【ADAMS中运动副添加摩擦因素】之 Revolute Joint Friction

yixiangren

　　借着回答网友的机会，把Revolute Joint Friction详细地看了一下。下面总结一下我的心得体会!
　　以下都是个人观点，水平有限，错误在所难免，主要的目的是抛砖引玉，大家一起讨论!
　　关于其他运动副中添加摩擦的详细讲解见：Adding Friction to Idealized Joints
　　转载请注明出处!
　　一、Revolute Joint Model
　　Joint reactions (Fa and Fr), bending moment (Tr), and torque preload (Tprfrc) determine the frictional torque in a revolute joint. You can turn off one or more of these force effects using switches SW1 through SW3. The joint reactions (Fa and Fr) (注：图3中FR21和Ff21) are converted into equivalent torques using the respective friction arm (Rn) (注：图3中p) and pin radius (Rp)(注：图3中r). The joint bending moment (Tr) is converted into an equivalent torque using pin radius (Rp) divided by bending reaction arm (Rb). The frictional torque (Tfrict) is applied along the axis of rotation in the direction that the FRD block computes.


　　图1 Revolute Joint Model



　　图2 Block Diagram of Revolute Joint
　　上述是帮助文件中关于转动副中摩擦的描述。这是一个较为复杂的模型，限于水平，这里将不对bending reaction 进行讨论。在模型中也将不对其进行设置。图2是ADAMS中如何将各个因素添加在一起，最终构成整个转动副的摩擦扭矩。





　　图3 轴颈摩擦力的确定
　　图3是常见的转动副轴颈摩擦模型(《机械原理》孙恒 陈作模 第六版 P91)。详细分析了轴颈摩擦的原理。设受径向载荷G作用下的轴颈1，在驱动力矩Md作用下，在轴承2中等速转动。轴承2对轴颈1的摩擦力Ff21=fv*G，其中fv=(1-0.5pi)f (注：对于配合进门且未经跑合的转动副取较大值，而对于有较大间隙的转动副去较小值)。摩擦力对轴颈的摩擦力矩为Mf=Ff21*r=fv*G*r。根据轴颈1受力平衡可得：G=-FR21，Md=-Mf，故可得：p=fv*r，其中r为转动副半径，红色圆为摩擦圆，p为摩擦圆半径。
　　更为复杂的模型这里不再讨论，如果是做摩擦方面研究的要更加深入的分析两者的不同。对于一般研究应注意对比图1和图3，有助于将我们所学的摩擦概念与ADAMS中使用的摩擦原理相对应。后面将通过ADAMS中的模型仿真来说明这些问题。
　　二、Friction Regime Determination (FRD)
　　下面讲解了ADAMS中的三种摩擦状态：动摩擦、静动摩擦转变、静摩擦。相信如果可以使用ADAMS软件，那么英语水平还是不错的，这里我就不再翻译，原文是对ADAMS的最好解释，大家自己琢磨吧!(注：红色字体是我觉得文中应该重点关注的内容)
　　Three friction regimes are allowed in ADAMS/View:
　　The regime:Means:
　　Dynamic friction A joint is in dynamic friction if its joint velocity magnitude exceeds 1.5 times the stiction transition velocity. The dynamic coefficient of friction (md) is used in the computation of frictional forces.
　　Transition between dynamic and static frictionIf the joint velocity magnitude is between 1 and 1.5 times the stiction transition velocity, the joint is considered to be transitioning between static and dynamic friction. A STEP function transitions the coefficient of friction between the dynamic (md) and static (ms) coefficients of friction.
　　Static frictionA joint is in static friction when the joint velocity magnitude falls below the stiction transition velocity. The effective coefficient of friction is computed using the joint creep, joint velocity, and static coefficient of friction ( ms ).
　　Revolute Joint Options：详细讲解了各个参数的含义。
　　For the option:Do the following:
　　Mu StaticDefine the coefficient of static friction in the joint. The magnitude of the frictional force is the product of Mu Static and the magnitude of the normal force in the joint, for example:
　　Friction Force Magnitude, F = μN
　　where μ = Mu Static and N = normal force
　　The static frictional force acts to oppose the net force or torque along the Degrees of freedom of the joint.
　　The range is > 0.
　　Mu DynamicDefine the coefficient of dynamic friction. The magnitude of the frictional force is the product of Mu Dynamic and the magnitude of the normal force in the joint, for example:
　　Friction force magnitude, F = μN
　　where μ = Mu Dynamic and N = normal force
　　The dynamic frictional force acts in the opposite direction of the velocity of the joint.
　　The range is > 0.
　　Friction ArmDefine the effective moment arm used to compute the axial component of the friction torque. The default is 1.0, and the range is > 0.
　　Bending Reaction ArmDefine the effective moment arm use to compute the contribution of the bending moment on the net friction torque in the revolute joint. The default is 1.0, and the range is > 0.
　　Pin RadiusDefines the radius of the pin.
　　The default is 1.0, and the range is > 0.
　　Stiction TransitionVelocityDefine the absolute velocity threshold for the transition from dynamic friction to static friction. If the absolute relative velocity of the joint marker is below the value, then static friction or stiction acts to make the joint stick.
　　The default is 0.1 length units/unit time on the surface of contact in the joint, and the range is > 0.
　　Max StictionDeformationDefine the maximum displacement that can occur in a joint once the frictional force in the joint enters the stiction regime. The slight deformation allows Adams/Solver to easily impose the Coulomb conditions for stiction or static friction, for example:
　　Friction force magnitude < static * normal force
　　Therefore, even at zero velocity, you can apply a finite stiction force if your system dynamics require it.
　　The default is 0.01 length units, and the range is > 0.
　　Friction Torque PreloadDefine the preload friction torque in the joint, which is usually caused by mechanical interference in the assembly of the joint.
　　The default is 0.0, and the Range is > 0.
　　EffectDefine the frictional effects included in the friction model, either Stiction and Sliding, Stiction, or Sliding. Stiction is static-friction effect, while Sliding is dynamic-friction effect. Excluding stiction in simulations that don't require it can greatly improve simulation speed. The default is Stiction and Sliding.
　　Input Forces to FrictionDefine the input forces to the friction model. By default, all user-defined preloads and joint-reaction force and moments are included. You can customize the friction-force model by limiting the input forces you specify. The inputs for a translational joint are:
　　&#8226; Preload
　　&#8226; Reaction Force
　　&#8226; Bending Moment
　　Friction Inactive DuringSpecify whether or not the frictional forces are to be calculated during a Static equilibrium or Quasi-static simulation.

