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12-1253 (SFD) Structural Calcs' a R1A STR1 UCr-�T U/fI ALE GIr - EER1Ir - G J JJJ � J J J •J ✓J� JJ�....�J�J ��J�J JJJ �...� f 78080 Calle Amigo, Suite 102 Telephone: (760) 771-9930 La Quinta, CA 92253 Cell: (760) 771-9998 Fax: (760) 808-9146 Structural Calculation For Larsen Residence At 78340 Birkdale Court Lot#78 Tradition a La Quinta, CA. Del CITY C)7�E— I (DIANTA D C BUILDING; OCT 2 2 2012 FPF OR Co By DA �l � X12 122, Type Of Proiect: Residential NSFR Designer: Fedderly and Associatesp�OFEss/o �© PSGHq,Q,o c I co O wN0. C 67613 cr_(EXP. 913 0 loi12/IZ/IZ Date: October 12, 2012 Design by: R.A. JN: 121094 ��i R1ATRrC�TR1AL E� ;�GIr�EER1IrJ�G TABLE OF CONTENTS 1. Design Criteria....................................................................................1 2. Beam Design.......................................................................................2-66 3. Lateral Design Loads (Seismic And Wind)...........................................67-72 4. Shear Wall Design.............................................................................73-82 5. Foundation Design..................................................................................83-91 18 0 0 DESIGN CRITERIA 2009 INTERNATIONAL BUILDING CODE 2010 CALIFORNIA BUILDING CODE SOILS BEARING PRESSURE = 1500 PSF(PER SOILS REPORT) EXTERIOR WALL = 25 PSF INTERIOR WALL = 15 PSF ROOF DEAD LOAD TILE ROOFING SHEATHING FRAMING DRYWALL INSULATION MISCELLANEOUS TOTAL DEAD LOAD TOTAL LIVE LOAD TOTAL LOAD . = 15.00 PSF = 2.00 PSF = 2.50 PSF = 2.00 PSF = 1.50 PSF = 7.00 PSF = 30.00 PSF = 20.00 PSF = 50.00 PSF E • 0 0 IA PROJECT: LARSEN RES., LOT -78. TRADITION PAGE: Structural CLIENT: BEAM41 ; HDR LEFT OF GUEST SUITE 3 DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY : R.A. INPUT DATA & DESIGN SUMMARY Yes MEMBER SIZE 6 x 8 Condition No. 1, Douglas Fir -Larch MEMBER SPAN L = 9.5 It ' UNIFORMLY DISTRIBUTED DEAD LOAD wo = 300 lbs UNIFORMLY DISTRIBUTED LIVE LOAD wL = 100 lbs / It CONCENTRATED DEAD LOADS Pot = 0 lbs I ! x ! l l (0 for no concentrated load) L, = 0 It 4 PO2 = 0 Ibs L2 = 0 It LIMIT OF LIVE LOAD LIMIT OF LONG-TERM :1L = L / 360 Camber => 0.27 inch A,,D., = 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 toad 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR wsa" „,„ = 9 lbs / It RLaft = 1.94 kips RRigM = 1.94 kips Vm. = 1.69 kips, at 7.5 inch from left end Mm. = 4.61 ft -kips, at 4.75 It 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 = NIA Fb = 1,350 psi F = FbE / Fb* = N/A A = 41.3 int I = 193 in" F„ = 170 psi Fb' = 1,350 psi S„ = 51.6 in' RB = NIA E' = 1,600 ksi F, = 170 psi /E = N/A CD Cr„r C, Ci CL CF Cv Cc C, 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 / S. = 1074 psi < Fb = 1350 psi [Satisfactory] f, = 1.5 Va„ / A = 61 psi < F, [Satisfactory] CHECK DEFLECTIONS AIL, u„xt = 0.06 in, at 4.750 It from left end, < "L = L / 360 [Satisfactory] A(Ca D . L. M.) = 0.24 in, at 4.750 ft from left end < 41110 . L = L / 240 [Satisfactory] Where Ka = 1.00 , (NDS 3.5.2) (DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.50, max) = 0.27 in, at 4.750 ft from left end Cq- 4ECK THE BEAM CAPACITY WITH AXIAL LOAD ;IAL LOAD F = 1 kips IE ALLOWABLE COMPRESSIVE STRE:S IS F.' = F. Co CP CF = 1172 psi Where F, = 925 psi Cp = 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 Le = Ke L = 11.01- = 114 in d = 7.5 in SF =slenderness ratio = 15.2 < F,,E = 0.822 E',,,;,, / SF = 2064 psi E'min = 580 -ksi F = Fa / F,* = 1.394 C = 0.8 IE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 24 psi < F,' ALLOWABLE FLEXURAL STRESS IS F; = 2160 psi, [ for CD = 1.6 1 ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1146 psi < Fp =CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f, / F.,)2 + fp / [Fp 0 - f, / Fes] = 0.537 0 . i 1 1 f t1[iz%i!it. 1 , 0.792 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] [Satisfactory] < 1 [Satisfactory] 0 • • Ll RA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: "BEAM #2 , HDR REAR OF GUEST SUITE 3 DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. Wood Beam�.Desian;Base�ori: NDS2005 �� : '�` : fi � _: '" _ 'UT DATA & DESIGN SUMMARY Ci CL CF Cv Cc C, 1.00 1.00 1.00 Camber => 0.14 inch MBER SIZE 6 x 8 psi < Fb = 1350 psi [Satisfactory] No. 1, Douglas Fir -Larch MBER SPAN L = 7 ft IFORMLY DISTRIBUTED DEAD LOAD wo = 510 lbs / ft A 1 IFORMLY DISTRIBUTED LIVE LOAD wL = 240 1 1 1 1 1 lbs / ft NCENTRATED DEAD LOADS Po, = 0 lbs (0 for no concentrated load) L, = 0 It :a PD2 = 0 lbs L2 = 0 It LECTION LIMIT OF LIVE LOAD "L = L / 360 Ci CL CF Cv Cc C, 1.00 1.00 1.00 Camber => 0.14 inch LECTION LIMIT OF LONG-TERM d Ka o . L = L / 240 fb = Mm" / Sx = 1082 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 VMax / A = 79 psi < F, [Satisfactory] CHECK DEFLECTIONS THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm in, at 3.500 ft from left end, < 4 L = L / 360 [Satisfactory] A tKv o . L . W4 = 0.13 in, at 3.500 It from left end < et K« o • L = L / 240 [Satisfactory] (1= yes, 0= no) 1 Yes , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D. mu) = 0.14 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. Southem Pine Choice => 2 Occupancy Live Load Choice => 2 4LYSIS (ERMINE REACTIONS, MOMENT, SHEAR wsen„,„ = 9 lbs / ft RLft = 2.66 kips RRill = 2.66 kips Vu„ = 2.18 kips, at 7.5 inch from left end Mm. = 4.65 ft -kips, at 3.50 ft from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = WA E = E.= 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Ft, = N/A A = 41.3 int I = 193 in° F„ = 170 psi Fe = 1,350 psi Sx = 51.6 in' Ra = N/A E' = 1,600 ksi F� = 170 psi /E = N/A CD CM C, Ci CL CF Cv Cc C, 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mm" / Sx = 1082 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 VMax / A = 79 psi < F, [Satisfactory] CHECK DEFLECTIONS A (L. Max) = 0.04 in, at 3.500 ft from left end, < 4 L = L / 360 [Satisfactory] A tKv o . L . W4 = 0.13 in, at 3.500 It from left end < et K« o • L = L / 240 [Satisfactory] Where K, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D. mu) = 0.14 in, at 3.500 ft from left end 0 ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F,' = F, Co CP CF = 1335 psi Where Fc = 925 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / cf's = 0.902 Fc = Fc Co CF = 1480 psi I, = Ke L = 1.0 L = 84 in d = 7.5 in SF = slenderness ratio = 11.2 < 50 FE = 0.822 E',;,, / SF = 3801 psi E',w, = 580 ksi F = FcE / Fc = 2.568 C = 0.8 ACTUAL COMPRESSIVE STRESS IS [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 24 psi < Fc' [Satisfactory] _ ALLOWABLE FLEXURAL STRESS IS Fo = 2160 psi, [ for Co = 1.6 ] _ ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1155 psi < Fp [Satisfactory] _CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )Z + f, / [F; (1 - f, / Fes] = 0.538 < 1 [Satisfactory] �61 1 1 ' 1 1 1 [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 24 psi < Fc' [Satisfactory] _ ALLOWABLE FLEXURAL STRESS IS Fo = 2160 psi, [ for Co = 1.6 ] _ ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1155 psi < Fp [Satisfactory] _CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / F.' )Z + f, / [F; (1 - f, / Fes] = 0.538 < 1 [Satisfactory] �61 • • fw PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: CLIENT: BEAM #3 , HDR REAR OF CART GRG DESIGN BY: R.A Structural JOB NO_ : DATE: 10/1/2012 REVIEW BY: R.A. UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD 4CENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 1, 6 x 6 No. 1, Douglas Fir -Larch i L = 3.5 It J wD = 510 lbs / ft wL = 300 lbs I ft . i PD, = 0 lbs L, = 0 It PDZ = 0 lbs LZ = 0 ft 41L = L / 360 Camber=> 0.02 inch AK«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 wse,,V„ = 7 lbs / ft RL ft = 1.43 kips V,,,,,,1 = 1.05 kips, at 5.5 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'mi„ = N/A d = 5.50 in FbE = N/A A = 30.3 int I = 76 in S„ = 27.7 in' Ra = N/A /E = N/A 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 R"m = 1.43 kips Mw.. = 1.25 ft -kips, at 1.75 ft from left end E = E. = 1600 ksi Fp = N/A Fb = 1,350 psi F = FbE / Fe = N/A F„ = 170 psi Fb' = 1,350 psi E' = 1,600 ksi F, = 170 psi CD CM Ct C, CL CF Cv C'. C, 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = Mmax / S,1= 541 psi < Fb = 1350 psi (Satisfactory] f, = 1.5 Vm,/ A = 52 psi < F, [Satisfactory] ECK DEFLECTIONS d (L,,) = 0.01 in, at 1.750 ft from left end, < AL = L / 360 [Satisfactory] e (Kcr D • L. W4 = 0.02 in, at 1.750 ft from left end < 11 Ka D • L = L / 240 [Satisfactory] Where Ku = 1.00 , (NDS 3.5.2) rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A o m. Mao = 0.02 in, at 1.750 ft from left end a i • 0 ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F. = F. Co Cp CF = 1420 psi Where Fc = 925 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2C)2 - F / cP = 0.960 F, = F, Co CF = 1480 psi Le = KL = 1.01- = 42 in d = 5.5 in SF = slenderness ratio = 7.6 < 50 FcE = 0.822 E',,,;,, / SF2 = 8176 psi E'min = 580 ksi F = FcE / F,* = 5.524 C = 0.8 ACTUAL COMPRESSIVE STRESS IS i 1 1 1 f. 1 1 (Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 33 psi < Fc' [Satisfactory] ALLOWABLE FLEXURAL STRESS IS F; = 2160 psi, [ for Co = 1.6 1 ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 640 psi < Fp [Satisfactory] =CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F,' )2 + fo / [F.' (1 - f. / FcE)1 = 0.298 < 1 [Satisfactory] 01 • LJ RA PROJECT: LARSEN RES., LOT 78, TRADITION- PAGE: Structural CLIENT: BEAM #4 , HDR SAT FRONT OF MEDIA DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. Ullniiil'Rraenr'LL.::ien;Base=en:M�SS2005 �' . ... . `� .,, , 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 6x 12 No. 1, Douglas Fir -Larch i L = 9.5 It ' wo = 540 lbs / It .,, WL = 320 lbs / It PD, = 0 lbs L, = 0 ft Q PD2 = 0 lbs f 1 LZ = 0 It "L = L / 360 Camber => 0.14 inch A Ka D - L = L / 240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes psi < Fb = f,; = 1.5 V,,., / A = 79 psi < Code Duration Factor, CD Condition A (L, Max) = 0.05 Code Designation 1 0.90 Dead Load (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) 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, Southem Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR ws„N, = 14 lbs / It Ruf, = 4.15 kips RR;ym = 4.15 kips V,,r, = 3.31 kips, at 11.5 inch from left end MM.. = 9.136 ft -kips, at 4.75 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'm,r, = N/A E = E.= 1600 ksi Fb = N/A d = 11.50 in FbE = NIA Fb = 1,350 psi F = FbE I Fb* = N/A A = 63.3 int 1 = 697 in F„ = 170 psi Fp = 1,350 psi SX = 121.2 in' Ra = N/A' E' = 1,600 ksi F, = 170 psi 'E= WA CD CM C, C, CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MMM / Sx = 976 psi < Fb = f,; = 1.5 V,,., / A = 79 psi < ECK DEFLECTIONS A (L, Max) = 0.05 in, at 4.750 ft from left end. A(Ku D . L . Max) = 0.14 in, at 4.750 ft from left end Where Ku = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A(,_5D,Max) = 0.14 in, at 4.750 It from left end 1350 psi F, [Satisfactory] [Satisfactory] -4L= L/ 360 [Satisfactory] A Kv D . L = L / 240 [Satisfactory] 0 IECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips _ ALLOWABLE COMPRESSIVE STRESS IS Fc = F, Co CP CF = 1372 psi Where F� = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c)os = Fc' = Fc Co CF = 1480 psi L8 = Ke L = 1.0 L = 114 in d = 11.5 in SF = slenderness ratio = 9.9 < FE = 0.822 E',,, / SF2 = 4852 psi E'm,,, = 580 ksi F = FE / Fc` = 3.278 C = 0.8 :-ACTUAL COMPRESSIVE STRESS IS I, = F / A = 16 psi < Fc r . b 0.927 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory) _ ALLOWABLE FLEXURAL STRESS IS F, = 2160 psi, [ for Cr = 1.6 ) ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1023 psi < Fo [Satisfactory) _CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] & / Fc )2 + % / [Fo' (1 - f, / Fes)] = 0.475 < 1 [Satisfactory) EO • • • Iw PROJECT: LARSEN.RES.,'LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #5 , FL BM AT HALLWAY DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 t REVIEW BY: R.A. •�__,�.w__�w=:Lim;�=:A_�_-f'-�nY�f1l�:.AAA!"+-;R _.. r..+;i� ... a. .rix i UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 6 x 6 -No. 1, Douglas Fir -Larch L = 55 ft i WD = 300 lbs/ft WL = 200 lbs / It Po, = 0 lbs L, = 0 It it PDZ = 0 lbs L2 = 0 It AL = L / 360 Camber => 10.08 inch -4Ka 0•L=L/240 ! 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 kLYSIS (ERMINE REACTIONS, MOMENT. SHEAR wse" VA = 7 lbs / ft Rum = 1.39 kips Vm., = 1.16 kips, at 5.5 inch from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 5.50 in FbE = N/A A = 30.3 int 1 = 76 in` Sx = 27.7 in' RB = N/A /E = N/A Co CM Ct C, CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mbtax / Sx = 829 psi < Fb = f" = 1.5 VMax / A = 58 psi < CHECK DEFLECTIONS 5 A (L M.x) = 0.03 in, at 2.750 ft from left end, A Mu D . L. Mx) = 0.09 in, at 2.750 It from left end Where K« = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A(l.So,Max) _ 0.08 in, at 2.750 ft from left end THE BEAM DESIGN IS ADEQUATE. i 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 RR;gt,t = 1.39 kips Mix = 1.92 ft -kips, at 2.75 It from left end E = Ex = 1600 ksi Fb = N/A Fb = 1,350 psi F. = FbE / Fp = N/A F„ = 170 psi Fb = 1,350 psi E' = 1,600 ksi F, = 170 psi « I Cv Cc Cr 1.00 1.00 1.00 1350 psi [Satisfactory] F, [Satisfactory] < 41L = L / 360 [Satisfactory] < .:t K« D , L = L/ 240 [Satisfactory] S • • • ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F� = F, Co CP CF = 1309 psi Where F, = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - ((1+F) / 2c)2 - F / c[o.5 = Fc = F, Co CF = 1480 psi L8 = KB L = 1.01- = 66 in d = 5.5 in SF = slenderness ratio = 12.0 < F, = 0.822 E'm, / SF2 = 3311 psi E'm7n = 580 ksi F = FF / F,* = 2.237 C = 0.8 E ACTUAL COMPRESSIVE STRESS IS f, = F / A = 33 psi < FZ ALLOWABLE FLEXURAL STRESS IS Fo = 2160 psi, I for Co = 1.6 ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 928 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2 (f. / F,' )2 + fo / [F.'(1 - fc / Fcc)) = 0.435 a 0.884 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] F; [Satisfactory] < 1 [Satisfactory] n �J • fw PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #6 , HDR AT REAR OF GARAGE DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. ��f_:�J'5���-.f\� ��.� .��.��►ff1C �IAAl Fx :ir `„ii_ -Y ,.: ',. _..�'., 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 = 3.5 It ` S WD = 510 lbs / It wL = 300 lbs / It Po, = 750 lbs L, = 0.5 It Poe = 2500 lbs L2 = 2 It r AL = L / 360 Camber => 0.03 inch AxaD•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 Designationr 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, Southem Pine Choice => 2 Occupancy Live Load Choice => 2 ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR ws,tr" = 9 lbs / It Rye„ = 3.15 kips RR;9M = 2.97 kips Vma„ = 2.64 kips, at 7.5 inch from left end Mm.. = 3.53 ft -kips, at 2.