Disc spring calculations

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This document provides specifications for two disc spring designs. It includes dimensions, material properties, calculations of theoretical spring force and stresses for different load cases. Key details include: - Design 1 has an outside diameter of 63mm, inside diameter of 31mm, thickness of 3.5mm, and is made of a material with Young's modulus of 206,000 N/mm^2. - Design 1 can withstand a maximum force of 15,030N over a stroke of 23.1mm when using 22 springs in series. - Design 2 has diameters of 28mm and 14.2mm, thickness of 1mm, and can provide 617.58N of force over a stroke of 0.

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DISC SPRING DESIGN:
Young’s modulus, E = 206000 N/mm2
Poisson’s ratio, ? = 0.3
Outside diameter of disc spring, De = 63 mm
Inside diameter of disc spring, Di = 31 mm
Ratio, De/Di = 2.032
? =
i
e
D
D
K1 = (?-1)
? x 1
(?+1) _ 2 ?
(?-1) ln (?)
K1 = 0.701
K2 = 6 x ? (?-1) _ 1 ?
ln (?)
? x ln (?)
K2 = 1.227
K3 = 6 x ? (?-1) _ 2 ?
ln (?)
? x ln (?)
K3 = 1.39
t = 3.5 mm
S = 1.05 mm
ho = 1.4 mm
F = 4 x E x S x t__ x (ho2 + t2 + 0.5S2 –1.5ho. S)
(1-?2) x K1 x De2

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F = 1.503x104 N
For S = ho The theoretical force
Spring force
Fc = 4 x E x hox t3
(1-?2 ) x K1 x De2
Fc = 1.955x104 N
Stresses
?1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t))
(1-?2) x K1 x De2 2
?1 = -2.03x103 N/mm2
?2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t))
(1-?2) x K1 x De2 2
?2 = 1.296x103 N/mm2
?3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di
(1-?2) x K1 x De2 2 De
?3 = 1.047x103 N/mm2
?4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di
(1-?2) x K1 x De2 2 De
?4 = -1.112x103 N/mm2
Spring are in series
Number of springs, n = 22
Available stroke, Sa = Sxn
Sa = 23.1 mm
Available spring force for Sa, F = 1.503x104 N
Total stroke, St = 6 mm

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5 of 7
Deflection per spring, S1 = St/n
S1 = 0.273 mm
E = 206000 N/mm2
? = 0.3
De = 28 mm
Di = 14.2 mm
De/Di = 1.972
? = De/Di
K1 = (?-1)
? x 1
(?+1) _ 2 ?
(?-1) ln (?)
K1 = 0.689
K2 = 6 x ? (?-1) _ 1 ?
ln (?)
? x ln (?)
K2 = 1.213
K3 = 6 x ? (?-1) _ 2 ?
ln (?)
? x ln (?)
K3 = 1.367
t = 1 mm
ho = 0.8 mm
F = 4 x E x S1 x t__ x (ho2 + t2 + 0.5S12 –1.5ho.S1)
(1-?2) x K1 x De2
F = 617.582 N

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For S1 = ho The theoretical force
Spring force
Fc = 4 x E x hox t3
(1-?2 ) x K1 x De2
Fc = 1.342x103 N
Stresses
?1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t))
(1-?2) x K1 x De2 2
?1 = -993.705 N/mm2
?2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t))
(1-?2) x K1 x De2 2
?2 = 256.939 N/mm2
?3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di
(1-?2) x K1 x De2 2 De
?3 = 551.22 N/mm2
?4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di
(1-?2) x K1 x De2 2 De
?4 = -319.965 N/mm2

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About Practice Head is assembled with Practice Torpedo intended for carrying out exercise firings. It is assembled with Homing Head in the forward section and oxygen flask in the rear section. Practice Head imparts positive buoyancy to the Torpedo at the end of run. The Practice Head is divided into two compartments viz. Ballast Compartment (Houses Light Device, Depth & Roll Recorder, Signal Flare Ejector, Discharge Valve, Stop Cock, Water discharge Valve, Bellow reducing Valve, Release Mechanism, Recess, Bypass Valve, Pressure Equalizer, Float, Sinking Plug etc.) which provides positive buoyancy at the end of run by discharging water (140 ltrs.) filled in the compartment and Instrument compartment (dry), houses (safety & recovery unit and its battery, combined homing and influence exploder equipment, noise maker, bollards & safety valve etc.) The recess in Ballast compartment houses the float which gets inflated at the end of run to provide floatation to the surfaced Torpedo. Several hand holes/recesses are provided on the casing/shell of Practice Head for assembly of the following components:- a) Signal Flare Ejector Assembly b) Depth and Roll Recorder Assembly c) Light Device d) Pressure equalizer e) Drain/Discharge Valve assembly f) Bollard Assembly g) Holding for Floater/Balloon Assembly h) Sinking Valve i) Safety Valve j) Inspection hand hole Technical Details: SrNo Items Specifications 1 Aluminum Alloy (AlMg5) Casing Body Material: AlMg5 ? Larger Outer Diameter of the Casing: 532.4 MM ? Smaller Outer Diameter of the Casing: 503.05 MM ? Total Length: 1204.20 MM ? Thickness: 6-8 mm ? Structural Details of Casing: The casing is of uniform outer dia for a certain distance from rear side and tapered from a definite distance to the front side. (Refer T-DAP-A1828-GADWG-PH- REV 00) ? Slope of the Tapered Portion: 1/8 ? Mass of Casing (Without components mounting, but including the ribs and collars on the body): 58.5 kg ? Maximum External Test Pressure: 12 kgf/cm2 ? Maximum Internal Test Pressure:- i. For Ballast Compartment: 2 kgf/cm2 ii. For Instrument Compartment: 1 kgf/cm2 ? Innerspace of casing assembly have 2 compartments:- i. Ballast Compartment and ii. Instrument Compartment ? Cut outs/ recesses shall be provided for the assembly of following components. a) Signal Flare Ejector Assembly b) Depth and Roll Recorder Assembly c) Light Device d) Pressure Equalizer e) Drain/ discharge valve assembly 2 Front Side Collar Material: AlMg5 ? Maximum Outer Diameter: 500 MM ? Pitch Circle Diameter: 468 MM ? All Dimensions as per drawing T-DAP-A1828-MDWG-C&R-REV-00 Application: In a torpedo, the ballast components and instrument compartment play crucial roles in maintaining stability, control, and overall operational effectiveness. The ballast system primarily manages buoyancy and trim, ensuring that the torpedo maintains a stable trajectory underwater.

