建筑结构竖向支撑结构体系

技术文章

高层建筑结构的竖向承重结构体系不但要承受和传递竖向荷载,还要抵抗侧向力的作用。竖向抗侧力结构的布置原则是不仅具有承受较大的水平剪力的能力还要使侧向力不至对基础产生非常大的变形。水平方向的荷载包括例如风荷载、偶然偏心或倾斜、水平地震作用以及撞击等。

在有限元软件 RFEM 中不仅可以计算结构的内力而且可以计算其竖向支撑结构体系。软件可以将结构内部所包含的全部承重构件、洞口等建立的模型尽可能的按照整体结构模型进行计算。

竖向支撑体系的预设几何尺寸除了可以按照在参考书目 [1] 中介绍的计算方法进行手算之外,用户使用 Dlubal 软件的独立程序 DUENQ 进行计算更为快速准确。结构工程师使用该软件不仅可以更加方便直观的了解结构体系的传力途径,还可以计算既有结构中各个竖向构件的最大承载力。

Distribution of Horizontal Forces

The horizontal load distribution for bending or torsional loading on the stiffening components can be calculated according to the following formulas.

Forces Caused by Bending
Vy,i = (Vy ∙ (Iz,i ∙ Iy - Iyz,i ∙ Iyz) - Vz ∙ (Iz,i ∙ Iyz - Iyz,i ∙ Iz)) / (Iy ∙ Iz - Iyz2)
Vz,i = (Vy ∙ (Iyz,i ∙ Iy - Iy,i ∙ Iyz) - Vz ∙ (Iyz,i ∙ Iyz - Iy,i ∙ Iz)) / (Iy ∙ Iz - Iyz2)
where
Vy,i, Vz,i is the shear force in the y- or z-direction, which affects the partial cross-section i
Vy, Vz is the shear force in the y- or z-direction, which affects the gross cross-section
Iy,i, Iz,i, Iyz,i are the moments of inertia of the partial cross-section i relating to the parallel axes Y and Z by the partial cross-section centroid Si
Iy, Iz are the total second moments of area relating to the overall centroid S

Forces Caused by Torsion
Vy,i = Mxs ∙ [Iyz,i ∙ (yM,i - yM) - Iz,i ∙ (zM,i - zM)] / Σ [Iω,i + Iy,i ∙ (yM,i - yM)2 - 2 ∙ Iyz,i ∙ (yM,i - yM) ∙ (zM,i - zM) + Iz,i ∙ (zM,i - zM)2]
Vz,i = Mxs ∙ [Iy,i ∙ (yM,i - yM) - Iyz,i ∙ (zM,i - zM)] / Σ [Iω,i + Iy,i ∙ (yM,i - yM)2 - 2 ∙ Iyz,i ∙ (yM,i - yM) ∙ (zM,i - zM) + Iz,i ∙ (zM,i - zM)2]
where
Vy,i, Vz,i is the shear force in the y- or z-direction, which affects the partial cross-section
Mxs is the secondary torsional moment, which affects the gross cross-section
Iy,i, Iz,i, Iyz,i are the moments of inertia of the partial cross-section i relating to the parallel axes Y and Z by the partial cross-section centroid Si
Iω,i is the warping constant relating to the shear center of the partial cross-section Mi
yM,i, zM,i is the coordinate of the shear center of the partial cross-section Mi
yM, zM is the coordinate of the overall shear center M

Example

The distribution of horizontal loads in the stiffening elements is explained on the system displayed in Figure 1.

Figure 01 - System

Wall thickness t = 30 cm

Cross-Section Properties

Partial Cross-Section 1
zS,1 = (2.15 ∙ 0.30 ∙ 0.30 / 2 + 4.70 ∙ 0.30 ∙ (4.70 / 2 + 0.30) + 2.15 ∙ 0.30 ∙ (0.30 + 4.70 + 0.30 / 2)) / (2.15 ∙ 0.30 ∙ 2 + 4.70 ∙ 0.30) = 2.65 m
yS,1 = (2.15 ∙ 0.30 ∙ 2.15 / 2 ∙ 2 + 4.70 ∙ 0.30 ∙ 0.30 / 2) / (2.15 ∙ 0.30 ∙ 2 + 4.70 ∙ 0.30) = 0.59 m
Iy,1 = 2.15 ∙ 0.303 / 12 ∙ 2 + 2.15 ∙ 0.30 ∙ (2.65 - 0.30 / 2)2 ∙ 2 + 0.30 ∙ 4.703 / 12 + 4.70 ∙ 0.30 ∙ (0.00)2 = 10.668 m4
Iz,1 = 0.30 ∙ 2.153 / 12 ∙ 2 + 2.15 ∙ 0.30 ∙ (2.15 / 2 - 0.59)2 ∙ 2 + 4.70 ∙ 0.303 / 12 + 4.70 ∙ 0.30 ∙ (0.59 - 0.30 / 2)2 = 1.084 m4

Partial Cross-Section 2
Iy,2 = 0.30 ∙ 4.003 / 12 = 1.600 m4
Iz,2 = 4.00 ∙ 0.303 / 12 = 0.009 m4

Gross Cross-Section
Iy = 10.668 + 1.600 = 12.268 m4
Iz = 1.084 + 0.009 = 1.093 m4

The cross-section properties determined in SHAPE-THIN 8 are displayed in Figure 2.

Figure 02 - Cross-Sectional Properties

Shear Forces of Partial Cross-Section
Vy,1 = 100 ∙ (1.084 ∙ 12.268) / (12.268 ∙ 1.093) = 99.18 kN
Vy,2 = 100 ∙ (0.009 ∙ 12.268) / (12.268 ∙ 1.093) = 0.823 kN

The shear forces of the partial cross-section determined in SHAPE-THIN 8 are displayed in Figure 3.

Figure 03 - Shear Forces of Partial Cross-Section

Reference

[1] Beck, H. & Schäfer, H. (1969). Die Berechnung von Hochhäusern durch Zusammenfassung aller aussteifenden Bauteile zu einem Balken. Der Bauingenieur, (Heft 3).

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