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Measurement Methods and Calculations to Determine Internal Deposit Stress Frank H. Leaman Specialty Testing & Development, Inc. York, PA.

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Presentation on theme: "Measurement Methods and Calculations to Determine Internal Deposit Stress Frank H. Leaman Specialty Testing & Development, Inc. York, PA."— Presentation transcript:

1 Measurement Methods and Calculations to Determine Internal Deposit Stress Frank H. Leaman Specialty Testing & Development, Inc. York, PA

2 Methods for Deposit Stress Determination Bent Strip (simple beam theory) Spiral Contractometer

3 Simple Beam Theory Where: N = Thickness of the plated coating (inches) T = Thickness of the test strip (inches) D = Deflection of the strip due to bending (inches) L = Length of the test section (inches) E = Modulus of elasticity of the test strip (lb/in 2 ) I = Moment of inertia of the cross section of test strip about its neutral axis S = Stress in plated layer (lb/in 2 ) Then: S = 4E (N+T) D 3NTL

4 Bent Strip Method (Initial Approach) During the application of a coating, one end of the test piece is held in a fixed position and the other end is free to move. It is difficult to measure the value for D.

5 Bent Strip Method (Different Approach) A test piece split into two legs spreads outward due to the deposit stress The deflection is easily read by placing the test piece over a scale Calculate the deposit stress value by using a simple formula

6 Simple Beam Tensile and Compressive Stress Tensile Compressive

7 Compressive and Tensile Stress CompressiveTensile

8 Stress Evaluation Using the Bent Strip Method Test Strip in a Plating Cell

9 In-Site 1 Plating Cell Ideal for small solution volumes and lab studies, particularly when working with precious metals

10 Bent Strip Test Piece Measuring Stand

11 Stress Evaluation Using the Bent Strip Method Bent Strip Plating Test Cell

12 Test Strip Plating Cell with Accessories

13 Typical Deposit Stress Evaluation Plating Set-Up

14 Deposit Stress Calculations for Test Strips The Stoney Formula:  = E T² M δ  3 L² t E = Modulus of elasticity of the substrate = 120,655 kg/cm². T = Thickness of the substrate in millimeters = 0.05077 mm. δ = 1/2 the distance between the test strip leg tips in mm. Example: 0.540 inch spread ÷ 2 x 25.385 mm/inch = 6.85 mm.  = Stress in megapascals, MPa. Note: MPa x 145 = PSI. L = Length of substrate on which the deposit is applied in mm. For Deposit Stress Analyzer test strips, this value is 76.2 mm. t = Deposit average thickness in millimeters. M = Correction for modulus of elasticity difference between the deposit and substrate: M = EDeposit ÷ ESubstrate = 206,900 ÷ 120,690 = 1.714  = E (.05077 mm) ² M ( δ mm) = mm³= MPa 3(76.2 mm)²(.002538 mm) 44.21 mm³ Deposit Stress in PSI = MPa x 145 = PSI Note: MPa is Megapascals, kg/cm.²

15 Spiral Contractometer Existing Design The test piece is a spiral. One end of the spiral is held, other end is free to move. As the free end moves, a dial registers the movement in degrees. The stress of the coating can be calculated.

16 Spiral on an Existing Contractometer

17 Spiral Plated on Existing Type Contractometer for Target Nickel Deposit Thickness of 500µ” in a Semi-bright Bath after 20 Seconds Wood’s Nickel Strike Deposit Location Thickness, µ” Outside Surface410 Inside Surface 85 Deposit stress over a 2 minute strike = 26.4% less than the New Design Contractometer result

18 New Spiral Properties New design spirals are constructed from 0.010 inch thick stainless steel and have a precise surface area of 13.57 in 2. Spirals mount on the contractometer in a way that the entire spiral plates from end to end and deposition of metal on the inside of spirals is minimal even if they are void of a masking material. The average test deposit thickness is 500 microinches.

19 Properties and Plating Conditions for Spiral Contractometer Tests Spiral MaterialStainless Steel Spiral Surface Area, in 2 13.57 Square Feet0.0942 Amps per square foot30 Amps2.90 Stock Thickness, inches0.010 Avg. Deposit Thickness, µ”500 Plating Time, Minutes21 Solution Temperature140° ± 1° F

20 A new geometry solves problems related to an exposed interior that allows deposition of the applied deposit to occur on the inside surface. Interior deposits reverse the type of stress and reduce calculated results as much as 30%. Interior masking is critical.

21 The new design provides masking of the interior surface by geometry and enables spirals to be plated tip to tip so the plated surface area is a constant value. Other advantages: Stainless steel inserts 30% glass filled nylon construction which prevents thread damage and spiral slipping More accurate results Saves time

22 Spiral Contractometer Equipment to Determine Internal Nickel Deposit Stress Spiral Contractometer with calibration weights, support stand and spiral test pieces. Container 4” diameter and 10” height for nickel strike anode basket and bath (1750 ml) Titanium Mesh Anode Basket 3.5” outside and 2.25” inside diameter, 8” high with support contact tabs and cover for Wood’s nickel strike Titanium Mesh Anode Basket 5” outside and 4” inside diameter with support contact tabs and cover for the plating bath Nickel anode buttons to fill the anode baskets Pyrex beaker 4000 ml for a nickel plating bath Support stand – designed to perfectly center over beaker Magnetic stirrer hot plate, 115 volt Digital temperature Controller pre-wired with probe to control ± 1 0 F Power Supply constant current, constant voltage, 0-5 amps, 0-30 volts Magnetic stirrer hot plate, 115 volt

23 Contractometer Stand, Anode Basket & Beaker

24 Contractometer Plating Set-Up

25 Data Recording for Spiral Contractometer Tests Deposit weight in grams:________ Kc degrees:________ Kt degrees:________ Degrees deflection caused by the deposit:________ Spiral weight in grams:________ Deposit weight in grams by subraction:________ Deposit thickness in microinches:________

