Plans : ECLIPSE ECLIPSE - ENGINEERING . COM
E N G I N E E R I N G
22nd, RECEII\IED
January 22 , 2014
JAN 2 9 2014
City of Tigard CVTYOF lG VON
Attn: Dan Nelson BUILDING
Re: American Eagle Outfitters—Store#2056
Washington Square mall
Plan Check Number=BUP2013-00308
Dan:
This letter was prepared in response to the comments recently issued for the American Eagle Store. Please
note that Eclipse Engineering, Inc. is only re-submitting structural details and calculations for the shelving racks
and storefront framing. All other submittal requirements shall be provided by the project architect or their
consultants.
Review Corrections:
1) Noted.
2) We have provided calculations for the seismic bracing of all metal stud walls. Interior walls shall run full
height to the roof structure so that diagonal bracing is not required. The studs have been sized accordingly.
t 3) We have provided calculations for the metal stud framing and bracing at the storefront per the seismic
requirements of ASCE 7.
4) We have attached the shelving rack calculations completed previously.
Eclipse Engineering, Inc. has only reviewed the structural components within our scope of services, and we do
not take any responsibility for any other portion of the project or plan submittal. Please feel free contact us with
any questions on this portion of the project—(541)389-9659.
Sincerely,
Eclipse Engineering, Inc. .��ckv PROI .
G I N�, /p
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.... — 76024P
Robert VanCamp, PE --
Project Engineer
OREGON
Enclosed: calculations cto 1,oPoots••e Oe
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City of Tigard Fpr vA��
A. •r.v-d Plans Expires 3O•2 4
By Air Date ( .t IA- Digitally signed by „
Roba
Robert �•�•°�
{/�� ON:cn•Pabert VanCamp
1 7 `7 O/ WmNbrvi Engineering.Inc,ou
Va n Ca m p 2011.0112,o-47:05�00
OFFICE COPY
MISSOULA WHITEFISH SPOKANE FEND
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5 EC Li PS E ECLIPSE - ENGINEERING . COM
E N G I N E E R I N G
RECEIVED
JAN 2 9 2014
Structural Calculations CITYOFTIGARD
BUILDING DIVISION
Light Gage Steel Storefront Framing
American Eagle Outfitters — Store #2056
Washington Square Mall
9767 SW Washington Square Rd — Sp. #D07
Tigard, Oregon 97223
PROFe
G I N� ,s/0
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OREGON
C 200
Prepared For: `1 ATV/04°'
'
American Eagle Outfitters
77 Hot Metal Street—6th Floor Expires (043e.2o114
Pittsburgh, PA 15203 Robert g
VanCamp
Please note: Eclipse Engineering, Inc. has reviewed only the adequacy of the storefront to
support the interior vertical and lateral loads of the above noted area. We neither take
responsibility for any other element nor the integrity of the structure as a whole.
MiS':�f'ai.A WHITEFISH
113 West P 'ISabalan .MT59602 1006BM1rMMMSeME MOW MT6O97 maw AMMO M,SO717 Spokane,WA90201 376 SW EU Dive.Suite&Bend,OR 97102
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i EC LI PS E AEO-Store#2056 1/16/2014
E N G I N E E R I N G Tigard, OR RVC
CALCULATE SEISMIC FORCE - Nonstructural Elements
psf:= lb-ft—2
Step 1 -Determine Seismic Use Group of Bldg. SUG:= 11 IE:= 1.0
Step 2-Determine Ss and S1 from maps SS:= 0.948 Si := 0.341
Step 3-Determine the Site Class Class D
Assume Site Class D unless established by the building official or a soils engineer
Step 4-Determine Fa and Fv Fa:= 1.121 F,:= 1.718
(use ICC program, see attached printout)
Step 5-Determine SMS and SM1 SMS:= Fa Ss SMI := FvS1
SMS= 1.063 SM1 = 0.586
Step 6 Determine S and SDI 2 2
p DS DI S DS:= 3'SMS SDI:= 3•SM1
SDS= 0.708 SDI = 0.391
Step 7- Determine Seismic Design Category SDC:= "D" From ASCE 7 Table 11.6-1
SDC:= "D" From ASCE 7 Table 11.6-2
Non-Structural Component Coefficients (ASCE 7 Table 13.5-1):
• 1. Interior Nonstrucural Walls and Partitions
A. All other walls and partitions ap:= 1.0 Rp:= 2.5
Weight of Interior Wall - W P:= 15-psf gypsum/studs/finishing
(conservative)
Total Height of Structure- h:= 304ft
Height to Wall Center of Gravity- z:= 15•ft
SEISMIC DESIGN IN ACCORDANCE WITH ASCE -7, Section 13.3.1
Finax:= 1.6-SDs IE.Wp Finax= 17.003-psf Fmin 0.3.SDs.IE.Wp Fmin= 3.188 psf
F:= 0.4 apR DS Wp I 1 + 2 hJ F= 3.4•psf
l
1E
Fp:= if(F>Finax,Finax,F)
Fp:= if(F<Fmin Fmin F) Fp= 3.4-psf
USE: MIN TRANSVERSE FORCE = 5psf FOR STOREFRONT WALLS
1/15
'i EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STOREFRONT CEILING JOISTS - Drywall Bulkhead
kp:= 1000 Ib ksi:= kp in 2 psf := lb ft–
t 2 plf:= lb•ft 1
Joist Design Data :
Joist Simple Span Length - Lei:= 611 Steel Yield Strength - F := 33ksi
Uniform Weight of Ceiling - pclg;= 5•psf
Modulus of Elasticity- E:= 29000ksi
Tributary Width of Ceiling - wclg:= 16•in
Uniform Load on Joist- Loci:= pclg we1g= 6.7•0f Cantilever Length - le:= 0.ft
Pt Load on Joist- pi := 0•lb
Location of Pt- a:= 0-ft
TRY: 362S162 -33 JOISTS @ 16" o.c.
Area of Joist- A:= 0.262in2 Radius of Gyration, y- ry:= 0.616in
Effective Section - Se:= 0.268in3 Radius of Gyration, x- rX:= 1.450in
Moment of Inertia - 1:= 0.551in 4 Thickness of t:= 0.0346in
Element-
Effective Width of
Element- w:= 1.625in Effective Length Factor- K:= 1.0
SIMPLE SPAN -LIGHT GAGE STEEL JOIST DESIGN
Maximum Design Shear Maximum Design Moment- Conservative
wcj•Lci Pi (Lcj+ 10 wej•Lej2
Vcj:_ + = 201b Mcj:= + p.1 = 30 ft-lb
2 Lcj 8
Allowable Bending Actual Bending Mcj
Stress - Fb:= 0.6.Fy= 19.8•ksi Stress- fb:= — = 1.343•ksi
Se
Allowable Shear Actual Shear Vcj
Stress - F„:= 0.4•Fy= 13.2•ksi Stress- fv:_ — = 0.076-ksi
A
5.wc..Lc4 P..1�Lc2
Load Deflection - Ac,;_ + 0.0642 = 0.012-in
384•E•I E•I
1 fv
c� = 5918 fb = 0.068 = 0.006
Ocj Fb F,
THE 362S162 - 33 JOISTS SPANNING 6'-0" ARE ADEQUATE
TO SUPPORT THE LOADS FOR THIS INTERIOR CEILING.