　　The joint velocity determines the instantaneous friction regime for a joint. The following is a block diagram of the friction regimes available in ADAMS/Solver.





　　图4 Block Diagram of Friction Regimes
　　三、ADAMS模型分析

　　模型是一个简单的圆柱体，与地面间添加转动副，并添加摩擦。本文采用柔性加载，只考虑稳态时的情况(时间足够长时)，具体的模型参数大家下载模型()查看，这里就不再一一论述。




　　图5 ADAMS模型(施加0.12N.m扭矩)




　　图6 ADAMS模型(施加0.121N.m扭矩)

　　当模型确定后，如果驱动力矩小于产生的摩擦阻力矩，那么系统仍保持静止，如果大于则圆柱体加速转动。通过Mf=Ff21*r=fv*G*r进行计算，通过 p=fv*r，由于转动副半径和摩擦圆半径已知，可计算出fv，则整个系统匀速运动的驱动力矩为0.12N.m附近，因此本文中加载了0.12N.m 和0.121N.m两种情况。通过对比发现当驱动力矩大于摩擦阻力矩时，圆柱体逐渐加速;驱动力矩小于摩擦阻力矩时，圆柱体保持静止(稳态)。注意对比图中曲线。


　　图7 ADAMS模型(圆柱体具有初始速度)
　　对比两中情况。左图中：开始时加载力矩小于摩擦阻力矩，转速逐渐减小;随着加载力矩的逐渐增大，当加载力矩大于摩擦阻力矩时，转速逐渐增加。右图中，由于加载力矩一直小于摩擦阻力矩，因此转速逐渐减小为0。
　　三、问题的引申
　　1、如何利用ADAMS仿真动静摩擦状态的转变;
　　2、如何利用ADAMS建立更为复杂的摩擦模型进行仿真分析
　　3、动摩擦系数、静摩擦系数、fv(选取原则)、转动副半径、摩擦圆半径之间的关系，有几个独立的变量。
　　ADAMS模型下载

　　限于水平，上述观点难免有误，旨在抛砖引玉，希望大家踊跃讨论!


转自：http://blog.sina.com.cn/s/blog_6203a6df0100nlrr.html