For calculation purpose wind load,

P_W = K₁ .K₂.P_w . X. D_o = 1000N/m²

D_o = 2.0 + 0.07 × 2 = 2.14 m.

K₁ = 0.7 and K₂ = 1

P_w = 0.7 × 1. 1000 .X . 2.14 = (1498 X) N

M_w = P_w . X/2 = (749 X² ) J

= 0.0138 X² MN/m²

 

Resultant tangential stress: F_tensile = F_a – F_d + F_bm

                                                  = 51.13 – 0.792X – 0.0671 +0.0138 X²

                                                  =0.0138X² – 0.792 X + 51.0629

F_{tensile max} = 98.1 × 0.85 = 83.0 MN/m²

0.0138X² – 0.792X – 31.93 = 0

X = 84 >> 18 m

F _{compressive maximum} = 0.125 E (t/D_o ) = 0.125 × 2 × 10⁵ (0.011/2.5)

                       = 110 MN/m²

F _{compressive stress} = F_d + F_bm - F_a = 0.0138 X² + 0.792 X + 0.0671 – 51.13

              110= 0.0138X² + 0.792X – 51.06

               0.0138X² + 0.792X – 161.06 = 0

 

X = 83 >> 16 m

If reinforcement of shell by tray support rings are also considered, X value will further increase.

If we consider longitudinal stress alone it is observed that hoop stress controlling the design and a uniform thickness of 16 mm is sufficient throughout the shell length.

References:

 

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