An acceptable alternative to the regular reinforcement requirements may be used subject to special limita- tions and other reinforcement requirements.
5.6.8.1 Limitations
(a) The reinforcement will have a circular cross section and be perpendicular to shell. (b) The reinforcement will have all integral construction using corner fillet radii.
5.6.8.2 Required Reinforcement Area, Ar. The required minimum reinforcement area related to
d/(Rtr)1/2is:
d/(Rtr)1/2 In Cylinders In Spheres & Heads
<0.20 None except r2required None except r2required
≥0.20 & <0.40 {4.05[d/(Rtr)1/2]1/2– 1.81}dtr {5.40[d/(Rtr)1/2]1/2– 2.41}dtr
≥0.40 0.75dtr dtrcosφ
where φ = sin–1(d/D)
5.6.8.3 Limits of Reinforcing Zone. Metal included in meeting the reinforcing area requirements shall be located in the zone boundary shown in Fig. 5.6.
Example 5.5 Problem
Using the rules of Article D-5 of VIII-2, determine the reinforcement requirements of an 8 in. I.D. nozzle which is centrally located in a 2:1 ellipsoidal head as shown in Fig. E5.5. The nozzle is inserted through the head and attached by a full penetration weld. The inside diameter of the head skirt is 41.75 in. The head material is SA-516 Gr. 70, and the nozzle material is SA-106 Gr. C. The design pressure is 700 psi, and the design temperature is 500°F. There is no corrosion allowance.
LIMITS OF REINFORCING ZONE FOR ALTERNATIVE NOZZLE DESIGN (ASME VIII-2)
FIG. E5.5
Solution
(1) The allowable stress intensity for SA-516 Gr. 70 is 20.5 ksi, and for SA-106 Gr. C it is 21.6 ksi. Since the nozzle material is stronger than the head material, no adjustment is required.
(2) Using Fig. AD-204.1, the minimum required thickness of a 2:1 ellipsoidal head is:
P/S = 700/20,500 = 0.034
which gives
t/L = 0.021
L = 0.9D = 0.9(41.75) = 37.575 in. t = 0.021(37.575) = 0.789 in.
Nominal thickness used is 1.0 in.
(3) Using AD-201(a), the minimum required nozzle thickness is:
tm= PR/(S – 0.5P)
= (700)(4)/[21,600 – 0.5(700)] = 0.132 in.
Nominal thickness used is 1.125 in.
(4) Limits parallel to head surface are
(a) For 100% of the required reinforcement:
X = d or (0.5d + t + tn), whichever is larger
= 8 in. or (4 + 1 + 1.25 = 6.125 in.)
Use X = 8 in.
(b) For 2/3 of required reinforcement:
X′ = r + 0.5(Rmt)1/2 or (0.5d + t + tn), whichever is larger
= [4 + 0.5(38.075 × 1)1/2= 7.085 in.] or 6.125 in.
Use X′ = 7.085 in.
(5) Limits perpendicular to the head surface are calculated as shown below. (a) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by:
(i) h < 2.5tn+ K or (ii) h≥ 2.5tn+ K (b) h = 2.50 in. 2.5tn= (2.5)(1.125) = 2.81 in. K = 0.25 in. 2.5tn+ K = 2.81 + 0.25 = 3.06 in.
(c) Determine the perpendicular limit as follows:
Y = 0.5(rmtn)1/2+ k or 1.73x + 2.5tp+ K, whichever is larger,
but not more than 2.5t nor (L + 2.5tp)
= 0.5[(4.5625)(1.125)]1/2+ 0.25 = 1.383 in. = 0 + 2.5(1.125) + 0.25 = 3.063 in. = 2.5(1) = 2.5 in.
= (4) + 2.5(1.125) = 6.813 in. = 2.5 in.
(6) Reinforcement area required according to AD-520 of VIII-2 is:
Ar= dtrF
= (8)(0.789)(1.0) = 6.312 in.2 2/3Ar= 2/3(6.312) = 4.208 in.2
(7) Using the 100% limit, reinforcement area available in the head is
A1= (t – tr)(2X – d)
= (1.0 – 0.789)(2 × 8 – 8) = 1.688 in.2
(8) Reinforcement area available in the nozzle is
A2= 2Y(tn– trn)
= 2(2.5)(1.125 – 0.132) = 4.965 in.2
(9) With the 100% limit, total reinforcement area available in the head and nozzle is:
AT= A1+ A2
= 1.688 + 4.965 = 6.653 in.2
Area available of 6.653 in.2is larger than area required of 6.312 in.2
(10) With the 2/3 limit, reinforcement area available in the head is:
A1= (t – tr)(2X′ – d)
= (1.0 – 0.789)(2 × 7.085 – 8) = 1.302 in.2
(11) With the 2/3 limit, total reinforcement area available in the head and nozzle is
Example 5.6 Problem
Using the rules of VIII-2, determine the reinforcement requirements for a 12 in. × 16 in. opening for the manway shown in Fig. E5.6. The manway forging is inserted through the vessel wall and attached by a full penetration weld. The 12 in. dimension lies along the longitudinal axis of the vessel. The manway cover seals against the outside surface of the manway forging. The I.D. of the shell is 41.875 in. The shell material is SA-516 Gr. 70, and the manway forging is SA-105. The design pressure is 700 psi, and the design temper- ature is 500°F. There is no corrosion allowance.
