Determining the Gasket Stiffness and Joint Constant Problem A pressure chamber is sealed by a gasketed cap fastened with eight preloaded bolts. Find the gasket stiffness and joint constants for two designs of the assembly as shown in Figure 15-33, one with a confined gasket and one with an

Question

Determining the Gasket Stiffness and Joint Constant

Problem A pressure chamber is sealed by a gasketed cap fastened with eight preloaded bolts. Find the gasket stiffness and joint constants for two designs of the assembly as shown in Figure 15-33, one with a confined gasket and one with an unconfined gasket. Also determine the loads felt by bolts and material.

Given The cylinder diameter [latex]D_{p}[/latex] = 4 in. Bolt circle diameter [latex]D_{bc}[/latex] = 5.5 in. Outside flange diameter [latex]D_{f}[/latex] = 7.25 in. The eight 3/8-16 UNC bolts are equispaced on the bolt circle. The flange on the steel chamber is 0.75 in thick. The aluminum cover thickness [latex]l_{c}[/latex] = 1.125 in. The clamped length [latex]l[/latex] of the joint is 1.875 in. The gasket thickness [latex]t[/latex] = 0.125 in. The pressure in the cylinder is 1 500 psi.

Assumptions The gasket material is rubber.

Accepted Answer

See Figure 15-33.

1 Figure 15-33 shows two alternate designs of gasket for the joint on the same view to save space. Whichever design is used, its gasket configuration will be present on both sides of the centerline. Don’t be confused by the depiction of different gaskets top and bottom; only one or the other will be used in the final assembly. We will deal first with the confined gasket configuration.

2 The force on each bolt can be found from the known pressure and cylinder dimensions, assuming that all bolts share the load equally. The total force on the end cap is

[latex] P_{ ext {total }}=p A=p \frac{\pi D_p^2}{4}=1500 \frac{\pi(4)^2}{4}=18850 lb [/latex] (a)

and the applied force [latex]P[/latex] on each bolt is

[latex] P=\frac{P_{ ext {total }}}{N_{ ext {bolts }}}=\frac{18850}{8}=2356 lb [/latex] (b)

3 First analyze the confined gasket case. A confined gasket allows the metal surfaces to contact just as if there were no gasket present. So the analysis of the material stiffness can ignore the confined gasket.

4 Determine the relevant ratios for this joint from equations 15-17. The joint aspect ratio is:

[latex] k_{b^{\prime}}=\left\lgroup 1+\frac{d}{l} ight group ^{-1} k_b \cong\left\lgroup 1+\frac{d}{l} ight group ^{-1} \frac{A_t A_b}{A_b l_t+A_t l_s} E_b [/latex] (15.17)

[latex] j=\frac{d}{l}=\frac{0.375}{1.875}=0.200 [/latex] (c)

5 There are two plate to bolt moduli of interest here since we have different materials in the joint. We will call them [latex]r_{H}[/latex] for the high modulus material (the steel cylinder flange) and [latex]r_{L}[/latex] for the low modulus aluminum cover plate.

[latex] \begin{array}{l} r_H=\frac{E_{ ext {material }}}{E_{ ext {bolt }}}=\frac{30 E 6}{30 E 6}=1.0 \ r_L=\frac{E_{ ext {material }}}{E_{ ext {bolt }}}=\frac{10.4 E 6}{30 E 6}=0.347 \end{array} [/latex] (d)

6 The plate thickness ratio relates the thicknesses of the low and high modulus materials in the joint.

[latex] t=\frac{T_L}{T_L+T_H}=\frac{1.125}{1.125+0.750}=0.600 [/latex] (e)

7 Calculate the Cr terms [latex]C_{H}[/latex] and [latex]C_{L} [/latex] for [latex]r_{H}[/latex] and [latex]r_{L}[/latex], respectively, using equation 15.19 and the coefficients [latex]p_i[/latex] from Table 15-8. For [latex]j[/latex] = 0.20, they are: [latex]p_{0}[/latex] = 0.6118, [latex]p_{1}[/latex] = –1.1715, [latex]p_{2}[/latex] = 1.0875, and [latex]p_{3}[/latex] = –0.3806.

[latex] C_r=p_3 r^3+p_2 r^2+p_1 r+p_0 [/latex] (15.19)

[latex] \begin{aligned} C_L &=C_r=p_3 r_L^3+p_2 r_L^2+p_1 r_L+p_0 \ &=-0.381(0.347)^3+1.088(0.347)^2-1.172(0.347)+0.612=0.321 \end{aligned} [/latex] (f)

Note that [latex]r_{H}[/latex] = 1 so [latex]C_{H}[/latex] is:

[latex] \begin{aligned} C_H &=C_r=p_3 r_H{ }^3+p_2 r_H{ }^2+p_1 r_H+p_0 \ &=-0.381+1.088-1.172+0.612=0.147 \end{aligned} [/latex] (g)

8 Because [latex]j[/latex] > 0.1 in this example, we need to use equation 15.20b to calculate [latex]C_{t}[/latex]. The coefficients [latex]q_{i}[/latex] are taken from Table 15-9 and for [latex]j[/latex] = 0.20, they are: [latex]q_{0}[/latex] = 0.101, [latex]q_{1}[/latex] = 8.547, [latex]q_{2}[/latex] = –24.166, and [latex]q_{3}[/latex] = 15.497.

[latex] \begin{array}{l} C_t=q_3 t^3+q_2 t^2+q_1 t+q_0 \ C_t=15.497(0.60)^3-24.166(0.60)^2+8.547(0.60)+0.101=-0.124 \end{array} [/latex] (h)

9 Calculate the amplitude of the correction factor to the linearized estimate using equation 15.21.

[latex] a=e^{0.0598(\ln j)^3+0.1385(\ln j)^2-0.4350(\ln j)-2.3516}=0.214 [/latex] (i)

10 The joint stiffness factor for the confined-gasket design is calculated with equation 15.22.

[latex] \begin{aligned} C &=C_H+\left(t+a C_t ight)\left(C_L-C_H ight) \ &=0.321+[0.60+0.214(-0.124)](0.321-0.147)=0.247 \end{aligned} [/latex] (j)

11 The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13.

[latex] \Delta \delta=\frac{P_b}{k_b}=\frac{P_m}{k_m} [/latex] (15.13a)

[latex] P_b=\frac{k_b}{k_m} P_m [/latex] (15.13b)

[latex] P_b=\frac{k_b}{k_m+k_b} P \ P_b=C P \quad ext { where } \quad C=\frac{k_b}{k_m+k_b} [/latex] (15.13c)

[latex] P_m=\frac{k_m}{k_b+k_m} P=(1-C) P [/latex] (15.13d)

[latex] \begin{array}{l} P_b=C P=0.247(2356) \cong 581.1 lb \ P_m=(1-C) P=(1-0.247)(2356) \cong 1775.1 lb \end{array} [/latex] (k)

12 We can estimate the bolt stiffness [latex] k_{b^{\prime}} [/latex] from equation 15.17 using its shank area = 0.110 in² and tensile stress area = 0.077 in² (Table 15-1), then estimate the material stiffness [latex] k_{m_1} [/latex] for the confined gasket case by using equation 15.13c, given [latex] k_{b^{\prime}} [/latex] and [latex]C[/latex].

[latex] k_{b^{\prime}}=\left\lgroup 1+\frac{d}{l} ight group ^{-1} k_b \cong\left\lgroup 1+\frac{d}{l} ight group ^{-1} \frac{A_t A_b}{A_b l_t+A_t l_s} E_b [/latex] (15.17)

[latex] P_b=\frac{k_b}{k_m+k_b} P \ P_b=C P \quad ext { where } \quad C=\frac{k_b}{k_m+k_b} [/latex] (15.13c)

[latex] \begin{aligned} ext {length of thread :} \quad l_{ ext {thd }} &=2 d+0.25=2(0.375)+0.25=1.0 ext { in } \ ext {length of shank :} \quad l_s &=l_{ ext {bolt }}-l_{ ext {thd }}=2.25-1.0=1.25 ext { in } \ ext {length of clamped thread :} \quad l_t &=l-l_s=1.875-1.25=0.625 ext { in } \end{aligned} [/latex]

[latex] \begin{aligned} k_{b^{\prime}} & \cong\left\lgroup 1+\frac{d}{l} ight group^{-1} \frac{A_t A_b}{A_b l_t+A_t l_s} E_b \ k_{b^{\prime}} & \cong\left\lgroup 1+\frac{0.375}{1.875} ight group^{-1} \frac{0.077(0.110)}{0.110(0.625)+0.077(1.25)} 30 E 6=1.290 E 6 lb / ext { in } \ C &=\frac{k_{b^{\prime}}}{k_{m_1}+k_{b^{\prime}}} \Rightarrow k_{m_1}=k_{b^{\prime}} \frac{1-C}{C}=1.29 E 6\left\lgroup \frac{1-0.247}{0.247} ight group=3.940 E 6 lb / in \end{aligned} [/latex] (l)

13 Now we will address the unconfined-gasket case. The bolt stiffness is not affected by the gasket but the material stiffness is. We now have two springs in series, the metal whose stiffness [latex]k_{m}[/latex] is defined in equation ([latex]l[/latex]) and the gasket which is calculated in equation ([latex]n[/latex]) below. These combine according to equation 14.2b (p. 787). The portion of the unconfined gasket subjected to the clamp force can be assumed to be from the outside diameter of the flange shown in Figure 15-33 to the inside diameter of the vessel. The bolt hole should be subtracted from the gasket area. The area of the clamped gasket around one bolt is:

