# Stiffening of Structures

### Technical Article

Buildings must be designed and dimensioned in such a way that both vertical and horizontal loads are conducted safely and without large deformations in the building. Examples of horizontal loads are wind, unintentional inclination, earthquakes, or a blast.

Finite element analysis programs such as RFEM allow you to determine internal forces and design stiffening structural elements. In this program, you can model a building including all structural components, openings, and other elements, and perform a calculation of the entire model.

Predimensioning of stiffening system can be performed using manual calculation according to the calculation method described in [1] or by using a program such as SHAPE‑THIN. This software provides engineers with a better understanding of the load transfer in a structure as well as the resistance contribution of the individual structural components.

#### 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
$$\begin{array}{l}{\mathrm V}_{\mathrm y,\mathrm i}\;=\;\frac{{\mathrm V}_\mathrm y\;\cdot\;({\mathrm I}_{\mathrm z,\mathrm i}\;\cdot\;{\mathrm I}_\mathrm y\;-\;{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;{\mathrm I}_\mathrm{yz})\;-\;{\mathrm V}_\mathrm z\;\cdot\;({\mathrm I}_{\mathrm z,\mathrm i}\;\cdot\;{\mathrm I}_\mathrm{yz}\;-\;{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;{\mathrm I}_\mathrm z)}{{\mathrm I}_\mathrm y\;\cdot\;{\mathrm I}_\mathrm z\;-\;{\mathrm I}_\mathrm{yz}²}\\{\mathrm V}_{\mathrm z,\mathrm i}\;=\;\frac{{\mathrm V}_\mathrm y\;\cdot\;({\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;{\mathrm I}_\mathrm y\;-\;{\mathrm I}_{\mathrm y,\mathrm i}\;\cdot\;{\mathrm I}_\mathrm{yz})\;-\;{\mathrm V}_\mathrm z\;\cdot\;({\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;{\mathrm I}_\mathrm{yz}\;-\;{\mathrm I}_{\mathrm y,\mathrm i}\;\cdot\;{\mathrm I}_\mathrm z)}{{\mathrm I}_\mathrm y\;\cdot\;{\mathrm I}_\mathrm z\;-\;{\mathrm I}_\mathrm{yz}²}\end{array}$$
where
Vy,i, Vz,i: shear force in the y- or z‑direction, which affects the partial cross‑section i
Vy, Vz: shear force in the y- or z‑direction, which affects the gross cross‑section
Iy,i, Iz,i, Iyz,i: 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: total second moments of area relating to the overall centroid S

Forces Caused by Torsion
$$\begin{array}{l}{\mathrm V}_{\mathrm y,\mathrm i}\;=\;\frac{{\mathrm M}_\mathrm{xs}\;\cdot\;\lbrack{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)\;-\;{\mathrm I}_{\mathrm z,\mathrm i}\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)\rbrack}{\mathrm\Sigma\;\lbrack{\mathrm I}_{\mathrm\omega,\mathrm i}\;+\;{\mathrm I}_{\mathrm y,\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)²\;-\;2\;\cdot\;{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)\;+\;{\mathrm I}_{\mathrm z,\mathrm i}\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)²\rbrack}\\{\mathrm V}_{\mathrm z,\mathrm i}\;=\;\frac{{\mathrm M}_\mathrm{xs}\;\cdot\;\lbrack{\mathrm I}_{\mathrm y,\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)\;-\;{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)\rbrack}{\mathrm\Sigma\;\lbrack{\mathrm I}_{\mathrm\omega,\mathrm i}\;+\;{\mathrm I}_{\mathrm y,\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)²\;-\;2\;\cdot\;{\mathrm I}_{\mathrm{yz},\mathrm i}\;\cdot\;({\mathrm y}_{\mathrm M,\mathrm i}\;-\;{\mathrm y}_\mathrm M)\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)\;+\;{\mathrm I}_{\mathrm z,\mathrm i}\;\cdot\;({\mathrm z}_{\mathrm M,\mathrm i}\;-\;{\mathrm z}_\mathrm M)²\rbrack}\end{array}$$
where
Vy,i, Vz,i: shear force in the y- or z‑direction, which affects the partial cross‑section
Mxs: secondary torsional moment, which affects the gross cross‑section
Iy,i, Iz,i, Iyz,i:  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: warping constant relating to the shear center of the partial cross‑section Mi
yM,i, zM,i: coordinate of the shear center of the partial cross‑section Mi
yM, zM: coordinate of the overall shear center M

