1

1.0 Introduction

1.1 Iron and Steel

Steel is the world's most important material. Without steel, the world as we

know it would not exist: from oil tankers to thumb tacks, from trucks to tin

cans, from transmission towers to toasters. Given the huge quantities of steel

produced, it is fortunate that the material is easy to recycle. Much of today's

steel is produced from scrap. Iron production requires these raw materials: iron

ore, coal and stone (LIMESTONE, dolomite). Steel production requires iron,

steel scrap and flux ("lime" - calcined limestone). The iron ore is smelted to

produce an impure metal called "hot metal" when liquid, or "pig iron" when

solid. The hot metal is refined to remove impurities and to develop the desired

composition. The liquid steel is continuously cast into blooms, slabs or billets,

and these semi-finished products are processed into the desired shapes by

rolling or forging.

1.2 Types of Structural Steel

The term structural steel refers to a number of steels that, because of their

economy and desirable mechanical properties, are suitable for load-carrying

members in structures. The customary way to specify a structural steel is to

use an ASTM ( American Society for testing and Materials ) designations. For

ferrous metals, the designation has a prefix letter “ A “ followed by two of three

numerical digits ( e.g., ASTM A36, ASTM A514 ).

There are 3 groups of hot-rolled structural steels for used in buildings;

1. Carbon steels use carbon as the chief strengthening element with

minimum yield stresses ranging from 220 MPa to 290 MPa. An increase

in carbon content raises the yield stress but reduces ductility, making

welding more difficult.

2. High- strength low- alloys steels ( HSLA ) have the yield stresses from

480 MPa to 840 MPa. In addition to carbon and manganese, these steels

contain one or more alloying elements such as columbium, vanadium,

chromium, silicon, copper, and nickel.

3. Quenched and tempered alloy steels have yield stresses of 480 MPa to

690 MPa. These steels of higher strength are obtained by heat-treating

2

low alloy steels. The heat treatment consist of quenching ( rapid cooling )

and tempering ( reheating ).

1.3 ASTM designations

Material conforming to one of the following standard specifications.

ASTM A36 – Structural steel

ASTM A53, Grade B – Pipe, Steel, Black and Hot-dipped, Zinc-coated Welded

and Seamless steel pipe.

ASTM A242 – High-strength Low- alloy structural steel

ASTM A709 – Structural Steel for Bridges

Certified mill test reports or certified reports of test made by the fabricator or

testing laboratory accordance with ASTM A6 or A568, as applicable and the

governing specification must constitute sufficient evidence of conformity with

one of the above ASTM standards.

1.4 Properties of Steel

Yield stress, Fy, is that unit tensile stress at which the stress- strain curve

exhibits a well-defined increase in strain ( deformation )

without increase in stress.

Tensile strength, Fu, is the largest unit stress that the material achieves in a

tension test.

Modulus of elasticity, E, is the slope of the initial straight-line portion of the

stress- strain diagram. It is usually taken as 200,000 MPa

for design calculation for all structural steel.

Ductility is the ability of the material to undergo large inelastic deformations

without fracture

3

Toughness is the ability of the material to absorb energy and is characterized

by the area under a stress- strain curve.

Weldability is the ability of steel to be welded without changing its basic

mechanical properties.

Poisson’s ratio is the ratio of the transverse strain to the longitudinal strain.

Poisson’s ratio is essentially the same for all structural

steels and has a value of 0.30 in the elastic range.

Shear modulus is the ratio of the shearing stress to shearing strain during the

initial elastic behavior.

Modulus of elasticity, E 200,000 MPa

Yield strength, Fy 248 MPa

Tensile strength, Fu 400 MPa

Endurance strength 207 MPa

Density, 7780 kg/m3

Poisson’s ratio, µ 0.30

Shear modulus, G 77,200 MPa

Coefficient of thermal expansion, 11.70 x 10-6/C

Table 1 - Typical Properties of A36 Steel

4

1.5 Structural Shapes

Structural steels are available of many shapes. The dimension and

weight must be added to the designation to uniquely identify the shape.

For example, W 40 x 436 refers to W- shape with an overall depth of

approximately 40 inches ( 1000mm ) that weighs 436 lb/ft 9 640 kg/m ).

Shape Designation

Wide flange beam W

American standard beam S

Bearing piles HP

Miscellaneous ( those that cannot be classified

as W, S, or HP ) M

Channel C

Angle L

Structural tee WT or ST

Structural tubing TS

Pipe pipe

Plate PL

Bar bar

Table 2 - Structural Shape Designation

Figure 1 Structural Shapes

5

Figure 2 – Combined Sections

1.6 Types of Construction

There are three basic types for construction and associated design

assumptions permitted, and each will govern in a specific manner the size of

members and the types and strength their connections:

Type 1, commonly designated as rigid frame ( continuous frame ), assumes that beam-column connections have sufficient rigidity to hold virtually

unchanged the original angles between intersecting members.

Type 2, commonly designated as simple framing ( unrestrained, free-ended ), assumes that, insofar as gravity loading is concerned, ends of beams and girders are connected for shear only and are free to rotate under

gravity load. Type 3, commonly designated as semi-rigid framing ( partially restrained ),

assumes that the connections of beams and girders posses a dependable and know moment capacity intermediate in degree between

the rigidity of Type 1 and the flexibility of Type 2.

