EG 1905 Pipe Racks

 

 

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ENGINEERING GUIDELINE

GROUP:

CIVIL/STRUCTURAL ENGINEERING

GUIDE NO.:

SUBJECT:   SUBJECT:

Pipe Supports Design

ISSUE DATE:  DATE: 

RESPONSIBLE ENGINEER: ENGINEER:   D. Mueller

Approved:   Approved:

EG-1905   EG-1905 16 Nov 01

10DEC99

TABLE OF CONTENTS

Page I.

General

2

II.

Design Loads

2

A. Vertical Loads B. Horizontal Loads C. Load Combinations

2 3 3

III.

Structur Structural al Design

4

IV.

Typical Piperack Configuration

7

Design Example

20 Pgs  Pgs 

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ENGINEERING GUIDELINE

General  General  This guide presents a general approach for the analysis and design of pipe supports, including main racks, miscellaneous and “T” supports.

II.

Design Loads A.

Vertical Loads

1.

Piping a. Main Rack – Vertical loads due to piping shall be the average unit loading at each level as determined from the piping layout drawings, but not less than 35 psf (22 psf dead load and 13 psf live load) per level. b. Miscellaneous Miscellaneous Racks and “T” Supports Supports - Vertical loads due due to piping shall the average unitbut loading at each level determined from and the pipingbe layout drawings, not less than 25 psfas(16 psf dead load 9 psf live load) per level. c. Pipes 14 inches and larger in diameter shall be considered as a concentrated load.

2.

Electrical Use a minimum of 20 psf (dead load) (for the width of the tray or area width of conduits) per level for conduits and single level cable trays. Use a minimum of 40 psf for double level cable trays.

3.

Equipment Loads should be obtained from vendor drawings for all equipment, such as air coolers with associated walkways and platforms.

4.

Live  Apply per the job specifications specifications to all walkways walkways,, platforms, platforms, stairs, and ladders. A 50 % reduction in live load is allowed when in combination with full wind or seismic. This reduction does not apply to piping live load (fluid load) or snow loads. Snow loading shall shall be included on platforms, platforms , cable trays, and equipment where applicable.

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Hydrotest Combine the hydrotest load of vapor lines with dead load only from the other piping lines.

6.

Structure Include the weight of all structural members, stairs, platforms and fireproofing in design.

B.

Horizontal Loads 1.

Wind  Apply wind loads to the structure, piping and equipment per EG-1901 . Use 8 in. diameter piping, minimum, if sizes are not known.

2.

Seismic Calculate seismic loads to the structure, piping and equipment per EG-1902. Use critical wind or seismic loads in design.

3.

Friction Use a longitudinal friction force equal to 10 % of the applied piping vertical load at each support level.

4.

Anchor and Guide  Anchor and guide forces, provided by the pipe stress group, may require horizontal bracing in the affected bay and may control the placement of vertical bracing.

C.

Load Combinations 1.

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Design pipe supports to accommodate the following followi ng load combinations: a.

Dead + Live

b.

Dead + Wind (or Seismic)

c.

Dead + Live + Wind (or Seismic)

d.

Dead + Live + Friction (or Anchor and Guide)

e.

Dead + Live + Anchor and Guide + Wind (or Seismic)

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III.

ENGINEERING GUIDELINE

Structural Structur al Design  A.

Design pipe racks as rigid frames f rames in the transverse transverse direction, with fixed bases. Struts transfer longitudinal loads to braced bays.

B.

Use computer program such as Staad-Pro to perform a rigid frame analysis. (See example for typical input file.)

C.

Make no allowance for future support levels unless specifically called for by the  job scope. Apply minimum loads on empty areas of partially loaded supports per section II of this guide to account for future pipe loading.

D.

Satisfy the applicable seismic requirements requirements also, when other loads control design.

E.

Limit lateral deflection deflection of piperack bent to L/100

F.

Limit lateral deflectionofatinertia anchor to ½when inch under anchor load 50% of the moment of points the beam the anchor load is only. along Use the weak axis of the beam.

G.

Following are some of the necessary parameters required for individual member design. Only those criteria, which are peculiar to pipe supports or have been known to cause confusion in their selection, are listed. 1.

Columns a.

Strong axis oriented perpendicular perpendicular to piping run.

b.

Effective Length Factor “K” :

TYPE OF SUPPORT Piperack with struts  And vertical vertical bracing Single frame “T” Support

X-X (STRONG)

Y-Y (WEAK)

1.2*

1.0

1.2* 2.1

1.6** 1.8** th

* For more accurate value use alignment charts, ASD 9  Ed., pp 5-137, Fig. C-C2.2 ** Assumes partial restraint/support from piping  

2.

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Transverse Transver se Beams a.

Reduce weak axis section modulus by 50 % when designing for Friction loads.

b.

