1 J.3b-N254 Structural Calcs-HVAC Anchorage.pdf

NASA N254 - HVAC Support

S T R U C T U R A L C A L C U L A T I O NS

Final Submission

July 1, 2024

Prepared by

AECOM

300 Lakeside Dr.

Suite 400 Oakland, CA 94612

AECOM Project No. 60667208

NARC Job No. K21007

Prepared for

NASA Ames Research Center.

Mountain View, CA 94025

© AECOM Restricted Calculation Cover Page Template (Q2[DCS]-351-FM5) Revision 2 December 6, 2021

1/2

DCS

NASA N254 HVAC Support Q2[DCS]-351-FM5

Project Job No.

N254 (K21007) 60667208

Client Department/Discipline

NASA Structural

Software Name

ETABS, TEDDS, Mathcad

Calculation Rev. No. Originator Self Check Reviewer/Checker Independent Peer

Reviewer Approver

0 Stephanie Ruybal Fabiola Macias-Wallis Tom Zimmerman

Add rows as required

Calculation Objective:

Evaluate the existing steel-framed structure for the proposed HVAC system using the existing construction drawings dated 2001 following provisions given by the IEBC. Design the support framing from the equipment base to the existing roof framing.

The proposed HVAC system includes the removal and replacement of existing rooftop HVAC equipment and ductwork. The new HVAC equipment includes 2 AC units, 2 exhaust fans, and ductwork that require the design of equipment support structures that are included in this calculation.

Calculation Methodology:

Evaluate the existing structure for the existing conditions Calculate the equipment reaction forces due to dead load and either wind or seismic loads Design the equipment support members for the equipment reaction forces Calculate the total design capacity ratio and deflection due to both the existing conditions and proposed HVAC equipment

References / Inputs/ Field Data:

References ─ American Institute of Steel Construction (AISC), Seismic Design Manual, 3rd Edition (AISC 341-16) ─ American Institute of Steel Construction (AISC), Steel Construction Manual, 15th Edition ─ American Institute of Steel Construction (AISC), Specification for Structural Steel Buildings (AISC 360-16) ─ American Society of Civil Engineers (ASCE), Minimum Design Load and Associated Criteria for Buildings and

Other Structures (ASCE 7-16) ─ American Society of Civil Engineers (ASCE), Wind Load Design for Petrochemical and Other Industrial

Facilities (ASCE 48303) ─ American Welding Society (AWS) Structural Welding – Steel (AWS D1.1 2015) ─ California Building Code (CBC) 2019 ─ International Code Council (ICC) International Existing Building Code (IEBC 2018)

C-1

© AECOM Restricted Calculation Cover Page Template (Q2[DCS]-351-FM5) Revision 2 December 6, 2021

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─ Original construction drawings 254-9704-G3, 254-9704-S1 thru 254-9704-S11, 254-9704-A1 thru 254-9704- A11

Design Criteria ─ Risk Category IV ─ Seismic Design Category D ─ Floor Live Load = N/A ─ Roof Live Load = N/A ─ Roof Snow Load = 0 psf ─ Basic Design Wind Speed = 102 mph, Exposure C ─ Earthquake Design Data:

▪ Seismic Importance Factor Ie = 1.5

▪ Mapped Spectral Response Acceleration Parameters Ss = 1.5, S1 = 0.6

▪ Site Class D

▪ Design Spectral Response Acceleration Parameters SDS = 1.2, SD1 = 0.68

▪ Horizontal Seismic Design Force FP = 0.311 kip

▪ Response Modification Coefficient R = 2.0

▪ Analysis Procedure Used = Equivalent lateral force procedure

─ Flood Design Data = N/A ─ Special Loads = N/A ─ Roof Rain Load = N/A

See Table of Contents for additional data in the Appendices.

Assumptions: (Include comments on need to revise calculations after more data is collected/confirmed and/or after assumptions have been verified.)

Existing steel beam strength = 36 ksi Existing dead loads = 16 psf (existing condition) that includes the following:

─ Metal deck = 2.7 psf ─ Roofing = 6.3 psf ─ Ceiling = 2.8 psf ─ MEP = 2 psf ─ Insulation = 2.1 psf

AC unit weight = 1800 lb each

Conclusions including confirmations to be obtained:

The existing structure is adequate to support the new HVAC loads.

This calculation is complete and ready for Discipline Review

Stephanie Ruybal 4/25/2024 Originator Name Signature Date

C-2

PROJECT TITLE:

NASA N254

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

Table of Contents

Description Page Number

Calculation Cover Page C-1 – C-2

Table of Contents 1

1. Sketches – see project drawings -

2. Design Equipment and Duct Support Members

2.1 Equipment and Duct Support Geometry

2.2 Duct Support Layout

2.3 Duct Support Design

2.4 Calculate AC Equipment Forces

2.5 Design New Wood Skid Members

2.6 Design Anchorage between AC Isolator and Sleeper

2.7 Design Anchorage between Fan and Sleeper

2.8 Design Anchorage between Sleepers and Existing Roof

Beams

2-1 2-3

2-4 – 2-20 2-21 – 2-27 2-28 – 2-40 2-41 – 2-46 2-47 – 2-50 2-51 – 2-60

3. Evaluate Existing Structure

3.1 Construction Document Review

3.2 Proposed HVAC Modifications

3.3 Code Discussion

3.4 Methodology of Evaluating the Existing Structure

3.5 Determine the Design Capacity Ratio (DCR)

3.6 Calculate Deflection

3.7 Evaluate the Existing Columns

3.8 Check that the Existing Structure is Adequate for Lateral

Loads

3-1 3-2 3-2 3-3 3-5 3-6 3-7 3-7

Appendix A – Drawings of Existing Structure

Appendix B – ETABS Model Report of Duct Portal Frames

Appendix C – Equipment Vendor Drawings

Appendix D – ETABS Model Report of Existing Structure

Appendicies are included as a separate document upon request.

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2. Design Equipment and Duct Support Members

The new HVAC system includes 2 identical AC units and 2 identical exhaust fans highlighted in yellow in the following figure

The new ductwork is highlighted in blue in the following figure

─ Bottom of duct elevations are outlined in blue; elevations for the 12” and 14” diameter ducts were measured from drawing 254-2101-M301

Figure 2.1. New HVAC Equipment Layout (see drawing 254-2101-M102)

2.1 Determine Equipment and Duct Support Geometry

Locate the new wood skids under the anchor locations of AC and Exhaust Fan units

The duct supports are designed in this calculation; however, the duct support locations will be by others spaced at 6’-0” max

─ An equipment and duct support layout was completed to determine the geometries and locations using a 3D model as shown in the following screenshot:

2-1

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

11/20/2023

Checked By:

F. Macias-Wallis

Date:

11/27/2023

Figure 2. New Wood Skid (green) Layout with New Equipment/Ducts (yellow) and Existing Wood Skids (brown) using Preliminary Duct Support Layout (cyan)

The new wood skids (green) will transfer the equipment and duct loads (yellow) into the existing roof framing members

Equipment support framing will be attached to 8x16 posts

─ The 8x16 posts will be attached (2) shaped blocks to fit the roof deck flutes

─ The shaped blocks will be bolted to the top flange of the existing roof framing members

2-2

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.2 Duct Support Layout

Figure 3. Duct Support Layout Numbering

2-3

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.3 Duct Support Design

2-4

CALCULATION SHEET

PROJECT TITLE: __NASA N254______________________________________

SUBJECT/FEATURE: __HVAC Support________________________________

AECOM PROJECT NO. __60667208____________________

NARC JOB NO.:__K21007______________________

CALC. BY:_S. Ruybal_ DATE :__11/14/23_

CHECKED BY:_F. Macias-Wallis_ DATE :_11/23/23_

Duct Support Design Calculate the duct reaction forces due to dead load, wind, & seismic loads and design the duct support frames.

