# Design of Fillet Welds According to EN 1993-1-8

### Technical Article

A fillet weld is the most common weld type in steel building construction. According to EN 1993‑1‑8, 4.3.2.1 (1) [1], fillet welds may be used for connecting structural parts where the fusion faces form an angle between 60° and 120°.

The effective weld thickness a of a fillet weld should be taken as the height of the largest triangle (with equal or unequal legs) that can be inscribed within the fusion faces and the weld surface, measured perpendicular to the outer side of this triangle, see Figure 01.

Image 01 - Fillet weld thickness a at normal penetration (a) and at deep penetration (b)

#### Design Resistance of Fillet Welds

According to 1993-1-8 [1], the design resistance of a fillet weld is usually determined using Directional Method or Simplified Method. The Directional Method is described below.

A uniform distribution of stress is assumed on the section of the weld, leading to the normal stresses and shear stresses shown in Figure 02, as follows:

- σ
_{⊥}normal stress perpendicular to weld axis - σ
_{||}normal stress parallel to weld axis - τ
_{⊥}shear stress (in plane of fillet weld surface) perpendicular to weld axis - τ
_{||}shear stress (in plane of fillet weld surface) parallel to weld axis

Image 02 - Weld Stresses on Throat Section of Fillet Weld

The normal stress σ_{||} parallel to the axis is not considered when verifying the design resistance of the fillet weld.

The design resistance of the fillet weld will be sufficient if the following conditions are met:

$$\begin{array}{l}\sqrt{{\mathrm{\sigma}}_{\perp}^{2}3\xb7({\mathrm{\tau}}_{\perp}^{2}{\mathrm{\tau}}_{\left|\right|}^{2})}\le \frac{{\mathrm{f}}_{\mathrm{u}}}{{\mathrm{\beta}}_{\mathrm{w}}\xb7{\mathrm{\gamma}}_{\mathrm{M}2}}\\ {\mathrm{\sigma}}_{\perp}\le 0.9\xb7\frac{{\mathrm{f}}_{\mathrm{u}}}{{\mathrm{\gamma}}_{\mathrm{M}2}}\end{array}$$

where

f_{u} is the nominal ultimate tensile stress of the weaker part joined

β_{w} is the appropriate correlation factor (see EN 1993-1-8, Table 4.1)

γ_{M2} is the partial safety factor for resistance of welds

#### Example

Design of a fillet weld of the beam displayed in Figure 03 from [2].

Material: S235, f_{u} = 36.0 kN/cm², β_{w} = 0.8

Internal forces: V_{z} = 350 kN

Centroid

$${\mathrm{z}}_{\mathrm{S}}=\frac{\mathrm{\Sigma}({\mathrm{A}}_{\mathrm{i}}\xb7{\mathrm{z}}_{\mathrm{Si}})}{{\mathrm{\Sigma A}}_{\mathrm{i}}}=\frac{91.48\xb743.7240.00\xb744.0048.00\xb723.0045.00\xb71.50}{224.48}=30.88\mathrm{cm}$$

**Moment of inertia**

With regard to the centroid, the moment of inertia is:

$\begin{array}{l}{\mathrm{I}}_{\mathrm{y}}=\sum ({\mathrm{I}}_{\mathrm{yi}}{\mathrm{A}}_{\mathrm{i}}\xb7{\mathrm{z}}_{\mathrm{si}}^{2})-\frac{{\left(\sum {\mathrm{A}}_{\mathrm{i}}\xb7{\mathrm{z}}_{\mathrm{Si}}\right)}^{2}}{{\mathrm{\Sigma A}}_{\mathrm{i}}}=\\ =850,88\frac{20,00\xb72,00\xb3}{12}\frac{1,20\xb740,00\xb3}{12}\frac{15,00\xb73,00\xb3}{12}91,48\xb743,72\xb240,00\xb744,00\xb248,00\xb723,00\xb245,00\xb71,50\xb2-\\ -\frac{(91,48\xb743,7240,00\xb744,0048,00\xb723,0045,00\xb71,50)\xb2}{224,48}=\\ =71.095{\mathrm{cm}}^{4}\end{array}$

**Static moments**

With regard to the centroid, the static moments of the cross-sections connected are calculated by using the welds ➀, ➁ and ➂:

S_{y,1} = A_{1} ∙ (z_{S,1} - z_{S}) = 91.48 ∙ (43.72- 30.88) = 1,175 cm³

S_{y,2} = S_{y,1} + A_{2} ∙ (z_{S,2} - z_{S})= 1175 + 40.00 ∙ (44.00 - 30.88) = 1,700 cm³