For calculating allowable strong axis bending (consider that the moment curve inflection points may be considered as locations of

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ENGINEERING GUIDELINE

lateral support for the compression flange, use ½ span as the unbraced length. Otherwise, consider the whole whole span to be laterally unbraced. unbraced. c.

Maintain as much uniformity uniformit y as is reasonable in member selection and moment connection details.

3.

Longitudinal Longitudinal Struts a.

4.

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Design the strut beam for the larger of : 1.

Actual loading from pipe.

2.

Pipe rack rack loadings as given in section II.A.1, II.A.1, multiplied by ½ of the piperack width. width. (ex. 35 psf psf x 0.5 x 20 ft for a 20 ft wide rack).

b.

Design all longitudinal longitudinal struts to carry applied axial forces.

c.

Strut loadoffrom (dead + live) is not want normally input into frame analysis bent,pipe but lead engineer may to consider inputting inputti ng 1 level on a multilevel strut rack.

d.

To isolate a portion of the rack from existing structure or another rack, slot connections at strut ends

e.

Use 2/3 of length for unbraced length when considering vertical beam load and/or axial load.

Vertical Vertica l Bracing (Longitudinal) a.

Bracing usually consists of WT or double angle sections in an inverted “V” (chevron bracing) configuration.

b.

Locate braced bays at approximately approximately 150 ft. intervals. intervals. Preferably provide two braced bays in each column row at every independent section of rack. But in no case should there be less than one braced bay in each column row, unless an alternate means of longitudinal support is provided.

c.

Size bracing for the largest load applied by the following : 1.

Accumulative force from friction.

2.

Longitudinal component of the wind acting at an angle of 45° to the pipe rack.

3.

Anchor Forces.

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ENGINEERING GUIDELINE

Longitudinal Longitudinal seismic forces.

Moment Connections a.

Provide end-plate moment connections in transverse beams.

b.

Consider using heavier column section in lieu of doubler plates when column web panel shear is exceeded.

c.

Design beam moment connections for the full beam capacity (.66Fy moment)

Air Cooler Support Girders a.

Analyze as continuous beam spanning across tops of columns.  Avoid lateral beams and horizontal bracing unless girder size becomes excessive.

b.

Space air coolers to allow piperack column to extend above the air coolers for header pipe support.

c.

Use 140 psf for preliminary frame design without vendor weights. Base final design of girder on actual vendor provided weights, or use from ‘d’.

d.

Typical Air Cooler Loads (For estimating only) (Loads shown on pipe support column) Piperack Width

20’

25’

30’

Dead Load

35 k/col

42 k/col

50 k/col

Live Load

3.5 k/col

4 k/col

5 k/col

Traverse Shear

5 k/col

5.5 k/col

6 k/col

Wind Couple, Vertical

±4.5 k/col

±4.5 k/col ±4.5 k/col

Longitudinal Shear (at braced bay only)

18 k/bay

18 k/bay

Wind Load

18 k/bay

Note: Wind loads are based on a design wind speed of 110 mph. mph. For other design wind speed, V, multiply wind loads above by V 2/1102.

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IV.

ENGINEERING GUIDELINE

Typical Piperack Configuration

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ENGINEERING GUIDELINE

TYPICAL PIPE BRIDGE Figure 1

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MISCELLANEOUS PIPE SUPPORTS  SUPPORTS  Figure 2

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DESIGN EXAMPLE

TABLE OF CONTENTS

Page 1. Sketches 2. Problem/Design Criteria 3. Dead Loads 4. Fluid Loads 5. Live Loads 6. Wind Loads 7. Friction Loads 8. Strut Design 9. Air Cooler Longitudinal Support Design 10. Longitudinal Bracing Design 11. Staad Model

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2 5 6 9 11 12 15 16 17 18 21

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VIII.

DESIGN EXAMPLE

Design Example

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DESIGN EXAMPLE

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DESIGN EXAMPLE

Problem:

Design the pipe rack illustrated in figures 1 and 2 using STAAD III. Figures 1 and 2 show typical information available at the beginning of a job, usually provided by the Piping Department.

Given Design Criteria:

a)

Wind ASCE 7-93 – 110 mph, Exposure C, Importance Factor 1.0.

b)

Seismic Zone 0

c)

Air Cooler Loads (obtained from Mech/Equip. vendor drawings) Dead Load Fluid Load Live Load

= 130 = 20 = 10

psf (based on plan area of rack) psf (based on plan area of rack) psf (based on plan area of rack)

* Note: Actual Air Cooler loads should be used to check the final design when they become available. d)

No future expansion is considered in design.

e)

All members are fireproofed except the longit longitudinal udinal struts, and the top most transverse beam, which is included for platform support.

f)

Engineering Guide EG-1905  

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DESIGN EXAMPLE

Load calculations for STAAD-III model: Dead Loads  Air Coolers: Coolers:  Air cooler cooler load pe perr colu column mn