Design Calculations:

A 3D model was created to visualize both the location and the geometry of each duct support Duct supports are located a maximum of 6'-0" spacing The table below summarizes the geometries of the duct supports:

Duct Support Number Height Width Type 1 2'-11" 3'-11" Tall 2 2'-11" 3'-11" Tall 3 7" 2'-7" Short 4 7" 2'-7" Short 5 7" 2'-7" Short 6 7" 1'-8" Short 7 7" 1'-8" Short 8 7" 3'-6" Short 9 1'-2" 3'-8" Tall

10 1'-2" 3'-8" Tall 13 7" 4'-0" Short 14 7" 4'-0" Short 15 7" 1'-8" Short 16 7" 1'-8" Short 17 7" 2'-7" Short 18 7" 2'-7" Short 19 7" 2'-7" Short 20 4'-0" 2'-5" Tall 21 4'-0" 2'-5" Tall 22 4'-7" 2'-1" Tall 23 5'-6" 4'-7" Tall

Not Used Not Used

Duct Supports.xmcd 12/5/2023 9:16 AM

2-5

SUBJECT/FEATURE: __HVAC Support________________________________

Duct Reaction Forces for Governing Tall Support:

Dead Load:

20 ga. duct = 1.5 psf 1" rigid insulation = 1.5 psf Duct sizes (see drawing 254-2101-M102):

Perimeter (in)

20 x 20 80 24 x 18 84 20 x 12 64 18 x 14 64 30 x 18 96

Duct Size (in)

DL 1.5 1.5( )psf 96 in 1.2 28.8 plf Dead load of duct with 20% contingency

2-6

SUBJECT/FEATURE: __HVAC Support________________________________

CALC. BY:_S. Ruybal_ DATE :

CHECKED BY:_F. Macias-Wallis_ DATE :

Seismic Load:

ap 2.5 Component amplification factor; ASCE 7 Table 13.6-1 for ducts

Rp 3 ASCE 7 Table 13.6-1 for ducts

SDS 1.2 Spectral acceleration, see stair tower calculation

Wp DL 6 ft 0.173 kip Duct weight per 6ft spacing

Ip 1.5 Component importance factor h 1.0 Roof height z h 1

Fp.

0.4 ap SDS

Rp Ip

1 2 z h

1.8 Horizontal seismic design force; ASCE 7 Eq. 13.3-1

Fp_min 0.3 SDS Ip 0.54

Fp_max 1.6 SDS Ip 2.88

Fp Fp. Wp 0.311 kip

Duct Supports.xmcd 4/23/2024 2:43 PM

2-7

4/23/24

5/1/24

SUBJECT/FEATURE: __HVAC Support________________________________

Calculate reaction forces due to seismic:

Ev 0.2 SDS Wp 0.041 kip Vertical seismic design force, ASCE 7 Sect 13.3.1.2

DR 0.8 Wp 0.138 kip Reliable dead load to resist seismic

MOT Fp 5.5ft( ) 1.711 kip ft Overturning moment due to seismic (conservatively used max height)

Dead

ASCE 7

2.3.1 Eq.

Vertical Force P

(kip)

Vertical Force P

(kip)

Horizontal Force V

(kip) Moment M

(kip-ft)

0.173 0.041 0.311 1.711

LC1 1.4D 1 0.242 0.000 0.000 0.000

LC2 1.2D+1.0E 6 0.208 0.041 0.311 1.711

LC3 0.9DR+1.0E 7 0.125 0.041 0.311 1.711

SesimicLRFD Load Combinations:

Ev 0.041 kip

Eh Fp 0.311 kip

M MOT 1.711 kip ft

Duct Supports.xmcd 4/23/2024 2:43 PM

2-8

4/23/24

SUBJECT/FEATURE: __HVAC Support________________________________

Summary Table:

Dead

ASCE 7

2.3.1 Eq.

Vertical Force P

(kip)

Vertical Force P

(kip)

Horizontal Force V

(kip) Moment M

(kip-ft)

0.173 0.372 0.471 3.517

LC1 1.4D 1 0.242 0.000 0.000 0.000

LC4 1.2D+1.0W 4 0.208 0.372 0.471 3.517

LC5 0.9DR+1.0W 6 0.125 0.372 0.471 3.517

WindLRFD Load Combinations:

Duct Framing Design:

An ETABS model was created to calculate the forces at the connections and to check the design capacity ratios of the duct portal frame members. Wind was conservatively applied in 3 directions simultaneously in the load combinations.

Check Vertical Framing Members:

Conservatively consider that the total lateral wind force is applied to the top of the vertical framing member and that member is fixed.

P Fh

0.235 kip Horizontal force on each vertical member l 5.5ft Controlling height of vertical member

E 29000ksi Steel modulus of elasticity

I 1.23in4 Moment of inertia for L3x3x1/4

Δ P l3

12 E I

0.158 in Deflection of vertical member l

Δ

417.576 Deflection of vertical member < L/200 OK

Duct Supports.xmcd 4/24/2024 8:35 AM

2-10

4/24/24

Project

NASA N254 HVAC

Job Ref.

Section

TYPE B SUPPORT: WOOD PERPENDICULAR TO ANGLE

Sheet no./rev.

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by Date App'd by Date

Total number of lag screw Ntotal = R C = 2

Applied load Applied load to the connection P = 627 lb

Dowel bearing length (main) (12.3.5) Size of tapered tip E = 0.406 in Actual penetration in main member Pact = L - W - ts = 4.625 in Dowel bearing length in main member lm = 3.500 in

Dowel bearing length (connected) (12.3.5) Dowel bearing length in connected member ls = ts = 0.250 in

Bending yield strength (lag screw) (Table 12J to 12K footnote no. 2) Bending yield strength of lag screw Fyb = 45000 psi

Dowel bearing strength (main member) (Table 12.3.3 footnote no. 2) Dowel bearing strength parallel to grain Fe_par = 11200 Gm 1 psi = 5152 psi

Dowel bearing strength perpendicular to grain Fe_perp = 6100 Gm1.45 1 psi / (D / 1 in) = 2503 psi

Dowel bearing strength for small dia. fasteners Fe = 16600 Gm1.84 1 psi = 3977 psi

Dowel bearing strength at an angle of load to grain Fem = (Fe_par Fe_perp) / ((Fe_par (sin(m))2) + (Fe_perp (cos(m))2)) Fem = 2503 psi Dowel bearing strength of main member Fem = 2503 psi

Dowel bearing strength (connected steel member) (Table 12K footnote no. 2) Dowel bearing strength of connected member Fes = 87000 psi

Preliminary yield limit equation coefficients (Table 12.3.1A notes) Dowel bearing strength ratio Re = Fem / Fes = 0.029 Dowel bearing length ratio Rt = lm / ls = 14.000 Preliminary yield limit equation coefficient k1 k1 = (((Re+(2Re2(1+Rt+Rt2))+(Rt2Re3)))-(Re(1+Rt)))/(1+Re) k1 = 0.182 Preliminary yield limit equation coefficient k2 k2 = -1+((2(1+Re))+((2Fyb(1+(2Re))Dr2))/(3Femlm2)) k2 = 0.512 Preliminary yield limit equation coefficient k3 k3 = -1+(((2(1+Re))/Re)+((2Fyb(2+Re)Dr2))/(3Femls2)) k3= 11.564 Angle of load to grain coefficient k k = 1 + (0.25 max (m, s) / 90) = 1.250 Reduction term (Table 12.3.1B) KD = 3.00