S_{y,3} = A_{3} ∙ (z_{S} - z_{S,3}) = 45.00 ∙ (30.88- 1.50) = 1,322 cm³

**Design of welds**

$\begin{array}{l}{\mathrm{\tau}}_{\left|\right|,\mathrm{Vz},\mathrm{i}}=\frac{-{\mathrm{V}}_{\mathrm{z}}\xb7{\mathrm{S}}_{\mathrm{y},\mathrm{i}}}{{\mathrm{I}}_{\mathrm{y}}\xb7{\mathrm{\Sigma a}}_{\mathrm{w},\mathrm{i}}}\le \frac{{\mathrm{f}}_{\mathrm{u}}}{\sqrt{3}\xb7{\mathrm{\beta}}_{\mathrm{w}}\xb7{\mathrm{\gamma}}_{\mathrm{M}2}}=\frac{36,0}{\sqrt{3}\xb70,8\xb71,25}=20,78\mathrm{kN}/\mathrm{cm}\xb2\\ {\mathrm{\tau}}_{\left|\right|,\mathrm{Vz},1}=\frac{-350\xb71.175}{71.095\xb72\xb70,4}=-7,23\mathrm{kN}/\mathrm{cm}\xb220,78\mathrm{kN}/\mathrm{cm}\xb2\\ {\mathrm{\tau}}_{\left|\right|,\mathrm{Vz},2}=\frac{-350\xb71.700}{71.095\xb72\xb70,5}=-8,37\mathrm{kN}/\mathrm{cm}\xb220,78\mathrm{kN}/\mathrm{cm}\xb2\\ {\mathrm{\tau}}_{\left|\right|,\mathrm{Vz},3}=\frac{-350\xb71.322}{71.095\xb72\xb70,4}=-8,13\mathrm{kN}/\mathrm{cm}\xb220,78\mathrm{kN}/\mathrm{cm}\xb2\end{array}$

**SHAPE-THIN**

In SHAPE-THIN, the shear stress (in the plane of the fillet weld surface) parallel to the weld axis τ_{||} can be calculated on fillet welds and the resistance can be designed. When modeling, the weld must be connected to the edges of two elements. One of these elements can also be a dummy element.

In Column H 'Continuous Element' of Table 1.6 Welds, you can define the continuous elements. No weld stresses are calculated on these elements. If there is no element specified in Column H, the weld stresses are determined on all elements to which the weld is connected. These elements can be taken from Column B 'Elements No.'.

Figure 04 shows the weld definition for the example described in this article.

Table 5.1 Welds displays the resulting stresses τ_{||} for the welds defined in Table 1.6 Welds. Figure 05 shows the weld stresses for the example described in this article.

#### Reference

#### Keywords

Fillet weld Weld Weld stress Weld seam design Fillet weld design

#### Downloads

#### Links

### Write Comment...

### Write Comment...

#### Contact us

Do you have questions or need advice?

Contact our free e-mail, chat, or forum support or find various suggested solutions and useful tips on our FAQ page.

###### Recommended Events

###### Videos

###### Models to Download

###### Knowledge Base Articles

#### Redistributing Shear Stresses from Null Elements

SHAPE-THIN allows you to calculate section properties and stresses of any cross‑sections. If a flange or a web is weakened by bolt holes, you can consider this by using null elements. The stresses are subsequently recalculated with the reduced cross‑section values. In this case, it is necessary to pay special attention to shear stresses. By default, these are set to zero in the area of the null elements. When recalculating shear stresses with the reduced cross‑section values and without further adaptation, it turns out that the integral of the shear stresses is no longer equal to the applied shear force. The following example shows in detail how to calculate the shear stress.

###### Screenshots

###### Product Features Articles

#### Material Database with Steels According to the Australian Standard AS/NZS 4600:2005

The material database in RFEM, RSTAB and SHAPE-THIN contains steels according to the Australian standard AS/NZS 4600:2005.###### Frequently Asked Questions (FAQ)

- What is the purpose of a fillet weld in SHAPE‑THIN?
- How can I display shear stresses on null elements in SHAPE‑THIN?
- Is it possible to model cold-formed sections in SHAPE‑THIN 8 and to design them in RF‑/STEEL Cold-Formed Sections?
- When calculating a cross-section, I get a message saying that the weld is not connected to two elements. What should I do?
- I would like to calculate the results, such as ordinates, static moments, stresses, at a specific location of a cross-section. What should I do?

- How can I calculate stresses in SHAPE‑THIN?
- When calculating a stiffened panel, I get the message that the boundary c/t-zone of the panel is not supported. What should I do?
- When calculating a longitudinally stiffened panel, I get a message saying that the stiffener cannot be at the end of the panel. What should I do?
- When selecting the elements for a stiffener of a stiffened panel, I get a message saying that at least one of the selected elements is a null element. What should I do?
- When calculating a buckling panel, I get a message saying that some elements are connected to the panel, but they are not defined as stiffeners. What should I do?