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