Yield limit equations (single shear) Mode Im (eq. 12.3-1) ZIm = (Dr lm Fem) / (4 k) = 825 lb

Mode Is (eq. 12.3-2) ZIs = (Dr ls Fes) / (4 k) = 2049 lb

Mode II (eq. 12.3-3) ZII = (k1 Dr ls Fes) / (3.6 k) = 414 lb

Mode IIIm (eq. 12.3-4) ZIIIm = (k2 Dr lm Fem) / ((1 + (2 Re)) 3.2 k) = 500 lb

Mode IIIs (eq. 12.3-5) ZIIIs = (k3 Dr ls Fem) / ((2 + Re) 3.2 k) = 420 lb

Mode IV (eq. 12.3-6) ZIV = (Dr 2 (((2 Fem Fyb) / (3 (1 + Re))))) / (3.2 k) = 474 lb Z = min (ZIm, ZIs,ZII, ZIIIm, ZIIIs, ZIV) = 414 lb Nominal capacity of single fastener Z = 414 lb

Actual load applied to the connection = 314 lb;

conservatively left load applied to the connection as 627 lb

2-14

Job Ref.

Section

TYPE B SUPPORT: WOOD PERPENDICULAR TO ANGLE

Sheet no./rev.

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by Date App'd by Date

Slenderness (Table 12.5.1C footnote no.1) Slenderness l / D = 5.600

Spacing requirements (perpendicular to grain loading)

End distance (Table 12.5.1A) End distance (full stregnth) aq_full = 4 D = 2.500 in End distance (minimum) aq_min = 2 D = 1.250 in End distance (actual) aq = 2.500 in

Edge distance (Table 12.5.1C) Loaded edge eq = 4 D = 2.500 in Unloaded edge ep = 1.5 D = 0.938 in

Center to center spacing (Table 12.5.1B) Center to center spacing (full strength) s_full = 4 D = 2.500 in Center to center spacing (minimum) s_min = 3 D = 1.875 in Center to center spacing (actual) s = 3.500 in

Row spacing (Table 12.5.1D) Row spacing srow = ((5 l) + (10 D)) / 8 = 2.969 in

Geometry factor C (12.5.1) End distance (actual) aq = 2.500 in End distance (full strength) aq_full = 2.500 in Geometry factor for end distance C = aq / aq_full = 1.00 Center to center spacing (actual) s = 3.500 in Center to center spacing (full strength) s_full = 2.500 in Geometry factor for spacing C = s / s_full = 1.40 Geometry factor C = min(1,C C) = 1.00

Adjustment factor Load duration factor (Table 2.3.2) CD = 1.00 Wet service factor (Table 11.3.3) CM = 1.0 Temperature factor (Table 11.3.4) Ct = 1.0 Group action factor (eq. 11.3-1) Cg = 1.0 Geometry factor (12.5.1) C = 1.00 End grain factor (12.5.2) Ceg = 1.0 Diaphragm factor (12.5.3) Cdi = 1.0 Toe nail factor (12.5.4) Ctn = 1.0

Total capacity of connection Capacity of connection Z’ = Z Ntotal Nint CD CM C= 828 lb P / Z' = 0.757

Design result PASS - Connection capacity exceeds applied load

2-15

SUBJECT/FEATURE: __HVAC Support________________________________

Design Duct Support 23/24:

This duct support is unique because it is not supported by an existing or new skid This support will be constructed on Durabloks that are set on an architectural pad The same frame design as the tall duct support designed in this calculation can be used This portion of the calculation will check the existing deck point loads, designing a brace to transfer lateral and any vertical forces, and design the connection to the Durablock

Check existing deck point load:

Rv 0.2076kip

0.342kip

2.898kip ft 4.583ft

0.907 kip Vertical reaction of 1 leg of support 23 due to wind

Apad 3.25ft 3.25 ft 10.563 ft2 Area of architectural support pad

Rv_area Rv

Apad

85.883 psf Vertical load spread over architectural support pad

Rv_all 92psf Allowable LRFD load of 22 gage roof deck with 8'-0" span

Rv_area Rv_all

0.934 CheckRv "OK"

Rv_area Rv_all

1.0if

"Decking NG" otherwise

"OK"

2-16

SUBJECT/FEATURE: __HVAC Support________________________________

Brace Design:

Fh 0.471 kip Horizontal design force

Rv 0.907 kip Vertical design force

By inspection, an L3x3x1/4" angle is adequate as a brace member and (2) 1/2" dia. bolts are adequate as calculated above. By inspection, a 3/8" thick gusset plate with 3/16" fillet weld is adequate.

Check withdrawal using same values as above except for design load and withdrawal angle Conservatively design top lag screws for entire design load Tedds is used to check the top screws for lateral forces and to calculate Z` for this connection Calculate the actual vertical design force accounting for structure dead load

Rv_24 Rv Wp 175lbf

0.733 kip Vertical reaction from wind counteracted by DL of duct and steel (considered to weigh 175 lbf) p24 4.625in Fastener penetration, 5" length

Z'24 1254lb Adjusted lateral design value per connection; see Tedds output next sheets

W' 1446.939 lb in

Adjusted withdrawal design value per connection; NDS αmin 22.5deg Minimum withdrawal angle

Z'α24 W' p24 Z'24

W' p24 cos αmin 2 Z'24 sin αmin 2

1423.39 lb NDS Eq 12.4-1

Zα24 Fh 2 Rv_24

871.287 lbf

871.287 1423.39

0.612 OK

2-17

Job Ref.

Section

Brace Support Sheet no./rev.

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by Date App'd by Date

LAG SCREW TIMBER TO STEEL CONNECTION DESIGN

In accordance with NDS 2018 Tedds calculation version 1.2.04

Design results summary Unit Required Provided Utilization Result

Connection capacity lbs 871 1254 0.695 PASS

P Main member

Connected member tm ts m

Main timber member details Species of main member Douglas Fir-South Size of main member (Table 1B) 6 x 8 Number of main member Nm = 1 Thickness of main member tm = 5.500 in Angle of load to grain of main member m = 25

Connected steel member details Number of connected steel member Ns = 1 Connected steel member thickness ts = 0.250 in Number of interfaces Nint = (Nm + Ns) – 1 = 1

Lag screw details Lag screw diameter (Table L2) 5/8'' Lag screw root diameter of threaded portion Dr = 0.471 in Length of lag screw L = 5.000 in Washer thickness W = 0.125 in Number of rows of lag screw R = 1 Number of columns of lag screw C = 2 Total number of lag screw Ntotal = R C = 2

2-18

Job Ref.

Section

Brace Support Sheet no./rev.

