Specifications RECFIVED
MAY 2 9 1.018
CITY OF T;GARD Igiff
BUILDING Dla,isioN
•
May 17,2018
Mr.Scott King
Sixty First Place Architects
7120 E. Indian School RD Suite l
Scottsdale,AZ 85251
Re: Din Tai Fung—Washington Square
Tigard,OR
Dear Scott:
Attached please find calculation sheets 1 through 8, dated May 17, 2018,which verify the structural
adequacy of the Din Tai Fung—Washington Square project,as shown on drawings 50.1 through 51.2
dated May 17, 2018. Design is based on the requirements of the 2014 Oregon Structural Specialty
Code, based on the 2012 International Building Code.
If you have any questions or need further information, please call me.
Sincerely,
; -
Travis C.Smith, PE
10 CT(44
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Project: Din Tel Fung By: TCS Sheet No.
laCiaff Location: Portland,OR Date: 1
Client: Revised: Job No.
Subject: Date:
Rooftop Mechanical Unit AS°
•
Roof Loads:
ASD Load Combinations: RTU Unit Name:AC-5
1.) 1.OD+1.0(LrorS) H= .' 54. '.. in
2.) 1.0D+0.75(LrorS)+0.75•(0.6W) W= ' 84 `,in
3.) 1.0D+0.75(LrorS)+0.75`(0.7E) L= 121 in
4.) 0.60+0.7E/0.6W Wt= 1900 lbs
Curb h= 14 in
Ladd'I= . .12 in
Gravi : Lateral:
Dead: Wind:
Deck= . `.:2.5 psf V= .`:.120 : :MPH
Rigid Insulation= 2 Kz= 0.7,;
Roofing/Waterproofing= 4 psf Kzt= 1
MEP/Ceiling= 4 ' psf Kd= 4.9
Framing= 4 ` `i psf
Mist= 3.5 . psf qz=.00256 x Kz x Kzt x Kd x V=2 ASCE 7-10\Eq.29.34
Total DL= 20 psf qz= 23.2 psf
GCr= ,T. 1.9':i ASCE 7-10.29.5.1
Live: Af= 45.4 ft2
Live Roof= 20 psf
F=qzxGxCrxAf
Total L= 20 psf
Fw= 3804.2 lbs
Snow: ASCE 7-10\Ch 7.7.1
Pg= 20. psf
Ct= `::`. 1
Ce= 0.4 Seismic:
I= 1 SDS= . 0.72 Per USGS
ap= 1
pf=0.7xCexCtx1xPg Rp= 2.5.
1= 1
Pf= 12.6 psf
Pf min= 20 psf Fp= 0.4 x ap x SOS x Wp '{1+2(z/h)j (ASCE 7-10,Ch 13)
(Rp/I) note:z=h
5= 20 psf
Local Minimum= 25 psf Fp= 656.6 lbs
Design Snow Load= 25 psf
D= 20 psf Ig-it Wind Controls
Lr= 20 psf F= 3804.2 lbs
5= 25 psf
{ F.
1.) 45 psf f ji
2.) 38.75 psf t P
3.) 38.75 psi z;
4.) 18 psf p�
l A2 Rt
eMECHANICAL UNIT ELEVATION VIEW
RoottopdachanicaLUoit Printed•S/ .1$at 10.35_an page lot 2
1<Pff Project: Din Tai Fung By: TCS Sheet No.
Location: Portland,OR Date: 01/00/00 2
Client: Revised: lob No.
Subject: Date: 0
Rooftop Mechanical Unit ASD
Roof Loads:
ASD Load Combinations: RTU Unit Name: Short Direction B controls
1.) 1.0 0+1.0(Is or S) H= 54 in
2.) 1.00+0.75(Lror5)+0.75*(0.6W) 8= 84.0 in I4-----"B------111
3.) 1.0 D+0.75(1r or S)+0.75*(0.7E) Depth= 121.0 in
4.) 0.60+0.7E/0.6W Wt= 1900 lbs
CurbH= 14 in z T +F
DISTRIBUTE GRAVITY LOADS L add'I= 12 in a
TO EACH OF FOUR SIDES -II L adds )4-1
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MAIN SUPPORTS MECHANICAL.UNIT ELEVATION VIEW
1
) MECHANICAL UNIT PLAN VIEW
Maximum Compression Reaction
R1=0.75*(0.7 or 0.6)F*(Curb H+H/2)*1/B+Wt/4
Controlled by LC 2,or 3
•
R1= 655.3 lbs When Applied at(2)corners
Controls--> R1= 130.0 plf When Applied on whole side
Maximum Tension Reaction
R2=(0.7 or 0.6)F*(Curb H+H/2)*1/B-0.6*Wt/4
Controlled by LC 4
R2= 414,5 lbs When Applied at(2)corners
R2= 82.2 plf When Applied on whole side
V=(0.7 or0.6)F/(2B+2Depth)
V= 66.8 plf When Applied on entire perimeter
V= 570.6 lbs When Applied at(4)corners
Side Member Demands: Gravity loads in addition to overturning
Tw= 1 ft
Mu= 12.40 k-in w1"2/8
Vu= 0.59 k wl/2
Main Member Demands:
Tw= 1 ft
Mu= 36.95 k-in wlA2/8
Vu= 1.02 k wi/2
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Project: Din Tai Fung By: TCS Sheet No. "
Location: Date:
1 Client: Revised: ` Job No.
Portland,OR Date:
Design Forces&Geometry:
Mu= 3.08. k-ft Lb= 7 ft Leg in Compression= Short
Section Size:
L-Shape: L4X4X5/16
d= 4 in lx= 3.67 in"4 ly= 3.67 in"4 lz= 1.46 in44
t= 0.313 in Sx= 1.27 in^3 Sy= 1.27 in"3 Sz= 0.936 in"3
b= 4 in Zx= 2.26 in^3 Zy= 2.26 inA3 rz= 0.781 in
A= 2.4 inA2 rx= 1.24 in ry= 1.24 in Ow= 0 in
Material Properties
Fy= 36 ksi = 0S
E= 29000 ksi
AISC 13th Edition
DESIGN Ref Section
Compactness Check: Table B4.1b
bit= 13 Case 10: Xp= 15 =0.54*(E/Fy)^0.5 Compact Section
Ar= 26 =0.91'(E/Fy)"0.5
Flexure Design
Lb= 84 in Cb= ; .1.
Yielding
Mn= 68.58 k-in =1.5'Fy'Sx Eq(F10-1)
5.715 k-ft
Lateral-Torsional Buckling
For single angless without continuous lateral-torsional restraint along the length
My= 45.72 k-in =Fy*Sx
Me= 248.935 k-in =(0.46•E*b"2*t^2`CbyLb-Equal Legs Eq(F10-4)
(b) When Me>My
Mn= 65 k-in =((1.92-1.17•(My/Me)"0.5J*My<1.5My Eq(F10-3)
5.40 k-ft
Leg Local Buckling
(a)For compact sections,the limit state of leg local buckling does not apply
Nominal Flexural Strength
Mn= 5A0 k-ft
pMn= 4.9 k-ft
DCR= 0.63 OK
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