Question 10.14: A spherically dished head is to be bolted to the welding-nec...
A spherically dished head is to be bolted to the welding-neck flange described in Example 10.9. The dished head is to be attached at the upper inner corner with the outer surface even with the ring’s outer surface (see Figure 10.25). What is the minimum required thickness of the flange ring when the spherical head is dished to a radius of L=2B?
From the geometry mentioned in Example 10.9,
A = 26.5, B = 10.75, L = 2B = 2(10.75) = 21.5.
Determine the minimum required head thickness as follows:
t=\frac{5 P L}{6 S}=\frac{5(2500)(21.5)}{6(17,500)} =2.560 in. use 2.625 in.
From geometry calculations,
L^{\prime}=L+\frac{t}{2}=21.5+\frac{2.625}{2}
L^{\prime}=22.813 in.
\cos \beta_{1}=\frac{\sqrt{4(22.813)^{2}-(10.75)^{2}}}{2(22.813)}=0.972\beta_{1}=13.626^{\circ}
The membrane force in the head due to pressure is
F^{\prime}=\frac{P L^{\prime}}{2}=\frac{(2500)(22.813)}{2}=28,520 lb / in.
Horizontal force = F^{\prime} \cos \beta_{1}
= (28,520)(0.9720) = 27,720 lb∕in.
Vertical force = F^{\prime} \sin \beta_{1} = (28,520)(0.2356)
= 6720 lb∕in.
Total horizontal force = π(10.75)(27,720)
= 936,150 lb.
Total vertical force = π(10.75)(6270)
= 226,900 lb.
The moment at the gasket seating condition is as follows:
| Load | Arm | Moment |
| H_{ G }=W_{ a } =700,800 | h_{ G }=0.5(C-G) =3.729 | M_{ G }=H_{ G } h_{ G } =2,613,000 |
| H_{ D }=H_{ v } =226,900 | h_{ D } 0.5(C-B) =5.875 | M_{ D }=H_{ D } h_{ D } =1,333,000 |
| H_{ G }=H_{ p } =250,600 | h_{ G }=0.5(C-G) =3.729 | M_{ G }=h_{ G } h_{ G } M_{ G }=h_{ G } h_{ G } =934,000 |
| H_{ T } =217,400 | h_{ T }=0.5(R+g_{1} +h_{ G }) =4.802 | M_{ T }=H_{ T } h_{ T } =1,044,000 |
| H_{ H } =936,150 | -h_{ H } = −0.5(T −t) = −0.5 T +1.313 | M_{ H }=H_{ H } h_{ H } = −468,100 T +1,229,200 |
| M_{0}=M_{ D }+M_{ G }+M_{ T }+M_{ H } =4,540,200–468,100 T | ||
The minimum thickness at the gasket seating condition is
F = 0 and J=\frac{(2,613,000)(26.5+10.75)}{(17,500)(10.75)(26.5-10.75)}
= 32.850
T=F \pm \sqrt{F^{2}+J}=\sqrt{32.850} =5.732 in.
The minimum thickness at the operating condition is
F=\frac{(2500)(10.75) \sqrt{4(22.813)^{2}-(10.75)^{2}}}{8(17,500)(26.5-10.75)}= 0.540.
If we assume T =5.75 and M_{0} =1,849,100,
J=32.850 \times \frac{1,849,100}{2,613,000}=23.246
T=0.540 \pm \sqrt{(0.540)^{2}+23.246}=5.392 in.
The minimum thickness of 5.75 in. based on the gasket seating condition is also satisfactory for the operating condition.