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Disc spring calculations

1. Multifacet Technical Center Pvt. Ltd. (MTCPL) #F810, “Vivekadhama”, Bharthnagar Off Magadi road, Bangalore – 91 Ph: 080-32946374 fax: 080-23588044 Email: Multifacet@dataone.in, mtc.bng@multifacet.net & mtc.bng@gmail.com Web: www.multifacet.net P.O No. - Report No. - Date: - Prepared by - Checked by KSM Approved by BVM Document Change Note: Total pages: -- Any previous copies of this document issued to the customer, must be destroyed if not of this current revision.Rev. No By Description Appvd Date - Customer: -

2. Title: - - Report No.: - Rev. No.: - 2 of 7 Disclaimer: ? The information contained is specific to the technical data mentioned in this report. ? No document will be provided to customer in any editable format. ? MTCPL is not responsible for any changes made to the report by any means and any consequences arising as a result of manipulation of data. ? Interpretation of the information contained herein should be based on complete report and not on parts thereof. ? This report is a property of MTCPL and should not be copied either directly or indirectly without a written consent from the company. ? Disputes, if any, will be within the Jurisdiction of Bangalore only.

3. Title: - - Report No.: - Rev. No.: - 3 of 7 DISC SPRING DESIGN: Young’s modulus, E = 206000 N/mm2 Poisson’s ratio, ? = 0.3 Outside diameter of disc spring, De = 63 mm Inside diameter of disc spring, Di = 31 mm Ratio, De/Di = 2.032 ? = i e D D K1 = (?-1) ? x 1 (?+1) _ 2 ? (?-1) ln (?) K1 = 0.701 K2 = 6 x ? (?-1) _ 1 ? ln (?) ? x ln (?) K2 = 1.227 K3 = 6 x ? (?-1) _ 2 ? ln (?) ? x ln (?) K3 = 1.39 t = 3.5 mm S = 1.05 mm ho = 1.4 mm F = 4 x E x S x t__ x (ho2 + t2 + 0.5S2 –1.5ho. S) (1-?2) x K1 x De2

4. Title: - - Report No.: - Rev. No.: - 4 of 7 F = 1.503x104 N For S = ho The theoretical force Spring force Fc = 4 x E x hox t3 (1-?2 ) x K1 x De2 Fc = 1.955x104 N Stresses ?1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t)) (1-?2) x K1 x De2 2 ?1 = -2.03x103 N/mm2 ?2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t)) (1-?2) x K1 x De2 2 ?2 = 1.296x103 N/mm2 ?3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di (1-?2) x K1 x De2 2 De ?3 = 1.047x103 N/mm2 ?4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di (1-?2) x K1 x De2 2 De ?4 = -1.112x103 N/mm2 Spring are in series Number of springs, n = 22 Available stroke, Sa = Sxn Sa = 23.1 mm Available spring force for Sa, F = 1.503x104 N Total stroke, St = 6 mm

5. Title: - - Report No.: - Rev. No.: - 5 of 7 Deflection per spring, S1 = St/n S1 = 0.273 mm E = 206000 N/mm2 ? = 0.3 De = 28 mm Di = 14.2 mm De/Di = 1.972 ? = De/Di K1 = (?-1) ? x 1 (?+1) _ 2 ? (?-1) ln (?) K1 = 0.689 K2 = 6 x ? (?-1) _ 1 ? ln (?) ? x ln (?) K2 = 1.213 K3 = 6 x ? (?-1) _ 2 ? ln (?) ? x ln (?) K3 = 1.367 t = 1 mm ho = 0.8 mm F = 4 x E x S1 x t__ x (ho2 + t2 + 0.5S12 –1.5ho.S1) (1-?2) x K1 x De2 F = 617.582 N

6. Title: - - Report No.: - Rev. No.: - 6 of 7 For S1 = ho The theoretical force Spring force Fc = 4 x E x hox t3 (1-?2 ) x K1 x De2 Fc = 1.342x103 N Stresses ?1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t)) (1-?2) x K1 x De2 2 ?1 = -993.705 N/mm2 ?2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t)) (1-?2) x K1 x De2 2 ?2 = 256.939 N/mm2 ?3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di (1-?2) x K1 x De2 2 De ?3 = 551.22 N/mm2 ?4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di (1-?2) x K1 x De2 2 De ?4 = -319.965 N/mm2

7. Title: - - Report No.: - Rev. No.: - 7 of 7

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Disc spring calculations

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