drive friction paper 1

5
solids handling Volume 18 Number 1 • January/March 1998 Belt Tension Around a Drive Drum Modelling Belt Tension Around a Drive Drum A. Harrison, USA Summary Analysis of the drive drum friction problem by classical calculus provides a slip test that allows a designer to determine a maxi- mum T^/T 2 for a conveyor. In the classical slip calculation, a drive friction coefficient of 0.25 to 0.35 is used. In this paper, a mechanical model is developed to simulate the starting of a drive drum with a distributed mass-spring system for the belt- ing. The model allows the belt to be pre-tensioned, then torque is applied to develop a 7 1 and T 2 tension. The torque is applied up to the point of drum slip. The model produces the tension distribution in the belt around the drum face. The model pro- vides a clearer understanding of belt tensions around a drive drum, including requirements for viscoelastic contraction when designing a drive system. 1. Introduction The problem of defining the way in which belt tension changes from its tight-side 7 1 tension to the outgoing slack-side 7 2 ten- sion has been contemplated by a number of authors. The basic relationship that defines the amount of 7 1 tension in relation to T 9 tension at the point of drive drum slip is n = e (D The derivation of this equation is based on a simple calculus for the problem, where small elements are integrated around the wrap angle 6 from 7" 2 to 7 r with an effective coefficient of fric- tion u defined at the point of limiting friction or the onset of bod- ily slip. The right hand side of the equation defines the amount of TJT Z that can be supported by a frictional drive drum against the belt. In the problem, u is the static friction of the belt on the drive drum. It should be remembered that the actual 7, and 7 2 tension at the drive drum is purely the result of conveyor line re- sistance and material lift, and not a function of the drive drum coupling to the belt by friction. Typical values quoted for the effective friction factor u between belt and pulley are 0.25 < u < 0.35 for steel pulley (lower value) and lagged pulley (higher value). However, actual measured val- ues of friction coefficient for rubber on steel and rubber on rub- Dr. ALEX HARRISON, President of Scientific Solutions Inc., Denver, Colorado, USA, 2200 Chambers Road, Unit J, Aurora, CO 80011, USA. Tel.: +1 303 344 90 24; Fax: +1 303 344 91 02. Details about the author on page 167. ber range from 0.7 to 1.2, much higher than the effective friction u used today for drive slip calculations. Clearly the u used above is considered an effective friction value over the entire drum wrap length. When 7", becomes large by comparison to T 2 , slip has a higher probability of occurring, and so for design purpose the relationship 7, (2) is applied as a test for the point of slip. For dual drive drums, other relationships have been developed to accommodate dif- ferent wrap angles [1], The basic model does not take into ac- count the following important parameters that are present in a real system: Contraction of the belting from 7, to 7 2 (the belt's elasticity). • The real value of u between drum and belting (0.7 to 1.2). Localised sliding as torque is applied to the drum. Variations in the normal force due to the contraction prob- lem. Variations in friction coefficient around the drum. Belt speed exiting the drum will be less than drum speed. A good deal of recent research has been conducted on the problem of drum friction, with two papers by HARRISON [2, 3] and another by ZEDIES [4], In references [2] and [3] the problem of surface roughness and friction tests are addressed. In reference [4], actual instrumentation is used to monitor the way the drive drum encounters the belting. Based on surface roughness and rubber hardness tests [2], a new equation was developed to de- scribe the manner in which 7/7 2 varies around the drive wrap, - =1 +AB T 2 (3) where A = 2.0 for rubber on steel, and 6 is the wrap angle [2]. The equivalent friction coefficient developed in this research is not constant but drops from about 0.86 to 0.56 between 100 degrees and 220 degrees of wrap, with ^ = 0.7 at 150 degrees of wrap. Therefore, n(6) is a curve that shows equivalent friction decreasing non-linearly as wrap increases, with the model de- scribed by Eq. (3). A conclusion of the research discussed above was that the mechanism that describes the way in which 7 1 evolves to 7 2 around a drive drum is complex and not well understood at the belt/drum interface. Friction coefficients are clearly much higher than used in design tests for slip. This paper describes another approach to the problem that involves physical modelling and simulation. 75