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by Date App'd by Date

Applied load Applied load to the connection P = 871 lb

Dowel bearing length (main) (12.3.5) Size of tapered tip E = 0.406 in Actual penetration in main member Pact = L - W - ts = 4.625 in Dowel bearing length in main member lm = 4.422 in

Dowel bearing length (connected) (12.3.5) Dowel bearing length in connected member ls = ts = 0.250 in

Bending yield strength (lag screw) (Table 12J to 12K footnote no. 2) Bending yield strength of lag screw Fyb = 45000 psi

Dowel bearing strength (main member) (Table 12.3.3 footnote no. 2) Dowel bearing strength parallel to grain Fe_par = 11200 Gm 1 psi = 5152 psi

Dowel bearing strength perpendicular to grain Fe_perp = 6100 Gm1.45 1 psi / (D / 1 in) = 2503 psi

Dowel bearing strength for small dia. fasteners Fe = 16600 Gm1.84 1 psi = 3977 psi

Dowel bearing strength at an angle of load to grain Fem = (Fe_par Fe_perp) / ((Fe_par (sin(m))2) + (Fe_perp (cos(m))2)) Fem = 4333 psi Dowel bearing strength of main member Fem = 4333 psi

Dowel bearing strength (connected steel member) (Table 12K footnote no. 2) Dowel bearing strength of connected member Fes = 87000 psi

Preliminary yield limit equation coefficients (Table 12.3.1A notes) Dowel bearing strength ratio Re = Fem / Fes = 0.050 Dowel bearing length ratio Rt = lm / ls = 17.688 Preliminary yield limit equation coefficient k1 k1 = (((Re+(2Re2(1+Rt+Rt2))+(Rt2Re3)))-(Re(1+Rt)))/(1+Re) k1 = 0.367 Preliminary yield limit equation coefficient k2 k2 = -1+((2(1+Re))+((2Fyb(1+(2Re))Dr2))/(3Femlm2)) k2 = 0.479 Preliminary yield limit equation coefficient k3 k3 = -1+(((2(1+Re))/Re)+((2Fyb(2+Re)Dr2))/(3Femls2)) k3= 8.620 Angle of load to grain coefficient k k = 1 + (0.25 max (m, s) / 90) = 1.069 Reduction term (Table 12.3.1B) KD = 3.00

Yield limit equations (single shear) Mode Im (eq. 12.3-1) ZIm = (Dr lm Fem) / (4 k) = 2109 lb

Mode Is (eq. 12.3-2) ZIs = (Dr ls Fes) / (4 k) = 2395 lb

Mode II (eq. 12.3-3) ZII = (k1 Dr ls Fes) / (3.6 k) = 978 lb

Mode IIIm (eq. 12.3-4) ZIIIm = (k2 Dr lm Fem) / ((1 + (2 Re)) 3.2 k) = 1147 lb

Mode IIIs (eq. 12.3-5) ZIIIs = (k3 Dr ls Fem) / ((2 + Re) 3.2 k) = 627 lb

Mode IV (eq. 12.3-6) ZIV = (Dr 2 (((2 Fem Fyb) / (3 (1 + Re))))) / (3.2 k) = 721 lb Z = min (ZIm, ZIs,ZII, ZIIIm, ZIIIs, ZIV) = 627 lb Nominal capacity of single fastener Z = 627 lb

2-19

Job Ref.

Section

Brace Support Sheet no./rev.

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by Date App'd by Date

Slenderness (Table 12.5.1C footnote no.1) Slenderness l / D = 7.075

Spacing requirements (perpendicular to grain loading)

End distance (Table 12.5.1A) End distance (full stregnth) aq_full = 4 D = 2.500 in End distance (minimum) aq_min = 2 D = 1.250 in End distance (actual) aq = 2.500 in

Edge distance (Table 12.5.1C) Loaded edge eq = 4 D = 2.500 in Unloaded edge ep = 1.5 D = 0.938 in

Center to center spacing (Table 12.5.1B) Center to center spacing (full strength) s_full = 4 D = 2.500 in Center to center spacing (minimum) s_min = 3 D = 1.875 in Center to center spacing (actual) s = 3.500 in

Row spacing (Table 12.5.1D) Row spacing srow = (5 D) = 3.125 in

Geometry factor C (12.5.1) End distance (actual) aq = 2.500 in End distance (full strength) aq_full = 2.500 in Geometry factor for end distance C = aq / aq_full = 1.00 Center to center spacing (actual) s = 3.500 in Center to center spacing (full strength) s_full = 2.500 in Geometry factor for spacing C = s / s_full = 1.40 Geometry factor C = min(1,C C) = 1.00

Adjustment factor Load duration factor (Table 2.3.2) CD = 1.00 Wet service factor (Table 11.3.3) CM = 1.0 Temperature factor (Table 11.3.4) Ct = 1.0 Group action factor (eq. 11.3-1) Cg = 1.0 Geometry factor (12.5.1) C = 1.00 End grain factor (12.5.2) Ceg = 1.0 Diaphragm factor (12.5.3) Cdi = 1.0 Toe nail factor (12.5.4) Ctn = 1.0

Total capacity of connection Capacity of connection Z’ = Z Ntotal Nint CD CM C= 1254 lb P / Z' = 0.695

Design result

2-20

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.4 Calculate AC Equipment Forces

2-21

SUBJECT/FEATURE: __HVAC Support________________________________

Calculate AC Equipment Forces

Equipment Dimensions:

Air Conditioner:

LAC 82.25in Length of AC

BAC 79in Width of AC hAC 44in Height of AC

Reference: RN Series A- Cabinet Standard ~ 6-10 Ton

Dead Loads:

DLAC 1500lbf Weight of AC unit, drawing 254-2101-M601 for AC-1A & AC-1B

DLACcurb 300lbf Weight of AC roof curb, drawing 254-2101-M601 for AC-1A & AC-1B

The dead load shown on drawing 254-2101-M601 for AC-1A & AC-1B is more than that on vendor documents showing equipment corner loads, RN-008-3-0-KB09-000:0000-00A-DCE-AGC-OOFAOOD-00-0000000VB. This may be because the weight estimate on the vendor document states that it does not account for an SPAs.

Calculate the corner loads of the equipment based on 1500 lbs instead of 1149 lbs by scaling the loads

Add roof curb as a point load to each corner

SF

1500lbf 1149lbf

1.305 Scale factor

Load Development.xmct 5/8/2024 3:41 PM

2-22

SUBJECT/FEATURE: __HVAC Support________________________________

Calculate Wind Forces on Equipment:

qh 16.514psf Velocity pressure: Velocity pressure calculated for duct supports

G 0.85 Gust effect factor: Gust effect factor from Section 26.11 of ASCE 7

B 6 4210 mm( ) 2 305 mm( ) 84.875 ft Width of building, see drawing 254-9704-S1 in Appendix A

L 2 7400 mm 48.556 ft Length of building, see drawing 254-9704-S1 in Appendix A h 4128 76( )mm 13.294 ft Height of building, see drawing 254-9704-S4 in Appendix A

Lateral Wind Force:

B h 1128.329 ft2 Vertical projected area of building

Af hAC max BAC LAC 25.132 ft2 Vertical projected area of equipment

GCr_1 1.9 Af 0.1B hif

"Interpolate per ASCE" otherwise

1.9 See ASCE 7 section 29.4.1

Fh qh GCr_1 Af 0.789 kip Lateral wind force, ASCE 7 equation 29.4-2

Vertical (uplift) Wind Force:

B L 4121.243 ft2 Horizontal projected area of building

Ar BAC LAC 45.123 ft2 Horizontal projected area of equipment

GCr_v 1.5 Ar 0.1B Lif

"Interpolate per ASCE" otherwise

1.5 See ASCE 7 section 29.4.1

Fv qh GCr_v Ar 1.118 kip Vertical upl ift force; ASCE 7 equation 29.4-3

Uplift DLAC DLACcurb Fv 0.682 kip No uplift exists

Load Development.xmct 12/5/2023 9:48 AM

2-24

SUBJECT/FEATURE: __HVAC Support________________________________

Calculate Seismic Forces on Equipment:

ap 2.5 Component amplification factor; ASCE 7 Table 13.6-1 for vibration-isolated components

SDS 1.2 Spectral acceleration

Ip 1.5 Component importance factor, ASCE 7 Section 13.1.3 for Risk Category IV

Wp DLAC DLACcurb 1.8 kip Component operating weight

Rp 2 ASCE 7 Table 13.6-1 z 1 Height of structure of point of attachment for roof-mount h 1 Average roof height of structure