= 30” Header Pipe

=

  (20 ft )2    130lbs   = 26kips   2   ft    2     118.8lbs

 

 ft 

(Ref. EG-1905 IV. Weight of Pipes)

Column load due to 30” Header Pipe 118.8lbs (20 ft ) = 2.  376kips   =  ft  Structure: Weight of each W10x26 Longitudinal Strut on each column

 26lbs   = 20 ft          = 0.52kips      ft    Weight of W16x67 A/C support beam plus fireproofing

 67 lbs + 251lbs   = 20 ft      ft       = 6.8kips   Column fireproofing load

=

244lbs  ft 

= 0.244

Traverse beam fireproofing load 252lbs W16x77 =  ft  W14x61

=

213lbs  ft 

kips  ft 

 

= 0.252

kips

= 0.213

kips

 ft   ft 

(Ref.????)

 

(Ref.????)

 

(Ref.????)  (Ref.????) 

Upper level piping (Elevation 132’-0”): Minimum piping load on first 12 ft

        = = 20 ft  22lbs 2      ft   

(Ref. EG-1905, pg.2, II.A.1.a) 0.44kips  ft 

 

Minimum load due cable tray on final 8 ft

 20lbs       = 20 ft     ft 2  =     142.8lbs  ft 

(Ref. EG-1905, pg.2, II.A.2)

0.40kips    ft 

Concentrated load due to 36” FLR

=

(Ref.????) (Ref.????)  

(Ref. EG-1905 IV. Weight of Pipes) (Ref. EG-1905, pg.2, II.A.1.c)

  0.44 kips     × 20 ft  −  3 ft ×  = 1.536kips    ft     

Mid level piping (Elevation 124’-0”): Minimum piping load

      = 20 ft  22lbs   = 2    ft     

0.44kips    ft 

Concentrated loads due to 30” 30” C CWS WS and 30“ CWR

=

118.8lbs  ft 

(Ref. EG-1905 IV. Weight of Pipes)

  0.44kips     × 20 ft  −  2.5 ft ×  = 1.276kips    ft     

Lower level piping (Elevation 118’-0”) Minimum piping load

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        = = 20 ft  22lbs 2    ft     

DESIGN EXAMPLE 0.44kips    ft 

Concentrated loads due to Insulated 16” STM

(Ref. EG-1905 IV. Weight of Pipes) (Ref. EG-1905 V. Weight of Piping Insulation)

 62.6lbs      440lbs   2  12lbs    3  ft    =    ft  + 1.01 ft         1.33 ft ×  ft      = 0.909kips    × 20 ft  −  

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Fluid Loads  Air Coolers: Coolers:  Air cooler cooler load pe perr colu column mn

= 30” Header Pipe

=

  (20 ft )2    20lbs   = 4kips   2   ft    2     291.2lbs  ft 

 

(Ref. EG-1905 IV. Weight of Pipes)

Column load due to 30” Header Pipe 291.2lbs (20 ft ) = 5. 824kips   =  ft  Upper level piping (Elevation 132’-0”): Minimum piping load on first 12 ft

 13lbs    0.26kips = 20 ft  2      =  ft       ft    Concentrated load due to 36” FLR

=

422.9lbs  ft 

(Ref. EG-1905 IV. Weight of Pipes)

  0.26 kips     × 20 ft  −  3 ft ×   = 7.678kips    ft     

Mid level piping (Elevation 124’-0”): Minimum piping load

 13lbs    0.26kips = 20 ft  2      =  ft     ft     

Concentrated loads due to 30” 30” C CWS WS and 30“ CWR

=

291 .2lbs  ft 

(Ref. EG-1905 IV. Weight of Pipes)

  0.26 kips     × 20 ft  −  2.5 ft ×   = 5.174kips    ft     

Lower level piping (Elevation 118’-0”) Minimum piping load

 13lbs    0.26kips = 20 ft  2      =  ft       ft    Concentrated loads due to Insulated 16” STM

=

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79.2lbs  ft 

(Ref. EG-1905 IV. Weight of Pipes)

  0.26kips     × 20 ft  − 1.33 ft ×   = 1.238kips    ft     

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Live Loads  Air Coolers: Coolers:  Air cooler cooler load pe perr colu column mn

=

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  (20 ft )2    10lbs   = 2kips     ft 2   2    

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DESIGN EXAMPLE

Wind Loads Note:  F  =  A f   × Gh  × C  f   × q  

Overall height

= 40 ft  + 19     ft  = 59 ft  , implies q =

0 to 20 ft

26.9lbs 2

Gh

Table le 17) = 1.20   (Ref. EG-1901, pg.15, Tab

 

(Ref. EG-1901, pg.14, Table 6)

 

(Ref. EG-1901, pg.14, Table 6)

 

