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V. ASCE 7-10 Wind Load Generation

To validate STAAD.Pro calculated equivalent joint loads for a closed structure subjected to Wind Loading. Wind Intensity is calculated as per ASCE 07 – 2010.

Details

When closed type structures are subjected to wind load, panels or closed surfaces are generated by the program based on the members in the ranges specified and their end joints. Wind load, in such way, is converted to equivalent joint loads and applied on the nodes of that closed surface. The area within each closed surface is determined and the share of this area (influence area) for each node in the list is calculated. Wind intensities have been calculated as per ASCE 07 – 2010. The detailed calculation of determination of equivalent joint loads is performed here in this document.

A two-story, single-bay frame is subjected to wind loading from +X direction. Assumptions made while calculating the wind intensities are tabulated in the Validation section of this document.

Verification

Calculation of Wind Intensity

Assumptions made in determination of wind intensities as per ASCE 07 – 2010

  1. Analysis Procedure – Directional Procedure for MWFRS of enclosed buildings, Chapter 27
  2. Type of roof – flat
  3. Building Classification Category = Category II (Table 1.5-1)
  4. Basic Wind Speed = 110 mph
  5. Exposure Category = Exposure B (Clause 26.7.3)
  6. Wind speed-up over Hills or Escarpment is not considered
  7. Building Height = 20 ft
  8. Building length along the direction of Wind (L) = 10 ft
  9. Building length normal the direction of Wind (B) = 20 ft
  10. Natural Frequency of the building = 2 Hz (Hence, as per clause 26.2, this is a rigid building)
  11. Building Damping Ratio = 0.05
  12. Enclosure Classification = Enclosed Building (Clause 26.10)
  13. Directionality Factor Kd = 0.85 (Table 26.6-1, for MWFRS)
  14. Topographic Factor, Kzt = 1 (Clause 26.8.2)
  15. Building Wall to generate Wind Load on – Windward side
  16. Gust Effect Factor, G = 0.85 (Clause 26.9.1)
  17. External pressure Coefficient Cp = 0.85 (Windward wall pressure coefficient, as per Table 27.4-1)
  18. Internal pressure coefficient GCpi = -0.18 (For Enclosed Buildings, Table 26.11-1)

Hence, the velocity pressure exposure coefficients (Kz) are calculated as per the formulae given in Table 27.3-1:

From Table 26.9.1, For Exposure B, α = 7 and zg = 1200 ft

Height (ft) Velocity Pressure Exposure Coefficient (Kz)
0 – 15 0.5747196698
16 0.5854155685
17 0.5956440691
18 0.6054513812
19 0.6148768735
20 0.6239543877

As per clause 27.3.2, velocity pressure qz = 0.00256x Kz x Kzt x Kd x V2 (lb/ft2)

Since Mean roof height h = Height of building = 20 ft

qh = 0.00256x (Kz)h=20 ft x Kzt x Kd x V2 = 16.42847 lb/ft2

Design wind pressure, p is calculated as per clause 27.4.1 as -

p = q x G x Cp - qi x (GCpi)

Values of qz and p for different heights are calculated using the above formulae and tabulated below.

Height (ft) Kz q z (lb/ft 2 ) p (lb/ft2)
15 0.57471967 15.13214 13.246979
16 0.58541557 15.41376 13.4384798
17 0.59564407 15.68307 13.6216122
18 0.60545138 15.94129 13.7972035
19 0.61487687 16.18946 13.9659587
20 0.62395439 16.42847 14.1284837

Procedure Used by STAAD.Pro for Calculating the Equivalent Joint Load from incident Wind Pressure

  1. Form closed panels. A closed panel is a region whose boundary consists entirely of members or of members and the ground surface.
  2. Find the center of gravity of each of the panels.
  3. For each panel, draw straight lines from the center of gravity (CG) to the midpoint of the members that form the panel boundary. So, the panel region will now contain several quadrilaterals whose two sides are made of portions of the respective members (or the ground) and the other two sides are lines going from the CG to the midpoint of the corresponding members.
  4. The area contained in any quadrilateral is allocated as the influence area for the node located at the meeting point of two members.
  5. Multiply the influence area by the average wind pressure contained inside the influence area and by the exposure factor for the node. This will yield the concentrated wind force for the joint.

Calculation of Equivalent Joint Loads

Evidently, members AB, BG and GH along with the ground surface, forms the closed panel ABGH. P is the CG of the panel, located at 5 ft away from line AH and 5 ft away from line AB. So, influence area for joint A is AJPI. Area of AJPI = 5 x 5 = 25 ft2.

Average intensity of wind pressure in this area AJPI = 13.246979 lb/ft2

Exposure factor = 1 (for all the nodes in the model)

So, equivalent joint load in joint A = Influence area x Avg. Intensity of wind pressure x Exposure

= 25 x 13.246979 x 1 = 331.174 lb = 0.331 kip

Joint B will be under influence of influence area BKQU and BKPJ. Area of each of those influence areas = 5 x 5 = 25 ft2. Intensity of wind pressure in each of those influence areas = 13.246979 lb/ft2. Hence, equivalent joint load for each of those areas = 0.331 kip

So, total joint load in joint B = 0.331 x 2 = 0.662 kip

Joint C will be under influence area CVQU. Average intensity of wind pressure in this influence area will be = (13.246979 + 14.1284837) / 2 = 13.68773135 lb/ft2

