Design Support for Timber Design – Serviceability Limit State, Compression Perpendicular to Grain, and Shear Force Reduction

In this article, the procedure for the deformation analysis, the compression perpendicular to grain design, and the shear force reduction based on Eurocode 5 will be presented in more detail. The assignment of design supports is described in the manual:

• Timber Design - Design Supports and Deflection

Serviceability

The segmentation for the deformation analysis is not discussed further here. It is detailed in this article:

• Deflection Check for Member Elements

Compression Perpendicular to Grain Design

Background

Design supports are assigned to a member or member set, not to a nodal support. While nodal supports provide clear support reactions, which can be used, for example, for the "compression perpendicular to grain" design, supports in spatial structures are often not represented by nodal supports. Typical examples are structures where a member rests on another member or on a surface. In such cases, no direct support reaction from a nodal support is available for the design. The required compressive force is therefore determined from the internal forces of the members connected at the node. This allows both classic supports and complex spatial support situations to be considered.

Support Situation

Due to the simplification when creating the structural system, the support situation at nodes with several members is not clearly defined. Thus, the program cannot automatically determine the compressive force without additional user input. The following image shows such a situation. Due to the model simplification, all members meet at one node.

Simplification of Structural System

This results in a variety of support situations. Four possible situations are shown in the next image. They will be discussed in more detail in this article.

Various Support Conditions of Multiple Members

Depending on which members cause compression perpendicular to grain forces, the user must clearly define the support situation.

Definition of the Support Situation in RFEM 6 and RSTAB 9

To define the support situation in the program, the design support must first be defined at the corresponding node. In the example of Case 1, Member 102 and 103 are each subjected to compression perpendicular to grain from the top side as well as the bottom side (+z and -z direction). Accordingly, a double-sided design support can be defined (see the following image). In Case 3 for Member 303, a design support is only required on the bottom side, and so on.

Definition of Design Support

The actual definition of the support situation is determined via the internal forces to be considered (see the next image).

For the following examples, the components that do not generate any compression perpendicular to grain forces are deactivated for a better overview, and all contact areas subjected to compression perpendicular to grain are examined.

Case 1

Member 103

• Top side (-z)

Member 104 generates compression perpendicular to grain on Member 103 via the axial force N. Accordingly, the checkbox 'N' for Member 104 is activated. All other checkboxes remain deactivated.

• Bottom side (+z)

Member 104 generates compression perpendicular to grain on Member 103 via the axial force N. Member 103 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, the checkbox 'N' for Member 104 is activated, as well as the checkbox 'V_z' for Member 103.

Member 102

• Top side (-z)

Member 104 generates compression perpendicular to grain on Member 102 via the axial force N. Member 103 generates compression perpendicular to grain on Member 102 via the shear force V_z. Accordingly, the checkbox 'N' for Member 104 is activated, as well as the checkbox 'V_z' for Member 103.

• Bottom side (+z)

Member 104 generates compression perpendicular to grain on Member 102 via the axial force N. Member 103 generates compression perpendicular to grain on Member 102 via the shear force V_z. Member 102 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, the checkbox 'N' for Member 104 is activated, as well as the checkbox 'V_z' for Member 103 and Member 102.

Definition of Internal Forces for Case 1

Case 2

Member 202

• Top side (-z)

Member 204 generates compression perpendicular to grain on Member 202 via the axial force N. Accordingly, the checkbox 'N' for Member 104 is activated. All other checkboxes remain deactivated.

• Bottom side (+z)

Member 204 generates compression perpendicular to grain on Member 202 via the axial force N. Member 202 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, the checkbox 'N' for Member 204 is activated, as well as the checkbox 'V_z' for Member 202.

Member 203

• Top side (-z)

Member 204 generates compression perpendicular to grain on Member 203 via the axial force N. Member 202 generates compression perpendicular to grain on Member 203 via the shear force V_z. Accordingly, the checkbox 'N' for Member 204 is activated, as well as the checkbox 'V_z' for Member 202.

• Bottom side (+z)

Member 204 generates compression perpendicular to grain on Member 203 via the axial force N. Member 202 generates compression perpendicular to grain on Member 203 via the shear force V_z. Member 203 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, the checkbox 'N' for Member 204 is activated, as well as the checkbox 'V_z' for Member 202 and Member 203.

Definition of Internal Forces for Case 2

Case 3

Member 303

• Top side (-z)

No design support defined

• Bottom side (+z)

Member 304 does not generate any compression perpendicular to grain on Member 303 via the axial force N. Member 303 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, no checkbox is activated for Member 304, but the checkbox 'V_z' is activated for Member 303.