#### Example

The distribution of horizontal loads on the stiffening elements is explained on the structural system shown in Figure 01.

Wall thickness t = 30 cm

Cross-Section Properties

Partial cross-section 1
$$\begin{array}{l}{\mathrm z}_{\mathrm S,1}\;=\;\frac{\displaystyle\frac{2.15\;\cdot\;0.30\;\cdot\;0.30}2\;+\;4.70\;\cdot\;0.30\;\cdot\;(\frac{4.70}2\;+\;0.30)\;+\;2.15\;\cdot\;0.30\;\cdot\;(0.30\;+\;4.70\;+\;\frac{0.30}2)}{2.15\;\cdot\;0.30\;\cdot\;2\;+\;4.70\;\cdot\;0.30}\;=\;2.65\;\mathrm m\\{\mathrm y}_{\mathrm S,1}\;=\;\frac{2.15\;\cdot\;0.30\;\cdot\;{\displaystyle\frac{2.15}2}\;\cdot\;2\;+\;4.70\;\cdot\;0.30\;\cdot\;{\displaystyle\frac{0.30}2}}{2.15\;\cdot\;0.30\;\cdot\;2\;+\;4.70\;\cdot\;0.30}\;=\;0.59\;\mathrm m\\{\mathrm I}_{\mathrm y,1}\;=\;2.15\;\cdot\;\frac{0.303}{12}\;\cdot\;2\;+\;2.15\;\cdot\;0.30\;\cdot\;(\frac{2.65\;-\;0.30}2)²\;\cdot\;2\;+\;0.30\;\cdot\;\frac{4.703}{12}\;+\;4.70\;\cdot\;0.30\;\cdot\;(0.00)²\;=\;10.668\;\mathrm m^4\\{\mathrm I}_{\mathrm z,1}\;=\;0.30\;\cdot\;\frac{2.153}{12}\;\cdot\;2\;+\;2.15\;\cdot\;0.30\;\cdot\;(\frac{2.15}2\;-\;0.59)²\;\cdot\;2\;+\;4.70\;\cdot\;\frac{0.303}{12}\;+\;4.70\;\cdot\;0.30\;\cdot\;(0.59\;-\;\frac{0.30}2)²\;=\;1.084\;\mathrm m^4\end{array}$$

Partial cross-section 2
$$\begin{array}{l}{\mathrm I}_{\mathrm y,2}\;=\;\frac{0.30\;\cdot\;4.003}{12}\;=\;1.600\;\mathrm m^4\\{\mathrm I}_{\mathrm z,2}\;=\;\frac{4.00\;\cdot\;0.303}{12}\;=\;0.009\;\mathrm m^4\end{array}$$

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 02.

Shear Forces of Partial Cross-Section
$$\begin{array}{l}{\mathrm V}_{\mathrm y,1}\;=\;\frac{100\;\cdot\;(1.084\;\cdot\;12.268)}{12.268\;\cdot\;1.093}\;=\;99.18\;\mathrm{kN}\\{\mathrm V}_{\mathrm y,2}\;=\;\frac{100\;\cdot\;(0.009\;\cdot\;12.268)}{12.268\;\cdot\;1.093}\;=\;0.823\;\mathrm{kN}\end{array}$$

The shear forces of the partial cross‑section determined in SHAPE‑THIN 8 are displayed in Figure 03.

#### Reference

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