Solution
(1) The allowable stress intensity for SA-516 Gr. 70 is 20.5 ksi, and for SA-105 it is 19.4 ksi. An adjustment of fr= 19.4/20.5 = 0.946 is required for A2.
(2) Using AD-201(a), the minimum required thickness of the shell is:
tr= (PR)/(SE – 0.5P)
= [(700)(20.9375)]/[(20500)(1.0) – 0.5(700)] = 0.728 in.
Nominal thickness used is 1.0 in.
(3) The manway forging is elliptical. Since there are no equations for determining the minimum required thickness of an elliptical shell in VIII-2, one needs to be located. For an elliptical shell, Eq. (2.41) for minimum required thickness is given in section 2.6.2. The maximum value of the minimum required thickness is used for all planes being examined as follows:
FIG. E5.6
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT
trn= Pa2b2/SE(a2sin2Φ + b2cos2Φ)3/2
= [(700)(8)2(6)2]/{(19,400)(1.0)[(8)2(1)2+ (6)2(0)2]3/2} = 0.385 in.
Nominal thickness used is 1.0 in.
(4) Examination of the longitudinal plane:
(a) Determine the limit parallel to the shell surface. (i) For 100% of reinforcement area required:
X = d or (0.5d + t + tn), whichever is larger
= 12 in. or (6 + 1 + 1 = 8.0 in.)
Use X = 12 in.
(ii) For 2/3 of reinforcement area required:
X′ = r + 0.5(Rmt)1/2 or (0.5d + t + tn), whichever is larger
= [6 + 0.5(21.4375 × 1)1/2] = 8.315 in. or 8.0 in.
Use 8.315 in.
(b) Determine the limit perpendicular to the shell surface.
(i) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by
(a) h < 2.5tn+ K or (b) h≥ 2.5tn+ K (ii) h = 2.50 in. 2.5tn= (2.5)(1.0) = 2.50 in. K = 0.25 in. 2.5tn+ K = 2.75 in.
Limit (5)(a)(i)(a) applies where h < 2.5tn+ K
(iii) Determine the perpendicular limits
Y = 0.5(rmtn)1/2+ K or 1.73x + 2.5tp+ K, whichever is larger,
but not more than 2.5t nor (L + 2.5tp)
= 0.5(6.625 × 1.0)1/2+ 0.25 = 1.537 in. = 0 + 2.5 × 1.0 + 0.25 = 2.75 in. = 2.5(1) = 2.50 in.
= 3 + 2.5(1.0) = 5.50 in. = 2.50 in.
(c) Reinforcement area required is For 100% reinforcement:
Ar= dtrF + 2tntr(1 – fr)
= (12)(0.728)(1) + 2(1.0)(0.728)(1 – 0.946) = 8.814 in.2
For 2/3 A: 2/3Ar= 5.876 in.
(d) Using the 100% limit, reinforcement area available in the shell is:
At= (t – tr)(2X – d)
= (1 – 0.728)(2 × 12 – 12) = 3.264 in.2
(e) Reinforcement area available in the nozzle wall is:
Outward: A21= 2(2.5)(1.0 – 0.385)(0.946) = 3.075 in.2 Inward: A22= 2(2.5)(1.0)(0.946) = 4.730 in.2
(f) Total reinforcement area available from the shell and nozzle is:
AT= A1+ A21+ A22
= 3.264 + 3.075 + 4.730 = 11.069 in.2
Area available of 11.069 in.2is larger than area required of 8.814 in.2
(g) Using the 2/3 limit, reinforcement area available in the shell is
A1= (1 – 0.728)(2 × 8.315 – 12) = 1.259 in.2
(h) With the 2/3 limit, total reinforcement area available from the shell and nozzle is
AT= 1.259 + 3.075 + 4.730 = 9.064 in.2
Area available of 9.064 in.2is larger than area required of 5.876 in.2
(5) Examination of circumferential plane:
The opening has a 16 in. dimension on this plane, but F = 0.5. (a) Reinforcement area required according to AD-520 of VIII-2 is
Ar= dtrF = (16)(0.728)(0.5) = 5.824 in.2
(b) Total reinforcement area available from the shell and nozzle is
AT= >9.064 in.2
Area available of 9.064 in.2is larger than area required of 5.824 in.2Since area is acceptable without any
adjustment of limits, the design is satisfactory.