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09110017 - CALCSFebruary 9, 2011 Building Department City of Cupertino 10300 Torre Avenue Cupertino, CA 95014 RE: A New Residence for Mrs. Martha Kang , C 9 20874H�0D g Cupef inoa James C. Chen, C. E., S. E. P.O. Box 20302 San Jose, CA 95160 Phone (408) 268 - 0612 rw- f _n ' ss Our Job No. 1101 Dear Inspector: I performed the Job Site framing review and check the field change for some members, all changes are checked OK. Please see the result of review as follows: 1. Dwg. S-4, use 2x10 rafters @ 24" o. c. in lieu of 2x8 rafters @ 16" o. c. as shown on Plan. For 2x10, 5=21.30 cu. in., for 2x8, S=13.14 cu. in.. 13.14x24/16 = 19.71 < 21.30. Check O. K. 2. Dwg. S-4, for Collector Simp. CS 16 is O. K. to nail to 2x10 rafter in lieu of 4x8 as shown on Plan, because the 1 '/2" width of 2x10 which is good to nail 1 '/4" wide CS 16. 3. Dwg. S-4, use 4x12 header in lieu of 8x8 as shown on Plan. For 4x12, 5=73.83 cu. in. which is larger than S=70.31 cu. in. for W. Check O. K. 4. Dwg. S-4, use a "A" Truss in lieu of TS 8x4x3/8 as shown on Plan. Please see attached calculation.. Check O. K. If you have any questions in regarding to this Very truly yours, 0. 2282 Exp IV311 �� James C. Chen, C. E., S. E. Copy to: Martha Kang, Owner Hesitate to call me. JAMES C. CHEN, C.E., S.E. P. O. BOX 20302 SAN JOSE, CA 95012 (408) 268 - 0612 TRUSS - 1 JOB NO. t (d 9 SHEET NO. �_ OF I DESIGN -d C -- DATE H2 M:= R -LI — _.w -L12 M= 3.01 x 103lbf-ft 2 TOP CHORD: TRY 4 X DF # 1 Fb := 1000• Ibf Fv := 95. Ibf . 2 M m m 2 Sreqd Sregd = 28.86 in3 1.25 -Fb 1.5-(R — w -ft) 2 Areqd := Aregd = 36.83 in 1.25-Fv USE 4 X 12 ( A:= 39.38-in2 SCF := 73.83.1.1•in3= 81.21-in3 1:= 415.28-in4 ) BOTTOM CHORD Areqd :_ T Ft. 1.25 USE 2-2X8 TRY 2 X DF NO.1 Areqd = 9.64 in A:= 10.88-in2 ) USE 5/8" � BOLTS p:= 2240-Ibf q:= 1410.1bf 0 := atan(s) Fn := p.q p-sin(0)2 + q•cos(0)2 T n:= Fri. 1.25 F1 Fn = 2.12 x 103 lbf n = 3.08 USE (3) — 5 -in Bolt 8 15 4 L:= -ft s:= H:= L -s H=2.5ft 2 12 L1 := 1 -ft L2:= L — Ll L2 = 6.5ft Hl := L1 -s H1 = 0.33 ft 1-12:= H — Hl H2 = 2.17 ft H L3:= 7 -ft Ibf l bf w:= (40.4.5)• w = 180 ft ft P:= (680 + 680)-Ibf see original calc. page 7 P = 1.36 x 1031bf P1 := (533 + 533)•Ibf P1 = 1.07 x 103Ibf R:= w -L + P + P1 R = 3.1 x 103Ibf 2 R -L — 0.5 -w -L2 — Pl-(L — L3) 3 T := T = 8.13 x 10 Ibf H2 M:= R -LI — _.w -L12 M= 3.01 x 103lbf-ft 2 TOP CHORD: TRY 4 X DF # 1 Fb := 1000• Ibf Fv := 95. Ibf . 2 M m m 2 Sreqd Sregd = 28.86 in3 1.25 -Fb 1.5-(R — w -ft) 2 Areqd := Aregd = 36.83 in 1.25-Fv USE 4 X 12 ( A:= 39.38-in2 SCF := 73.83.1.1•in3= 81.21-in3 1:= 415.28-in4 ) BOTTOM CHORD Areqd :_ T Ft. 1.25 USE 2-2X8 TRY 2 X DF NO.1 Areqd = 9.64 in A:= 10.88-in2 ) USE 5/8" � BOLTS p:= 2240-Ibf q:= 1410.1bf 0 := atan(s) Fn := p.q p-sin(0)2 + q•cos(0)2 T n:= Fri. 1.25 F1 Fn = 2.12 x 103 lbf n = 3.08 USE (3) — 5 -in Bolt 8