(Ref. EG-1901, pg.14, Table 6)

 ft 

20 to 40 ft  

q =

30.3lbs

40 to 60 ft  

q =

35.0lbs

 ft 2

 ft 2

 Air Coo Coolers: lers: For Wind on Equipment C  f   = 1.3  

(Ref. EG-1901, pg.23, Table 11)

Effective Height of air cooler he1  = Header Pipe Diameter = 30 in  he2   = Actual Bundle height = 6 ft   he3  = 50% x Actual height of the coolers base support = 50% x 8 ft = 4 ft   H eff = total effective height of the air cooler = he1+he2 +h +he3 = 12.5 ft   eff Horizontal wind load per bent  F  = 12.5 ft × 20 ft × 1.20   × 1.3 × 35lbs  ft 2

= 13.650kips  

Value applied to the tops of each column. kips 13.650kips = = 6.825   column 2 columns Vertical wind load per column  Assume the hor horizontal izontal w wind ind for force ce acts 2/3 2/3 of th the e effe effective ctive he height ight above the top most traverse beam. 13.650kips × (2 3)12.5 ft  = = ±5.688kips   20 ft  Value is positive (+) on the windward column and negative (-) on the leeward column. Structure: For Wind on Windward Steel C  f   = 1.8   (Ref. EG-1901, pg.23, Table 11) (Ref. EG-1901, pg.25, Note 3) For Wind on Leeward Steel C  f   = 1.5  

(Ref. EG-1901, pg.23, Table 11)

(Ref. EG-1901, pg.25, Note 3) Columns up to 20 ft Uniform load on the windward column: 26.9lbs kips (10in + 2in × 2)  ft  × 1  .20 × 1.8 ×   = 0.068  F  =   2  ft  12in  ft  Uniform load on the leeward column: 26.9lbs (10in + 2in × 2)  ft  × 1  .20 × 1.5 ×    F  = 12in  ft 2 Columns 20 ft to 40 ft: Uniform load on the windward column: 30.3lbs (10in + 2in × 2)  ft  × 1  .20 × 1.8 ×    F  = 12in  ft 2

= 0.056

kips

= 0.076

kips

 ft 

 ft 

 

 

Uniform load on the leeward column:  F  = (10in + 2in × 2)  ft  × 1  .20 × 1.5 × 30.32lbs  12in  ft 

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= 0.064 kips    ft 

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DESIGN EXAMPLE

Lower Strut (Elevation 115’-0”): Concentrated load on the windward column at the strut elevation: 26.9lbs (10in)  ft  × 20 ft × 1. 20 × 1.8 ×   = 0.968kips    F  = 12in  ft 2 Concentrated load on the leeward column at the strut elevation: 26.9lbs (10in)  ft   F  = 12in × 20 ft × 1. 20 × 1.5 ×  ft 2   = 0.807kips   Mid and Upper Strut (Elevation 121’-0” and 128’-0”): Concentrated load on the windward column at the strut elevation: 30.3lbs (10in) ft  × 20 ft × 1. 20 × 1.8 × 2   = 1.091kips    F  = 12in  ft  Concentrated load on the leeward column at the strut elevation: 30.3lbs (10in)  ft  × 20 ft × 1. 20 × 1.5 ×   = 0.909kips    F  = 12in  ft 2  Air Coo Cooler ler Suppo Support rt Bea Beams ms (Eleva (Elevation tion 140 140’-0”): ’-0”): Concentrated load on the windward column at the beam elevation: 30.3lbs (16in)  ft  × 20 ft × 1. 20 × 1.8 ×   = 1.745kips    F  = 12in  ft 2 Concentrated load on the leeward column at the beam elevation: 30.3lbs (16in)  ft   F  = 12in × 20 ft × 1. 20 × 1.5 ×  ft 2   = 1.454kips   Piping: For Wind on Piping C  f   = 0.7  

(Ref. EG-1901, pg.23, Table 11)

The projected area for piping shall be based on the diameter of the largest pipe plus 10 percent of the width of the piperack times the lingth of the pipes. (Ref. EG-1901, pg.25, Note J) Lower Level Piping: Concentrated load split between the windward and leeward column applied at the supporting beam elevation:

   16in + 2in × 2 26 .9lbs × 20 ft × 1.20 × 0.7 × = 1.657 kips   + 10% × ( 20 ft )       ft      12in  ft 

 F  = 

Mid Level Piping: Concentrated load split between the windward and leeward column applied at the supporting beam elevation:

 

 F  = 

  30.3lbs + 10% × ( 20 ft )   = 2.727 kips    ×  20 ft × 1.20 × 0.7 ×  ft   ft   

30in

 12in

Upper Level Piping: Concentrated load split between the windward and leeward column applied at the supporting beam elevation:

    36in 30.3lbs ×  20 ft × 1.20 × 0.7 × = 2.545kips   + 10% × ( 20 ft )      ft     12in  ft 