Fp_1

0.4 ap SDS

2.7 Seismic design force, equation 13.3-1 ASCE 7

Fp_max 1.6 SDS Ip 2.88 Max seismic design force, equation 13.3-2 ASCE 7

Fp_min 0.3 SDS Ip 0.54 Min seismic design force, equation 13.3-2 ASCE 7

Fp Fp_1 Wp 4.86 kip Seismic horizontal design force

Ev 0.2 SDS Wp 0.432 kip Vertical seismic force, section 13.3.1.2 of ASCE 7

Load Development.xmct 4/24/2024 4:04 PM

2-25

SUBJECT/FEATURE: __HVAC Support________________________________

LAC (ft) = 6.85 DL Equipment Vertical Seismic

Ev Horizontal Seismic Fp

0.629 0.159 1.793

0.184 0.046 0.523

Point Load from Isolator (kip) Uniform Load (klf)

Calculate moment from AC:

d1 6 4 4.5( )in 14.5 in Distance from AC unit to curb support d2 16 8( )in 24 in Distance from curb to sleeper

Meq hAC d1 d2

Fp 27.472 kip ft Moment about horizontal axis due to seismic, assumes vertical center of gravity of AC unit is 2/3 the height of the unit

LRFD Load Combinations for controlling support (front left corner):

Dead Vertical Force P

(klf) Vertical Force Ev

(klf) Horizontal Force

V (klf) Moment M (kip-ft)

0.184 0.046 0.523 27.472

LC 1 1.4D 0.258 0.000 0.000 1.000

LC 2 1.2D+E 0.221 0.046 0.523 27.472

LC 3 0.9DR+E 0.132 0.046 0.523 27.472

Seismic

DR 0.8( ) 0.184 klf 0.147 klf Reliable dead load to resist seismic; reduction of dead load that resists seismic that is equal to 80% of the total dead load

Load Development.xmct 5/8/2024 3:41 PM

2-27

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.5 Design New Wood Skid Members

2-28

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Node Force Moment Fx Fz My

(kips) (kips) (kip_ft) 2 0 0.095 0 3 0 0.029 0

Load case: Dead Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0 0.004 0 2 0 1.236 0 3 0 0.236 0

Load case: Seismic Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 -0.245 0.323 0 2 -2.895 2.992 0 3 -0.883 0.932 0

Load combination: 1.0D (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0 0.033 0 2 0 1.33 0 3 0 0.265 0

Load combination: 1.0D + 0.7E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 -0.172 0.259 0 2 -2.026 3.425 0 3 -0.618 0.917 0

Load combination: 0.6D + 0.7E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 -0.172 0.246 0 2 -2.026 2.892 0 3 -0.618 0.811 0

2-31

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Load combination: 1.4D (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0 0.046 0 2 0 1.862 0 3 0 0.371 0

Load combination: 1.2D + 1.0E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 -0.245 0.362 0 2 -2.895 4.588 0 3 -0.883 1.25 0

Load combination: 0.9D + 1.0E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 -0.245 0.352 0 2 -2.895 4.189 0 3 -0.883 1.17 0

Load combination: 1.0D - 0.7E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0.172 -0.193 0 2 2.026 -0.764 0 3 0.618 -0.388 0

Load combination: 0.6D - 0.7E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0.172 -0.206 0 2 2.026 -1.296 0 3 0.618 -0.493 0

Load combination: 1.2D -1.0E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0.245 -0.283 0 2 2.895 -1.396 0 3 0.883 -0.614 0

2-32

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Load combination: 0.9D - 1.0E (Strength) Node Force Moment

Fx Fz My (kips) (kips) (kip_ft)

1 0.245 -0.293 0 2 2.895 -1.795 0 3 0.883 -0.694 0

Beam - Span 1

Member details Service condition Wet Elevated temperature 100 degF to 125 degF

Sawn lumber section details Number of sections in member N = 1 Nominal breadth of sections bnom = 6 in Breadth of sections b = 5.5 in Nominal depth of sections dnom = 8 in Depth of sections d = 7.5 in Material Douglas Fir-South, Posts and timbers, No.1 grade

7.

5"

5.5"

6''x8'' sawn lumber section Cross-sectional area, A, 41.25 in 2

Section modulus, S x, 51.6 in3

Section modulus, S y, 37.8 in3

Second moment of area, I x, 193.4 in4

Second moment of area, I y, 104 in4

Radius of gyration, r x, 2.165 in Radius of gyration, r y, 1.588 in Douglas Fir-South, Posts and timbers, No.1 grade Bending, Fb, 1150 psi Shear parallel to grain, F v, 165 psi Compression parallel to grain, F c, 925 psi Compression perpendicular to grain, F c_perp, 520 psi Tension parallel to grain, F t, 775 psi Modulus of elasticity, E, 1200000 psi Minimum modulus of elasticity, Emin, 440000 psi Density, , 33.193 lbm/ft 3

Specific gravity, G, 0.46

Span details Bearing length Lb = 7.5 in

Consider Combination 5 - 1.2D + 1.0E (Strength)

Adjustment factors - Table 4.3.1 Wet service factor for bending - Table 4D CMb = 1 Wet service factor for shear - Table 4D CMv = 1 Wet service factor for tension - Table 4D CMt = 1 Wet service factor for perp.comp. - Table 4D CMc_perp = 0.67 Wet service factor for mod.of elasticity - Table 4D CME = 1

2-33

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Temperature factor - Table 2.3.3 Ct = 0.7 Temp.factor for modulus of elasticity - Table 2.3.3 CtE = 0.9 Depth-to-breadth ratio dnom / bnom = 1.33 Slenderness ratio - eq.3.3-5 RB = (Le,y d / b2) = 5.140 Reference bending design value Fb = Fb Ct KFb b = 1390 lb/in2

Adjusted modulus of elasticity Emin' = Emin CtE KFE s = 592416 lb/in2

Critical buckling design value - cl.3.3.3.8 FbE = 1.2 Emin' / RB2 = 26908 lb/in2

Beam stability factor - eq.3.3-6 CL = [1 + (FbE / Fb)] / 1.9 - [([1 + (FbE / Fb)] / 1.9)2 - (FbE / Fb) / 0.95] = 0.997

Time dependent deformation factor - cl.3.5.2 Kcr = 2 Format conversion factor for bending - Table 2.3.5 KFb = 2.54 Format conversion factor for shear - Table 2.3.5 KFv = 2.88 Format conversion factor for tension - Table 2.3.5 KFt = 2.7 Format conv.factor for perp.comp. - Table 2.3.5 KFc_perp = 1.67 Form.conv.factor for mod.of elasticity - Table 2.3.5 KFE = 1.76 Resistance factor for bending - Table 2.3.6 b = 0.85 Resistance factor for shear - Table 2.3.6 v = 0.75 Resistance factor for compression - Table 2.3.6 c = 0.9 Resistance factor for tension - Table 2.3.6 t = 0.8 Resistance factor for mod.of elasticity - Table 2.3.6 s = 0.85 Time effect factor - Table N.3 = 0.8

Check design at end of span

Compression members - General - cl.3.6 Design axial compression force P = 1091 lb Design compression parallel to grain - Table 4.3.1 Fc' = Fc CMc Ct CP KFc c = 835 lb/in2

Actual compression parallel to grain fc = P / (b d) = 26 lb/in2 fc / Fc' = 0.032 PASS - Design compression stress exceeds actual compression stress

Bending members - Flexure - cl.3.3 Design bending moment Mx = 3376 lb_ft Design bending stress - Table 4.3.1 Fb,x' = Fb Ct CL KFb b = 1387 lb/in2