2/15
•
i EC LI PS E AEO-Store#2056 1/23/2014
N G I N E E R I N G Tigard, OR RVC
LT GAUGE STL STORE FRONT CEILING HEADER - Sect. A.1.501
Header Design Data :
Header Length - Lht := 21.5ft
Uniform Weight of Tributary Width 8-ft
Ceiling - Pcig:= 5•psf of Ceiling - Wei:= 2 = 4 ft
Uniform Weight of
Soffit- Psft:= 5•psf Height of Soffit- Hsft:= 4.ft
Uniform Load on Header- Loh' := Prig Wei+ PstrHsft+ 2•(1.83•plf) = 43.7•plf
TRY: DOUBLE 800S162 - 43
Area of Header- A:= 2•(0.537in2) Radius of Gyration, y- ry:= 0.546in
Effective Section - Se:= 2•(1.019in3) Radius of Gyration, X- rX:= 2.397in
Moment of Inertia - I:= 2•(4.500in4) Thickness of Element- t:= 0.0451in
Effective Width of-
Element W:= 3.25in Effective Length Factor- K:= 1.0
SIMPLE SPAN - LIGHT GAGE STEEL BEAM DESIGN
Maximum Design Shear (at Maximum Design Moment-
Base of Wall) -
2
�"?hl'Lhf Wh1.LhI
Vhl := 2 = 469 lb Mbl := 8 = 2523 ft•lb
Allowable Bending Actual Bending Mhi
Stress - Fb:= 0.6.Fy= 19.8•ksi Stress - fb:_ — = 14.854•ksi
Se
Allowable Shear Actual Shear VhI
:
Stress- F�,:= 0.4•Fy= 13.2•ksi Stress- f�, _ — = 0.437•ksi
A
5w L 4
h'
Transverse Load Deflection - Ah1 := h' = 0.804•in
384•E•I
Lh 1 fb fv
— = 321 — = 0.75 — = 0.033
Oh t Fb Fv
THE DOUBLE 800S162 - 43 HEADER SPANNING 21'6" IS
ADEQUATE TO SUPPORT THE LOADS FOR THIS INTERIOR
SOFFIT AND CEILING.
3/15
. 5 EC Ll PS E AEO-Store#2056 1/23/2014
E N G I N E E R I N G Tigard, OR RVC
LT GAUGE STL STORE FRONT CEILING HEADER - Sect. A.1.500
Header Design Data :
Header Length - i,1,2 :- sil
Uniform Weight of Tributary Width 8•ft
Ceiling - Pelg 5•psf of Ceiling - WcIg 2 = 4 ft
Uniform Weight of
Soffit- Psft:= 5•psf Height of Soffit- Hsft:= 4-ft
Uniform Load on Header- Wh2:= Pclg Wclg+ Psft•Hsft+ 2•(1.08•plf) = 42.2.pIf
TRY: DOUBLE 600S137 - 33
Area of Header- A:= 2•(0.318in2) Radius of Gyration, y- ry:= 0.464in
Effective Section - Se:= 2•(0.455in3) Radius of Gyration, x- rx:= 2.229in
Moment of Inertia - i:= 2•(1.548in4) Thickness of Element- t:= 0.0346in
. Effective Width of
Element- w:= 2.75in Effective Length Factor- K:= 1.0
SIMPLE SPAN - LIGHT GAGE STEEL BEAM DESIGN
Maximum Design Shear (at Maximum Design Moment-
Base of Wall) -
L L
X2 2 2
Vh2:_ h2 = 169 lb Mh2:_ 8 h = 337 ft.lb
2 Allowable Bending Actual Bending Mh2
Stress- Fb:= 0.6.Fy= 19.8•ksi Stress- fb:= — = 4.448•ksi
Se
Allowable Shear Actual Shear Vh2
Stress- F�,:= 0.4•Fy= 13.2•ksi Stress- f�,:_ — = 0.265•ksi
A
5•W •Lh24
h2
Transverse Load Deflection - Ah1 :_ = 0.043•in
384•E•I
Lhl = 59(2 fb = 0.225 4 = 0.02
Oh I Fb Fv
THE DOUBLE 600S137 - 33 HEADER SPANNING 8'0" IS
ADEQUATE TO SUPPORT THE LOADS FOR THIS INTERIOR
SOFFIT AND CEILING.
4,15
ECLI PS E AEO-Store#2056 1/23/2014
E N G I N E E R I N G Tigard, OR RVC
LIGHT GAGE STEEL CONNECTION ANALYSES
No. 8 Screws through 0.033 in Materal - V833:= 164.1b T833:= 72•lb
No. 10 Screws through 0.033 in Materal - V1033 := 177•1b T1033 := 84.11)
MAXIMUM END SHEAR v V 201b V`' V
= 0.122 ' = 0.1 9
FOR JOISTS- ci - ci=-
V833 2•T833
FASTEN CEILING JOISTS TO 362T125 - 33 TRACK WITH (2) # 8
SCREWS & FASTEN TRACK TO LIGHT GAUGE STEEL HEADER
ON WITH (2) #8 SCREW @ 16" o.c.
MAXIMUM END SHEAR FOR HEADERS- max(Vhl,V1i2) = 469 lb
MAXIMUM BRACE REACTION - Ptb:= 0•Ib
NUMBER OF SCREWS USED - Ns:= 3
max(Vhl,vh2) max(Vhl,Vh2) Ptb Ptb
= 0.954 = 0.884 = 0 = 0
3•V833 3•V1033 2'�'ti3; 2•T833
FASTEN CEILING HEADERS TO 362T125 - 33 TRACK WITH
(3) # 8 SCREW PER SIDE OF HEADER & FASTEN TRACK TO
SUPPORTING STRUCTURE WITH (6) #8 SCREWS PER END.
5%15
Wei EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL STUD BRACES - Header/Glass Support
Brace Design Data -4ft max spacing: ((( \\ psf := lb•ft 2
Total Hor¢ontal Load - P:_ (5psf)•(4ft)•I 1- ) = 160 lb plf:= lb•ft 1
Unbraced Length, y- Ly:= 20ft l J ksi:= 1000.1b.in—2
Unbraced Length, x- Lx:= 20ft k:= 1000•Ib
TRY: Double 362S162-33
Area of Brace- A:= 2.0.262in2 Radius of Gyration, y- ry:= 2.0.616in
Effective Section - Se:= 2.0.268in3 Radius of Gyration, x- rx:= 2.1.45in
Moment of Inertia - i:= 2.0.551 in4
Effective Width of Element- w:= 2.1.625•in
Thickness of Element- t:= 0.0346in
Number of Studs- N:= 1
Modulus of Elasticity- E:= 29000ksi
Steel Yield Strength - F Y:= 33ksi
Effective Length Factor- K:= 1.0
Capacity of Column
z E 2
Buckling Stresses - Fey:_ = 7.5•ksi F„:_ i E = 41.8•ksi
KLy112
C )
(K.L„)2
rx
Fe:= if(Fey<FeX,Fey,Fex) Fe= 7.5•ksi
Allowable Stress - Fn:= if Fe> Fy,Fyl l — Fy}Fe F„= 7.5•ksi
2 4•F,J
Effective Area - Ae:= w-t-N Ae= 0.11•in2
Nominal Axial Strength - P„:= Ae•Fn Pn= 848.1 lb
Pn2:= A�2 E Pn2= 661.4 lb
25.7• wl 2
t
Factor of Safety- 52 := 1.92 ft,= 1.92
Pn Axial Load - Pa:_ " Pa= 442-lb
6/15
ECLI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
SIMPLE SPAN - LIGHT GAGE STEEL BRACE DESIGN
Assume Simple Span
Span - L:= Lx= 20ft
Dead Load on Brace - wd:= 0.89p1f wd= 0.9•plf
Axial Load - P:= P•cos(45deg) P= 113.1 lb
Shear Load - Pv:= P P,= 113.1 lb
2
Maximum Design Moment- wd
M :_ M = 44.5 ft lb
8
wd•L
Maximum Design Shear- V:_ + Pv V= 1221b
2
Allowable Bending Stress- Fb:= 0.6•F Fb= 19.8•ksi
Actual Bending Stress- fb:= M fb= 1•ksi
Se
Allowable Moment- M„,:= Se-Fb Maxo= 884.4 if lb
Allowable Shear Stress- F,:= 0.4.Fy F„= 13.2•ksi
Actual Shear Stress - fv:= V A f,= 0.2•ksi
5•wd•L4 L
Dead Load Deflection - O:= 0= 0.1.in — = 2393.9
384•E•I
Combined Axial and Bending Stresses :
M P
+ — = 0.31
Maxo Pa
Capacity of#8 TEK screw Ve:= 164-lb
Number of Screws Required - N:= ceil( P) N= 1
Ve
USE: (2) 362S162-33 LIGHT GAGE STEEL STUD BRACE @ 4'-0"
