Products ## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 10.9

What is the minimum required thickness of a welding neck flange as shown in Fig. 10.14a with the following design data? (Note: These data are the same as those used for the blind flange in Example 10.8. In Fig. 10.15 is a sample calculation of a welding neck flange.

Design pressure, p, = 2,500 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.  ## Verified Solution

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.

$b_{o}=N / 2=0.5 \ in$.              and              $b=0.5 \sqrt{b_{0}}=0.3535$

G = 13.75 + (2 × 1) – (2 × 0.3535) = 15.043 in.

4 . Determine bolt loadings and sizing of bolts with N = 1 ; b = 0.3535; y = 10,000 ; m = 3.0.

$H=\frac{\pi}{4} G^{2} p=\frac{\pi}{4}(15.043)^{2}(2500)=444,323$

$H_{p}=2 b \pi G m p=2(0.3535) \pi(15.043)(3.0)(2500)=250,591$

$W_{m 1}=H+H_{p}=(444,323)+(250,591)=694,914$

$W_{m 2}=\pi b G y=\pi(0.3535)(15.043)(10,000)=167,060$

$A_{m}=\text { the greater of } W_{m 1} / S_{b h}=(694,914) /(19,200)=36.2 \text { in.}^{2}$

$\text { or } W_{m 2} / S_{b c}=(167,060) /(19,200)=8.7 \text { in. }^{2}$

$A_{b}$ = actual bolt area = 36.8 in.²                    16 bolts at 2-in. diameter

$W_{a}=0.5\left(A_{m}+A_{b}\right) S_{b c}=0.5(36.2+36.8)(19,200)=700,800$

$W_{0}=W_{m 1}=694,914$

5 . Calculate total flange moment for the design condition.

$H_{D}=\frac{\pi}{4} B^{2} p=\frac{\pi}{4}(10.75)^{2}(2500)=226,906$

$H_{G}=H_{p}=250,591$

$H_{T}=H-H_{D}=(444,323)-(226,906)=217,417$

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

$M_{D}=H_{D} \times h_{D}=(226,910)(4.1875)=950,170$

$M_{G}=H_{G} \times h_{G}=(250,590)(3.7285)=934,330$

$M_{T}=H_{T} \times h_{T}=(217,420)(4.8018)=1,043,990$

$M_{d e}=M_{D}+M_{G}+M_{T}=2,928,490$

6 . Calculate total flange moment for bolt-up condition.

$H_{G}=W_{a}=700,800$

Lever Arm

$h_{G}=0.5(C-G)=3.7285$

Flange Moment

$M_{b u}=H_{G} \times h_{G}=(700,800)(3.7285)=2,612,930$

7 . Use the greater of $M_{d e}$ or $M_{bu}\left(S_{h} / S_{c}\right) ; M_{0}=2,928,490$.

8 . Shape constants from the ASME Code, VIII-1, Appendix 2: K = A/B = (26.5)/(10.75) = 2.465. From Fig. 2-7.1, Section VIII-1;

T = 1.35                  Z = 1.39                Y = 2.29              U = 2.51.

$g_{1} / g_{0}=3.375 / 1.0=3.375$

$h_{0}=\sqrt{B g_{0}}=\sqrt{(10.75)(1.0)}=3.279$

From Fig. 2-7.2, Section VIII-1, F = 0.57.

From Fig. 2-7.3, Section VIII-1, V = 0.04.

From Fig. 2-7.6, Section VIII-1; f = 1.0.

$e=F / h_{0}=(.57) /(3.279)=0.1738$

$d=(U / V) h_{0} g_{0}^{2}=(2.51 / .04)(3.279)(1)^{2}=205.76$

9 . Calculate stresses. Assume a flange thickness t = 4.5 in.

L = (te + 1) / T + t³ / d = (1.320) + (0.443) = 1.763

Longitudinal Hub Stress

$S_{H}=f M_{0} / \operatorname{Lg}_{1}^{2} B=(1)(2,928,490) /(1.763)(3.375)^{2}(10.75)$

$S_{H}=13,570 \ psi$

$S_{R}=\left(\frac{4}{3} t e+1\right) M_{0} / L t^{2} B=(2.0428)(2,928,490) /(1.763)(4.5)^{2}(10.75)$

$S_{R}=15,590 \ psi$

Tangential Flange Stress

\begin{aligned}S_{T} &=\left(Y M_{0} / t^{2} B\right)-Z S_{R} \\&=(2.29)(2,928,490) /(4.5)^{2}(10.75)-(1.39)(15,590)\end{aligned}

$S_{T}=9140 \ psi$

10 . Allowable stresses

$S_{H} \leq 1.5 S_{f}:(1.5)(17,500)=26,250>13,570 \ psi$

$S_{R} \leq S_{f}: 17,500>15,590 \ psi$

$S_{T} \leq S_{f}: 17,500>9140 \ psi$