pobieranie: Receptances_App
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APPENDIX 1: Addition of a system at a remote co-ordinate
When the main system B is a real machine it is rarely possible to attach another system C at the point on B where the exciting force is applied. Also the relevant displacement will often be at the same position but in a different direction. It follows that the cross receptance b1 2 is the required receptance where 1 and 2 refer to the same position but a different direction. Also 1 and 2 may be defined as a relative displacement and a relative force (ie. equal and opposite forces) without changing the theory of system addition. It is thus necessary to derive the equation for α1 2 for a combined system A when the additional system C has been attached away from the cutting position, ie. at some co-ordinate 3. This situation is shown in Figure 3 with the associated separation into individual systems to enable the analysis to be carried out. The forces and displacements introduced at 3 are such that the separated systems behave in the same way as when joined.
For system C by definition X c3 = γ 33Fc3 ...................................................... (9.27)
For system B since two forces are applied the displacement at any co-ordinate will be the sum of the displacements caused by each force. ie. superposition is assumed to apply. Thus X b3 = β33F b3 + β 32F2 .............................................. (9.28) and X1 = β 12F2 + β13F b3 ................................................. (9.29)
Now for the systems to be identical when separated as when joined X b3 = Xc3 ......................................................... (9.30) and since there is no external force at 3. F b3 + F c3 = 0 ...................................................... (9.31) Substitute Fc3 = − Fb3 from (9.31) and for Xb3 from (9.28) and Xc3 from (9.27) in equation (9.30). Thus therefore β33F b3 + β32 F2 = − γ 33Fb3 F b3 = β 32 F2 β33 + γ 33
β β F and substituting for Fb 3 in (9.29) gives X1 = β 12F2 + 13 32 2 β +γ
33 33
therefore X1 β β = α1 2 = β1 2 + 1 3 3 2 F2 β3 3 + γ 3 3 .............................................. (9.32)
It is possible to measure β12, β13, β 32 and β33 as functions of frequency and predict the effect on α12 of adding a particular system C.
APPENDIX 2: Axial Vibration of a stepped bar
Find the first natural frequency of the stepped bar shown. Each section has the same length L but the large bar has a diameter √2 times the smaller bar. First divide in to two sub-systems,
The natural frequency equation is given by, β11 + γ 11 = 0 where γ 11 = and β11 = − cosλ bL A bE b λ bsinλ bL sinλ c L Ac E c λ c cosλ c L
If the bars are made from the same material E b = E c = E and also ρ b = ρ c = ρ ρ so that λ b = λ c = λ = ω E However because of the different diameters Ab ≠ Ac . πD2 π2D2 c = b = 2A Ac = b 4 4 thus sinλL γ 11 = 2A bEλcosλL and β11 = − cosλL AE bλsinλL
The natural frequency equation is thus sinλL cosλL − =0 2A bEλcosλL Ab EλsinλL so that sinλL cosλL = and hence 2cosλL sinλL tan 2 λL = 2 and tan λL = ± 2 Note the negative sign is important when higher natural frequencies are to be found. For the first natural frequency, tan λL = + 2 λL = 0.9553 since λ2 = ω 2ρ / E 0.9553 E ω n1 = L ρ and for the second natural frequency tan λL = − 2 λL = 2.186 since λ2 = ω 2ρ / E 2.186 E ω n2 = L ρ 9.7.1 Classical Mode Shapes If the system has no damping then it will have undamped natural frequencies. If we wish to determine the deflected shapes (mode shapes) at these frequencies we cannot have an external excitation as at an undamped natural frequency the response will tend to infinity. Hence we apply no external excitation. If any point in the system is assumed to have an amplitude of unity then we find the forces at the joins, all of which will have zero external excitation. Hence to find the associated mode shapes, divide the system into its sub-systems and assume an amplitude of unity at the join.
