Herman 
星期三, 2009/09/09 at 10:03
RecurDyn求解精度实验2 (40)
RecurDyn Virtual Lab 2: Simple Pendulum
Problem
AB is a uniform bar with a mass of 2 kg and a length of 450 mm. Bar AB swings in a vertical plane about the pivot at A. The angular velocity (theta dot) = 3 rad/s when theta = 30 degrees. Compute the force supported by the pin at A at that instant.
Theoretical Solution
Total reaction force is 14.4926 N.
| Given: |
Symbol |
Value |
Units |
| Bar Mass |
M |
2 |
kg |
| Bar Length |
L |
0.45 |
m |
| Gravity |
G |
9.80655 |
m/s |
| Initial angle from horizontal |
theta0 |
30 |
deg |
| Initial angular velocity |
thetadot0 |
3 |
rad/s |
If you want to know more information of this process.please reading the attachment file of this article.
Tracy 
星期一, 2009/08/31 at 14:42
Herman 
星期三, 2009/08/26 at 17:31
RecurDyn Virtual Lab 1: Spring-Mass Model
Problem
A 10-kg cylinder is released from rest with x=1 m, where the spring is not stretched. Determine the following:
(a) Maximum velocity v of the cylinder and the corresponding value of x (a)
(b) Maximum value of x during the motion (b)
The stiffness of the spring is 450 N/m.
Theoretical Solution
(a) maximum v=1.4619 m/s occurring at x=1.2179 m
(b) maximum x=1.4358 m
| Given: |
Symbol |
Value |
Units |
| Cylinder Mass |
m |
10 |
kg |
| Spring Stiffness |
k |
450 |
N/m |
| Gravity |
g |
9.80655 |
m/s |
| Initial Displacement |
x0 |
1 |
m |


RecurDyn Solution
To start RecurDyn and create a model:
- 1. On your Desktop, double-click the RecurDyn tool.
- 2. Create a new model:
To create the cylinder part (10 kg):
- 1. From the Toolkit bar, click the Cylinder tool.
- 2. Select centers of the two ends of the cylinder as (0,0.9,0) and (0,1.1,0) so the center of the cylinder is exactly 1 m away from the origin in the –y direction.
- 3. Double-click the cylinder to enter the Body edit mode. Then, right-click the cylinder and click Properties.
- 4. Change the radius to 0.05 m.
- 5. Exit Body edit mode.
- 1. Right-click the cylinder and select Properties.
- 2. Set Material Input Type to User Input.
- 3. Change Mass to 10 kg. Because the cylinder will not be rotating, the moments of inertia can remain unchanged.
- 1. From the Force menu, choose Spring.
- 2. Using the default creation method (Point, Point), select the origin and then the center of the cylinder.
To change the mass to 10 kg:
To create a spring part (spring-damper, k=450 N/m, no damping):
- 1. Right-click the spring and select Properties to set the spring constant and the damping coefficient as shown in the dialog box on the right.
- 2. Click OK.
- 1. From the Analysis menu, choose Dynamic/Kinematic.
- 2. Turn off numerical damping by clicking the Parameters tab, and setting the numerical damping coefficient to 0.
- 3. Click the General tab and change the Plot Multiplier Step Factor to 10.
- 4. Run the simulation by clicking Simulate.
To run an analysis:
RecurDyn prompts you to save your file if you have not already done so.
To create plots and determine velocity and distance:
- Enter the plotting environment by clicking the Plot Result tool.
- In the Database window, expand Body1 and double-click Pos_TY and Vel_TY to plot the position and velocity in the y-direction.
- Select the Trace Data tool to find the point at which the maximum velocity occurs (frame 347, 1.4622 m/s). Then, find the position that corresponds to that velocity (1.2175 m). You can also see that the maximum displacement is 1.4359 m.
If you would like to increase the precision of the plots, you can increase the number of steps, as well as the plot multiplier factor.
Comparison of Results
Results are summarized in the following table:
| Results |
RecurDyn |
Theoretical |
Percent Error |
| Maximum velocity (m/s) |
1.4622 |
1.4619 |
0.0223% |
| Position at vmax (m) |
1.2175 |
1.2179 |
-0.0348% |
| Maximum spring length (m) |
1.4359 |
1.4358 |
0.0037% |
Tracy 
星期五, 2009/07/17 at 16:52
force 就是力,也就是 F. 在RecurDyn 中力可以用公式或者方程来表示。force 中有各种各样的力的,例如spring ,bushing, 等
Joint 就是约束, 限制物体的自由度。
contact 是力的一种。物体接触时候产生的力。 例如小球掉到地面上。
force 和 Joint 在一些极端的条件下可以等价的。 例如,绞结,你就可以用bushing 来代替,把绞结相应的钢都定义的非常大就可以了。
例如球绞结,他限制了物体连接处的 x,y,z 三个方向不能运动。如果用bushing 来代替, 你可以把bushing 的X,Y,Z 的刚度都定义到很大(10E9),这样,物体在x,y,z 方向的位移非常小了-〉0 。相当于把x,y,z 都给限制住了。
Tracy 
星期四, 2009/07/16 at 16:52
多体动力学仿真最早起源于二十世纪七十年代早期,当时局限于完全刚体的仿真。之后的几年,许多应用领域明显提出了分析柔性体的必要性。二十世纪八十年代,第一个多体 动力学软件在市面上出现,该软件采用所谓的“模态缩减”方法求解有限元结构问题,通过有限元程序预先求解结构的特征值,进而得到模态缩减 的柔性体结构。该方法已经满足了许多应用的需要,但是对于一些重要的应用呈现出了巨大的局限性。一方面这些线性化的柔性体不允许出现大变形,更为重要的是不具备足够的刚度信息,考虑接触非常辛苦,甚至不能实现。因为结果的好坏直接取决于从有限元程序倒入特征值,评价组件的应力显得非常重要。
新技术不是为了取代模态缩减法,而是采用“节点法”对它补充扩展,应用真实的有限元结构,在一个系统中合并两种方法的优点。
目前为止,采用MFBD结构,不再需要一个单独的有限元工具预先对柔性体求解,在软件进行仿真时,生成完整的刚度矩阵和质量矩阵。仿真过程中通过反复计算结构矩阵,可以进行精确的应力分析、考虑柔性体结构间的碰撞和接触以及考虑非线性变形,计算过程中软件能够自动判断结构间的接触。
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