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Topic: Steel Connection The steel connection in building construction is mainly divided into two types: (a) Welding (b) Bolting In this lecture, both types of steel connection will be discussed through the calculation and real-life example. -----------------------------------------------------------------------------------------

PART I: Welding Welding is the process of joining metal parts by fusing them and filling in with molten metal from the electrode. However, due to the high temperature of the welding process, it should be carried out under close supervision. (Figure 1.1)

Figure 1.1:- Welding

And welding is divided into two main types

(1) Butt weld Butt weld is named after edge preparation. It also means a weld made in a groove (gap) between two members to be jointed (Figure 1.2)

Figure 1.2:- Butt Weld

(2) Fillet weld Fillet weld is normally done without edge preparation. It also means a weld or nearly triangular cross section joining two surfaces approximately at right angles to each other in a lap joint. (Figure 1.3)

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Figure 1.3:- Fillet weld

(3) Weld Terminology

Base metal – the metal to be welded or cut Bevel angle – angle formed between the prepared edge of a member and a plane perpendicular to the surface of the member Root – where the members approach Weld leg – distance from root to the toe of the fillet weld Actual throat – minimum distance from the root of a weld to its face Effective throat – minimum distance from the roof of a weld to its face minus any reinforcement. Root preparation – the depth that a weld extends into the roof of a joint Joint preparation – minimum depth a groove weld extends from its face into a joint, exclusive of reinforcement For example,

Figure 1.4:- Butt Weld

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Figure 1.5:-Fillet Weld General Consideration in Design in Welding (a) Butt weld design

Clause 6.6.6. of BS 5950: Part I This clause states the design strength should be taken as equal to that of the parent metal provided that the weld metal is not less than that of the parent metal And full penetration depth is required.

(b) Fillet weld design

Clause 6.6.2 of BS5950: Part I - Fillet welds should be returned around corners for twice the leg length (Figure 1.6(a))

(a) (b) Figure 1.6:- Fillet weld design

- In lap joints the lap length should not be less than four times the thickness of

the thinner plate. (Figure 1.6(b)). - The spacing between intermittent welds should not exceed 30mm nor 16t for

parts in compression nor 24t for parts in tension (when t is thickness of the plate) (Figure 1.7)

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Figure 1.7:- Fillet weld design Common defects found in welding (a) Over-reinforcement or undercutting (b) Incomplete penetration (c) Porosity and slag inclusion (d) Residual stresses after suddenly cooling (e) Distortion and twisting (f) Surface cracking, for example, lack of surface preparation

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Part II: Bolting (or Rivet)

It is achieved by inserting high strength bolts into connecting holes between plates and then tightened to a percentage of the allowable bolt tensile strength. On the other hand, in a riveted joint, a heated rivet is forced into a hole connecting two “plates”. As the rivet cool, a tension will be developed in the rivet and the plates are forced together.

Figure 2.1:-Lap joint for bolt and rivet / Butt joint for bolt and rivet

Bolted Connections Bolted connections are increasingly used instead of rivets and more often than welds It is mainly because (1) rivets may be too labor intensive, for example, it require heating up at the bolt; (2) welds may cause secondary cracks if not properly performed. Types of the bolts (1) Ordinary Bolts “Black” hexagonal head bolt with nut and washer are commonly used. The clamping force resulted from the tightened of bolt-nut system is only sufficient to prevent movement in the axial direction of the bolt. Slipping to bearing occurs.

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Figure 2.2 Ordinary Bolts It is specified in BS4190

- Two grades - Grade 4.6 mild steel – yield strength 235MPa - Grade 8.8 high yield steel 0 yield strength 627MPa - Common diameter: 16, 20, 22, 24, 27 and 30mm

(2) Friction Grip Bolts

- high yield steel is preformed so that a high tension can be provided - high strength friction grip bolt is specified in BS4395 with three grades:

general grade (similar to grade 8.8 ordinary bolts), higher grade (parallel shank) and higher grade (waisted shank)

- the bolt must be used with hardened steel washers to prevent damage to the connected parts

- Care must be taken to ensure that bolts are tightened up to the required tension and no slip will be resulted.