 F  = 

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Friction Load Note: The load is doubled to account for reducing Sy by half. (Ref. EG-1905, pg.2, III.C.2.a) Use a longitudinal force equal to 10% of the applied piping vertical load at each support level. At points of concentrated pipe loads use 30% of applied vertical load. (Ref. EG-1905, II.B.3) Uniform load on each transverse beam: ω

0.26 kips ) × 10% = 0.140 kips = 2 × (0.44 kips  ft  +    ft   ft 

 

Concentrated Load due to 36” FLR line on upper beam (no water):  F 1 = 2 × (1.536 kips   ) × 30 % = 0.922kips   Concentrated Load due to 30” CWS & 30” CWR line on middle beam:  F 2

= 2 × (1.276kips +  5 .174kips) × 30% = 3.87 kips  

Concentrated Load due to 16” STM line on upper beam (no water):  F 3 = 2 × (0.909 kips   ) × 30% = 0.545kips  

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DESIGN EXAMPLE

Design Calculations: Verify Strut Design: Strut size: W10x26 Properties: d   A f  

r t  =  1.54in  

 F  y =  36ksi  

= 1.0  

C b

S   xx  = 27.9   in3  

= 4.07  

S  y   in 3    y   = 4.89

l  =  20 ft   

Loads: Vertical load Self weight of the beam = 26lbs/ft   Weight of pipe = 36 psf x 20 ft x 0.5 = 350 lbs/ft Total vertical load = 26 lbs/ft + 350 lbs/ft = 376 lbs/ft   Horizontal load Friction Load = 10% x 350 lbs/ft = 35 lbs/ft   Calculations: l  r t 

=

20 ft  1.54in

×

 

12in

= 155.844 ≥

510 × 103 C b

 ft   

F  y

3

 F bx

=

 F bx

=

12 × 10 C b     ld   A f  

 F bx

 F by

=  12.285ksi   ….Controls = 0.75   F     y = 27ksi  

 M  x

=

 M  y

=

)

=

(35lbs /  ft ) ×  (20 ft )2

=

 F bx

=

 F bx

=

170 × 10 (119.  02)2 =  7 ksi  

 

 

(Ref. AISC-ASD F1-7)

3

12 × 10     12in    20 ft ×  × 4.07  ft       

= 12.285ksi  

(Ref. AISC-ASD F1-8)

(Ref. AISC-ASD F2-1)

= 18.8kip ⋅  ft    = 1.75kip ⋅  ft   

12in

   ft  S  x  M  y ×

 f  by

  

8 8

= 119.02  

 

≤ 0  .6F y  

(376lbs /  ft ) ×  (20 ft )2

 M  x ×  f  bx

≤  0.6F y  

3

(

(Ref. ????)

3

170 × 10 C  b (l  r  t )  2  

(Ref. EG-1905, pg.2, III.G.3)

= 8.086ksi  

12in  

 ft 

S  y

Stress Ratio  =

= 4.294ksi    f  bx  F bx

+

 f  by  F by

=

8.086 ksi 12.285 ksi

+

4.294ksi 27 ksi

= 0.82 ≤ 1.0  

…. O.K. (Ref. AISC-ASD H1-3)

Preliminary Design Size W10x26 for Struts *Note: When the piping arrangement is finalized, individual struts may require further inspection, especially for axial loads from wind.

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 MUS TA NG E NG IN INE EERS & CONSTR UCTOR UCTOR S , INC.

DESIGN EXAMPLE

Verify Air Cooler Longitudinal Support Beam Design: Beam size: W16x67 Properties: d  r t  =  2.75in   = 2.4   S  x  = 117.0   in3    A f   C b

 F  y =  36ksi  

= 1.0  

3

S  y   = 23.2   in  

l  =  20 ft   

Loads: Horizontal load Load due to the wind on the Air Cooler  F  = 12.5 ft × 20 ft × 1.20   × 1.3 × 35lbs  ft 2

= 13.650kips  

(Ref. Previous calculation)

Note: For the 20 ft span between bents the Air Coolers have 3 contact points along each longitudinal support beam The component applied at the mid-span of the beam is 13.650kips 2 = 3.413kips   = 2 beams Vertical load Self weight of the beam plus the fireproofing = 67 lbs/ft + 261 lbs/ft = 328 lbs/ft   Load due to the wind on the Air Cooler 13.650kips × (2 3)12.5 ft  (Ref. Previous calculation) calculation)   20 ft  = = ±5.688kips   The component applied at the mid-span of the beam is 5.688kips = = 2.844kips   2 beams Weight of the Air Coolers (Dead+Fluid+Live) applied at mid-span of the support beam is lbs 20 ft  20 ft    =   ×   × (130 + 20 + 10) 2   = 16kips   2    ft    2 Calculations: 102 × 103 C b  F  y

l  r t 

=

20 ft   2.75in

 F bx

2 2   F  y (l  r t  )  ≤ 0.6 F  y       = −  3 1530 × 103 C b 

 F bx

= 12 × 10    C b ≤ 0  .6F y  

 F bx

 F by

a)

