Learning Module: Bolts loaded in shear (EN)

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Learning Module: Bolts loaded in shear (EN)

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This tutorial is intended for students. It aims to explain how shear forces are distributed between bolts and what checks are necessary for bolts loaded predominantly in shear. Hand calculations are compared to IDEA StatiCa Connection.

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.

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Geometry

d_0=22\textrm{ mm}

Solution

V=50\textrm{ kN}

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:

$V_{Ed}$
$n_b$

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:

$r_i$
$p$

F_{1,M}

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:

$M_{Ed}$
$r_1$
$r_i$

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:

V_{Ed}

Exactly the same shear forces are in IDEA StatiCa.

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:

$Shear resistance$
$Add more bolts with a spacing 70 mm:$

Virtual Lab – Bolts in Shear

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

The whole member consisting of two plates transfers 100 kN. Each plate transfers the same magnitude 100 kN / 2 = 50 kN .

Each bolt transfers the same force, 100 kN / 2 bolts = 50 kN .

Note that for long bolted connections (EN 1993-1-8 – 3.8), the shear force in bolts is not uniformly distributed, but the first and last bolts are loaded more than the bolts in the middle. Refer to this article for more information. This is why reduction factor \(\beta_{Lf}=1-\frac{L_j-15d}{200d}\le1.0\) is used for long bolted connections.

For each bolt, there are 2 shear planes. Every shear plane is loaded by 100 kN / 2 bolts / 2 shear planes = 25 kN .

15\cdot d = 15\cdot 20 = 300\textrm{ mm}

Add more bolts with a spacing 70 mm:

We can estimate the force acting on the most-loaded bolt by calculating reduction factor \(\beta_{Lf}\), although it is not possible to clearly determine the force in each bolt. The resistance of the bolt remains the same, but the load varies. The bolt group can safely transfer the force:

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

where:

$\(\beta_{Lf} = 1-\frac{L_j-15d}{200d}=1-\frac{420-15\cdot 20}{200\cdot 20}=0.97\) – reduction factor for long bolted connections$
$\(N\) – number of bolts$
$\(n\) – number of shear planes$

The load acting on the first bolt assuming the uniform force distribution would be \(\frac{F}{N} = \frac{100 \textrm{ kN}}{7} = 14.29\textrm{ kN}\), and \(7.14 \textrm{ kN}\) per each shear plane.

We may assume the force in the first bolt is higher, probably close to \(\frac{14.29\textrm{ kN}}{\beta_{Lf}}=14.72\textrm{ kN}\), i.e., \(7.36\textrm{ kN}\) at one shear plane.

The most loaded bolt is B7, which is indeed located at the beginning of the bolt group. The load acting at each shear plane is 9 kN, i.e., 18 kN at bolt B7. This is more than the estimated 14.72 kN. It seems IDEA StatiCa provides a more conservative bolt force distribution; however, it may change with consecutive loading due to nonlinear calculation and nonlinear load-deformation diagram of the bolt in shear.