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EXPIRES: •
2 c(.l s i one 1 2-_(,,oc4NEc.?' By <`.
• Date re,/27Jc0
GROMACKENZIE ' '°b# Z�oZIS
0690 SW Bancroft St/PO Box 69039 Portland,OR 97201-0039 Sht. Z of Z
Tel: 503.224.9560 Net:infoegromack.com Fax: 503.228.1285 22000 GROUP $ACrEHZIE. AL, FIGHTS RESERVED
G R O U P
• MACKE NZ I E
0690 SW BANCROFT STREET-P.O.BOX 69039
PORTLAND,OREGON 97201-0039
TEL(503)224-9560-FAX(503)228-1285
MECHANICAL UNIT SUPPORT
PROJECT NAME: Precision Interconnect Building 10 PROJECT NO.: 2000215 DATE: 6/27/00
ELEMENT ID:
DESIGNER: SKS
W := 1200•lb TOTAL WEIGHT OF MECHANICAL UNIT
P corner 350-lb MAXIMUM CORNER WEIGHT (use W/3 if not specified)
P curb 50•Ib assumes 200 lb curb (200/4=50)
I meth 8.5•ft mechanical unit length
`"mech 6 ft mechanical unit width
h mech 5 ft mechanical unit height
s :=4•ft PURLIN SPACING 4,AINt+,
I pur:=31 ft PURLIN SPAN yy
L SC
I EXPIRES: (p-3o•CU
P. lofS
Roof Dead Load:
roofing '=2.5•psf
plywood :=2.5•psf
insulation := 1.O.psf
sprinkler :=1.5•psf
electrical :=1.O.psf
mechanical =0.5•psf
ceiling :=1.O'psf
purlin :=2.5•psf
DL 1 :=roofing+ceiling+insulation--sprinkler-I-electrical-F mechanical t plywood
DL 1 = l0opsf
DL 2 :=DL 1 l-purlin
DL 2 = 12.5 opsf
Roof Live Load
snow :=25-psf
LL :=snow
LL =25 opsf
`consider unbalanced snow condition for cantilevered roof system
I'• 2oF 8
FRAMING
P mech P corner+ P curb P meth =400 lb
P11 :=LL
mech �"mech
P11 =318.75 lb
Find maximum shear and moment:
P mech+Pll •(241'2•ft)
M max s M max=R.625.103ain•lh
V max := P mech+P11
Vm =718.75 lb
M max
S REQ 900•psi•1.15.1.3 S REQ =6.41.in3
3 V max ,
A REQ '-2 95•psi•1.15 A REQ =9.868°in`
S 4x6 17.65-in3 Define Actual Section Modulus Here
A 4x6 =19.25-in` Define Actual Area Here
S xx :=if S REQ<S 4x6,"OK","NO GOOD"
S xx ="OK"
A :=if A REQ<A 4x6,"OK","NO GOOD"
A="OK" Use 4x6 No. 2 DFL
HANGER
R max :=V max R max =718.75 lb
R allow =1150-lb Define Allowable Hanger Value Here
HANGER :=if R max<R allow,"OK","NO GOOD"
Use Simpson LUS46 Hanger
or similar
HANGER ="OK"
P. 3 of
PURLINS
3-1/8x15 Glulam:
S x :=117.2•in3
A:=46.88•in2
1pur=31 ft
w := DL2+LL s w = 150°plf
w 2 :=DL 2•s+LL 2 w 2 = 100opIf (below mechanical unit)
mech•1 mech 2 LL-w mech'I mech
P mech P corner+P curb+LL' 4 g• P comer+P curb l- 4
P mech =898.438 lb
M max 252.72•k in (see attached)
Vmax .=2.61 •k
(see attached)
f
M
b := fb =2.156.103°psi
7
F b :=2400•psi•1.15 Tc F b =2.692.103^psi
3 Vmax
fv :=2— fv =83.511opsi
Fv :=190•psi•1.15 F =218.5°psi
Bending:=if F b>f b,"OK","NO GOOD" Bending="OK"
Shear :=if F v>f v,"OK","NO GOOD" Shear ="OK"
p. 4oF9
Choose "Title Block" menu item
on Settings Screen to change
• these five lines to your own
special title information
&company logo
Date:06127100 Page:
SINGLE SPAN BEAM ANALYSIS
SPAN DATA -- MEMBER DATA
Center Span - 31.00 ft I:Inertia - 1.000 in4 All distances for load locations refer
Left Cantilever - 0.00 ft E:Elastic Modulus 1 psi to left support.Neg 1'1 distances
Right Cantilever = 0.00 ft End Fixity Pin/Pin means load is on left cantilever
UNIFORM LOADS TRAPEZOIDAL LOADS
#1: 0.100k/ft @ Left, 0.100k/ft @ Right,Start @ 0.00ft > 8.50ft
#2: 0.150k/ft @ Left, 0.150k/ft @ Right,Start @ 8.50ft-> 31,OOft
CONCENTRATED LOADS APPLIED MOMENTS
#1 - 0.900k at 8.50ft
SUMMARY
Moments Shears Deflections
Max.Span Moment - 21.06 k-ft Left Support 2.61 k Max.@ Span = -3654173206.796 in at 15.25 ft
Max.Mom.Location - 14.26 ft Right Support - 2.51 k
Min.Span Moment = -0.00 k-ft Reactions
Min.Morn.Location - 31.00 ft Left Support - 2.61 k
Max @ Left Support - 0.00 k-ft Right Support - 2.51 k
Max @ Right Support 0.00 k-ft Query Values