26 Average Deposit Thickness Calculation in Inches T = _________W____________ = Inches D (87.55 cm 2 ) (2.54 cm/inch) W = Grams of nickel D = Density of nickel = 8.90 g/cm 3, and T = Deposit thickness in inches For the new spirals plated on the new design contractometers, the constant spiral plated surface area is 13.57 in 2 and the following shortened formula applies: T = W = Inches 1979.2

27 Calculating Deposit Stress Stress = 13.02 (D) (M) ÷ w x d = PSI D = Degrees caused by the deposit, M = Modulus of Elasticity of the deposit ÷ that of the substrate = 206,897 ÷ 198,186 = 1.044 for nickel deposits over new spirals that are 0.010 inch thick, w = degrees K t from spiral calibration if the stress is tensile or degrees Kc if the stress is compressive, and d = Deposit thickness in inches. Calculation Example: S = 13.02 (26) (1.04897) ÷ 33 (0.000536) = 20,073 PSI

28 Modulus of Elasticity Values Stock Material Cu-Fe Alloy Ni –Fe Alloy Ni-Fe Alloy Pure Ni ES*120,690 144,830179,310 206,900 Stock Thickness, in0.0020 0.0015 0.0010 0.0010 Metal ED** Values for M*** Cadmium 31,7200.263 0.219 0.177 0.153 Chromium 248,280 2.06 1.71 1.39 1.20 Cobalt 206,8971.72 1.43 1.15 1.00 Copper 117,2400.971 0.810 0.654 0.567 Gold 74,4800.617 0.514 0.415 0.360 Nickel 206,9001.71 1.42 1.14 1.00 Platinum 146,9001.22 1.02 0.819 0.710 Rhodium 289,6502.40 2.00 1.62 1.400 Silver 75,860 0.629 0.524 0.423 0.367 Zinc 82,7600.686 0.571 0.462 0.400 ES*, modulus of elasticity of substrate material in the Stoney Formula. ED**, Modulus of elasticity of deposit for use in modified Deposit Stress Analyzer and Stoney formulas. M***, modulus of elasticity of deposit ÷ modulus of elasticity of substrate for deposit stress determinations using the modified Deposit Stress Analyzer and Stoney Formulas.

29 A Frequent Mistake in Test Procedure 1 2 3 Deposit Thickness To Stock Ratio1:201:201:5 Stock Thickness, Inches0.0100.0020.002 Deposit Thickness, µ Inches500500100 Minutes Plated20420 Current Density, ASF303030 Deposit Stress, PSI14,06014,1276,865 Note: Extra thick deposits of the harder metals increases the degree of stiffness which results in lower proportional test strip spread. Spiral Test Strips

30 Formulas for Bent Strip with One End Stationary* Bent Strip Stress Curve For the comparison of equations that follow that apply to calculating the internal deposit stress of applied metallic coatings over various substrate materials, the value of U = 8.5 units = 0.780 inch will consistently be used as a basis. It will be noted that the calculated internal deposit stress values vary from equation to equation, particularly where the equation fails to address Modulus of Elasticity differences between the substrate and the deposit. Relationship between δ and Z. Example: For a given test strip, U = 8.5 units = 0.780 inch, and δ = U in inches x 25.385 mm/inch ÷ 2, so in this case δ = 9.90 mm. δ = 4Z Z = δ ÷ 4 L = 76.155 mm Using δ = 9.900 mm, Z = 9.90 mm ÷ 4 = 2.475 mm R = L² + 4Z² = 5824.1 = 303.34 mm 8Z19.2 *Note: These formulas only work for bent strip applications and are not applicable for the spiral contractometer method.

31 Stoney Formula Without and With Correction for Modulus of Elasticity Differences Between the Deposit and the Substrate Example: For a Cu-Fe test strip, U = 8.5 units = 0.780 inch δ = U in inches x 25.385 mm/ inch ÷ 2 = 9.900 mm WITHOUT σ = 4ET²Z = ET² δ = 91.137 MPa = 13,214.9 PSI 3L²t3L²t L = test strip plating length = 76.2mm, T = Stock thickness = 0.05077mm and t = Deposit thickness = 0.000075 inch = 0.001904mm WITH M = Edeposit ÷ Esubstrate = 206900 ÷ 120690 = 1.714 σ = ET² δM = 120690(0.05077)²(9.900mm)(1.715) = 156.30 MPa 3 L² 3(76.2mm)²(0.00194mm) σ = MPa (145 PSI/MPa) σ = 22,663.5 PSI

32 Other Bent Strip Formulas to Determine Internal Deposit Stress in Applied Metallic Coatings Barklie and Davies Formula σ = ET² 6Rt (1 – t/T) Heussner, Balden and Morse Formula σ = 4ET²Z 3t (T + t) L Brenner and Senderoff Formulas σ = ET(T+ t) = Edeposit ÷ Esubstrate 6Rt σ = E (t + T)³ 3Rt (2T + t)

33 Brenner and Senderoff Formula for Bent Strip Applications Brenner and Senderoff Formula σ = ET(T+ t) = Edeposit ÷ Esubstrate = 1.714 6Rt σ = 120690 MPa (.05077mm)(.05077mm +1.714(.001904mm) = 95.538 MPa 6(303.34mm)(.001904mm) 95.538 MPa x 145PSI/MPa = 13,853 PSI Note: This formula doesn’t correct for large differences in Modulus of Elasticity values. The uncorrected Stoney result was 13,215 PSI. To be correct, this Brenner and Senderoff formula requires modification as follows: σ = ET ² = 22,310 PSI where = 1.714 6Rt

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