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## Q. 10.14

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 inside corner with the outside surface even with the ring’s outside (see Fig. 10.23 ). What is the minimum required thickness of the flange ring when the spherical head is dished to a radius of L = 2B ?

## Verified Solution

From the geometry of Example 10.9,

A = 26.5;              B = 10.75;              L = 2B = 2(10.75) = 21.5

Determine the minimum required head thickness as follows:

$t=\frac{5 P L}{6 S}=\frac{5(2500)(21.5)}{6(17,500)}=2.560 \text { in. }$;              use 2.625 in.

From geometry calculations

$L^{\prime}=L+\frac{t}{2}=21.5+\frac{2.625}{2}$

$L^{\prime}=22.813 in .$

$\cos \beta_{1}=\frac{\sqrt{4(22.813)^{2}-(10.75)^{2}}}{2(22.813)}=0.972$;                  $\beta_{1}=13.626^{\circ}$

Membrane force in the head due to pressure is

$F^{\prime}=\frac{p L}{2 T}=\frac{(2500)(21.5)}{2(2.625)}=10,240 \ lb$

Horizontal force =$F^{\prime} \cos \beta_{1}=(10,240)(0.972)=9950 \ lb$

Vertical force =$F^{\prime} \sin \beta_{1}=(10,240)(.236)=2420 \ lb$

Total horizontal force =$\pi(10.75)(9950)=336,000$

Total vertical force =$\pi(10.75)(2420)=81,700$

Moment at gasket seating condition is

 Load Arm Moment $H_{G}=W_{a}=700,800$ $h_{G}=0.5(C-G)=3.729$ $M_{G}=H_{G} h_{G}=2,613,000$

Moment at operating condition is

 Load Arm Moment \begin{aligned}H_{D} &=H_{v}\\&=81,700\end{aligned} $h_{D}=0.5(C-B)=5.875$ $M_{D}=H_{D} h_{D}=480,000$ \begin{aligned}H_{G}&=H_{p} \\&=250,600\end{aligned} $h_{G}=0.5(C-G)=3.729$ $M_{G}=H_{G} h_{G}=934,000$ $H_{T}=336,100$ \begin{aligned}h_{T}&=0.5\left(R+g_{1}+h_{G}\right) \\&=4.802\end{aligned} $M_{T}=H_{T} h_{T}=1,614,000$ $H_{H}=336,100$ \begin{aligned}-h_{H} &=-0.5(T-t) \\&=-0.5T+1.313\end{aligned} \begin{aligned}M_{H}=H_{H} h_{H}=&-168,050 T \\&+441,300\end{aligned}
$M_{0}=M_{D}+M_{G}+M_{T}+M_{H}=3,469,000-168,050 \ T$

The minimum thickness at gasket seating condition is

F = 0          and        $J=\frac{(2,613,000)(26.5+10.75)}{(17,500)(10.75)(26.5-10.75)}=32.850$

$T=F \pm \sqrt{F^{2}+J}=\sqrt{32.850}=5.732 \text { in. }$

The minimum thickness at 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 $M_{0}=2,503,000$ ,

$J=32.850 \times \frac{2,503,000}{2,613,000}=31.467$

$T=0.540 \pm \sqrt{(0.540)^{2}+(31.467)}=6.175 \text { in. }$

If we assume T = 6.25 and $M_{0}=2,419,000$ ,

$J=32.850 \times \frac{2,419,000}{2,613,000}=30.411$

$T=0.540 \pm \sqrt{(0.540)^{2}+(30.411)}=6.081 \text { in. }$

The minimum thickness is approximately 6.108 in. Although exact thickness can be determined, T = 6.25 in. is satisfactory.