THE CALCULUS OF THE EQUIVALENT RIGIDITY COEFFICIENTS FOR THE SHAFTS OF THE (/$67,&$/ SYSTEMS

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1 MECHAICA EGIEERIG, ISS THE CACUUS OF THE EQUIVAET RIGIDITY COEFFICIETS FOR THE SHAFTS OF THE (/$67,&$/ SYSTEMS $VVRF3URI'U(QJ1FXúRU'5 *$1 Assistant Eng. Aura POTÂRICHE MECMET The Research Center o Machines, Mechanic and Technological Equipments "Dunarea de Jos" University o Galati ABSTRACT The components o the mechanical systems like machines and/ technological equipment have a certain elasticity rigidity. Knowing the rigidity o each component is very imptant the studies whose goal is to establish the dynamic ets and the stresses in each shat, gear wheel, steel structure, aso. The rigidity can be an imptant act the dynamic load estimation process. The components with high elasticity are the most imptant inducers o elastical ces and couples o ce; we can enumerate: shats, coupling gears, gears, elastical couplings, springs, long steel structures, some wking devices, aso. This article presents a method and an equation involving the rigidities o the elastical shats o the mechanical transmissions with gears in any point o the system, so that the dynamic analysis should become easier. KEYWORDS: elastical systems, equivalent rigidity, gearing, shat stress 1. Introduction The equivalent coeicient o rigidity is the mechanical eature o an equivalent elastical element (generally named spring), which replaces the real element on the basic principle o the equation o the potential energy, [1]. This means that the demation potential energy o the equivalent element 9 is equal to the demation potential energy o the actual element Calculus o the equivalent rigidity o the shats with one step gearing In der to describe the rigidity equation method, it is considered a simple mechanical driving system as in ig.1, where is the driving mot moment, :' is the moment o wking device, 2 and 3 are the wheels o the one step gearing, and are the rigidity coeicients o the shat I (driving shat) respectively shat II (driven shat). 29,, Fig. 1 The model to calculate the equivalent rigidities o the shats It is considered that the mechanical eiciency o the gearing is and the ratio is

2 J, (1) J where J and J are the angular speeds o the wheel gear 2 respectively 3. The instantaneous real angular rotations o the shats terminations are: - the shat I -, - the shat II -, The equivalent inertia moments o the wking device and o wheel gear 3 can be calculated accding to [2] Calculus o the equivalent rigidity on the driving shat I the needs o equation is to be done on the shat I, the ig. 2 shows the calculus model, where the signiicance o the notations is as - the equivalent rigidity coeicient o the driven shat, - the equivalent angular rotations o the driven shat s terminations - the average angular speed o the driving shat (in steady-state conditions), The demation potential energy o the shat II real system (ig. 1) can be written as: 9 ( ) (2) F the same shat II, the potential energy, on the basis o the equivalent model rom ig. 2, has the expression: ( ) Fig. 2 The model to calculate the equivalent rigidity o the driven shat 9 (3) ( ) ( ) (4) Taking into consideration that, in steady-state conditions, the wking device moment (o resistance) is equal to the elastical moment rom the driven shat, it may be written as IRUUHDOV\VWHP ( ) (5) IRUHTXYDOHQWV\VWHP ( ) :' (6) Dividing the relations (5) and (6), it is obtained: :' (7) Considering the relation (4), it may be written (8) (9) :' From (9), we can write: (1) :' In der to estimate the raction between wking device moments rom (1), it has to be written the wking device power both real system and equivalent system. a)ideal step gearing ( ) I there are no mechanical losses in the gearing 2-3, all power rom the mot goes to the wking device, that s why it may be written 3 :' (11) Equating the expressions (2) and (3) o the potential energy o the shat II, we obtain: 3 From the relation (11), we can write the raction between the wking device equivalent moment and the wking device real moment as

3 (12) Since, in steady-state conditions, the average angular speed o the wking device is equal to angular speed o the wheel gear 3 ( ) and the average angular speed o the mot is equal to the angular speed o the wheel gear 2 ( ), the relation (12) may be written: (13) Taking into consideration relation (13), the calculus mula the rigidity coeicient o the shat II on the mot shat is: (14) b)step gearing with mechanical losses ( < ) I there are mechanical losses in the gearing 2-3, the power rom the mot goes partially to the wking device and the dierence is dissipated in the gearing. In this case, we may write 3 + 3ORVV, (15) where 3 ORVV is the power losses in the gearing. I the loss o power is written unction o as 3ORVV ORVV, (16) where ORVV is the equivalent loss o moment, in the steady-state conditions (, ), the power balance done by (15) can be written like ( ORVV ) ORVV, (17) (18) Consequently, the relation (1) becomes: (19) (2) 2.2. Calculus o the equivalent rigidity on the driven shat Figure 3 shows the calculus model o the rigidity equation on the axle o the driven shat II. The signiicance o the notations is as - the equivalent rigidity coeicient o the driving shat, - the equivalent angular rotations o the driving shat s terminations - the average angular speed o the driving shat (in steady-state conditions) As in 2.1, the demation potential energy o the driving shat I can be written like this 9 ( ) (21) F the same shat I, the potential energy calculated with the equivalent rigidity and angular delections, is as ollows ( ) 9 (22),, Fig. 3 The model to calculate the equivalent rigidity o the driving shat Since, the let side o the relation (18) is the mechanical eiciency o the gearing 2-3 and the raction o the right side is the inverse o the gearing ratio (1), we may write 31 Since the potential energy o the shat I has to remain the same ater the process o equation, rom relations (21) and (22) it can be written the raction between the rigidities as

