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# Design of Bolted Connections | Concept and Formulas with Example

Bolted connections are a type of structural joint used to join two or more structural components in a steel structure using bolts. Bolts are a form of threaded fasteners which has a male thread and preformed matching female thread, such as nuts. Concepts of bolt value, the strength of bolted joints are important to design a bolted connection, and the same are discussed further.

## Types of Joints in Bolted Connections

There are two predominant types of joints in a bolted connection namely, lap joint and butt joint. There are sub-types within these two types i.e., eccentric connections, pure moment connections etc. but are beyond the scope of this blog. The nature of the joints and sub-types within these joints are explained as follows.

### 1. Lap Joint

In a lap joint, the main members to be connected are placed over one another to form an overlap between the members, and then the bolting is done on the overlapped portion. Because of the very nature of the connection, an eccentricity is produced.

Lap joint

### 2. Butt Joint

In this type of joint, a cover plate is used to join two members. Based on the number of cover plates there are two types of butt joint namely, single cover butt joint and double cover butt joint.

Butt joint types

To know a little more about bolts, types of bolts, other different types of butt and lap joints click here.

### Most Preferred Joint in a Bolted Connection

Out of all the above-mentioned joints, the double cover butt joint is the most preferred for the following reasons.

• The shear capacity of the bolt is more than in a lap joint

• No eccentricity is present in the connection

## Types of Failure in Bolted Connections

In a bolted connection either the connecting plate might fail or the bolt might fail. Therefore, it becomes important to consider the "Limit States" or failure modes of both bolt and the plate. Possible limit states by which a bolted connection might fail are mentioned below.

### Failure Modes of Bolts in a Bolted Connection

• Shear Failure of Bolts

• Bearing Failure of Bolts

• Tensile Failure of Bolts

### Failure Modes of Plate in a Bolted Connection

• Shear Failure of Plate

• Bearing Failure of Plate

• Tensile Failure of Plate

### 1. Shear Failure of Bolts

As the name suggests, this failure occurs due to shear force at the interface of surfaces in a joint. Depending on the number of shear surfaces there are two types of shear failure that could occur in a bolted connection, namely, the single shear failure and double shear failure.

#### Single Shear Failure

Here, the bolt is subjected to a single shear force which could cause the failure of the bolt. This type of failure occurs in single cover butt joint and lap joint.

Single shear failure of bolt in a lap joint

#### Double Shear Failure

Here, the bolt is subjected to two shear forces at two separate shear planes. This type of failure occurs in the double cover butt joint.

Double shear failure of bolt in double cover butt joint

### 2. Bearing Failure of Bolts

In this failure, the bolt fails in bearing due to contact with the plates. This type of failure occurs in cases where a low-strength bolt is used with a plate of very high grade, which usually doesn't occur in practice.

Bearing failure of bolt

### 3. Tensile Failure of Bolts

The tensile strength of the bolt is the amount of pull the bolt can withstand in the perpendicular direction to the plane of loading. If the pull on this perpendicular axis exceeds the tensile strength of the bolt, then the bolt will fail in tension.

Tensile failure of bolt

### 4. Shear Failure of Plate

Shear failure of plate

### 5. Bearing Failure of Plate

Both shear and bearing failure of the plate can be avoided by providing sufficient centre to centre distances between the bolts as mentioned in section 10 of IS 800: 2007.

### 6. Tension Failure of Plate

Due to a reduction in the net area (i.e., due to bolt holes) of the plate along the bolt line, the tensile strength of the plate will be lesser than the actual value at this section. Because of this, the plate might fail under tension. Therefore, it becomes important to calculate the least net area among different bolt lines to find the least tensile strength of the plate and check it for safety for the applied load.

Tensile failure of plate

## Assumptions in Designing a Bolted Connection

• All bolts in a connection are stressed equally

• Friction between the plates in a connection is neglected

• The shear stress distribution is uniform for the bolt

• The bearing stress distribution is uniform for the bolt

• The bending of the bolt is neglected in case of smaller grip length, in case of larger grip lengths reduction factors must be considered for bending of the bolt

## Basic Terminologies in Designing a Bolted Connection

• Pitch distance (p) - centre to centre distance between two adjacent bolt holes in the direction of the applied load

• Gauge distance (g) - centre to centre distance between two adjacent bolt holes in the perpendicular direction of the applied load

• Edge distance (e) - the distance between the edge of the plate to the nearest centerline of a bolt hole in the perpendicular direction to the applied load

• End distance (e') - the distance between the end of the plate to the nearest centerline of the bolt hole in the direction of the applied load

Terminologies used in a bolted connection

For detailed specifications regarding all the above-mentioned terms refer to IS 800: 2007, Section 10 Connections.