o.c. WITH (3) #8 TEK SCREWS AT EACH END - MAX UNBRACED
LENGTH = 20'
7/15
. EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - 6" Studs
Stud Design Data : kp:= 1000.1b ksi:= kp in 2 psf lb ft 2 plf:= lb•ft
Unbraced Length, y- Ly:= 4ft Unbraced Length, x- Lx:= 29ft
Uniform Weight of Spacing of Studs
Wall on Studs- Pd 10 psf in Wall - ss:= 24 in
Transverse Pressure Uniform Transverse
on Studs- P, 5•psf Load on Studs - we Pc sS= 10 plf
TRY: 600S200 -43
Area of Brace - A:= 0.492in2 Radius of Gyration, y- ry:= 0.739 in
Effective Section - Se:= 0.873in3 Radius of Gyration, x- rx:= 2.335in
Moment of Inertia - I:= 2.683 in4
Effective Width of Thickness of
Element- w:= 2.00in Element- t:= 0.0451in
Number of Studs- N:= 1 Modulus of Elasticity- E:= 29000ksi
Steel Yield Strength - Fy:= 33ksi Effective Length Factor- K:= 1.0
Capacity of Column
2.E Tr2.E
Buckling Stresses - Fey:_ Fex
- (K:1/2 K Lx 2
y ( rx
Fey= 67.8•ksi FeX= 12.9•ksi
Fe:= if Fey<FeX,Fey,Fex) Fe= 12.9•ksi
Allowable Stress - Fa:= if Fe>Fy,Ft,. 1 — Fy ,F, F„= 12.9•ksi
2 l 4.Fc
Effective Area - Ae:= w•t•N Ae= 0.09•in2
Nominal Axial Strength - Pa:— Ae F„ Pa= 1162.3 lb
A
P„2:= �2 E
P„2= 2786.3 lb
2
25.7.1 w
t
Factor of Safety- Si,:= 1.92 Ste= 1.92
Allowable Axial Load - Pa:= F° = 605 lb Fa:= Pa = 6.711.ksi
52e A
8/15
. 'i EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - Cont...
Stud Height- Hs:= 29•ft Ceiling Height- Hg b:= 12•ft
Axial Load on Actual Axial Pd
Studs- Pd pd.Hg,b•Ss= 240 lb Stress- fa:= = 2.661•ksi—
Ac
Maximum Design Shear(at Maximum Design Moment-
Base of Wall) -
2"
w Vmax
Vmax —Hgwb)= 951b Mmax:= 2 w = 453 ft•lb
t ,
Design Shear at Top of Wall - Vt0p:= wt•Hgwb—Vmax= 25 lb
Allowable Bending Actual Mmax
Stress - Fb:= 0.6•Fy= 19.8.ksi Stress- b:= = 6.225•ksi
Se
Allowable Shear Actual Shear Vmax
Stress- F„:= 0.4•Fy= 13.2•ksi Stress - f„:_ = 1.055•ksi
At
Transverse Load Deflection -
A:_ (wt 0.5 Hs
1 [Hg,,,b2•(2•Lx— Hg,,b)2— (2•Hg,,,b)•(0.5.HS)2•(2•Lx— Hg,,b) + Hs.(0.5•Hs)3] = 0.745•in
I1 24•E•I•HS L
f b f f f
= 467 = 0.314 - = 0.08 b + a = 0.711
0 Fb F„ Fb Fa
9/15
• 'i EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - 3-5/8" Studs
Stud Design Data : kp:= 1000.1b ksi:= kp in 2 psf:= lb ft 2 plf:= lb•ft 1
Unbraced Length, y- Ly:= 4ft Unbraced Length, x- Lx:= 29ft
Uniform Weight of Spacing of Studs
Wall on Studs - Pa 10•psf in Wall - ss:= 12•in
Transverse Pressure Uniform Transverse
on Studs - Pt:= 5 psf Load on Studs - wt= Pt ss= 5 plf
TRY: 362S200 -43
Area of Brace - A:= 0.385in2 Radius of Gyration, y- ry:= 0.767in
Effective Section - Se:= 0.427in3 Radius of Gyration, x- rx:= 1.474in
Moment of Inertia - i:= 0.836in4
Effective Width of Thickness of
Element- w'= 2'o0tn Element- t:= 0.0451in
Number of Studs- N:= 1 Modulus of Elasticity- E:= 29000ksi
Steel Yield Strength - Fy:= 33ksi Effective Length Factor- K:= 1.0
• Capacity of Column
2 E z E
Buckling Stresses - Fey:_ Fex it
. K•Ly12
Cry / (K.Lx)2
rx
Fey= 73.1•ksi Fex= 5.1•ksi
Fe:= if Fey<Fex,Fey,Fex Fe= 5.1•ksi
Allowable Stress - F„:= if Fe>Py,Fyr1— Fy 1,Fe F„= 5.1•ksi
2 l 4•F,J
Effective Area - Ae:= w•t•N Ae= 0.09•in2
Nominal Axial Strength - P„:= Ae F„ P„= 463.2 lb
A
P„2:= �2 E
Piz= 2180.3 lb
25.7•(w 2
t
Factor of Safety- ft,:= 1.92 Ste= 1.92
Allowable Axial Load - Pa:= PO = 2411b Fa:= Pa = 2.674-ksi
lie Ac
10/15
• "i EC LI PS E AEO-Store#2056 1/23/2014
E N G I N E E R I N G Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - Cont...
Stud Height- HS:= 29•ft Ceiling Height- Hob:= 12•ft
Axial Load on Actual Axial Pd
Studs - Pd Pd'Hgwb•Ss= 120 lb Stress - fa:= Ae = 1.33•ksi
Maximum Design Shear (at Maximum Design Moment-
Base of Wall) -
2
Vmax:= HI(2'Lx — Hgb)= 48lb Mmax:= 2 = 226 t•lb.Hmb
t
Design Shear at Top of Wall - Vtop:= wt•Hgb— Vmax= 12 lb
Allowable Bending Actual Bending Mmax
Stress- Fb:= 0.6•Fy= 19.8•ksi Stress- fb:= 3 = 6.364•ksi
Se
Allowable Shear Actual Shear Vmax
Stress - F�:= 0.4.Fy= 13.2•ksi Stress - fv:_ = 0.528•ksi
Ac
Transverse Load Deflection -
Q:_ Cwt 0.5 HS [Hb2•(2•Lx— Hgwb12— (2•Hgb)•(0.5•HS)2•(2.Lx— Hgb) + Hs.(0.5.HS)3] = 1.195•in
24.E•I•HS JI /
f v f v
f f
- = 291 = 0.321 = 0.04 b + a = 0.819
A Fb Fv, Fb Fa
11/15
. i EC Ll PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - 3-5/8" Studs
Stud Design Data : kp:= 1000.1b ksi:= kp in 2 psf:= lb-ft-
b ft 2 plf:= lb•ft 1
Unbraced Length, y- Ly.= 4ft Unbraced Length, x- Lx:= 29ft
Uniform Weight of Spacing of Studs
Wall on Studs - pd 10•psf in Wall - ss:= 16•in
Transverse Pressure Uniform Transverse
on Studs - Pt:= 5•psf Load on Studs - wt Pt ss= 6.7 plf
TRY: 3625200 -54
Area of Brace - A:= 0.479in2 Radius of Gyration, y- ry:= 0.761 in
Effective Section - Se:= 0.490in3 Radius of Gyration, x- rx:= 1.467in
Moment of Inertia - I:= 1.030in4
Effective Width of Thickness of
Element- w:= 2.00in Element- t:= 0.0566in
Number of Studs- N:= 1 Modulus of Elasticity- E:= 29000ksi
Steel Yield Strength - Fy:= 33ksi Effective Length Factor- K:= 1.0
- Capacity of Column
Tr 2 E 7r2
.E Stresses - Fey F„= E
. K Lyl2 K Lx 2
Cry ) \ r,� )
Fey= 71.9•ksi FeX= 5.1•1csi
Fe:= if(Fey<Fex,Fey,Fex) Fe= 5.1•ksi
Allowable Stress- F,:= i Fe>2,Fy1 1 — 4 FF 1,Fe1 Fn= 5.1•ksi
F,
J
Effective Area - Ae:= w•t•N Ae= 0.11•in2
Nominal Axial Strength - Pn:= Ae•Fn Pn= 575.8 lb
Pn2 A�2 E Pn2= 4272.41b
25.7.(w)
2
t
Factor of Safety- f2e:= 1.92 52,= 1.92
Allowable Axial Load - Pa:= pn = 300 lb Fa:= Pa = 2.649-ksi
S2c Ae
12/15
. i EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - Cont...