Exposure Factor = 1

Area of CVQU = 5 x 5 = 25 ft2

Hence, equivalent joint load at joint C = 342.193 lb ≈ 0.343 kip

Joint W will be under influence of area WXRZ and WVQZ. Area of both of these = 25 ft2. Average intensity of wind pressure in both of the influence areas = Average intensity of wind in the influence area CVQU = 13.68773135 lb/ft2. Hence, equivalent joint load on joint W = 2 x 0.343 kip = 0.686 kip

Joint G will be under the influence of area GZRM, GZQK, GMSL and GKPL. Area of each of these influence areas = 5 x 5 = 25 ft2. Average intensity of wind pressure in each of these areas = average intensity of wind pressure in area BKQU = 13.246979 lb/ft2. Hence, equivalent joint load on joint G = 4 x 0.331 = 1.324 kip

Joint H will be under the influence of areas HLPI and HTSL. Area of each of those influence areas = 5 x 5 = 25 ft2. Intensity of wind pressure in each of those influence areas = 13.246979 lb/ft2. Hence, equivalent joint load on joint H = 2 x 0.331 = 0.662 kip

Since the structure is symmetric about the line WH, the equivalent joint load at D = equivalent joint load at C = 0.343 kip, equivalent joint load at B = equivalent joint load at E = 0.662 kip and equivalent joint load at A = equivalent joint load at F = 0.331 kip.

Comparison

Table 1. Comparison of results
Parameter Hand Calculation STAAD.Pro Difference Comments
Joint load at node (kips) 1 0.331 0.331 none
2 0.662 0.662 none
5 0.662 0.662 none
6 0.324 0.325 negligible
9 0.331 0.331 none
10 0.662 0.662 none
13 0.343 0.343 None
15 0.686 0.687 negligible
17 0.343 0.343 none

STAAD Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\06 Loading\Wind Load\ASCE 7-10 Wind Load Generation.STD is typically installed with the program.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 05-Sep-18
END JOB INFORMATION
*****************************************************************
*This problem has been created to verify the program calculated
*intensities of wind pressure as per ASCE 07-2010 and
*verify equivalent joint loads in the joints of a
*closed structure subjected to wind
*****************************************************************
INPUT WIDTH 79
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 0 10 0; 3 10 10 0; 4 10 0 0; 5 0 0 10; 6 0 10 10; 7 10 10 10;
8 10 0 10; 9 0 0 20; 10 0 10 20; 11 10 10 20; 12 10 0 20; 13 0 20 0;
14 10 20 0; 15 0 20 10; 16 10 20 10; 17 0 20 20; 18 10 20 20;
MEMBER INCIDENCES
1 1 2; 2 2 3; 3 3 4; 4 2 6; 5 3 7; 6 5 6; 7 6 7; 8 7 8; 9 6 10; 10 7 11;
11 9 10; 12 10 11; 13 11 12; 14 2 13; 15 3 14; 16 6 15; 17 7 16; 18 10 17;
19 11 18; 20 13 14; 21 13 15; 22 14 16; 23 15 16; 24 15 17; 25 16 18; 26 17 18;
DEFINE MATERIAL START
ISOTROPIC STEEL_50_KSI
E 4.176e+06
POISSON 0.3
DENSITY 0.489024
ALPHA 6.5e-06
DAMP 0.03
TYPE STEEL
STRENGTH FY 7200 FU 8928 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 TO 26 TABLE ST W14X873
CONSTANTS
MATERIAL STEEL_50_KSI ALL
SUPPORTS
1 4 5 8 9 12 FIXED
DEFINE WIND LOAD
TYPE 1 WIND 1
<! STAAD PRO GENERATED DATA DO NOT MODIFY !!!
ASCE-7-2010:PARAMS 110.000 MPH 0 1 1 0 0.000 FT 30.000 FT 10.000 FT 1 2 -
20.000 FT 10.000 FT 10.000 FT 2.000 0.050 0 0 0 0 1 0.624 1.000 1.000 0.850 -
0 1 1 1 0.850 0.800 -0.180
!> END GENERATED DATA BLOCK
INT 0.013247 0.013247 0.0133217 0.0133951 0.0134672 0.0135381 0.0136078 -
0.0136764 0.0137439 0.0138104 0.0138759 0.0139404 0.014004 0.0140667 -
0.0141285 0.0169213 HEIG 0 15 15.3846 15.7692 16.1539 16.5385 16.9231 -
17.3077 17.6923 18.0769 18.4615 18.8461 19.2308 19.6154 20 20
EXP 1 JOINT 1 TO 12
LOAD 1 LOADTYPE Wind  TITLE LOAD CASE 1
WIND LOAD X 1 TYPE 1 YR 0 40
PERFORM ANALYSIS PRINT LOAD DATA
FINISH

STAAD Output

   LOADING     1  LOADTYPE WIND  TITLE LOAD CASE 1                            
   -----------
   JOINT LOAD - UNIT KIP  FEET
   JOINT   FORCE-X   FORCE-Y     FORCE-Z     MOM-X     MOM-Y     MOM-Z
       1      0.33      0.00        0.00      0.00      0.00      0.00
       2      0.66      0.00        0.00      0.00      0.00      0.00
       5      0.66      0.00        0.00      0.00      0.00      0.00
       6      1.32      0.00        0.00      0.00      0.00      0.00
       9      0.33      0.00        0.00      0.00      0.00      0.00
      10      0.66      0.00        0.00      0.00      0.00      0.00
      13      0.34      0.00        0.00      0.00      0.00      0.00
      15      0.69      0.00        0.00      0.00      0.00      0.00
      17      0.34      0.00        0.00      0.00      0.00      0.00
   ************ END OF DATA FROM INTERNAL STORAGE ************