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'R,i„ = N/A E = Ex = 1600 ksi Fb = N/A d = 7.50 in Fbe = N/A Fb = 1,350 psi F; = FbE I Fb = N/A A = 41.3 int I = 193 in' F„ = 170 psi Fe = 1,350 psi SX = 51.6 in' Ra = N/A E' = 1,600 ksi F, = 170 psi /E = N/A CD CM C, C, 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 = Mma. / S. = 822 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 Vmu / A = 96 psi < F, [Satisfactory] (CHECK DEFLECTIONS A(,_ Max) = 0.00 in, at 1.750 ft from left end, < A L = L / 360 [Satisfactory] d (cu o . L. Max) = 0.02 in, at 1.775 ft from left end < /1110 - L = L / 240 [Satisfactory] Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) (1.50. uax) = 0.03 in, at 1.775 It from left end • • ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips = ALLOWABLE COMPRESSIVE STRESS IS Fc = Fc CD CP CF = 1449 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber onty) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cf's = F, = Fc CO CF = 1480 psi Le = KB L = 1.01- = 42 in d = 7.5 in SF = slenderness ratio = 5.6 < F, = 0.822 E'm„, / SF2 - 15203 psi E'mi„ = 580 ksi F = FcE / Fc' = 10.272 C = 0.8 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 psi < Fc = ALLOWABLE FLEXURAL STRESS IS Fe = 2160 psi, [ for CD = 1.6 ACTUAL FLEXURAL STRESS IS f, = (M + Fe) / S = 895 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2 (fc / Fc )2 + fo / [Fu (1 - fc / FcF)I = 0.415 0.979 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory) F; [Satisfactory) < 1 [Satisfactory) I t 1 � 1 0.979 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory) F; [Satisfactory) < 1 [Satisfactory) I • fw PROJECT: LARSEN RES -LOT 78, TRADITION PAGE: CLIENT: BEAM #7 , HDR AT 1 CAR GARAGE DESIGN BY: R:A Structural JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. 'UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN IFORMLY DISTRIBUTED DEAD LOAD IFORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 6x 12 No. 1 • Douglas Fir -Larch i L = 10.5 It wo = 600 Itis / It wL = 300 lbs / ft P0, = 2500 Itis L, = 2.25 It PD7 = 0 Itis 1 L2 = 0 It AL = L / 360 Camber => 0.31 inch .1KuO•L=L/240 6 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 => 4 Construction Load kLYSIS rERMINE REACTIONS, MOMENT, SHEAR wsdfw = 14 lbs / ft RLO = 6.76 kips Vm, x = 5.89 kips, at 11.5 inch from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 11.50 in FbE = N/A A = 63.3 int 1 = 697 in` S„ = 121.2 in' R8 = N/A /.E= N/A CD CIN C, Ci CL CF 1.25 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1540 psi < Fb = f,' = 1.5 VMM / A = 140 psi < CHECK DEFLECTIONS 5 (L, m,4 = 0.07 in, at 5.250 ft from left end • A a« D . L. MU) = 0.28 in, at 5.250 It from left end Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.50, re,,,) _ 1 0.31 in, at 4.950 It 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, Southam Pine 6 No. 2. Southern Pine Choice => 2 RR;ym = 5.33 kips Mma„ = 15.56 ft -kips, at 4.65 It from left end E = E: = 1600 ksi Fb = N/A Fb = 1.350 psi F'= FbE / Fb = N/A F„ = 170 psi Fb' = 1,688 psi E' = 1.600 ksi F, = 213 psi Cv Cc Cr 1.00 1.00 1.00 1688 psi F, [Satisfactory] [satisfactory] /11. = L / 360 [Satisfactory] '11111),L= L/ 240 [Satisfactory] 3 • • r-] IECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F.' = F. CD CP CF = 1343 psi Where Fc = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2C)2 - F l c)o.s = F,' = F, CD CF = 1480 psi La = Ke L = 1.0 L = 126 in d = 11.5 in SF = slenderness ratio = 11.0 < Fc,e = 0:822 E'min / SF2 = 3971 psi E'min = 580 ksi F = FcE / F,* = 2.683 C = 0.8 ACTUAL COMPRESSIVE STRESS IS f, = F / A = 16 psi < Fe' : ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, [ for Co = 1.6 : ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1588 psi < :CK COMBINED STRESS (NDS 2005 Sec. 3.9.21 (f. / F.* )Z + fb / [Fp (1 - fc / Fes] = 0.738 0.907 50 [Satisfies NDS 2005 Sec. 3.7.1.41; [Satisfactory] F; [Satisfactory] < 1 (Satisfactory] i 0 • PA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: CLIENT: BEAM #8-; HDR AT 2 CAR GARAGE DESIGN BY: R.A Structural JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. UT DATA IL DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/2 x 19 1/2 Glulam 24F -1.8E L = 20.5 ft Code Duration Factor, Cr, Condition WD = 600 lbs / It wL = 300 lbs / ft PD, = 2500 lbs L, = 14 It P,, = 0 lbs 3 1.15 Snow Load L2 = 0 It /1L = L / 360 4 1.25 Construction Load Camber => 0.77 inch A11D.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 wsetr a = 26 lbs/ft RLee = 10.28 kips RR;gm = 11.19 kips V,,,,,, = 9.69 kips, at 19.5 inch from right end MM. = 57.08 ft -kips, at 11.00 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E',,,i„ = N/A E = Ex = 1800 ksi Fb = N/A d = 19.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fp = N/A A = 107.3 inz I = 3,398 in' F„ = 265 psi Fe = 2,276 psi S. = 348.6 in' Re = N/A E' = 1,800 ksi F,; = 265 psi 'E= N/A CD Cti1 C, 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 = Mma. / Sx = 1965 psi < Fb = 2276 psi [Satisfactory] k = 1.5 Vm,� / A = 136 psi < F, [Satisfactory] CHECK DEFLECTIONS n (L. Ltax) = 0.19 in. at 10.250 It from left end, < -4L. = L / 360 [Satisfactory] A (Kc, D . L. MB4 = 0.71 in, at 10.250 It from left end < =1 Kp D • L = L / 240 [Satisfactory] Where K-, = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.50. M") = 0.77 in, at 10.625 It from left end r� LJ • U 'HECK THE BEAM CAPACITY WITH AXIAL LOAD KIAL LOAD F = . 1 kips HE ALLOWABLE COMPRESSIVE STRESS IS F,' = F. CD CP CF = 2338 psi Where F, = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+f) / 2c)'- F / cjo.s = Fc = Fr CD CF = 2560 psi LB = K0 L = 1.0 L = 246 in d = 19.5 in SF = slenderness ratio = 12.6 < Fce = 0.822 E-mj, / SF2 = 4803 psi E'min = 930 ksi F = Fct / Fc' = 1.876 C = 0.9 HE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 9 psi < F,' ALLOWABLE FLEXURAL STRESS IS Fti = 3641 psi, [ for Co = 1.6 ] ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1993 psi < F, °CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )2+ fo / [Fp (1 - fc / F.E)] = 0.548 I # 1 1 a c� 1 � 0.913 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] [Satisfactory] < 1 [Satisfactory] PROJECT: LARSEN RES., LOT 78. TRADITION PAGE: CLIENT: BEAM #9, HEADER AT REAR OF KITCHEN DESIGN BY: R.A Structural JOB NO.: DATE: 10/10/2012 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 6 x 12 No, 1, Douglas Fir -Larch L = 13.5 ft wo = 360 Ibs / ft WL = 140 lbs / ft Po, = 0 lbs ' L, = 0 ft PD2 = 0 lbs LZ = 0 It :1,. = L / 360 Camber => 0.38 inch i1KuD•�=L/240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES Code Duration Factor, CD Condition f, = 1.5 VMa% / A = Code Designation 1 0.90 Dead Load t (Ku D - L. Max) = 0.34 1 Select Structural. Douglas Fir -Larch 2 1.00 Occupancy Live Load in, at 6.750 ft from left end 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 wglp M = 14 lbs / ft RLIft = 3.47 kips RR,ah, = 3.47 kips VMa, = 2.98 kips. at 11.5 inch from left end MMax = 11.70 ft -kips, at 6.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b • = 5.50 in E•n„n = N/A E = E,= 1600 ksi FD = N/A d = 11.50 in FbE = N/A Fp = 1,350 psi F = FbE / Fp = N/A A = 63.3 in' I = 697 in F„ = 170 psi Fp' = 1,350 psi % = 121.2 in' RB= N/A E' = 1,600 ksi F, = 170 psi /E= N/A Co CM C, C, CL CF Cv C, C,. 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MM8, / Sx = 1158 psi < Fb = 1350 psi [Satisfactory] f, = 1.5 VMa% / A = 71 psi < F, (Satisfactory] ECK DEFLECTIONS . (L. Max) = 0.09 in, at 6.750 ft from left end, < a = L/ 360 (Satisfactory] t (Ku D - L. Max) = 0.34 in, at 6.750. ft from left end < : t Kcr D . L = L / 240 (Satisfactory] Where Ka = 1.00 . (NDS 3.5.2) rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) n (+:eo. Maa) = 0.38 in, at 6.750 ft from left end 0 0 0 CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc- = F, Co CP CF = 1225 psi Where F� = 925 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) 12c - [(1+F) / 2c)2 - F / cI°-5 = 0.828 Fc = Ft Co CF = 1480 psi Le = KB L = 1.01- = 162 in d = 11.5 in SF = slenderness ratio = 14.1 < 50 F,,E = 0.822E -m, /SF 2 = 2403 psi E'mm = 580 ksi F = FCE / F,' = 1.623 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS [Satisfies NDS 2005 Sec. 3.7.1.41 fc = F / A = 16 psi < F,' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS F,, = 2160 psi, [ for CL = 1.6 1 THE ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 1206 psi < Fo [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2) (fc / F,* )2+ fp / [F,; (1 - fc / F.E)j = 0.562 < 1 [Satisfactory) 0 1 1 •L 1 l [Satisfies NDS 2005 Sec. 3.7.1.41 fc = F / A = 16 psi < F,' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS F,, = 2160 psi, [ for CL = 1.6 1 THE ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 1206 psi < Fo [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2) (fc / F,* )2+ fp / [F,; (1 - fc / F.E)j = 0.562 < 1 [Satisfactory) 0 • • fw PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: CLIENT: BEAM #10, FL BM AT FRONT OF KITCHEN DESIGN BY : R.A Structural JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/8 x 15 Glulam 24F -1.8E , L = 19 ft I +; wD = 360 lbs / It 1.00 wL = 200 lbs / It .! PD, = 0 lbs "° i� i l I t L, = 0 It li PD2 = 0 lbs 2.00 L2 = 0 It => .aL = L / 360 ANALYSIS Camber=:* 0.64 inch AK«D,L=L/240 DETERMINE REACTIONS, i 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 wseu" = 18 lbs / ft Rif, = 5.49 kips V,. = 4.77 kips, at 15 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E',,,i„ = N/A d = 15.00 in FbE = N/A A = 76.9 int 1 = 1,441 in S„ = 192.2 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 CHECK BENDING AND SHEAR CAPACITIES fb = MMM / SX = 1629 psi < Fb = f, = 1.5 V&t, / A = 93 psi < CHECK DEFLECTIONS psi A (L. w.) = 0.23 in, at 9.500 ft from left end, A ow D • L . Mal = 0.65 in, at 9.500 ft from left end Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD t SELF WEIGHT) A ow. Maxi = 0.64 in, at 9.500 ft from left end THE BEAM DESIGN IS ADEQUATE. i RRi9M = 5.49 kips MMU = 26.10 ft -kips, at 9.50 It from left end E = E, = 1800 ksi Fb = N/A Fb = 2,400 psi F = FbE / Fb* = N/A F„ = 265 psi Fp = 2,371 psi E' = 1.800 ksi F, = 265 psi Cv Cc Cr 0.99 1.00 1.00 2371 psi F, [Satisfactory] [Satisfactory] a L = L / 360 [Satisfactory] ft Kcr o . L = L / 240 [Satisfactory] IECK THE BEAM CAPACITY WITH AXIAL LOAD IAL LOAD F = 1 kips E ALLOWABLE COMPRESSIVE STRESS IS Fc = F, CD CP CF = 2156 psi Where F, = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - 1(1+F) / 2c)2 - F / cf. . = 0.842 F. = F. CO CF = 2560 psi I, = K0 L = 1.0 L = 228 in d = 15 in SF = slenderness ratio = 15.2 < 50 Fc�E = 0.822 E'mm / SF2 = 3309 psi E'min = 930 ksi F = FcE / F,* = 1.292 C = 0.9 E ACTUAL COMPRESSIVE STRESS IS [Satisfies NDS 2005 Sec. 3.7.1.41 I, = F / A = 13 psi < Fc' [Satisfactory] E ALLOWABLE FLEXURAL STRESS IS F; = 3793 psi, I for Co = 1.6 ] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1668 psi < Fo [Satisfactory] ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fo / Fc )2 + % / [F.'(1 - f, / F,F)] = 0.442 < 1 [Satisfactory] • • 0 1 G 1 i t � fa [Satisfies NDS 2005 Sec. 3.7.1.41 I, = F / A = 13 psi < Fc' [Satisfactory] E ALLOWABLE FLEXURAL STRESS IS F; = 3793 psi, I for Co = 1.6 ] E ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 1668 psi < Fo [Satisfactory] ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fo / Fc )2 + % / [F.'(1 - f, / F,F)] = 0.442 < 1 [Satisfactory] • • 0 • • fw PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: StructuralCLIENT : BEAM #11 FL BM AT FRONT OF NOOK DESIGN BY: RAJOB NO.: DATE: "10/1/2012 REVIEW BY: R.A, WondlBeairi'Desiatn<Base'on_NDS�2005:r?_ '" ;�.s;:>�; 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 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes GLB 5 1/8 x 15 Glulam 24F -1.8E Code Duration Factor. Cn Condition L = 17.5 It j 'wD = 360 lbs / ft wL = 200 lbs / It PD, = 5500 lbs`') .l �, t i. .'t 1 l I .l i L, = 16 It I PD2 = 0 lbs L2 = 0 ft "L = L / 360 Camber => 0.62 inch AX,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 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 rERMINE REACTIONS, MOMENT, SHEAR wsa,,,,,„ = 18 lbs / ft RLen = 5.53 kips RR;9,,, = 10.09 kips Vwa„ = 9.37 kips, at 15 inch from right end M,,,ax = 26.45 ft -kips, at 9.48 ft from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E',,,i„ = N/A E = E, = 1800 ksi Fb = N/A d = 15.00 in FbE = N/A Fb = 2,400 psi F = FbE I Fb = N/A A = 76.9 ins 1 = 1,441 in° F„ = 265 psi Fb' = 2,390 psi S„ = 192.2 in' RR = N/A E' = 1,800 ksi F,; = 265 psi /E = N/A Co CM C, Ci CL CF Cv Cc Cr 1.00 1.00 • 1.00 1.00 1.00 1.00 1.00 1.00 1.00 _CK BENDING AND SHEAR CAPACITIES fb = MMM / S. = 1652 psi < Fb = 2390 psi [Satisfactory] f„=1.5Vm,„/A= 183 psi < F' [Satisfactory] I _CK DEFLECTIONS A (L. u aa,q = 0.16 in, at 8.750 ft from left end, < '4L = L / 360 [Satisfactory] A"D . L . � = 0.57 in, at 8.750 ft from left end < A K� D . L = L / 240 [Satisfactory] Where K�. = 1.00 , (NDS 3.5.2) GERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (+.5D. ua4 = 0.62 in, at 8.750 It from left end i ( ZZ CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = Fc CD CP CF = 2252 psi Where Fc = 1600 psi Cp = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)z - F / c]o.s = F, = Fc CD CF = 2560 psi L, KL = 1.01 = 210 in d = 15 in SF = slenderness ratio = 14.0 < Fc = 0.822 E',,;,, / SF = 3900 psi E'min = 930 ksi F = FF / Fc* = 1.524 C = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 13 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fe = 3824 psi. [ for Co = 1.6 ] E ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1691 psi < Fo ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fc / Fc )2+ fp / IF.* (1 - fc / Fce)] = 0.444 • • } i , f] a{ 1, t f r / � l r 0.880 1 50 (Satisfies NDS 2005 Sec. 3.7.1.41 i [Satisfactory] , [Satisfactory] r < 1 [Satisfactory] • • :7 Iw PROJECT: LARSEN RES.,'LOT `78, TRADITION PAGE: Structural CLIENT: BEAM #12, FL BM AT REAR OF WET BAR DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. Wnrii1rRPam"[>aien'RiitiP,�n 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 t 6 x 12 No. 1. Douglas Fir -Larch L= 8 It „ WD = 690 Ibs / ft wL = 360 lbs/ft PD, = 0 lbs t l I l I L, = 0 ft y PD2 = 0 lbs L2 = 0 ft AL = L / 360 Camber => 0.09 inch A110.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, Lural, 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 wsefftyr = 14 lbs / ft R1e,r = 4.25 kips RRtgm = 4.25 kips Vmax = 3.24 kips, at 11.5 inch from left end Mm. = 8.51 ft -kips, at 4.00 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E',,,i„ = WA E = Ex= 1600 ksi Ft, = N/A d = 11.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fe = N/A A = 63.3 int I = 697 in` F„ = 170 psi Fe = 1,350 psi Sx = 121.2 in' Ra = N/A E' = 1,600 ksi F, = 170 psi /E = N/A Co CM Cr Ci CL Cr Cv C� C, 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 = 842 psi < Fb = 1350 psi [Satisfactory] fv, = 1.5 Vmx I A = 77 psi < F� [Satisfactory] CHECK DEFLECTIONS A (L, max) = 0.03 in, at 4.000 It from left end, < AL = L / 360 [Satisfactory] A (Ku D . L. Max) = 0.09 in, at 4.000 ft from left end < n Kcr D . L = L / 240 [Satisfactory] Where K. = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) n t1.s0, ma4 = 0.09 in, at 4.000 It from left end • • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = F. Co CP CF = 1407 psi Where Fc = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(l +F) / 2c)2 - F / clos = Fc = Fc Co CF = 1480 psi L, = K, L = 1.0 L = 96 in d = 11.5 in SF = slenderness ratio = 8.3 < FcE, = 0.822 E',r,,,, / SF2 = 6842 psi F,in = 580 ksi F = FcF / Fj = 4.623 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 16 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, I for Co = 1.6 ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 890 psi < :CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F,' )Z + % / IF.