simpack_系统动力学

Multi-body System Dynamics 方向华 总监 GET集团北京分公司 Model Setup in SIMPACK Bodies Joints Force Elements Constraints Excitations Sensors ... Mass, Center of ~, I-Tensor, Marker, 3D- Primitive from Marker, to Marker, Type from Marker, to Marker, Type from Marker, to Marker, Type Type, Paramete r, u-Vectors from Marker, to Marker, Type Draw Topology Separate into Bodies, Joints, Force Elements, ... FEMCAD ... External Data Real System 物理系统抽象 Open MBS Tree Structure Closed Loop MBS Tree Structure 开环、闭环拓扑结构 Inertial Frame Joint 1 (1 DOF) Joint 2 Joint 3 Body 1 Body 2 One of the possible Solutions in SIMPACK (relative Kinematics): Joints 1,2: define the Topology ⇒ in SIMPACK: joints Joint 3: defines closed Loop ⇒ in SIMPACK: constraints joints: give system degrees of freedom constraints: lock motion - close kinematic chains - make closed loops - reduce number of degrees of freedom Z Y X Application Example: Simple Crank 正弦机构 Joints/Constraints α β γ x y z α 1313 αα Body name Reference Frame: Constraint: Bodies: Joints: Force Elements: with: name of body with: - from Marker --> to Marker - locked direction of motion (in Coordinates of from Marker) with: - from Marker --> to Marker - joint state position (in Coordinates of from Marker) with: - SIMPACK force element type (cmp calculated w.r.t from Marker) Isys ∑ DOFsystem = ? ∑ FOSsystem = ? joint; α - 1 DOF - Revolution around x-Axis Body 2 joint; α -1 DOF - Revolution around x-Axis constraint; L: z (locked Transl. in z, reduce Number of DOF) Body 1 Task: How do you calculate the Number of Degrees of Freedom (DOF) of closed Loop Systems and Number of first order States (FOS)? Z Y X 正弦机构 Task: How do you calculate the Number of Degrees of Freedom of closed Loop Systems (DOF) and Number of first order States (FOS)? ∑ DOFsystem = ∑ DOFjoint - ∑ constraint ∑ FOSsystem = 2•∑DOFjoint + ∑ constraint joint; α - 1 DOF - Revolution around x-Axis Body 2 joint; α -1 DOF - Revolution around x-Axis constraint; L: z (locked Transl. in z, reduce Number of DOF) Body 1 Number of differential Equations Number of algebraic Equations Z Y X 正弦机构 DOF system: ∑ DOFsystem = ∑ DOFjoint - ∑ constraint FOS system: ∑ FOSsystem = 2•∑ DOFjoint + ∑ constraint Joint 4 Joint 3 Joint 2 Joint 1 Body 3 Body 2 Body 1 αα Body 3 α Body 2 Body 1 L: y,z Solution 1: Solution 2: ∑ DOFsystem = 18 - 17 = 1 ∑ FOSsystem = 36 + 17 = 53 ∑ DOFsystem = 3 - 2 = 1 ∑ FOSsystem = 6 + 2 = 8 Body 3 Body 2 Body 1 6 DOF 6 DOF L: x,y,z, β,γ 6 DOF L: x,y,z, β,γ L: y,z L: x,y,z,β,γ Z Y X „Smart“ SIMPACK Model „Stupid“ SIMPACK Model Independent and Dependent Joints N_Independent Joints = ∑ DOFjoint (= N_Joint States) - ∑ constraint = ∑ DOFsystem Solution II β-dep α-indep Solution I β-indep α-dep non-linear implicit equations solved by Newton-Iteration ---> solution depending on initial joint states of body a and b! Problem β α a b α-indep Multiple Solutions are possible β-dep Z Y X Isys x y Draw Topologie of McPherson Suspension System: ∑DOFsystem = ∑DOFjoint - ∑constraint = ? ∑FOSsystem = 2•∑DOFjoint + ∑constraint = ? Isys x y body Reference Frame: Constraints: Bodies: Joints: Force Elements: dummy wheel_platearm steering rod damper_upper damper_lower wheel rackdummy 0DOF α Draw Topologie of McPherson Suspension System: body Reference Frame: Constraints: Bodies: Joints: Force Elements: Isys x y 0 DOF α, γ βα, β, γ 0DOF dummy wheel_platearm steering rod damper_upper damper_lower x, y, z x, y, z z spring damper ‚help‘ (tyre) wheel ∑DOFsystem = ∑DOFjoint - ∑constraint = 8 - 6 = 2 ∑FOSsystem = 2•∑DOFjoint + ∑constraint = 18 + 6 = 24 rackdummy Exercise 4 (2): Set Up of a Complex Complete Vehicle Model track_joint_19 (vehicle reference system) Topologie Automotive_Plus_Models before definition of Substruktur Intersections (e.g. Substructures “Steering” and “Front Axle Left”): 6 DOF (Type 19) Sub: STEER dummy 0 DOF Sub: FRONT AXLE LEFT dummy_ chassis 0 DOF dummy_ steering 0 DOF Exercise 4 (3): Set Up of a Complex Complete Vehicle Model track_joint_19 (vehicle reference system) Topologie Automotive_Plus_Models after definition of Substruktur Intersections (e.g. Substructures “Steering” and “Front Axle Left”): 6 DOF (Type 19) Sub: STEER dummy 0 DOF Sub: FRONT AXLE LEFT dummy_ chassis 0 DOF dummy_ steering 0 DOF Exercise 4 (4): Set Up of a Complex Complete Vehicle Model Sub: REAR AXLE RIGHT tyre fl (49) Topologie Automotive_Plus_Models e.g Simple Complete Car 6 DOF (Type 19) Sub: STEER dummy 0 DOF 0 DOF track_joint_19 (vehicle reference system) Sub: REAR AXLE LEFT dummy_ chassis Sub: FRONT AXLE RIGHT dummy_ chassis dummy_ steering 0 DOF 0 DOF Sub: FRONT AXLE LEFT dummy_ chassis dummy_ steering 0 DOF 0 DOF dummy_ chassis tyre rl (49) tyre rr (49) tyre fr (49) The model consists of the following functional bodies: - ENG: engine block - CRS: crankshaft - CNR: conrod (4x) - PIN: pin (4x) - PIS: piston (4x) - TSP: torsional damper primary - TSS: torsional damper secondary - FLP: flywheel primary - FLS: flywheel secondary - CLU: clutch - LOD: engine load brake x z y >> Draw Topology of the Complete Model and Check DOF and FOS: ∑DOFsystem = ∑DOFjoint - ∑constraint = ? ∑FOSsystem = 2•∑DOFjoint + ∑constraint = ? CHA CRS CNR PIN PIS FLP FLSTSS CLU ENG LOD TSP body Reference Frame: Constraints: Bodies: Joints: Force Elements: Connector: Body 0 DOF Connector Model Substructure: The engine model should consist out of four substructures: • Engine block with crankshaft ………………….. to be created • Crank train with gas forces…………………….. to be created • Additional masses (flywheel, clutch, torsional damper)…………… predefined • Engine load (brake torque)…………………..….predefined Create substructure: engine block with crankshaft • Review topology of the substructure • Review DOFs of the substructure ENGConnector DOF 0 DOF 6 CRS α FEL 43 (4x) ∑ =+= 716DOFENG Create substructure: crank train CNR CRS Connector DOF 0 α PIN α ∑ =−= 022DOFCTR ENG Connector DOF 0 PIS y,α DOF 0 FEL 50: Gas force 0 DOF 6 DOF Assemble the main model (1) CHA FE 43 (4x) CNR PINFE 50 PIS CTR (4x) FLP FLS 0 DOF TSS 0 DOF 0 DOF CLU 0 DOF α, y � � � 0 DOF αd αd αd ADMADM LOD Excitation u(t) � 0 DOF TSP ENB_01 CRB_01 FE 13 LOD ( ) ( )∑ =⋅−⋅+= 724247DOF GAS CRS ENG αd αd αd ~ ~ ~ • Every body in the MBS model has it‘s own BFRF • The BFRF is always located at (0,0,0) by definition and can not be deleted • All body fixed marker coordinates are given with respect to the BFRF (except marker coordianates given relatively to a reference marker) Even if the BFRF is an ordinary marker, it is not recommended to use it for modelling purposes. In order to keep a clear model structure it is better to create a new marker at (0,0,0) and assign an appropriate name to it. ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = 0 0 0 BFRF P1BFRF P2BFRF P1 P2 The position of a body in space is given by its relation to the inertia system (Isys) or another body: Example: Body1 is fixed to the inertia system with its BFRF at P1 Body2 is rotating around P3 on Body1. The position of a body in space results from the joint definition of this body (= assignment of joint coupling markers) ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = 1 1 1 1Isys 1Body1 z y x BFRF BFRF 0 0 0 ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = 0 0 0 22 BFRFBody 23BodyBFRF P 24BodyBFRF P 12BodyBFRF P 13BodyBFRF P Isys IsysIsys P1= Graphical elements (3D primitives) are visualisation elements without any physical meaning (except functional primitves like e.g. gearwheels) to the MBS system. The position of any 3D primitive on a body is given with respect to a marker located on the body (most commonly the BFRF). P1 belongs to Body1, even if no 3D graphic is visible at it‘s location. The cuboid is defined with respect to BFRF with primitve built in coordinates of P2. The center of the sphere is defined with respect to P2 with additional primitve built in coordinates. Therefore no marker is needed in it‘s center.Isys ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = center center center sphereP2 z y x rr ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = center center center cuboidBFRF z y x rr P2 BFRF P1 Changing the built-in positions of 3D primitives will not change the position of the body in space. Even if the shape of the body has changed, all marker positions will stay at the same location. The body did not move at all. Isys ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = center center center sphereP2 z y x rr ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = center center center cuboidBFRF z y x rrP2 BFRF P1 Marker coordinates from an assembly drawing only: SIMPACK body coordinates given in the Assembly Coordinate System = Isys = BFRF (e.g. Vehicle coordinate system). Example (1) (body joint definition at position P2 between marker P2 in Isys and marker P2 on body ) BFRFIsys = ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ == P2 P2 P2 IsysBFRF z y x P2P2 P2 P1 CG ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ == CG CG CG IsysBFRF z y x CGCG ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ == P1 P1 P1 IsysBFRF z y x P1P1 Example (2) (body joint definition at position P2 between marker P2 in Isys and marker P2 on body ) Marker coordinates from a part drawing only: SIMPACK body coordinates given in the Part’s Coordinate System = BFRF, located „somewhere“ on the part. PLEASE KEEP IN MIND: BFRF position in space results from joint definition between marker P2 in Isys and marker P2 position relative to BFRF in this case ! Isys ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = 0 0 LP2BFRF P2 P1BFRF = ⎟⎟ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎜⎜ ⎝ ⎛ = 0 2 0 LCGBFRF ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ = 0 0 0 P1BFRF CG Example (3) (body joint definition at position P2 between marker P2 in Isys and marker P2 on body ) BFRFIsys = ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ == P2 P2 P2 IsysBFRF z y x P2P2 P2 P1 CG ⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ == P1 P1 P1 IsysBFRF z y x P1P1 ⎟⎟ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎜⎜ ⎝ ⎛ = 0 2 0 LCGBFRF Marker coordinates combined from an assembly drawing and from an part drawing. P1 and P2 coordinates given in the assembly coordinate system (e.g. vehicle coordinate system). CG coordinates given in the Part coordinate system (e.g. P1 system). --> use marker relative to marker and/or CG relative to marker functionality in SIMPACK Thank you! E-mail to : xianghua.fang@bj-esp.com