 # V.Wind On Closed Structure 2

Evaluate the equivalent joint loads for a closed structure subjected to wind loads.

## Details

A structure lying in YZ plane is composed of three members and is subjected to wind load from (+X) direction (i.e., out-of-plane). Those three members, along with the ground surface, form a closed panel. The wind load incident on that panel is converted to equivalent joint loads incident on the nodes by the program.

The exposure factor, e, for the structure is 0.85.

Note: This structure is similar to that in V.Wind On Closed Structure 1 with an added node at the mid-point of the left column (Point E). Table 1. Wind intensity variation with height above ground
Height (m) Intensity of wind load (kg/m2)
≤ 5m 1.5
5m - 10 m 2
10 m - 20 m 3

## Validation

Determine CG of the Wind Area

Naming the nodal points A, B, C, D, and E, divide the area into two sub-areas: a rectangle and a triangle. The division is taken at nodes 5 and 3 (points E and B).

Area of the triangle ABE = 1/2 × base × height = 1/2 ×(10m) ×(10m) = 50 m2

CG of triangle ABE: ,

Area of rectangle BCDE = 10 m × 10 m = 100 m2

CG of rectangle BCDE: ,

CG of total area ABCD: ,

The CG is labeled point P.

The influence area of each joint is then take as the midpoint of the connecting members to this CG. Label the mid-points of each member R, N, Q, and J. The point at support level between nodes 1 and 2 is labeled K. Equivalent Joint Load on Node 2 (A)

The influence area for node 2 is the shape formed by points A, Q, P, and N. Notice that this area can be decomposed into a trapezoid and two triangles. Also notice that this area has two different wind pressures, above and below y = 10m (which coincides with line EB).

Determine the Z coordinate of where line EB intersects line NP and label this point S.

slope of NP = (15 - 7.77)/(4.44) = 1.625

z = 4.44 - (10 - 7.77)/1.625 = 4.615 m

Determine the Z coordinate of where line EB intersects line PQ and label this point M.

slope of PQ = (15 - 7.77)/(5 - 4.44) = 13.0

z = 4.44 + (10 - 7.77)/13 = 3.077 m ### Influence area of point A

The sub-areas of AQPN:

ASMP = 0.5×(10 - 7.77)×(4.615 - 3.077) = 1.71 m2

ANQMS = 0.5×(5 + 4.615 - 3.077)×5 = 16.35 m2

AANQ = 0.5×5×5 = 12.5 m2

The equivalent joint load, F, is calculated as Influence area, A, × avg. wind pressure, p × exposure factor, e

F = 0.85×[2×(1.71) + 3×(16.35 + 12.5)] = 76.47 kN

Equivalent Joint Load on Node 3 (B)

The influence area for node 3 is the shape formed by points B, J, P and Q. Label a point on line BJ with the same Y coordinate as P as point V. ### Influence area of point B

The sub-areas of BJPQ:

ABMQ = 0.5×(10 - 4.615)×5 = 13.46 m2

ABVPM = 0.5×[(10 - 4.615) + (10 - 4.44)]×(10 - 7.77) = 12.2 m2

AVJP = 0.5×(10 - 4.44)×(7.77 - 5) = 7.72 m2

F = 0.85×[2×(12.2 + 7.72) + 3×(13.46)] = 67.85 kN

Equivalent Joint Load on Node 4 (C)

The influence area for node 4 is the shape formed by points C, K, P, and J. The change in pressure magnitude is at the same Y coordinate as point J. Label the point on line ED with this same Y coordinate as point R. Determine the Z coordinate of where line RJ intersects line PK and label this point W.

slope of PK = (7.77)/(5 - 4.44) = 14.0

z = 4.44 + (7.77 - 5)/14 = 4.643 m ### Influence area of point C

The sub-areas of CKPJ:

AJWP = 0.5×(7.77 - 5)×(10 - 4.643) = 7.44 m2

ABVPM = 0.5×[(10 - 4.643) + 5]×5 = 25.89 m2

F = 0.85×[1.5×(25.89) + 2×(7.44)] = 45.66 kN

Equivalent Joint Load on Node 1 (D)

The influence area for node 1 is the shape formed by points D, R, P, and K. ### Influence area of point D

The sub-areas of DRPK:

ARPW = 0.5×(7.77 - 5)×4.643 = 6.45 m2

ADRWK = 0.5×[4.643 + 5]×5 = 24.11 m2

F = 0.85×[1.5×(24.11) + 2×(6.17)] = 41.71 kN

Equivalent Joint Load on Node 5 (E)

The influence area for node 5 is the shape formed by points N, P, and R. Label a point on line RE with the same Y coordinate as P as point L. ### Influence area of point E

The sub-areas of NPR:

ANSE = 0.5×(5)×3.077 = 7.69 m2

AESPL = 0.5×[3.077 + 4.44]×(10 - 7.77) = 8.36 m2

ALPR = 0.5×(4.44)×(7.77 - 5) = 6.17 m2

F = 0.85×[2×(8.36 + 6.17) + 3×(7.69)] = 44.31 kN

## Results

Table 2. Comparison of results
Joint load at node 1 (kN) 41.71 41.70 negligible
Joint load at node 2 (kN) 76.47 76.46 negligible
Joint load at node 3 (kN) 67.85 68.11 negligible
Joint load at node 4 (kN) 45.66 45.66 none
Joint load at node 5 (kN) 44.31 44.32 negligible

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 30-Aug-18
END JOB INFORMATION
*******************************************************************************
* This problem is created to verify the calculations of equivalent -
* joint load on the joints of a closed structure subjected to wind load
*******************************************************************************
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 20 0; 3 0 10 10; 4 0 0 10; 5 0 10 0;
MEMBER INCIDENCES
1 1 5; 2 2 3; 3 3 4; 4 5 2;
DEFINE MATERIAL START
ISOTROPIC CONCRETE
E 2.17185e+07
POISSON 0.17
DENSITY 23.5616
ALPHA 1e-05
DAMP 0.05
TYPE CONCRETE
STRENGTH FCU 27579
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 TO 4 PRIS YD 0.4 ZD 0.4
CONSTANTS
MATERIAL CONCRETE ALL
SUPPORTS
1 4 FIXED
TYPE 1 WIND 1
INT 1.5 2 3 HEIG  5  10  20
EXP 0.85 JOINT 1 TO 5
WIND LOAD X 1 TYPE 1 YR 0 20
FINISH


   LOADING     1  LOADTYPE WIND  TITLE LOAD CASE 1