- (1 - fc / FcE)l = 0.413 0.951 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] F; [Satisfactory] < 1 [Satisfactory] I • • • RA i PROJECT: LARSEN RES., LOT 78, TRADITION PAGE Structural CLIENT: BEAM #13, FL BM AT LEFT OF DINING i DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY Duration Factor, C. Condition 1 MEMBER SIZE GLB 8 3/4 x 30 Glulam 24F -1.8E MEMBER SPAN L = 34 ft UNIFORMLY DISTRIBUTED DEAD LOAD wD = 350 lbs / ft UNIFORMLY DISTRIBUTED LIVE LOAD WL = 140 lbs/ft CONCENTRATED DEAD LOADS PD1 = 4250 lbs (0 for no concentrated load) L, = 6 It => PD2 = 10000 lbs L2 = 15 ft DEFLECTION LIMIT OF LIVE LOAD "'L = L / 360 lbs / It RL,,, = 18.48 kips DEFLECTION LIMIT OF LONG-TERM LIK1D, L = L / 240 kips, at 30 inch from left end 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 Constriction Load 5 1.60 Wind/Earthquake Load 6 2.00 Impact Load Choice => 2 Occupancy Live Load ANALYSIS DETERMINE REACTIONS, MOMENT, SHEAR w$e,," = 63 lbs / It RL,,, = 18.48 kips V,.,,, = 17.10 kips, at 30 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 8.75 in E'min = NIA d = 30.00 in FbE = N/A A = 262.5 int I = 19,688 in' Sx = 1312.5 in' RB = NIA /E = N/A Camber => 1.24 inch THE BEAM DESIGN IS ADEQUATE. RRi9rd = 14.56 kips Mm. = 208.75 ft -kips, at 26.07 It from left end E = E, = 1800 ksi Fb = N/A r' 2,400 psi F = FbE / Fe = N/A F„ = 265 psi Fe = 1,978 psi E' = 1,800 ksi F, = 265 psi ' I'• t 1 � � I 1 I 1 / 1 Camber => 1.24 inch THE BEAM DESIGN IS ADEQUATE. RRi9rd = 14.56 kips Mm. = 208.75 ft -kips, at 26.07 It from left end E = E, = 1800 ksi Fb = N/A F„ = 2,400 psi F = FbE / Fe = N/A F„ = 265 psi Fe = 1,978 psi E' = 1,800 ksi F, = 265 psi CD Cm C, Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.82 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mmax / Sx = 1909 psi < Fb = 1978 psi f,; = 1.5 V,,ax / A = 98 psi < F„ [Satisfactory] CHECK DEFLECTIONS A(L ,,,x) = 0.12 in, at 17.000 It from left end, < A(Ku D . L, MM) = 0.95 in, at 17.000 It from left end < Where K. = 1.00 . , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) Am5O,,,,a4 = 1.24 in, at 17.000 It from left end [Satisfactory] -'L = L / 360 [Satisfactory] A Kc D . L = L / 240 [Satisfactory] • • • ECK.THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F,' = Fc CD CP CF = 2280 psi Where Fc = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / cjo.s = Fc = Fc Co CF = 2560 psi Le = KB L = 1.0 L = 408 in d = 30 in SF = slenderness, ratio = 13.6 < F, = 0.822 E'mi„ / SFZ = 4133 psi E'mm = 930 ksi F = Fce / Fc* = 1.614 C = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 4 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS F; = 3165 psi, [ for Cc) = 1.6 THE ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1920 psi < CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F. )Z + fp / [Fo (1 - fc / FcE)j = 0.607 i � r r i E / � 1 0.890 50 [Satisfies NDS 2005 Sec. 3.7.1.41 (Satisfactory) Fp [Satisfactory) < 1 [Satisfactory] 9 • RA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #14, HDR AT REAR OF NOOK DESIGN BY : R.A JOB NO.: DATE: 10/1/2012 ? REVIEW BY: R.A. W66d Bttaailli=06sicn gase:on-NDS2005,!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 GLB 5112 x 18 Glulam 24F -1.8E i L = 12.5 ft wD = 100 lbs / It wL = 20 IbS / ft 4_ } POI = 18500 Ibs L, = 2.5 ft i PD2 = 0 lbs L2 = 0 It n. L = L / 360 Camber => 0.26 inch AK«O.L=L/240 in, at 5.500 ft from left end THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes " 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES Code Duration Factor, C„ Condition psi < Fb = f,' = 1.5 VMax / A = 235 psi < CHECK DEFLECTIONS 1 0.90 Dead Load A (L. M.* = 0.00 in, at 6.250 ft from left end, A (K« D . L . MU) = 0.17 in, at 5.500 ft from left end 2 1.00 Occupancy Live Load , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A(1,50. Max) = 0.26 in, at 5.500 ft from left end 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 I ALYSIS ) ' (ERMINE REACTIONS, MOMENT, SHEAR wseffM = 24 lbs / It RLeft = 15.70 kips RR;gm = 4.60 kips Vm. = 15.48 kips, at 18 inch from left end Mme„ = 38.79 ft -kips, at 2.50 ft from left end (ERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'rr,;r, = N/A E= E.= 1800 ksi Fb = N/A d = 18.00 in FbE = N/A Fb = 2,400 psi F = FbE / Fb* = N/A A = 99.0 in2 I = 2,673 in F„ = 265 psi Fe = 2,400 psi S„ = 297.0 in' Ra = N/A E' = 1,800 ksi F, = 265 psi lE = N/A Co CM C, C. CL. CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / S. = 1567 psi < Fb = f,' = 1.5 VMax / A = 235 psi < CHECK DEFLECTIONS A (L. M.* = 0.00 in, at 6.250 ft from left end, A (K« D . L . MU) = 0.17 in, at 5.500 ft from left end Where Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A(1,50. Max) = 0.26 in, at 5.500 ft from left end Cv CC Cr 1.00 1.00 1.00 2400 psi F, [Satisfactory] [Satisfactory] n L = L / 360 [Satisfactory] Zt Ku D - L = L / 240 [Satisfactory] 1] lu CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = F, CD CP CF = 2487 psi Where F� = 1600 psi CO = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - ((1+F) /2c)2 - F / c)°-' = 0.972 F, = F,. Co CF = 2560 psi Le = K0 L = 1.0 L = 150 in d = 18 in SF = slenderness ratio = 8.3 < 50 F,,F = 0.822 E'„„� / SFZ = 11008 psi E'min = 930 ksi F = Fe / F�- = 4.300 C = 0.9 THEACTUAL COMPRESSIVE STRESS IS [Satisfies NOS 2005 Sec. 3.7.1.41 f, = F / A = 10 psi < Fc- [Satisfactory) THE ALLOWABLE FLEXURAL STRESS IS Ft, = 3840 psi, [ for Co = 1.6 ) THE ACTUAL FLEXURAL STRESS IS f, = (M + Fe) / S = 1598 psi < Fp [Satisfactory) CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )Z + fo / [F.'(1 - f. / F.E)) = 0.416 < 1 [Satisfactory) e • • «A PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #15, HDR AT.REAR OF VERANDA DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMARARY 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 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes GLB 5 1/8 x 16 1/2 Glulam 24F -1.8E i. L = 17 ft 1 0.90 Dead Load WD = .. 600' lbs / ft wL = 300 lbs / ItI. i I, t I ( ! . , P PD1 = 0 lbs L, = 0 ft 4 1.25 Construction Load PD2 = 0 lbs L2 = 0 ft -4L = L / 360 6 2.00 Impact Load Camber => 0.51 inch AKGD,L=L/240 Choice => 2 Occupancy Live Load 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 4LYSIS (ERMINE REACTIONS, MOMENT, SHEAR wsetf m = 20 lbs / ft R1efl = 7.82 kips RRi9,„ = 7.82 kips Vu�x = 6.56 kips, at 16.5 inch from left end Mm, = 33.24 ft -kips, at 8.50 ft from left end (ERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A E = E, = 1800 ksi Fp = N/A d = 16.50 in FbE = N/A Fb = 2,400 psi F = FbE I Fb' = N/A A = 84.6 int 1 = 1,919 in F„ = 265 psi Fp = 2,374 psi S„ = 232.5 in Rg = N/A E' = 1,800 ksi F, = 265 psi /.E = N/A CD Coq C, Ci CL CF Cv C, C, 1.00 1.00 1,00 1.00 1.00 1.00 0.99 1.00 1.00 _CK BENDING AND SHEAR CAPACITIES fb = Mµ. / SX = 1715 psi < Fb = 2374 psi [Satisfactory] f,; = 1.5 Vm,/ A = 116 psi < F,; [Satisfactory] _CK DEFLECTIONS A (L- Max )= 0.16 in, at 8.500 ft from left end, < :1 L = L / 360 [Satisfactory] A tKa D . L. M.4 = 0.50 in, at 8.500 ft from left end < d KD . L = L / 240 [Satisfactory] Where K. = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A 11,5D, ma4 = 0.51 in, at 8.500 ft from left end • • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. Co CP CF = 2351 psi Where F, = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2C)2 - F / cr = 0.918 'Fc* = Fc Co CF = 2560 psi Le = Ke L = 1.01- = 204 in d = 16.5 in SF = slenderness ratio = 12.4 < 50 Fc = 0.822 E'mjn / SF2 = 5001 psi E'min = 930 ksi F = F, / Fj = 1.954 C = 0.9 THE ACTUAL COMPRESSIVE STRESS IS f 1 i I � �.. 'i 1 1 [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 12 psi < F� [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fo = 3799 psi, (for CD = 1.6 1 THE ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1751 psi < Fo [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 & / Fc )2 + % / [Fo' 0 - fc / FcE)) = 0.462 < 1 [Satisfactory] • • • fw PROJECT: LARSEN RES., ,LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #16, HDR.AT REAR OF GREAT ROOM DESIGN BY: RA JOB NO:: DATE: -.10/1/2012 REVIEW BY: R.A. ,UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD IFORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/8 x 19 1/2 Glulam 24F -1.8E L = 20.5 ft wo = .600 lbs / It wL = 300 lbs / It Po, = 0 lbs L, = 0 ft PD2 = 0 lbs L2 = 0 ft A L = L 1360 dKaD.L= L /'240 Does member have continuous lateral support by top diaphragm ? Camber - 0.65 inch THE BEAM DESIGN IS ADEQUATE. (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 wse,r,,,,, = 24 lbs / ft RLea = 9.47 kips RRgM = 9.47 kips VMax = 7.97 kips, at 19.5 inch from left end Mm. 48.53 ft -kips, at 10.25 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 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 / Fe = N/A A = 99.9 int I = 3,167 in F„ = 265 psi :Fti = 2,292 psi SX = 324.8 in' RB= N/A E' = 1,800 ksi F,; = 265 psi /E= N/A Co CM C( Ci CL CF Cv Cc C, 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 = 1793 psi < Fb = 2292 psi [Satisfactory] f,;= 1.5 VMax / A = 120 psi < F,; [Satisfactory] ECK DEFLECTIONS A (L. M-4 = 0.21 in, at 10.250 ft from left end, < d (Ku D+ L. Ma4 = 0.64 in, at 10.250 ft from left end < Where Ku = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D. Mao = 0.65 in, at 10.250 ft from left end 0 AL = L / 360 [Satisfactory) d Ka D . L = L / 240 [Satisfactory) e • • IECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS Fe = F, CO CP CF = 2338 psi Where Fe = 1600 psi --FT —,_T ._.LIT.T __. ` ` CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - 1(1+F) / 2c)'- F / c]°-5 = i 0.913 Fc = F, CD CF = 2560 psi LB = Ka L = 1.0 L = 246 in d = 19.5 in SF = slenderness ratio = 12.6 < 50 [Satisfies NDS 2005 Sec. 3.7.1.4] Fc.E = 0.822 E',n;,, / SF = 4803 psi E'min = 930 ksi F = FcE / Fc* = 1.876 C = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 10 psi < Fc' [Satisfactory] : ALLOWABLE FLEXURAL STRESS IS Ft, = 3667 psi. [ for CD = 1.6 ] : ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1823 psi < Fp [Satisfactory] :CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f. / F.' )2 + fp / [Fp (1 - fc / F&J] = 0.498 < 1 [Satisfactory] r (9 • • Iw PROJECT: LARSEN RES., LOT -78,T RADITION PAGE: Structural CLIENT: BEAM'#17, HDR AT REAR OF GREAT. ROOM DESIGN BY: R.A JOB NO.: .. DATE: 10/1/2012 REVIEW BY: R.A. UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM GLB 5 1/8 x 19 1/2 Glulam 24F -1.8E L =, .20.5 ft wo = 600 . lbs/ft wL = 300 lbs / It PD1 = 0 lbs L1 = 0 ft Poe = 0 lbs L2 = 0, ft AL=L/360 1.25 /JKcrD.L=L/240 5 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 _YSIS ERMINE REACTIONS, MOMENT, SHEAR wSeg M = 24 lbs / ft RLen = 9.47 kips Vm. = 7.97 kips, at'19.5 inch from left end MINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 19.50 in FbE = N/A A = 99.9 int 1 = 3,167 in S. = 324.8 in Ra = N/A E = N/A ksi Fv' = 265 psi wp 1-T-1 --I - F1.7-1. Camber - 0.65 inch THE BEAM DESIGN IS ADEQUATE. CD CM Ct Ci CL Rkigm = 9.47 kips Mm. = 48.53 ft -kips, at 10.25 ft from left end E = Ex = 1800 ksi Fb N/A Fb = 2,400 psi F = FbE / Fe = N/A F„ = 265 psi Fe = 2,292 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 0.95 1.00 1:00 CHECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1793 psi < Fb = 2292 psi f, = 1.5 Vmax / A = 120 psi < F� [Satisfactory] CHECK DEFLECTIONS A a ma�) = 0.21 in, at 10.250 ft from left end, < (Kcr o. L. M1a�) = 0.64 in, at 10.250 ft from left end < Where Ka = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.51), Max) = 0.65 in, at 10.250 ft from left end [Satisfactory] -JL = L / 360 [Satisfactory] 4 Ka o . L = L / 240 [Satisfactory] 1 • • • IECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = .1 . kips E ALLOWABLE COMPRESSIVE STRESS IS F.* = Fc Co CP CF = 2338 psi Where Fc = 1600 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c -.1(1+F) / 2C)2 - F / c)0.5 = Fc. = Fc CD CF = 2560 psi Le = Ke L = 1.0 L = 246 in d = 19.5; in SF = slenderness ratio = 12.6 < FcE = 0.822 E'm;,, / SF2 = 4803 psi E'min = 930 ksi F = FcE / Fc = 1.876 C = 0.9 ACTUAL COMPRESSIVE STRESS IS fc = F / A = 10 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS F; = 3667 psi, [ for Co = 1.6 THE ACTUAL FLEXURAL STRESS IS It, = (M + Fe) / S = 1823 • psi < CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2. (fc / Fc )2 + fo / [Fu (1 - fc / FcE)] = 0.498 / 1 0.913 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] Fe [Satisfactory] < 1 [Satisfactory] n r� J • is RA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM 918, HDR AT FRONT OF GREAT ROOM DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY : RA- 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 5 1/8 x 16 1/2 Glulam 24F -1.8E L = 16.5 ft WD = 600 lbs / ft wL = 300 lbs / It PD1 = 0 lbs L, = 0 ft PD2 = 0 lbs L2 = 0 It AL=L/360 Construction Load /a111D•L=L/240 1.60 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 w8e,r M = 20 lbs / ft RLee = 7.59 kips Vmm = 6.33 kips, at 16.5 inch from left end iAINE SECTION PROPERTIES& ALLOWABLE STRESSES 1, I T in 1 1 'nz yr 1-1 1-17F I. 1. 1. 1 / Camber => 0.45 inch THE BEAM DESIGN IS ADEQUATE. RR;9m = 7.59 kips Mm„, = 31.31 ft -kips, at 8.25 ft from left end b = 5.13 in E'm;,, = N/A E= Ex= 1800 ksi Fb = N/A d = 16.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fb' = N/A A = 84.6 int 1 = 1,919 in F„ = 265 psi Fb' = 2,382 psi S„ = 232.5 in Ra= N/A E' = 1,800 ksi Fv' = 265 psi 1E = WA CD CM Cl Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb= MMU /Sx= 1616 psi < Fb= f,=1.5Vmw/A= 112 psi < CHECK DEFLECTIONS A IL, M,x, = 0.14 in, at 8.250 ft from left end, e (KC, D . L. Max) = 0.44 in, at 8.250 ft from left end Where Ku = 1.00 (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d t1.5D, ym = 0.45 in, at 8.250 ft from left end Cv C, C, 0.99 1.00 1.00 2382 psi F, [Satisfactory] [Satisfactory] d L = L/ 360 [Satisfactory] d Ka D • L = L1 240 [Satisfactory] �iJ • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = F, Cl) CP CF = 2369 psi Where Fc = 1600 psi CD = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) 12C)2 - F / c]°-5 = 0.925 Fc' = Fc. CD CF = 2560 psi I, = Ke L = 1.0 L = 198 in d = 16.5 in SF = slendemess ratio = 12.0 < 50 Fc, = 0.822 E'mj„ / SF2 = 5309 psi E'min = 930. ksi F = FCE / F j = 2.074 C = 0.9 THE ACTUAL COMPRESSIVE STRESS IS 0 [Satisfies NDS 2005 Sec. 3.7.1.4] fr = F / A = 12 psi < Fc' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fp = 3810 psi, [ for Co = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fo = (M + Fe) / S = 1651 psi < Fp [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f, / F. )2 + fp / [Fp 0 - f, / FcE)] = 0.434 < 1 [Satisfactory] • 0 RA PROJECT: LARSEN RES., LOT 78; TRADITION 4 PAGE Structural CLIENT: BEAM #19, HDR AT RIGkT OF GREAT ROOM DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. UT DATA & DESIGN SUMMARY WBER SIZE WBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD VCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM t 6 X 8 No. 1, Douglas Fir -Larch , -,;c i, L = 6.5 ft -- ---Z----r wD = 450 lbs / ft wL = 200 lbs / It Por = 0 lbs v° a-1 .1_ ( f -f-1 .t 1. 