“Cole County Law Enforcement Center” Jefferson City, Missouri
Design Calculations for Glazed Aluminum Curtain Walls
STRUCTURAL DESIGN CRITERIA Wind Load (nominal) 22.6 psf (typical zone) 27.8 psf (corner zone) Dead Load 8.0 psf (aluminum frame with 1” I.G. unit infill)
REFERENCE STANDARDS
1. The Aluminum Association, Aluminum Design Manual - Specifications and Guidelines for Aluminum Structures; Sixth Ed., October 1994.
2. American Institute of Steel Construction, Manual of Steel
Construction; Ninth Edition, 1989. 3. American Forest & Paper Association/ANSI, National Design Specification for Wood Construction; Revised Edition, 1997.
ALLOWABLE DEFLECTIONS All framing members - normal to wall - L/175 where L < 13’-6” All framing members - normal to wall - L/240 + ¼” where L > 13’-6” All framing members - parallel to wall - 25% of glass edge bite MATERIALS (UNLESS OTHERWISE NOTED)
Structural Aluminum Extrusions 6063-T6 Sheet 3003-H14
Structural Steel Rolled Shapes ASTM A36 Sheet and Bar Sizes AISI M1020
Stainless Steel (AS NOTED)
Jason E. Barrick, P.E. 2850 Valley Vista Drive Sedona, AZ 86351 (928) 284-9710 [phone] [email protected] [e-mail]
• Formal design calculations by Jason E. Barrick, P.E. dated 10/17/10 • Shop drawings by Mark’s Mobile Glass with comments and sketches by Jason E. Barrick, P.E.
(forwarded separately via e-mail)
Jason E. Barrick, P.E.
2850 Valley Vista Drive Da
Sedona, AZ 86351
(928) 284-9710 Cole County Law Enforcement Designed by
Nominal wind load calculations (eff. wind area = 10 ft^2) Sheet
WIND LOAD CALCULATION PER ASCE 7-98 (h ≤ 60 ft)
Wind speed (3-sec gust) = 90 mph
Mean roof height = 60 ft Importance factors for
Effective wind area = 10 sq-ft non-hurricane prone regions
Exposure category = B Category Factors
Roof slope reduction = 1 I 0.87
II 1
17.626 psf III 1.15
IV 1.15
where:
0.85 (Velocity pressure exposure coeff.)
1 (Topographic factor) Values for K_z (Ref. Table 6-5)
1 (Wind directionality factor) z (or h) B' (Case 1) C' (Case '1')
V = 90 mph 15 0.7 0.85
I = 1 (Importance factor) 20 0.7 0.9
25 0.7 0.94
30 0.7 0.98
40 0.76 1.04
50 0.81 1.09
60 0.85 1.13
1 (Zone 4, positive) 70 0.89 1.17
-1.1 (Zone 4, negative) 80 0.93 1.21
-1.4 (Zone 5, negative) 90 0.96 1.24
100 0.99 1.26
0.18 (Ref. Table 6-7)
-0.18GC_p (Ref. Figure 6-5A)
Trib. Area GC_p (4, posi.) GC_p (4, neg.)
1 1 -1.1
10 1 -1.1
15 0.967 -1.084
20 0.933 -1.067
35 0.900 -1.034
50 0.867 -1
=·····= IVKKK00256.0q2
dztzh
=zK
=ztK
=dK
��
��
�
=pGC
���
=piGC
70 0.853 -0.973100 0.833 -0.933
145 0.818 -0.918
500 0.7 -0.8
Note: Italicized values are interpolated
20.8 psf (Zone 4)
-22.6 psf (Zone 4)
-27.8 psf (Zone 5)
ASSUMPTIONS
1) Building is ENCLOSED (therefore, GC_pi = ± 0.18)
2) No topographic effects are present
3) Structure is not located in "Special Wind Region"
( ) ( )[ ]��
��
�
=−·=piph GCGCqp
Jason E. Barrick, P.E. D
2850 Valley Vista Drive
Sedona, AZ 86351 Cole County Law Enforcement Designed(928) 284-9710
[email protected] Check int. mull. @ Mark 'W1' (WL=27.8 psf) Shee
Wind load = 27.8 psfDead load = 8 psf
0.823 in
Span length (L) = 144 inLeft DLO = 32 in
Right DLO = 32 inSite-line = 2.5 inModule = 34.5 in
6.660 lbf/in
4.532 < OK4
in=∆····
=allow
4
W Lreq,x x
E384
Lw5I
=·���
����
�+
+=
144
W L.L.S
2
DLODLOw
rightlef t
W L
==∆ 175/Lallow
x xI
17264 lbf-in
5126 psi Section properties for:12.693.36
9500 psi 2.301.83
0.54 < 1.00 OKWind load reaction = 480Dead load reaction = 276
Ext. #257 is adequate at intermed. mullion where L = 144"and T.W. = 34.5".
