Specifications (13) RECETITED
Approved plans .
shall be on job site.
JUL I , ?OP
STRUCTURAL CALCULATIONS
BUILDING
CIITYF TIGAR(y
_DI DIVISION
To OFFICE COPY
Washington County
For CITY OPTIGARD
REVIR D FOR CODE COMPLIANCE
10824 SW 118th Ct.,Tigard,OR Approved: I
OTC: I
Bearing wall demo •
Permit#: - 00 A_--7
Submitted Address: A a u Sw f g
Suite#:
5/22/17 By: Date: 7
TABLE OF CONTENTS
Calculations 1-7
Plan markup SK1
Framing detail SK2
The bearing wall between the dining room and living room is being demoed. A glulam beam is being added to
support the floor framing. Due to an adequate stem wall foundation no foundation modifications/additions are
required.
If there are any questions please contact David Linton, 541-510-7200. p PROFfiss
0AGINEER
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5/22/2017
Page I 1
Wood Rafters
PROJECT NAME: 10824 SW 118th Tigard DATE: 5/22/17
E Unit Definitions
Q FLOOR BEAM
Basic Input Input OutputVeri
! i f iN ttAtast ` is
Cleft)il ;a e.�
3i.•
: ," . ' - / bearin2 _ i, 4-
) .., v. ,, • -. 4i or 1
:4n
3
(15.67ft 10.33ft wa= 156•plf
)
wd= 12psf 2 + 2
40psf wL=520 plf
wL wd 12psf
w=676•plf
w:= wd+ wL
1 := 19ft
M =3.1 x 104 ft-lb
VF,. -—, V=6.4 x 103 lb
2ti
I ptot=0.95 in
Otot 240
1 OL =0.633 in
OL _ 360
Wood Properties -24F-V4 Glulam
Fb:= 2400psi F,,:= 265psi E:= 1800000psi
CD:= 1.0 Cfu:= 1.0 CM:= 1.0 Ce= 1.0
Ct:= 1.0 CL:= 1.0
C;:= 1.0
b =.5$'n d'•= 15in
1
3
S:= b d2 =206.25 in3 A:= b•d= 82.5-in2 I:= b•12 = 1.547 x 103 in4
6
21ft10'1 12in10.1 5.125in10'1 Cv=0.98
C�= min 1.0, 1 I \ / b J
3
Fb, := bD' M' t-CmF •C C C C C C Cc Fb, =2.354x 10 -psi
FV Fv•CD•CMCtC, Fv=265•psi
E'•— ECM C, E`=1.8 x 106-psi
M SCeq= 155.505•in3 SCeq =0.754
SCeq F S
b
1.5 V ACeq=36.351 int ACeq =0.441
Aieq:= ":1,
A
5./ 4 5 w l4
1„q:= max384E'•0tot 384E'•0Lj 'req= 1.338x 103-In 1Ceq =0.865
\
GLB 5.5x15 1
0 FLOOR BEAM
2
WOOD COLUMN
PROJECT NAME: 10824 SW 118th Tigard DATE: 1/31/17
Unit Definitions
U FLOOR COLUMN
Column Loading and Properties
B := 1 1 Solid Sawn Member
2 Glued Laminated Member
3 Round Timber Pile
P:= 64001b P=6.4 x 103 lb
ht:= 8-ft
b := 5.5-in If round timber pile, b is the diameter
d:= 5.5-in
Ke:= 1.0 Column Factor Refer to NDS Appendix G
le:= Ke-ht le=8 ft
(1
le = 17.455 < 50 CheckO:= if e <_ 50,1,"Not Good"�d / Gheck0= 1
d
lb,:= ht Unbraced length for bending
c':= 0.8 if B = 1 0.8 for sawn lumber
0.85 if B =3 0.85 for round timber piles
0.9 for glued-laminated timbers
0.9 if B =2
c' =0.8
KcE:= 0.3 0.3 for visually graded lumber Kbe := 0.439 0.439 for visually graded lumber
0.418 for glued-laminated products and 0.610 for glued-laminated products and
machine stress rated sawn lumber machine stress rated sawn lumber
3
Material Properties:
Emin 580ksi
E:= 1300-ksi CM:= 1.0 Wet Service Factor, Table 4A
Fb:= 1200-psiCD:= 1.0
Load Duration Factor, See Table 2.3.2
Ft:= 825-psi Ct:= 1.0 Temperature Factor, See Table 2.3.3
Fc:= 1000-psi
CT:= 1.0 Buckling Stiffness Factor, Section 4.4.2
Cr:= 1.0 Repetitive use Factor(1.15 for members used
1 as studs. See Table 4A.
_ 12 to 9
if B = 1 Size Factor, See Section 4.3.6
CF' d For Sizes less than 12 deep
1 see Below
10 CF= 1.091
1291.5iin2•ft if B =2
d b ht Cfu:= 1 Flat use factor, see Table 4A
Cf:= 1 Form factor, See Section 3.3.4
CF:= 0.74 Input Size factor for smaller Members if Applies, See Table 4A
F"c:= FC.CD.Cm.Ct.CF F"c=740•psi
Emin= Emin'Cm.Ct-CT Emin=580•ksi
4
Column Compression w/o Bending
0.822•E'min
FcE
�1el2
FcE ( FcE.\2 FcE
1 + - 1 + — —
F"c F"c F"c CP =0.876
:— 2 c' — \ 2 c' ) c'
F'c=648.294•psi
Fc:= F �Cp
f :—_ P fc=211.6 psi Checkl := if(fc<_F'c,"OK" ,"Not Good") Checkl ="OK"
b•d
5
11111111111111111111111111111111.11.11.1........... -
COMBINED LOADING, SECTION 3.9
Flexure and Axial Compression
M:= P•1.5in M =800 ft.lb
2
S:= b d S=27.729•in3
6
F't:= Ft.CD CM Ct CF
Fbs:= Fb-CD-CM-Ct-CF.Cfu"Cr'Cf
lbT d�
5
RB : 2
\ b
Kbe•E
FbE:= 2
RB
FbE FbE 2 FbE
1 + — 1 + — --
Fbs Fbs Fbs
CL. 1.9 1.9 j .95 _ CL=0.999
M
fbl := s f b2:= 0
F'bl Fb"CD"CM"Ct-CF•Cr Cf-CL
F'b2= Fb.CD-CM-Ct-CF•Cfu Cr Cf
let := le
KcE"
E'min KcE"E'min Kbe"E!min
FbEI :=
FcEI FcE2 2 2
l 2 lee RB
d)
(-1)
fe\2 fbi fb2
+ + =0.727
F', f / f \2
F, 1 c — bl
bl FcEI F�b2' 1 — FcE2 \FbEl _
/fc
2 fbt fb2 Check < 1, OKNG:= if — + + - 2 Check4="OK"
4 F'c fe fc \ fb I
F'bl 1 -- -
FcE1 F b2" 1 — FcE2 \FbE1 _
6
Foundation
qa:= 1500psf fe:= 2000psi fy := 60ksi
P=6.4x 103 lb P,,:= P•1.6
Areq=4.267 ft2
Aieq._ _
q
W= 1.25 ft
W:= 15in
Areq 1 req=3.413 ft
LCeq ._
Existing stem wall and foundation are adequate for point load from column
Fl FLOOR COLUMN
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