1 F I. L, = 0 ft Poz = 0 lbs L2 = 0 ft A L = L 1 360 Camber => 0.09 inch d 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 ANALYSIS S. = 51.6 DETERMINE REACTIONS, MOMENT, SHEAR Wseffv„ = 9 lbs / ft RLefl = 2.14 kips Vm. = 1.73 kips, at 7.5 inch 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 RR19M = 2.14 kips Mm. = 3.48 ft -kips, at 3.25 It from left end RINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in Cmin = N/A E= E,= 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fe = N/A A = 41.3 int 1 = 193 in F„ = 170 psi Fe = 1,350 psi S. = 51.6 in' RB = N/A C = 1,600 ksi F,; = 170 psi /.E = N/A CD CM C, 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 = 810 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 VMm / A = 63 psi < Fv [Satisfactory] CHECK DEFLECTIONS A (L. Max) = 0.03 in, at 3.250 ft from left end, < d L = L / 360 [Satisfactory] d (a« D . L. Max) = 0.09 in, at 3.250 It from left end < n K« o • L = L / 240 [Satisfactory] Where Ku = 1.00 , , (NDS 3.5.2) t DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A t1.50, ma4 = 0.09 • in, at 3.250 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1. kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc = F, Co 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]o.5 = Fc` = F, Co CF = 1480 psi 1, = Ke L = 1.01- = 78 in d = 7.5 in SF = slenderness ratio = 10.4 < F, = 0.822 E'm,, / SFZ = 4408 psi E',I„ = 580 ksi F = F�E / F j = 2.978 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 24 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Ft, = 2160 psi, [ for Co = 1.6 ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 883 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc/Fi )2+fb/[Fn'(1 -f./Fes] = 0.411 • 0 FFT 117f TT I 1 T 1 1 1 0.918 50 [Satisfies NDS 2005 Sec. 3.7.1.4] [Satisfactory] Fp [Satisfactory] < 1 [Satisfactory] I �J • RA PROJECT: LARSEN RES., LOT 78;_TRADITION PAGE: Structural CLIENT: BEAM #20; HDR AT LEFT.M. BDRM DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY 1 Yes 1.00 1.00 1.00 t MEMBER SIZE 6 z 8 No. 1, Douglas Fir -Larch f,; = 1.5 Vmu / A = Code MEMBER SPAN L= 8.5 ft Dead Load d (KID . L • Max) = 0.16 UNIFORMLY DISTRIBUTED DEAD LOAD WD= 300, lbs / It 1.00 1 UNIFORMLY DISTRIBUTED LIVE LOAD wL = 100 lbs / ft -7'r'' llY � `r i• (T I I -6 1 FITT CONCENTRATED DEAD LOADS PD1 = 0 .lbs "h (0 for no concentrated load) L, = 0 It Construction Load Poe = 0 lbs , .. 1.60 L2 = 0 It 5 DEFLECTION LIMIT OF LIVE LOAD dL = L / 360 Camber=> 0.18 inch DEFLECTION LIMIT OF LONG-TERM d KID, L = L / 240 6 No. 2, Southern Pine THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES Code Duration Factor. Cn Condition f,; = 1.5 Vmu / A = Code Designation 1 0.90 Dead Load d (KID . L • Max) = 0.16 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load in, at 4.250 It from left end 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 kLYSIS GERMINE REACTIONS, MOMENT, SHEAR Wse11 iM = 9 lbs / ft Rlea = 1.74 kips RR;9t,t = 1.74 kips Vmax = 1.48 kips, at 7.5 inch from left end M'. = 3.69 ft -kips, at 4.25 It from left end FERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E= Ex= 1600 ksi Fb = N/A d = 7.50 in FDE = N/A Fb = 1,350 psi F = FbE / Fp' = N/A A = 41.3 int -1 = 193 in° F„ = 170 psi Fe = 1,350 psi Sx = 51.6 in' RB = NA E' = 1,600 ksi F, = 170 psi /E = WA 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 = 860 psi < Fb = f,; = 1.5 Vmu / A = 54 psi < CHECK DEFLECTIONS 11(L.M-) = 0.04 in, at 4.250 ft from left end, d (KID . L • Max) = 0.16 in, at 4.250 It from left end Where KI = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.5D. Max) = 0.18 in, at 4.250 It from left end Cv C, Cr 1.00 1.00 1.00 1350 psi F, [Satisfactory] [Satisfactory] n L = L/ 360 [Satisfactory] d KI o . L = L/ 240 [Satisfactory] • • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc = Fc Co CP CF = 1247 psi Where Fc = 925 psi Co = 1.60: CF = 1.00 (Lumber only) Cp = (1+F) / 2c - ((1+F) / 2c)2 - F / cf-5 = F. = Fc CD CF = 1480 psi I, = KL = 1.01 = 102 in d = 7.5 in SF = slenderness ratio = 13.6 < F, = 0.822 E'mi,, / SF = 2578 psi Fmin = 580 ksi F = F�F / F j = 1.742 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 24 , psi < F.' THE ALLOWABLE FLEXURAL STRESS IS Fb = 2160 psi, [ for Co = 1.6 THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 932 psi < CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2 (fc / F,')2 + fb / [Fe (1 - fc / FctJ} = 0.436 I I 1 1 0.842 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] Fe [Satisfactory] 1 < 1 [Satisfactory] • • fw PROJECT: LARSEN RES:, LOT 78, TRADITION Structural CLIENT: BEAM #21, HDR AT REAR OF M: BORM JOB NO.: ... DATE: 10/1/2012 UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) DEFLECTION LIMIT OF LIVE LOAD DEFLECTION LIMIT OF LONG-TERM PAGE: t DESIGN BY: R.A REVIEW BY: R.A Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes r, 6 x 10 No. 1, Douglas Fir -Larch - _ _ T` L.y - - - L= 11 ft I )b, 1 r''I wD = 360 lbs / ft 1.15 wL = - 140 lbs / ft ,� �i - - � (--I A4 T trr f I PD1 = 0 lbs 1.60 L, = 0 ft i PDz = 0 lbs => LZ = 0 ft A L = L /.360 Camber => 0.29 inch d Kcr D+ L = L 1240 11 lbs/ft RUH = 2.81 kips 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 kLYSIS FERMINE REACTIONS, MOMENT, SHEAR wSP,r „1„ = 11 lbs/ft RUH = 2.81 kips V� = 2.41 kips, at 9.5 inch from left end FERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'mi„ = N/A d = 9.50 in FbE = NA A = 52.3 in' I = 393 in° Sx = 82.7 in' RB = N/A LE = 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 = MMm / Sx = 1122 psi < Fb = f,;=1.5Vmax/A= 69 psi < CHECK DEFLECTIONS 5 d (L, Ma.) = 0.07 in, at 5.500 ft from left end, d(Ka D+L, Max) = 0.27 in, at 5.500 It from left end Where KQ = 1.00. , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) Id (1.51), Max) = 0.29 in, at 5.500 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 RRigM = 2.81 kips M' = 7.73 ft -kips, at 5.50 It from left end E = Ex = 1600 ksi Fb = N/A Fb = 1,350 psi F = FbE / Ft = N/A Fv = 170 psi Fe = 1,350 psi E' = 1,600 ksi F,; = 170 psi Cv C, Cr 1.00 1.00 1.00 1350 psi F,; [Satisfactory] [Satisfactory] 11L= L l 360 [Satisfactory] d KU o + L = L/ 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = ' 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = Fe Co CP CF = 1234 psi Where Fc = 925 psi Co = 1.60 CF = 1.00. (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]°-5 = 0.834 F, = Fc Co CF 1480 psi LB = K0 L = 1.0 L = 132 in d = 9.5 in SF = slendemess ratio = 13.9 < 50 FcE = 0.822 E',,,;,, / SF = 2469 psi Fmin = 580 ksi F = FE / F j = 1.669 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 19 psi < F,' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fp = 1 2160 psi, [ for Co = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1179 psi < Fe [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F.' )Z + % / [Fo (1 - f. / Fes] = 0.550 < 1 [Satisfactory] J • U • 1� u RA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BM 22, CANT HDR REAR OF M. BDRM DESIGN BY: RA JOB NO.: DATE: 9/24/2012 REVIEW BY: R.A. Wood Seam Dmilth Base on NDS 2005 DATA & DESIGN SUMMARY SECTION 6 x 10 No. 1, Douglas Fir -Larch BEAM SPAN L j = 7 ft CANTILEVER L2= 1 ft, (0 for no cantilever) SLOPED DEAD LOADS wog,, = 0.3 kips / ft WDL,Z = 0.3 kips / ft PROJECTED LIVE LOADS wu,, = 0 kips / ft wu,Z = 0.1 kips / ft CONCENTRATED LOADS Poi = 1.8 kips 12 PLL = 1 kips SLOPE 0 :12(0= 0.00 0 ) DEFLECTION LIMIT OF LIVE LOAD M.„. du=L/ 360 LONG-TERM DEFLECTION (NDS 3.5.2) d111.L=L/ 240 Code Duration Factor, C„ Condition 1 0.90 Dead Load to— / 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 n.j. „ 2 No. 1, Douglas Fir -Larch 3 No. 2, Douglas Fir -Larch 4 Select Structural, Southern Pine r 5 No. 1, Southern Pine 6 No. 2, Southern Pine Choice => 2 THE BEAM DESIGN IS ADEQUATE. ANALYSIS DETERMINE REACTIONS, MOMIlENTS & SHEARS R2=0.5(coso W-.,+w,,.,JLi+(IV— +w,,.,)(L,+0.5L2)L2+P L;+L2= 4.68 kips ll/ 1 L, L, A= w,<.,., /.i+ L2+P-R2= 0.62 kips cosocoso MM;,,=0.5(ces0+w,,.,)L2 PL2= 3.0 ft -kips X, = 2.07 ft X2 = 2.07 ft W"-'(CosB+w��) \2 (x,+ X') Xg= 2.86 ft M.u,.= - 8 0.6 ft -kips V n,ax = 4.2811 kips, at R2 left. MINE SECTION PROPERTIES AND DESIGN FACTORS L„ = A4-03 , L2) = 2.9 ft, (NDS 2005 Table 3.3.3) 6 x 10 Properties b = 5.50 in Fb = 1,350 psi 1,E = 5.9 ft, (Tab 3.3.3 footnote 1) d = 9.50 in F„ = 170 psi, (NDS 97 CH included) RB = 4.7 < 50 A = 52.3 int E' = 1,600 ksi E'y = 1,600 ksi r4q)_ r� • S. = 82.7 in Fe = 1,347 psi FbF = 31378 psi = 393 in F,' = 170 psi Fe = 1350 psi E = E% = 1600 ksi E',,,;, = 580 ksi F = FeE / FD' = 23.24 Co CM Cr Ci CL CF Cv Cc C, 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 .1.00 IECK BENDING AND SHEAR CAPACITIES Cantilever. f b' = M min / S X = 435 psi < F b = 1347 psi [Satisfactory) Middle Span: ft, = M,ya., /S,t = 93 psi < Fb = 1350 psi [Satisfactory] Shear. f„' = 1.5 Viya., /A = 123 psi < F„' [Satisfactory] (neglected d offset conservatively) IECK DEFLECTION AT LIVE LOAD CONDITION L = L 1/cos 0 = 7.00 ft, beam sloped span a = L 2/ cos 0 = 1.00 ft, beam sloped cantilever length P= PLL COS 0 = 1.00 kips, perpendicular to beam w, = WL/ I COS2 0 = 0.00 klf, perpendicular to beam W2 = wLL,2 COS2 0 = 0.10 kif, perpendicular to beam w,L 3a+ w:a3(4L +3a)lcos0 A "d _CPa2(L+a) + _ 3 E/ 24 E/ 24 E/ J = 0.01 in, downward to vertical direction. < 2 L2 / 360 = 0.07 in [Satisfactory) 4 /'aL2 Sw,/,w:LZa' OM id _ — + — ]COSO= -0.01 in, uplift to vertical direction. 16E/ 384E/ 320 < L, / 360 = 0.23 in [Satisfactory) IECK DEFLECTION AT LONG-TERM LOAD, Ka DL + LL, CONDITION P = PKCrDL+LL COs 0 = 2.80 kips, perpendicular to beam K« = 1.00 , (NDS 3.5.2) w, = K,r wm, cos 0 + wLL.� cost 0 = 0.30 klf, perpendicular to beam W2 = Ku w01.2 COS 0 + wLL 2 COS2 0 = 0.40 klf, perpendicular to beam Pa (La) w,.L3a,+ w:a� (4L +3a)lcosf> Ar"`� 3E/ 24E1: 24F_/ J _ 0.01 in, downward to vertical direction. < 2 L2 / 240 = 0.10 in [Satisfactory] PaLZ 5w, L4 w:L2a2 A'r"r _ + _ — 16E/ 384E/ 32E/ cos0 = 0.00 in, downward to vertical direction. < L, / 240 = 0.35 in [Satisfactory] • • FLA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE StructuralCLIENT : ; BEAM #23; FL BM LEFT. OF M. BDRM DESIGN BY: R.A JOB NO.: : DATE: 10/1/2012 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) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/8 x 16 1/2 Glulam 24F -1.8E L. -_'- �.1., : Duration Factor. Cr, Condition L = 17 It , 1� (-� r• f wD = 600 Ibs / ft 3 1.15 wL = 300 lbs / ft ��" w �• S fr 1i I -D2T _r 7 8�-T� PD1 = 0 lbs Wind/Earthquake Load 6 L, = 0 ft Choice => PDZ = 0 lbs - -- - - ------ -.. �_ L2 = 0 it 20 lbs / It RLe„ = 7.82 kips dL= L! 360 6.56 Camber--> 0.51 inch DETERMINE SECTION d Ktt D.,, = L / 240 b = 5.13 in E'min = N/A 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 Wsetfm = 20 lbs / It RLe„ = 7.82 kips VMS = 6.56 kips, at 16.5 inch from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.13 in E'min = N/A d = 16.50 in FbE = N/A A = 84.6 int I = 1,919 in' S, = 232.5 in' RB = N/A E = N/A CD CI,,, C, C; CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mmax / SX = 1715 • psi < Fb = f,;=1.5VMa,,/A= 116 psi < CHECK DEFLECTIONS d(L,,,,.4 = 0.16 1 in, at 8.500 it from left end, d (Ka D. L, Max) = 0.50 in, at 8.500 It from left end Where KU = 1.00 -, (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) 41(1.5D. ma)o = 0.51 in, at 8.500 It from left end THE BEAM DESIGN IS ADEQUATE. RRigm = 7.82 kips MMS„ = 33.24 ft -kips, at 8.50 it from left end E = E, = 1800 ksi Fb = N/A Fb = 2,400 psi F = FbE / Fe = N/A F„ = 265 psi Fb' = 2,374 psi E' = 1,800 ksi F, = 265 psi Cv Cc Cr 0.99 1.00 1.00 2374 psi F,; [Satisfactory] [Satisfactory] d , = L / 360 [Satisfactory] d Kv D . I = L/ 240 [Satisfactory] e • } CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = • 1 . - kips ` , 1 THE ALLOWABLE COMPRESSIVE STRESS IST II r I i� TTF! � Fc = F, Co CP CF = 2351 psi v$ Where Fc = 1600' psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)'- F / c]" = 0.918 Fj = Fc Co CF = 2560 psi Le = Ke L = 1.0 L = 204 in d = 16.5 ' in SF = slenderness ratio = 12.4 < 50 [Satisfies NDS 2005 Sec. 3.7.1.41 F,= 0.822 E'min / SF Z = 5001 psi E'min = 930 ksi F = FcE / Fc' _ 1.954 C = 0.9 THE ACTUAL COMPRESSIVE STRESS IS fe, = F/ A = 12 psi < F,' [Satisfactory] I THE ALLOWABLE FLEXURAL STRESS IS FD = 3799 psi, [ for CD = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1751 psi < Fp [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (f, / F.' )Z + fp / [Fe (1 - f� / Fes) = 0.462 < 1 [Satisfactory] Iw PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: CLIENT: BEAM #24, HDR RIGHT OF HER OFFICE DESIGN BY: R.A Structural JOB NO.: DATE: 110/11/2012 REVIEW BY: R.A. • UT DATA & DESIGN SUMMARY ABER SIZE iABER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD VCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 6 x 8 No. 1, Douglas Fir -Larch t. i- L_Z f,; = 1.5 VMax / A = L = 6.5 It ^: R>1 wo = 390 lbs / It 1 0.90 wL = 160 lbs / ft 1 Select Structural, Douglas Fir -Larch PD1 = 0 lbs N, .l"�-�--- T- I- 1•_ra_ _ -1 L, _ 0 It No. 1, Douglas Fir -Larch 3 PoZ = 0 lbs 3 LZ = 0 ft 1.25 Construction Load dL = L / 360 Camber=> 0.08 inch Select Structural, Southern Pine AWD.L=L/.240 1.60 Wind/Earthquake Load THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes psi < Fb = f,; = 1.5 VMax / A = 53 psi < Code Duration Factor, C„ Condition A (L, �yax) = 0.02 Code Designation 1 0.90 Dead Load rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) 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, Sout hem Pine 6 2.00 Impact Load 6 No. 2, Southern Pine Choice => 2 Occupancy Live Load Choice => 2 ALYSIS rERMINE REACTIONS, MOMENT, SHEAR Wseif 1 = 9 lbs / ft PLO = 1.82 kips RRi9h, = 1.82 kips V,,,a„ = 1.47 kips, at 7.5 inch from left end Mm. = 2.95 ft -kips, at 3.25 It from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E= Ex= 1600 ksi Fb = N/A d = 7.50 in FbE = NIA Fb = 1,350 psi F = FbE / Fe = N/A A = 41.3 int I = 193 in' F„ = 170 psi Fe = 1,350 psi Sx = 51.6 in3 RB = N/A E' = 1,600 ksi F, = 170 psi /,E = N/A CD CM C, Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 687 psi < Fb = f,; = 1.5 VMax / A = 53 psi < ECK DEFLECTIONS A (L, �yax) = 0.02 in, at 3.250 ft from left end, A (Ka D. L. eeax) = 0.07 in, at 3.250 ft from left end Where Ka = 1.00 1 , (NDS 3.5.2) rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.5D. Mao = 0.08 in, at 3.250 ft from left end r -1 L J 1350 psi F, [Satisfactory] [Satisfactory] A L = L / 360 [Satisfactory] A Ka D * L = L / 240 [Satisfactory] CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 . kips THE ALLOWABLE COMPRESSIVE STRESS IS F� = Fc Co CP CF = 1359 psi Where Fc = 925 psi CO = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c -`[(1+F) / 2c)2 - F / c]o.5 = F, = F, Co CF = 1480 psi I, = Ke L = 1.01- = 78 in d = 7.5 in SF = slenderness ratio = 10.4 < Fcrz = 0.822 E'min / SF2 = 4408 psi E'min = 580 , ksi F = Fcr= / Fc* = 2.978 c = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 24 psi < Fc' ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, (for Co = 1.6 j ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 760 psi < F; =CK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fc / F.' )2 + fb / [Fp (1 - fc / Fes)[ = 0.354 0 I • d I / 1 0.918 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] [Satisfactory] < 1 [Satisfactory] Iw PROJECT: LARSEN RES.,' LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #25, FL. BM AT REAR'OF HER BATH DESIGN BY: R JOB NO.: . ' DATE: 10/1/2012 REVIEW BY: R.A. UT DATA & DESIGN SUMMARY LIBER SIZE ABER SPAN FORMLY DISTRIBUTED DEAD LOAD FORMLY DISTRIBUTED LIVE LOAD 4CENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM GLB 5 1/$ x19 1/2 Glulam 24F -1.8E L= 20.5 ft wD = 570 lbs / ft wL = 280 lbs / It PD1 = 0 lbs L, = 0 It PD2 = 0 lbs LZ = 0 ft AL=L/.360 AKcr D+L' L / 240 Does member have continuous lateral support by top diaphragm ? z Camber => 0.62 inch 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 wse,,,,,u, = 24 lbs i ft RLee = 8.96 kips RRigM = 8.96 kips Vm� = 7.54 kips, at 19.5 inch from left end Mm. = 45.90 ft -kips, at 10.25 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 = 19.50 in FbE = N/A Fb = 2,400 psi F = FbE / Fp' = N/A A = 99.9 int I = 3,167 in F„ = 265 psi Fp = 2,292 psi S„ = 324.8 in3 RB = N/A E' = 1,800 ksi F, = 265 psi i,E = N/A CD CM Ct Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 0.95 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mmax / S, = 1696 psi < Fb = 2292 psi [Satisfactory] f,'= 1.5 VNm/ A = 113 psi < F,; [Satisfactory] CHECK DEFLECTIONS A a. r,,, 4 = 0.20 in, at 10.250 ft from left end, < A (KU o + L. Max) = 0.61 in, at 10.250 ft from left end < Where Ku = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d (1.51). Max) = 0.62 in, at 10.250 It from left end • A L = L / 360 [Satisfactory] a Ku D+ L= L 1 240 [Satisfactory] So ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = . 1 kips ALLOWABLE COMPRESSIVE STRESS IS F.' = Fc CD CP CF = 2338 psi Where F, = 1600 , psi Co = 1.60 CF = 1.00. (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c10.5 = 0.913 Fc* = FCD CF = 2560 psi I = K8 L = 1.0 L = 246 in d = 19.5 in SF = slenderness ratio = 12.6 < 50 Fc = 0.822 E',,,;,, / SF = 4803 psi E',,,;n = 930 ksi F = FcE / F j = 1.876 C = 0.9 ACTUAL COMPRESSIVE STRESS IS I F-FITTI-TTI-71 - I. [Satisfies NDS 2005 Sec. 3.7.1.41 fr = F / A = 10 psi < Fc' [Satisfactory] ALLOWABLE FLEXURAL STRESS IS Fp = 3667 psi, [ for Co = 1.6 ] ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1726 psi < Fp [Satisfactory] :-CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fo / F.' )2 + % / [Fe' (1 - f. / Fes] = 0.472 < 1 [Satisfactory] • C] LJ • • fw PROJECT: LARSEN RES., LOT 78, TRADITION ; PAGE: Structural CLIENT: BEAM #26, HDR AT FRONT OF HER BATH DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY Yes Code i MEMBER SIZE GLB 5 1/8 x 12' Glulam 24F -1.8E MEMBER SPAN L= 9.25 ft 3 UNIFORMLY DISTRIBUTED DEAD LOAD wo = 150 lbs / It 1.25 UNIFORMLY DISTRIBUTED LIVE LOAD wL = '20 lbs / It Wind/Earthquake Load CONCENTRATED DEAD LOADS PD, = 9000 lbs Choice (0 for no concentrated load) L, = 1.25 ft 12.00 in FbE = N/A PD2 = 0 lbs psi F = FbE / Fb' = N/A L2 = 0 ft int 1 = 738 in DEFLECTION LIMIT OF LIVE LOAD A L = L 1360 _ psi Camber => 0.15 inch DEFLECTION LIMIT OF LONG-TERM 4110+1- = L / 240 in' RB = NIA E' = 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 wsen M = 15 lbs/ft R,R„ = 8.64 kips 1.00 1.00 1.00 1.00 1.00 1.00 RR;g„ t = 2.07 kips Vm. = 8.45 kips, at 12 inch from left end M'. = 10.65 ft -kips, at 1.25 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES in, at 3.950 ft from left end < XuD+L = L / 240 [Satisfactory] Where Kir = 1.00 + , (NDS 3.5.2) b = 5.13 in Umi„ = 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 int 1 = 738 in Fv = 265 psi Fe = 2,400 psi Sx = 123.0 in' RB = NIA E' = 1,800 ksi F,,' = 265 psi E = N/A CD CM C, Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = MMax / Sx = 1039 psi < Fb = 2400 psi [Satisfactory] fv = 1.5 Vm x / A = 206 psi < F,; [Satisfactory] ECK DEFLECTIONS A (L, M..) = 0.00 in, at 4.625 it from left end, < A L = L / 360 [Satisfactory] A(KerD+L,Max) = 0.10 in, at 3.950 ft from left end < XuD+L = L / 240 [Satisfactory] Where Kir = 1.00 + , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) 4 (1.5D, Max) = 0.15 - in, at 3.950 ft from left end (5D • • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = F. Co CP CF = 2466 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 Co CF = 2560 psi Le = Ke L = 1.01- = 111 in d = 12 in SF = slenderness ratio = 9.3 < F, = 0.822 E'min / SF = 8935 psi E'min = 930 ksi F = FcE / Fc* = i 3.490 c 0.9 THE ACTUAL COMPRESSIVE STRESS IS fc = F / A = 16 psi < Fc' E ALLOWABLE FLEXURAL STRESS IS Fe = 3840 psi, [ for Co = 1.6 E ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 1088 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / Fc' )Z + fo / [Fo (1 - f. / FcE)] = 0.284 0.963 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory] Fp [Satisfactory] < 1 [Satisfactory] • • IA PROJECT: LARSEN RES., LOT 78, TRADITION f PAGE: Structural CLIENT: BEAM #27, HDR AT FRONT OF HIS TOILET,, : DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY 1 Yes i MEMBER SIZE 6 x 8 No. 1, Douglas Fir -Larch MEMBER SPAN L = . 6 ft 0.90 UNIFORMLY DISTRIBUTED DEAD LOAD wD = 570 lbs / ft Select Structural, Douglas Fir -Larch UNIFORMLY DISTRIBUTED LIVE LOAD wL = 240 lbs / It I I�'' .! _ 8 T-1-11 CONCENTRATED DEAD LOADS PD, _' _ 0 lbs 1.15 (0 for no concentrated load) L, = 0 ft No. 2, Douglas Fir -Larch 4 PD2 =' 0 lbs 4 L2 = 0 ft 1.60 DEFLECTION LIMIT OF LIVE LOAD d L = L / 360 Camber => 0.08 inch DEFLECTION LIMIT OF LONG-TERM d Kerb • L = L / 240 2.00 Impact Load 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 Wseff , = 9 lbs/ft RLea = 2.46 kips RR -MM = 2.46 kips VLja� = 1.94 kips, at 7.5 inch from left end Mrs = 3.69 ft -kips, at 3.00 It from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E= E,= 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fe = N/A A = 41.3 int 1 = 193 in F„ = 170 psi Fe = 1,350 psi SX = 51.6 in3 Re = N/A E' = 1,600 ksi Fv'= 170 psi /.E= N/A Co CM C, Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = M&4M / SX = 858 psi < Fb = f,;=1.5Vmx/A= 71 psi < ECK DEFLECTIONS d (L, Max) = 0.02 in, at 3.000 It from left end, A pca D . L. Ma,q = 0.08 in, at 3.000 ft from left end Where Ka = 1.00 � , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) d o.so. Max) = 0.08 in, at 3.000 ft from left end Cv C, Cr 1.00 1.00 1.00 1350 psi F,; [Satisfactory) [Satisfactory] d L = L / 360 [Satisfactory] d Ku D • 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 = 1 1380 psi Where Fc = 925 psi Co = 1.60 CF = 1.00 (Lumber only) Cp = (1+F) / 2c - [(1+F) / 2c)'- F / c]° -s = 0.932 Fc' = Fc CD CF = 1480 psi LB = Ke L = 1.01 = 72 in d - 7.5 in SF = slenderness ratio = 9.6 < 50 Fc = 0.822 E'mtri / SF2 = 5173 psi E'u �n = 580 ksi F = FcE / F j = 3.495 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 24 psi < Fc' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fo = 2160 psi, [ for Co = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 930 psi < Fo [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (% / F� )2 + % / [Fe (1 - f, / Fes] _ 0.433 < 1 [Satisfactory] 1 i 9 3 • • IRA PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #28, HDR AT FRONT OF HIS OFFICE.. DESIGN BY: R.A JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. 'UT DATA & DESIGN SUMMARY MBER SIZE MBER SPAN IFORMLY DISTRIBUTED DEAD LOAD IFORMLY DISTRIBUTED LIVE LOAD NCENTRATED DEAD LOADS (0 for no concentrated load) LIMIT OF LIVE LOAD LIMIT OF LONG-TERM 6x8 No. 1, Douglas Fir -Larch - L - 7 ft wo = 300 lbs / ft � wL = 200 lbs / ft f .j --J_ 1 TTI_1 7 ,t -i _ � -I PD1 = 0 Ibs lbl (-F- J) �T -) ]. L,= 0 ft -- _._ ___J PD2 = 0 lbs L2 = 0 ft d L = L / 360 Camber => 0.08 inch dKID•L=L/240 THE BEAM DESIGN IS ADEQUATE. Does member have continuous lateral support by top diaphragm ? (1= yes, 0= no) 1 Yes 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES Code Duration Factor, C, Condition f,; = 1.5 Vmx / A = Code Designation 1 0.90 Dead Load d (Ker D • L. Max) = 0.09 1 Select Structural, Douglas Fir -Larch 2 1.00 Occupancy Live Load in, at 3.500 ft from left end 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 ALYSIS rERMINE REACTIONS, MOMENT, SHEAR wseff" = 9 lbs / It RLe = 1.78 kips RRigM = 1.78 kips Vm. = 1.46 kips, at 7.5 inch from left end Mm. = 3.12 ft -kips, at 3.50 ft from left end rERMINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A E= Ez = 1600 ksi Fp = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fp = N/A A = 41.3 int I = 193 in F„ = 170 psi Fe = 1,350 psi S„ = 51.6 in3 RB = N/A E' = 1,600 ksi F, = 170 psi IE= WA CD CM C, 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 = Mma. / S. = 725 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 Vmx / A = 53 psi < F,; [Satisfactory] CHECK DEFLECTIONS A (L. Mx) = 0.03 in, at 3.500 ft from left end, < d L = L / 360 [Satisfactory] d (Ker D • L. Max) = 0.09 in, at 3.500 ft from left end < d Ka o . L = L / 240 [Satisfactory] When: Ku = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A 11,5D, ma -4 = 0.08 in, at 3.500 ft from left end • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS Fc' = F. CD CP CF 1335 psi Where F, = 925 psi Co = 1.60 CF = 1.00 (Lumber only) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cf-5 = Fc* = Fc CD CF = 1480 psi I, = KB L = 1.0 L = 84 in d = 7.5 in SF =slenderness ratio = 11.2 < FcE = 0.822 E'm , / SF = 3801 psi E'min = 580 ksi F = FE/Fj = 2.568 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f, = F / A = 24 psi < Fc' THE ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, [ for Co = 1.6 ) .ACTUAL FLEXURAL STRESS IS fp = (M + Fe) / S = 798 psi < ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fc / F.)2 + fb / (Fo (1 - f. / Fc.E)1 = 0.372 IF f.-FFF j7T_Fj- -f 0.902 50 [Satisfies NDS 2005 Sec. 3.7.1 A) [Satisfactory) F; [Satisfactory] < 1 [Satisfactory] 0 Iw PROJECT: LARSENRES., LOT;78, TRADITION Structural CLIENT: =BEAM #29, HDR AT LEFT OF HIS OFFICE JOB NO.: DATE: 10/1/2012 PAGE: DESIGN BY: R.A INPUT DATA & DESIGN SUMMARY 1 Yes Code Duration Factor, C„ Condition MEMBER SIZE 6 x 10 No. 1, Douglas Fir -Larch 2 MEMBER SPAN L = 10.5 It 1.15 UNIFORMLY DISTRIBUTED DEAD LOAD wo = 360 lbs / ft Construction Load UNIFORMLY DISTRIBUTED LIVE LOAD wL = 140 lbs / ft 6 CONCENTRATED DEAD LOADS PD1 = 0 lbs => (0 for no concentrated load) L, = 0 ft i PDz = 0. lbs � -- --- --- - --------_- � VMax = LZ = 0 ft psi F = FbE / Fb = N/A DEFLECTION LIMIT OF LIVE LOAD A L = L ! 360 Camber => 0.24 inch DEFLECTION LIMIT OF LONG-TERM AXID.L = L / 240 ksi F, = 170 psi 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 2.68 kips DETERMINE REACTIONS, MOMENT, SHEAR wsetf M = 11 lbs / ft RLeB = 2.68 kips VMax = 2.28 kips, at 9.5 inch from left end RINE SECTION PROPERTIES& ALLOWABLE STRESSES b = 5.50 in E'min = N/A d = 9.50 in FbE = N/A A = 52.3 int 1 = 393 in S,, = 82.7 in3 RB = N/A /E = WA => 2 RRigM = CD CM Ct Ci CL CF 1.00 1.00 1.00 1.00 1.00 1.00 CHECK BENDING AND SHEAR CAPACITIES fb = Mma. / SX = 1022 psi < Fb = f,=1.5Vma,,/A= 65 psi < CHECK DEFLECTIONS 3 A (L, Max) = 0.06 in, at 5.250 ft from left end, 4 (Ka D . L. Mal) = 0.22 in, at 5.250 ft from left end Where K« = 1.00 , (NDS 3.5.2) DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A 11.50, Maxi = 0.24 in, at 5.250 ft from left end THE BEAM DESIGN IS ADEQUATE. Cv Cc Cr 1.00 1.00 1.00 1350 psi F,; [Satisfactory] [Satisfactory] d L = L/ 360 [Satisfactory] d II D . L = L/ 240 [Satisfactory] 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 RRigM = 2.68 kips 7.05 ft -kips, at 5.25 ft from left end E = E. = 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] d L = L/ 360 [Satisfactory] d II D . L = L/ 240 [Satisfactory] • :7 r �J ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = I 1 kips ALLOWABLE COMPRESSIVE STRESS IS F.' = F. Co CP CF = 1261 psi Where Fc = 925 psi Co = 1.60 CF = 1.00 (Lumber onty) CP = (1+F) / 2c - [(1+F) / 2c)2 - F / clo" = 0.852 Fc* = Fc Co CF = 1480 psi Le = K0I- = 1.01- = 126 in d = 9.5 in SF = slenderness ratio = 13.3 < 50 F, = 0.822 E'min / SF = 2710 psi E'min = 580 ksi F = F,_. / F,* = 1.831 C = 0.8 ACTUAL COMPRESSIVE STRESS IS i [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 19 psi < Fc' (Satisfactory) ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, [ for CD = 1.6 ] ACTUAL FLEXURAL STRESS IS fe = (M + Fe) / S = 1080 psi < Fp [Satisfactory) =CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (f. / F..)2 + fp / [Fp 0 - f. / Fes] = 0.504 < 1 [Satisfactory) 40 • • 0 Iw PROJECT: LARSEN RES., LOT.78, TRADITION PAGE: Structural CLIENT: BEAM #30, HDR RIGHT OF;GUEST SUITE 2' DESIGN BY: R.A . JOB NO.: DATE: 10/1/2012 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY 1 Yes Code Duration Factor, Cn Condition 1 �. Dead Load MEMBER SIZE 6 x 8 Occupancy Live Load No. 1, Douglas Fir -Larch 1.15 Snow load 4 MEMBER SPAN L = 3.5 It Wind/Earthquake Load 6 2.00 UNIFORMLY DISTRIBUTED DEAD LOAD wD = .600 lbs / ft ANALYSIS 170 psi UNIFORMLY DISTRIBUTED LIVE LOAD wL = . 300 lbs / ft VMex = 2.09 kips, at 7.5 inch from left end CONCENTRATED DEAD LOADS PD1 = 1500 lbs (0 for no concentrated load) L, = 1 ft PO2 = 0 lbs LZ = 0 It DEFLECTION LIMIT OF LIVE LOAD d L = L / 360 Camber => 0.02 inch DEFLECTION LIMIT OF LONG-TERM dK«o+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 170 psi DETERMINE REACTIONS, MOMENT, SHEAR wse,,,M = 9 lbs / ft RLef, = 2.66 kips VMex = 2.09 kips, at 7.5 inch 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 R1jght = 2.02 kips Mm. = 2.24 ft -kips, at 1.30 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES Ci CL CF Cv C, Cr 1.00 1.00 1:00 1.00 1.00 1.00 1.00 1.00 1.00 =CK BENDING AND SHEAR CAPACITIES b = 5.50 in E'mi„ = N/A E= E.= 1600 ksi Fb = N/A d = 7.50 in FbE = N/A Fb = 1,350 psi F = FbE / Fe = N/A A = 41.3 int 1 = 193 in F„ = 170 psi Fe = 1,350 psi S. = 51.6 in' RB = N/A E' = 1,600 ksi F,; = 170 psi 1,E = 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 =CK BENDING AND SHEAR CAPACITIES fb = MMax / S. = 522 psi < Fb = 1350 psi [Satisfactory] f,; = 1.5 VMa„ / A = 76 psi < F,; [Satisfactory] =CK DEFLECTIONS A h Mei = 0.00 in, at 1.750 ft from left end, < AL = L / 360 [Satisfactory] d (KID + L, MW = 0.02 in, at 1.675 It from left end < d Ku o + L = L / 240 [Satisfactory] Where K« = 1.00 , (NDS 3.5.2) (ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (1.SD, Mail = 0.02 in, at 1.675 ft from left end 0 r� • • ECK THE BEAM CAPACITY WITH AXIAL LOAD AL LOAD F = 1 kips ALLOWABLE COMPRESSIVE STRESS IS F� = F. Co CP CF = 1449 psi Where Fe = 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,31- = 1.01- = 42 in d = 7.5 in SF = stendemess ratio = 5.6 < Fc, = 0.822E',,,,/SF 2 = 15203 psi E',,,;,, = 580 ksi F = F�E / Fj = 10.272 C = 0.8 ACTUAL COMPRESSIVE STRESS IS f, = F / A = 24 psi < Fc- ALLOWABLE FLEXURAL STRESS IS Fo = 2160 psi, [ for CD = 1.6 ) THE ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 595 psi < Fb CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.2) (fc / Fc )2 + fb / [Fe (1 - f. / Fes] = 0.276 1 � T 0.979 50 [Satisfies NDS 2005 Sec. 3.7.1.41 [Satisfactory) [Satisfactory) < 1 [Satisfactory] 60, • r� RA PROJECT: LARSEN RES.,;LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #31,.HDR REAR OF GUEST SUITE 2 DESIGN BY : R.A JOB NO.: DATE:10/1/2012 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) 6 x 8' No. 1, Douglas Fir -Larch L = 6.5 ft wD = 330 lbs / It wL = 120 lbs / It PD1 = -0 lbs L, = 0. ft P02 = 0 lbs L2 = 0 It L, 'oa I -17 -1 -T -Tb ?T1 d-1 1.' I. T .T. T T-1 7. F 17T 1 1 DEFLECTION LIMIT OF LIVE LOAD AL = L / 360 Camber - 0.07 inch DEFLECTION LIMIT OF LONG-TERM 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 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 wsetr w = 9 lbs/ft RLafl = 1.49 kips RRight = 1.49 kips VMax = 1.20 kips, at 7.5 inch from left end MM.. = 2.42 ft -kips, at 3.25 It 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 / Fy = N/A A = 41.3 int 1 = 193 in F„ = 170 psi Fp' = 1,350 psi SX = 51.6 in RB = N/A E' = 1,600 ksi F, = 170 psi 1,E = N/A CD CM C( 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 / S. = 564 psi < Fb = 1350 psi [Satisfactory) f. = 1.5 VMax / A = 44 psi < F� [Satisfactory] CHECK DEFLECTIONS A (L, Max) = 0.02 in, at 3.250 It from left end, < d L = L / 360 [Satisfactory) d (Ka D • L. Max) - 0-06 in, at 3.250 It from left end < d Ka D . L = L / 240 [Satisfactory] Where Ka = 1.00 , (NDS 3.5.2) rEERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) A (,,SD, Max) = 0.07 in, at 3.250 ft from left end 062- • • • CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 - kips THE ALLOWABLE COMPRESSIVE STRESS ISI F.'