Thus we have X b1 = Xc1 = 1.0 and we have already shown,
β11 = − Thus β11 =
cosλL A bEλsinλL
Xb1 1 cosλL = =− Fb1 F b1 Ab EλsinλL Ab EλsinλL ∴F b1 = − cosλL
Now consider the amplitudes along the bar B at some position x from the left hand end, X cosλ(L - x) β x1 = bx = − F b1 A bEλsinλL and substituting for F b1 For the first mode, x cos0.9553 1 − L = 1.733cos0.95531 − x λL = 0.9553 and thus X bx = L cos0.9553 and for the second mode x cos2.186 1− L x λL = 2.186 and thus X bx = = −1.733cos2.186 1 − cos2.186 L Now consider the sub system C. There is no external force at coordinate 1 so that A EλsinλL F c1 = −F b1 = b cosλL Now consider the amplitudes along the bar C at some position y from the left hand end, (Note: it is not wise to use x again for position as confusion may arise with system C) X cy sinλy = γ 11 = Fc1 Ac EλcosλL and substituting for F c1 sinλy A bEλsinλL U cy = Ac EλcosλL cosλL Since A c = 2Ab sinλy tan λL θ cy = 2 cosλL For the first mode, sin0.9553 λL = 0.9553 and thus X cy = and for the second mode sin2.186 λL = 2.186 and thus X cy = y tan 2.186 y L = 1.226sin2.186 2cos2.186 L y tan 0.9553 y L = 1.225sin0.9553 2cos0.9553 L
To plot the mode shape take ten points on each bar. For bar C y y mode 1 X cy = 1.225sin0.9553 mode 2 X cy = 1.226sin2.186 L L y/L=0 y/L=0.1 y/L=0.2 y/L=0.3 y/L=0.4 y/L=0.5 y/L=0.6 y/L=0.7 y/L=0.8 y/L=0.9 y/L=1.0 For bar B x mode 1 X b x = 1.733cos0.9553 1− L x/L=0 Ub x = 1.0 x/L=0.1 x/L=0.2 x/L=0.3 x/L=0.4 x/L=0.5 x/L=0.6 x/L=0.7 x/L=0.8 x/L=0.9 x/L=1.0 x mode 2 X b x = −1.733cos2.186 1− L Ub x = 1.0 Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = 1.1310 1.2510 1.3598 1.4560 1.5390 1.6080 1.6623 1.7015 1.7251 1.7330 Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = 0.6694 0.3069 -0.0703 -0.4442 -0.7969 -1.1117 -1.3735 -1.5700 -1.6918 -1.7330 Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy = = = = = = = = = = = 0 0.1169 0.2326 0.3463 0.4568 0.5631 0.6643 0.7595 0.8480 0.9282 1.0 Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy =0 = 0.2659 = 0.5191 = 7476 = 0.9405 = 1.0887 = 1.1851 = 1.2250 = 1.2066 = 1.1308 = 1.0
We may present these mode shapes in graphical form
Mode 1
Mode 2.
APPENDIX 3: Receptance Addition with two coupling coordinates
5 6
For system B U1 = β 11F1 + β12F2 + β13F b3 + β 14Fb4 ......................................... U2 = β 12F1 + β 22F2 + β 23Fb3 + β 24Fb4 ........................................ U b3 = β13F1 + β23F2 + β 33Fb3 + β 34F b4 ....................................... U b4 = β14F1 + β 24F2 + β 34F b3 + β 44F b4 ...................................... For system C U c3 = γ 33Fc3 + γ 34F c4 + γ 35F5 + γ 36 F6 U c4 = γ 34Fc3 + γ 44F c4 + γ 45F5 + γ 46 F6 U5 = γ 35Fc3 + γ 45Fc4 + γ 55F5 + γ 56 F6 U6 = γ 36 Fc3 + γ 46Fc4 + γ 56 F5 + γ 66 F6 At boundary, for compatability, U3 = U b3 = Uc3 ......................................................... (9.41) U 4 = U b4 = Uc4 ........................................................ (9.42) At boundary, for force equivalence, F b3 + F c3 = F3 ........................................................... (9.43) F b4 + F c4 = F 4 .......................................................... (9.44) Substituting in (9.41) and (19.42) from (9.35), (9.36), (9.37) and (9.38), β13F1 + β 23F2 + β33F b3 + β 34F b4 = γ 33Fc3 + γ 34F c4 + γ 35F5 + γ 36 F6 ................... (9.45) β14F1 + β 24F2 + β34 Fb3 + β 44F b4 = γ 34Fc3 + γ 44F c4 + γ 45F5 + γ 46 F6 .................. (9.46) Substituting for F c3 and F c4 from (9.43) and (9.344) in (9.45)and (9.46) β13F1 + β 23F2 + β33F b3 + β 34F b4 = γ 33(F3 − Fb3 ) + γ 34 (F4 − Fb4 ) + γ 35F5 + γ 36 F6 β14F1 + β 24F2 + β34 Fb3 + β 44F b4 = γ 34 (F3 − F b3 ) + γ 44 (F4 − Fb4 ) + γ 45F5 + γ 46 F6 rearranging, (β33 + γ 33 )Fb3 + (β34 + γ 34 )Fb4 = −β13F1 − β 23F2 + γ 33F3 + γ 34 F4 + γ 35F5 + γ 36F6 (β34 + γ 34 )Fb3 + (β 44 + γ 44 )Fb4 = −β14 F1 − β24 F2 + γ 34F3 + γ 44 F4 + γ 45F5 + γ 46F6 and substituting, P = −β13F1 − β 23F2 + γ 33F3 + γ 34F 4 + γ 35F5 + γ 36 F6 ......................... (9.47) Q = −β 14F1 − β 24F2 + γ 34 F3 + γ 44F4 + γ 45F5 + γ 46F6 ......................... (9.48) gives, (β33 + γ 33 )Fb3 + (β34 + γ 34 )Fb4 = P ....................................... ...................................... ........................................ ....................................... (9.37) (9.38) (9.39) (9.40) (9.33) (9.34) (9.35) (9.36)
3 C 4 B
1 2
(β34 + γ 34 )Fb3 + (β 44 + γ 44 )Fb4 = Q Solving for F b3 and F b4 gives, P(β44 + γ 44 ) − Q(β34 + γ 34 ) F b3 = (β33 + γ 33 )(β 44 + γ 44 ) − (β34 + γ 34 )2 and F b4 = and substituting, yields, P(β44 + γ 44 ) − Q(β34 + γ 34 ) ............................................. (9.50) Δ Q(β33 + γ 33 ) − P(β34 + γ 34 ) F b4 = .............................................. (9.51) Δ F b3 = All six responses U1, U2 , U3, U4 , U 5 and U6 may be obtained using equations (9.33) to (9.51), U1 = β 11F1 + β12F2 + β13F b3 + β 14Fb4 ......................................... U2 = β 12F1 + β 22F2 + β 23Fb3 + β 24Fb4 ........................................ U3 = β13F1 + β 23F2 + β33F b3 + β 34 Fb4 ......................................... U 4 = β14F1 + β24 F2 + β34 Fb3 + β 44 Fb4 ........................................ U5 = γ 35(F3 − Fb3) + γ 45(F4 − Fb4 ) + γ 55F5 + γ 56F6 ............................... U6 = γ 36 (F3 − Fb3) + γ 46 (F 4 − F b4 ) + γ 56F5 + γ 66F6 .............................. (9.33) (9.34) (9.35) (9.36) (9.39) (9.40) Q(β33 + γ 33) − P(β34 + γ 34 ) (β33 + γ 33 )(β 44 + γ 44 ) − (β34 + γ 34 )2
Δ = (β33 + γ 33)(β44 + γ 44 ) − (β34 + γ 34 )2 ....................................... (9.49)
For any particular receptance we require the response to just one force. Thus for example, α 1 3 is, by definition, given by α13 = therefore P(β44 + γ 44 ) − Q(β34 + γ 34 ) Q(β33 + γ 33) − P(β34 + γ 34 ) U1 = β 13 +β14 Δ Δ and P and Q reduce to, P = γ 33F3 and Q = γ 34F3 so that γ F (β + γ 44 ) − γ 34F3 (β34 + γ 34 ) γ F (β + γ 33) − γ 33F3(β34 + γ 34 ) U1 = β 13 33 3 44 + β14 34 3 33 Δ Δ therefore γ (β + γ 44 ) − γ 34 (β34 + γ 34 ) γ (β + γ 33 ) − γ 33(β34 + γ 34 ) U α13 = 1 = β13 33 44 + β14 34 33 F3 Δ Δ or rearranging, U1 with all other forces than F3 zero. F3
α13 =
U1 β13[γ 33(β44 + γ 44 ) − γ 34 (β34 + γ 34 )] + β14 [γ 34(β33 + γ 33) − γ 33 (β34 + γ 34 ) ] = F3 Δ
The same result could have been obtained using, U α13 = 3 with all other forces than F1 zero. F1 This only applies for linear conservative systems when Maxwell's reciprocal Theorem holds.
.