Upload: yashodhan-karkare

Post on 06-Nov-2015

27 views

Category:

Documents


0 download

DESCRIPTION

Drive Friction

TRANSCRIPT

  • solidshandling Volume 18 Number 1 January/March 1998 Belt Tension Around a Drive Drum

    Modelling Belt TensionAround a Drive Drum

    A. Harrison, USA

    SummaryAnalysis of the drive drum friction problem by classical calculusprovides a slip test that allows a designer to determine a maxi-mum T^/T2 for a conveyor. In the classical slip calculation, adrive friction coefficient of 0.25 to 0.35 is used. In this paper, amechanical model is developed to simulate the starting of adrive drum with a distributed mass-spring system for the belt-ing. The model allows the belt to be pre-tensioned, then torqueis applied to develop a 71 and T2 tension. The torque is appliedup to the point of drum slip. The model produces the tensiondistribution in the belt around the drum face. The model pro-vides a clearer understanding of belt tensions around a drivedrum, including requirements for viscoelastic contraction whendesigning a drive system.

    1. IntroductionThe problem of defining the way in which belt tension changesfrom its tight-side 71 tension to the outgoing slack-side 72 ten-sion has been contemplated by a number of authors. The basicrelationship that defines the amount of 71 tension in relation toT9 tension at the point of drive drum slip is

    n = e (D

    The derivation of this equation is based on a simple calculus forthe problem, where small elements are integrated around thewrap angle 6 from 7"2 to 7r with an effective coefficient of fric-tion u defined at the point of limiting friction or the onset of bod-ily slip. The right hand side of the equation defines the amountof TJTZ that can be supported by a frictional drive drum againstthe belt. In the problem, u is the static friction of the belt on thedrive drum. It should be remembered that the actual 7, and 72tension at the drive drum is purely the result of conveyor line re-sistance and material lift, and not a function of the drive drumcoupling to the belt by friction.Typical values quoted for the effective friction factor u betweenbelt and pulley are 0.25 < u < 0.35 for steel pulley (lower value)and lagged pulley (higher value). However, actual measured val-ues of friction coefficient for rubber on steel and rubber on rub-

    Dr. ALEX HARRISON, President of Scientific Solutions Inc., Denver, Colorado,USA, 2200 Chambers Road, Unit J, Aurora, CO 80011, USA. Tel.: +1 303344 90 24; Fax: +1 303 344 91 02.

    Details about the author on page 167.

    ber range from 0.7 to 1.2, much higher than the effective frictionu used today for drive slip calculations.Clearly the u used above is considered an effective frictionvalue over the entire drum wrap length. When 7", becomes largeby comparison to T2, slip has a higher probability of occurring,and so for design purpose the relationship

    7,(2)

    is applied as a test for the point of slip. For dual drive drums,other relationships have been developed to accommodate dif-ferent wrap angles [1], The basic model does not take into ac-count the following important parameters that are present in areal system: Contraction of the belting from 7, to 72 (the belt's elasticity). The real value of u between drum and belting (0.7 to 1.2). Localised sliding as torque is applied to the drum. Variations in the normal force due to the contraction prob-

    lem. Variations in friction coefficient around the drum. Belt speed exiting the drum will be less than drum speed.A good deal of recent research has been conducted on theproblem of drum friction, with two papers by HARRISON [2, 3] andanother by ZEDIES [4], In references [2] and [3] the problem ofsurface roughness and friction tests are addressed. In reference[4], actual instrumentation is used to monitor the way the drivedrum encounters the belting. Based on surface roughness andrubber hardness tests [2], a new equation was developed to de-scribe the manner in which 7/72 varies around the drive wrap,

