# Chapter 11. Bolted Connections – Design of Steel Structures

## Bolted Connections

#### 11.1 Introduction

The fundamentals of bolted connections are presented in Chapter 2. In this chapter, bolted connections subjected to combined actions such as shear, twisting or bending are considered. The types of connections considered include bracket connections, beam-to-column end connections and splicing of beams. Bolted connections have certain advantages over welded connections even though they require holes to be made in the parts to be connected. The foremost of these advantages is the ease with which bolted connections can be made especially in the field i.e., during erection since doing quality welding in the field is very difficult. The problems of misalignment and misfit between the connecting parts during erection can be easily overcome with bolted connections, especially in the case of high strength friction grip bolts, since tolerance on the hole size is more.

#### 11.2 Bracket Connections

Brackets are used to support beams at some eccentricity from the column face. A bracket may be connected to the flange of a column in either of the following ways

1. at right angle to the flange of a column
2. in the plane of the flange of a column

#### 11.2.1 Bracket Connected at Right Angle to the Flange of a Column

When a bracket is connected at a right angle to the flange of a column with the help of angles as shown in Figure 11.1, the bolts ‘A’ connecting the angles to the column flange are subjected to direct shear and bending moment. Alternatively, a T-stub may also be used in which case no angles are needed. It may be assumed that the neutral axis exists at h/7 from the bottom of the bracket where h is the height of the bracket.

Figure 11.1 Bracket connected at right angle to the flange of a column

The factored moment M = Pe is resisted by the tensions developed in the bolts above the neutral axis x. For equilibrium, the moment of the tensions in the bolts above the neutral axis should be equal to Pe.

where T1, T2, …. Tn are the tensions in each of the bolts at distances y1, y2 …. yn from the neutral axis, respectively. If it is assumed that the tension in a bolt is proportional to the distance (y1, y2 ….. yn) from the neutral axis,

then it may be written as

Substituting these in Equation (11.1)

or

is the maximum tension developed in the bolt of the topmost row.

The shear force in each bolt due to the direct shear is given by

where N is the total number of bolts in the connection.

As per IS800:2007, a bolt subjected to tension and shear should satisfy the interaction equation

where Tb = Tn; Vdb and Tdb are shear and tension capacities of bolt, respectively.

For design purpose, the approximate number of bolts required in a vertical line may be known from

where m is the number of vertical lines of bolts in the connection, p is the pitch of bolts in the vertical direction.

#### 11.2.2 Bracket Connected in the Plane of Flange of a Column

When bracket is connected to the flange of a column in its plane as shown in Figure 11.2, bolts are subjected to direct shear and torsion. In this type of connection, a bolt is subjected to a resultant shear force which is due to direct shear and twisting moment.

Figure 11.2 Bracket connected in the plane of the flange of a column

It is assumed that the direct shear is resisted equally by all the bolts. Therefore, the vertical shear force V' on each bolt is given by

V′ = P/N     (11.6)

where N is the total number of bolts.

The twisting moment M = Pe is resisted by the shear forces V1″, V2″, …… VN″ developed in the bolts (Figure 11.3). These shear forces act perpendicular to the lines joining the bolts with the centre of gravity of the bolt group. If the applied moment is clockwise, the directions of these shear forces are such that they produce an anti-clockwise moment. For equilibrium,

Figure 11.3 Shear forces on a bolt that is in critical condition

V1r1 + V2r2 + ................ + VNrN = M

or

Viri = M     (11.7)

It is assumed that the shear forces V1″, V2″, ……, VN″ are proportional to their distances r1, r2, ……., rN from the centre of gravity of the bolts C, respectively, then

or

where V1″ is the maximum shear force in the extreme bolt due to twisting moment Pe. Substituting Equation (11.8) in Equation (11.7),

or

or

or

The horizontal and vertical components of V1″ may be expressed as

The horizontal and vertical components of resultant shear force V1 on the extreme bolt are given by

Therefore, the resultant shear force V1 on the extreme bolt is expressed as

which should be less than or equal to the design strength of bolt Vdb, i.e. the least value of design strength of bolt in shear Vdsb and design strength of bolt in bearing Vdpb.

For design purpose, the approximate number of bolts required in a vertical line may be known from

where m is the number of vertical lines of bolts in the connection, p is the pitch of bolts in vertical direction.

Example 11.1

Re-do Example 10.1 using M20 bolts of product Grade C and property class 4.6 if 12 mm thick bracket is connected by two ∠100 100 × 10.