= F. CD CP CF = 1359 psi��� �_� Where F. = 925 psi Co = 1.60 - �- CF = 1.00 (Lumber only) Cp = (1+F) / 2c - 1(1+F) / 2c)2 - F / cr.s = 0.918 Fj = F, CD CF = 1480 psi LB = KB L = 1.01- = 78 in d = 7.5 in SF = slenderness ratio = 10.4 < 50 [Satisfies NDS 2005 Sec. 3.7.1.4] F�, = 0.822 E'min / SF2 = 4408 psi E'min = 580 ksi F = FcE / Fj = 2.978 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS f,. = F/ A = 24 psi < F� [Satisfactory] ALLOWABLE FLEXURAL STRESS IS FW = 2160 psi, [ for CD = 1.6 ] :ACTUAL FLEXURAL STRESS IS fb = (M + Fe) / S = 637 psi < Fo [Satisfactory] :CK COMBINED STRESS [NDS 2005 Sec. 3.9.2] (fe / Fc )2 + fb / [F.'(1 - I. / F.E)] = 0.297 < 1 [Satisfactory] Ll • PROJECT: LARSEN RES., LOT 78, TRADITION PAGE: Structural CLIENT: BEAM #32, FL, SM AT ENTRY GATE DESIGN BY: R.A JOB NO.: DATE: 10/1/2b12 REVIEW BY: R.A. INPUT DATA & DESIGN SUMMARY Yes Code Duration Factor, MEMBER SIZE 6 x 10 No. 1, Douglas Fir -Larch Dead Load MEMBER SPAN L = ft -ittD1 A/ UNIFORMLY DISTRIBUTED DEAD LOAD .9:5. wD = 330. lbs / ft �'"'r�I_ UNIFORMLY DISTRIBUTED LIVE LOAD wL = 120 lbs / ft T1 _ CONCENTRATED DEAD LOADS PD, = 0 lbs Choice (0 for no concentrated load) L, = 0 ft d = 9.50 PD2 = 0 lbs r psi L2 = 0 ft 52.3 DEFLECTION LIMIT OF LIVE LOAD AL = L / 360 Camber => 0.15 inch DEFLECTION LIMIT OF LONG-TERM dKaD. L = L 1240 SX = 82.7 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 d = 9.50 DETERMINE REACTIONS, MOMENT, SHEAR 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 wSelfw = 11 lbs / ft RLea = 2.19 kips 1.00 1.00 1.00 1.00 1.00 1.00 RRigm = 2.19 kips Vm. = 1.83 kips, at 9.5 inch from left end MM.. = 5.20 ft -kips, at 4.75 ft from left end DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES , (NDS 3.5.2) TERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) dow. ea4 = 0.15 in, at 4.750 ft from left end 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 / Fe = N/A A = 52.3 int I = 393 in F„ = 170 psi Fb' = 1,350 psi SX = 82.7 - in 3 RB - N/A E' = 1,600 ksi F,; = 170 psi 1,E = N/A CD CM C, Ci CL CF Cv Cc Cr 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ECK BENDING AND SHEAR CAPACITIES fb = Mma. / Sx = 755 psi < Fb = 1350 psi (Satisfactory) f,, = 1.5 Vm,„ / A = 52 psi < F, (Satisfactory) ECK DEFLECTIONS d (L. Ma) = 0.03 in, at 4.750 ft from left end, < d L = L 1360 [Satisfactory] d (W D . L , Ma4= 0.13 in, at 4.750 ft from left end < d Kcr D • L = L / 240 [Satisfactory] Where Kp = 1.00' , (NDS 3.5.2) TERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT) dow. ea4 = 0.15 in, at 4.750 ft from left end CHECK THE BEAM CAPACITY WITH AXIAL LOAD AXIAL LOAD F = 1 kips THE ALLOWABLE COMPRESSIVE STRESS IS F.' = Fc Co CP CF = 1309 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 0.884 Fc' = Fc CD CF = 1480 psi I, = Ke L = 1.0L = 114 in d = 9.5 in SF = slenderness ratio = 12.0 < 50 F, = 0.822 E'min / SF2 = 3311 psi E'min = 580 ksi F = FcE / Fc* = 2.237 C = 0.8 THE ACTUAL COMPRESSIVE STRESS IS � 1 �r_rj[Z.-T—f — :c. i [Satisfies NDS 2005 Sec. 3.7.1.41 f, = F / A = 19 psi < Fc' [Satisfactory] THE ALLOWABLE FLEXURAL STRESS IS Fp = 2160 psi, [ for Co = 1.6 ] THE ACTUAL FLEXURAL STRESS IS fe = (M + Fe) / S = 812 psi < Fp [Satisfactory] CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21 (fo / F.' )2+ fp / [Fp (1 - fc / Fes] = 0.378 < 1 [Satisfactory] • 49 • 0 Reza PROJECT: Wood Post Design PAGE: CLIENT : Larsen Residence DESIGN BY: R.A. Asgharpour JOB NO.: 121094 DATE : 10/9/2012 REVIEW BY: R.A. ibtes for Wood Post Design Based on NDS 2005 IRATION FACTOR (1.0, 1.15, 1.25, 1.6) Cp = 1.00 , (NDS 2.3.2) )MMERCIAL GRADE (# 1 or # 2) # 2 Pnct Axial Canar_ity for nnuaias Fir -Larch It 2_ lkinsl Height Section Size ft 4x4 4x6 4 x 8 4x10 4x12 6x6 6x8 6x10 6x12 8x8 8x10 6 10.89 16.85 21.84 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 8.04 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 12.83 16.25 19.67 26.62 33.72 16 1 1.90 2.98 3.92 5.00 6.08 1 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 1 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 1 0.93 1.46 1.92 2.45 2.98 1 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 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 1.79 2.17 3.25 4.43 5.61 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 1.55 1.89 2.83 3.86 4.89 5.92 9.49 12.02 30 1 0.55 0.86 1.14 1.45 1 1.76 1 2.65 1 3.62 1 4.58 5.54 8.91 11.29 Post Axial Canacitv for Southern Pine # 2_ lkinsl Height Section Size ft 4x4 4 x 6 4x8 4x10 4x12 6 x 6 6x8 6x10 6x12 8x8 8x10 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 17.98 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 23.24 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 1 8.09 10.30 12.50 1 11.90 16.22 1 20.55 24.87 26.09 1 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 13.76 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 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 1 3.12 3.98 4.84 1 6.27 8.56 1 10.84 13.12 18.42 23.33 19 1.36 2.13 2.81 3.58 4.35 5.72 7.80 9.88 11.96 17.21 21.80 20 1.23 1.93 2.54 3.24 3.93 5.23 7.13 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 14.95 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 1 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 1.63 2.08 2.53 1 3.48 4.75 1 6.01 7.28 11.31 1 14.33 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 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 1.55 1.89 2.63 3.58 4.54 5.50 8.72 11.04 1 30 1 0.55 0.86 1.14 1 1.45 1.76 2.46 3.36 4.25 1 5.15 8.20 10.39 Note: 1. The bold values require steel bearing plate based on, Fc_, 625 psi. 2. The table values are from Wood Column software at www.Engineering-Intemational.com . 0 r' :zf I L' l/ I 0 ft, LOCAnUft LA Wil, iA ( ) 88 = 1.51 &600 - Sm = LOO &a _aMO S1Si1BCDMIGN C;ATEC(3tRYD _,.. 1613.S-6) SIM CLASS B (A ) (IBS 2613.5.2) OCCUPANCY'CA7IIGORY-H ;(B g 1GQ45) C FACIMI= LQO. = (A 11.5-1) RESPONSE :R="... (ASCD 7-S1:TAWE 12.2-1) ArHAP.DYBUA dW AAAM Z 8=6.5' - "t) AT -STS40OLDUNR= LS ( %K TABIE 122-1) R£sDUNDANCYFACTMp*- I39 :- 77-65, SWT_ 12-3-4.2) BASESHRAR V p g T!(L4 x" a W -: (ASM?-nS;II 12`8-2 x.128-2) BASE SHEAR V =130 s [1.010 a LQ/(L4 x GS)] =i l45w . GTI C2t1',�[i —WpmI G ASSUbM iSDs'f' CONSEWAMM ANWE=20* pm =17.8 PS.F n _ kAS� 7-0�� CRHtB 6-2) W • II�ID ESPOSi1R CA7 ORY C4 r -W. SECT 6,5.63) WRWAMMIFfACTO1t1= Loo (ASCE 7-M% TAX 6-1) TOPOGRAPEIFCFACTOR $a - LAW (ASE : 7--W. S E'6-1_72) WIINDFAC!'ORpb-7lxFaaIa 2 (AS=7-45� EQU 64) MA7 UMEEl6H3r15$(1=121)M9iFACrMpg=2L54psf �fM� 20 S CX--- L29). WBSb = 23-01 psf MAXIMUM 25*0,-1..35 jWfMit lbT24ASpe �1A�JMI3FdG�T 39$(1.-=1_40j;�WIIB}2�ACR�[1Sas=2+L9Z � • - 68 017 LATERAL ANALYSIS SECTION 1 LONGITUDINAL ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF WIND LOAD = 22.97 x ( 18.00 - 10.00/2 ) = 299.00 PLF SEISMIC LOAD = 0.145 x ( 30.00x60.0,(Y+ 25.00x2x5 + I O.00x4x5) = 326.00 PLF SEISMIC GOVERNS = 326.00 PLF MAX. SHEAR = 326.00 x 34.00 / 2 x 60.00 = 93.00 PLF CHORD FORCE = 326.00 x 34.00 x 34.00 / 8 x 60.00 = 785.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D'S PER TOP PLATE SPLICE SECTION 2 LONGITUDINAL ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = t22.9g7�- WIND LOAD = 22.97 x ( 18.00 - 10.00/2) LF SEISMIC LOAD = 0.145 x ( 30.00x26.00 + P7 149.00 PLF SEISMIC GOVERNS = 149.00 PLF MAX. SHEAR = 149.00 x 40.00 / 2 x 26.00 = 115.00 PLF CHORD FORCE = 149.00 x 40.00 x 40.00 / 8 x 26.00 = 1146.00 LBS • USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (10) 16D'S PER TOP PLATE SPLICE SECTION 3 LONGITUDINAL ROOF AVERAGE HEIGHT = 24.00 FT TOP PLATE AVERAGE HEIGHT = 16.00 FT 25 FT. MAX ROOF HEIGHT WIND FORCp,-7l8 WIND LOAD = 24.03 x ( 24.00 - 16.00/2 ) 0 SEISMIC LOAD = 0.145 x ( 30.0004.00 + 25.00x2x8 + I O.00x2x8) = 229.00 PLF SEISMIC GOVERNS = 229.00 PLF MAX. SHEAR = 229.00 x 39.00 / 2 x 34.00 = 131.00 PLF CHORD FORCE = 229.00 x 39.00 x 39.00 / 8 x 34.00 = 1281.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (10) 16D'S PER TOP PLATE SPLICE SECTION 4 LONGITUDINAL ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = 2 - 299. WIND LOAD = 22.97 x ( 18.00 - 10.00/2 ) 00 SEISMIC LOAD = 0.145 x ( 30.00x50.00 + 25.00x2x5 + IO.00x2x5) = 268.00 PLF SEISMIC GOVERNS = 268.00 PLF MAX. SHEAR = 268.00 x 23.00 / 2 x 50.00 = 62.00 PLF CHORD FORCE = 268.00 x 23.00 x 23.00 / 8 x 50.00 = 355.00 LBS • USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D'S PER TOP PLATE SPLICE • LATERAL ANALYSIS SECTION 5 LONGITUDINAL ROOF AVERAGE HEIGHT = 15.00 FT TOP PLATE AVERAGE HEIGHT = 12.00 FT 15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF WIND LOAD = 21.54 x ( 15.00 - 12.00/2 ) = 1.94.00 PLF SEISMIC LOAD = 0.145 x ( 30.00x20.00 + 25.00x2x6) = 1.3 1.00 PLF WIND GOVERNS = 194.00 PLF .MAX. SHEAR = 194.00 x 14.00 / 2 x 14.00 = 97.00 PLF CHORD FORCE = 194.00 x 14.00 x 14.00 / 8 x 14.00 = 340.00 LBS USE 1/2" CDX UNBLOCKED WITH 8D'S AT 6", 12" O/C USE (6) 16D'S PER TOP PLATE SPLICE��--L-- • • LATERAL ANALYSIS SECTION 5 LONGITUDIN ROOF AVERAGE HEIGHT = 15.00 FT TOP PLATE A RAC-vk HEIGHT = 12.00 FT 15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 P WIND LOAD= 21.54x ( 15.00 -12.00/2) =c4.00 SEISMIC LOAD = 0.145 x ( 30.00x20.00 + 25.00x2x = I.A.00 PLF SEISMIC GOVERNS = 131.00 PLF MAX. SHEAR= 131.00x 14.00/2x 14.00=66.0 LF CHORD FORCE = 13 l .00 x 14.00 x 14.00 / 8 x l 00 = 29.00 LB S USE '/z" CDX UNBLOCKED WITH 8D' S AT 6' 12" 6/C USE (6) 16D'S PER TOP PLATE SPLICE • LATERAL ANALYSIS SECTION 1 TRANSVERSE ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF WIND LOAD = 22.97 x ( 18.00 - 10.00/2) _ P F SEISMIC LOAD = 0.145 x ( 30.00x40.00 + 2 5 + 10.00x6x5) = 23 2. 00 PLF SEISMIC GOVERNS = 232.00 PLF MAX. SHEAR = 23 2. 00 x 22.00 / 2 x 40.00 = 64.00 PLF CHORD FORCE = 23 2. 00 x 22.00 x 22.00 / 8 x 40.00 = 351.00 LBS USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 2 TRANSVERSE ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF WIND LOAD = 22.97 x ( 18.00- 10.00/2) F SEISMIC LOAD = 0.145 x ( 30.00x50.00 + 25.OGx2x5 + 10.00x2x5) = 268.00 PLF SEISMIC GOVERNS = 268.00 PLF MAX. SHEAR = 268.00 x 26.00 / 2 x 50.00 = 70.00 PLF CHORD FORCE = 268.00 x 26.00 x 26.00 / 8 x 50.00 = 453.00 LBS • USE '/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 3 TRANSVERSE ROOF AVERAGE HEIGHT = 24.00 FT TOP PLATE AVERAGE HEIGHT = 16.00 FT 25 FT. MAX ROOF HEIGHT WIND FORCE = 03 PSF WIND LOAD = 24.03 x ( 24.00 - 16.00/2) 85.00 LF SEISMIC LOAD = 0.145 x (30 OOx55.00 + Ox2x8 + 10.00x2x8) = 321.00 PLF SEISMIC GOVERNS = 321.00 PLF MAX. SHEAR = 321.00 x 27.00 / 2 x 55.00 = 79.00 PLF CHORD FORCE = 321.00 x 27.00 x 27.00 / 8 x 55.00 = 532.00 LBS USE 1/2" CDX UNBLOCKED WITH 8D'S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE SECTION 4 TRANSVERSE ROOF AVERAGE HEIGHT = 18.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT 20 FT. MAX ROOF HEIGHT WIND FORCE = PSF WIND LOAD = 22.97 x ( 18.00 - 10.00/2 ) 99.0 LF - SEISMIC LOAD =0.145 x (30.00x40.00 +. Ox2x5 + 10.00x2x5) = 225.00 PLF SEISMIC GOVERNS = 225.00 PLF MAX. SHEAR = 225.00 x 18.00 / 2 x 40.00 = 51.00 PLF CHORD FORCE = 225.00 x 18.00 x 18.00 / 8 x 40.00 = 228.00 LBS • USE '/2" CDX UNBLOCKED WITH. 8D' S AT 6", 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE 0 • LATERAL ANALYSIS SECTION 5 TRANSVERSE .ROOF AVERAGE HEIGHT = 15.00 FT TOP PLATE AVERAGE HEIGHT = 12.00 FT 15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF WIND LOAD = 21.54 x ( 15.00 - 12.00/2 ) = 194.00 PLF SEISMIC LOAD = 0.145 x ( 30.00x20.00 + 25.00x2x6) = 131.00 PLF WIND GOVERNS = 194.00 PLF MAX. SHEAR = 194.00 x 14.00 / 2 x 14.00 = 97.00 PLF /\ CHORD FORCE = 194.00 x 14.00 x 14.00 / 8 x 14.00 = 340.00 LBS �\ USE 1/2" CDX UNBLOCKED WITH 8D'S AT 6", 12" O/C USE (6) 16D'S PER TOP.PLATE SPLICE • • 0 • LATERAL ANALYSIS SECTION 5 TRANSI ROOF AVERAGE HEIGHT = 15.00 FT TOP PLA 15 FT. MAX ROOF HEIGHT WIND FORCE WIND LOAD= 21.54 x( 15.00- 12.00/2) =®194.0 SEISMIC LOAD = 0.145 x ( 30.00x20.00 .0 SEISMIC GOVERNS = 13 1. 00 PLF AVERAGE HEIGHT = 12.00 FT PSF LF i) = 131.00 PLF MAX. SHEAR = 13 1. 00 x 14.00 / 2 x 14.00 = 66. 0 PLF CHORD FORCE = 131.00 x 14.00 x 14.00 / 8 x .00 = 229.00 LBS USE'/s" CDX UNBLOCKED WITH 8D'S AT 6 , 12" O/C USE (6) 16D' S PER TOP PLATE SPLICE (7D • SHEAR WALL DESIGN SW #1 RIGHT ELEVATION TOTAL LOAD = 326.00 x 35.00/2 = 5705.00 LBS. SHEAR WALL LENGTH = 4.00 + 7.00 + 8.00 = 19.00 FT. SHEAR WALL = 5707.00 / 19.00 = 300.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 1588.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2548.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW 92 LEFT OF HER CLOSET TOTAL LOAD = 326.00 x 41.00/2 = 6683.00 LBS. SHEAR WALL LENGTH = 16.00 FT. SHEAR WALL = 6683.00 / 16.00 = 418.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1682.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2830.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW 93 RIGHT OF GREAT ROOM FULL HT TOTAL LOAD = 326.00 x 6.00/2 + 229.00 x 40.00/2 = 5558.00 LBS. SHEAR WALL LENGTH = 16.00 FT. • SHEAR WALL = 5558.00 / 16.00 = 348.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG = 1390.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 3794.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #4 LEFT OF POOL BATH TOTAL LOAD = 149 x 50.00/2 = 3425.00 LBS. SHEAR WALL LENGTH=8.00 SHEAR WALL = 3425.00 / 8.00 = 466.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG = 1765.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4747.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #5 LEFT OF VERANDA TOTAL LOAD = 149 x 50.00/2 = 3425.00 LBS. SHEAR WALL LENGTH=8.00 SHEAR WALL = 3425.00 / 8.00 = 466.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG = 1765.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4747.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END • 703 • SHEAR WALL DESIGN SW #6 STEEL COLUMN LEFT OF DINING TOTAL LOAD = 229 x 60.00/2 = 6870.00 LBS. STEEL COLUMN DESIGN CANTILEVER STEEL COLUMN R=1.5 LOAD AT STEEL COLUMN = 6870.00 x 6.5/1.5 = 29.750 K USE SHEAR WALL TYPE 12 AT BEAM TO TOP PLATE SEE STEEL COLUMN AND FLAG POLE FOOTING DESIGN SW #7 RIGHT AND LEFT OF ENTRY GATE TOTAL LOAD = 194.00 x 26.00/2 = 2522.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 2522.00 / 8.00 = 315.00 PLF USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0 .MAX. DRAG = 1347.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2547.00 LBS USE SLNI:PSON H.DU5 HOLDOWN EACH END SW 98 LEFT OF PANTRY AND NOOK TOTAL LOAD = 229.00 x 26.00/2 = 2938.00 LBS. SHEAR WALL LENGTH. = 6.00 + 7.00 = 13). 00 FT. SHEAR WALL = 2938.00 / 13.00 = 226.00 PLF • USE SHEAR WALL TYPE I 1 WITH 5/8" x 12" A.B. AT 32" O.0 MAX. DRAG = 1202.