[latex] \frac{1}{k_{ ext {total }}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots+\frac{1}{k_n} [/latex] (14.2b)

[latex] A_g=\frac{\pi}{4}\left[\frac{\left(D_f^2-D_p^2 ight)}{N_{b o l t s}}-d^2 ight]=\frac{\pi}{4}\left[\frac{\left(7.25^2-4^2 ight)}{8}-0.375^2 ight]=3.479 in ^2 [/latex] (m)

14 The stiffness of this piece of gasket is found from equation 15.11c (p. 885).

[latex] k_m=\frac{A_m E_m}{l} [/latex] (15.11c)

[latex] k_{m_2}=k_g=\frac{A_g E_g}{t}=\frac{3.479(10 E 3)}{0.125} \cong 2.783 E 5 lb / in [/latex] (n)

The modulus of elasticity [latex]E_{g}[/latex] of the gasket material is found in Table 15-10 (p. 900).

15 The combined stiffness of the gasketed joint (from equation 14.2b, p. 787) is

[latex] \frac{1}{k_{ ext {total }}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots+\frac{1}{k_n} [/latex] (14.2b)

[latex] k_m=\frac{1}{\frac{1}{k_{m_1}}+\frac{1}{k_{m_2}}}=\frac{1}{\frac{1}{3.94 E 6}+\frac{1}{2.783 E 5}} \cong 2.600 E 5 lb / in [/latex] (o)

Note that the combined stiffness is essentially the same as that of the soft gasket alone, since it dominates the equation. We could have used the gasket stiffness [latex]k_{g}[/latex] to represent the joint stiffness [latex]k_{m}[/latex] with little error.

16 The joint constant with the unconfined gasket is now

[latex] \begin{array}{c} C=\frac{k_{b^{\prime}}}{k_m+k_{b^{\prime}}}=\frac{1.290 E 6}{2.600 E 5+1.290 E 6}=0.832 \ ext {and} \quad (1-C)=0.168 \end{array} [/latex] (p)

17 The portions of the applied load [latex]P[/latex] felt by the bolt and the material with a soft, unconfined gasket in the joint can now be found from equations 15.13 (p. 886).

[latex] \Delta \delta=\frac{P_b}{k_b}=\frac{P_m}{k_m} [/latex] (15.13a)

[latex] P_b=\frac{k_b}{k_m} P_m [/latex] (15.13b)

[latex] P_b=\frac{k_b}{k_m+k_b} P \ P_b=C P \quad ext { where } \quad C=\frac{k_b}{k_m+k_b} [/latex] (15.13c)

[latex] P_m=\frac{k_m}{k_b+k_m} P=(1-C) P [/latex] (15.13d)

[latex] \begin{array}{l} P_b=C P=0.832(2356) \cong 1961 lb \ P_m=(1-C) P=(1-0.832)(2356) \cong 395 lb \end{array} [/latex] (q)

18 See what has happened as a result of introducing an unconfined soft gasket. Compare the value of [latex]C[/latex] in equation ([latex]j[/latex]) to [latex]C[/latex] in equation ([latex]p[/latex]). The bolt has gone from feeling only 25% of the applied load with no gasket (or a confined gasket) to feeling 83% of the applied load with a soft, unconfined gasket. In effect, the roles of the bolt and the material have been reversed by the introduction of the soft gasket. An unconfined, soft gasket severely limits the ability of the bolt to accommodate fatigue loads, as was accomplished in the previous example. The files EX15-04 can be found on the CD-ROM.

Table 15-8 Parameters for Equation 15.19
[latex]j [/latex]	[latex]p_{0}[/latex]	[latex]p_{1}[/latex]	[latex]p_{2}[/latex]	[latex]p_{3}[/latex]
0.10	0.4389	–0.9197	0.8901	–0.3187
0.20	0.6118	–1.1715	1.0875	–0.3806
0.30	0.6932	–1.2426	1.1177	–0.3845
0.40	0.7351	–1.2612	1.1111	–0.3779
0.50	0.7580	–1.2632	1.0979	–0.3708
0.60	0.7709	–1.2600	1.0851	–0.3647
0.70	0.7773	–1.2543	1.0735	–0.3595
0.80	0.7800	–1.2503	1.0672	–0.3571
0.90	0.7797	–1.2458	1.0620	–0.3552
1.00	0.7774	–1.2413	1.0577	–0.3537
1.25	0.7667	–1.2333	1.0548	–0.3535
1.50	0.7518	–1.2264	1.0554	–0.3550
1.75	0.7350	–1.2202	1.0581	–0.3574
2.00	0.7175	–1.2133	1.0604	–0.3596

