Study of strain energy based shear model for single lap bolt
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摘要: 为了简化复杂结构在冲击数值分析中的大量螺栓连接,可用等效的载荷位移模型代替复杂的螺栓连接关系,本文中针对单搭接螺栓连接在剪切载荷下建立了连接本构关系。首先通过对有预紧力的单搭接螺栓进行实验和精细有限元模拟,揭示了螺栓剪切载荷位移曲线的特征并针对不同特征阶段进行了相应的物理机理分析。在此基础上对于载荷位移曲线的界面黏结、部分滑移、整体滑移阶段提出了连接本构模型的基本形式和各阶段的参数估算方法。在部分滑移阶段考虑了4个方面的刚度贡献,其中部件对螺栓的支撑刚度是三维非轴对称变形问题,理论求解非常困难,本文中通过应力分布研究,采用应变能法解决了螺栓的支撑刚度的估算问题。提出的单搭接螺栓剪切模型物理含义明确,参数估算简单,准确度高。Abstract: In simplifying large numbers of bolt connections in simulation to substitute the construction of the whole joint structure, it is essential to establish a joint constitutive model. In this paper, a shear constitutive model of the single lap bolt was established. Firstly, the curves of the load-displacement of the single lap shear bolt were obtained from experiments, and the corresponding physical mechanism was analyzed using the finite-element method. Then, the curve of the load-displacement was divided into several phases according to the physical mechanism. Based on the first three phases, the basic form of the shear constitutive model was proposed and the related parameters were determined in detail. In the second phase of the shear model, the support stiffness was solved by the strain energy method based on reasonable stress distribution hypothesis, which was a step most difficult to take. Finally, this constitutive model was verified by numerous examples. The results show that this shear model for the single lap bolt has the advantages of providing clear physical meaning, easy parameter estimation, and high accuracy.
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Key words:
- solid mechanics /
- shear stiffness /
- constitutive model /
- bolt joint /
- strain energy
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表 1 不同结构选材k1验证
Table 1. Verification of k1 for different material
螺栓 部件 k1/(N·mm-1) 误差/% 预测值 有限元值 钢 铝 1.83×104 1.68×104 8.8 铝 钢 9.69×103 8.88×103 9.1 铝 铝 8.44×103 7.70×103 9.5 表 2 不同螺栓直径k1验证
Table 2. Verification of k1 for different bolt diameters
螺栓直径/mm 长细比 k1/(N·mm-1) 误差/% 预测值 有限元值 6 5.0 4.29×103 4.13×103 3.8 10 3.0 2.53×104 2.30×104 9.9 16 1.9 1.17×105 9.14×104 28.3 表 3 不同部件厚度k1验证
Table 3. Verification of k1 for different thicknesses
部件厚度/mm 长细比 k1/(N·mm-1) 误差/% 预测值 有限元值 20 2.0 6.57×104 5.76×104 14.0 40 4.0 1.23×104 1.03×104 8.3 70 7.0 2.76×103 2.64×103 4.7 -
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