Form Determination - Tensile Fabric Structures: Design, Analysis, and Construction - page 82

An equilibrium configuration has been identified. The bar forces are closer to the
target of 10.0; however, additional accuracy is desired. One way to select a force
density factor to increase the force in members 1 & 2 and decrease the force in
member 3 is to scale the previously used force density values by the desired change
in the resulting member forces.

For the next iteration try:
g 1 = 0.00572(10.0/8.59) = 0.00666 (ref 5.5.2-2)

g 2 = 0.00629(10.0/7.97) = 0.00789
g 3 = 0.00375(10.0/11.44) = 0.00327

g X  g X  g X  P

X C =

1 A 2 B 3 D CX

(ref 5.5.2-3)

g  g  g

1 2 3

= 0.00666(0.0) + 0.00789(0.0) + 0.00327(4000) -0.0 = 735
(0.00666 + 0.00789 + 0.00327)

g Y  g Y  g Y  P

1 A 2 B 3 D CY

Y C = (ref 5.5.2-4)

g  g  g

1 2 3

= 0.00666(1000) + 0.00789(-1000) + 0.00327(0.0) -2.0 = -182
(0.00666 + 0.00789 + 0.00327)

Check for equilibrium, recalling F i = g i L i

(ref 5.5.2-2)

2 2

L 1 = 735  =1392

F 1 = 0.00666(1392) = 9.26 kN
F 1X = 4.89 kN
F 1Y = 7.86 kN

1182

2 2

L 2 = 735  = 1100

F 2 = 0.0789(1100) = 8.68 kN
F 2X = 5.80 kN
F 2Y = 6.46 kN

818

L 3 = 3265  = 3270

F 3 = 0.00327(3270) = 10.71 kN
F 3X = 10.69 kN
F 3Y = 0.62 kN

2

182

2

At the new location for node C:

 F

= -4.89 - 5.80 + 10.69 = 0.0 (ref 5.1.3-1)

 F = 7.86 - 6.46 + 0.60 = 2.0 (ref 5.1.3-2)