    - = 1 + A BT2

    (3)

    where A = 2.0 for rubber on steel, and 6 is the wrap angle [2].The equivalent friction coefficient developed in this research isnot constant but drops from about 0.86 to 0.56 between 100degrees and 220 degrees of wrap, with ^ = 0.7 at 150 degreesof wrap. Therefore, n(6) is a curve that shows equivalent frictiondecreasing non-linearly as wrap increases, with the model de-scribed by Eq. (3).A conclusion of the research discussed above was that themechanism that describes the way in which 71 evolves to 72around a drive drum is complex and not well understood at thebelt/drum interface. Friction coefficients are clearly much higherthan used in design tests for slip. This paper describes anotherapproach to the problem that involves physical modelling andsimulation.

    75

  • Belt Tension Around a Drive Drumbulk

    Volume 18 Number 1 January/March 1998 SOlMS

    2. Physical ModellingDetermining the way that 71 tension evolves to a lower 7~2 valuearound a drive drum is one of the subjects of this paper. A phys-ical model using masses and damped springs has been devel-oped to simulate a belt wrapped around a drive drum. Themodel may be generally applied to a drive design review oraudit.

    Distributed mass and elasticity elements are developed thatform the input to a simulation engine [5], The simulation engineallows the model to run under a set of program control func-tions, written so that the time when torque is applied can beuser-input.For example, the rate of pre-tension applied to the model canbe controlled so that initial conditions are stabilised. After initialpre-tensioning oscillations are stabilised, the rate of torque ap-plication is controlled to prevent drum slip while the model dif-ferentially tensions all the springs on the drum face.Fig. 1 illustrates the model of a drive drum, constructed so thatreal-world parameters are applied. For example, the followingset of parameters are used in the model that will be described inthis paper:Belt mass = 25.5 kg/m (fabric belt) - 7 masses, 7 kg/massBelt mass = 36.5 kg/m (steel cord belt) - 5 masses, 14 kg/massDrive drum = 1.0 m diameter, wrap angle 210 degreesStatic friction = 0.9, kinetic friction = 0.7 (rubber on steel)Spring stiffness k = belt stiffness, length L/Q per element spring

    6k, L/6

    k , L 6k

    k , L

    Fig. 1: Physical model of a drive drum for torque application and simulation ofspring tensions

    Fig. 2: Procedural functions for the simulation

    Physical Model Inputs \Control Parameters for

    Simulation J

    Working Model Inputs ]Simulation Engine I

    Drum and BeltData

    Control RampAlgorithms

    Runge-KuttaAuto Time StopCollision Detection

    /SimulationNI Outputs J

    Fig. 2 shows the interface between the physical model and theworking model simulator [5]. All mass positions and spring ten-sions are tracked during the simulation. All inputs to the physi-cal model need to be correct so that the simulation will be sta-ble and converge to real solutions.In establishing the model, the following procedure is required:a) Each spring is attached to a mass element at a location that

    will not cause significant over turning moments.b) Each spring has the same length L/6, where L is the length of

    the unsupported spring.c) Every mass contacts the pulley face with even bearing pres-

    sure so that sliding forces are rapid to compute and so thatbouncing is not induced when the drum torque is applied.

    d) Damping is applied to each spring to reduce dynamic inter-actions while pre-tension and torque is being applied.

    e) At time t = 0, a displacement is applied to the drum along itshorizontal axis, to pre-tension the springs to a typical belttension.

    f) At a later time after all initial oscillations die away (the systemis in static equilibrium and tensioned) the torque is rampedup slowly to create a 7^ and 7"2 tension. This process contin-ues until the drum rotates beneath the masses (slip). Eachmass adjusts its position by sliding on the drumface until allforces are balanced. Spring forces then represent belt ten-sions.

    3. Low Pre-Tension ExampleTension in a fabric belt is generally lower than in higher stiffnesssteel cord belts, and therefore spring stiffness has to be set ac-cordingly. A model was constructed for a 6 element spring sys-tem, as shown in Fig. 1. Spring lengths were of the order of0.28 m and the drive drum diameter was 1.0 m. The exactlength of the spring lengths determines the initial strain, howeverthese initial conditions are no longer relevant once the torque isapplied to the drum and spring tensions start to deviate fromeach other. Static friction was set at 0.9 and sliding or kineticfriction was set at 0.7.