## Design Strength of Bolts

Black bolts are the least expensive bolts available and are also called ordinary, unfinished, or common bolts. They are primarily used in light structures under static loading such as small trusses, bracings, etc.

The grade of a bolt indicates the properties of the bolt such as nominal ultimate tensile strength, and yield strength. It varies from 4.6 to 10.9. Generally, a bolt of grade 4.6 is used in construction. A grade of 4.6 means, 4 indicates 1/100 the of the nominal ultimate tensile strength of the bolt in N/mm^2, and 0.6 indicates that the yield strength of the bolt is 60% of the ultimate tensile strength of the bolt. Thus a bolt with grade 4.6 has an ultimate tensile strength of 400 N/mm^2 and a yield strength of 240 N/mm^2. The same is indicated in the picture below.

Ultimate and yield strength of bolt

### 1. Bearing Capacity of the Bolt

As per IS 800: 2007, the Design Bearing Strength of the Bolt (Vdpb) is given by,

Vdpb = Vnpb/γmb,

where,

γmb - partial safety factor;

Vnpb = Nominal Bearing Strength of the Bolt = 2.5 * Kb * d * t * fu;

d - nominal diameter of the bolt;

t - the sum of the thickness of the connected plates experiencing bearing stress in the same direction;

fu - ultimate tensile strength of the plate (i.e., for Fe410, fu=410N/mm^2);

Kb - smaller of the following values:

• e/(3 * do),

• (p/(3 * do)) - 0.25,

• fub/fu, and,

• 1

where,

e,p - end and pitch distances of the fastener along bearing direction;

e = 1.5 * do; p = 2.5 * d;

do - diameter of the hole (refer Table 19 (clause 10.2.1) of IS 800: 2007, a simplified version is provided below) ;

fub - ultimate tensile strength of the bolt;

fu - ultimate tensile strength of the plate.

 Diameter of the bolt (d) 12 14 16 20 24 28 30 Diameter of the hole (do) 13 15 18 22 26 31 33

### 2. Shear Capacity of the Bolt

As per IS 800: 2007, the Design Shear Strength of the Bolt (Vdsb) is given by,

Vdsb = Vnsb/γmb,

where,

γmb - partial safety factor;

Vnsb = Nominal Shear Capacity of the Bolt = (fub/√3) * ((nn * Anb) + (ns * Asb));

fub - ultimate tensile strength of a bolt;

nn - number of shear planes with threads intercepting the shear plane;

ns - number of shear planes without thread intercepting the shear plane (i.e., shank area intercepts the shear plane), usually taken as zero (0);

Anb - net shear area of the bolt at threads, taken as the area corresponding to root diameter at the thread;

Anb = 0.78 * ((π*d^2)/4); d - diameter of the bolt; and

Asb - nominal plain shank area of the bolt.

#### Reduction Factors in Design Shear Strength of the Bolt

• If the joint is too long - when the distance between the first and last rows of bolts in the joint, measured in the direction of load transfer exceeds 15 times the diameter of the bolt (d);

• If the grip length is large - when the grip length (equal to the thickness of the connected plates) exceeds 5 times the diameter of the bolt (d);

• If a packing plate of more than 6 mm in thickness is provided

Refer to section 10.3.3 shear Capacity of Bolt of IS 800: 2007 to get the formulas for reduction factors of each of the above-mentioned cases.

### Bolt Value or Design Strength of the Bolt

Bolt value is the least of Design Bearing Strength (Vdpb) and Design Shear Strength (Vdsb) of the bolt. Let us understand the same using an example problem.

### Bolt Value Example Problem

Question: Calculate the bolt value of a 20mm diameter bolt of grade 4.6 to join main plates of 12mm thick under the following cases. Assume Fe410 for all plates.

1. Lap joint

2. Single cover butt joint with cover plate = 10mm thickness

3. Double cover butt joint with cover plate = 8mm thickness

Solution:

Given data,

Diameter od the bolt (d) = 20mm

for Fe410 grade plates, fu = 410 N/mm^2

for bolt grade 4.6, fub = 400 N/mm^2

Common data,

Diameter of the bolt hole (from table 19 of IS 800: 2007) = 22mm

γmb = 1.25 (from Table 5 of IS 800: 2007)

It is assumed that the threaded portion alone is crossing the shear plane (i.e., ns = 0).