The moment due to eccentricity, M = Pe = 200 × 300/1,000 = 60 kNm

Yield strength of the bolt material = 0.6 × 400 = 240 MPa

Design of bolts ‘A’

Bolts ‘A’ are subjected to single shear and the assuming shear plane lies in the shank of bolts

From Equations (2.7) and (2.8),

The number of bolts, N = 2 × 7 = 14

The bolts are provided as shown in Figure 11.5.

Figure 11.5

Figure 11.4

The depth of the bracket, h = 6 × 60 + 2 × 40 = 440 mm

The neutral axis is at h/7 = 440/7 = 63 mm from the bottom of the bracket.

The number of bolts may be reduced to 12 and the calculations may be repeated.

h = 5 × 60 + 2 × 40 = 380 mm

h/7 = 54 mm

y1 = 46 mm, y2 = 106 mm, y3 = 166 mm, y4 = 226 mm, y5 = yn = 286 mm

yi2 = 2(462 + 1062 + 1662 + 2262 + 2862) = 3,46,565mm2

Check for bearing

kb = 0.6

Vnpb = 2.5 ×0.6 × 20 × 10 × 410 = 125kN

Vdpb = 125/1.25 = 100kN > 16.7kN OK

Design of bolts ‘B’

These bolts are subjected to direct shear and twisting.

The twisting moment, M = 200 × 240/1,000 = 48 kNm

As these bolts are in double shear, Vdsb = 2 × 58 = 116 kN

Since there is only one vertical line of bolts, m = 1

7 bolts may be tried as shown in Figure 11.6.

Figure 11.6

Check for bearing

Vnpb = 2.5 ×0.6 × 20 × 12 × = 148kN > 91kN

Hence, the depth of angles / bracket may be 440 mm as assumed initially.

Example 11.2

Re-do Example 10.3 using high strength M20 bolts of property class 8.8 if

1. no slip is permitted at the ultimate load,
2. a slip is permitted at the ultimate load but not at the working load.

Since the width of the flange of the column is 200 mm (Figure 10.6), e = 100 + 300 = 400 mm

M = 200 × 400/1,000 = 80 kNm

ƒ0 = 0.7 × 800 = 560 MPa

(a) if no slip is permitted at the ultimate load

Vnf = μ f ne Kh F0 where F0 = Anb f0 = 245 ×560 = 1,37,225N

= 0.55 ×1 × 1 × 1 ×1,37,225 = 75,474kN

Vdsf = 75,474/1.25 = 60kN

Eighteen bolts may be provided as shown in Figure 11.7.

Figure 11.7

ri2 = 4[(502 + 2402) + 502 + 120)2 + 502 + 602) = 4,72,000mm2

(b) if slip is permitted at the ultimate load but not at the working load

If slip may occur at the ultimate load, bolts are designed for bearing and shear at the ultimate load.

Twelve bolts may be provided as shown in Figure 11.8.

Figure 11.8

ri2 = 4[(502 + 1502) + (502 + 902) + (502 + 302)] = 1,56,000mm2

The bolts provided should be checked for no slip at the working load.

Working load = 200/1.5 = 133 kN

Hence, no slip occurs at the working load.

#### 11.3 Simple Beam End Connections

These are the connections which are used to transfer only the end shear of a beam to a column or any other supporting member. It is assumed that the beam ends can rotate freely at these connections even though some amount of rigidity exists in these type of connections. Some of the simple beam end connections are described here.

#### 11.3.1 Web-angle Connection

As its welded counterpart, two angles are provided, each on either side of the web of the beam. One set of angle legs are connected to the web of the beam by bolts ‘A’ which are in double shear condition. The other set of angle legs are connected to a column or any other supporting member by bolts ‘B’ which are in single shear condition. It is assumed that the end shear V from the beam is transferred to the column at its face as shown in Figure 11.9. Due to this, bolts ‘A’ are subjected to shear, twisting and bearing. The twisting moment on these bolts is equal to Ve1 where e1 is the eccentricity of V from the centre line of bolts ‘A’. The maximum depth of the angles is limited to 0.75 times the depth of the beam. Usually, the angles are provided near the compression flange of the beam to provide a lateral restraint to the beam end. Bolts ‘B’ are assumed to be subjected to shear and bearing.