    At time t = 0, a slow rising pre-tension of 5 kN was applied tothe belting for 2 seconds, and then held at this level until 5 sec-onds. The control logic for the simulation sets a true conditionby the first value and the false condition by the second value, asfollows :

    time steppretension Force Fxtorque

    0.025 s

    if (f < 2,5000 * f/2,5000)if (f > 5,50 * (f - 5), 0)

    Each mass was set at 25 kg and each spring had a stiffness of120 kN/m. Damping for each spring was set to produce smoothsteady-state conditions. Setting the damping too large results indynamic stiffness and possible instabilities due to the produc-tion of high frequencies in the model.Fig. 3 shows the result of the model simulation. Soon aftertorque is applied, motion of the drum is observed in this partic-ular model.

    Tensions in the spring elements between 5 and 10 seconds arebasically locked because there is no relative sliding of themasses at a low value of applied torque. The upper 2 springsnear 7^ (the highest rising curves in the left hand side (Ihs) box ofFig. 3) experience a rise in tension before others around thedrum. The T2 side spring tension starts to fall at the same timethe 71 spring tension rises (see right hand side (rhs) box ofFig. 3).

    76

  • bulksolidshandling Volume 18 Number 1 January/March 1998 Belt Tension Around a Drive Drum

    Fig. 3: Results of a low tension simulation

    In general, tensions for springs shown in the Ins box are orderedfrom top to bottom in the same way they are drawn in Fig. 1,namely the top spring of length L contains the tight side tension7^, and so on. In the rhs box, the lowest spring tension is alsothe last long spring, i.e. at 7~2 belt tension.This model slips early in the simulation, allowing the sliding ofthe drum to drag the masses to a point of static equilibrium. Thispoint is reached when the masses locate at a point on the drumsurface where spring forces across each element is balancedby the drag force at its particular normal load. The simulation inthis particular model describes constant creep or sliding of thebelt relative to the drive drum, typical of many operating fabricbelt drives that show up to 5% slip when running.Analysis of the data shows that all masses slide to a point of sta-tic equilibrium, with an active dynamic friction of uk = 0.7 foreach mass. The drum takes about 1 minute to reach a pointwhere the system is at static equilibrium. At this time, no morespring extensions occur, and the tensions are measured:

    = 791 7 N 72 = 2411 N.The ratio T^/T2 = 3.2837. From the general equation for slip, theeffective friction coefficient u, that supports this tension ratio atslip, with 210 degrees of wrap, is

    M, = {In (7~/72)} / 9 = 0.3257.This result is very interesting in that the static friction coefficientof each mass is set at near measured values of \a = 0.9, the slid-ing friction is uk = 0.7, and the equivalent or effective friction isH = 0.3257.The simulation has produced a result for \a that the simple the-ory described by Eq. (1) would require, namely a ^ in the range0.25 < u< 0.35. The value of the effective friction for the systemis about half the sliding friction value.By taking the data from Fig. 3, it is possible to plot the way inwhich the spring element tensions vary around the drive drum.Fig. 4 shows the graph of the evolution of spring forces at dif-ferent levels of K = T^/T2, up to the point of static equilibrium.This graph shows that the tension in a belt around a drivingdrum evolves with an approximate "S" curve. There may be

    some error in the curve near 7T, just before system static equi-librium (slip) due to dynamic oscillations in the mass locations(and hence spring tensions) at the point where the systemchanges to static equilibrium.