Receptances_App
APPENDIX 1: Addition of a system at a remote co-ordinate
When the main system B is a real machine it is rarely possible to attach another system C at the point on B where the exciting force is applied. Also the relevant displacement will often be at the same position but in a different direction. It follows that the cross receptance b1 2 is the required receptance where 1 and 2 refer to the same position but a different direction. Also 1 and 2 may be defined as a relative displacement and a relative force (ie. equal and opposite forces) without changing the theory of system addition. It is thus necessary to derive the equation for α1 2 for a combined system A when the additional system C has been attached away from the cutting position, ie. at some co-ordinate 3. This situation is shown in Figure 3 with the associated separation into individual systems to enable the analysis to be carried out. The forces and displacements introduced at 3 are such that the separated systems behave in the same way as when joined.
For system C by definition X c3 = γ 33Fc3 ...................................................... (9.27)
For system B since two forces are applied the displacement at any co-ordinate will be the sum of the displacements caused by each force. ie. superposition is assumed to apply. Thus X b3 = β33F b3 + β 32F2 .............................................. (9.28) and X1 = β 12F2 + β13F b3 ................................................. (9.29)
Now for the systems to be identical when separated as when joined X b3 = Xc3 ......................................................... (9.30) and since there is no external force at 3. F b3 + F c3 = 0 ...................................................... (9.31) Substitute Fc3 = − Fb3 from (9.31) and for Xb3 from (9.28) and Xc3 from (9.27) in equation (9.30). Thus therefore β33F b3 + β32 F2 = − γ 33Fb3 F b3 = β 32 F2 β33 + γ 33
β β F and substituting for Fb 3 in (9.29) gives X1 = β 12F2 + 13 32 2 β +γ
33 33
therefore X1 β β = α1 2 = β1 2 + 1 3 3 2 F2 β3 3 + γ 3 3 .............................................. (9.32)
It is possible to measure β12, β13, β 32 and β33 as functions of frequency and predict the effect on α12 of adding a particular system C.
APPENDIX 2: Axial Vibration of a stepped bar
Find the first natural frequency of the stepped bar shown. Each section has the same length L but the large bar has a diameter √2 times the smaller bar. First divide in to two sub-systems,
The natural frequency equation is given by, β11 + γ 11 = 0 where γ 11 = and β11 = − cosλ bL A bE b λ bsinλ bL sinλ c L Ac E c λ c cosλ c L
If the bars are made from the same material E b = E c = E and also ρ b = ρ c = ρ ρ so that λ b = λ c = λ = ω E However because of the different diameters Ab ≠ Ac . πD2 π2D2 c = b = 2A Ac = b 4 4 thus sinλL γ 11 = 2A bEλcosλL and β11 = − cosλL AE bλsinλL
The natural frequency equation is thus sinλL cosλL − =0 2A bEλcosλL Ab EλsinλL so that sinλL cosλL = and hence 2cosλL sinλL tan 2 λL = 2 and tan λL = ± 2 Note the negative sign is important when higher natural frequencies are to be found. For the first natural frequency, tan λL = + 2 λL = 0.9553 since λ2 = ω 2ρ / E 0.9553 E ω n1 = L ρ and for the second natural frequency tan λL = − 2 λL = 2.186 since λ2 = ω 2ρ / E 2.186 E ω n2 = L ρ 9.7.1 Classical Mode Shapes If the system has no damping then it will have undamped natural frequencies. If we wish to determine the deflected shapes (mode shapes) at these frequencies we cannot have an external excitation as at an undamped natural frequency the response will tend to infinity. Hence we apply no external excitation. If any point in the system is assumed to have an amplitude of unity then we find the forces at the joins, all of which will have zero external excitation. Hence to find the associated mode shapes, divide the system into its sub-systems and assume an amplitude of unity at the join.