Stud Height- Hs:= 29.0 Ceiling Height- H - 12.ft
gwb�-
Axial Load on Actual Axial Pd
Studs - Pd PdrHgwb•Ss= 160 lb Stress -
fa:= = 1.413 ksi
Maximum Design Shear (at Maximum Design Moment-
Base of Wall) -
wt•H b1 Vmax2
Vmax C 2 L j(2Lx_ Hb) = 631b Mmax 2 = 302 ft•]b
t
Design Shear at Top of Wall - viol,:= wt•Hg..•b- Vmax= 171b
Allowable Bending Fb — 0.6 F 19.8 ksi Actual Bending f - Mmax = 7 394•ksi
Stress- b y= Stress- b Se
Allowable Shear Actual Shear Vmax
Stress- F,:= 0.4.Fy= 13.2•ksi Stress - f,, = 0.56•ksi
_ Ae
Transverse Load Deflection -
A:= C Wt 0.5 HS1 [Hgwb2•(2.Lx— H b)2— (2•Hgb)•(0.5•HS)2•(2•Lx— Hs�,b) + HS•(0.5•HS)3] = 1.294•in
24•E•1•HS JI L l
f f f f
. = 269 b = 0.373 v = 0.042 b + a = 0.907
A Fb Fs, Fb Fa
13/15
. i EC LI PS E AEO-Store#2056 1/23/2014
ENGINEERING Tigard, OR RVC
•
LIGHT GAUGE STEEL WALL STUDS - 2-1/2" Studs
Stud Design Data : kp:= 1000•1b ksi:= kp•in 2 psf:= lb•ft 2 plf:= lb•ft 1
Unbraced Length, y- Ly:= 4ft Unbraced Length, x- Lx:= 29ft
Uniform Weight of Spacing of Studs
Wall on Studs - Pd 10•psf in Wall- ss:= 12•in
Transverse Pressure Uniform Transverse
on Studs- Pt:= 5-psf Load on Studs- Wt pc s,= 5-plf
TRY: 250S162-43
Area of Brace - A:= 2 0.289in2 Radius of Gyration, y- ry:= 2.0.620in
Effective Section - Se:= 2.0.240in3 Radius of Gyration, X- rx:= 2.1.022in
Moment of Inertia - I:= 2.0.302in4
Effective Width of Thickness of
Element- w:= 2.1.625in Element- t:= 0.0451in
Number of Studs- N:= 1 Modulus of Elasticity- E:= 29000ksi
Steel Yield Strength - F := 33ksi Effective Length Factor- K:= 1.0
- Capacity of Column
2•E -n-2
Buckling Stresses- Fey:= 'r F„:-
E
K•I,y2 (K.1..x2
( rY ) rx
Fey= 191•ksi FeX= 9.9•ksi
F,:= if(Fey<FeX,Fey,Fex) Fe= 9.9•ksi
Allowable Stress- F,,:= i Fe>2y,Fy(1 — 4 FF ),Fe Fn= 9.9•ksi
e
Effective Area - Ae:= w•t•N Ae= 0.15•in2
Nominal Axial Strength - Pn:= Ae•Fn P„= 1447.3 lb
Pn2:= A�2 E
Pn2= 1239.6 lb
25.7•(r )2
` t
Factor of Safety- S2e:= 1.92 SZ,= 1.92
P a
Allowable Axial Load - Pa:_ —n = 754 lb Fa:= = 5.143•ksi
. fic Ac
• -: EC LI PSE AEO-Store#2056 1/23/2014
. ENGINEERING Tigard, OR RVC
LIGHT GAUGE STEEL WALL STUDS - Cont...
Stud Height- 1-1,:= 29.ft Ceiling Height- H -= 12.ft
gwb•
Axial Load on Actual Pd
Studs- Pd Pd.Hgwb•ss= 120 lb Stress - a:= �e = 0.819•ksi
Maximum Design Shear (at Maximum Design Moment-
Base of Wall) -
2
Vmax I w2 L b)•(2'Lx– Hg b) = 481b Mmax 2 w = 226 ft-lb
x t
Design Shear at Top of Wall- Vtop:= wc•Hg,,,b— Vmax= 12 lb
Allowable Bending Actual Bending Mmax
Stress- Fb:= 0.6•Fy= 19.8•ksi Stress - fb:= = 5.661•ksi
Se
Allowable Shear Actual Shear Vmax
Stress - Fv:= 0.4•Fy= 13.2•ksi Stress- fv:_ = 0.325•ksi Ae
Transverse Load Deflection -
. 0.– (wt 0.5 HS [Hg,„b2 (2 Lx– Hg,,,b)2– (2•Hgwb)•(0.5•Hs)2•(2•Lx– H8,,,,b) + HS•(0.5•HS)3] = 1.654•in
24•E•1•I ) LL
Hs fb fv fb fa
— = 210 — = 0.286 — = 0.025 — + — = 0.445
A Fb Fv Fb Fa
ECLIPSE ECLIPSE - ENGINEERING . COM
E N G I N E E R I N G
RECEIVED
JAN 29 2014
Structural Calculations gUIDINGTIGARD
Steel Storage Racks
Pipp PC Manual Carriage with Retail Shelving
for Pipp Mobile Storage Systems, Inc.
American Eagle Outfitters - Store #2056
Washington Square Mall
9767 SW Washington Square Rd - Sp. #D07
Tigard, Oregon 97223
.\C�`cO PROFFss
.441 76024PE
/�4
�GIN� ' :
y 76024PE
OREGON
O'C. 13.X' 0
Prepared For: 6FAT VP
American Eagle Outfitters Expires . ("36.2°14 ,
r
77 Hot Metal Street— 6 Floor
Pittsburgh, PA 15203
Robert r
EKa=
VanCamp
Please note: Eclipse Engineering, Inc. has reviewed only the adequacy of the storefront to
support the interior vertical and lateral loads of the above noted area. We neither take
responsibility for any other element nor the integrity of the structure as a whole.
