = 5.00 ft COLUMN DEPTH c2 = 0 in FOOTING LENGTH L = 5.00 ft BASE PLATE WIDTH b, = 6 in FOOTING THICKNESS T = 18 in BASE PLATE DEPTH bZ = 6 in LONGITUDINAL REINF. 5 # 5 @ 13 in o.c. FOOTING CONCRETE STRENGTH fc' = 2.5 ksi TRANSVERSE REINF. 5 # 5 @ 13 in ox, REBAR YIELD STRESS fy = 40 ksi F AXIAL DEAD LOAD Poi = 18 k AXIAL LIVE LOAD P, = 18 k _ LATERAL LOAD (O=WIND, 1=SEISMIC) = 1 Se;smlc.SD .._ 777777 SEISMIC AXIAL LOAD PLAT = 0 k, SD o SURCHARGE qs = 0 ksf SOIL WEIGHT ws = 0.11 kcf FOOTING EMBEDMENT DEPTH Df = 2 ft FOOTING THICKNESS T = 18 in-f c I ALLOW SOIL PRESSURE Qa = 1.5 ksf �q FOOTING WIDTH B = 5 ft FOOTING LENGTH L = 5 ft �I i BOTTOM REINFORCING # 5 i THE PAD DESIGN IS ADEQUATE. i_ — Y ANALYSIS DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1) CASE 1 DL+ LL P = 36 kips 1.2 DL+ 1 6 LL Pu = 50 kips CASE 2: DL + LL + E / 1.4 P = 36 kips 1.2 DL + 1 0 LL + 1.0 E Pu - 40 kips CASE 3 0.9 DL + E / 1.4 P = 16 kips 0.9 DL + 1.0 E Pu = 16 kips CHECK SOIL BEARING CAPACITY (ACI 318-05 SECA 5.2 2) P CASE 1 CASE 2 CASE 3 qWX = BL +g s+(0.15 — �,,,.)7' = 1.50 ksf, 1 50 ksf, 0.71 ksf q MAX < k Q a , [Satisfactory] where k = 1 for gravity loads, 4/3 for lateral loads. DESIGN FOR FLEXURE (ACI 318-05 SECA 5.4.2. 10.2, 10.3.5, 10.5 4, 7.12.2. 12.2. & 12.5) 085.f, I- P�ci.c — ' e—�� Cie / T 4 l p- S, *Lr Pin.c=M/NI0.0018' 3P /y rI LONGITUDINAL TRANSVERSE d 14.59 14.36 b 50 60 q u,max 2A2 2.62 Mu 28.43 28.43 P 0,001 0.001 Pmin 0.001 0.001 As 0.67 0.88 RegD 3 # 5 3 :/ 5 Max_. Spacing 18 in o.c. 18 in o.c. USE 5 # 5 @ 13 in o.c. 5 # 5 @ 13 in o.c. Pmax 0.019 0.019 Check Pnrud < Pmax [Satisfactory] [Satisfactory] CHECK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3. 15.5.2. 11 1 3.1, & 11.3) 20bd fc LONGITUDINAL TRANSVERSE Vu 11.60 11.87 0.75 0.75 Wn 66.1 64.7 Check VL, < Wn [Satisfactory] [Satisfactory] PUNCHING SHEAR (ACI 318-05 SEC.15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2) �I n—(2+Y)� %1.41, = 15285 kips where m = 0.75 (ACI 318-05, Section 9.3.2.3) (tc = ratio of long side to short side of concentrated load = 1.00 bo = c1 + c2 + bi + b2 + 4d = 70.1 in Ap = bo d = 1019.0 in2 y = MIN(2 , 4 / pc , 40 d / bo) = 2.0 l bz+c•z 1l f 'u = Pu, mar I — BL o (b�+,, + d +d JJ = 46.10 kips < 0 V „ [Satisfactory] (cont'd) FOUNDATION DESIGN SOIL BEARING PRESUURE = 1500 PSF PER SOILS REPORT MAX LOAD = 45.00x41.00/2 + 15.00x18.00/2 + 150xl2xl2/144 = 1383 PLF FOUNDATION WIDTH REQUIRED = (1383.00/1500)xl2 = 9.7" USE IT'WIDE x IT'DEEP FOOTINGS WITH (2) # 5 TOP AND BOTTOM FOOTING CAPACITY = (l500x4Oxl2)/(12xl2) = 5000.00 LBS SOIL BEARING PRESUURE = 1500 PSF PER SOILS REPORT MAX LOAD = 65.00x24.00/2 + 10.00x18.00/2 + 150xl5xl8/144 = 1686 PLF FOUNDATION WIDTH REQUIRED = (1686.00/1500)xl2 = 13.25" USE 15" WIDE x 18" DEEP FOOTINGS WITH (2) # 5 TOP AND BOTTOM FOOTING CAPACITY = (l500x52xl5)/(12xl2) = 8125.00 LBS PADS DESIGN BM # 14 = 7800.00 LBS USE 2'-6" SQ. x 18" DEEP W/ (3) 94 BARS ECH WAY BM # 16 = 12000.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 17 + BM # 18 = 26850.00 LBS USE 4'-6" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 24 = 16550.00 LBS USE 4'-0" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 25 + BM #25A = 18000.00 LBS USE 4'-0" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 30 LEFT + BM # 31= 14300.00 LBS USE 3'-6" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 32 HDR= 19000.00 LBS USE 4'-0" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 35 = 9000.00 LBS USE 3'-0", SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 36 RIGHT= 11000.00 LBS USE 3'-6" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 40 LEFT=10500.00 LBS USE 3'-6" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY a FOUNDATION DESIGN BM # 41 LEFT = 13500.00 LBS USE 3'-6" SQ. x 18" DEEP W/ (4) #4 BARS ECH WAY BM # 45 LEFT = 11750.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM#46LEFT =9550.00LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 50 = 9625.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY j