3.7 Conical Sections, VIII-1
3.7.1 Internal Pressure
For internal pressure, the design equation for a conical section is given by
t = PD/[2 cos α (SE – 0.6P)], where α < 30° (3.33)
where
t = required thickness, in. P = internal pressure, psi
D = inside diameter of conical section under consideration, in. S = allowable tensile stress, psi
E = Joint Efficiency Factor
Equation (3.33) can be expressed in terms of internal pressure as
P = 2SEt cos α/(D + 1.2t cos α) (3.34)
Equations (3.33) and (3.34) can also be expressed in terms of outside diameter as FIG. 3.6
t = PDo/[2SE cos α + P(2 – 1.2 cos α)] (3.35)
P = 2SEt cos α/[Do– t(2 – 1.2 cos α)] (3.36)
Equations (3.33) to (3.36), which are applicable at any angle α, are limited by VIII-1 to α ≤ 30°. When the angle α exceeds 30°, then VIII-1 requires a knuckle at the large end, as shown in Fig. 3.6(c) and (e). This type of construction will be discussed later in this section.
After determining the thickness of the cone for internal pressure, the designer must evaluate the cone-to- shell junction. The cone-to-shell junction at the large end of the cone is in compression due to internal pres- sure, in most cases. The designer must check the junction for required reinforcement needed to contain the unbalanced forces in accordance with Paragraph 1-5 of Appendix 1 of VIII-1. The required area is obtained from
ArL= (k QLRL/SsE1) (1 – ∆/α) tan α (3.37)
where,
ArL= required area at the large end of the cone, in.
E1= Joint Efficiency Factor of the longitudinal joint in the cylinder
Ec= modulus of elasticity of the cone, psi
Er= modulus of elasticity of the reinforcing ring, psi
Es= modulus of elasticity of the cylinder, psi
k = 1 when additional area of reinforcement is not required y/SrErbut not less than 1.0 when a stiffening ring is required
QL= axial load at the large end, lb/in., including pressure end-load
RL= large radius of the cone, in.
Sc= allowable stress in the cone, psi
Sr= allowable stress in the reinforcing ring, psi
Ss= allowable stress in the cylinder, psi
y = SsEsfor the reinforcing ring on the shell
ScEcfor the reinforcing ring on the cone
∆ = angle obtained from Table 3.2
The area calculated from Eq. (3.37) must be furnished at the junction. Part of this area may be available at the junction as excess area. This excess area can be calculated from the equation
AeL= (ts– t)(RLts)1/2+ tc– tr)(RLtc/cos α)1/2 (3.38)
where
AeL= available area at the junction, in.2
t = minimum required thickness of the shell, in. tc= nominal cone thickness, in.
TABLE 3.2
VALUES OF ∆ FOR JUNCTIONS AT THE LARGE CYLINDER
DUE TO INTERNAL PRESSURE
P/SsE1 0.001 0.002 0.003 0.004 0.005
∆, deg. 11 15 18 21 23
P/SsE1 0.006 0.007 0.008 0.0091
∆, deg. 25 27 28.5 30
NOTE:
tr= minimum required thickness of the cone, in.
ts= nominal shell thickness, in.
If this excess area is less than that calculated from Eq. (3.37), then additional area in the form of stiffening rings must be added.
The cone-to-shell junction at the small end of the cone is in tension due to internal pressure, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1-5 of Appendix 1 of VIII-1. The required area at the small end of the cone is obtained from
Ars= (k Qs, Rs/SsE1)(1 – ∆/α) tan α (3.39)
where
Ars= required area at the small end of the cone, in.2
Qs= axial load (including pressure end load) at small end, lb/in.
Rs= small radius of the cone, in.
∆ = angle obtained from Table 3.3
The area calculated from Eq. (3.39) must be furnished at the junction. Part of this area may be available at the junction as excess area. This excess area can be calculated from the equation
Aes= 0.78(Rsts)1/2[(ts– t) + (tc– tr)/cos α] (3.40)
If this excess area is less than that calculated from Eq. (3.39), then additional area in the form of stiffening rings must be added.
When the angle α exceeds 30°, VIII-1 requires a knuckle at the large end, as shown in Fig. 3.6(c) and (e). The required thickness for the knuckle (called a flange) at the large end of the cone is obtained from the equation
t = PLM/(2SE – 0.2P) (3.41)
where
M = (1/4)[3 + (L/r)1/2]
L = Di/2 cos α
Di= inside diameter at the knuckle-to-cone junction
= D – 2r (1 – cos α)
r = inside knuckle radius, in.
Equation (3.41) can be expressed in terms of the required pressure
P = 2SEt/(LM + 0.2t) (3.42)
Equations (3.41) and (3.42) can also be written in terms of the outside diameter, Do, as
t = PLoM/[2SE + P(M – 0.2)] (3.43)
TABLE 3.3
VALUES OF ∆ FOR JUNCTIONS AT THE SMALL CYLINDER
DUE TO INTERNAL PRESSURE
P/SsE1 0.002 0.005 0.010 0.02 0.04
∆, deg. 4 6 9 12.5 17.5
P/SsE1 0.08 0.10 0.1251
∆, deg. 24 27 30
NOTE:
or in terms of required pressure,
P = 2SEt/[MLo– t(M – 0.2)] (3.44)
When a knuckle is used at the cone-to-shell junction, the diameter at the large end of the cone is slightly less than the diameter of the cone without a knuckle, as shown in Fig. 3.6. Thus, the design of the cone as given by Eq. 3.33 is based on diameter Dirather than on the shell diameter.
ASME VIII-1 does not give rules for the design of knuckles (flues) at the small end of cones. One design method uses the pressure-area procedure (Zick and Germain, 1963) to obtain the required thickness. Referring to Fig. 3.7 for terminology, we can determine the required thickness based on membrane forces in the flue and adjacent cone and shell areas from
tf= (180/απr)[P(K1+ K2+ K3)/1.5SE – K4– K5] (3.45)
where
E = Joint Efficiency Factor
K1= 0.125 (2r + D1)2tan α – απr2/360
K2= 0.28D1(D1ts) 1/2
K3= 0.78K6(K6tc) 1/2
K4= 0.78tc(K6tc) 1/2
K5= 0.55ts(D1ts) 1/2
K6= [D1+ 2r(1 – cos α)]/2 cos α
P = internal pressure, psi S = allowable stress, psi tc= thickness of the cone, in.
tf= thickness of the flue, in.
ts= thickness of the shell, in.
α = flue angle, deg. The flue angle is normally the same as the cone angle. Example 3.11
Problem
Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions shown in Fig. E3.11. Let axial compressive load at cone vicinity from mounted equipment = 50 kips.
Small Cylinder Cone Large Cylinder Reinforcing Ring
Allowable stress, psi 15,000 16,000 17,500 13,000
Joint Efficiency Factor 0.85 1.0 0.85 —
Modulus of elasticity, ksi 27,000 29,000 25,000 30,000
Pressure, psi 100 100 100 —
Solution
Small Shell
The required thickness from Eq. (2.1) is
t = 100 × 60/(15,000 × 0.85 – 0.6 × 100)
= 0.47 in.
Use t = 1/2 in.
Cone
From Eq. (3.33), the cone thickness is calculated as
t = 100 × 2 × 7 × 12/[2 cos 28(16,000 × 1.0 – 0.6 × 100)]
= 0.60 in.
Use t = 5/8 in.
Large Shell
Again, using Eq. (2.1), we get
t = 100 × 7 × 12/(17,500 × 0.85 – 0.6 × 100)
= 0.57 in.
Use t = 5/8 in.
Large Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to the shell. Then from Eq. (3.37), we calculate the stiffness ratio, k, as
k = 17,500 × 25,000,000/(13,000 × 30,000,000)
= 1.12
The axial loads are given by
QL= PRL/2 – axial equipment load
= 100 × 84/2 – [50,000/(2π84)] = 4105 lb/in.
Next, we need to calculate the need for reinforcement in accordance with Table 3.2.
P/SsE1= 100/17,500 × 0.85 = 0.0067
From Table 3.2,∆ = 26.4°. Reinforcement is needed since α = 28°. The amount of reinforcement is calculated from Eq. (3.37):
ArL= (1.12 × 4105 × 84/17,500 × 0.85)(1 – 26.4/28) tan 28
= 0.79 in.2
The available area in the shell and cone due to excess thickness is calculated from Eq. (3.38):
AeL= (ts– t)(RLts)1/2+ (tc– tr)(RLtc/cos α)1/2
= (0.625 – 0.57)(84 × 0.625)1/2+ (0.625 – 0.60)(84 × 0.625/cos 28)1/2 = 0.040 + 0.193
= 0.59 in.2
The additional area needed at the large junction = 0.79 – 0.59 = 0.20 in.2Use a 2 in. × 1/4 in. bar rolled the
hard way.
Small Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from
k = 15,000 × 27,000,000/(13,000 × 30,000,000)
= 1.04
The axial loads are equal to
Qs= PRs/2 – axial equipment load
= 100 × 60/2 – [50,000/(2π60)] = 2867 lb/in.
The need for reinforcement is obtained from Table 3.3.
P/SsE1= 100/15,000 × 0.85 = 0.0078
From Table 3.3,∆ = 7.68°. Since this is less than 28°, reinforcement is required in accordance with Eq. (3.39),
Ars= (1.04 × 2867 × 60/15,000 × 0.85)(1 – 7.68/28) tan 28
= 5.41 in.2
In order to determine what excess area, if any, is available at the cone-to-shell junction, we must calculate the required thickness of the cone at the small junction. This information is needed because the cone thick- ness used so far is based on the large diameter rather than on the small one. From Eq. (3.33), the minimum cone thickness at the small end is
t = 100 × 2 × 5 × 12/[2 cos 28(16,000 × 1.0 – 0.6 × 100)]
= 0.43 in.
Aes= 0.78(60 × 0.50)1/2[(0.50 – 0.47) + (0.625 – 0.43)/cos 28]
= 1.07 in.2
The additional area needed at the small junction = 5.41 – 1.07 = 4.34 in.2Therefore, use a 4.5 in. × 1 in. bar
rolled the hard way. Example 3.12 Problem
Determine the required thickness of the cone, knuckle, flue, and the two cylinders shown in Fig. E3.12. Let
P = 100 psi, S = 16,000 psi, and joint efficiency = 0.85.
Solution Small Shell From Eq. (2.1), t = 100 × 60/(16,000 × 0.85 – 0.6 × 100) = 0.44 in. Use t = 1/2 in. FIG. E3.12
Large Shell
From Eq. (2.1),
t = 100 × 7 × 12/(16,000 × 0.85 – 0.6 × 100)
= 0.62 in.
Use t = 5/8 in.
Knuckle at Large End
The thickness of the knuckle is obtained from Eq. (3.41).
Di= 7 × 2 × 12 – 2 × 10(1 – cos 25) = 166.13 in. L = 166.13/(2 × cos 25) = 91.65 in. M = (1/4)[3 + (91.65/10)1/2] = 1.51 t = 100 × 91.65 × 1.51/(2 × 16,000 × 0.85 – 0.2 × 100) = 0.51 in. Use t = 9/16 in. Cone
From Eq. (3.33), with D = Di= 166.13 in.,
t = 100 × 166.13/[2 cos 25(16,000 × 0.85 – 0.6 × 100)]
= 0.67 in.
Use t = 11/16 in.
Flue
From Eq. (3.45), the required thickness of the flue is
K1= 0.125(2 × 4 + 120)2tan 25 – 25 π42/360 = 951.51 K2= 0.28 × 120(120 × 0.50)1/2 = 260.26 K6= [120 + 2 × 4(1 – cos 25)]/2 cos 25 = 66.62
K3= 0.78 × 66.62(66.62 × 0.6875)1/2 = 351.67 K4= 0.78 × 0.6875(66.62 × 0.6875)1/2 = 3.62 K5= 0.55 × 0.50(120 × 0.50)1/2 = 2.13 t = (180/25π4)[100(951.51 + 260.26 + 351.67)/1.5 × 16,000 × 0.85 – 3.62 – 2.13] = 1.10 in. 3.7.2 External Pressure
The design of conical shells for external pressure follows the same procedure as that for cylindrical shells given in sections 2.4.1 and 2.4.2, with the following exceptions:
Item Cylinder Cone
Thickness t of cylinder te= (t of cone)(cos α) (3.46)
Diameter Doof cylinder DL= outside large diameter of the cone (3.47)
Length L of cylinder Le= (L/2)(1 + Ds/DL), where L is obtained from Fig. 3.6. (3.48)
After designing the cone for external pressure, the cone-to-shell junctions must be evaluated. Due to exter- nal pressure, the cone-to-shell junction at the large end of the cone is tension, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1-8 of Appendix 1 of VIII- 1. The required area is obtained from
ArL= (kQLRL/SsE1){1 – (0.25)[(PRL– QL)/QL](∆/α)} tan α (3.49)
where all terms are the same as those in Eq. (3.37) and ∆ is obtained from Table 3.4.
The area calculated from Eq. (3.49) must be furnished at the junction. Some of this area may be available as excess area at the junction. This excess area can be calculated from the equation
AeL= 0.55(DLts)1/2(ts+ tc/cos α) (3.50)
TABLE 3.4
VALUES OF ∆ FOR JUNCTIONS AT THE LARGE CYLINDER
DUE TO EXTERNAL PRESSURE
P/SsE1 0 0.002 0.005 0.010 0.02 ∆, deg. 0 5 7 10 15 P/SsE1 0.04 0.08 0.10 0.125 0.15 ∆, deg. 21 29 33 37 40 P/SsE1 0.20 0.25 0.30 0.351 ∆, deg. 47 52 57 60 NOTE:
If this excess area is less than that calculated from Eq. (3.49), then additional area in the form of stiffening rings must be added.
In addition to having a sufficient reinforcement area, the cone-to-shell junction must have an adequate moment of inertia to resist external pressure forces when the junction is considered as a line of support. The required moment of inertia is calculated as follows:
1. Determine the quantity ATLfrom the equation
ATL= LLts/2 + Lctc/2 + As (3.51)
where
As= area of the stiffening ring
LL= effective length of the shell
Lc= effective length of the cone = [L2+ (RL– Rs)2]1/2
L = axial length of the cone
2. Calculate the quantities
M = (R2
L– R2S/(3RLtan α) + LL/2 – (RLtan α)/2 (3.52)
and
FL= PM + (axial forces other than pressure)(tan α) (3.53)
3. Calculate B from the equation
B = 0.75(FLDL/ATL) (3.54)
4. Enter the appropriate EPC and determine and A value.
5. If B falls below the left end of the temperature line, calculate A from the equation
A = 2B/Ex (3.55)
where Exis the smaller of Ec, Er,or Es.
6. Calculate the moment of inertia from one of the following equations:
Is= AD2LATL/14 (3.56)
or
I′s= AD2LATL/10.9 (3.57)
where
Is= required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in.4
I′s= required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about
their combined neutral axis, in.4
7. The required moment of inertia must be greater than the furnished one.
The cone-to-shell junction at the small end of the cone due to external pressure is in compression, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1–8 of Appendix 1 of VIII-1. The required area is obtained from
The area calculated from Eq. (3.58) must be furnished at the junction. Some of this area may be available at the junction as excess area. This excess area can be calculated from the equation
Aes= 0.55(Dsts)1/2[(ts– t) + (tc– tr)/cos α] (3.59)
If this excess area is less than that calculated from Eq. (3.58), then additional area in the form of stiffening rings must be added.
In addition to having a sufficient area, the cone-to-shell junction must have an adequate moment of iner- tia to resist external pressure forces when the junction is considered as line of support. The required moment of inertia is calculated as follows:
1. Determine the quantity ATSfrom the equation
ATs= LSts/2 + Lctc/2 + As (3.60)
where
As= area of the stiffening ring
LS= effective length of the shell
Lc= effective length of the cone
= [L2+ (R
L– Rs)2]1/2
L = axial length of the cone
2. Calculate the quantities
N = (R2
L– R2S)/(6Rstan α) + Ls/2 + (Rstan α)/2 (3.61)
and
Fs= PN + (axial forces other than pressure)(tan α) (3.62)
3. Calculate B from the equation
B = 0.75(FsDS/ATS) (3.63)
4. Enter the appropriate EPC and determine an A value.
5. If B falls below the left end of the temperature line, calculate A from the equation
A = 2B/Ex (3.64)
where Exis the smaller of Ec, Er, or Es.
6. Calculate the moment of inertia from one of the following equations:
Is= AD2SATS/14 (3.65)
or
I′s= AD2sATS/10.9 (3.66)
where
Is= required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in.4
I′s= required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about
their combined neutral axis, in.4
When the cone is flanged and flued, then the required thickness of the cone is determined as before, except that Eq. (3.48) is replaced by
Le= r1sin α + (Lc/2)[(Ds+ DL)/DLs] for sketch (c) in Fig. 3.6 (3.67)
Le= r2(Dss/DL) sin α + (Lc/2)[(Ds+ DL)/DL] for sketch (d) in Fig. 3.6 (3.68)
Le= [r1+ r2(Dss/DLs)] sin α + (Lc/2)[(Ds+ DL)/DLs] for sketch (d) in Fig. 3.6 (3.69)
Example 3.13 Problem
Check the calculated thicknesses in Example 3.11 due to an external pressure of 15 psi. The axial com- pressive load at cone vicinity from mounted equipment = 50 kips. Figure 2.4 applies for external pres- sure. Notice that the modulus of elasticity, shown in Fig. 2.4, for external pressure calculations must be used and is different from the values listed below for junction reinforcement. The design temperature is 100°F.