Actual bending stress - eq.3.3-2 fb,x = Mx / Sx = 786 lb/in2 fb,x / Fb,x' = 0.567 PASS - Design bending stress exceeds actual bending stress

Bending members - Shear - cl.3.4 Design shear force Vx = 2058 lb Design shear stress - Table 4.3.1 Fv,x' = Fv Ct KFv v = 200 lb/in2

Actual shear stress - eq.3.4-2 fv,x = 3 Vx / (2 b d) = 75 lb/in2 fv,x / Fv,x' = 0.375 PASS - Design shear stress exceeds actual shear stress

Design for bearing - cl.3.10 Design perpendicular compression Rx = 4588 lb

2-34

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Design bearing stress - Table 4.3.1 Fc_perp,x' = Fc_perp CMc_perp Ct KFc_perp c = 367 lb/in2

Actual bearing stress fc_perp,x = Rx / (b Lb) = 111 lb/in2 fc_perp,x / Fc_perp,x' = 0.303 PASS - Design bearing stress exceeds actual bearing stress perpendicular to grain

Combined bending and axial loading - cl.3.9 Critical buckling design value in x-axis FcE1 = 0.822 Emin' / (Lb,x / d)2 = 2972 lb/in2

Critical buckling design value in y-axis FcE2 = 0.822 Emin' / (Lb,y / b)2 = 1598 lb/in2

Bending and compression check - eqs.3.9-3 and 3.9-4 max((fc / Fc')2 + fb,x / (Fb,x' (1 - (fc / FcE1))), (fc / FcE2)) = 0.573 < 1.0

PASS - Combined bending and compressive stresses are within permissible limits

Consider Combination 6 - 0.9D + 1.0E (Strength)

Check design 3ft 9.601in along span

Bending members - Deflection - cl.3.5 Instantaneous deflection x' = x / CtE = 0.052 in Final deflection x,Final = Kcr x' = 0.104 in Allowable deflection x,Allowable = Lm1_s1 / 250 = 0.384 in x,Final / x,Allowable = 0.27

PASS - Allowable deflection exceeds final deflection

Beam - Span 2

Member details Service condition Wet Elevated temperature 100 degF to 125 degF

Sawn lumber section details Number of sections in member N = 1 Nominal breadth of sections bnom = 6 in Breadth of sections b = 5.5 in Nominal depth of sections dnom = 8 in Depth of sections d = 7.5 in Material Douglas Fir-South, Posts and timbers, No.1 grade

2-35

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

7.

5"

5.5"

6''x8'' sawn lumber section Cross-sectional area, A, 41.25 in 2

Section modulus, S x, 51.6 in3

Section modulus, S y, 37.8 in3

Second moment of area, I x, 193.4 in4

Second moment of area, I y, 104 in4

Radius of gyration, r x, 2.165 in Radius of gyration, r y, 1.588 in Douglas Fir-South, Posts and timbers, No.1 grade Bending, Fb, 1150 psi Shear parallel to grain, F v, 165 psi Compression parallel to grain, F c, 925 psi Compression perpendicular to grain, F c_perp, 520 psi Tension parallel to grain, F t, 775 psi Modulus of elasticity, E, 1200000 psi Minimum modulus of elasticity, Emin, 440000 psi Density, , 33.193 lbm/ft 3

Specific gravity, G, 0.46

Span details Bearing length Lb = 7.5 in

Consider Combination 6 - 0.9D + 1.0E (Strength)

Adjustment factors - Table 4.3.1 Wet service factor for bending - Table 4D CMb = 1 Wet service factor for shear - Table 4D CMv = 1 Wet service factor for tension - Table 4D CMt = 1 Wet service factor for perp.comp. - Table 4D CMc_perp = 0.67 Wet service factor for mod.of elasticity - Table 4D CME = 1 Temperature factor - Table 2.3.3 Ct = 0.7 Temp.factor for modulus of elasticity - Table 2.3.3 CtE = 0.9 Depth-to-breadth ratio dnom / bnom = 1.33 Slenderness ratio - eq.3.3-5 RB = (Le,y d / b2) = 5.140 Reference bending design value Fb = Fb Ct KFb b = 1738 lb/in2

Adjusted modulus of elasticity Emin' = Emin CtE KFE s = 592416 lb/in2

Critical buckling design value - cl.3.3.3.8 FbE = 1.2 Emin' / RB2 = 26908 lb/in2

Beam stability factor - eq.3.3-6 CL = [1 + (FbE / Fb)] / 1.9 - [([1 + (FbE / Fb)] / 1.9)2 - (FbE / Fb) / 0.95] = 0.997

Time dependent deformation factor - cl.3.5.2 Kcr = 2 Format conversion factor for bending - Table 2.3.5 KFb = 2.54 Format conversion factor for shear - Table 2.3.5 KFv = 2.88 Format conversion factor for tension - Table 2.3.5 KFt = 2.7 Format conv.factor for perp.comp. - Table 2.3.5 KFc_perp = 1.67 Form.conv.factor for mod.of elasticity - Table 2.3.5 KFE = 1.76 Resistance factor for bending - Table 2.3.6 b = 0.85 Resistance factor for shear - Table 2.3.6 v = 0.75 Resistance factor for compression - Table 2.3.6 c = 0.9 Resistance factor for tension - Table 2.3.6 t = 0.8

2-36

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/22/2023 Chk'd by Date App'd by Date

Resistance factor for mod.of elasticity - Table 2.3.6 s = 0.85 Time effect factor - Table N.3 = 1

Check design 4ft 9.331in along span

Compression members - General - cl.3.6 Design axial compression force P = 695 lb Design compression parallel to grain - Table 4.3.1 Fc' = Fc CMc Ct CP KFc c = 972 lb/in2

Actual compression parallel to grain fc = P / (b d) = 17 lb/in2 fc / Fc' = 0.017 PASS - Design compression stress exceeds actual compression stress

Bending members - Flexure - cl.3.3 Design bending moment Mx = 2359 lb_ft Design bending stress - Table 4.3.1 Fb,x' = Fb Ct CL KFb b = 1732 lb/in2

Actual bending stress - eq.3.3-2 fb,x = Mx / Sx = 549 lb/in2 fb,x / Fb,x' = 0.317 PASS - Design bending stress exceeds actual bending stress

Combined bending and axial loading - cl.3.9 Critical buckling design value in x-axis FcE1 = 0.822 Emin' / (Lb,x / d)2 = 2972 lb/in2

Critical buckling design value in y-axis FcE2 = 0.822 Emin' / (Lb,y / b)2 = 1598 lb/in2

Bending and compression check - eqs.3.9-3 and 3.9-4 max((fc / Fc')2 + fb,x / (Fb,x' (1 - (fc / FcE1))), (fc / FcE2)) = 0.319 < 1.0

PASS - Combined bending and compressive stresses are within permissible limits

Check design 4ft 4.876in along span

Bending members - Deflection - cl.3.5 Instantaneous deflection x' = x / CtE = 0.128 in Final deflection x,Final = Kcr x' = 0.255 in Allowable deflection x,Allowable = Lm1_s2 / 250 = 0.384 in x,Final / x,Allowable = 0.664

PASS - Allowable deflection exceeds final deflection

2-37

Job Ref.

Section

Wood Beam A - Vertical & Axial Sheet no./rev.