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1887.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #9 RIGHT OF 2 CAR GARAGE TOTAL LOAD = 149.00 x 40.00/2 = 2980.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 2980.00 / 8.00 = 373.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1155.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE .MAX. UPLIFT = 2974.00 LBS USE SI:MPSON HDU5 .HOLDOWN EACH END SW #10 LEFT OF CART GARAGE TOTAL LOAD = 149.00 x 40.00/2 = 2980.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR. WALL = 2980.00 / 8.00 = 373.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1755.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2974.00 LBS USE SIM:PSON' HDU5 HOLDOWN EACH. END • r ' nsxwyef' • • :7 SHEAR WALL DESIGN SW #6 STEEL COLUMN LEFT OF DINING TOTAL LOAD = 229 x 60.00/2 = 6870.00 LBS. STEEL COLUMN DESIGN CANTILEVER STEEL COLUMN R=1.5 LOAD AT STEEL COLUMN = 6870.00 x 6.5/1.5 29.750 K USE SHEAR WALL TYPE 12 AT BEAM TO T P PLATE SEE STEEL COLUMN AND FLAG POLE FO � ING DESIGN SW #7 RIGHT AND LEFT OF ENTRY i TOTAL LOAD = 131.00 x 26.00/2 = 1703. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 1706.00 / 8.00 = 213.00 USE SHEAR WALL TYPE 11 WITH 5/8", MAX. DRAG = 910.00 LBS. USE (8) l6 J MAX. UPLIFT = 1720.00 LBS USE SIMM SW #8 LEFT OF PANTRY AND N TOTAL LOAD = 229.00 x 26.00/2 = SHEAR WALL LENGTH = 6.00 + 7 SHEAR WALL = 2938.00 / 13.00 = USE SHEAR WALL TYPE 11 WITI MAX. DRAG = 1202.00 LBS. USE MAX. UPLIFT = 1887.00 LBS USE, SW #9 RIGHT OF 2 CAR GAR) TOTAL LOAD = 149.00 x 40.00/2 SHEAR WALL LENGTH = 8.00 F SHEAR WALL = 2980.00 J 8.00 = USE SHEAR WALL TYPE 12 W MAX. DRAG = 1155.00 LBS. U MAX. UPLIFT = 2974.00 LBS SW #10 LEFT OF CART G) TOTAL LOAD = 149.00 x 40 SHEAR WALL LENGTH = 8 SHEAR WALL = 2980.00 / 8. USE SHEAR WALL TYPE 1: MAX. DRAG = 1755.00 LBS. MAX. UPLIFT = 2974.00 LB1 S. k 12" A.B. AT 32" O.0 S PER TOP PLATE SPLICE ;ON HDU5 HOLDOWN EACH END 18.00 LBS. = 13.00 FT. 00 PLF 8"x12"A.B.AT 32" O.0 16D' S PER TOP PLATE SPLICE IPSON HDU5 HOLDOWN EACH END 2980.00 LBS. 73.00 PLF H 5/8" x 12" A.B. AT 12" O.0 (10) 16D' S PER TOP PLATE SPLICE SIMPSON HDU5 HOLDOWN EACH END = 2980.00 LBS. FT. = 373.00 PLF WITH 5/8" x 12" A.B. AT 12" O.0 JSE (12) 16D' S PER TOP PLATE SPLICE USE SIMPSON HDU5 HOLDOWN EACH END (5) 0 � 0 • RA PROJECT t ARSENRESIDENQE CQ778; TRADITION, :'' PAGE: CLIENT SHEAVVALL6L'EFTOF..DINtNG DESIGN BY: ;RA Structural JOB NO (DATE .ionnol2 REVIEW BY: 'RA INPUT DATA & DESIGN SUMMARY kips Mrx = 327.25 COLUMN SECTION (Tube, Pipe, or WF) +IN12X120 . _ W Shape COLUMN YIELD STRESS Fy = '16.,-% ksi CANTILEVER HEIGHT H = ". '11 ft COLUMN TOP. LATERAL LOAD F = '.29.75 kips, ASD (Strong Axis Bending only) > Mrx [Satisfactory] COLUMN TOP GRAVITY LOAD P = '15 'kips, ASD DIAMETER OF POLE FOOTING b = ... '5. ft ALLOW SOIL PRESSURE Qa = '. 1>5 ksf LATERAL SOIL CAPACITY P p = ksf / It RESTRAINED @ GRADE 7(1yes,0=no) 1: '� Yes Use 5 ft dia x 7.74 ft deep footing restrained @ ground level THE DESIGN IS ADEQUATE. YSIS X COMBINED COMPRESSION AND BENDING CAPACITY OF COLUMN (AISC 360-05, H1) P.+gAt +MPrz ry for 0.2 Pc 9(Ma I Pc = 0.99 < 1.0 [Satisfactory] Pr +L +Mry for P, <0.2 12Pc (M.. MCy I Pc Where Pr = 15.00 kips Mrx = 327.25 ft -kips Mry = 0. ft -kips KL y = 22. ft, weak axis unbraced axial length PC = Pn / tic = 873 / 1.67 = 522.78 kips, (AISC 360-05 Chapter E) > Pr [Satisfactory] Mcx = Mn / Dib = 558.00 / 1.67 = 334.13 ft -kips, (AISC 360-05 Chapter F) > Mrx [Satisfactory] Mcy = Mn / Ob = 256.20 11.67 = 153.41 ft -kips, (AISC 360-05 Chapter F) > Mry [Satisfactory] N POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8) By trials, use pole depth, d = 7.743 It Lateral bearing @ bottom, S3 = 2 Pp Min(d , 12') = 4.65 ksf Lateral bearing @ dl 3, S r = 2 Pp Min(d / 3, 12') = 1.55 ksf Require Depth is given by [VI1+ +4. A ] ./or nonconstrained d — JJ = 7.743 ft [Satisfactory] 4.25Pit for constrained hS•, Where P= F= 29.75 kips A= 2.34P/(bSl)= 5.88 h= Mmax / F = 11.00 ft ECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2) 9soil = P / (a b214) = 0.76 ksf, (net weight of pole footing included.) < Qa [Satisfactory] ECK STRONG AXIS LATERAL DEFLECTION Q = 3E1 = 0.74 in < 2 H/ 240 = 1.10 in [Satisfactory] SHEAR WALL DESIGN SW #11 RIGHT OF LAUNDRY TOTAL LOAD = 268.00 x 24.00/2 = 3216.00 LBS. SHEAR WALL LENGTH= 14.00 FT. SHEAR WALL = 3216.00 / 14.00 = 230.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG =1187.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1363.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #12 FRONT OF CART GARAGE TOTAL LOAD = 268.00 x 8.00/2 = 1072.00 LBS. USE HARDY FRAME HFX-18x10 1 1/8 STD CAPACITY = 1855.00 LBS SEE FOUNDATIO CALCULATIONS SW 413 RIGHT OF GUEST SUITE 3 TOTAL LOAD = 268.00 x 34.00/2 = 4556.00 LBS. SHEAR WALL LENGTH = 16.00 FT. SHEAR WALL = 4556.00 / 16.000 = 285.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG = 1708.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE • MAX. UPLIFT = 2364.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #14 LEFT OF GUEST SUITE 3 AND MEDIA TOTAL LOAD = 268.00 x 17.00/2 = 2278.00 LBS. SHEAR WALL LENGTH = 7.75 + 6.25 += 14.00 FT. SHEAR WALL = 2278.00 / 14.00 = 163.00 PLF " "llZl- USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 4'0.0 MAX. DRAG = 725.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1479.00 LBS USE SIMPSON HDU8 HOLDOWN EACH END SW #15 FRONT OF BDRM 3 AND CART TOTAL LOAD = 225.00 x 13.00/2 = 1463.00 LBS. SHEAR WALL LENGTH = 6.25 (4.75) + 5.50 = 10.25 FT. SHEAR WALL = 1463.00 / 10.25 = 143.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 372.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1003.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END 0 706 n U SHEAR WALL DESIGN SW #16 REAR OF BDRM 3 AND CART TOTAL LOAD = 225.00 x 30.00/2 = 3375.00 LBS. SHEAR WALL LENGTH = 11.50 + 10.00 = 21.50 FT. SHEAR WALL = 3375.00 / 21.50 = 157.00 PLF USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 48" O.0 MAX. DRAG = 888.00 LBS. USE (8) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 625.00 LBS NO HOLDOWN REQUIRED SW #17 FRONT OF MEDIA TOTAL LOAD = 225.00 x 34.00/2 = 3825.00 LBS, SHEAR WALL LENGTH = 6:25 FT. SHEAR WALL = 73825.00 / 6.25= 606.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.C. MAX. DRAG = 1352.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4461.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW 918 FRONT OF 2 -CAR GARAGE TOTAL LOAD = 268.00 x 25.00/2 = 3350.00 LBS. USE HARDY FRAME HFX-18x10 1 1/8 STD CAPACITY = 1855.00 LBS. • USE (2) HFX-18x10 1 1/8 STD TOTAL CAPACITY = 1855.00 x 2 = 3710.00 LBS SEE FOUNDATIO CALCULATIONS SW # 19 'REAR OF GARAGE TOTAL LOAD = 268.00x24.00/2 + 321.00x28.00/2 + 232.0008.00/2 = 121 18.00 LBS. SHEAR WALL LENGTH = 13.00(10.00) + 7.50 + 17.00 = 34.50 FT. SHEAR WALL = 12118.00 / 34.50 = 352.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.C. .MAX. DRAG = 1753.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4863.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #20 FRONT OF KITCHEN AND PANTRY TOTAL LOAD = 225.00 x 50.00/2 = 5625.00 LBS. SHEAR WALL LENGTH = 15.00 + 11.00 = 26.00 FT. .SHEAR WALL = 5625.00 / 26.00 = 216.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C. MAX. DRAG = 862.00 LBS. USE (8) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1238.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END P� 77 • SHEAR WALL DESIGN SW #21 REAR OF GREAT ROOM, TOTAL LOAD = 321.00 x 28.00/2 = 4494.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 4424.00 / 8.00 = 562.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 3482.00 LBS. USE (28) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4020.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW 422 REAR OF GREAT ROOM TOTAL LOAD = 268.00 x 26.00/2 = 3484.00 LBS. SHEAR WALL LENGTH = 4.00 + 4.50 = 8.50 FT. SHEAR WALL = 3484.00 / 8.50 = 410.00 PLF REDUCE CAPACITY FOR HT TO WDTH RATIO 610 x4x2/10 = 488.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0 MAX. DRAG = 1582.00 LBS. USE (12) 16D'S PER TOP :PLATE SPLICE .MAX. UPLIFT = 4.953.00 LBS USE SIMPSON H:DU5 HOLDOWN EACH END SW #23 REAR OF VERANDA TOTAL LOAD = 268.00 x 26.00/2 = 3484.00 LBS. SHEAR WALL LENGTH = 4.00 + 4.00 + 4.00 + 4.00 = 16.00 FT. • SHEAR WALL = 3484.00 / 16.00 = 218.00 PLF REDUCE CAPACITY FOR HT TO WDTH RATIO 610 x4x2/1.3 = 375.00 PLF USE SHEAR WALL TYPE 16 WITH 5/8" x 12" A.B. AT 8" O.C. MAX. DRAG = 694.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2651.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #24 RIGHT OF M. SUITE 2 TOTAL LOAD = 194.00 x 26.00/2 = 2522.00 LBS. SHEAR WALL LENGTH=6.00.FT. SHEAR WALL = 2522.00 / 6.00 = 420.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 1.2" A.B. AT 12" O.C. MAX. DRAG = 1333.00 LBS. USE (1.2) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4386.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #25 TYP ATGUEST SUITES i A.ND 2 TOTAL LOAD = 194.00 x 15.00/2 = 1455.00 LBS. SHEAR WALL LENGTH= 7.00 FT. SHEAR WALL =1 455.00 / 7.00 = 208.00 PLF USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 48" O.C. MAX. DRAG = 1655.00 LBS. USE (12) 16D'SPER TOP PLATE SPLICE MAX. UPLIFT = 1435.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END • • SHEAR WALL DESIGN SW #21 REAR OF GREAT ROOM TOTAL LOAD = 321.00 x 28.00/2 = 4494.00 LBS. SHEAR WALL LENGTH = 8.00 FT. SHEAR WALL = 4424.00 / 8.00 = 562.00 PLF USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A MAX. DRAG = 3482.00 LBS. USE (28) 16D'S IAC MAX. UPLIFT = 4020.00 LBS USE SIMPSON SW 922 REAR OF GREAT ROOM TOTAL LOAD = 268.00 x 26.00/2 = 3484.00 SHEAR WALL LENGTH = 4.00 + 4.50 = 8.' SHEAR WALL = 3484.00 J 8.50 = 410JU REDUCE CAPACITY FOR HT TO W USE SHEAR WALL TYPE 13 WITH MAX. DRAG = 1582.00 LBS. USE (1MAX. UPLIFT = 4953.00 LBS USE S SW #23 REAR OF VERANDA TOTAL LOAD = 268.00 x 26.00/2 = 3 SHEAR WALL LENGTH = 4.00 + 4.a SHEAR WALL = 3484.00 / 16.00 = 21 REDUCE CAPACITY FOR HT TO WUSE SHEAR WALL TYPE i3 WITH MAX. DRAG = 694.00 LBS. USE (6)MAX. UPLIFT = 2651.00 LBS USE SW #24 RIGHT OF M. SUITE 2 TOTAL LOAD = 131.00 x 26.00/2SHEAR WALL LENGTH=6.00 FT. SHEAR WALL= 1703.00 / 6.00 =USE SHEAR WALL TYPE 12 WIMAX. DRAG = 935.00 LBS. USEMAX. UPLIFT = 3007.00 LBS USf SW #25 TYP ATGUEST SUITI TOTAL LOAD = 131.00 x 15.00SHEAR WALL LENGTH= 7.001.SHEAR WALL = 983.00 / 7.00 = USE SHEAR WALL TYPE 10 A MAX. DRAG= 1135.00 LBS. U`. MAX. UPLIFT = 984.00 LBS US BS. FT. 13. AT8"O.0 TOP PLATE SPLICE U5 HOLDOWN EACH END IO 610 x4x2/IO = 488.00 PLF 12" A.B. AT 8" O.0 'S PER TOP PLATE SPLICE N HDU5 HOLDOWN EACH END 4.00 LBS. + 4.00 + 4.00 = 16.00 FT. .00 PLF )TH RATIO 61.0 x4x2/13 = 375.00 PLF /8" x 1.2" A.B. AT 8" O.C. 6D' S PER TOP PLATE SPLICE VIPSON HDU5 HOLDOWN EACH END 703.00 LBS. 34.00 PLF 15/8" x 12" A.B. AT 12" O.C. ) 16D' S PER TOP PLATE SPLICE SIMPSON HDU5 HOLDOWN EACH END 1 AND 2 = 983.00 LBS. 10.00 PLF 'H 5/8" x 12" A.B. AT 48" O.C. (10) 16D' S PER TOP PLATE SPLICE SIMPSON HDU5 HOLDOWN EACH END J78( • SHEAR WALL DESIGN SW #26 RIGHT OF GUEST BATH I \ TOTAL LOAD = 194.00 x 26.00/2 = 2522.00 LBS. \� SHEAR WALL LENGTH= 6.00 FT SHEAR WALL = 2522.-00 / 6.00 = 420.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1278.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4068.00 LBS USE SIMPSON .HDU5 HOLDOWN EACH END SW #27 LEFT OF GUEST BATH l TOTAL LOAD = 194.00 x 12.00/2 = 1164.00 LBS. /r SHEAR WALL LENGTH= 8.00 FT ; SHEAR WALL = 1.164.00 / 8.00 = 146.00 PLF / USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 48" O.0 MAX. DRAG = 560.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 429.00 LBS NO HOLDOWNS SW #28 REAR OF GUEST BATH 1 TOTAL LOAD = 194.00 x 15.00/2 = 1455.00 LBS. • SHEAR WALL LENGTH= 4.25 FT SHEAR WALL = 1455.00 / 4.25 ='342.00 PLF USE SHEAR WALL TYPE la. WITH 5/8" x 12" A.B. AT f2" O.0 MAX.. DRAG = 971.00 LBS. USE (8) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2856.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END SW #29 FRONT OF GUEST HOUSE TOTAL LOAD = 194.00 x 22.00/2 = 2134.00 LBS. SHEAR WALL LENGTH= 4.75 FT SHEAR WALL = 2134.00 / 4.75 = 449.00 PLF USE SHEAR WALL TYPE 1.2 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 1278.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT ='-')9')0.00 LBS USE SIMPSON HDU5 HOLDOWN' EACH END SW 430 REAR OF GUEST SUITE 2 --"� TOTAL LOAD = 232.00 x 28.00/2 = 3248.00 LBS. SHEAR WALL .LENGTH= 7.50 FT SHEAR WALL = 3248.00 / 7.50 = 433.00 PLF USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 666.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1.958.00 LBS USE SIMPSON H:DU5 .HOLDOWN EACH END • SHEAR WALL DESIGN SW #26 RIGHT OF GUEST BATH 1 TOTAL LOAD = 13 1. 00 x 26.00/2 = 1703.00 LBS. SHEAR WALL LENGTH= 6.00 FT SHEAR WALL = 1703.00 / 6.00 = 284.00 PLF . USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A. MAX. DRAG = 876.00 LBS. USE (6) 16D'S PER I MAX. UPLIFT = 2694.00 LBS USE SIMPSON HD SW #27 LEFT OF GUEST BATH 1 TOTAL LOAD = 131.00 x 12.00/2 = 786.00 LB,' SHEAR WALL LENGTH= 8.00 FT SHEAR WALL = 786.00 / 8.00 = 98.00 PLF USE SHEAR WALL TYPE 10 WITH 5/8" x 1 MAX. DRAG = 326.00 LBS. USE (6) 16D'S P MAX. UPLIFT = 294.00 LBS NO HOLDO S SW #28 REAR OF GUEST BATH 1 TOTAL LOAD = 131.00 x 15.00/2 = 983.0 • SHEAR WALL LENGTH= 4.25 FT SHEAR WALL = 983.00 / 4.25 = 232.00 P USE SHEAR WALL TYPE 11 WITH 5/8" MAX. DRAG = 666.00 LBS. USE (6) 16D, MAX. UPLIFT = 1958.00 LBS USE SDa SW #29 FRONT OF GUEST HOUSE TOTAL LOAD = 131.00 x 22.00/2 = 1 SHEAR WALL LENGTH= 4.75 FT SHEAR WALL = 1441.00 / 4.75 = 304 USE SHEAR WALL TYPE 11 WITH' MAX. DRAG = 876.00 LBS. USE (6) MAX. UPLIFT = 2694.00 LBS USE SI ;AT 12" O.0 0 PLATE SPLICE 5 HOLDOWN EACH END A.B. AT 48" O.0 L TOP PLATE SPLICE 12" A.B. AT 32" O.0 PER TOP PLATE SPLICE IN HDU5 HOLDOWN EACH END 00 LBS. PLF x 12" A.B. AT 32",O.0 l' S PER TOP PLATE SPLICE SON HDUS HOLDOWN EACH END SW 930 REAR OF GUEST SUITE Z TOTAL LOAD = 232.00 x 28.00/2 = 248.00 LB.S. SHEAR WALL LENGTH= 7.50 FT SHEAR WALL = 3248.00 / 7.50 = 433.00 PLF USE SHEAR WALL TYPE 12 WITq 5/8" x 12" A.B. AT 12" O.0 MAX. DRAG = 666.00 LBS. USE 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 1958.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END • • SHEAR WALL DESIGN SW #31 FRONT OF GUEST SUITE 2 TOTAL LOAD = 131.00 x 11.00/2 = 721.00 LBS. SHEAR WALL LENGTH= 7.00 FT SHEAR WALL = 721.00 / 7.00 = 103.00 PLF USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 48" O.0 MAX. DRAG = 53.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 294.00 LBS NO HOLDOWNS SW #32 FRONT OF M. BDRM TOTAL LOAD = 232.00 x 42.00/2 = 4872.00 LBS. SHEAR WALL LENGTH= 12.00 FT SHEAR WALL = 4872.00 / 12.00 = 406.00 PLF 8t� USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 4" O.0 MAX. DRAG =1 266. 00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 4658.00 LBS USE SHVIPSON HDU5 HOLDOWN EACH END SW #33 REAR OF M. SUITE TOTAL LOAD = 232.00 x 22.00/2 = 2552.00 LBS. SHEAR WALL LENGTH= 10.00 FT SHEAR WALL = 2552.00 / 10.00 = 253.00 PLF USE SHEAR WALL TYPE I I WITH 5/8" x 12" A.B. AT 32" O.0 • MAX. DRAG = 128 1. 00 LB S. USE (10) 16D'S PER TOP PLATE SPLICE MAX. UPLIFT = 2752.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END • • • • Reza PROJECT: Sfiear Wail #12;AND 18;: HARDYFRAME PAGE: WENT:DESIGN BY: R A, Asgharpour JOB NO 4 DATE: #I #. REVIEW BY: R.A. Foo>t�ng: P, UT DATA ' L LENGTH L,,, = 1.8 It 0 -� L HEIGHT h = ; .10 It r 7- L THICKNESS I= .8 in M TING LENGTH L = 9 It PW L, = 315 ft In TING WIDTH B = ' 3.5 It TING THICKNESS T= 18 in Pi TING EMBEDMENT DEPTH D = .1.5 ftI-- ---- )WABLE SOIL PRESSURE qa = ; . 