113 West Main.Sate B.Mssed&MT 59802 1005 Rake(Ave,Stile E.Me h.MT 59337 4211MeM Rmrsde Ave_Sue 717 Wane.WA 99201 376 SW Bd Dnee,State B.Rend,OR 97702
Phone'(e46)7214733•Fax(406)72144988 Fe761e:(406)862.3715•Fax 4054623718 Phase:(509)921.7731•Fax(509)921.5704 P111ne:(541)389-9859•Fax(541)312-8706
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Pipp Mobile STEEL STORAGE RACK DESIGN
2009 IBC & 2010 OSSC &ASCE 7-05 - 15.5.3
kips:= 1000•Ib
Design Vertical Steel Posts at Each Corner : 1
Shelving Dimensions: plf:= lb.ft-
2
psf:= Ib•ft
Total Height of Shelving Unit- ht:= 10.ft pcf:= Ib ft 3
Width of Shelving Unit- w:= 4-ft ksi:= 1000.1b-in—2
Depth of Shelving Unit- d:= 1.33•ft
Number of Shelves- N := 11
Vertical Shelf Spacing - S:= 12.0•in
Shelving Loads:
Maximum Live Load From Jeans (stacked 12 pairs tall and 3 wide):
Weight of(1) Pair of Jeans- W. .== 1.40•Ib
Weight of Jeans per shelf- Wti:= 3.12.Wi Wtj= 50.4 lb
psf in sf- W.LLB:_ LLB = 9.4737•psf
w•d
ote: With 11 shelves, the floor
Design Live Load on Shelf- LL:= 10.psf oading will be max 100 psf, which
Dead Load on Shelf- DL:= 2.0.psf s OK on 100 psf retail floor
Section Properties of 3/4" x 1 9/16" x 16 Gauge Steel Channel :
Modulus of Elasticity of Steel - E:= 29000.ksi
Steel Yield Stress - F := 33.ksi
Physical Dimensions of Channel :
Channel Width- out-to-out- b:= 1.5625.in
Channel Depth - out-to-out- h:= 0.75•in
Radius at Corners- Rc:= 0.188•in
Channel Thickness- t:= 0.0593.in 16 Gauge
Channel Width- CL-to- CL- be:= b— 0.51 be= 1.5329.in
Channel Height- CL- to-CL- h�:= h—t h = 0.6907-in
Radius of Gyration in x and y- rx:= 0.4982•in ry•= 0.3239 in
Section Modulus in x and y- SSx:= 0.0614.in3 Sy:= 0.0628 in3
Moment of Inertia in x and y- Ix:= 0.0557.in4 Iy:= 0.0236.in4
Full Cross Sectional Area - AP:= 0.2246.1n2
Length of Unbraced Post- Lx:= S= 12•in 1.1:= S= 12.in Lt:= S= 12.in
Effective Length Factor- Kx:= 1.0 K := 1.0 Kt:= 1.0
1
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Section Properties Continued:
Density of Steel - psteel:= 490•pcf
Weight of Post- Wp:= psteel•Ap•ht Wp = 7.6426.1b
Vertical DL on Post- Pd := DL•w•.25•d•N + Wp Pd = 36.9026 lb
Vertical LL on Post- P1:= LL•w•.25•d•N Pi= 146.3 lb
Total Vertical Load on Post- Pp := Pd + Pl Pp= 183.2026•lb
Floor Load Calculations :
Weight of Mobile Carriage: We:= 50•lb
Total Load on Each Unit: W:= 4-Pp+ We W= 782.811b
Area of Each Shelf Unit: Au := w•(d + 8.in) A„= 7.9867 ft 2
Floor Load under Shelf: PSF:= A PSF= 98.0147•psf
NOTE: SHELVING LIVE LOAD IS LESS THAN 100 psf REQUIRED FOR RETAIL FLOOR LOADING
This is considering the aisle area next to the shelf to help disburse the shelf load
Find the Seismic Load using Full Design Live Load :
ASCE-7 Seismic Design Procedure:
Importance Factor- IE:= 1.0
Determine SS and S1 from maps- Ss:= 0.948 S1:= 0.341
Determine the Site Class - Class D
Determine Fa and Fv - Fa:= 1.121 Fv:= 1.718
Determine SMS and SM1 SMS Pa-Ss SM1 Fv•S1
SMS= 1.0627 SM1= 0.5858
Determine SDS and SDI_ SDS 3•SMS SDI 3•SMl
SDS= 0.708 SDI= 0.391
Structural System - Section 15.5.3 ASCE-7:
4. Steel Storage Racks R:= 4.0 110:= 2 Cd := 3.5
Rp:= R ap•= 2.5 Ip := 1.0
Total Vertical LL Load on Shelf- W1:= LL•w•d W1= 53.21b
W
Total Vertical DL Load on Shelf- Wd := DL•w•d+ 4•
WP Wd = 13.4191 lb
Seismic Analysis Procedure per ASCE-7 Section 13.3.1:
Average Roof Height- hr:= 24•ft
Height of Rack Attachment- z:= 0•ft For Ground Floor
2
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
0.4•ap•SDS z
Seismic Base Shear Factor- Vt:= Rp h 1 + 2• Vt= 0.1771
r
Ip
Shear Factor Boundaries- Vtm11,:= 0.3•Sps•Ip Vtmin = 0.2125
Vtmax 1.6•SDS•Ip Vtmax= 1.1336
Vt:= if(Vt>Vtmax,Vtmax,Vt)
Vt:= if(Vt<Vtmin,Vtmin,Vt) Vt= 0.213
Seismic Loads Continued :
v
For ASD, Shear may be reduced - V := t V = 0.1518
p 1.4 p
Seismic DL Base Shear- Vtd:= Vp•Wd•N Vtd = 22.41 lb
DL Force per Shelf: Fd := Vp•Wd Fd = 2.04 lb
Seismic LL Base Shear- Vti:= V •WI•N Vti = 88.84 lb
LL Force per Shelf: F1:= Vp•W1 FI= 8.08 lb
0.67* LL Force per Shelf: F1.67:= 0.67•Vp'WI F1.67= 5.41 lb
Force Distribution per ASCE-7 Section 15.5.3.3:
Operating Weight is one of Two Loading Conditions :
Condition #1: Each Shelf Loaded to 67% of Live Weight
Cumulative Heights of Shelves- H1:= .5•S+ 1.5•S+ 2.5.S+ 3.5.S+ 4.5•S+ 5.5•S+ 6.5.S+ 7.5.S
H2:= 8.5.S+ 9.5.S+ 10.5.S H := H1+ H2
Total Moment at Shelf Base- Mt:= H•Wd + H 0.67•Wl Mt= 2968.3 ft•lb
Vertical Distribution Factors for Each Shelf-
Total Base Shear- Vtotal:= Vtd+ 0.67.Vti Vtotal = 81.93 lb
Wd•0.5•S+ Wi•0.67.0.5•S
C1:= C1= 0.0083 F1 Ci-(Vtotai) F1= 0.68 lb
Mt
Wd•1.5.S+ W1•0.67.1.5•S
C2 - C2= 0.0248 F2:= C2•(Vt0te1) F2= 2.03 lb
Mt
Wd•2.5•S+ W1.0.67.2.5.S
C3:= C3= 0.0413 F3:= C3•(Vtotal) F3= 3.39 lb
Mt
Wd 3.5•S+ W1.0.67.3.5.S
C4 C4= 0.0579 F4:= C4•(Vtotal) F4= 4.741b
Mt
3
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Wd•4.5•S+ W1.0.67.4.5•S
C5:= C5= 0.0744 F5:= C5•(Vtotal) F5= 6.09 lb
Mt
Wd•5.5.5+ WI 0.67.5.5 S
C6:= C6= 0.0909 F6:= C6•(Vtotai) F6= 7.45 lb
Mt
Wd•6.5•S+ W1.0.67.6.5•S
C7:= C7= 0.1074 F7:= C7•(Vt0tal) F7= 8.8 lb
Mt
Wd•7.5•S+ W1.0.67-7.5-S
Cg:= Mr Cg= 0.124 Fg:= C8•(Vtotal) F8= 10.16lb
Wd•8.5•S+ WI.0.67.8.5•S
Cg:= M - C9= 0.1405 Fg:= Cg•(Vtotal) Fg= 11.511b
t
Wd•9.5•S+ WI.0.67.9.5•S
C10:= M C10= 0.157 Flo:= C1o•(Vtotal) Flo= 12.87 lb
t
Wd•10.5.S+ WI.0.67.10.5•S
C11:= C11 = 0.1736 F11:= Cil.(Vtotal) F11 = 14.22 lb
Mt
C1+ C2+ C3+ C4+ C5+ C6+ C7+ Cg+ C9+ C10+ C11 = 1 Coefficients Should total 1.0
Condition #2: Top Shelf Only Loaded to 100% of Live Weight
Total Moment at Base of Shelf- Mta := (N — 1)•S•Wd + (N — 1)•S•W1= 666.19 ft-lb
Total Base Shear- Vtota12:= Vtd+ FI= 30.49 lb
Wd•0.5•S+ 0•W1.0.5•S
Cla := M Cia = 0 Fla:= C1a•(Vtotal2) Fla = 0.31 lb
to
Wd•(N — 1)•S+ W1-(N — 1).S
Cga:= Cga= 1 F8a:= Cga (Vtotal2) Fga = 30.49 lb
M
to
Condition #1 Controls for total base shear.
