Question 10.9: What is the minimum required thickness of a welding neck fla...
What is the minimum required thickness of a welding neck flange as shown in Figure 10.15 with the following design data? (Note: These data are the same as those used for the blind flange in Example 10.8. In Figure 10.16 is a sample calculation for a welding neck flange.)
Design pressure P =2500 psi
Design temperature=250 °F
Bolt-up and gasket seating temperature=70 °F
Flange material is SA-105
Bolting material is SA-325 Grade 1
Gasket details are spiral-wound metal, fiber-filled, stainless steel; inside diameter is 13.75 in., and width is 1.0 in.
Figure 10.15 Flange dimensions mentioned in Example 10.9.
1) Allowable bolt stress at design and seating temperatures= S_{b} =19,200 psi
2) Allowable flange stress at design and seating temperatures= S_{f} =17,500 psi
3) Gasket dimensions are as follows:
b_{0}=N / 2=0.5 in and b=0.5 \sqrt{b_{0}}=0.3535
G = 13.75 + (2 × 1) − (2 × 0.3535) = 15.043in.
4) Determine the bolt loadings and sizing of bolts with
N =1, b=0.3535, y=10,000, m =3.0:
H=\frac{\pi}{4} G^{2} P=\frac{\pi}{4}(15.043)^{2}(2500)=444,323H_{ p }=2 b \pi G m P=2(0.3535)
π(15.043)(3.0)(2500) = 250,591
W_{ m 1}=H+H_{ p }=(444,323)+(250,591)= 694,914
W_{ m 2} = πbGy = π(0.3535)(15.043)(10,000)
= 167,060
A_{ m }=\text { the greater of } W_{ m 1} / S_{ bh }=(694,914) /(19,200) = 36.2 in.^{2} or
W_{ m 2} / S_{ bc }=(167,060) /(19,200)=8.7 in ^{2}A_{b} = actual bolt area = 36.8 in.^{2}
16 bolds at 2in.diameter
W_{ a }=0.5\left(A_{ m }+A_{ b }\right) S_{ bc }=0.5(36.2+36.8)(19,200) = 700,800
W_{ b }=W_{ m 1}=694,9145) Calculate the total flange moment for the design condition. Flange loads:
H_{ D }=\frac{\pi}{4} B^{2} p=\frac{\pi}{4}(10.75)^{2}(2500)=226,906H_{ G }=H_{ p }=250,591
H_{ T }=H-H_{ D }=(444,322)-(226,906)
= 217,417.
Lever arms:
h_{ D }=R+0.5 g_{1} =(2.5)+0.5(3.375)=4.1875
h_{ G }=0.5(C-G) =0.5(22.5-15.043)=3.7285
h_{ T }=0.5\left(R+g_{1}+h_{ G }\right) =0.5(2.5+3.375+3.7285)
= 4.8018.
Flange moments:
M_{ D }=H_{ D } \times h_{ D } =(226,910)(4.1875)=950,170
M_{ G }=H_{ G } \times h_{ G } =(250,590)(3.7285)=934,330
M_{ T }=H_{ T } \times h_{ T } =(217,417)(4.8018)=1,043,990
M_{ de }=M_{ D }+M_{ G }+M_{ T } =2,928,490
6) Calculate the total flange moment for bolt-up condition. Flange load:
H_{ G }=W_{ a } =700,800
Lever arm:
h_{ G }=0.5(C-G) =3.7285
Flange moment:
M_{ bu }=H_{ G } \times h_{ G } =(700,800)(3.7285)=2,612,930 .
7) Use the greater of M_{de} or M_{bu}(S_{h}/S_{c}); M_{0} =2,928,490.
8) Shape constants from the ASME Code, VIII-1, Appendix 2 : K =A/B=26.5/10.75=2.465. From Figure 2-7.1, Section VIII-1,
T = 1.35, Z = 1.39, Y = 2.29, U = 2.51
g_{1} / g_{0} =3.375 / 1.0=3.375
h_{0}=\sqrt{B g_{0}}=\sqrt{(10.75)(1.0)} =3.279
h / h_{0} =6.25 / 3.279=1.906
From Figure 2-7.2, Section VIII-1, F =0.57.
From Figure 2-7.3, Section VIII-1, V =0.04.
From Figure 2-7.6, Section VIII-1, f =1.0
e=F / h_{0} =0.57 / 3.279=0.1738
d=\frac{U}{V} h_{0} g_{0}^{2}=\frac{2.51}{0.04}(3.279)(1)^{2} =205.76
9) Calculate the stresses. Assume a flange thickness t =4.5 in.
L=\frac{t e+1}{T}+\frac{t^{3}}{d}=1.320+0.443=1.763Longitudinal hub stress:
S_{ H }=f M_{0} / L g_{1}^{2} B=(1)(2,928,490) /\times(1.763)(3.375)^{2}(10.75)
S_{ H }=13,570 psi
Radial flange stress:
S_{ R }=\left(\frac{4}{3} t e+1\right) M_{0} / L t^{2} B= (2.0428)(2,928,490) /(1.763)(4.5)^{2}(10.75)
= 15,590 psi.
Tangential flange stress:
S_{ T }=\frac{Y M_{0}}{t^{2} B}-Z S_{ R }= \frac{(2.29)(2,928,490)}{(4.5)^{2}(10.75)}-(1.39)(15,590)
= 9140 psi.
10) Allowable stresses:
S_{ H } \leq 1.5 S_{ f } : (1.5)(17,500)=26,250
>13,570 psi
S_{ R } \leq S_{ f } : 17,500>15,590 psi
S_{ T } \leq S_{ f } : 17,500>9140 psi
