枪钻杆自由振动的稳定性分析
1College of Energy and Mining Engineering, Shandong University of Science and Technology, Qingdao 266590, China.
Materials (Basel, Switzerland)
|March 27, 2025
概括
这项研究模拟了枪钻杆,揭示了由于其不对称的设计而具有独特的不稳定的速度范围. 振动分析和稳定性对于提高深孔加工精度和效率至关重要.
科学领域:
- 机械工程 机械工程
- 振动分析 振动分析
- 机械加工动力学 机械加工动力学
背景情况:
- 枪钻对于深孔加工至关重要,但面临着振动的挑战,影响准确性和效率.
- 枪钻杆的独特,环形不对称的横截面,由于内部的切割流体通道,与标准的圆形旋转机相比,使它们的动态行为复杂化.
- 了解这些复杂的动态对于优化枪支演练性能至关重要.
研究的目的:
- 为了开发一个动态模型的枪钻杆.
- 分析枪钻杆的自由振动特性和稳定性.
- 为了研究尺寸参数对枪支演练动态的影响.
主要方法:
- 雷利旋转机模型的开发用于枪支钻杆.
- 应用Galerkin方法来评估自由振动和稳定性.
- 用ANSYS模拟进行比较分析以验证.
主要成果:
- 枪钻杆表现出明显的不稳定的速度范围,而不是在圆形旋转机中观察到的.
- 尺寸因素显著影响枪钻管的振动特性和稳定性.
- 理论预测得到了ANSYS模拟的验证,证实了模型的可靠性.
结论:
- 枪钻杆的不对称设计导致了独特的动态行为和稳定性限制.
- 准确的建模和分析对于预测和减轻深孔加工中的振动问题至关重要.
- 这项研究为提高枪支演习的设计和操作效率提供了基础.
更多相关视频
00:09Visualization of Flow Field Around a Vibrating Pipeline Within an Equilibrium Scour Hole
Published on: August 26, 2019
5.5K
06:45Design and Application of a Fault Detection Method Based on Adaptive Filters and Rotational Speed Estimation for an Electro-Hydrostatic Actuator
Published on: October 28, 2022
1.5K
相关概念视频
Stability of structures
149
In mechanical engineering, the stability of systems under various forces is critical for designing durable and efficient structures. One fundamental way to explore these concepts is by analyzing systems like two rods connected at a pivot point, O, with a torsional spring of spring constant k at the pivot point. This system is similar in appearance to a scissor jack used to change tires on a car. In this case, the arms of the linkage (equivalent to the rods in this system) are entirely vertical,...
149
Applications of Stress
236
Consider a structure made of a boom and a rod designed to support a load. These two components are connected by a pin and stabilized by brackets and pins. The boom and the rod are detached from their supports to assess the different stresses imposed on this structure, and a free-body diagram is drawn. Then, all the forces applied, including the load acting on the structure, are identified. The reaction forces exerted on both the boom and the rod are computed using the equilibrium equations.
The...
The...
236
Temperature Dependent Deformation
134
In a nonhomogeneous rod made up of steel and brass, restrained at both ends and subjected to a temperature change, several steps are involved in calculating the stress and compressive load. Due to the problem's static indeterminacy, one end support is disconnected, allowing the rod to experience the temperature change freely. Next, an unknown force is applied at the free end, triggering deformations in the rod's steel and brass portions. These deformations are then calculated and added...
134
Statically Indeterminate Problem Solving
349
Statically indeterminate problems are those where statics alone can not determine the internal forces or reactions. Consider a structure comprising two cylindrical rods made of steel and brass. These rods are joined at point B and restrained by rigid supports at points A and C. Now, the reactions at points A and C and the deflection at point B are to be determined. This rod structure is classified as statically indeterminate as the structure has more supports than are necessary for maintaining...
349
Stability of Equilibrium Configuration: Problem Solving
558
The stability of equilibrium configurations is an important concept in physics, engineering, and other related fields. In simple terms, it refers to the tendency of an object or system to return to its equilibrium position after being disturbed. The stability of an equilibrium configuration can be analyzed by considering the potential energy function of the system and examining its behavior near the equilibrium point.
Problem-solving in the context of the stability of equilibrium configuration...
Problem-solving in the context of the stability of equilibrium configuration...
558
Rigid Body Equilibrium Problems - II
6.9K
A rigid body is in static equilibrium when the net force and the net torque acting on the system are equal to zero.
Consider two children sitting on a seesaw, which has negligible mass. The first child has a mass (m1) of 26 kg and sits at point A, which is 1.6 meters (r1) from the pivot point B; the second child has a mass (m2) of 32 kg and sits at point C. How far from the pivot point B should the second child sit (r2) to balance the seesaw?
Consider two children sitting on a seesaw, which has negligible mass. The first child has a mass (m1) of 26 kg and sits at point A, which is 1.6 meters (r1) from the pivot point B; the second child has a mass (m2) of 32 kg and sits at point C. How far from the pivot point B should the second child sit (r2) to balance the seesaw?
6.9K