Calc. by

SSR

Date

11/21/2023 Chk'd by Date App'd by Date

WOOD BEAM A - LATERAL MOMENT CALCULATION

In accordance with the ANSI/AF&PA NDS 2018 using the LRFD method Tedds calculation version 2.2.20

ANALYSIS

Tedds calculation version 1.0.37

Results

Forces

Strength combinations - Moment envelope (kip_ft)

0.1

-1.1

0.9

-1.1

Strength combinations - Shear envelope (kips)

-0.7

-0.3

Member results

Envelope - Strength combinations Member Position Shear force Moment

(ft) (kips) (kip_ft) Beam 8 1.026 (max abs) -0.718 0 -1.096 (min)

11.95 0 0.928 (max) 0

2-38

SUBJECT/FEATURE: __HVAC Support________________________________

CHECKED BY:_F. Macias-Wallis_ DATE :_11/27/23_

Design Wood Beam Members for Combined Bending and Axial Loads Purpose:

Apply the NDS combined bending and axial compression equation 3.9-3 to design the wood beam Only Beam A will be designed since it experiences the largest loads from the equipment

Results from Tedds:

fb1 786psi Actual bending stress, see Tedds output fc 44ps Actual compression parallel to grain, see Tedds output

Sy 37.8in3 Section modulus, see Tedds output

FbE 26908ps Critical buckling design value, see Tedds output

FcE2 1598psi Critical buckling design value, see Tedds output

My 1.096kip ft Design bending moment, see Tedds output (note that Tedds was used only to calculate the moment due to the seismic lateral loads)

Design Calculations:

Minor Axis Bending Capacity:

λ 1 Time effect factor for 1.4D=0.6, can be increased to 1.0 for other load combinations; NDS Table N3 ϕb 0.85 Resistance factor, NDS Table N2

KF_Fb 2.16 ϕb

2.541 Format conversion factor, NDS Table N1

Fb 1000psi Bending; NDS Supplement Table 4A for Douglas Fir-Larch No. 1

CF_Fc 1.05 Size factor for compression; NDS Supplement Cf Table

CM_Fb 0.85 Fb CF_Fc 1150psiif

1.0 otherwise

0.85 Wet Service Factor, NDS Supplement Wet Service Factor

Ct 0.7 Temperature factor, NDS 2.3.3

Wood Beam Interaction 11/22/2023 8:41 AM

2-39

SUBJECT/FEATURE: __HVAC Support________________________________

CHECKED BY:_F. Macias-Wallis_ DATE :_11/27/23_

CF_Fb 1.3 Size factor for bending; NDS Supplement Cf Table

Ci 1.0 Insising factor

Cr 1.0 Repetitive member factor; NDS Supplemental Repetitive Member Factor

Cfu 1.05 Flat Use Design Factor, NDS Supplement

F'b2 λ KF_Fb ϕb Fb CM_Fb Ct CF_Fb Ci Cr Cfu 1754.298 psi fb2 My Sy

0.348 ksi

Combined Interaction Equation:

Axial Capacity Term:

fc F'c

2 fb1

F'b1 1

FcE1

= 0.601 See Tedds output

Minor Axis Bending Term:

fb2

F'b2 1

FcE2 fb1 FbE

0.204

Combined Interaction Equation:

0.601 fb2

F'b2 1

FcE2 fb1 FbE

0.805 NDS Equation 3.9-3

A single 6x8 No. 1 Douglas Fir (south) is acceptable for the design forces for all wood skid beams.

Wood Beam Interaction 11/22/2023 8:41 AM

2-40

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.6 Design Anchorage between AC Isolator and Sleeper

2-41

SUBJECT/FEATURE: __HVAC Support________________________________

AC Curb to 6x8 Anchorage Design Caluclate the size and minimum number of anchors between the AC equipment curb and the new wood 6x8 skid members.

Design Parameters:

AC Curb Reactions:

The factored max shear = 0.523 klf; factored max tension = 0.046 klf + moment calculated previously that is spread along a 6.85ft curb

Vu 0.523klf Design shear per foot of curb

Tu 0.046klf( ) 6.85ft( )[ ] 27.472kip ft

79in

4.488 kip Design tension along 1 new wood skid member

Design Calculations:

Use Tedds to check lateral forces on wood and bolts per foot of curb, see next sheets Check combined lateral and withdrawal n 2 Number of lag screws per foot of curb nmin ceil n ft

6.85 ft

14 Minimum number of lag screws per wood skid member

Vu_s Vu 6.85 ft nmin

0.256 kip Design shear per bolt

Tu_s Tu nmin

0.321 kip Design tension per bolt

Curb to 6x8.xmcd 4/24/2024 4:42 PM

2-42

4/24/24

Job Ref.

60667208 Section

AC Curb to 6x8

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by

F. Macias- Date

11/27/2023

LAG SCREW TIMBER TO STEEL CONNECTION DESIGN

In accordance with NDS 2018 Tedds calculation version 1.2.04

Design results summary Unit Required Provided Utilization Result

Connection capacity lbs 341 420 0.812 PASS

P

Main member

Connected member tm ts

Main timber member details Species of main member Douglas Fir-South Size of main member (Table 1B) 6 x 8 Number of main member Nm = 1 Thickness of main member tm = 5.500 in Angle of load to grain of main member m = 90

Connected steel member details Number of connected steel member Ns = 1 Connected steel member thickness ts = 0.250 in Number of interfaces Nint = (Nm + Ns) – 1 = 1

Lag screw details Lag screw diameter (Table L2) 5/8'' Lag screw root diameter of threaded portion Dr = 0.471 in Length of lag screw L = 5.000 in Washer thickness W = 0.125 in Number of rows of lag screw R = 1 Number of columns of lag screw C = 1

2-43

Job Ref.

60667208 Section

AC Curb to 6x8

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by

F. Macias- Date

11/27/2023

Total number of lag screw Ntotal = R C = 1

Applied load Applied load to the connection P = 341 lb

Dowel bearing length (main) (12.3.5) Size of tapered tip E = 0.406 in Actual penetration in main member Pact = L - W - ts = 4.625 in Dowel bearing length in main member lm = 4.422 in

Dowel bearing length (connected) (12.3.5) Dowel bearing length in connected member ls = ts = 0.250 in

Bending yield strength (lag screw) (Table 12J to 12K footnote no. 2) Bending yield strength of lag screw Fyb = 45000 psi

Dowel bearing strength (main member) (Table 12.3.3 footnote no. 2) Dowel bearing strength parallel to grain Fe_par = 11200 Gm 1 psi = 5152 psi

Dowel bearing strength perpendicular to grain Fe_perp = 6100 Gm1.45 1 psi / (D / 1 in) = 2503 psi

Dowel bearing strength for small dia. fasteners Fe = 16600 Gm1.84 1 psi = 3977 psi

Dowel bearing strength at an angle of load to grain Fem = (Fe_par Fe_perp) / ((Fe_par (sin(m))2) + (Fe_perp (cos(m))2)) Fem = 2503 psi Dowel bearing strength of main member Fem = 2503 psi

Dowel bearing strength (connected steel member) (Table 12K footnote no. 2) Dowel bearing strength of connected member Fes = 87000 psi

Preliminary yield limit equation coefficients (Table 12.3.1A notes) Dowel bearing strength ratio Re = Fem / Fes = 0.029 Dowel bearing length ratio Rt = lm / ls = 17.688 Preliminary yield limit equation coefficient k1 k1 = (((Re+(2Re2(1+Rt+Rt2))+(Rt2Re3)))-(Re(1+Rt)))/(1+Re) k1 = 0.221 Preliminary yield limit equation coefficient k2 k2 = -1+((2(1+Re))+((2Fyb(1+(2Re))Dr2))/(3Femlm2)) k2 = 0.484 Preliminary yield limit equation coefficient k3 k3 = -1+(((2(1+Re))/Re)+((2Fyb(2+Re)Dr2))/(3Femls2)) k3= 11.564 Angle of load to grain coefficient k k = 1 + (0.25 max (m, s) / 90) = 1.250 Reduction term (Table 12.3.1B) KD = 3.00

Yield limit equations (single shear) Mode Im (eq. 12.3-1) ZIm = (Dr lm Fem) / (4 k) = 1042 lb

Mode Is (eq. 12.3-2) ZIs = (Dr ls Fes) / (4 k) = 2049 lb

Mode II (eq. 12.3-3) ZII = (k1 Dr ls Fes) / (3.6 k) = 503 lb

Mode IIIm (eq. 12.3-4) ZIIIm = (k2 Dr lm Fem) / ((1 + (2 Re)) 3.2 k) = 596 lb

Mode IIIs (eq. 12.3-5) ZIIIs = (k3 Dr ls Fem) / ((2 + Re) 3.2 k) = 420 lb

Mode IV (eq. 12.3-6) ZIV = (Dr 2 (((2 Fem Fyb) / (3 (1 + Re))))) / (3.2 k) = 474 lb Z = min (ZIm, ZIs,ZII, ZIIIm, ZIIIs, ZIV) = 420 lb Nominal capacity of single fastener Z = 420 lb

2-44

Job Ref.