=·
=8
LwM
2
W L
m ax,x x
==xx
m ax,x x
x x,bS
Mf
=x xI
=x xS
=y yI
=y yS
=xx,bF
=x x,b
x x,b
F
f
∴
Jason E. Barrick, P.E. D
2850 Valley Vista Drive
Sedona, AZ 86351 Cole County Law Enforcement Designed(928) 284-9710
[email protected] Check int. mull. @ Mark 'W3' (WL=27.8 psf) Shee
Wind load = 27.8 psfDead load = 8 psf
0.500 in
Span length (L) = 72 inLeft DLO = 32 in
==∆ 175/Lallow
Right DLO = 32 inSite-line = 2.5 inModule = 34.5 in
6.660 lbf/in
0.466 < OK
4316 lbf-in
1281 psi Section properties for:12.693.36
9500 psi 2.301.83
0.13 < 1.00 OKWind load reaction = 240Dead load reaction = 138
Ext. #257 is adequate at intermed. mullion where L = 72"and T.W. = 34.5".
4in
=·
=8
LwM
2
W L
m ax,x x
==xx
m ax,x x
x x,bS
Mf
=∆····
=allow
4
W Lreq,x x
E384
Lw5I
=·���
����
�+
+=
144
W L.L.S
2
DLODLOw
rightlef t
W L
=x xI
=x xS
=y yI
=y yS
x xI
=xx,bF
=x x,b
x x,b
F
f
∴
See note below
NOTE: Allowable bending stress for T6 is 15000 psi, so analysis above is conservati For Ext. #257, allowable bending moment is (15000 psi)(3.368 in^3) = 50520 lb-in.
Jason E. Barrick, P.E. D
2850 Valley Vista Drive
Sedona, AZ 86351 Cole County Law Enforcement Designed(928) 284-9710
[email protected] Check int. mull. @ Mark 'W5' (WL=27.8 psf) Shee
Wind load = 27.8 psfDead load = 8 psf
0.333 in
Span length (L) = 48 inLeft DLO = 32 in
Right DLO = 32 inSite-line = 2.5 inModule = 34.5 in
6.660 lbf/in
0.138 < OK
1918 lbf-in
570 psi Section properties for:12.693.36
9500 psi 2.301.83
0.06 < 1.00 OKWind load reaction = 160Dead load reaction = 92
Ext. #257 is adequate at intermed. mullion where L = 48"and T.W. = 34.5".
4in
=·
=8
LwM
2
W L
m ax,x x
==xx
m ax,x x
x x,bS
Mf
=∆····
=allow
4
W Lreq,x x
E384
Lw5I
=·���
����
�+
+=
144
W L.L.S
2
DLODLOw
rightlef t
W L
==∆ 175/Lallow
=x xI
=x xS
=y yI
=y yS
x xI
=xx,bF
=x x,b
x x,b
F
f
∴
NOTE: Allowable bending stress for T6 is 15000 psi, so analysis above is conservati For Ext. #257, allowable bending moment is (15000 psi)(3.368 in^3) = 50520 lb-in.
Cole County Law EnforcementCbeam 2000
...10/14/2010 11:25 File: W4-jamb
Mark 'W4' - Right-jamb vert. analysis (22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
Max. Span Deflection = -0.0678" (Span 2, @ 90.00")Max. Positive Moment = 5263"# (Span 1, @ 46.50")Max. Negative Moment = -9043"# (Span 3, @ 42.00")
1 124.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 4.708 4.708 0.000 124.000 2 4.708 4.708 0.000 90.000 3 4.708 4.708 0.000 42.000 4 4.708 4.708 0.000 104.000
1 223 2 668 4 646
3
45
4.708 lb/in
Cole County Law EnforcementCbeam 2000
...10/14/2010 11:25 File: W4-jamb
Mark 'W4' - Right-jamb vert. analysis (22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
Max. Span Deflection = -0.0678" (Span 2, @ 90.00")Max. Positive Moment = 5263"# (Span 1, @ 46.50")Max. Negative Moment = -9043"# (Span 3, @ 42.00")
1 124.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 4.708 4.708 0.000 124.000 2 4.708 4.708 0.000 90.000 3 4.708 4.708 0.000 42.000 4 4.708 4.708 0.000 104.000
1 223 2 668 4 646
3
45
4.708 lb/in
1
5 158
12
Maximum distributed load vshown only, see distributetable for detailed informa
1
5 158
12
Maximum distributed load vshown only, see distributetable for detailed informa
w_WL = (30")(22.6 psf/144) = 4.708 lb/in P_DL,1 = (4.708)(8/22.6)(214") = 357 lb P_DL,3 = (4.708)(8/22.6)(146") = 243 lb
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:14 File: W4-int
Mark 'W4' - Intermed. vert analysis (1st from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Max. Span Deflection = -0.1356" (Span 2, @ 90.00")Max. Positive Moment = 10526"# (Span 1, @ 46.50")Max. Negative Moment = -18087"# (Span 3, @ 42.00")
1 124.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 124.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
3
45
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:14 File: W4-int
Mark 'W4' - Intermed. vert analysis (1st from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Max. Span Deflection = -0.1356" (Span 2, @ 90.00")Max. Positive Moment = 10526"# (Span 1, @ 46.50")Max. Negative Moment = -18087"# (Span 3, @ 42.00")
1 124.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 124.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
3
45
1
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
1 445 2 1337 4 1292 5 316
12
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
1
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
1 445 2 1337 4 1292 5 316
12