Applications of the bolting (1) Internal truss joint (2) Brackets (3) Joint in built-up members (4) Beam to beam connections (5) Beam to column connections (6) Column to foundation connections

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For example:

Design in Bolt connection (1) Direct shear joint

- Refer to Section 6.2 of BS5950, part I - Spacing, edge and end distance is

strictly defined (a) Minimum spacing = 2.5d (b) Minimum edge and end distance is 1.25D (c) Maximum edge distance is εt×11

where d = bolt diameter D = hole diameter

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t = thickness of thinner plate connected

5.0)/275( yp=ε

Py = steel design strength in MPa (d) shear capacity of the bolt = sss ApP ×= …..Eq(1)

ps = shear strength of the bolts As = shear stress area

(2) Direct tension Joints

When bolts are in direct tension only, the tension capacity of an ordinary bolts is:

ttt ApP ×= ……Eq(2) pt = tensile strength of the

bolts At = tensile stress area

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(3) Eccentric Connections: Bolts in direct shear and Tension

Moment is applied to the plane of connection The bolt group rotates about its centre of bolt group

Considering the Torsion force Let FT be the force due to Torsion on bolt B M = FT x r1 +FT(r2/r1)r2+… = (FT/r1) (r1

2+r22+….)

then

( ) ∑×= 21/ rrFM T

( )∑∑ +=× 221 )/( yxrFeP T

∑∑ +

××=∴

22

1

yx

rePFT …..Eq(3)

On the other hand, force due to shear on each bolt = PFS = / (nos. of bolts)……Eq(4)

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Then try to find out the maximum force on the bolt Consider Bolt B, Vertical force on bolt B =

φcosTs FF + …..Eq(5) Horizontal force on bolt B =

φsinTF ……Eq(6) Resultant force on bolt B,

( ) ( )[ ] 2122 sincos φφ TTSR FFFF ++= …..Eq(7)

where Fs = shear force, FT = Torsion force

The bolt size can then be determined from the maximum force on the bolt.

(4) Eccentric Connections: Bolts in direct shear and tension - Occurs in bracket types

connection - Applied shear force Fs

must not exceed the shear capacity Ps.

- Applied tension force FT must not exceed the tension capacity Pt.

- Additional requirement: 4.1// ≤+ TTSS PFPF

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Moment

( )∑=

=⋅

++=

++=

n

iit

t

tt

yyFeP

yyyF

yyFyFM

1

21

22

211

1221

/2

))(/(2

)/(2

K

K

where n is number of bolts Maximum bolt tension (Force due to tension),

∑=

⋅⋅= n

ii

t

y

yePF

1

2

1

2…..Eq(8)

Shear Force on each bolt, Fs = P / no of bolts…..Eq(9)

2 sides

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Example 1 In Figure Q.1, Calculate the Maximum bolt tension and the shear force on each bolt.

Figure Q.1

Solution: y1 = 400mm

000,300)400300200100( 22224

1=+++=∑

=

=

n

iiy

Maximum bolt tension (Force due to tension) By Eq(8),

kNy

yePF n

ii

t 45000,3002400450150

21

2

1 =×××=⋅⋅=

∑=

By Eq(9), the shear force on each bolt = kN1510150

=

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Example 2 In Figure Q.2, Calculate the tension force on the bolt A and shear force on each bolt.

Figure Q.2

Solution:

From Eq.(3), 71.16715075 221 =+=r

000,145)15050(4)75(8)7550(4)15075(4 2222222222 =++=+++=+=∑ ∑ ∑ yxr

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The tension force on bolt A is

From Eq.(3), kNyx

rePFT 05.52000,145

71.16730015022

1 =××

=+

××=

∑∑

The shear force on each bolts From Eq.(4), PFS = / (nos. of bolts) = 150 / 8 =18.75kN

-END OF THE LECTURE 7-