Plans • MAA /13/2008/THU 05:25 PM P. 002
IvICGINNIS & ASSOCIAlES 1110 WESTMARK ST. LOUIS, MO 63131 -1735
CONSULTING ENGINEERS, INC. PHONE; (314) 469 - 6460 FAx: (314) 469 - 0319
March 13 2008 RECEIVED
Mr. Jim Gross MAR 1 4 Z008
Tuff Shed Store 160 CITY OF TIGARD
6500 NE Halsey St. #A BIJILDINGDItJISION
Portland, OR 97213
RE: Modifications to Plans for Monte Sellers's Accessory Building
10600 SW Fairhaven, Tigard, OR
Tuff Shed Project 160 - 541851
McGinnis & Associates Job #15685
Dear Mr. Gross,
The following comments are regarding plans provided by McGinnis and Associates for the above
referenced project dated January 7 2008. From my understanding, you wish to modify the plans such that the
walls are framed using 2x4 studs rather than 2x6's as specified on the schedule on sheet A2. Hem Fir grade #1 will
be used for the lumber. As the attached calculations indicate, the studs can resist the required gravity and lateral
loading as specified on the general notes of the plans. It is therefore permissible to make the modification.
if you have any questions or need any further information, please do not hesitate to call.
Sincerely,
MCGINNIS AND ASSOCIATES,
CONSULTING ENGINEERS, INC.
;
Jeffrey S. Austin, P.E. s o PROF
r 1 73 frg
• oREGO • .`y
Daniel W. McGi nis, P 4 �FC W, M�Q
• xrasa
CJ � ,z0 OOI
City of Tigard
Approved Plans
Byc.s• Date /s/.4,,
/ 6Ka. sw 1G .4 kk•Itn
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- MAR/13/2008/THU 05:26 PM P. 003
McGINNIS & ASSOCIATES JOB: TS Accessory Building
Consulting Engineers, inc. #160-541851 NO. 15685
1110 Westmark Drive SHEET NO. 7 OF 12
Saint Louis, Missouri 63131-1735 CALC. BY JSA DATE 3/13/2008
Consider 2x Wood Studs- Ott No of Members: Ir.:11
... . , 80 . , . '
Commercial # of b d Area Sx l E
Species Grade Member Size Mem. (in) (in) (in (in (1n (kel)
Hem-Fir No. 1 2x4 1 1.5 3.5 5.25 3.06 5.36 1500
Plate Height: ilitri4;1906V4* Unbraced Length iggro Unbraced Length: kl := k.ht iklSd.90*
Factor:
Compression Parallel to Grain
Non-Adjusted Design Value: Ifft
PROp
Grading Adjustment Factor: istigA (See NDS Section 3.7.1)
e:comge - .gi b
4,10
Coefficient Used for Column RFC (See NDS Section 3.7.1) 8 3PE
Stability Factor
Maximum Bending giTtat4 Effective Length Factor: k := 1.84 Le := k itigl 84' lir
Unbraced Length: . — (Ref NDS Table 3.3.3) IIP A oREGON . k co
f / I.e
'' 1: le• ell+ r3(
Determine the allowable axial Compressive stress per NDS Section 3.6 $ w.
.x IE
1 KciluetmtriT*Opitsi.4 '':,, :!
CM
C, C C1 C CT
1.00 1,00 1.00 1.00 1.15 1.00
Modified Modulus of Elasticity: E' := E-Cm-Ct E' = 1500000 psi
KoE.E
Stress used for column FOE := . FcE - 413.22 psi
stability factor: kj )2
( ddressed (id))
1 +
FcE ( FcE 2 FcE
Column stability Fc F Fc
factor: CP := 2-c 2.c c Cp = 0.28
Allowable axial compression
design stress w/ no lateral load: _
0,00: F :.= Fc lakgalta4,1
Maximum Allowable
PallowNo_LL := A (id) F 151147214414M
Axial Load w/no lateral load:
Allowable axial compression
design stress with lateral load: IORM Pc_w_LL :- Fc r!" kk4*.P01
. . __
I p;‘0_pRomorsvis000visacrokisseursisejansi
• •MAr, /13 /2008 /THU 05:26 PM P. 004
McGINNIS & ASSOCIATES JOB: TS Accessory 13uildina
Consulting Engineers, Inc. #160-541851 N0. 1 685
1110 Westmark Drive SHEET NO. 6 OF 12
Saint Louis, Missouri 63131 - 1735 CALC. BY JSA DATE 311312008
Determine the allowable bending stress per NDS Section 3.3
Determine the slenderness RB := if L Oft, ,1 RB = 5.86
bdressed (id)
Coefficient used for beam KbE := 0.439 Stress value used for beam KbE' El
stability factor: stability factor: FbE := 2 FbE = 19.1 ksi
R
B
Tabulated bending design Fb_materlal = 975 psi
value from Table 4A:
Tabulated bending design
value multiplied by applicable Fbstar Fb material'CM'Ct
adjustment factors: _
FbE FbE 2 FbE
1+
Fbstar Fbstar Fbstar
Beam stability factor. CL := If Lu > 0 •ft, 1.9 1.9 0.95 ' 1.0 CL = 0.99
Allowable bending stress: Fb allow := Fb material'CM'Ct'Ci'CD'CL'Cr ! °laj}ovii` � '�0
Lateral pressure: P w 14.17 ps Tributary width of member: logot4
Maximum moment due 2
to lateral pressure; Mmax Pwhtw-twidth•ht •0.125 Mmax = 2191b
Mmax fb = 541 psi
Maximum bending stress: fb :=
S ■
2
Find the maximum compressive stress to f4 fb
satisfy the bending and axial compression ,c + fc S 1.0 (NDS equation 3.9 -3)
interaction equation. Fb_allow 1 — ')
F OE
5 compressive y f:l::' 9401
Solving for the com ressive �` ��a •,,,,,,,., ;,t,
stress results in:
Maximum allowable Pallow w LL A(ld)•fc P'al'ow, 4v; = =L)4E', Z'
axial load wd lateral load:
Controlling allowable axial load; ' ' :149p lid l I .. ; ... Aid00 g0WiiiteraLlaad. goyarrrs ,. .
Rroof DL+ . LL �.-
Required axial load capacity: Prequired = f 4' Ilb"
rtspacing - twidth �' ` Pr iliii* "
5 • Pwhtw'twldth• ht _E 1,
Stud deflection @ mldheight: Alateral °lateral = 0.45 in T 5 _ ; ,AP ', {
384E 1 x(id) .44te. 44':.E: :u::,.
P: \0 PROJECTS115000115600115685 75190_541851