Introduction

Bolts are commonly used as fasteners connecting steel elements in assembly joints. They facilitate the rapid construction of steel structures. Contractors typically require all connections at the construction site to be bolted; in other words, field welding should be avoided despite the fact that bolted connections may be much more complicated than welded connections and require more material.

The aim of this learning module is to make students familiar with and confident when designing simple bolted connections through interaction with connection design software providing visual feedback. Students should be familiar with the basics of design according to Eurocode – for more information, refer e.g. ESDEP lecture notes.

Example: Bolts in shear

Calculate the load on bolts and check the maximum loaded bolt.

Geometry

A plate with a thickness of 20 mm is connected to a cantilever consisting of two plates with a thickness of 8 mm each by two bolts M20 8.8. A cantilever is loaded by a force 50 kN at a distance 100 mm from the plate edge. Bolt holes are standard (\(d_0=22\textrm{ mm}\)) and the shear force passes through the threads.

Solution

Crucial is the determination of the center of rotation and the loading on each bolt. The center of rotation for a bolt group loaded in shear is assumed to be in the center. The bolt connection is loaded by a shear force and bending moment. The shear force is \(V=50\textrm{ kN}\) and the bending moment is \(M=50\cdot 0.175=8.75 \textrm{kNm}\). In IDEA StatiCa, the shear force position must be correctly set or the connection loaded by a combination of shear force and bending moment.

The shear force is transferred via bolts uniformly, i.e. each bolt transfers the same portion of the shear load:

\[F_{1,V}=V_{Ed}/n_b=50/2=25\textrm{ kN}\]

where:

• \(V_{Ed}\) – set shear force
• \(n_b\) – number of bolts

In IDEA StatiCa, if the shear force position is set to Bolts, the bolt group is loaded purely by shear:

The forces are indeed identical and each bolt is loaded by 12.5/12.5, i.e. 12.5 kN in each shear plane.

The bending moment is transferred via the bolt group as well. Each bolt is loaded proportionally with the distance of the bolt to the center of rotation. In this example, there are only two bolts with identical distance:

\[r_i=p/2=70/2=35\textrm{ mm}\]

where:

• \(r_i\) – distance from bolt to the center of rotation
• \(p\) – bolt spacing

The force acting on each bolt, \(F_{1,M}\), is calculated:

\[F_{1,M}=M_{Ed}\frac{r_1}{\Sigma r_i^2}=8.75\frac{0.035}{0.035^2+0.035^2}=125\textrm{ kN}\]

where:

• \(M_{Ed}\) – bending moment acting on the connection
• \(r_1\) – distance from the investigated bolt to the center of rotation
• \(r_i\) – distance from each bolt to the center of rotation

Despite the point of application of the load being quite close, just 100 mm from the plate edge, the force in the bolt resulting from the bending moment is very large.

In IDEA StatiCa, the connection may be loaded only by bending moment:

Now, we must do a vector sum of both effects – shear force and bending moment. The direction of forces is crucial here. The forces in bolts from shear force \(V_{Ed}\) acts downwards while the forces from bending moment rotate around the center of rotation. That means one goes upwards and the other downwards. The force in one bolt is subtracted: \(F_{1,v,Ed} = 25 - 125 = - 100\textrm{ kN}\), the force in the other bolt is summed: \(F_{2,v,Ed} = 25 + 125 = 150\textrm{ kN}\).

Exactly the same shear forces are in IDEA StatiCa.

The bigger force determines the design, \(F_{v,Ed}=F_{2,v,Ed}=150\textrm{ kN}\).

Detailed checks are provided for bolt B2. Bolts loaded in shear must be checked for:

• Shear resistance
• Bearing resistance

Virtual Lab – Bolts in Shear

Change the load effect to tensile load only, 100 kN.

Create a long-bolted connection. Long bolted connection is longer than \(15\cdot d = 15\cdot 20 = 300\textrm{ mm}\). Extend the overlap of members:

Add more bolts with a spacing 70 mm:

Qual é a força que atua no parafuso mais solicitado?

Podemos estimar a força que atua no parafuso mais solicitado calculando o fator de redução \(\beta_{Lf}\), embora não seja possível determinar claramente a força em cada parafuso. A resistência do parafuso mantém-se igual, mas a carga varia. O grupo de parafusos pode transferir com segurança a força:

\[F_{v,Rd} = \beta_{Lf} \cdot N \cdot \frac{n \cdot \alpha_v \cdot f_{ub} \cdot A}{\gamma_{M2}}\]

onde:

• \(\beta_{Lf} = 1-\frac{L_j-15d}{200d}=1-\frac{420-15\cdot 20}{200\cdot 20}=0.97\) – fator de redução para ligações aparafusadas longas
• \(N\) – número de parafusos
• \(n\) – número de planos de corte

A carga que atua no primeiro parafuso, assumindo uma distribuição uniforme de forças, seria \(\frac{F}{N} = \frac{100 \textrm{ kN}}{7} = 14.29\textrm{ kN}\), e \(7.14 \textrm{ kN}\) por cada plano de corte.

Podemos assumir que a força no primeiro parafuso é superior, provavelmente próxima de \(\frac{14.29\textrm{ kN}}{\beta_{Lf}}=14.72\textrm{ kN}\), ou seja, \(7.36\textrm{ kN}\) num plano de corte.

Compare a sua força estimada com os resultados do IDEA StatiCa

O parafuso mais carregado é o B7, que está de facto localizado no início do grupo de parafusos. A carga que atua em cada plano de corte é de 9 kN, ou seja, 18 kN no parafuso B7. Este valor é superior aos 14,72 kN estimados. Parece que o IDEA StatiCa fornece uma distribuição de forças nos parafusos mais conservadora; no entanto, pode variar com o carregamento consecutivo devido ao cálculo não linear e ao diagrama não linear força-deformação do parafuso ao corte.