Member 302

• Top side (-z)

Member 304 does not generate any compression perpendicular to grain on Member 303 via the axial force N. Member 303 generates compression perpendicular to grain on Member 302 via the shear force V_z. Accordingly, no checkbox is activated for Member 304, but the checkbox 'V_z' is activated for Member 303.

• Bottom side (+z)

Member 304 does not generate any compression perpendicular to grain on Member 302 via the axial force N. Member 303 generates compression perpendicular to grain on Member 302 via the shear force V_z. Member 302 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, no checkbox is activated for Member 304, but the checkbox 'V_z' is activated for Member 303 and Member 302.

Definition of Internal Forces for Case 3

Case 4

Member 403

• Top side (-z)

No design support defined

• Bottom side (+z)

Member 404 does not generate any compression perpendicular to grain on Member 403 via the axial force N. Member 403 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Member 402 does not generate any compression perpendicular to grain on Member 403 via the shear force V_z. Accordingly, no checkbox is activated for Member 404 and 402, but the checkbox 'V_z' is activated for Member 403.

Member 402

• Top side (-z)

No design support defined

• Bottom side (+z)

Member 404 does not generate any compression perpendicular to grain on Member 402 via the axial force N. Member 403 does not generate any compression perpendicular to grain on Member 402 via the shear force V_z. Member 402 generates compression perpendicular to grain on its bottom side (+z) via the shear force V_z. Accordingly, no checkbox is activated for Member 404 and 403, but the checkbox 'V_z' is activated for Member 402.

Definition of Internal Forces for Case 4

Case 1 - Alternative

Member 102

• Bottom side (+z)

It is also possible to reverse the input. This will be shown using the example of Case 1 for Member 102 on the bottom side (+z). Here, all checkboxes of Members 102, 103, and 104 are to be deactivated. The support force can most easily be taken here from the axial force N of Member 101. For this, this checkbox must be activated.

Alternative Definition of Internal Forces for Case 4

Results

The cross-sections are 100/100 mm, as are the contact areas. The compression perpendicular to grain factor k_c,90 is simplified to 1.0. The calculation is performed with 5 kN vertical load each at the member ends of members x02, x03, and x04. For the aforementioned conditions, with a characteristic compression perpendicular to grain strength of 2.5 N/mm², a k_mod of 0.6, a partial safety factor of 1.3, and considering the hanging effect, the following design ratios result, which correspond to the ratio of the total force (15 kN).

Case 1:

Case 2:

Case 3:

Case 4:

Design Ratios in Timber Design for Various Members and Cases

Design Ratios in Timber Design for Various Members and Cases

Compression Perpendicular to Grain Reinforcements

If the load-bearing capacity of an unreinforced bearing is insufficient to transfer the applied force, the bearing can be reinforced by using fully threaded screws driven in perpendicular to the grain. It must be ensured that the compressive force is evenly distributed among all screws and that the forces resulting in the screw heads can be transferred into the support. To achieve this, a steel plate can be used, which directs the forces in the screw heads into the support. The screw heads must be flush with the timber surface. The following failure modes must be investigated:

1. Pressing of the screw into the timber (analogous to the withdrawal resistance)
2. Buckling of the screw in the timber component
3. Compression perpendicular to grain failure at the level of the screw tip

The reinforcement elements can be activated as shown in the image.

Activating Transversal Compression Reinforcements in Design Supports

The input of the relevant screw parameters currently still has to be done manually. The values can be taken from the corresponding approvals and product data sheets.

Definition of Fully Threaded Screw for Support Reinforcement

Example

The beam shown in the next image is to be reinforced with the fasteners from the previous image. The following parameters are defined:

Beam on Column

With a linear load distribution, the following results are obtained:

Unreinforced Support

\( \mathrm{f_{c,90,z,d}} = \mathrm{k_{mod}} \cdot \frac{\mathrm{f_{c,90,z,k}}}{\gamma_{M}} = 0.60 \cdot \frac{2.50\, \mathrm{N/mm^2}}{1.30} = 1.15\, \mathrm{N/mm^2} \)

\( \mathrm{l_{ef}} = \mathrm{l} + 30\, \mathrm{mm} = 200\, \mathrm{mm} + 30\, \mathrm{mm} = 230\, \mathrm{mm} \)

\( \mathrm{A_{ef}} = \mathrm{b} \cdot \mathrm{l_{ef}} = 100\, \mathrm{mm} \cdot 230\, \mathrm{mm} = 0.023\, \mathrm{m^2} \)

\( \mathrm{\sigma_{c,90,d}} = \frac{\mathrm{F_{c,90,d}}}{\mathrm{A_{ef}}} = \frac{80.00\, \mathrm{kN}}{0.023\, \mathrm{m^2}} = 3.48\, \mathrm{N/mm^2} \)

\( \mathrm{\eta_{1}} = \frac{\mathrm{\sigma_{c,90,d}}}{\mathrm{k_{c,90}} \cdot \mathrm{f_{c,90,z,d}}} = \frac{3.48\, \mathrm{N/mm^2}}{1.75 \cdot 1.15\, \mathrm{N/mm^2}} = 1.72 \)

→ The support must be reinforced.