60667208 Section

AC Curb to 6x8

Calc. by

S. Ruybal Date

11/21/2023 Chk'd by

F. Macias- Date

11/27/2023

Slenderness (Table 12.5.1C footnote no.1) Slenderness l / D = 7.075

Spacing requirements (perpendicular to grain loading)

End distance (Table 12.5.1A) End distance (full stregnth) aq_full = 4 D = 2.500 in End distance (minimum) aq_min = 2 D = 1.250 in End distance (actual) aq = 3.500 in

Edge distance (Table 12.5.1C) Loaded edge eq = 4 D = 2.500 in Unloaded edge ep = 1.5 D = 0.938 in

Geometry factor C (12.5.1) End distance (actual) aq = 3.500 in End distance (full strength) aq_full = 2.500 in Geometry factor for end distance C = min(1,aq / aq_full) = 1.00

Adjustment factor Load duration factor (Table 2.3.2) CD = 1.00 Wet service factor (Table 11.3.3) CM = 1.0 Temperature factor (Table 11.3.4) Ct = 1.0 Group action factor (eq. 11.3-1) Cg = 1.0 Geometry factor (12.5.1) C = 1.00 End grain factor (12.5.2) Ceg = 1.0 Diaphragm factor (12.5.3) Cdi = 1.0 Toe nail factor (12.5.4) Ctn = 1.0

Total capacity of connection Capacity of connection Z’ = Z Ntotal Nint CD CM C= 420 lb P / Z' = 0.812

Design result

2-45

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.7 Design Anchorage between Fan and Sleeper

2-47

SUBJECT/FEATURE: __HVAC Support________________________________

CALC. BY:_S. Ruybal_ DATE :__4/24/24

CHECKED BY:_F. Macias-Wallis_ DATE :_5/1/24 _

Exhaust Fan Anchorage Equipment Dimensions:

LEF 35.5625in Length of Exhaust Fan

BEF 22.3125in Width of Exhaust Fan hEF 30.5in Height of Exhaust Fan

Reference: Flat Blade Centrifugal Blower Clockwise Top Horizontal Arrangement 10, Class I page 1 of 8

Dead Loads:

wtEF 250lbf Weight of each exhaust fan unit, drawing 254-2101-M601

DLEF

wtEF

0.063 kip DL on each corner of each exhaust fan

Calculate Wind Forces on Equipment:

qz 15.3psf Velocity pressure: Velocity pressure calculated for stair tower design

G 0.85 Gust effect factor: Gust effect factor from Section 26.11 of ASCE 7

Cf_AC_check hEF min LEF BEF 1.367

Cf 1.0 Force Coefficient, ASCE 7 Figure 29.4-1

Af hEF max LEF BEF 7.532 ft2

Fw qz G Cf Af 0.098 kip Design wind force, Equation 29.4-1 ASCE 7

WindEF Fw

0.024 kip Design wind force at each corner of each equipment

Exhaust Fan Loads.xmcd 4/24/2024 5:02 PM

2-48

SUBJECT/FEATURE: __HVAC Support________________________________

Calculate Seismic Forces on Equipment:

ap 2.5 Component amplification factor; ASCE 7 Table 13.6-1 for air conditioning units

SDS 1.0 Spectral acceleration, ATC Hazards for Seismic in Appendix of stair tower calculation

Ip 1.0 Component importance factor, ASCE 7 Section 13.1.3

Wp wtEF 0.25 kip Component operating weight

Rp 6 ASCE 7 Table 13.6-1 for air conditioning units z 1 Height of structure of point of attachment for roof-mount h 1 Average roof height of structure

Fp_1

0.4 ap SDS

0.5 Seismic design force, equation 13.3-1 ASCE 7

Fp_max 1.6 SDS Ip 1.6 Max seismic design force, equation 13.3-2 ASCE 7

Fp_min 0.3 SDS Ip 0.3 Min seismic design force, equation 13.3-2 ASCE 7

Fp Fp_1 Wp 0.125 kip Seismic horizontal design force

Fp_EF Fp

0.031 kip Seismic horizontal design force at each corner of each exhaust fan

Meq hEF Fp 0.212 kip ft Moment about horizontal axis due to seismic, assumes vertical center of gravity of equipment is 2/3 the height of the unit

Feq Meq min LEF BEF

0.114 kip Vertical force due to seismic moment

Exhaust Fan Loads.xmcd 4/24/2024 5:02 PM

2-49

SUBJECT/FEATURE: __HVAC Support________________________________

Feq_EF Feq

0.028 kip Vertical force due to seismic moment at each corner of each exhaust fan

Ev 0.2 SDS Wp 0.05 kip Vertical seismic force, section 13.3.1.2 of ASCE 7

Ev_EF Ev

0.013 kip Vertical seismic force at each corner of each exhaust fan

Load Summary:

DL Equipment (kip)

Vertical Seismic Ev (kip)

Horizontal Seismic Fp (kip)

Vertical Seismic Feq (kip)

Each Corner of Exhaust Fan 0.063 0.013 0.031 0.028

Design Anchorage to 6x8

Vu Fp_EF 0.031 kip Design shear per bolt

Tu Feq_EF Ev_EF 0.041 kip Design tension per bolt

Because the design shear and tension values are << than those of the AC bolt design shear and tension, by inspection, (1) 1/2" dia x 5" long lag screw is OK at each corner

Exhaust Fan Loads.xmcd 4/24/2024 5:07 PM

2-50

AECOM PROJECT NO.

60667208

SUBJECT/FEATURE:

HVAC SUPPORT

NARC JOB NO.

K21007

Calc By:

S.Ruybal

Date:

4/25/2024

Checked By:

F. Macias-Wallis

Date:

5/1/2024

2.8 Design Anchorage between Sleepers and Existing Roof Beams

2-51

Job Ref.

60667208 Section

Sleeper to 8x16 Anchorage

Calc. by

S. Ruybal Date

4/24/2024 Chk'd by

F. Macias- Date

5/1/2024

LAG SCREW TIMBER TO STEEL CONNECTION DESIGN

In accordance with NDS 2018 Tedds calculation version 1.2.04

Design results summary Unit Required Provided Utilization Result

Connection capacity lbs 3000 3413 0.879 PASS

P Main member

Connected member tm ts m

Main timber member details Species of main member Douglas Fir-South Size of main member (Table 1B) 8 x 16 Number of main member Nm = 1 Thickness of main member tm = 7.500 in Angle of load to grain of main member m = 45

Connected steel member details Number of connected steel member Ns = 2…