By Inspection, Force Distribution for intermediate shelves without LL are negligible.
Column Design in Short Direction : MS:= 4-S-0/td +Va) MS= 27.813 ft-lb
Bending Stress on Column- fbx:= M5 S 1 fbx= 5.4358-ksi OK
Allowable Bending Stress- Fb := 0.6.Fy Fb= 19.8.ksi
4
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Find Allowable Axial Load for Channel Column
Allowable Buckling Stresses-
'Tr2•E ir2•E
0ex.x 2 crex.x= 493.3359•ksi oex.y \2 °ex.y= 208.5245•ksi
K • K •Ly
rx ry i
crex if(o ex.x<0ex.y,crex.x cex.y) 0ex= 208.5245•ksi
Distance from Shear Center- t'hc2•bc2
to CL of Web via X-axis ec:=
4-1x ec= 0.2983-in
Distance From CL Web to Centroid- xc:= 0.649•in– 0.5•t xc= 0.6193.in
Distance From Shear Center x0:= xc+ ec xo= 0.9177.in
to Centroid-
Polar Radius of Gyration - ro:=V rx2 + ry2 + xo2 ro= 1.0933-in
Torsion Constant- 3:= 1•(2•b•t3 + h•t3) J= 0.00027•in4
in Constant- C t•b •h 3•b•t+ 2•h•t
Warping g w:_ J Cw= 0.00648-in 6
12 6-b-t+ h•t
Shear Modulus- G := 11300•ksi
1 7c2•E'Cw
vt:= • G•3+ / 2 Qt= 59.3144•ksi
Ap•ro 2 (Kt•4)
2
p:= 1– (— p= 0.2954
ro
Fet 21a to-ex+ at) – (Qex+ Qt�2 – 4'0'Qex'6t] Fet= 48.8063•ksi
Elastic Flexural Buckling Stress- Fe:= if(Fet<c'ex,Fet•°ex) Fe= 48.8063•ksi
Allowable Compressive Stress- Fo:= if Fe> 2,Fy• 1 – 4 F ,Fe Fn = 27.4218•ksi
e
Factor of Safety for Axial Comp. - no:= 1.92
5
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Find Effective Area -
Determine the Effective Width of Flange -
Flat width of Channel Flange- wf:= b— Rc— 0.5.t wf= 1.3449-in
Flange Plate Buckling Coefficient- kf:= 0.43
1.052 f E
w F
Flange Slenderness Factor- xf t Xf= 1.1188
0.22 1
Pf 1 �f J Xf pf= 0.7181
Effective Flange Width- be:= if(Xf>0.673,pf•wf,wf) be= 0.9657.in
Determine Effective Width of Web-
Flat width of Channel Web- ww:= h— 2.R. — t ww= 0.3147•in
Web Plate Buckling Coefficient- kw:= 4.0
w F
Web Slenderness Factor- aw:= 105 w n �w= 0.0858
kw t E
pw Ii — 0.221 1 Pw= —18.2086
\ xw J Xw
Effective Web Width- he:= if(xw>0.673,pw•ww,ww) he= 0.3147-in
Effective Column Area - Ae:= t•(he+ 2•be) Ae= 0.1332.in2
Nominal Column Capacity- Pr, := Ae•Fn Pr, = 3652 lb
P
Allowable Column Capacity- Pa:_ Pa= 1902 lb
Check Combined Stresses -
7r2•E-Ix zr2 E I
Pcrx Pcrx= 1 x 105 lb Pcry:_ Pcry= 46908 lb
r
Kx•Lx 12 Ky.�Y)2
Pcr:= if(Pcrx< Pcry,Pcrx.Pcry) Pcr= 46908 lb
Magnification Factor- := 1 — SZo Pp a= 0.9925 C m:= 0.85
Pcr
Combined Stress:
Pp + Cm•f bx = 0.331 MUST BE LESS THAN 1.0
Pa Fb•a
Final Design: 3/4" x 1 9/16" x 16ga CHANNEL IS ADEQUATE FOR
REQD COMBINED AXIAL AND BENDING LOADS
NOTE: Pp is the total vertical load on post, not 67% live load, so the design is conservative
6
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
STEEL STORAGE RACK DESIGN
PER 2009 IBC & 2010 OSSC & ASCE 7-05 - 15.5.3
Find Overturning Forces
Total Height of Shelving Unit- Ht:= ht= /Oft Width of Shelving Unit- w = 4ft
Depth of Shelving Unit- d= 1.33 ft
Number of Shelves- N= 11 Vertical Shelf Spacing- S= 12.in
Height to Top Shelf Center of G - htop Ht htop= /Oft
Height to Shelf Center of G - he:= (N 2 1) S h�= 6•ft
From Vertical Distribution of Seismic Force previously calculated -
Controlling Load Cases -
Weight of Rack and 67% of LL- W:= (Wd + 0.67.N•N W= 539.6946lb
Seismic Rack and 67% of LL- V:= Vtd+ 0.67•Vtl V= 81.934 lb
M1:= F1.0.5•S+ F2.1.5.S+ F3.2.5.S+ F4.3.5.S+ F5.4.5•S+ F6.5.5•S+ F7•6.5•S+ F8.7.5•S
M2:= F9.8.5.S+ F10•9.5•S+ F11.10.5•S
Overturning Rack and 67% of LL- M := M1+ M2= 599.6ft•Ib
Weight of Rack and 100%Top Shelf- Wa:= Wd•N + W1 Wa= 200.8106 lb
Seismic Rack and 100%Top Shelf- Va:= Vtd+ Fl Va= 30.48611b
Overturning Rack and 100%Top Shelf- Ma:= Vtd•hc+ Fi•htop Ma= 215.2ft•Ib
Controlling Weight- We:= if(W>Wp,W,Wp) We= 539.695 lb
Controlling Shear- Vc:= if(V>Va,V,Va) = 81.934 lb
Controlling Moment- Mot:= if(M > Ma,M,Ma) Mot= 599.61 ft•Ib
vvc
Tension Force on Column Anchor- T:= Mot – 0.60. 2 T= 288.92 lb
per side of shelving unit d 2
T:= if(T <0•Ib,0•Ib,T) T= 288.92441b
V`
Shear Force on Column Anchor- V:= — V= 411b
2
USE: HILTI KWIK BOLT TZ ANCHOR (or equivalent) -
USE 3/8"(1) x 2" embed installed per the requirements of Hilti
Allowable Tension Force- Ta:= 1006•Ib For 2500psi Concrete
Allowable Shear Force - Va:= 999•Ib
Combined Loading - p•1.0.T + p•1.0•V 0 427 p:= 1.3 For Concrete
Ta Va MUST BE LESS THAN 1.0
7
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Strap Bracing on Backside of Shelving Units :
Total Seismic Base Shear- Vc= 81.934 lb
V
Shear Force on Backside into Braces- Vb:= 2 Vb = 40.967 lb
Number of Shelf Bays Supported by Brace- Nb := 1
Each Shelf is Individually Braced
2 (Vb.hc12
Tension Force into Brace- Tb := Nb• Vb + J Tb= 73.8543 lb
w
USE 0.75" x 16ga STRAPS
Width of Strap- b5:= 0.75-in
Thickness of Strap- is:= 0.0545.in
Area of Strap- A5:= b5•t5 A5= 0.0409•in2
T
Tension Stress in Strap- ft:= ft= 1.81•ksi
s
Allowable Tension Stress- Ft:= 0.6.33•ksi Ft= 19.8•ksi
USE #10 SCREWS or RIVETS TO CONNECT STRAP TO FRAME :
Shear Capacity of#10 Screw - Vc:= 293-lb
Connecting to 16ga metal
T b
Number of Screws Required - N5:= N5= 0.2521
V
c
USE: min (1) #10 SCREW or RIVET AT EACH END OF STRAP
