EM DRIVE TESTING in CAVENDISH PENDULUM:

PROOF that CHANGE in WIRE's CROSS-SECTIONAL AREA is NEGLIGIBLE for CAVENDISH TORSIONAL PENDULUM'S MUSIC WIRE FOR LOADS that RESULT in STRESS EQUAL TO OR BELOW YIELD STRESS



I had promised to post this proof that the concern expressed by some that the change in area was relevant is not borne by the facts.



Of course, for those knowledgeable about mechanics of materials, this is obvious, since it is well-known to Engineers that design mechanical and aerospace structures with metals, that elastic strains in metals are infinitesimal and hence such changes in diameter due to longitudinal strain are negligible. I proceed to prove this for this case, and provide specific numerical examples for Monomorphic's rig.



Since some people expressed a concern with change in area, this is a concern about the strains not being infinitesimal, hence we should use finite strain and finite stress measures to explore the validity of this concern. Therefore I use the "true strain" (Hencky or Logarithmic measure) and the "true stress" (Cauchy measure: force divided by actual present cross sectional area).



Transverse strain



Since the torsional wire is made of steel, the material is isotropic. Using Cartesian coordinates, with z being the longitudinal direction of the wire, and x and y are the transverse directions, the transverse (Hencky or logarithmic measure of) strains ε x in the x direction and ε y in the y direction are:



ε x = ε y = ln (D/Do)



where Do is the undeformed diameter of the wire and D is the deformed diameter of the wire



therefore the ratio of the deformed to the undeformed diameter of the wire



D/Do = eε x = eε y



Area ratio



The ratio of the present cross-sectional area "A" of the wire to the original cross-sectional area "Ao" of the wire is:



A/Ao = (D/Do)2



substituting the expression for the diameter ratio in terms of the transverse strains:



A/Ao = e2ε x = e2ε y



Longitudinal strain



From the Poisson's ratio "ν" (Greek letter nu) definition, the longitudinal strain is:



ε x = ε y = - ν ε z



substituting this equation into the expression for the area ratio, one gets an expression for the ratio A/Ao of the deformed to undeformed cross-sectional area as a function of the longitudinal strain ε z :



A/Ao = e- 2 ν ε z



Yield stress



Using the "true stress" (Cauchy stress: ratio of force to present cross-sectional area) value of the yield stress:



σ z = E ε z



where E=modulus of elasticity. Therefore the yield stress is related to yield strain as follows:



σ z YIELD = E ε z YIELD



and therefore the ratio of the cross-sectional area deformed at yield, to the undeformed cross-sectional area is:



A/Ao = e- 2 ν σ z YIELD /E





Torsional stiffness



The torsional stiffness "k" of the wire is defined as follows, as the constant relating the torsional moment to the angle of rotation of the torsional pendulum:



T = k θ



T=Torsional moment (Newton-meter or pound force-inch)



θ= angle of rotation (radians)



k = torsional spring stiffness (Newton-meter per radian or inch-pound force per radian)



where



k = (Pi * D4/(32 L)) E/(2(1+ν))



L= length of torsional wire between support and hanging mass



Therefore, expressing this in terms of the stress-free diameter Do of the wire:



k = (Pi * Do4/(32 L)) E/(2(1+ν)) (D/Do)4



and since it was previously shown that:



A/Ao = e- 2 ν σ z YIELD /E

= (D/Do)2



it follows that



k = (Pi * Do4/(32 L)) E/(2(1+ν)) (D/Do)4

= ko e- 4 ν σ z YIELD /E



where "ko" is the torsional stiffness based on the original stress-free diameter: the torsional stiffness for infinitesimally small load



ko= (Pi * Do4/(32 L)) E/(2(1+ν))



and the ratio of the actual torsional stiffness at yield to the torsional stiffness under no load is simply:



k/ ko = e- 4 ν σ z YIELD /E



FREQUENCY



The angular frequency is



ω = √(k/I)



and therefore the natural frequency of the torsional pendulum (analyzed as a single degree of freedom) is:



f =(1/(2 π)) √(k/I)



where



I = moment of inertia of suspended mass



Expressing this in terms of the torsional stiffness ko for the undeformed wire (under no load):



f =(1/(2 π)) (√(ko/I)) ( √(k/ko) )



since the natural frequency of the torsional pendulum, calculated on the basis of the undeformed wire is:



fo = (1/(2 π)) (√(ko/I) )



using this, and expressing the ratio of the torsional stiffness k under load to the torsional stiffness ko for the undeformed wire (under no load), the following follows:



f =fo e- 2 ν σ z YIELD /E



where it is assumed that the moment of inertia of the wire is negligibly small compared with the moment of inertia of the suspended mass (assumed to remain constant during the test)



Numerical Example



Numerical values are shown first in US units instead of SI units because Monomorphic is in the USA and has expressed the weight of his rig in pounds force. Values are converted to SI units, for international purposes.



