1 - Acta Technica Corviniensis

The Euler ecuation about the condition of unslip of theband on motor drum is:28μ ⋅αk f ⋅ S 4 = S 1 ⋅ e [N] (7)where k f [‐] is the safety coefficient for unslip onmotor drum (k f =1,2…1,3), e is the base of naturalslogarithms, μ [‐] is the friction coefficient between theband and the driven drum (μ=0,25…0,35) and α[rad]it is the angle of wrap up for the band on motor drum.With relations (1),…, (4) and (6) may be get:⋅[ qb⋅( 1+kî) + kî⋅qrg+ qî+ qrp][ q ⋅( 1−k) + q ]⎧w⋅cosβ⎫kf⋅L⋅⎨⎬sin b î îS⎩+β ⋅⎭1 = [N] (8)μ⋅αe −k⋅kSS +fic4 = k î ⋅S1[N] (9)k fThe reduce resistant moments for motor shaft may beobtioned with:( S −S)4 1 DTMrr= ⋅ [Nm] (10)η ⋅i2Rwhere the tensions in band S 1 and S 2 are fromexpressions (8) and (9).For horizontal transport conveyers, without deviationdrums when β=0 and from expressions (8) and (9)result:k f ⋅L⋅w⋅[ q b ⋅( 1+k î ) + k î ⋅q rg + q î + q rp ]S1 =[N] (11)eμ ⋅α− k f ⋅k î( 1+k )⎪⎧⎡qb⋅ î ⎤⎪⎫S4 = L ⋅⎨kî ⋅S1+ w ⋅ ⎢⎥⎬[N] (12)⎪⎩ ⎣+ kî⋅qrg+ qî+ qrp⎦⎪⎭The reduce resistance moments at motor shaft, maybe determinate with the formula (10) where S 1 and S 2are given by expression (11) and (12).The reduce resistance moments at shaft of drivenmotor, may be determinate for the belt transportconveyer with deviation drum is presented in fig.2.Fig.2. Belt inclinated transport conveyerwith deviation drumACTA TECHNICA CORVINIENSIS – Bulletin of EngineeringThe tensions in band, in points 1,…,8, for β>0 are:S 1 = S x [N] (13)SS 2 ≈ S 1 [N] (14)3 k î 1 ⋅ S 2= [N] (15)( q + q )S4= S3+ b rg ⋅L2⋅w⋅cosβ[N] (16)−q⋅L⋅sinβS = S +87b2S = î ⋅ [N] (17)5 k 1 S4S 6 ≈ S 5 [N] (18)S = î ⋅ [N] (19)7 k 2 S6( q + q + q )+b( q + q ) ⋅L⋅sinβbîîrp1⋅L1⋅wcos⋅ β[N] (20)where k î1 [‐] it is the band loading coefficient becausethe band passing over deviation drums (k î1 =1,02…1,03)and k î2 [‐] it is the band loading coefficient over returndrum (k î2 =1,07…1,09).The band loading that passing over deviation drumshave some coefficients values, because the angles ofwrap up of band on these drums are equal.The condition by unslip of band on motor drum is:μ ⋅αk f ⋅ S 8 = S 1 ⋅e(21)With these expressions (13),…,(21) may be get thenext ecuations:Sk f ⋅S A1 =e⋅α−k2⋅k î ⋅k î 1 2 fμ [N] (22)S = 28 S1⋅ k î 1 ⋅ k î 2 + S A [N] (23)where S A has this formula:⎛qb⋅( L1+ kî1⋅kî2 ⋅L2) ⎞SA= w⋅cosβ⋅⎜⎟qrgkî1kî2 L2L1( qîqrp)⎝+ ⋅ ⋅ ⋅ + ⋅ +⎠ (24)+ sinβ⋅ L ⋅ q + q −k⋅k⋅q⋅L( ( ))1bîî1For horizontal transport conveyer (β=0) with banddeviation drums, the tensions S 1 and S 8 from the bandare:k f ⋅SBS1 = [N] (25)μ ⋅α2e − k ⋅k⋅kî1î2bî 2S = 2S1⋅ k î 1 ⋅ k î 2 + S B8 [N] (26)where S B may be calculate with:S B⎡⎤⎢q b ⋅( L1+ k î 1⋅k î 2 ⋅L2)= w ⋅ ⎢ + q rg ⋅k î 1⋅k î 2 ⋅L2⎢(27)⎣+ L1⋅2( ) ⎥ ⎥⎥ q î + q rp ⎦For these two situations (β>0 and β=0) the reducedresistant moments at motor shaft may be computewith:( S8− S1) DTMrr= ⋅ [Nm] (28)η ⋅i2Rf2012. Fascicule 3 [July–September]