VERIFY HILTI BOLT CONNECTION FROM STRAP :
h
Tension Force from Brace- T„:= Nb•Vb•- Tv= 61.45lb
(Vertical Component) w
Shear Force from Brace - Vh:= Nb•Vb Vh = 40.967 lb
(Horizontal Component)
1.0•Tv (1.0-Vh)
Combined Loading - + = 0.1 MUST BE LESS THAN 1.0
Ta Va
8
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard, OR RVC
STEEL ANIT-TIP CLIP AND ANTI-TIP TRACK DESIGN
Tension (Uplift) Force on each side - T — 288.92443 lb
Connection from Shelf to Anti-Tip track:
Capacity of 1/4" diameter bolt in 16 ga steel - Zc:= 312.Ib
if(T < 2�Z ,"(2) 1/4" Bolts are Adequate" ."No Good") = "(2) 1/4" Bolts are Adequate"
Use 3/16" Diameter anti-tip device -
Yield Stress of Angle Steel - Fy:= 36.ksi
Thickness of Anti-tip Head - to:= 0.090•in
Width of Anti-tip Rod + Radius- br:= 0.25.in
Width of Anti-tip Head - ba:= 0.490-in
Width of Anti-tip Flange - La:= ba— br La= 0.12•in
2
Tension Force per Flange leg- Ti:= 0.5.T T1= 144.4622 lb
TI-La
Bending Moment on Leg - MI:= 2 M1= 0.722311-ft•Ib
ba•tat
Section Modulus of Leg - Si:= 6 S1= 0.0007 in3
Mi
Bending Stress on Leg - fb :-= S fb = 13.1031 ksi
i
f b
Ratio of Allowable Loads - = 0.485 MUST BE LESS THAN 1.00
0.75-Fy
Width of Anti-Tip track- L:= 5.1.in
Thickness of Aluminum Track- tt:= 0.25.in Average Thickness
Spacing of Bolts - Stb:= 24.in
L tt2
Section Modulus of Track- St:= 6 St= 0.0531-in3
T.Stb
Design Moment on Track- M :— 8 M = 72.2ft-lb
for continuous track section
Bending Stress on Track- fb : fb = 16.3157-ksi
St
Allowable Stress of Aluminum - Fb:= 21-ksi
Ratio of Allowable Loads fb Fb— 1 = 0.777 MUST BE LESS THAN 1.00
ANTI - CLIP STEEL CONNECTION AND TRACK ARE ADEQUATE
9
•
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Connection from Steel Racks to Wall
Seismic Analysis Procedure per ASCE-7 Section 13.3.1:
Average Roof Height- hr:= 24.0•ft
Height of Rack Attachments- zb:= 10.ft At Top for fixed racks connected to walls
0.4.ap•SDS zb
Seismic Base Shear Factor- Vt:- • 1 + 2.— Vt= 0.3247
r
Ip
Shear Factor Boundaries- Vtmin:= 0.3•SDs•Ip Vtmin= 0.2125
Vtmax 1.6•SDS•Ip Vtmax= 1.1336
Vt:= if(Vt>Vtmax,Vtmax
Vt:= if(Vt<Vtmin.Vtmin,Vt) Vt= 0.325
Seismic Coefficient- Vt= 0.3247
Number of Shelves- N = 11
Weight per Shelf- Wtf := 50-lb
Total Weight on Rack- WT:= 4.Pp WT= 732.8106 lb
0.7•Vt•WT
Seismic Force at top and bottom - T„:= 2 T„= 83.2844 lb
Connection at Top:
Standard Stud Spacing - Sttud := 16.in
Width of Rack- w = 4ft
Number of Connection Points - Nc:= floor( 1 Nc= 3
on each rack Sttud)
Tv
Force on each connection point- Fc:= Fc= 27.7615 lb
N
c
Capacity per inch of embedment- Ws:= 135.lb
in
Fc
Required Embedment- ds:= W ds= 0.2056•in
For Steel Studs: 5
Pullout Capacity in 20 ga studs T20:= 83-lb For#10 Screw - Per Scafco
MIN #10 SCREW ATTACHED EXISTING WALL STUD IS ADEQUATE
TO RESIST SEISMIC FORCES ON SHELVING UNITS. EXPANSION
BOLT IS ADEQUATE BY INSPECTION AT THE BASE
10
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
STEEL STORAGE RACK DESIGN
STEEL CHANNEL SECTION PROPERTIES W/O LIPS
Channel Dimensions 3/4 x 1 9/16 x 16 gauge -
Channel Flange Length- df:= 1.5625•in
Channel Web Width- bw:= 0.75•in
Channel Thickness- t:= 0.0598•in
Radius at Corners- R:= 0.188•in
Area of one Flange- Af:= (df- t)•t Af= 0.0899.in2
Area of Web- Aw:= bw•t Aw= 0.0448•in2
Total Area of Channel- At:= 2.Af+ A, At= 0.2246•in2
X-X Section Properties:
df- t
Distance to Flange CL- yfx:= t+ 2 yfx= 0.8111-in
Distance to Web CL ywx= 2 ywx= 0.0299•in
Centroid of Channel - Yc:= 2 Af Yfx+ Aw Ywx yc= 0.6551•in
2-Af+ Aw
Dist- Flange CL to Centroid - Yxi:= Yfx-Yc Yxi= 0.156.in
Dist- Web CL to Centroid- Yx2 Yc— Ywx Yx2= 0.6252-in
t (df- 03
Moment of Inertia of one Flange•Ifx:= 12 Ifx= 0.0169.in4
Moment of Inertia of Web- I, := 12 = 1.3365 x 10 5.in4
Using Parallel Axis Theorem -
Moment of Inertia of Channel - Ix:= 2•(Ifx+ Af•Yx12) + �Iwx+ Aw.Yx22) Ix= 0.0557•in4
Distance from Centroid to edge- cx:= if(yc>df— Yc,Yc,df—Yc) cx= 0.9074•in
I
Section Modulus of Channel- Sx:= X Sx= 0.0614•in3
cx
I
Radius of Gyration- rx:= A rx= 0.4982.in
At
11
Eclipse Engineering, Inc. American Eagle Oufitters-Store#2056 11/25/2013
Consulting Engineers Tigard,OR RVC
Y - Y Section Properties :
Distance to Outer Flange CL- xf1:= bw— 2 xf1 = 0.7201•in
Distance to Near Flange CL- xfz:= Z xf2= 0.0299•in
b
Centroid of Channel - := 2 xc= 0.375•in
Dist- Flanges to Centroid- x1:= Xf1 — xc x1= 0.3451.in
ldf— t)'t3
Moment of Inertia of Flange- Ify:= 12 I fy= 0.000027•in4
t bw3
Moment of Inertia of Web- Iwy:= 12 I,,,,y= 0.0021•in4
Using Parallel Axis Theorem-
Moment of Inertia of Channel - Iy:= IWy+ 2•(Ify+ Af.x12) Iy= 0.0236.in4
Distance from Centroid to edge- cy:= - cy= 0.375.in
Section Modulus of Channel- Sy:= Iy Sy= 0.0628•in3
w
Radius of Gyration- ry:= A ry = 0.3239.in
At
12
Eclipse Engineering, Inc American Eagle Outfitters-Store#2056 11/25/2013
Consulting Engineers Tigard, OR RVC
Project Name = WASHINGTON SQUARE MALL
Conterminous 48 States
2005 ASCE 7 Standard
Latitude = 45.450774
Longitude = -122.780703
Spectral Response Accelerations Ss and S1
Ss and S1 = Mapped Spectral Acceleration Values