1.5 ksf D J LOAD AT TOP WALL Pr.DL = : 1 kips LOAD AT TOP WALLPr,LI = ' 1. .• kips L w - LOAD LOCATION a = ' ; 0:1 . -.ft L SELF WEIGHT PW = . = 0.1 " kips :RAL LOAD TYPE (0=wind,1=seismic) -0 :..'wind LOADS AT WALL TOP F = : ;1.875. kips THE FOOTING DESIGN IS ADEQUATE - M = ", ' 0 ft -kips 3RETE STRENGTH ksi 4R YIELD STRESS fy = 60 ksi BARS, LONGITUDINAL 3 # 5 FOM BARS, LONGITUDINAL .4 # 5 - FOM BARS, TRANSVERSE # .5. @ 12 in o.c. (LYSIS X OVERTURNING FACTOR (IBC 06 1605.2.1, 1801.2.1, 8 ASCE 7-05 12.13.4) F = MR / Mo = 1.82 > 1.6/0.9 for wind [Satisfactory] Where Pf = 6.85125 kips (footing self weight) Mo = F (h + 0) + M = 19 ft -kips (overturning moment) MR = (Pr,00 (1-1 + a) + Pf (0.5 L) + PW (1-1 + 0.51-W) = 35 ft -kips (resisting moment without live load) SOIL CAPACITY (ALLOWABLE STRESS DESIGN) Ps = 4.725 kips (soil weight in footing size) P = (Pr,DL + Pr.1-0 + PW + (Pr - Ps) = 4.23 kips (total vertical net load) MR = (Pr,DI + Pr, LL) (1-1 + a) + Pf (0.5 L) + PW (1-1 + 0.51-W) = 39 e = 0.5 L - (MR - Mo) / P = l -0.17 It (eccentricity from middle of footing) P(1+ G I L ll /// for e < - 4mur = BL 6 2P for e > L = 0.12 ksf - 3B(0.5L -e)' 6 Where e = -0.17 ft, < (L / 6) FOOTING CAPACITY (STRENGTH DESIGN) M.,R = 1.2 [Pr,0L (1-1 + a) + Pf (0.5 L) + PW (1-1 + 0.51-W)) + 0.5 Pr, LL(Lt + a) _ Mu.o = 1.6 [F(h + D) + M) = 31 ft -kips Pu = 1.2 (Pr,DL + Pf + PW) + 0.5 Pr, LL = 10 kips eu=0.5L-(Mu,R-Mu,o)/Pu= 3.18 It 1+ 6e„ P. I L /// , for e„ < L BL 6 = 1.45 ksf 2P„ L for e,. > - 3B(0.5L-e„)' 6 It -kips (resisting moment with live load) c 4/3qa [Satisfactory) 44 ft -kips 0 ' ple s. • RENDING MOMENT S SHEAR AT EACH FOOTING SECTION Section 0 1/10 L 2/10 L 3/10 L 4/10 L 5/10 L 6/10 L 7/10 L 8/10 L 9/10 L L Xu (ft) 0 0.90 1.80 2.70 3.60 4.50 5.40 6.30 7.20 8.10 9.00 Pu,w (Ido 0.0 0.0 0.0 0.0 0.0 10.9 -48.6 0.0 0.0 0.0 0.0 Muw(ft-k) 0 0 0 0 0 -12 -33 35 -37 -38 -40 Vu,w (kips) 0 0 0 0 0 -27 -10 -2 -2 -2 -2 Pu.t(kst) 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 Mut (ft -k) 0 0 -1 -3 -0 -9 -13 -18 -24 -30 -37 Vu,t (kips) 0 -1 -2 -2 3 -4 -5 -6 -7 -7 -8 qu (ksQ -1.4 -1.1 -0.8 -0.5 -0.1 0.0 0.0 0.0 0.0 0.0 0.0 MU,q (ft -k) 0 2 7 14 23 32 41 50 59 68 77 Vu.q (kips) 0 4 7 9 10 10 10 10 10 10 10 L" Mu (ft -k) 0 2 5 11 17 10 -6 3 -1 0 0 L' Vu (kips) 0 3 6 7 7 -21 -6 2 2 1 0 20 15 10 5 0 -5 -10 20 0 -20 -40 Location Mu.rrmx d (in) PregO Pprowo Vujnax �Vc = 2 0 b d (f� )o.s Top Longitudinal -5 ft -k 14.69 0.0001 0.0015 21 kips 52 kips Bottom Longitudinal 17 ft -k 14.69 0.0018 0.0020 21 kips 52 kips .Bottom Transverse 0 ft -k / ft 14.06 0.0018 0.0018 0 kips / It 14 kips / ft 0.85 1- 1- �u o f` 0.383bd'f' Where P = Pmm = 0.0018 f 0.85Q I PA,U = rc 6u = 0.0129 [Satisfactory) fy Ft!+-, 0M OV (conrd) �01 0 FOUNDATION DESIGN SOIL BEARING PRESUURE = 1800 PSF PER SOILS REPORT MAX LOAD=50.00x46.00/2 + 25.00x17.00 + 150xl5xl8/144=1856.00 PLF FOUNDATION WIDTH REQUIRED = (1856.00/1500)x12 = 15" USE 15" WIDE x 18" DEEP FOOTINGS WITH (2) # 5 TOP AND BOTTOM FOOTING CAPACITY = (1500x52xl5)/(12x12) = 8125.00 LBS PADS DESIGN BM # 8 = 11200.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 14 = 15700.00lBS USE 3'-6" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 16 = 9500.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 17 = 11000.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY • BM # 18 = 8000.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY • BM # 23 = 8000.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 25 = 9000.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY BM # 26= 8650.00 LBS USE 3'-0" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY ri • 16 Reza PROJECT: ;Maximum Load, For 2 sq.. ft. Pad Footing (I500psf) PAGE: As harpour CLIENT: DESIGN BY: R.A. M JOB NO.: DATE: REVIEW BY: R.A. Pad `F.00tin D'esl nrBased=o' A '111 INPUT DATA LONGITUDINAL TRANSVERSE d DESIGN SUMMARY 8.50 b 24 COLUMN WIDTH c, = 0 in FOOTING WIDTH 8 = 2.00 ft COLUMN DEPTH c, = 0 in FOOTING LENGTH L = 200 ft BASE PLATE WIDTH b; = 4 in FOOTING THICKNESS T = 12 in BASE PLATE DEPTH b, = 4 in LONGITUDINAL REINF 3 # 4 @ 9 :n o c FOOTING CONCRETE STRENGTH ft = 2.5 ksi TRANSVERSE REINF 3 # 4 @ 9 ul o c REBAR YIELD STRESS fy = 40 ksi AXIAL DEAD LOAD Pn; = 2.5 k 1 AXIAL LIVE LOAD P;,F. = 2.5 k LATERAL LOAD (O=WIND, 1=SEISMIC) = 1 Seismic.SD SEISMIC AXIAL LOAD Pl.tir = 0 k. SD ' SURCHARGE qs = 0 ksf SOIL WEIGHT wy = 0.11 kcf FOOTING EMBEDMENT DEPTH D, = 2 ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Qa = 1:5 ksf •. FOOTING WIDTH B = 2 ft t "•` "� •: FOOTING LENGTH L = 2 ft BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. kLYSIS IGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9,2.1) E 1: DL + LLP = 5 kips 1.2 DL - 1.6 LL Pu = E2: DL+LL+E/1.4 P = 5 kips 12 OL+1.0LL-1.0E Pu = E3: 0.9 DL -E/1.4 P = 2 kips 0.9 DL+1.0E P:: = CK SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2) CASE 1 CASE 2 CASE 3 (U, 15 - WO/" = 1.29 ksf. 1,29 ksf. 0.60 ksf q MAX < k Q a • [Satisfactory] where k = 1 for gravity loads, 413 for lateral loads. IGN FOR FLEXURE (ACI 318.05 SECA5.4.2. 10.2, 10.3.5, 10.5.4, 7.12.2, 12 2. & 12.5) O.RS j" . I — 1 0 1R3h,/ / q.85 j> .%�• r: r, ( % 4 "fi P- , / :! /-::n — �r'r, !':a•,=.l//�\\ll.11l)I\C/ �1)! i 7 kips 6 kips 2 kips LONGITUDINAL TRANSVERSE d 8.75 8.50 b 24 24 q u.max 1.75 1 75 Mu 1.47 1.47 P 0.000 0.000 ('mm 0.000 0.000 k 0.07 0.08 RegO 1 # 4 1 # 4 Max. Spacing 18 in o.c. 18 in o.c. USE 3 # 4 @ gin o.c. 3 # 4 @ 9 in o c ('Max 0.019 0.019 Check pgrod ` ('max [Satisfactory] [Satisfactory] 7 kips 6 kips 2 kips FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1. & 11.3) PUNCHING SHEAR (ACI 318-05 S,EC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2) (2-I"y)�hJ.l=1p = 54.98 kips where G = 0.75 (ACI 318-05. Section 9.3.2.3) (1, = ratio of long side to short side of concentrated load = 1.00 bo = Cl + C2 + b1 + 02 + 4d = 42.5 in Ap = bo d = 366.6 W y - MIN(2 . 4 / ftp. 40 d /\bo) = 2.0 I' - P I - I )'` cq + d �� �'—�r ' - �l I = 5.63 kips < 0 V n [Satisfactory] • (cont'd) £SS LONGITUDINAL TRANSVERSE V„ 0.66 0.73 0 0.75 0.75 OV„ 15.8 15.3 Check V„ < OVA [Satisfactory) [Satisfactory] PUNCHING SHEAR (ACI 318-05 S,EC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2) (2-I"y)�hJ.l=1p = 54.98 kips where G = 0.75 (ACI 318-05. Section 9.3.2.3) (1, = ratio of long side to short side of concentrated load = 1.00 bo = Cl + C2 + b1 + 02 + 4d = 42.5 in Ap = bo d = 366.6 W y - MIN(2 . 4 / ftp. 40 d /\bo) = 2.0 I' - P I - I )'` cq + d �� �'—�r ' - �l I = 5.63 kips < 0 V n [Satisfactory] • (cont'd) £SS 7 r� 10 Reza LONGITUDINAL TRANSVERSE d 8.75 8.50 PROJECT: Max. Load For 2.5 sq. ft. Pad Footing (1500psO 2.02 PAGE: As har CLIENT: 3.43 a 0.000 0.001 DESIGN BY: R.A. 9 p Our JOB NO.: A, 0.17 DATE: RegD REVIEW BY: R.A. Pid"166 in' _Desi -'rr'S ed dK ACl 318-05. Max. Spacing 18 in o.c 18 in o.c. USE 3 # 4 @ 12 in o.c. 3 # 4 @ 12 in o.c Pmax 0.019 0.019 INPUT DATA [Satisfactory] [Satisfactory] DESIGN SUMMARY COLUMN WIDTH c; = 0 in FOOTING WIDTH B = 250 ft COLUMN DEPTH c, = 0 in FOOTING LENGTH L - 250 ft BASE PLATE WIDTH b, = 4 in FOOTING THICKNESS T = 12 in BASE PLATE DEPTH b, = 4 in LONGITUDINAL REINF 3 # 4 @ 12 rn o c FOOTING CONCRETE STRENGTH fc = 215 ksi TRANSVERSE REINF 3 # 4 @ 12 m o c REBAR YIELD STRESS fy = 40 ksi AXIAL DEAD LOAD Pct = 4.5 k ' AXIAL LIVE LOAD P'_ = 4.5" k LATERAL LOAD (O=WIND, 1=SEISMIC) = 1 Seismic.SD . SEISMIC AXIAL LOAD P", = 0 k. SD SURCHARGE qs = 0 ksf ;,:•:: - u" SOIL WEIGHT W, = 0.11 kcf FOOTING EMBEDMENT DEPTH D1 = 2 ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Oa = 1.5 ksf FOOTING WIDTH B = 2:5 ft ". • vl ,. FOOTING LENGTH L = 2.5 ft "" • �s...�i BOTTOM REINFORCING # 4 A� �.. THE PAD DESIGN IS ADEQUATE. ANALYSIS DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318.05 SEC.9.2.1) CASE 1: DL + LLP = 9 kips 1.2 DL + 1.6 LL CASE 2: DL+LL+E/1.4 P = 9 kips 1.20L* 1.0 LL+1.OE CASE 3: 0.9 DL+E/1.4 P = 4 lops 0.9 DL+1.OE CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2) p CASE 1 CASE 2 CASE 3 q,,m, v _ - + y , + (U. 15 - 1.48 ksf, 1.46 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 316-05 SECA5.4.2. 10.2, 10.3.5. 10.5.4, 7.12.2, 12.2. & 12.5) I t/ +r 11 o>:SI' (I - I- 11 .< Pu = 13 kips Pu = 10 kips Pu = 4 kips //A* 11.0018- -!) 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 I1min 0.001 0 001 A, 0.17 0.18 RegD 1 # 4 1 # 4 Max. Spacing 18 in o.c 18 in o.c. USE 3 # 4 @ 12 in o.c. 3 # 4 @ 12 in o.c Pmax 0.019 0.019 Check Pprod ' Pmax [Satisfactory] [Satisfactory] • r� • CK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3. 15.5.2, 11.1.3.1. 6 11.3) Or, = 20hd f 1. ECK PUNCHING SHEAR (ACI 318.05 S,EC.15.5.2. 11.12.1.2. 11.12.6, F 13.5.3.2) 01"n=(2+.Y)o.lr•A = 54.98 kips where o = 0.75 (ACI 318-05, Section 9.3.2.3 ) N = ratio of long side to short side of concentrated load = 1.00 b0 = C1+C2+b1+b2+4d = 42.5 in AP = bb d = 366.6 in - y = MIN(2 , 4 / (ic. 40 d / bb) = 2.0 f'u - Pu. max I lf/ i I>; + c, 4r/l I ' h' `''" d )� = 11 02 kips < c5 V [Satisfactory] J (cont'd) 97 LONGITUDINAL TRANSVERSE Vu 2.21 2.31 0.75 0.75 Wn 19.7 19.1 Check V. < Wn [Satisfactory] [Satisfactory] ECK PUNCHING SHEAR (ACI 318.05 S,EC.15.5.2. 11.12.1.2. 11.12.6, F 13.5.3.2) 01"n=(2+.Y)o.lr•A = 54.98 kips where o = 0.75 (ACI 318-05, Section 9.3.2.3 ) N = ratio of long side to short side of concentrated load = 1.00 b0 = C1+C2+b1+b2+4d = 42.5 in AP = bb d = 366.6 in - y = MIN(2 , 4 / (ic. 40 d / bb) = 2.0 f'u - Pu. max I lf/ i I>; + c, 4r/l I ' h' `''" d )� = 11 02 kips < c5 V [Satisfactory] J (cont'd) 97 r • � 0 • Reza PROJECT: Max. Load For 3.0 sq. ft. Pad Footing (1500psf) PAGE: As har1pour CLIENT: DESIGN BY: R.A. JOB NO.: DATE: REVIEW BY: R.A. PAd:Footiri ";Desi' li,Btised-on:ACI.318-05..:�r. LONGITUDINAL TRANSVERSE d 8.75 8.50 b 36 36 q u.max INPUT DATA 2.02 Mu 6.09 DESIGN SUMMARY p 0.001 0.001 COLUMN WIDTH C: = 0 in FOOTING WIDTH B = 300 It COLUMN DEPTH Ca = 0 in FOOTING LENGTH L = 300 h BASE PLATE WIDTH b; = 4 in FOOTING THICKNESS T = 12 ,n BASE PLATE DEPTH b•_ = 4 in LONGITUDINAL REINF 3 # 4 @ 15 .n o c FOOTING CONCRETE STRENGTH fc' = 2.5 ks, TRANSVERSE REINF 3 # 4 @ 15 ,n o c REBAR YIELD STRESS fy = 40 ksf AXIAL DEAD LOAD Pr,. = 6.5 k ' AXIAL LIVE LOAD PL, = 6.5 k LATERAL LOAD (O=WIND, 1=SEISMIC) = 1 Seismic.SD SEISMIC AXIAL LOAD PLN, = 0 k, SD SURCHARGE Cls = 0 ksf SOIL WEIGHT ws = 0.11 kcf FOOTING EMBEDMENT DEPTH D1 = 2 ft FOOTING THICKNESS T = 12 in ALLOW SOIL PRESSURE Oa = 1.5 ksf FOOTING WIDTH B = 3 ft l' FOOTING LENGTH L = 3 ft BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. 4LYSIS IGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1) E 1' DL + LL P = 13 kips 1.2 DL + 1 6 LL E2: DL+ILL +E/1.4 P = 13 kips 1.2 DL+10ILL +1.0E E3: 0.9 DL+E/1.4 IP = 6 kips 0.9 DL+1.0E CK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2) 1, CASE 1 CASE 2 CASE 3 rl.%IAA' Bl. +'7A (() 15 - n, )1`= 1,48 ksf, 1-48 ksf, 0.69 ksf q MA. < k O 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— II- 1`/rr II RS Pu = 18 kips Pu = 14 kips Pu = 6 kips LONGITUDINAL TRANSVERSE d 8.75 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 A, 0.31 0.32 RegD 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 Pproo < pmax [Satisfactory] [Satisfactory] i ry. • lie :CK FLEXURE SHEAR (ACI 318.05 SEC.9.3.2.3. 15.5.2, 11.1.3.1, & 11.3) 41'11-2ohcl f. CHECK PUNCHING SHEAR (ACI 318-05 SEC.15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2) Or', = � � + !) O ./ r..4 = 54.98 kips where 0 - 0.75 (ACI 318-05, Section 9.3.2.3 ) li, ratio of long side to short side of concentrated load bo - c, + c2 + b, + b2 + 4d = 42.5 in AP = bo d' = 366.6 y = MIN(2 . 4 / [ic . 40 d / bo) = 2.0 1` it ^ Nu. mai �I - / I , r/ 1 = 16.61 kips HL\ 2 )I a 1.00 < ya V [Satisfactory] (cont'd) LONGITUDINAL TRANSVERSE VU 4.17 4.30 p 0.75 0.75 On 23.6 23.0 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) Or', = � � + !) O ./ r..4 = 54.98 kips where 0 - 0.75 (ACI 318-05, Section 9.3.2.3 ) li, ratio of long side to short side of concentrated load bo - c, + c2 + b, + b2 + 4d = 42.5 in AP = bo d' = 366.6 y = MIN(2 . 4 / [ic . 40 d / bo) = 2.0 1` it ^ Nu. mai �I - / I , r/ 1 = 16.61 kips HL\ 2 )I a 1.00 < ya V [Satisfactory] (cont'd) �J 0 Reza LONGITUDINAL TRANSVERSE d 8.75 8.50 b PROJECT: Max. Load For 3.5 sq. ft. Pad Footing (1500psf) 1 94 PAGE: 9.44 As har CLIENT: 0.001 0.001 PmIn 0.001 DESIGN BY: R.A. g p OUr JOB NO.: RegD 3 # 4 DATE: Max. Spacing REVIEW BY: R.A. .Pad Footirt�.Desi-n:Based:on ACI�31:8-05, 4 # 4 @ 12 in o.c. 4 # 4 @ 12 in o.c. Pmax 0.019 0.019 Check PProd < Pmax [Satisfactory] [Satisfactory] INPUT DATA DESIGN SUMMARY COLUMN WIDTH C. = 0 in FOOTING WIDTH B = 350 ft COLUMN DEPTH C, = 0 in FOOTING LENGTH L = 3.50 h BASE PLATE WIDTH b; = 4 in FOOTING THICKNESS T = 12 In BASE PLATE DEPTH b, = 4 in LONGITUDINAL REINF 4 # 4 @ 12 in o c FOOTING CONCRETE STRENGTH fc' = 2.5 ksi TRANSVERSE REINF 4 # 4 @ 12 in o c REBAR YIELD STRESS fy = 40 ksi AXIAL DEAD LOAD Pa = 8.5 k ' AXIAL LIVE LOAD PLL = 8.5 k LATERAL LOAD (O=WIND. 1=SEISMIC) = 1 Seismic•SD SEISMIC AXIAL LOAD PLA] = 0 k. SD SURCHARGE qs = 0. ksf SOIL WEIGHT ws = 0.11 kcf FOOTING EMBEDMENT DEPTH DI = 2. ft FOOTING THICKNESS T = 12 in 7 ALLOW SOIL PRESSURE Oa = 1.5 ksf FOOTING WIDTH B = 3.5 h FOOTING LENGTH L = 3.5 h :d '•; - -%^• � f,. .. , BOTTOM REINFORCING # 4 THE PAD DESIGN IS ADEQUATE. 1LYSIS IGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1) E 1: DL + LL P = 17 kips 1.2 DL + 1.6 LL Pu = E2: DL+LL+E/1.4 P = 17 kips 1.2 DL + 1.0 LL + 1.0 E Pu = E3: 09 DL+E/1.4 P = 8 kips 0.9 DL+1.OE Pu = CK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2) /, CASE 1 CASE 2 CASE 3 -- (0- t 5 - w,j 7' = 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. GN FOR FLEXURE (ACI 318-05 SEC.15.4,2. 10.2. 10.3.5. 10.5.4, 7.12.2. 12.2. 8 12.5) 0.95!!1-I- 1/� ~� 0.958 +. `1 1! 0.383h,121*' c _ `� /- - d /� 1 24 kips 19 kips 8 kips LONGITUDINAL TRANSVERSE d 8.75 8.50 b 42 42 q u,max 1.94 1 94 Mu 9.44 9.44 P 0.001 0.001 PmIn 0.001 0.001 A, 0.48 0.50 RegD 3 # 4 3 # 4 Max. Spacing 18 in o.c. 18 in o.c. USE 4 # 4 @ 12 in o.c. 4 # 4 @ 12 in o.c. Pmax 0.019 0.019 Check PProd < Pmax [Satisfactory] [Satisfactory] 24 kips 19 kips 8 kips • • (conva) CHECK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3. 15.5.2. 11.1.3.1. & 11.3) 0i 20hdF CHECK PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2. 11.12.6. & 13.5-3.2) Y�I'II 't2`} 54.98 kips where o = 0.75 (ACI 318.05. Section 9.3.2.3 ) PC = ratio of long Side to short side of Concentrated load = 1.00 bo = C1 + CZ + b, + by + 4d = 42.5 in Ap = bo d = 366.6 y = MIN(2. 4 / lic . 40 d I bo) = 2.0 I' it ' !'u, masI I ` /ij 2—=-7--- r d ).1 22.28 kips < 0 V n [Satisfactory] LONGITUDINAL TRANSVERSE V„ 6.38 6.52 6 0.75 0.75 pV„ 27.6 26.8 Check V. < it V„ (Satisfactory] [Satisfactory] CHECK PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2. 11.12.6. & 13.5-3.2) Y�I'II 't2`} 54.98 kips where o = 0.75 (ACI 318.05. Section 9.3.2.3 ) PC = ratio of long Side to short side of Concentrated load = 1.00 bo = C1 + CZ + b, + by + 4d = 42.5 in Ap = bo d = 366.6 y = MIN(2. 4 / lic . 40 d I bo) = 2.0 I' it ' !'u, masI I ` /ij 2—=-7--- r d ).1 22.28 kips < 0 V n [Satisfactory]