Figure 11.9 Web-angle connection

#### 11.3.2 Seat-angle Connection

This is similar to its welded counterpart described in Sec. 10.3.2 except that bolts are used to connect seat and cleat angles to the supporting member. The expressions given there (Equations 10.17, 10.19 and 10.20) for the bearing length b1 and the bending moment in the seat and the moment capacity of the seat remain the same. The bolts connecting the vertical leg of the seat-angle to the supporting member are subjected to shear and bearing. The beam is connected to the seat by two bolts to keep the beam in position. The cleat-angle is connected to the beam and the supporting member by at least two bolts as shown in Figure 11.10 to provide lateral support to the compression flange.

Figure 11.10 Seat-angle connection

#### 11.3.3 Stiffened Seat Connection

When the end shear from a beam is more, the seat-angle with the required thickness may not be available or the vertical leg of the seat-angle may not accommodate the required number of bolts. In such cases, the horizontal leg of the seat-angle is stiffened, usually, with a pair of angles whose vertical legs support the horizontal leg of the seat-angle as shown in Figure 11.11. The width of the outstanding legs of stiffener angles is governed by the local buckling and the web crippling of the beam. The width of the outstanding legs of stiffener angles should be greater than the bearing length b1 and is limited to 14εt (as per Clause 8.7.1.2 of IS800:2007) where t is the thickness of the stiffener angle. The outstanding legs of the stiffener angles are tack bolted and their combined thickness should be greater than the thickness of the web of the beam.

Figure 11.11 Stiffened seat connection

The bolts connecting the vertical legs of the stiffener angles to a column are subjected to vertical shear and bending moment. They should be designed as in the case of the bracket connection described in Sec. 11.2.1.

Example 11.3

Design a web-angle connection for a beam MB350 @ 52.4 kg/m which transfers a factored end shear of 200 kN to the flange of the column HB 300 @ 63 kg/m. Use M20 bolts of product Grade C and property class 5.6.

ƒub = 500 MPa

Design of bolts ‘A’

The bolts are in double shear.

kb is the least of

kb = 0.61

The thickness of the web of the beam = 8.1 mm

The design strength of the bolt, Vdb = 80.5 kN

Using 2 ∠90 90 × 8, e1 = 50 mm (gauge for 90 mm leg, Appendix A)

The twisting moment, M = 200 × 50/1,000 = 10 kNm

For a single vertical line of bolts, m = 1

Four bolts may be provided as shown in Figure 11.12.

Figure 11.12

ri2 = 2(902 + 302 ) = 18,000mm2

Design of bolts ‘B’

These bolts are in single shear.

The number of bolts = 200/73 = 2.8

Four bolts are provided, 2 in each leg of angle as shown in Figure 11.13.

Figure 11.13

The edge distance, e = 70 mm

kb = 1.0

Vnpb = 2.5 ×1.0 × 20 × 8 × 410 = 164kN

Vdpb = 164/1.25 = 131kN > 73kN OK

Example 11.4

Re-do Example 11.3 as a seat-angle connection using M20 bolts of product Grade C and the property class 4.6.

∠ 130 130 × 16 may be used as seat-angle as in Example 10.5.

The thickness of the flange of the column HB300 @ 63 kg/m = 10.6 mm

The bolts connecting the seat-angle to the column are in single shear.

Vdsb = 72.5/1.25 = 58 kN

kb is the least of

kb = 0.38

Vnpb = 2.5 × 0.38 × 20 × 10.6 × 400 = 80.6 kN

Vdpb = 80.6/1.25 = 64 kN

∴ Design strength of bolt = 58 kN

The number of bolts = 200/58 = 3.4

Four bolts may be provided as shown in Figure 11.14.

Figure 11.14

∠100 100 × 10 may be provided as cleat-angle to connect the upper flange of the beam to the column.

Example 11.5

Design a stiffened seat connection to transfer a factored end shear of 400 kN from a beam MB550 @ 104 kg/m to a column HB 450 @ 87.2 kg/m using M16 bolts of product Grade C and the property class 4.6.

n2 = 2.5 (18 + 19.3) = 93.25 mm

With an erection clearance of 10 mm, the minimum width of the outstanding legs of the stiffener angles = 64 + 10 = 74 mm

The width of the outstanding legs of the stiffener angles may be 80 mm

The required bearing area of the stiffener angles =

The thickness of the stiffener angles =

Hence, 2 ∠80 80 × 12 may be used as the stiffener angles with ∠100 100 × 10 as the seat-angle as shown in Figure 11.15(a).