Small Cylinder Cone Large Cylinder Reinforcing Ring
Allowable stress, psi 15,000 16,000 17,500 13,000
Joint Efficiency Factor 0.85 1.0 0.85 —
Modulus of elasticity, ksi 27,000 29,000 25,000 30,000
External P, psi 15 15 15 —
Effective L 10 ft — 20 ft —
Solution
Small Shell
From Example 3.11, use t = 1/2 in.
Do= 2(60 + 0.5) = 121 in.
Then L/Do= 0.99 and Do/t = 242.
From Fig. 2.6, A = 0.00038. From Fig. 2.4, B = 5500 psi. From Eq. (2.26),
P = (4/3)(5500)/(242) = 30.3 psi > 15 psi
By trial and error, it can be shown that for t = 0.40 in., P = 15 psi.
Large Shell
From Example 3.11, try t = 5/8 in.
Do= 2(84 + 0.625) = 169.3 in.
Then L/Do= 1.42 and Do/t = 271.
From Fig. 2.6, A = 0.00021. From Fig. 2.4, B = 3050 psi. From Eq. (2.26),
P = (4/3)(3050)/(271) = 15.0 psi
Cone
From Example 3.11, use t = 5/8 in.
Do= 2(84 + 0.625) = 169.3 in.
From Eq. (3.46), te= 0.625 cos 28 = 0.552 in.
From Eq. (3.47), DL= 169.3 in.
From Fig. E3.11, L = (7 – 5)/tan 28 = 3.76 ft. From Eq. (3.48),
Le= (3.76 × 12/2)(1 + 121/169.3) = 38.68 in.
Then Le/DL= 0.23 and DL/te= 307.
From Fig. 2.6, A = 0.00121. From Fig. 2.4, B = 13,000 psi. From Eq. (2.26),
P = (4/3)(13,000)/(307) = 56.5 psi > 15 psi
By trial and error, it can be shown that for t = 0.33 in., P = 15 psi.
Large Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to the shell. Then from Eq. (3.49), we calculate the stiffness ratio, k, as
k = 17,500 × 25,000,000/(13,000 × 30,000,000)
= 1.12
The axial loads are given by
QL= PRL/2 – axial equipment load
= –15 × 84.625/2 – [50,000/(2π84.625)] = – 728.7 lb/in.
Next, we must calculate the need for reinforcement in accordance with Table 3.4.
P/SsE1= 15/(17,500 × 0.85) = 0.001
From Table 3.4,∆ = 2.5°. Hence, reinforcement is needed. The amount of reinforcement is calculated from Eq. (3.49):
ArL= (1.12 × 728.7 × 84.625/17,500 × 0.85) {1 – (0.25)[15 × 84.625 + 728.7)/728.7](2.5/28)} tan 28
= 2.32 in.2
AeL= 0.55(84.625 × 0.625)1/2(0.625 + 0.625/cos 28)
= 5.33 in.2
Therefore, the junction is inherently reinforced and no additional rings are required.
Next, determine the required moment of inertia at the large junction needed for external pressure. From Eq. (3.51), and assuming a ring area of 1.0 in2,
ATL= (240)(0.625)/2 + (3.67 × 12)(0.625)/2 + 1.0 = 89.8 in. From Eq. (3.52), M = (84.6252– 60.52)/(3 × 84.625 tan 28) + (240/2) – (84.625 tan 28)/2 = 123.44 From Eq. (3.53), FL= 15 × 123.44 + (50,000/2π84.625)(tan α) = 1901.6 From Eq. (3.54), B = 0.75(1901.6 × 84.625/89.8) = 1344 From Eq. (3.55), A = 2 × 1344/25,000,000 = 1.08 × 10–4 From Eq. (3.57), I′s= 1.08 × 10–4× 84.6252× 89.8/10.9 = 6.37 in.4
A trial run indicates that the available moment of inertia is inadequate without a stiffening ring. Assume that a 2 in. × 1/2 in. ring rolled the hard way will be used. The available moment of inertia is obtained from Fig. E3.13(a) and (b). The neutral axis is at
x1= [(5.64 × 0.6252/2) + (5.64 × 0.625 × 2.65/2)
+ [2 × 0.5(– 1.0)]/[(5.64 × 0.625) + (5.64 × 0.625) + (2 × 0.5)] = (1.10 + 4.67 – 1.00)/(3.53 + 3.53 + 1.0)
x2= 2.65/2 – 0.59 = 0.74 in.
x3= 1 + 0.59 = 1.59 in.
