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专题报告

专题报告

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  • 回答

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  • 回答

    附加模块RS-Knick没有考虑弯曲扭转屈曲。
  • 回答

    RSKNICK和RF-STABIL使用给定的法向力状态对整个模型执行特征值分析。结果,取决于所需的特征值的数量,关键负载因子与相关的屈曲数字一起输出,并且对于每个本征形状的每个条,输出围绕强轴和弱轴的屈曲长度。

    由于通常每个载荷工况在元件中具有不同的法向力状态,因此对于每个载荷情况产生单独的相关的屈曲长度结果。屈曲长度,其屈曲形状,支撑在框架平面中弯曲,是用于证明相应载荷情况的正确长度。

    由于不同的负载情况,每次分析的结果可能不同,因此假设所有计算分析的最长屈曲长度在所有负载情况下都是安全的。

  • 回答

    RSKNICK不输出杆组的屈曲长度系数。您只能从单个条形的结果开始。作为一组杆的规则,您可以查看杆,其中输出最小的临界屈曲载荷N cr

    考虑单个杆中的法向力也可能是有帮助的。如果它们在横截面不变的横杆组中是相同的,则屈曲长度系数是相同的。然后,该值也可用于该组条。
  • 回答

    RSBUCK uses a momentary representation of the axial force distribution in the respective load state. These axial forces are increased iteratively until the buckling load case occurs.

    In the numerical analysis, the stability load is indicated by the fact that the determinant of the stiffness matrix becomes zero.

    If the effective length factor is known, the program uses it to determine the buckling load and buckling mode. For this lowest buckling load, all effective lengths and effective length factirs are determined.

    Example:
    Hinged-hinged column with a length of 20 m, cross-section: HE-B 500, load: only self-weight

    For the first buckling mode, you obtain an effective length factor of kcr,y = 2.92 for the buckling about the major axis.

    For the buckling about the minor axis with a buckling load of 651.3 kN, you obtain an effective length factor of 1.00.

    If we solve the expression used for the buckling load Ncr = π² x E x I / Lcr² for Lcr and plug in Ncr = 651.3 kN and Iy = 107,200 cm4, we obtain an Lcr,y of 58.4 m, which results in an effective length factor kcr,y of 2.92.

    In RSBUCK, we obtain the two effective length factors for each buckling mode and buckling load.

    If you want to determine the correct effective length factor for the deflection perpendicular to the y-axis (buckling about major axis), it is necessary to calculate several buckling modes and choose the appropriate one.

    In this case, it is the third buckling mode with a buckling load of 5485.5 kN. From this load, the effective lengths and effective length factors are determined, so that we obtain kcr,y = 1.0 und kcr,z = 0.345. For a quadratic cross-section, you obtain two equal effective lengths, because the stiffnesses in both directions are the same. Therefore, to obtain the correct values for the effective length factors in RSBUCK, it is necessary to look at several buckling modes.

  • 回答

    Lcr,y is too great in the first buckling mode because of the following reason.

    In RSBUCK, the lowest critical buckling load is calculated first.

    For example, the buckling load is obtained for a hinged column (Euler buckling mode 1 with HE-B 300) for the buckling about the z-axis. With this buckling load, the effective length Lcr,y is determined retroactively.

    If you also want to obtain the correct effective length for Lcr,y, you must also look at the second buckling mode. To do this, set the number of buckling modes to be determined to 2.

    In the second buckling mode, you obtain a higher buckling load (sway about the y-axis), from which you obtain the correct buckling load Lcr,y.

    In RSBUCK, it is necessary to calculate several buckling modes to obtain results for the individual directions.

  • 回答

    Check if the settings for considering the favorable effect by tension forces are the same in RSTAB and RSBUCK.

    RSTAB determines the critical load factor according to a nonlinear calculation method: The loading is increased gradually by the value of the load factor increase until the system becomes unstable. RSBUCK, however, carries out a linear eigenvalue analysis. Therefore, elements acting nonlinearly like failing members or supports may have different effects in RSTAB and RSBUCK.

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