    4. High Pre-Tension ExampleThe previous example showed a relative low tension belt model,typical of those used in underground coal mines. It showed thateven though friction of a belt against a drive drum is measuredat about jx = 0.9 at limiting friction, the effective friction as seenat the point of slip (T,/T2 = 3.2837) is \JL = 0.3257. This simula-tion confirmed the typical value applied in Eq. (1).Another drive drum model with higher stiffness springs(k = 1000 kN/m) was constructed to simulate a steel cord belt.A 4-spring element with 5 masses was used. Each mass had avalue of M = 14 kg and the same drum parameters were usedas in the previous example. The friction used was also the sameas in the previous example, namely us = 0.9 and uk = 0.7.

    Fig. 4: Tension in the belt between 7, and 72 for various K ratios

    0 30 60 90 120 150 180 210Degrees

  • Belt Tension Around a Drive Drum Volume 1 8 Number 1 January/March 1 998bulk

    Various Tensions around Druml 1TUT2 Tensions!(NJOe+005(N)(N)6e+005(N)""

    11.8026+005

    1.600 2.400

    (N)0e+005(N);1.966e+005

    1.802e+005

    1.6386+005

    i^ TSangefe'Olle+005

    1.147e+005

    9.830e+004

    8.192e+004

    4.915s+0V

    3.27?e+004

    1.638e+OQ4!

    t^C

    x^

    1.600

    i J""' >i^

    W^C

    N.2.400

    A-xAft

    a200 (a)Fig. 5: Simulation to the point of slip for a 100 kN model pre-tension

    For a model pre-tension of 85 kN, 71 and T2 tensions at drivedrum were 172 kN and 14.1 kN respectively, at the point ofdrum slip. The slip ratio T^/T2 = 12.226, which gives an effectivefriction factor of u = 0.654. As pre-tension in the model in-creases to 165 kN, the effective friction increases u = 0.8 be-fore the drum physically slides beneath the distributed masses.Fig. 5 shows the evolution of spring tensions around the drumwith 4 spring elements and 5 masses. The model was pre-ten-sioned to 100 kN, and the above simulation techniques wereapplied. Tensions achieved prior to bodily slip of the drum were7., = 202 kN and T2 = 8 kN. This gives an effective friction coef-ficient of u = 0.84. The mass-drum interface friction was 0.9.

    Fig. 6: Full time history of the simulation in Fig. 5

    -16.000 i =; * i

    0.400 1.200

    ; ^Y

    2000 2.800 \ It]

    This test reveals that a significant torque can be applied to thedrum prior to slip, well above the values traditionally used forEq. (1). Spring tensions stabilise to constant values in this modelat 3.2 seconds after torque is applied. This is where constantdrive slip is modelled. The pre-tension is so high that the modelmaintains stability even though the drum slips. This is not thecase with lower pre-tensioned models. Increasing the numberof springs to 6 (as before) results in very high frequencies due tothe spring stiffness and so the model becomes unstable duringsimulation.Fig. 6 shows a full time history simulation of the model. The ini-tial pre-tension causes dynamic overshoot in all spring tensions.

    After the system settles down,torque is applied gently and thetensions split as shown, up to thepoint the drum slips. The lowergraph shows the rotation of thedrum and total slip at about 2.8seconds.A number of stable simulationswere conducted for the drive withthe same mechanical conditionsas above. The model's pre-ten-sion value was increased for eachsimulation so that K increased.Fig. 7 shows the way the modelsimulates an effective friction co-efficient as K = T^/T2 increases.The effective friction u was calcu-lated for each simulation. Clearlyu increases as K is increased,

    showing that the pre-tension levelaffects the effective friction coeffi-cient at a given wrap angle.

    78

  • bulkVolume 18 Number 1 January/March 1998 Belt Tension Around a Drive Drum

    MuEff-

    0.90.80.70.6

    0.30.20.1

    010 15

    K = T1/T2 Ratio20 25

    Fig. 7: Variation in effective friction M with 7/72 (drum diameter = 1 m)

