A vertical vessel made with a cylindrical shell and hemispherical heads is to be installed out of doors near Corpus Christi, Texas. The shell has 5 ft. 0 in. inside diameter 1.0 in. nominal wall thickness, and 100 ft 0 in. length, tangent to tangent. The contract specification requires the vessel

Question

A vertical vessel made with a cylindrical shell and hemispherical heads is to be installed out of doors near Corpus Christi, Texas. The shell has 5 ft. 0 in. inside diameter 1.0 in. nominal wall thickness, and 100 ft 0 in. length, tangent to tangent. The contract specification requires the vessel to be designed according to the ASCE 7-10 Standard. What are the lateral wind forces to be used for design. Use [latex] K_{zt}[/latex] =1.0 and Exposure D and Category III.

Accepted Answer

Figure 16.8 shows that Corpus Christi, Texas, is located in a 160 mph wind zone. Therefore, the wind forces at various locations using the ASD method are calculated as follows:

The wind pressure is obtained from Eq. (16.9) and Figure 16.8 as

[latex] q_{ z }=(0.6)\left[0.00256(0.95) K_{ z }(1.0)(160)^{2}\right]=37.4 K_{ z }[/latex]

The hemispherical top head projection can be approximated as an equivalent cylinder with length [latex] L^{\prime}[/latex] as follows:

Outside diameter [latex] D_{o}[/latex] =inside diameter+2t =5+ 2(1/12)=5.167 ft.

Projected area of the hemispherical head=projected area of the equivalent cylinder

[latex] \pi D_{ o }^{2} /(4)(2)=D_{ o } L^{\prime}[/latex]

or, [latex] L^{\prime}=\pi D_{ o } / 8=2.03 ft[/latex]

Overall length of the cylinder=100+2.03=102.03 ft.

The values of [latex] K_{z}[/latex] are obtained from Table 16.3 as follows.

For z < 15 ft,

[latex] K_{ z }=2.01(15 / 700)^{2 / 11.5} [/latex] =1.030

[latex] q_{z}[/latex] = 37.4(1.03) = 38.5 psf.

The force is

[latex] F_{1}=38.5 D_{ o } L[/latex] = 38.5[(5.0 + 2(1)∕12)](15)

= 2980 lbs.

For 15 ft ≤ z ≤ 102.03 ft,

[latex] K_{ z }=2.01(z / 700)^{2 / 11.5}=0.643 z^{0.174}[/latex]

[latex] q_{z}=37.4\left(0.643 z^{0.174}\right)=24.0 z^{0.174}[/latex] psf

The distribution of this pressure is shown in Figure 16.10. The distribution is nonlinear. The force due to this pressure can be obtained in one of two ways. The first is to approximate the distribution by a series of straight line segments and then calculate the area of each of these segments. The second way is more direct and consists of integrating the aforementioned equation over the length. Hence,

[latex] F_{2}=5.167 \int_{15}^{102.03} 24.0 z^{0.174} d z[/latex] =21,560 lbs .

The total wind force is

F = 2980 + 21 560 ≈ 24 540 lbs.

Table 16.3 Values of 𝛼 and [latex] z_{g}[/latex] .

Exposure[latex] ^{a)}[/latex]	[latex] \alpha[/latex]	[latex] Z_{g}[/latex] ( ft )
B	7.0	1200
C	9.5	900
D	11.5	700
a) For the definition of the exposure categories, see Section 26.7.3 of ASCE 7-10.

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