= 53.23 ≤

 

3

 F bx

(ld   A f  )

=

×

12in  ft   F bx

= 87.273 ≤

510 × 103 C b  F  y

= 119.02  

 2 36(87.237)2  = −    = 17.548ksi   3  3 1530 × 10 

  3 12 × 10     12in    × 2.4  20 ft ×  ft       

(Ref. AISC-ASD F1-6)

= 20.833ksi  (Ref. AISC-ASD F1-8)

=  20.833ksi   ….Controls = 0.75   F     y = 27ksi  

(Ref. AISC-ASD F2-1)

Operating

 M  x

=

16,000lbs × 20  ft  4  M  x ×

 f  bx

EG1905+REVB.DOC

=

+

  2 (328 lbs /  ft ) × (20 ft )

 

 

8

= 96.4kip ⋅  ft   

12in  

S  x

 ft 

= 9.877ksi  …less than F bbx x   O.K.

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 MUS TA NG E NG IN INE EERS & CONSTR UCTOR UCTOR S , INC. b)

DESIGN EXAMPLE

Operating + Wind 2.844lbs × 20 ft      4

 M  x

=  M  x +

 M  y

= 3.413kips  × 20  ft  = 17.07 kips   4

 M  x ×  f  bx

=

12in  

 ft 

= 11.35ksi  

S  x  M  y ×

 f  by

= 110.62kip ⋅  ft   

=

12in

 ft    S  y 2

Stress Ratio  =

= 17.65ksi    f  bx  F bx

+

 f  by  F by

=

11.35ksi 20.833ksi

+

17.65ksi 27 ksi

= 1.20 ≤ 1.33  

…. O.K. (Ref. AISC-ASD H1-3)

Preliminary Design Size W16x67 for Longitudinal Air Cooler Support Beams Design the Longitudinal Bracing To develop the quarter wind factor, assume the total wind on one bent is unity. Take the longitudinal component of the wind at 45 degrees to the rack.

 Lateral  Wind   = 1kip / bent    (Unity)  H  prime

EG1905+REVB.DOC

  °  1kip / bent      =    × (20 ft  × Cos 45 ) = 0.707  kips / bent    20  ft     

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 MUS TA NG E NG IN INE EERS & CONSTR UCTOR UCTOR S , INC.  H 1

DESIGN EXAMPLE

=  H  prime  × Cos 45°   = 0.5kips / bent   Longitudinal component of the quarter wind

The horizontal wind at each column line is  H  =

 H 1 2

=  0.25kips / bent   

The quarter wind factor for the total wind per braced bay is  F  =  H  ×

number  of   bents − 1   number  of   braced  bays

= 0.25 kips × (9 − 1)bents = 1 bent 

2braced  bays

kip   braced  bay

This is the force per kip of lateral wind. Brace Lengths  L1

 2 2   ) + (10  ft ) = 15   .62 ft    = (140 ft  − 128   ft 

 L2

 2 2 = (128 ft  − 121    ft    ) + (10 ft ) = 12  .207 ft   

 L3

= (121 ft  − 115   ft     )2 + (10  ft )2 = 11  .662 ft   

 L4

 2 2 = (115 ft  − 101    ft    ) + (10 ft ) = 17  .205 ft   

EG1905+REVB.DOC

(Ref. Figure 1,2, and 4)

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DESIGN EXAMPLE

Brace Loads

(Ref. Wind Load Calculations)

 Load  =  Hor    . Wind  on the  Air  Cooler  + Wind    on the  Air  Cooler  Support  Beams + Wind  on the Columns  

 

 Load  = 13.650kips + (1.745kips + 1.454kips )  +  0.076

kips  ft 

+ 0.064 

kips    ft 

 

 Load × F  15.62 ft  = 13.82kips    Brace  Load 1 =    × 2 braces 10 ft 

   × 6 ft  = 17.69kips    

 Load  =  Load  + Wind  on the Upper  Level   Piping    + Wind  on the Upper  Level  Struts + Wind  on the Columns  

   

 Load  = 17.69 kips + 2.545kips + (1.091kips + 0.909kips   ) +  0.076

 Brace  Load 2

=

kips  ft 

+ 0.064 

kips    ft 

   × 9.5 ft  = 23.56kips    

 Load × F  12.207 ft    × = 14.38kips   2 braces 10 ft 

 Load  =  Load  + Wind  on the  Mid   Level  Piping    + Wind  on the  Mid   Level  Struts + Wind  on the Columns  

   

 Load  = 23.56kips + 2.727 kips + (1.091kips + 0.909 kips    ) +  0.076

 Brace  Load 3

=

kips  ft 

+ 0.064 

kips    ft 

   × 6.5 ft  = 29.20kips    

 Load × F  11.662 ft  = 17.03kips     × 2 braces 10 ft 

 Load  =  Load  + Wind  on the  Lower  Level  Piping    + Wind  on the  Lower  Level  Struts + Wind  on the Columns  