The moments of inertia of the cone about its two major axes are
I_ x= 5.643× 0.625/12 = 9.34 in.4 I_ y= 0.6253× 5.64/12 = 0.12 in.4 I_ x_y= 0.0
The moment of inertia of the cone around an axis y through its centroid is
Iy= Ix_sin2α + Iy_cos2α + 2I_x_ysin αcos α
= 9.34 sin228 + 0.12 cos228 + 0.0 = 2.15 in.4
The total moment of inertia of the composite section = (I of shell about its neutral axis) + (area of shell) (dis- tance to composite section neutral axis)2+ (I of cone about the y axis through its centroid) + (area of cone)
(distance to composite section neutral axis)2+ (I of the stiffener about its neutral axis) + (area of stiffener)
(distance to composite section neutral axis).2
I = 0.6253× 5.64/12 + (5.64 × 0.625)(0.59 – 0.625/2)2+ 2.15
+ (5.64 × 0.625)(0.74)2+ 2.03× 0.5/12 + (2.0 × 0.5)(1.59)2 = 0.12 + 0.27 + 2.15 + 1.93 + 0.33 + 2.53
= 7.33 in.4
Hence, use 2 in. × 1/2 in. ring at the junction.
Small Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from
k = 15,000 × 27,000,000/(13,000 × 30,000,000)
= 1.04
The axial loads are equal to
Qs= PRs/2 – axial equipment load
= –15 × 60.5/2 – [50,000/(2π60.5)] = –585.3 lb/in.
Reinforcement is required in accordance with Eq. (3.58).
Ars= [1.04 × 585.3 × 60.5/(15,000 × 0.85)] tan 28
= 1.54 in.2
From Eq. (3.59), the available area is
Aes= 0.55(2 × 60.5 × 0.5)1/2[(0.5 – 0.4) + (0.625 – 0.33)/cos28]
= 1.86 in.2
Hence, the small junction is inherently reinforced and no additional rings are needed.
Next, determine the required moment of inertia at the small junction needed for external pressure. From Eq. (3.60),
ATs= (120)(0.5)/2 + (3.76 × 12)(0.625)/2 + 0.0
From Eq. (3.61), N = (84.6252– 60.52)/(6 × 60.5 tan 28) + 120/2 – (60.5 tan 28)/2 = 62.06 in. From Eq. (3.62), Fs= 15 × 62.06 + (50,000/2π60.5)(tan 28) = 1000.8 lbs/in. From Eq. (3.63), B = 0.75(1000.8 × 60.5/44.1) = 1030 psi From Eq. (3.64), A = 2 × 1030/27,000,000 = 7.63 × 10–5 From Eq. (3.66), I′s= 7.63 × 10–5× 60.52× 44.1/10.9 = 1.13 in.4
The available moment of inertia is obtained from Fig. E3.13(c) and (d). The neutral axis is at
x1= [(4.26 × 0.52/2) + (4.76 × 0.625 × 2.23/2)]/[(4.26 × 0.5) + (4.76 × 0.625)] = 0.75 in.
x2= 0.365 in.
The moments of inertia of the cone about its two major axes are
I_ x= 4.763× 0.625/12 = 5.62 in.4 I_ y= 0.6253× 4.76/12 = 0.10 in.4 I_ x_y= 0.0
The moment of inertia of the cone around an axis y through its centroid is
Iy= I_xsin2α + I_y_cos2α + 2I_x_ysin α cos α
= 5.62 sin228 + 0.10 cos228 + 0.0 = 1.32 in.4
I = 0.53× 4.26/12 + (4.26 × 0.5)(0.75 – 0.5/2)2+ 1.32 + (4.76 × 0.625)(0.365)2 = 0.04 + 0.53 + 1.32 + 0.4
= 2.29 in.4
Hence, an adequate moment of inertia is available at the junction.