Stainless steel fittings and valve body dimensions
Release Date:
2026-06-18
Stainless steel fittings and valve body dimensions Operating conditions: 1. The material is 304 stainless steel round bar, cold-drawn structural steel, or forgings; 2. The shell is filled with fluid at a pressure of P; 3. The outer surface of the shell is cylindrical.
1. Tensile force F = PπD²/4 2. Effective engagement length of the thread (excluding chamfer): L = 1 + F / (πD4[τ] × 0.35) (L ≥ 3 × pitch) L2 = 1 + F / (πD5[τ] × 0.35) (L2 ≥ 3 × pitch) 3. Optimal thread pitch P = L/5 to L/9 4. Mid-section shell wall thickness S1: Select based on the pressure P and the inner diameter D1, referring to the pressure‑resistant wall thickness of steel pipes. (See page 333 for details.) When the pressure P is ≤160 bar, the minimum wall thickness shall be selected based on P = 160 bar. To ensure safety under external forces. 5. End-shell wall thickness S2: S2 = 1.35 × S1 (for a circular shell) 6. Under the same pressure, the minimum wall thickness of a square or hexagonal tube is smaller than that of a circular tube.
P(MPa)
Fluid pressure inside the housing and fittings
F(N)
Tensile force on the housing and fittings caused by fluid pressure
Allowable tensile stress (see valve material property data for details)
[σy](MPa)
Allowable compressive stress (see valve material property data for details)
[τ](MPa)
Allowable shear stress (see valve material property data for details)
K=A/D
K=1.2～1.4
Square tube and hexagonal tube Cold-drawn steel processing
Square tube
Hexagonal tube
Wall thickness ratio S4/S2 =
0.8
0.9
7. Wall thickness S3 of the thread relief groove: Refer to the housing dimension table. 8. Calculate the compression width A of the step; the compressive stress σy in the step shall satisfy σy = F / compression area ≤ 0.8 × [σy]. 4F/(π(D6-0.5)²-πD3²)≤0.8×[σy] (0.5 mm is the safety margin) A=(D6-D3)/2 9. Calculate the step thickness B (with a minimum value of 3.5 mm), ensuring that the shear stress in the step, τ = F / shear area, does not exceed 0.6 × [τ]. F/(πD³×(B-0.5))≤0.6×[τ] (0.5 mm is the safety margin) 10. Calculate the nut‑clamping thickness C; the bending section modulus is W = πD⁵(C − 0.5)²/6 (with 0.5 mm as a safety margin); the bending moment is M = F × (D₇ − D₃)/2, and M/W ≤ [σL].
Housing Dimensions Table (Unit: mm)
M
M2
Medium and low pressure: P ≤ 160 bar (when the pressure is ≥ 120 bar, the wall thickness in the casting table shall be increased by 20%)