   

 Load  = 29.20kips + 1.657 kips + (0.968kips + 0.807  kips) +  0.068

 Brace  Load 4

=

kips  ft 

+ 0.056 

kips    ft 

   × 10 ft  = 33.87 kips    

 Load × F  17.205 ft     × = 29.14kips   2 braces 10 ft 

To size braces, reference AISC Columns – Allowable Axial Loads, p. 3-85 (1/3 increase allowable for wind) For lengths 1,2, and 3 select WT4x9 For length 4 select WT5x11

EG1905+REVB.DOC

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DESIGN EXAMPLE

STAAD JOINT and MEMBER NUMBERS

EG1905+REVB.DOC

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DESIGN EXAMPLE

STAAD 2D INPUT FILE

STAAD SPACE EG-1905 STANDARD PIPERACK DESIGN EXAMPLE INPUT WIDTH 72 *MUSTANG ENGINEERS AND CONSTRUCTORS, INC. *FILE .... EG1905.STD OUTPUT WIDTH 72 UNIT FEET KIP JOINT COORDINATES 1 .000 .000 .000 2 .000 13.500 .000 3 .000 16.500 .000 4 .000 19.500 .000 5 .000 22.500 .000 6 .000 26.500 .000 7 .000 30.500 .000 8 .000 38.500 .000 9 20.000 .000 .000 10 20.000 13.500 11 20.000 16.500 12 20.000 19.500 13 20.000 22.500 14 20.000 26.500 15 20.000 30.500 16 20.000 38.500 MEMBER INCIDENCES 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 9 10 9 10 11

.000 .000 .000 .000 .000 .000 .000

10 11 12 11 12 13 12 13 14 13 14 15 14 15 16 15 3 11 16 5 13 17 7 15 18 8 16 MEMBER PROPERTY AMERICAN * TRAVERSE BEAMS 18 TABLE ST W10X26 16 17 TABLE ST W14X61 15 TABLE ST W16X77 * COLUMNS 1 TO 14 TABLE ST W14X82 MEMBER RELEASE *15RELEASE WEAK AXIS MOMENT ON TRAVERSE BEAMS TO 17 START MY

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DESIGN EXAMPLE

15 TO 17 END MY * RELEASE STRONG AND WEAK AXIS MOMENTS ON TOP MOST BEAM 18 START MY MZ 18 END MY MZ SUPPORT * BASES ARE FIXED 1 9 FIXED * PROVIDE SUPPORT IN THE Z DIRECTION WHERE STRUTS ARE LOCATED 2 4 6 8 10 12 14 16 FIXED BUT FX FY MX MY MZ CONSTANT E STEEL ALL DENSITY STEEL ALL POISSON STEEL ALL LOAD 1 DEAD LOAD SELFWEIGHT Y -1. JOINT LOAD *AIR COOLERS 8 16 FY -26. * 30" HEADER ON AIR COOLER 16 FY -2.376 *2 STRUTS 4 6 10 12 14 FY -.52 * AIR COOLER LONGITUDINAL SUPPORT BEAMS 8 16 FY -6.8 MEMBER LOAD * FIREPROOFING - COLUMNS 1 TO 14 UNI GY -.244 * FIREPROOFING - TRAVERSE BEAMS 15 UNI GY -.252 16 17 UNI GY -.213 * UPPER LEVEL PIPING AND ELECTRICAL 17 UNI GY -.44 0. 12. 17 UNI GY -.4 12. 20. 17 CON GY -1.536 3. * MID LEVEL PIPING 16 UNI GY -.44 16 CON GY -1.276 2.5 16 CON GY -1.276PIPING 5.5 * LOWER LEVEL 15 UNI GY -.44 15 CON GY -.909 17.5 LOAD 2 FLUID LOAD JOINT LOAD * AIR COOLERS 8 16 FY -4. *30" HEADER ON AIR COOLER 16 FY -5.82 * UPPER PIPING LEVEL MEMBER LOAD 17 UNI GY -.26 0. 12. 17 CON GY -7.67 3. * MID PIPING LEVEL 16 UNI GY -.26 16 CON GY -5.174 2.5 16 CON GY -5.174 5.5 * LOWER LEVEL PIPING

