Specifications (28) - Design Code: Oregon State Specialty Code
Referenced Standards: Hilti Kwik HUS-EZ (ESR-3027)
Hilti Kwik Pro self drilling screws (ESR-2196)
Shelving Data: Height of shelving Unit (feet) = ht = 12.00
Width of Shelving Unit (feet)= w= 4.00
Depth of Shelving Unit (feet) = d = 2.00
Total Number of Shelves per Unit = 6
Vertical Shelf Spacing (inches) =S = 24.00
Shelf Loads: Maximum Live Load per Shelf(lbs.) = 80
Actual Dead Load per Shelf(lbs.) = 25
Design Load Combinations: Design per ASCE 7-10
Shelving units are to be designed per ASCE-7 ASD load combinations
as well as RMI ASD load combinations.
A review of RMI ASD load combinations indicates the following:
Combinations 1 & 2: DL (only) and DL+ LL:These load
combinations are covered in the ASCE 7-10 load combinations.
Combination 3: .75(DL— .67EL+ Plapp): This load combination is
for determining maximum leg uplift. This combination produces
Dead Load that is greater than that required by ASCE 7 and
produces Seismic Load that is less than ASCE 7. Therefore, combining
the two will produce less leg uplift than the similar ASCE 7 combination.
Combination 4: .75(DL+LL+.67EL): This load combination is to find
maximum gravity plus seismic effect on the unit. Again, the ASCE 7
load combinations with gravity plus seismic produces larger load and
will therefore control the design.
Combination 5: DL+LL+Impact Load: This combination is
required at beam design only.
A review of ASCE 7 ASD load combinations indicates the following:
Combinations 1, 2, 3,and 4: Are basically equivalent.The shelf and its
contents are conservatively considered Dead Load for purposes of
these load combinations.
Combinations 5 and 8: Are nearly equivalent, however combination 8
will produce less deal load with an equivalent seismic load to
combination 5. The lower DL decreases the shelving unit's
resistance to overturning and will thus produce greater leg uplift and
and downward loads while producing the greatest lateral seismic effect.
Therefore combination 8 controls the design of the units. (0.6DL+ 0.7EL)
Combination 6 produces less seismic effect than combination 8.
Combination 7 is dead load only and will not control the design.
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IN MMII
Seismic Loads: Per ASCE 7-10
From attached USGS Data Sheet: Ss = 0.983 Si = 0.422
Assumed Site Class = D
Fa= 1.107
Fv = 1.578
SMs= (Fa)(Ss) = 1.088
SMi = (Fv )(Si) = 0.666
SDS = (2/3)(SMs) = 0.725
Sol = (2/3)(SMi) = 10.444
Occupancy Category = II
Seismic Design Category= D
IE = 1.0 (racks in Storage Rooms NOT accessible to the public)
hn = 12.00
R = 4
Oo= 2
Cd = 3.5
Ct = 0.02
x = 0.75
T= Ct(hn)x= 0.129
Cs = (SDs)/(R/I) = x0.181
Cs(max) =Sds/(T)(R/I) = 0.861 OK
Cs(min) = 0.044(SDs)(I) = 0.032 OK
If Si > 0.6, then Cs(min)= (0.5)(Si)/(R/I) = N/A FALSE
Therefore, Seismic Base Shear, V= C5W
Distribution of Seismic Forces:
Consider two cases per ASCE 7-10:
Case 1 =Weight of unit plus all levels loaded to 67% capacity
Case 2 =Weight of Unit plus top shelf only loaded to 100% capacity
CHECK FREESTANDING SHELVING - 2'-0" WIDTH
Page 3
Case 1:
Total Dead Load = 150 lbs.
Total Live Load = 321.6 lbs.
Total Load = 471.6 lbs.
E =V= CsW = 86 lbs. (Seismic Base Shear)
Seismic force V must be distributed laterally to each level as follows:
k
wx hx
_CCvx n k = 1
Fx vx V
EKi k
=1
From this we get the following lateral load distributions:
h1 = 4 inches Fl = 0.89 lbs.
h2= 28.00 inches F2 = 6.24 lbs.
h3 = 52.00 inches F3 = 11.58 lbs.
h4 = 76.00 inches F4 = 16.92 lbs.
h5 = 100.00 inches F5 = 22.27 lbs.
h6 = 124.00 inches F6 = 27.61 lbs.
z•-a"
{ Rys = 0.9(D+L)(d�2) - (E(Fxhx)) _� -218.711lbs.
l§ Rye = D+L- Ry1. = 690.31i lbs.
1,,
4 - - A THEREFORE NET UPLIFT
i - -,
Vertical Seismic load effect per ASCE 7-10:
Ev= 0.2DdsD = �� 68.41!Ibs.
j > 1::,
i t 1. it Therefore, maximum downward load = 758.72i lbs.
IL 1 1 > I Therefore, maximum uplift load = i -287 121Ibs.
s
F
hi3
ii
f
I
`a i >1 ,
A
V
Ryl Ry2
Page 4
IIIIIIIII
Case 2: Total Dead Load = 150 lbs.
Total Live Load = 80 lbs.
Total Load = 230 lbs.
E =V= CSW= 42 lbs. (Seismic Base Shear)
Seismic force V must be distributed laterally to each level as follows:
k
_ wx hx
F `v vx rs
x vx � k k = 1
WI h2
I=1
From this we get the following lateral load distributions:
h1 = 6 inches Fl = 0.41 lbs.
h2= 30.00 inches F2 = 2.06 lbs.
h3 = 54.00 inches F3 = 3.71 lbs.
h4 = 78.00 inches F4 = 5.36 lbs.
h5 = 102.00 inches F5 = 7.01 lbs.
h6 = 126.00 inches F6 = 8.65 lbs.
