Chemical Engineering - Apllied Mechanics

Belt transmissions

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1. BELT TRANSMISSIONS 1.1 FLAT BELTS Figure 1 shows a belt transmission in which the drive torque is applied to the smaller pulley. Figure 2 shows the isolated driving pulley. The drive torque is denoted M d, while the tensions in the two sides of the= belt are denoted= 1T =and 2T , respectively. The radius of the pulley is 1R , while the component of the= reaction force acting at the centre of the pulley, along=the straight line passing through the centers of the two= pulleys, is denoted= Q .= = = = Figure 1 Belt transmission = = = = = Figure 2 Isolated driving pulley=Figure 3 Wrap angle and independent variable ϕ The balance of the mome nts about the axis passing through point O 1, orthogonal to the lateral side of the pulley, gives: ( ) 121 M T TR d= − =(1)= In accordance with eq.(1) the tension 1T =must be higher than the tension 2T . The belt side subjected to= the tension 1T =is denoted=tight side, while the opposite side is denoted slack side. The balance of the forces acting along the straight line passing through the centers of the two pulleys gives: 12T cos T cos Q αα += (2) where α is the inclination angle of the belt sides with respect to the axis passing through the centers of the pulley (Figure ). In order to study the law with which the tension of the belt changes from the lower value 2T , in the= slack side, to the higher value 1T , in=the=tight side, it is possible to isolate an infinitesimal=portion of the belt,= which is in contact with the smallest pulley and subtended by an infinitesimal angle= ≤ϕ . The independent variable= ϕ =is measured starting from the first contact point between belt and pulley, on the slack side= (Figure=3). Owing to the tension of the belt, a pressure distribution is generate≤=on the contact surface between belt and pulley.=As a result, an infinitesimal normal force, n dF , is exerted on the infinitesimal portion of the belt that has been considered. The friction between belt and pulley and the presence of=the normal force= n dF , mentioned above, cause an infinitesimal tangential force= t dF . The versus of this force is coherent with that of the angular velocity of the pulley (Figures=2,=4).=In fact, in absence of=any slippage= between belt and pulley, the tangential velocity of the pulley, which should coincide with the belt velocity, must be coherent with the versus of the rotary motion of the pulley.= = = = Figure 4 Isolated driving pulley= = If no slippage occurs=between belt and pulley it is possible to write:= tn dF dF sf < = (3)= where sf=is the static friction coefficient.=In order to reduce the risk of slippage between belt and pulley it is possible to set up the computation of the maximum torque that can be transmitted by the belt,=considering the limit force that can cause an impending slippage. In this case, the following relationship between the normal and tangential infinitesimal forces exists:= tn dF dFf = = (4)= Let us denote q=the linear mass density of the belt. The length of the infinitesimal arc of belt that has been considered is: Rd ϕ .=This very short portion of the belt is subjected to a centrifugal inertia force given= by:= () n 2 2 iv dF = R ≤ = v ≤ R qq ϕϕ = (5 )= where v=is the tangential velocity of the pulley. If the velocity v=is not very high, as it happens in most operating conditions, this radial force=can be neglected.= The tension of the belt changes form the slack side to the tight side, therefore, it depends on the variable= ϕ =defined above. With reference to Figure=4, if= ( ) T=T ϕ =is the tension on the right=side of the infinitesimal portion of the belt, the tension on the left side, evaluated at an angle= () d ϕϕ + =is:= ()T d T + dT ϕϕ += = (6)= In accordance with the assumptions mentioned above, the infinitesimal radial force= n dF =is given by:= ( ) n dF R d p ϕ = = (7)= Then, the corresponding frictionϑinduced tangential force is:= ( ) t dF R d fp ϕ = = (8)= = = = Figure 5 Balance of forces, exerted on a short portion of belt, in the radial and tangential direction= = The balance of the forces exerted in the radial direction can be expressed as follows=(Figure=4, 5):= () () dd T sin T + d T sin R d 0 22pϕϕ ϕ  + −=     =(9)= Eq.(7) can be rewritten as follows, after removing the higherϑorder infinitesimal terms:= ( ) dd 2T dTR d 22 p ϕϕ ϕ   +=    = (10)= In the end, we obtain:= ( ) T d Rd p ϕϕ = = (11)= That is:= TR p = =(12)= The balance of the forces exerted in the tangential direction can be expressed as follows:= ( ) t dd T cos T + d T cos d F 0 22 ϕϕ   −+=    (13) For infinitesimal angles dϕ =it is possible to write:= t T T dT dF 0 −− + = = (14)= Therefore, we obtain:= t dT dF= = (15)= Based on eqs.(2) and (5)=it is possible to write:= dT Rd Td fp f ϕϕ = = = (16)= That is:= dT≤ Tf ϕ = = (17)= Eq.(17) must be integrated over the wrap angle, 1ϕ , of the belt. The limits of integration for the left term= of eq.(17) are,= 2T =and 1T , respectively. Therefore, we have:= 11 2 T T0 dT ≤ T f ϕ ϕ = ∫∫ = (18)= As a result, we obtain:= 1 21 ln T ln T fϕ −= = (19)= Eq.(17) can be also rewritten as:= 1 1 2T T fe ϕ = or 1 12TT fe ϕ = =(20)= In order to ensure a higher security level,=that is to avoid any slippage, the following condition should be= met:= 1 12TT fe ϕ ≤ = (21)= In order to start a system that contains a belt transmission the belt must be preϑtensioned. That is, the tension in the belt must be not null also when the pulleys are stopped. In general, the hinge of one pulley is fixed to a frame while the other one can be moved along the straight line passing through the centers of the= two pulleys. That is, the span between the two pulleys can be adjusted. Therefore, the belt preϑtension can be= obtained=increasing the original span that is associated with the configuration in which the belt is at rest (subjected to a null tension). Let us denote= Q =the reaction force, directed as the straight=line passing through= the centers of the two pulleys, acting on the centre of the pulley whose position can be adjusted (Figure=1).= The computation performed considering the maximum torque that can be transmitted without any= slippage between belt and pulley gives:= () max 121 M T TR d = − = (22)= 1 1 2 T T fe ϕ = 12T cos T cos Q αα += = In order to reduce the risk of possible slippages between belt and pulley, the following condition should be met:= max MM dd ≤ = (23)= Obviously, the condition (23) is in accordance with the condition (21).= When the pulleys are stopped, that is, when the drive torque= ϕ d=is null, the tension in the slack side is equal to that of the tight side:= 12 0TT T = = = (24)= The tension 0T =must not be null in order to ensure a not null pressure p=between belt and pulley.= In=fact, the radial force= n dF , from which the tangential force t dF =depends, nullifies for a null pressure= p.= In accordance with eq.(20), that is case of=impending slippage between belt and pulley, the tension of the= belt increases with an exponential law from the slack side to the tight side=(Figure=6=ϑ=a). a) b) Figure 6 Increase of the belt tension from the slack side to the tight side However, in order to ensure correct operating conditions, the inequalities (21) and (23) must be met. Therefore, the tension of the belt increases with an exponential law from 2T =to= 1T , over a portion of the actual wrap angle. That is:= * 12TT fe ϕ = with * 1ϕϕ < =(25) = Then, the tension remains=constant over the arc subtended by the angles ϕ =contained in the range= (Figure=6=ϑ=b): * 1ϕ ϕϕ