Figure 11.15

Check for the local buckling of the outstanding legs of the stiffener angles

From Equations (2.7) and (2.8),

Twelve bolts may be provided as shown in Figure 11.15(b).

h = 5 × 60 + 2 × 40 = 380 mm

h/7 = 380/7 = 54 mm

y1 = 46 mm, y2 = 106 mm, y3 = 166 mm, y4 = 226 mm, y5 = yn = 286 mm

Check for bearing

kb is the least of

kb = 0.74

Vnpb = 2.5 × 0.74 × 16 × 12 × 410 = 145.6 kN

#### 11.4 Moment Resistant Beam End Connection

A typical moment resistant beam end connection is shown in Figure 11.16. T-stubs cut from I sections can be effectively used to connect the flanges of the beam to a column to transfer end moment. To transfer end shear, web-angle connection may be used. For the moment direction shown, bolts ‘A’ in the upper T-stub are subjected to tension. Bolts ‘B’ are subjected to shear and bearing.

Figure 11.16 Typical moment resistant connection

Bolts ‘A’ connecting the upper T-stub to the column are subjected to an additional tension Q due to “prying effect” which is illustrated in Figure 11.17. Due to end moment tension is produced in the uppre flange of the beam. Due to this tension, the T-stub is pulled by force P because of which couples of magnitude Ql1 are formed in the flange of T-stub. That is, an additional tension Q is developed in the bolts ‘A’ and forces Q are produced at the edges of flange of T-stub as they are pressed against the flange of the column.

Figure 11.17 Prying effect

∴     Ql1 = Pl2/2

or

IS800:2007 recommends the following expression for calculating the prying force Q.

where β = 2 for a non-pre-tensioned bolt

= 1 for a pre-tensioned bolt

η = 1.5

be = effective width of the flange per pair of bolts

f0 = proof stress = 0.7 fub

t = thickness of the flange of the T-stub

lv = distance from the bolt to the toe of the fillet weld or half the root radius (Figure 11.18)

Figure 11.18 Lengths lv and le

le = distance between the line of action of the prying force and the centre of the line of the bolt (Figure 11.18) or 1.1 whichever is less

The thickness of the flange of the T-stub should be such that no yielding occurs. At yielding, the flange of the T-stub is subjected to a bending moment of 0.5 Telv. The moment capacity of the flange of the T-stub should be equal to or more than the factored bending moment.

Figure 11.19

Example 11.6

Re-do Example 10.7 using M20 bolts of product Grade C and property class 4.6.

From Equations (2.7) and (2.8),

The tension in the bolts ‘A’ due to the applied moment =

Since the bolts ‘A’ are provided in two horizontal lines in the flange of the T-stub, the force in the bolts in each horizontal line

Te = 143/2 = 71.5 kN

T-stubs cut from MB450 may be tried to connect the flanges of the beam to the column. From Appendix A, bf = 150 mm, tf = 17.4 mm, tw = 9.4 mm, g = 90 mm and R1 = 15 mm.

le is minimum of (150 − 90)/2

∴ le = 28.6 mm

be = length of T-stub = 140 mm

The bending moment in the flange of the T-stub =

The moment capacity of the flange of the T-stub = 1.2 Ze ƒym0

The prying force,

The prying force in each bolt = 33.6/2 = 16.8 kN

The total tensile force in each of the bolts ‘A

Design of bolts ‘B’

kb = 0.61
Vnpb = 2.5 ×0.61 × 20 × 9.4 × 410 = 117.6kN
Vdpb = 117/1.25 = 94kN OK
Vdb = 72.6kN

The number of bolts required to connect the flange of the beam to the web of the

T-stub

Two M20 bolts may be provided to connect the flange of the beam to the web of the T-stub.

Check for tensile strength of web of T-stub

The design strength due to the yielding of the gross section, Tdg = Ag fym0

= (140 × 9.4) × 250/1.1 = 299 kN

The design strength due to the rupture at the net section, Tdn = 0.9 An fum1

= 0.9 × (140 − 2 × 22) × 9.4 × 410/1.25
= 266 kN

Both the strengths Tdg and Tdn are more than 143 kN. Hence it is OK.

To transfer an end shear of 200 kN, the web-angle connection designed in Example 11.3 may be provided.