Thus we have X b1 = Xc1 = 1.0 and we have already shown,
β11 = − Thus β11 =
cosλL A bEλsinλL
Xb1 1 cosλL = =− Fb1 F b1 Ab EλsinλL Ab EλsinλL ∴F b1 = − cosλL
Now consider the amplitudes along the bar B at some position x from the left hand end, X cosλ(L - x) β x1 = bx = − F b1 A bEλsinλL and substituting for F b1 For the first mode, x cos0.9553 1 − L = 1.733cos0.95531 − x λL = 0.9553 and thus X bx = L cos0.9553 and for the second mode x cos2.186 1− L x λL = 2.186 and thus X bx = = −1.733cos2.186 1 − cos2.186 L Now consider the sub system C. There is no external force at coordinate 1 so that A EλsinλL F c1 = −F b1 = b cosλL Now consider the amplitudes along the bar C at some position y from the left hand end, (Note: it is not wise to use x again for position as confusion may arise with system C) X cy sinλy = γ 11 = Fc1 Ac EλcosλL and substituting for F c1 sinλy A bEλsinλL U cy = Ac EλcosλL cosλL Since A c = 2Ab sinλy tan λL θ cy = 2 cosλL For the first mode, sin0.9553 λL = 0.9553 and thus X cy = and for the second mode sin2.186 λL = 2.186 and thus X cy = y tan 2.186 y L = 1.226sin2.186 2cos2.186 L y tan 0.9553 y L = 1.225sin0.9553 2cos0.9553 L
To plot the mode shape take ten points on each bar. For bar C y y mode 1 X cy = 1.225sin0.9553 mode 2 X cy = 1.226sin2.186 L L y/L=0 y/L=0.1 y/L=0.2 y/L=0.3 y/L=0.4 y/L=0.5 y/L=0.6 y/L=0.7 y/L=0.8 y/L=0.9 y/L=1.0 For bar B x mode 1 X b x = 1.733cos0.9553 1− L x/L=0 Ub x = 1.0 x/L=0.1 x/L=0.2 x/L=0.3 x/L=0.4 x/L=0.5 x/L=0.6 x/L=0.7 x/L=0.8 x/L=0.9 x/L=1.0 x mode 2 X b x = −1.733cos2.186 1− L Ub x = 1.0 Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = 1.1310 1.2510 1.3598 1.4560 1.5390 1.6080 1.6623 1.7015 1.7251 1.7330 Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = Ub x = 0.6694 0.3069 -0.0703 -0.4442 -0.7969 -1.1117 -1.3735 -1.5700 -1.6918 -1.7330 Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy = = = = = = = = = = = 0 0.1169 0.2326 0.3463 0.4568 0.5631 0.6643 0.7595 0.8480 0.9282 1.0 Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy Ucy =0 = 0.2659 = 0.5191 = 7476 = 0.9405 = 1.0887 = 1.1851 = 1.2250 = 1.2066 = 1.1308 = 1.0
We may present these mode shapes in graphical form
Mode 1
Mode 2.
APPENDIX 3: Receptance Addition with two coupling coordinates
5 6
For system B U1 = β 11F1 + β12F2 + β13F b3 + β 14Fb4 ......................................... U2 = β 12F1 + β 22F2 + β 23Fb3 + β 24Fb4 ........................................ U b3 = β13F1 + β23F2 + β 33Fb3 + β 34F b4 ....................................... U b4 = β14F1 + β 24F2 + β 34F b3 + β 44F b4 ...................................... For system C U c3 = γ 33Fc3 + γ 34F c4 + γ 35F5 + γ 36 F6 U c4 = γ 34Fc3 + γ 44F c4 + γ 45F5 + γ 46 F6 U5 = γ 35Fc3 + γ 45Fc4 + γ 55F5 + γ 56 F6 U6 = γ 36 Fc3 + γ 46Fc4 + γ 56 F5 + γ 66 F6 At boundary, for compatability, U3 = U b3 = Uc3 ......................................................... (9.41) U 4 = U b4 = Uc4 ........................................................ (9.42) At boundary, for force equivalence, F b3 + F c3 = F3 ........................................................... (9.43) F b4 + F c4 = F 4 .......................................................... (9.44) Substituting in (9.41) and (19.42) from (9.35), (9.36), (9.37) and (9.38), β13F1 + β 23F2 + β33F b3 + β 34F b4 = γ 33Fc3 + γ 34F c4 + γ 35F5 + γ 36 F6 ................... (9.45) β14F1 + β 24F2 + β34 Fb3 + β 44F b4 = γ 34Fc3 + γ 44F c4 + γ 45F5 + γ 46 F6 .................. (9.46) Substituting for F c3 and F c4 from (9.43) and (9.344) in (9.45)and (9.46) β13F1 + β 23F2 + β33F b3 + β 34F b4 = γ 33(F3 − Fb3 ) + γ 34 (F4 − Fb4 ) + γ 35F5 + γ 36 F6 β14F1 + β 24F2 + β34 Fb3 + β 44F b4 = γ 34 (F3 − F b3 ) + γ 44 (F4 − Fb4 ) + γ 45F5 + γ 46 F6 rearranging, (β33 + γ 33 )Fb3 + (β34 + γ 34 )Fb4 = −β13F1 − β 23F2 + γ 33F3 + γ 34 F4 + γ 35F5 + γ 36F6 (β34 + γ 34 )Fb3 + (β 44 + γ 44 )Fb4 = −β14 F1 − β24 F2 + γ 34F3 + γ 44 F4 + γ 45F5 + γ 46F6 and substituting, P = −β13F1 − β 23F2 + γ 33F3 + γ 34F 4 + γ 35F5 + γ 36 F6 ......................... (9.47) Q = −β 14F1 − β 24F2 + γ 34 F3 + γ 44F4 + γ 45F5 + γ 46F6 ......................... (9.48) gives, (β33 + γ 33 )Fb3 + (β34 + γ 34 )Fb4 = P ....................................... ...................................... ........................................ ....................................... (9.37) (9.38) (9.39) (9.40) (9.33) (9.34) (9.35) (9.36)
3 C 4 B
1 2
(β34 + γ 34 )Fb3 + (β 44 + γ 44 )Fb4 = Q Solving for F b3 and F b4 gives, P(β44 + γ 44 ) − Q(β34 + γ 34 ) F b3 = (β33 + γ 33 )(β 44 + γ 44 ) − (β34 + γ 34 )2 and F b4 = and substituting, yields, P(β44 + γ 44 ) − Q(β34 + γ 34 ) ............................................. (9.50) Δ Q(β33 + γ 33 ) − P(β34 + γ 34 ) F b4 = .............................................. (9.51) Δ F b3 = All six responses U1, U2 , U3, U4 , U 5 and U6 may be obtained using equations (9.33) to (9.51), U1 = β 11F1 + β12F2 + β13F b3 + β 14Fb4 ......................................... U2 = β 12F1 + β 22F2 + β 23Fb3 + β 24Fb4 ........................................ U3 = β13F1 + β 23F2 + β33F b3 + β 34 Fb4 ......................................... U 4 = β14F1 + β24 F2 + β34 Fb3 + β 44 Fb4 ........................................ U5 = γ 35(F3 − Fb3) + γ 45(F4 − Fb4 ) + γ 55F5 + γ 56F6 ............................... U6 = γ 36 (F3 − Fb3) + γ 46 (F 4 − F b4 ) + γ 56F5 + γ 66F6 .............................. (9.33) (9.34) (9.35) (9.36) (9.39) (9.40) Q(β33 + γ 33) − P(β34 + γ 34 ) (β33 + γ 33 )(β 44 + γ 44 ) − (β34 + γ 34 )2
Δ = (β33 + γ 33)(β44 + γ 44 ) − (β34 + γ 34 )2 ....................................... (9.49)
For any particular receptance we require the response to just one force. Thus for example, α 1 3 is, by definition, given by α13 = therefore P(β44 + γ 44 ) − Q(β34 + γ 34 ) Q(β33 + γ 33) − P(β34 + γ 34 ) U1 = β 13 +β14 Δ Δ and P and Q reduce to, P = γ 33F3 and Q = γ 34F3 so that γ F (β + γ 44 ) − γ 34F3 (β34 + γ 34 ) γ F (β + γ 33) − γ 33F3(β34 + γ 34 ) U1 = β 13 33 3 44 + β14 34 3 33 Δ Δ therefore γ (β + γ 44 ) − γ 34 (β34 + γ 34 ) γ (β + γ 33 ) − γ 33(β34 + γ 34 ) U α13 = 1 = β13 33 44 + β14 34 33 F3 Δ Δ or rearranging, U1 with all other forces than F3 zero. F3
α13 =
U1 β13[γ 33(β44 + γ 44 ) − γ 34 (β34 + γ 34 )] + β14 [γ 34(β33 + γ 33) − γ 33 (β34 + γ 34 ) ] = F3 Δ
The same result could have been obtained using, U α13 = 3 with all other forces than F1 zero. F1 This only applies for linear conservative systems when Maxwell's reciprocal Theorem holds.
.