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DESIGN EXAMPLE

15 UNI GY -.26 15 CON GY -1.238 17.5 LOAD 3 LIVE LOADS *AIR COOLER JOINT LOAD PLATFORMS 8 16 FY -2.0 LOAD 4 NORTH WIND *AIR COOLER JOINT LOAD JOINT LOAD 8 FY -5.688 16 FY 5.688 8 16 FX -6.825 MEMBER LOAD * COLUMNS LEEWARD 1 2 UNI GX -.057 3 UNI GX -.057 0. 2.5 3 UNI GX -.064 2.5 3. 4 TO 7 UNI GX -.064 * COLUMNS WINDWARD 8 9 UNI GX -.068 10 10 UNI UNI GX GX -.068 -.076 0. 2.52.5 3. 11 TO 14 UNI GX -.076 JOINT LOAD * LOWER STRUT LEEWARD 2 FX -.807 * LOWER STRUT WINDWARD 10 FX -.968 * MID AND UPPER STRUT LEEWARD 4 6 FX -.909 * MID AND UPPER STRUT WINDWARD 12 14 FX -1.091 * AIR COOLER LONGITUDINAL SUPPORT BEAM LEEWARD 8 FX -1.454 * AIR COOLER LONGITUDINAL SUPPORT BEAM WINDWARD 16 FX -1.745 * LOWER LEVEL PIPING 3 11 FX -.829 PIPING * MID LEVEL 5 13 FX -1.364 * UPPER LEVEL PIPING 7 15 FX -1.273 LOAD 5 SOUTH WIND *AIR COOLER JOINT LOAD JOINT LOAD 8 FY 5.688 16 FY -5.688 8 16 FX 6.825 MEMBER LOAD * COLUMNS LEEWARD 8 9 UNI GX .057 10 UNI GX .057 0. 2.5 10 UNI GX .064 2.5 3. 11 TO 14 UNI GX .064 * COLUMNS WINDWARD 1 2 UNI GX .068

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DESIGN EXAMPLE

3 UNI GX .068 0. 2.5 3 UNI GX .076 2.5 3. 4 TO 7 UNI GX .076 JOINT LOAD * LOWER STRUT LEEWARD 10 FX .807 * LOWER STRUT WINDWARD 2 FX .968 * MID AND UPPER STRUT LEEWARD 12 14 FX .909 * MID AND UPPER STRUT WINDWARD 4 6 FX 1.091 * AIR COOLER LONGITUDINAL SUPPORT BEAM LEEWARD 16 FX 1.454 * AIR COOLER LONGITUDINAL SUPPORT BEAM WINDWARD 8 FX 1.745 * LOWER LEVEL PIPING 3 11 FX .829 * MID LEVEL PIPING 5 13 FX 1.364 *7 UPPER LEVEL PIPING 15 FX 1.273 LOAD 6 FRICTION * DOUBLE LOAD TO ACCOUNT FOR REDUCING SY OF THE BEAM BY HALF. * REMEMBER TO REDUCE BEAM END REACTION BY HALF IF LOADS ARE USED FOR * VERTICAL BRACING DESIGN. MEMBER LOAD * UNIFORM PIPE FRICTION 10% OF OPERATING PIPE LOAD 15 16 UNI GZ .14 17 UNI GZ .14 0. 12. * CONCENTRATED LOAD USE 30% * 36" DIA. FLR 17 CON GZ .922 3. * 30" DIA. CWS 16 CON GZ 3.87 2.5 * 30" DIA. CWR 16 CON GZ 3.87 5.5 *1516" DIA.GZ STM CON .545 17.5 LOAD COMB 7 DEAD + LIVE 1 1. 2 1. 3 1. LOAD COMB 8 DEAD + NORTH WIND 1 .75 4 .75 LOAD COMB 9 DEAD + SOUTH WIND 1 .75 5 .75 LOAD COMB 10 DEAD + LIVE + NORTH WIND 1 .75 2 .75 3 .75 4 .75 LOAD COMB 11 DEAD + LIVE + SOUTH WIND 1 .75 2 .75 3 .75 5 .75 LOAD COMB 12 DEAD + LIVE + FRICTION 1 1. 2 1. 3 1. 6 1. PERFORM ANALYSIS SECTION 0. .25 .5 .75 1. ALL PARAMETER CODE AISC * NOTE: IT IS IMPORTANT TO SUPPLY ALL REQUIRED PARAMETERS IF A

EG1905+REVB.DOC

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DESIGN EXAMPLE

* CORRECT DESIGN IS EXPECTED. * COLUMNS KZ 1.3 MEMB 1 2 8 9 KZ 2. MEMB MEMB 2 73 149 10 LY 6. LY 6. MEMB 4 5 11 12 LY 12. MEMB 6 7 13 14 LZ 16.5 MEMB 1 2 8 9 LZ 6. MEMB 3 4 10 11 LZ 8. MEMB 5 6 12 13 UNL 6. MEMB 2 3 9 10 UNL 7. MEMB 4 5 11 12 UNL 12. MEMB 6 7 13 14 * BEAMS CMY 1. ALL * CMZ DEFAULTS TO 0.85 & CB DEFAULTS TO 1.0 FOR ALL MEMBERS. LOAD LIST 1 6 TO 12 PRINT SUPPORT REACTIONS CHECK CODE ALL FINISH

EG1905+REVB.DOC

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