    5. Application to DesignIn conveyor design various tension calculation methods areused to find the effective tension 7"e around the conveyor. Thesemethods include CEMA [1], ISO 5048 or DIN 22121. Whilsteach of these methods have drawbacks they are widely used inindustry to produce an effective tension Te around a particularconveyor profile. Commercial models may be more accuratethan ISO or CEMA.Using the simulation results for a typical 1 m diameter drivedrum, Eq. (1) may be re-written to include the effect of belt ten-sion on effective friction. For the above conditions of actual fric-tion, namely us = 0.7 and uk = 0.9, the equation for u(K) is

    \i(K) -aK-e~ (4)

    where a = 6 x 10"3 and the equation for the slip test becomes

    < e (5)

    A design slip test may be summarised with a 5 point procedureas follows :1. Compute the effective tension 7e for the conveyor.2. Select a 7~2 tension based on lowest profile tension for sag.3. Compute the value of 71 based on a best assessment for Te.4. Compute drive friction coefficient using Eq. (4)5. Determine whether 7^ is too great, using Eqs. (5 or 6).There are many instances where the drive slips beneath thebelt, contrary to design calculations [1]. Each particular con-veyor drive design may be analysed using simulation, so that theinfluence of drive diameter and lagging type may be taken intoaccount. Viscoelastic considerations of the belting can now beaccommodated since the tension evolution between 7^ and 7~2is known up to the point of slip [6]. This process is particularlyimportant since there are new lagging materials in the marketthat claim a larger coefficient of driving friction to prevent slip.For example, ceramic lagging is reported to exhibit an effectivefriction value of M =0.45 to 0.5. A simulation of ceramic againstthe rubber belting will give the actual values of friction for the sliptest. Mechanical impression of the lagging nodules in the caseof ceramic lagging may be the cause of the apparently higher ef-fective friction factor, and this problem is being researched.

    6. ConclusionsSimulation of a drive drum against a rubber belt with static andsliding friction values near 0.9 and 0.7 respectively shows thatthe effective friction coefficient u varies between 0.32 and 0.84,depending on the belt tension. At low belt tensions in relation tobelt stiffness the slip predicting Eq. (2) is valid, however as theratio T^/T2 increases, a new effective friction coefficient that istension dependent needs to be applied to prove the point ofslip. Another outcome of the modelling is the tension distributionin the belt around the drive drum. This result, together with vis-coelastic relaxation rates, may lead to improved belting materi-als.Although a specific simulation example is described in thispaper, along with assumptions and conclusions, every drivedrum system will require individual examination to determine thereal point of slip.

    During design iterations, the tension at some point along a beltneeds to be set, and usually a value is selected for T2 or tail ten-sion (in head drive conveyors) to ensure that sag at the loadpoint is kept within design limits. This may change for tail drivesystems. In general, setting 7~2 or tail tension is relatively simpleif gravity take-ups are used.Supposing that 72 is set in a design to control the sag at theload point, the relationship for the slip test becomes

    where the effective tension due to conveyor resistance isTe = (T,-T2).The research shows that the slip test depends on the effectivetension due to conveyor resistance. Larger effective drive drumfriction values are predicted with this research compared to tra-ditionally used drive friction values of between 0.25 and 0.35. Ina traditional application, Te is not one of the parameters that af-fects the effective friction coefficient.

    References[1] CEMA: Belt Conveyors for Bulk Materials Handling; 3rd

    Edition, 1995.[2] HARRISON, A. and ROBERTS, A.W.: Mechanisms of force

    transfer on conveyor belt drive drums; bulk solids handlingVol. 12 (1992) No. 4, pp. 581-584.

    [3] HARRISON, A. and BARFOOT, G.: Load sharing between mul-tiple drive conveyors; Procs. SME Conference, Reno,Nevada, 1993.

    [4] ZEDIES, H.: Investigation of the strain on drum-laggingsaiming the optimization of the lagging; Doctoral thesis, Uni-versity of Hannover, Germany, 1987.

    [5] Working Model 2D for Windows 95, Users Manual, Knowl-edge Rev., Summit software 1989-1996

    [6] HARRISON, A.: Lagging and belting dynamics in conveyordrives; bulk solids handling Vol. 16 (1996) No. 2, pp. 353-359.

    79