Ryi = 0.9(D+L)(d/2) - (F(Fxhx)) = 1 -207.291 lbs.
• Ry2 = D+L- Rys = I 437.291Ibs.
- --> F6
! I
THEREFORE NET UPLIFT
r_ I
Fs Vertical Seismic load effect per ASCE 7-10:
Ev= 0.2Dd5D = I 33.36�Ibs.
1
n
N ._. F4
g D±L Therefore, maximum downward load = 470.66!Ibs.
Therefore, maximum uplift load = -240.66Ibs.
Therefore, CASE 1 Governs
--\--- --j 1-2
Pup = -287.12 lbs.
Pdown = 1 758.72 lbs.
_ Plateral = € 86 -j lbs.
_> }1
l
- Ry.l Rye
Page 5
Anchorage To Concrete Floor:
Base Plate/Clip: Check Plate Thickness of: 14 ga. = 0.0713 in.
Pa= 758.72 lbs.
Fy= 36,000 psi
B = 3.5 in.
N = 1.75 in.
bf = 3 in.
d = 1.5 in.
m = (N-0.95d)/2 = 0.1625 in.
n = (B-0.8bf)/2 = 0.55 in. Governs
n' _V(dbf)/4 = 0.53 in.
l = max. of m, n, and n'
` 3
.3 ?
1 =1 II
-- "
miry \ 1' BA'
= 0.059 inches < 0.07 in. OK
Plate Connection to Post: (2) 1/4" dia. TEK screws
Maximum uplift = -287.12 lbs.
Uplift resisted by (2) screws, therefore -143.56 lbs. per screw
Maximum lateral load = 86 lbs.
Lateral Load resisited by (2) screws, therefore 42.8 lbs. per screw
Vtot =V(Pup^2+Plat^2) = 150 lbs. per screw
Vallow = 645 lbs. from ICC ESR report
Therefore, (2) 1/4" dia. TEK screws are OK
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•
Check Wedge Anchors: (2) 3/8" dia.x 2 1/2" embedment Hilti Kwik HUS-EZ
Normal weight concrete, f'c = 2500 psi
Plat=V= 42.8 lbs.
Pup=T= -143.56 lbs.
Tallow= 1334 lbs.
Vallow= 905 lbs.
Tapplied ± Vapplled ,C
Tallowable,ASD Vallowable,ASD 1.2
0.15 OK_
Check Concrete Slab Puncture:
Pdown = 758.72 lbs.
Vu= (1.7)Pdown = 1289.83 lbs.
Slab t= 3.5 in.
Base Plate Dimensions: Bi = 3.5 in.
B2= 1.75 in.
Concrete shear area = 2(Bi+ 82)t = 36.75 in^2
DVn = (0.85)2dfc(Ac) = 3123.75 lbs. > 758.72 lbs. I OK
Check Shelving Unit Members:
Double Rivet Beam Type 1 (per ARCO Industries)
t = 14 ga. = 0.0747 in. Sx = 0.119 inA3
Fy= 36,000 psi lx = 0.200 inA4
= + =
Load per level 40 lbs. 20 lbs. (self weight) 60 lbs.
Load per beam = 30 lbs.
Max. beam span = 4 ft.
M =wL^2/8 = 180 in.-lbs.
b/t = 14.6419 < 155VFy= 25.83, therefore Qs = 0.94
Fb = 0.6QsFy = 20304 psi
Sreq = M/Fb = 0.10638 inA3 OK I
deflection = 0.00745 in. OK
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Double Rivet Beam Type 2 (per ARCO Industries)
t = 14 ga. = 0.0747 in. Sx= 0.105 in^3
Fy= 36,000 psi lx = 0.055 in^4
Load per level = 40 lbs. + 20 lbs. (self weight) = 60 lbs.
Load per beam = 30 lbs.
Max. beam span = 1 ft.
M =wL^2/8 = 45 in.-lbs.
b/t = 12.55 < 155VFy= 25.83, therefore Qs = 0.94
Fb = 0.6QsFy = 20304 psi
Sreq = M/Fb = 0.0266 in^3 OK
deflection = 0.00011 in. OK
Rivet Beam Connection to Column
Connection must be designed for 1000 lb. upward force
Each beam/column connection is made with (2) 1/4" dia. Rivets
Arivet= 0.049 inA2
Fv= 17.5 ksi = 17,500 psi
Vallow = FvA = 858 lbs. per rivet
Total Vallow = 1715 lbs. > 1000 lbs. ; OK I
Check Moment Connection:
A502-1 Rivet per AISC Table I-D "Shear" page 4-5
See above for Vallow Calculation
Rivets are 1.5 inches apart, therefore allowable maximum moment
= (858 lbs)(1.5") = 1286.25 in.-lbs.
From Finite element analysis, max. moment = 1124 ft-lbs. OK
Page 8
T-Shape Dhelving Unit Upright: Per ARCO Industries
t=14 ga.= 0.747 in. A = 0.386 in^2
Q=Qs= 0.75 S = 0.099 in^3
Rmin = 0.054 in. I = 0.100 in^4
Fy= 36 ksi
L= 12 ft. (top 1'-8" is not loaded)
Lateral Load = 43 lbs. per leg
Vertical Load = 758.72 lbs.
Check Bending:
Distributed lateral load = 4.14 plf
Mmax = 45.0389 ft-lbs.
fb=M/S= 5.45927 ksi
Fb = 16.2 ksi L OK
Check Axial Load:
fa = P/A = 1.97 ksi
Fa = 19.37 ksi (from AISC equation E2-1 and A-B5-11)
Combined Axial and Bending:
Unity Check = 0.43847 < 1.0 OK
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