4 ( ) ( ) (23) Considering 2-3 as ideal, all power rom the mot goes to the wking device, that s why we may write: 3 (3) Assuming that, in steady-state conditions, the mot has the same average angular speed like the wheel gear 2 and the wking device has the same average angular speed like the wheel gear 3, that meaning and, the mot moment has to be equal to the elastical tsion moment rom the shat I. In consequence, it can be written IRUWKHUHDOV\VWHP ( ) (24) IRUWKHV\VWHPWKHTXYDOHQWUJGW\ ( ) (25) Dividing the relations (24) and (25) it is obtained (26) Considering the raction o the rigidities done by (23), the relation (26) becomes, (27) (28) From the relation (28),we may say that the equivalent rigidity o the shat I is unction o the raction o the mot moments as From (3), we can write: (31) Since and, the raction between the mot moments can be written unction o the gear ratio as (32) In this case, the calculus mula o the equivalent rigidity o the driving shat on the driven shat axle is: (33) a)gearing with power losses ( < ) Taking into consideration the mechanical losses rom the gearing 2-3, the power rom the mot goes partially to the wking device and the dierence is dissipated in the gearing. In this case, the balance o the power being can be written 3 3ORVV, (34) where 3 ORVV is the power losses in the gearing. Writing the loss o power as a unction o like 3ORVV ORVV (35) where ORVV is the equivalent loss o moment, in the steady-state conditions, the power balance done by (34) can be written like (29) The raction between mot moments rom (29) can be determined by writing the mot power both real system and equivalent system. a)gearing with no power losses ( ) 32 ( ORVV ), (36) ORVV (37) Since, in the let side is the mechanical eiciency o the gearing and the raction between angular speeds rom right side is the gearing ratio, the relation (37) becomes

5 (38) With the determinated raction o the moments (38), we can write the calculus mula the equivalent rigidity o the driving shat with real gearing like a unction o the ratio and the mechanical eiciency as 3. The equivalent rigidities o the mechanism s shats with gearings To exempliy the method o the rigidity equation the mechanism with multiple gearing, it is considered the driving system a belt convey rom ig. 4. The skeleton diagram o the acting device is shown in ig. 5, where 2, 3, 4, 5, 6 and 7 are the gearing wheels o the mechanical transmission. It considers as known the mechanical eiciencies and the ratio o the gearings as (,& * '3 5* 2& %& & Fig. 4 The principle model o a belt convey egend: EM-electromot, IC-inside coupling, OC-outside coupling, G-gear reducer unit, C-case, RG-reducing gear, BC-belt convey, DP-drive pulley Fig. 5 The skeleton diagram the belt convey (39) 33 -gearing 2-3, -gearing 4-5, -gearing 6-7, Using the calculus relationships

6 ~ V O v k ~ w O O P V T T U T v v \ [ \ [ 5 a a b` ` ` ` a ` ` ` a g g & determinated in 2, we will equate the shats rigidities both on the electromot axle and on the drive pulley axle Rigidities on the electromot axle Figure 6 shows the calculus diagram o the equivalent rigidities on electromot axle, where the mulae the equivalent moments o inertia can be taken rom [2]. The calculus relationships used in this case are (14) ideal gearings and (2) real gearings. ƒ ƒ ƒ ƒ ƒ ƒ ~ } w@x y{z x y{z } x y{z Fig. 6 The calculus diagram the equivalent rigidities on the electromot axle shat The equivalent rigidities on the electromot axle the shats 2, 3 and 4 are shown in the table 1 both cases (without and with mechanical losses). Table 1 Equivalent rigidities on the driving axle shat Ideal gearings ( ) Real gearings ( < ) Q!R%S WX1Y ]^1_ 3.2. Rigidities on the drive pulley axle Figure 7 shows the calculus diagram o the equivalent rigidities on the drive axle shat, where, like 3.1, the equivalent moments o inertia can be taken rom [2]. The calculus relationships used in this case are (33) ideal gearings and (39) real gearings. q l8m%n k1l8m%n o1l8m%n p v v v q o p r u s Fig. 7 The calculus diagram the equivalent rigidities on the drive pulley axle shat Table 2 shows the equivalent rigidities on the drive pulley axle the shats 1, 2 and 3, both in the case when the mechanical losses are taken into consideration and in case they aren t. 5. Conclusions The methods and calculus mulae presented in this study are useul both to design engineers and to dynamics experts, as well as to students,, candidates master s and doct s degree. Regarding the ratio o the gearings, we may draw some conclusions about its inluence on the equivalent rigidities: 1 i (gearing changing the sense o rotation only), the equivalent rigidities stay unchanged; 2 i > (reduced step gearing), the rigidity o the driving shat (on the driven shat axle) is ampliied by and the rigidity o the driven shat (on the driving shat axle) is divided by ; 3 i < (ampliier step gearing) the rigidity o the driving shat (on the driven shat axle) is divided by and the rigidity o the driven shat (on the driving shat axle) is ampliied by. Taking into consideration the losses o power in the gearings trough their mechanical coeicients <, the equivalent rigidities o the shats are always increased by multiplying with. Table 2 Equivalent rigidities on the driven axle shat Ideal gearings ( ) Real gearings ( < ) cd%e ` h!i%j 5. Reerences [1]Bratu, P. P. -!" # %$&'%(!) * +!, -/.1 23) 4 +!4+!, 6 * ) &87999 [2]Debeleac, C.., :/; <!>@?1AB The dynamic modelling o the mechanical systems. Calculus o the equivalent mass and equivalent mass inertia &C.12 D33- E FHGIKJ+!3, - (! MGF" University o Galati, Fascicle XIV Mechanical Engineering, Galati, 27 34