Withdrawal Resistance of a Screw \( \mathrm{F_{ax,90,Rk}} = \mathrm{f_{ax,k}} \cdot \mathrm{d} \cdot \mathrm{l_{g}} \cdot \left( \frac{\rho_{k}}{\rho_{a}} \right)^{0.8} = 12.00\, \mathrm{N/mm^2} \cdot 8\, \mathrm{mm} \cdot 545\, \mathrm{mm} \cdot \left( \frac{385.00\, \mathrm{kg/m^3}}{350.00\, \mathrm{kg/m^3}} \right)^{0.8} = 56.47\, \mathrm{kN} \)

\( \mathrm{F_{ax,90,Rd}} = \frac{\mathrm{k_{mod}} \cdot \mathrm{F_{ax,90,Rk}}}{\gamma_{M}} = \frac{0.60 \cdot 56.47\, \mathrm{kN}}{1.30} = 26.06\, \mathrm{kN} \)

Stability Resistance of a Screw \( \mathrm{N_{pl,k}} = \pi \cdot \frac{(d_{1})^{2}}{4} \cdot \mathrm{f_{y,k}} = \pi \cdot \frac{(5\, \mathrm{mm})^{2}}{4} \cdot 900.00\, \mathrm{N/mm^2} = 17.67\, \mathrm{kN} \)

\( \mathrm{c_{h}} = \frac{(0.22 + 0.014 \cdot \mathrm{d}) \cdot \rho_{k}}{1.17} = \frac{(0.22 + 0.014 \cdot 8\, \mathrm{mm}) \cdot 385.00\, \mathrm{kg/m^3}}{1.17} = 109.25\, \mathrm{N/mm^2} \)

\( \mathrm{I_{S}} = \frac{\pi \cdot (d_{1})^{4}}{64} = \frac{\pi \cdot (5\, \mathrm{mm})^{4}}{64} = 30.68\, \mathrm{mm^4} \)

\( \mathrm{N_{Ki,k}} = \sqrt{c_{h} \cdot E_{S} \cdot I_{S}} = \sqrt{109.25\, \mathrm{N/mm^2} \cdot 210000.00\, \mathrm{N/mm^2} \cdot 30.68\, \mathrm{mm^4}} = 26.53\, \mathrm{kN} \)

\( \overline{\mathrm{\lambda}}_{k} = \sqrt{\frac{\mathrm{N_{pl,k}}}{\mathrm{N_{Ki,k}}}} = \sqrt{\frac{17.67\, \mathrm{kN}}{26.53\, \mathrm{kN}}} = 0.82 \)

\( \mathrm{k} = 0.5 \cdot \left[ 1 + 0.49 \cdot \left( \overline{\mathrm{\lambda}}_{k} - 0.2 \right) + \left( \overline{\mathrm{\lambda}}_{k} \right)^{2} \right] = 0.5 \cdot \left[ 1 + 0.49 \cdot \left( 0.82 - 0.2 \right) + \left( 0.82 \right)^{2} \right] = 0.98 \)

\( \mathrm{κ_{c}} = \frac{1}{\mathrm{k} + \sqrt{(\mathrm{k})^{2} - (\overline{\mathrm{\lambda}}_{k})^{2}}} = \frac{1}{0.98 + \sqrt{(0.98)^{2} - (0.82)^{2}}} = 0.65 \)

\( \mathrm{F_{c,Rk}} = \mathrm{κ_{c}} \cdot \mathrm{N_{pl,k}} = 0.65 \cdot 17.67\, \mathrm{kN} = 11.52\, \mathrm{kN} \)

\( \mathrm{F_{c,Rd}} = \frac{\mathrm{F_{c,Rk}}}{\gamma_{M1}} = \frac{11.52\, \mathrm{kN}}{1.10} = 10.47\, \mathrm{kN} \)

Design of the Fully Threaded Screw \( \mathrm{F_{S,90,Rd}} = \min\left( \mathrm{F_{ax,90,Rd}}, \, \mathrm{F_{c,Rd}} \right) = \min\left( 26.06\, \mathrm{kN}, \, 10.47\, \mathrm{kN} \right) = 10.47\, \mathrm{kN} \)