Table 15-9 Parameters for Equation 15.20
[latex]j[/latex]	[latex]q_{0}[/latex]	[latex]q_{1}[/latex]	[latex]q_{2}[/latex]	[latex]q_{3}[/latex]	[latex]q_{4}[/latex]	[latex]q_{5}[/latex]
0.10	0.0079	17.0400	–92.832	202.44	–209.38	82.726
0.20	0.1010	8.5465	–24.166	15.497
0.30	0.0861	8.2344	–22.274	13.963
0.40	0.0695	8.0297	–20.727	12.646
0.50	0.0533	7.8676	–19.357	11.457
0.60	0.0372	7.6705	–17.951	10.262
0.70	0.0197	7.3030	–16.235	8.9273
0.80	0.0029	6.9893	–14.737	7.7545
0.90	–0.0123	6.7006	–13.363	6.6784
1.00	–0.0265	6.4643	–12.188	5.7481
1.25	–0.0524	5.7363	–9.3326	3.6348
1.50	–0.0678	5.0674	–7.0322	2.0107
1.75	–0.0763	4.5187	–5.1590	0.6861
2.00	–0.0784	3.9617	–3.5248	–0.3956

Table 15-1 Principal Dimensions of Unified National Standard Screw Threads Data Calculated from Equations 15.1—See Reference 3 for More Information
		Coarse Threads—UNC			Fine Threads—UNF
Size	Major Diameter [latex]d[/latex] (in)	Threads per inch	Minor Diameter [latex]d_{r}[/latex] (in)	Tensile Stress Area [latex]A_{t} (in^2)[/latex]	Threads per inch	Minor Diameter [latex]d_{r}[/latex] (in)	Tensile Stress Area [latex]A_{t} (in^2)[/latex]
0	0.0600	–	–	–	80	0.0438	0.0018
1	0.0730	64	0.0527	0.0026	72	0.0550	0.0028
2	0.0860	56	0.0628	0.0037	64	0.0657	0.0039
3	0.0990	48	0.0719	0.0049	56	0.0758	0.0052
4	0.1120	40	0.0795	0.0060	48	0.0849	0.0066
5	0.1250	40	0.0925	0.0080	44	0.0955	0.0083
6	0.1380	32	0.0974	0.0091	40	0.1055	0.0101
8	0.1640	32	0.1234	0.0140	36	0.1279	0.0147
10	0.1900	24	0.1359	0.0175	32	0.1494	0.0200
12	0.2160	24	0.1619	0.0242	28	0.1696	0.0258
1/4	0.2500	20	0.1850	0.0318	28	0.2036	0.0364
5/16	0.3125	18	0.2403	0.0524	24	0.2584	0.0581
3/8	0.3750	16	0.2938	0.0775	24	0.3209	0.0878
7/16	0.4375	14	0.3447	0.1063	20	0.3725	0.1187
1/2	0.5000	13	0.4001	0.1419	20	0.4350	0.1600
9/16	0.5625	12	0.4542	0.1819	18	0.4903	0.2030
5/8	0.6250	11	0.5069	0.2260	18	0.5528	0.2560
3/4	0.7500	10	0.6201	0.3345	16	0.6688	0.3730
7/8	0.8750	9	0.7307	0.4617	14	0.7822	0.5095
1	1.0000	8	0.8376	0.6057	12	0.8917	0.6630
1 1/8	1.1250	7	0.9394	0.7633	12	1.0167	0.8557
1 1/4	1.2500	7	1.0644	0.9691	12	1.1417	1.0729
1 3/8	1.3750	6	1.1585	1.1549	12	1.2667	1.3147
1 1/2	1.5000	6	1.2835	1.4053	12	1.3917	1.5810
1 3/4	1.7500	5	1.4902	1.8995
2	2.0000	4.5	1.7113	2.4982
2 1/4	2.2500	4.5	1.9613	3.2477
2 1/2	2.5000	4	2.1752	3.9988
2 3/4	2.7500	4	2.4252	4.9340
3	3.0000	4	2.6752	5.9674
3 1/4	3.2500	4	2.9252	7.0989
3 1/2	3.5000	4	3.1752	8.3286
3 3/4	3.7500	4	3.4252	9.6565
4	4.0000	4	3.6752	11.0826

Table 15-10 Young's Modulus for Some Gasket Materials Source: Reference 12 with permission of McGraw-Hill, Inc., New York
Material	Modulus of Elasticity
Material	psi	MPa
Cork	12.5E3	86
Compressed asbestos	70E3	480
Copper-asbestos	13.5E6	93E3
Copper (pure)	17.5E6	121E3
Plain rubber	10E3	69
Spiral wound	41E3	280
Teflon	35E3	240
Vegetable fiber	17E3	120

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