MAXIMUM MOMENT = 21.06 k-ft Between Supports @ X = 0.00 ft,M - 0.00 k-ft, V - 2.611 k, Defl. - 0.000 in
V 1 OS'
0
0.151 111111111 1111 III I .1 I II I I 1 III I I 11 0.15 Mmex-21.05ft-k 14.26 fl
Mmtn--0.00 ft-k•31.00 ft -0.00
010=Eau 0 2.61
Vmex-2.61 kips•0.00 ft
Y m t n--2 51 kips 4.31 00 ft V
-2.51
0rnex a 0.00 In•0.00 ft
Drnln -3654173206 79 In a 15 25 110
-3654173206.79
31.00 ft I I I I 1 I
1111 0.0 5.1 10.3 15.5 20 7 25.8 51.0
V4.4B(c) 1983-95 ENERCALC Mackenzie Engineering,Inc.,KW-0602071
GIRDER
6 3/4x31 1/2 Glulam Beam:
S x := II 16•in3
A :=212.6•in2
wt :=46•plf
pur=31 8
12
F b :=2400•psi•1.15• 3
F b =2.179.10 °psi
F v := 190•psi•1.15
F v =218.5°psi
P purlin_DL DL 2•s•1 pur P purlin_DL = 1.55°k
P purlin_LL LL•s•I pur P 3.1°k
purlin_LL =
Calculate Reactions from end spans and use loads to determine maximum moment. (see attached)
For Balanced and Unbalanced Snow Conditions:
M max :=2601240•in•Ib
V max 29640•Ib
fb := MS ax fb =2.331.103°psi / Fb =2.479.103°psi
x
3 V max
fv2 A fv =209.125 opsi / Fv =218.5°psi
Bending:=if f b<F b,"OK","No Good"
Bending= "OK"
Shear :=if fv<Fv,"OK",'No Good"
Shear="OK"
. GoFS
Choose "Title Block" menu item
on Settings Screen to change
e these five lines to your own
special title information
& company logo
Date:06127/00 Page:
SINGLE SPAN BEAM ANALYSIS / 3/1,t k 311k- Ga-• g•
woStS1 c.4SE l.a rDll - : Fvv4 SNow 4- M641 vNCr oN UtNT1LtI6J.
SPAN DATA MEMBER DATA
Center Span = 25.00 ft I:Inertia - 1.000 in4 All distances for load locations refer
Left Cantilever = 0.00 ft E:Elastic Modulus - 1 psi to left support.Neg(1 distances
Right Cantilever = 8.50 ft End Fixity Pin/Pin means load is on left cantilever
UNIFORM LOADS TRAPEZOIDAL LOADS
Center Span#1 = 0.05 klft
Right Cant#1 = 0.05 klft
CONCENTRATED LOADS APPLIED MOMENTS
# 1 - 19.260k at 33.50ft
#2 - 4.650k at 4.00ft
N 3 - 4.650k at 8.00ft
#4 - 4.650k at 12.00ft
#5 - 4.650k at 16.00ft
II 6 - 4.650k at 20.00ft
#7 - 4.650k at 24.00ft
N 8 - 4.650k at 28.00ft
II 9 - 4.650k at 32.00ft
#10 - 0.700k at 32.00ft
- - - SUMMARY - -Moments - Shears Deflections
Max.Span Moment = 16.34 k-ft Lett Support 4.18 k Max.@ Span - 5361496299.249 in at 17.76 ft
Max.Mom.Location = 4.02 ft Right Support = 29.65 k
Min.Span Moment = -216.77 k-ft Reactions Right Cant = -23452901511.887 in at 33.50 ft
Min.Mom.Location = 25.00 ft Left Support - 4.18 k
Max @ Left Support = 0.00 k-ft Right Support - 54.52 k
Max @ Right Support = -216.77 k-ft Query Values
MAXIMUM MOMENT = 216.77 k-ft Between Supports @ X - 0.00 ft.M - 0.00 k-ft, V - 4.180 k, Defl. - 0.000 in
At Right Cantilever @ X = 0.00 ft,M - 0.00 k-ft, V - 0.000 k, Defl. = 0.000 in
Mmax.16 33ft-k•4 02 f1 16 33
N Mmin--216.77 11-k•24 99 ft
-Z16 77
29 64
I 1 111 111 1 1.1110.04 1111 11l1.1 1 1H0.04 1
Vmax-29.64 kips•25.05 ff
Ymin--24.86 kips•24.99 ft Y
-24.86
299.
Dmax-5361496299.24155n 6�17.149675 ft 24
Dmin--23452901511 88 in•33 3011
-23452901511.88
25.00 8.50 "I I I I i I
0.0 5.5 1 1 1 16.7 22.3 27.9 35 5
p. 7oFS
V4.4B lc)1983-95 ENERCALC Mackenzie Engineering,Inc.,KW-0602071
ANCHORAGE
I
seismic
ap = 1 Ca :=0.36 I = 1 Rp :=3
hx :=34ft W:=1250•Ib
h r:=34•ft
ap•Ca•1p h
F P .= R • 1 +3.,-- •W Fp =600 lb
F
F s • 1.4 Fs =428.571 lb
overturning
h mech =5 ft mech =6 ft
h mech
M OT =F' , MOT= 1.071 103 lb-ft
w mech
M RES :=W 2 M RES =3.75.103 lb•ft
M RES
FS OT:= MOT FS OT=3.5
screws
h mech I mech
Fs• 2 — W• 2
Tension :_
I mech Tension =-498.95 lb
Fs
Shear :=4 Shear = 107.143 lb
#12 screw with minimum 1-1/2-inch embedment:
T allow 231 lb
:= 163.1b
V allow
check combined
Tension Shear
combined :=if + <1.33,"OK" ,"NO GOOD"
T allow V allow
combined ="OK"
p.8 °F8