Using the actual value for music wire chosen by Monomorphic's torsional pendulum:



Music wire ASTM A228

σ z YIELD = 420 MPa = 60,916 psi (Yield stress)

E = 207 GPa = 30*106 psi (Young's modulus of elasticity)

ε z YIELD =σ z YIELD / E = 0.00203 = 0.203% (Yield strain)

ν = 0.3 (Poisson's ratio)



Substituting these values into the expression for the area ratio at yield:



A/Ao = e- 2 * 0.3 * 0.00203



= 0.99878242 The change in cross-sectional area at yield being only - 0.12%



natural frequency at yield



Using the previous expression, it is obvious that the frequency decreases in direct proportion to the decrease in cross-sectional area



f =fo (A/Ao)



=fo e- 2 ν σ z YIELD /E



and therefore, at yield:



f =fo e- 2 * 0.3 * 0.00203



= fo 0.99878242 The natural frequency at yield is lower by - 0.12%



where it is assumed that the moment of inertia of the wire is negligibly small compared with the moment of inertia of the suspended mass (assumed to remain constant during the test)



torsional stiffness ratio at yield



k/ ko = e- 4 ν σ z YIELD /E

= e - 4 * 0.3 * 0.00203

= 0.99756696 The torsional stiffness at yield is lower by - 0.24%



change in diameter at yield:



D/Do = √ (A/Ao) = 0.99939119



ε x = ε y = ln (D/Do) = - 0.00060900 = -0.06%



(D - Do)/Do = - 0.00060897 = - 0.06 %



Notice: since the transverse strain measure on the diameter (using the Engineering Strain measure) is only 0.06%, this strain is infinitesimal, and hence it was not necessary to use finite strain or finite stress measures, as the strain on the cross-section is so small that at these values the stress measures are equivalent (True Stress ~ Engineering Stress and True Strain ~ Engineering Strain).



Actual Diameter of music wire used by Monomorphic



Do = 0.033 inches = 0.838 mm



Change in diameter:

D - Do = - 0.00060897 * 0.033 inches = - 0.00002010 inches

= - 0.5105 μm





Tolerance on diameter of music wire used by Monomorphic



Music Wire ASTM A228

Minimum tolerance on diameter = 8 μm = 0.0003 inches (another supplier had a greater tolerance of 0.0004 inches)



Therefore, the ratio of the change in diameter to the wire's tolerance in diameter, for a load high enough to result in yield stress (permanent deformation threshold) is:



(Change in diameter)/(Tolerance on diameter) = - (0.00002010 inches)/(0.0003 inches) = 6.7%



Therefore, the change in diameter of the wire, even at a stress high enough to result in incipient yield of the material, is only 6.7% of the tolerance in diameter. Hence the change in diameter, and the change in cross-sectional area, for stresses below the yield limit is negligible, and concerns about change in diameter are ill-founded, since the dimensional tolerance changes of the wire itself are 15 times greater.



Calculation of Maximum Load for Monomorphic's wire



σ z YIELD = 420 MPa = 60,916 psi



Longitudinal Force = F z



σ z YIELD = 60,916 psi = F z / A = ( F z / Ao ) *(Ao/A)



60,916 psi = ( F z / (Pi*(Do/2)2) ) *(Ao/A)



= ( F z / (Pi*(0.033 inches/2)2) ) /0.99878242



F z = 60,916 psi *0.99878242 * (Pi*(0.033 inches/2)2)

= 52.038 lbf

= 231.476 Newtons

= 23.6040 kg force



The maximum load, the load that will result in permanent yield, is 52.038 pound force (23.604 kgf)



Due to the fact that the change in diameter, and hence the change in cross-sectional area is negligible, this load could also have been calculated for engineering purposes, based on the initial, unloaded, cross-sectional area:



F z = 60,916 psi * (Pi*(0.033 inches/2)2)

= 52.101 lbf

=231.758 Newtons

=23.6328 kg force



The difference due to the change in cross-sectional area, being only 0.12%



Since Monomorphic's rig is only 25 lbf (111 Newtons or 11.3 kg force), it is less than 1/2 of the weight that would result in permanent deformation of the torsional wire



The change in diameter of the music wire is completely negligible, particularly in comparison with the manufacturing tolerance of the music wire itself and with respect to the variation in yield stress of the wire itself.