\( \mathrm{n} = \mathrm{n_{0}} \cdot \mathrm{n_{90}} = 4 \cdot 1 = 4 \)

\( \mathrm{\eta_{2}} = \frac{\mathrm{F_{c,90,d}}}{\mathrm{n} \cdot \mathrm{F_{S,90,Rd}} + \mathrm{k_{c,90}} \cdot \mathrm{A_{ef}} \cdot \mathrm{f_{c,90,z,d}}} = \frac{80.00\, \mathrm{kN}}{4 \cdot 10.47\, \mathrm{kN} + 1.75 \cdot 0.023\, \mathrm{m^2} \cdot 1.15\, \mathrm{N/mm^2}} = 0.91 \)

Design of the Compression Perpendicular to Grain Stress at the Level of the Screw Tip (Linear) \( \mathrm{a_{1} = a_{1c}} = 40\, \mathrm{mm} \) \( \mathrm{l_{ef,2}} = \mathrm{a_{1c}} + (\mathrm{n_{0}} - 1) \cdot \mathrm{a_{1}} + \mathrm{l_{g}} \) \( = 40\, \mathrm{mm} + (4 - 1) \cdot 40\, \mathrm{mm} + 545\, \mathrm{mm} = 705\, \mathrm{mm} \)

\( \mathrm{A_{ef,2}} = \mathrm{b} \cdot \mathrm{l_{ef,2}} = 100\, \mathrm{mm} \cdot 705\, \mathrm{mm} = 0.071\, \mathrm{m^2} \)

\( \mathrm{\eta_{3}} = \frac{\mathrm{F_{c,90,d}}}{\mathrm{A_{ef,2}} \cdot \mathrm{f_{c,90,z,d}}} = \frac{80.00\, \mathrm{kN}}{0.071\, \mathrm{m^2} \cdot 1.15\, \mathrm{N/mm^2}} = 0.98 \)

Governing Design \( \mathrm{\eta} = \max\left( \mathrm{\eta_{2}}, \, \mathrm{\eta_{3}} \right) = \max\left( 0.91, 0.98 \right) = 0.98 \)

\( \mathrm{\eta} = 0.98 \leq 1 \)

Design Check Details of Transverse Compression Reinforcement Elements

The design for compression perpendicular to grain failure at the level of the screw tip becomes governing in this example. However, the meaningfulness of this design can be questioned, as the design is carried out at the screw tip – 20 mm below the upper edge of the beam. At this point, however, there are almost no compression perpendicular to grain stresses left, because these stresses are already transferred as shear stresses into the support.

Alternatively, the load distribution can also be considered nonlinearly. Further information can be found in [1].

Linear (Left) and Nonlinear (Right) Load Distribution

Ensuring Load Transfer

In order for the timber and fully threaded screws to act together effectively, the acting compressive force must be distributed as evenly as possible across all screws. Furthermore, it must be ensured that the compressive stresses introduced via the screw heads can be absorbed by the bearing material. These requirements can generally only be met with a flat and sufficiently rigid bearing, which is often implemented using an adequately thick steel plate. The required steel plate thickness in [mm] can be approximately determined according to [2] as follows:

For beam supports, an elastomeric layer is often additionally provided beneath the steel plate. This allows the bearing to rotate better, which promotes more uniform load transfer.

Shear Force Reduction

With the shear force reduction option in the design supports, the shear design at the support is performed using the governing shear force. In this case, the shear force at a certain distance from the support edge is considered in the design. The distance depends on the selected standard. This assumes that the force acts on the opposite side of the bearing, i.e., usually on the top side of the beam. The resulting compression perpendicular to grain stresses increase the shear strength. In the current Eurocode, the shear strength is not increased but, as already mentioned, the design is carried out with the reduced shear force. An interaction, as is the case, for example, in the current SIA 265, is planned for the second generation of Eurocode 5.

Activating Shear Force Reduction in Design Supports

For the example in the next image, the governing shear force at a distance h from the support edge can be read as 39 kN. Although the maximum shear force over the support is 60 kN, due to the aforementioned reasons, the shear force may be reduced in the design from 60 kN to 39 kN.

Shear Force Distribution Under Uniformly Distributed Load with Marked Governing Shear Force

Definition of Internal Forces for Shear Force Reduction

With the settings from the previous image, the input is executed correctly and the shear force reduction is considered in the design.

Shear Design Ratio

Design with Reduced Shear Force

Without shear force reduction, the shear design is not fulfilled.

Shear Design Ratio Without Shear Force Reduction