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
w_WL = (30")(22.6 psf/144) = 9.417 lb/in P_DL,1 = (9.417)(8/22.6)(214") = 713 lb P_DL,4 = (9.417)(8/22.6)(146") = 487 lb
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:33 File: W4-int2
Mark 'W4' - Intermed. vert analysis (2nd from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Max. Span Deflection = -0.1651" (Span 2, @ 90.00")Max. Positive Moment = 4719"# (Span 4, @ 72.80")Max. Negative Moment = -19920"# (Span 4, @ 0.00")
1 88.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:33 File: W4-int2
Mark 'W4' - Intermed. vert analysis (2nd from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Max. Span Deflection = -0.1651" (Span 2, @ 90.00")Max. Positive Moment = 4719"# (Span 4, @ 72.80")Max. Negative Moment = -19920"# (Span 4, @ 0.00")
1 88.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice
1
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 88.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
1 264 2 1136 4 1353 5 298
12
3
45
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
1
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 88.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
1 264 2 1136 4 1353 5 298
12
3
45
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
w_WL = (30")(22.6 psf/144) = 9.417 lb/in P_DL,1 = (9.417)(8/22.6)(178") = 593 lb P_DL,4 = (9.417)(8/22.6)(146") = 487 lb
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:44 File: W4-int3
Mark 'W4' - Intermed. vert analysis (3rd from right, 22.6 psf)
By: jeb
Beam Results
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:44 File: W4-int3
Mark 'W4' - Intermed. vert analysis (3rd from right, 22.6 psf)
By: jeb
Beam Results
1
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
Max. Span Deflection = -0.1720" (Span 2, @ 90.00")Max. Positive Moment = 4590"# (Span 4, @ 72.80")Max. Negative Moment = -20348"# (Span 4, @ 0.00")
1 52.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 52.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
1 8 2 1043 4 1368 5 294
12
3
45
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
1
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 3
Support Reactions Joint Pounds
Max. Span Deflection = -0.1720" (Span 2, @ 90.00")Max. Positive Moment = 4590"# (Span 4, @ 72.80")Max. Negative Moment = -20348"# (Span 4, @ 0.00")
1 52.000 12.695 3.368 1.0e+007 2 90.000 12.695 3.368 1.0e+007Splice 3 42.000 12.695 3.368 1.0e+007 4 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 52.000 2 9.417 9.417 0.000 90.000 3 9.417 9.417 0.000 42.000 4 9.417 9.417 0.000 104.000
1 8 2 1043 4 1368 5 294
12
3
45
9.417 lb/in
Maximum distributed load vshown only, see distributetable for detailed informa
w_WL = (30")(22.6 psf/144) = 9.417 lb/in P_DL,1 = (9.417)(8/22.6)(142") = 473 lb P_DL,4 = (9.417)(8/22.6)(146") = 487 lb
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:52 File: W4-int4
Mark 'W4' - Intermed. vert analysis (4th from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 2
Support Reactions Joint Pounds
Max. Span Deflection = -0.3356" (Span 1, @ 83.70")Max. Positive Moment = 13730"# (Span 1, @ 54.00")Max. Negative Moment = -29664"# (Span 3, @ 0.00")
1 108.000 12.695 3.368 1.0e+007Splice 2 42.000 12.695 3.368 1.0e+007 3 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 108.000 2 9.417 9.417 0.000 42.000 3 9.417 9.417 0.000 104.000
1 509 3 1679 4 204
1
2
34
9.417 lb/in
Maximum distributed load vshown only, see distribute
Cole County Law EnforcementCbeam 2000
...10/14/2010 10:52 File: W4-int4
Mark 'W4' - Intermed. vert analysis (4th from right, 22.6 psf)
By: jeb
Beam Results
Member Information
Span Length(in) I(in^4) S(in^3) E(psi)
Distributed Load InformationSpan W1(#/in) W2(#/in) X1(in) X2(in)
Joints Free to Displace
Free Joints - 2
Support Reactions Joint Pounds
Max. Span Deflection = -0.3356" (Span 1, @ 83.70")Max. Positive Moment = 13730"# (Span 1, @ 54.00")Max. Negative Moment = -29664"# (Span 3, @ 0.00")
1 108.000 12.695 3.368 1.0e+007Splice 2 42.000 12.695 3.368 1.0e+007 3 104.000 12.695 3.368 1.0e+007
1 9.417 9.417 0.000 108.000 2 9.417 9.417 0.000 42.000 3 9.417 9.417 0.000 104.000
1 509 3 1679 4 204
1
2
34
9.417 lb/in
Maximum distributed load vshown only, see distribute
w_WL = (30")(22.6 psf/144) = 9.417 lb/in P_DL,1 = (9.417)(8/22.6)(108") = 360 lb P_DL,3 = (9.417)(8/22.6)(146") = 487 lb
1
table for detailed informa
1
table for detailed informa
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