Anb = 0.78 * (π*d^2)/4)

Anb = 0.78 * ((π*20^2)/4) = 245 mm^2

e = 1.5 * do = 1.5 * 22 = 33 mm

p = 2.5 * d = 2.5 * 20 = 50 mm

Case 1: Lap joint (nn =1)

1. Design bearing strength of the bolt

Vdpb = Vnpb/γmb,

Vnpb = 2.5 * Kb * d * t * fu,

Kb least of:

• e/(3 * do) = 33/(3*22) = 0.5,

• (p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,

• fub/fu, = 400/410 = 0.97, and

• 1

Therefore Kb = 0.5.

t = 12 mm (thickness of the main plate)

Vnpb = 2.5 * 0.5 * 20 * 12 * 410 = 123 kN

Vdpb = 123/1.25 = 98.4 kN

2. Design shear strength of the bolt

Vdsb = Vnsb/γmb,

Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),

Vnsb = (400/√3) * (1 * 245) = 56.58 kN

Vdsb = 56.58/1.25 = 45.26 kN

Therefore, the bolt value or design strength of the bolt in a lap joint is = 45.26 kN (least).

Case 2: Single cover butt joint (nn =1)

1. Design bearing strength of the bolt

Vdpb = Vnpb/γmb,

Vnpb = 2.5 * Kb * d * t * fu,

Kb least of:

• e/(3 * do) = 33/(3*22) = 0.5,

• (p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,

• fub/fu, = 400/410 = 0.97, and

• 1

Therefore Kb = 0.5.

t = 10 mm (thickness of the cover plate)

Note:

• if the thickness of the cover plate < thickness of the main plate, then t = the thickness of cover plate;

• if the thickness of the cover plate > thickness of the main plate, then t = the thickness of the main plate;

• if two cover plates are used, t = summation of the thicknesses of both the cover plates provided that the sum is less than the thickness of the main plate.

Vnpb = 2.5 * 0.5 * 20 * 10 * 410 = 102.5 kN

Vdpb = 123/1.25 = 82 kN

2. Design shear strength of the bolt

Vdsb = Vnsb/γmb,

Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),

Vnsb = (400/√3) * (1 * 245) = 56.58 kN

Vdsb = 56.58/1.25 = 45.26 kN

Therefore, the bolt value or design strength of the bolt in a single cover butt joint is = 45.26 kN (least).

Case 3: Lap joint (nn =2)

1. Design bearing strength of the bolt

Vdpb = Vnpb/γmb,

Vnpb = 2.5 * Kb * d * t * fu,

Kb least of:

• e/(3 * do) = 33/(3*22) = 0.5,

• (p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,

• fub/fu, = 400/410 = 0.97, and

• 1

Therefore Kb = 0.5.

As the sum of the thickness of the two cover plates is 16mm (8+8), the t value is taken as the thickness of the main plate (a lesser value must be taken).

t = 12 mm (thickness of the main plate)

Vnpb = 2.5 * 0.5 * 20 * 12 * 410 = 123 kN

Vdpb = 123/1.25 = 98.4 kN

2. Design shear strength of the bolt

Vdsb = Vnsb/γmb,

Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),

Vnsb = (400/√3) * (2 * 245) = 113.16 kN

Vdsb = 56.58/1.25 = 90.52 kN

Therefore, the bolt value or design strength of the bolt in a double cover butt joint is = 90.52 kN (least).

## Design Tensile Strength of the Plate

As per IS 800: 2007, Section 6 Design of Tension Members, the design strength in tension of a plate, Tdn as governed by rupture of net cross-sectional area (An) is given by,

Tdn = (0.9 * fu * An)/γm1,

where,

An - the net effective area of the plate = (b - n*do)*t, for a chain of holes in a row;

Design tensile strength of plate governed by net-section

An = (b - n*do + Σ[(psi^2)/(4*gi)])*t, for staggered arrangement;

fu- ultimate tensile strength of the plate;

γm1 = 1.25, refer table 5 of IS800: 2007.

## Strength of a Bolted Connection

The strength of a bolted connection is the least of bolt value (design strength of the bolt) and the design tensile strength of the plate.

## Practise Problem

Refer to the below attached video lectures on Introduction to bolted connection and codal provisions for bolted connection.

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