#### 11.5 Splicing of Beams/Plate Girders

Usually beams or plate girders are spliced where the absolute value of the bending moment or shear force is minimum. Two separate splices may be provided to transfer the bending moment and the shear force. It may be assumed that the web splice carries the shear force whereas the flange splice carries the bending moment. The web may be spliced by providing splice plates on either side of the web as shown in Figure 11.20. Even though the web splice is designed for shear force only, the bolts are subjected to shear force and twisting moment due to the eccentricity ‘e’ of the shear force (V) from the centre line of the bolts. The web splice plates are designed for the direct shear force V.

Figure 11.20

Flanges may be spliced by providing splice plates for each flange as shown in Figure 11.20. For the direction of moments shown in Figure 11.20, the splice plate connecting the upper flanges is subjected to compression F and the splice plate connecting the lower flanges is subjected to tension F. The cross sectional area of each splice plate should be in excess of at least 5% of the cross sectional area of each flange. The strength of the bolts provided for the splicing flanges should not be less than 5% in excess of the strength of the flanges. In no case, the strength of the bolts should be less than 50% of the effective strength of the flange. It is preferable to use high strength bolts since it involves fewer bolts thereby fewer holes need to be made.

Example 11.7

Design a splice for a beam WB 450 at a section where the factored bending moment is 205 kNm and the factored shear force is 61 kN. The grade of the steel is E250. Use high strength bolts of the product Grade 8.8. Assume that no slip is permitted at the working load.

Design of flange splice

The force in the splice plate at the ultimate load =

1. At working load

Suppose M20 bolts are used.

Vnf = µƒneKhF0 = 0.55 × 1 × 1.0 × 245 × 0.7 × 800 = 75.5 kN

Vdsf = 75,474/1.1 = 68.6 kN

Six bolts may be provided, two in each row, at a pitch of 60 mm.

2. At ultimate load

kb is the least of

∴  kb = 0.61

Vnpb = 2.5 × 0.61 × 20 × 15.4 × 410 = 191 kN

Vdpb = 191/1.25 = 153 kN

Design strength of the bolt at the ultimate load = 116 kN

Design strength of 6 bolts at the ultimate load = 6 × 116 = 696 kN > 456 kN

OK

Width of the splice plate = the width of the flange of WB450 = 200 mm

The design strength of the flange splice plate due to yielding

The design strength of the flange splice plate at the net section = 0.9 An ƒum1

= 0.9 × (200 − 2 × 22)t × 410/1.25

= 39,557 t

∴     39,557t = 456 × 103

or     t = 11.5 mm

Since the thickness of the flange of WB 450 is 15.4 mm, the thickness of the splice plate which should be 5% in excess, may be 16 mm.

Design of web splice

To splice a web, 2 bolts in a vertical line may be tried.

1. At working load

The direct shear force on each bolt =

The shear force on each bolt due to the twisting moment =

The resultant shear force on each bolt =

Vnf = μf ne Kh F0 = 0.55 × 2 × 1.0 × 157 × 0.7 × 800 = 97 kN

Vdsf = 97/1.1 = 88 kN > 26 kN

OK

2. At ultimate load

The resultant force on each bolt = 1.5 × 26 = 39 kN > (Vdsf = 88 kN)

Therefore, no slip occurs at the ultimate load also.

Providing 180 mm depth of the splice plates on either side of the web as shown in Figure 11.21, the thickness of each plate ‘t’ may be calculated as shown below.

Figure 11.21

The shear capacity of the splice plates =

∴ 47,238 t = 61 × 103

or t = 1.3 mm

Two 6 mm thick plates may be provided.

#### Problems

For the following problems, consider the grade of the steel as E250

1. A T-stub cut from MB 600 is connected to the column HB400 using 12 M16 bolts of the product Grade C as shown in Figure 11.22. Find the design load P that can be applied at an eccentricity 150 mm.
2. If two bracket plates are connected to the flanges of the column SC 250 as shown in Figure 11.23, find the design load P that can be applied at an eccentricity of 300 mm.
3. Re-do Problem 10.3 using M16 bolts of the property class 8.8.
4. Re-do Problem 10.4 using bolts of the product Grade C.
5. Re-do Problem 10.5 using bolts of the product Grade C.
6. Re-do Problem10.6 using high strength bolts.
7. Design a splice for a welded plate girder in Example 7.1 using high strength bolts where the factored BM is 5,000 kNm and the factored shear force is 800 kN.

Figure 11.22

Figure 11.23