Site Class B - Fa = 1.0 ,Fv = 1.0
Data are based on a 0.05 deg grid spacing
Period Sa
(sec) (g)
0.2 0.948 (Ss, Site Class B)
1.0 0.341 (S1, Site Class B)
Conterminous 48 States
2005 ASCE 7 Standard
Latitude = 45.450774
Longitude = -122.780703
Spectral Response Accelerations SMs and SM1
SMs = Fa x Ss and SM1 = Fv x S1
Site Class D - Fa = 1.121 ,Fv = 1.718
Period Sa
(sec) (g)
0.2 1.063 (SMs, Site Class D)
1 .0 0.586 (SM1, Site Class D)
Conterminous 48 States
2005 ASCE 7 Standard
Latitude = 45.450774
Longitude = -122.780703
Design Spectral Response Accelerations SDs and 501
SDs = 2/3 x SMs and SD1 = 2/3 x SM1
Site Class D - Fa = 1.121 ,Fv = 1.718
Period Sa
(sec) (g)
iis 0.2 0.709 (SDs, Site Class D)
1 .0 0.390 (SD1, Site Class D)
,a
/ ,
Eclipse Engineering, Inc American Eagle Outfitters-Store#2056 11/25/2013
Consulting Engineers ineers Ti ard, OR RVC
9
-! and Welds) SCAFC0
Fasteners (Screws a Steel Scud Manufacturing Lo.
Screw Table Notes
1. Screw spacing and edge distance shall not be less than 3 x D. (D = Nominal screw diameter)
l2. The allowable screw values are based on the steel properties of the members being connected, per AISI section
E4.
3. When connecting materials of different metal thicknesses or yield strength, the lowest applicable values should be
used.
4. The nominal strength of the screw must be at least 3.75 times the allowable loads.
5. Values include a 3.0 factor of safety.
6. Applied loads may be multiplied by 0.75 for seismic or wind loading, per AISI A 5.1.3.
7. Penetration of screws through joined materials should not be less than 3 exposed threads. Screws should be
installed and tightened in accordance with screw manufacturer's recommendations.
Allowable Loads for Screw Connections (lbs/screw)
No.12 ( . No.10 i No.8 No.6
Steel Thickness Steel Properties Die.=0.216(in) ( Dia.=0.190(in) Dia,=0.164(in) Dia.=0.138(in)
Mils Design(in) Fy(ksi) Fu(ksi) Shear Pullout Shear Pullout Shear Pullout Shear Pullout
18 0.0188 33 45 66 39 60 33
27 0.0283 33 45 121 59 111 50
30 0.0312 33 45 151 76 141 65 129 55
33 0.0346 33 45 177 84 164 72 151 61
43 0.0451 33 45 280 124 263 109 244 94 224 79
54 0.0566 33 45 394 156 370 137 344 118
68 0.0713 33 45 557 156 523 173
C Weld Table Notes
1. Weld capacities based on AISI, section E2.
2. When connecting materials of different metal thickness or tensile strength (Fu), the lowest applicable values
should be used.
3. Values include a 2.5 factor of safety.
4. Based on the minimum allowance load for fillet or flare groove welds, longitudinal or transverse loads.
5. Allowable loads based on E60xx electrodes
6. For material less than or equal to .1242" thick, drawings show nominal weld size. For such material, the effective
throat of the weld shall not be less than the thickness of the thinnest connected part.
Allowable Loads For Fillet Welds And Flare Groove Welds
Design 1 Steel Properties
Thickness Yield Tensile E60XX Electrodes
Mil In. ksl ksi Ihs/in
43 0.0451 33 45 609
54 0.0566 33 45 764
68 0.0713 33 45 963
97 0.1017 33 45 1373
118 0.1242 33 45 1677
54 0.0566 50 65 1104
68 0.0713 50 65 1390
97 0.1017 50 65 1983
118 0.1242 50 65 _ 2422
.
48
• Eclipse Engineering, Inc American Eagle Outfitters-Store#2056 11/25/2013
Consulting Engineers Tigard, OR RVC
• Page 11 of 14 ESR-1917
TABLE 9-KB-TZ CARBON AND STAINLESS STEEL ALLOWABLE SEISMIC TENSION(ASD),NORMAL-WEIGHT
CRACKED CONCRETE,CONDITION B(pounds)''•' _
Concrete Compressive Strength=
Nominal Embedment re+e 2,500 psi Pc•=3,000 psi Pc=4,000 psi re=8,000 psi
Anchor Depth he, _ ,
Diameter (In.) Carbon Stainless Carbon Stainless Carbon Stainless Carbon Stainless
steel steel steel steel steel steel steel steel
3/8 2 1,006 1,037 1,102 1,136 1,273 1,312 1,559 1,607
1/2 2 1,065 1212 1,167 1,328 1,348 1,533 1,651 1,878
3 1/4 2,178 2,207 2,386 2,418 2,755 2,792 3,375 3,419
31/8 2,081 2,081 2,280 2,280 2,632 2,632 3,224 3,224
5/8
4 3,014 2,588 3,301 2,835 3,812 3,274 4,669 4,010
314 3 3/4 2,736 3,594 2,997 3,937 3,460 4,546 4,238 5,568
4 3/4 3,900 3,900 4,272 4,272 4,933 4,933 6,042 6,042
For SI: 1 lbf=4.45 N,1 psi=0.00689 MPa For pound-inch units:1 mm=0.03937 inches
'Values are for single anchors with no edge distance or spacing reduction.For other cases,calculation of Rd as per ACT 318-05 and conversion
to ASD in accordance with Section 4.2.1 Eq.(5)is required.
2Values are for normal weight concrete.For sand-lightweight concrete,multiply values by 0.60.
'Condition 8 applies where supplementary reinforcement in conformance with ACI 318-05 Section D.4.4 is not provided,or where pullout or
pryout strength governs.For cases where the presence of supplementary reinforcement can be verified,the strength reduction factors
associated with Condition A may be used.
TABLE 10-KB-TZ CARBON AND STAINLESS
STEEL ALLOWABLE SEISMIC SHEAR LOAD(ASD),
(pounds)'
Nominal Allowable Steel Capacity,Seismic Shear
. Anchor
Diameter Carbon Steel Stainless Steel
3/8 999 1,252
1/2 2,839 3,049
5/8 4,678 5,245
3/4 6,313 6,477
For SI:1 lb(=4.45 N
'Values are for single anchors with no edge distance or
spacing reduction due to concrete failure.