6

2.0 Basic Structural Engineering

Understanding Load Flow All structures are subjected to forces that are imposed by gravity, wind and

seismic events (see Figure 3). The combination and anticipated severity of these

forces will determine the maximum design force the member can sustain. The

structural engineer will then select a member that meets all of the strength as

well as serviceability issues such as deflection and/or vibration criteria for any

specific project. The following is a brief discussion on each of the types of loads

and how these loads are transferred to the other structural components.

2.1 Gravity Loads Gravity loads include all forces that are acting in the vertical plane (see Figure

2). These types of forces are commonly broken down into dead loads and live

loads in a uniform pounds per square foot loading nomenclature. Dead loads

account for the anticipated weight of objects that are expected to remain in

place permanently. Dead loads include roofing materials, mechanical

equipment, ceilings, floor finishes, metal decking, floor slabs, structural

materials, cladding, facades and parapets. Live loads are those loads that are

anticipated to be mobile or transient in nature. Live loads include occupancy

loading, office equipment and furnishings.

The support of gravity loads starts with beams and purlins. Purlins generally

refer to the roof while beams generally refer to floor members. Beams and

purlins support no other structural members directly. That is to say, these ele-

ments carry vertical loads that are uniform over an area and transfer the uniform

loads into end reactions carried by girders.

Girders generally support other members, typically beams and/or purlins, and span

column to column or are supported by other primary structural members. Girders may

support a series of beams or purlins or they may support other girders. Forces

imposed on girders from beams, purlins, or other girders are most often

transferred to the structural columns. The structural column carries the ver-

tical loads from all floors and roof areas above to the foundation elements.

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leeward wind (suction )

wind

Figure 3 - Forces experienced by Structures

8

2.2 Horizontal Loads

Forces created by wind or seismic activity are considered to act in the

horizontal plane. While seismic activity is capable of including vertical forces,

this discussion will be based only on horizontal forces. The majority of this

section will address wind forces and how they are transferred to the primary

structural systems of the building (see Figure 4).

Wind pressures act on the building's vertical surfaces and create varying forces

across the surface of the façade. The exterior façade elements, as well as the

primary lateral load resisting system, are subjected to the calculated wind

pressures stipulated by code requirements. This variation accounts for façade

elements being exposed to isolated concentrations of wind pressures that may

be redistributed throughout the structural system. Design wind pressures can

be calculated using a documented and statistical history of wind speeds and

pressure in conjunction with the building type and shape. Calculated wind

pressures act as a pushing force on the windward side of a building. On the

leeward (trailing) side of the building, the wind pressures act as a pulling or

suction force. As a result, the exterior façade of the entire building must be

capable of resisting both inward and outward pressures.

Roof structures made up of very light material may be subjected to net upward

or suction pressures from wind as well. Roofs typically constructed of metal

decking, thin insulation and a membrane roof material without ballast

have the potential to encounter net upward forces. Roof shape may also

determine the net uplift pressures caused by wind. Curved roofs will actually

exhibit a combination of downward pressures on the top portion of the curve

and upward pressure on the lower portion of the curve. This distribution of

downward and upward pressures caused by the curve is similar to the

principles of air pressure and lift acting on an airplane wing.

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Figure 4 - Gravity and wind loads Figure 5 - Loads on Column and Beams

As the wind pressures are applied to the exterior of the building, the façade

(actually a structural element to some degree), transfers the horizontal

pressures to the adjacent floor or roof. As these pressures are transferred, the

floor and roof systems must have a means to distribute the forces to the lateral

load resisting systems. Floors and roofs that are generally solid or without large

openings or discontinuities may behave as a diaphragm. A diaphragm is a

structural element that acts as a single plane with the connecting beams and

columns. When experiencing a force, this single plane causes the beams and

columns to displace horizontally the same amount as the diaphragm. This can

be exemplified by a sheet of paper or cardboard that is supported by a series of

columns. Should the paper, a flexible diaphragm, be pushed horizontally, all

points in contact with the paper will move laterally by the same amount. The

metal roof decking on most projects will behave as a flexible diaphragm.

Substituting a piece of cardboard for paper in our example, the paper will

behave more like a rigid diaphragm. A typical floor decking and composite

structural slab are examples of a rigid diaphragm.

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2. 3 Seismic

Seismic activity induces horizontal forces, and at times, vertical loads. The

discussions in this publication will focus on horizontal forces imposed during

seismic activity. Forces created during a seismic event are directly related to

weight or mass of the various levels on a specific building. During seismic

activity horizontal diaphragms behave like wind load transfers with respect to

the primary lateral load resisting systems. However, the induced forces are

much more sensitive to the shape of the building and the positioning of the

lateral load resisting systems. It is advantageous to consider a very regular

building plan in areas of the country with significant seismic activity.

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3.0 Centroid of an Area

The centroid of an area is analogous to the center of gravity of a homogenous

body. The centroid is often described as the point at which a thin homogenous

plate would balance. The centroid of a complex area can be found by dividing

the area into basic shapes ( rectangles, triangles, circles ).

Centroid of Plane Area

AT Xc = a x = a1 x1 + a2 x2 + a3 x3 + ……………

AT Yc = a y = a1 y1 + a2 y2 + a3 y3 + ……………

12

Table 3 - Properties of common Geometric Shapes

13

14

Problem 1.

Locate the centroid of the shaded area shown.

Solution

15

2

1 1 1

2

2 2 2

2 2

3 3 3

8 4 32 0 2

2 2 4 2 3

4 111 1.57 2 4 3.576

2 3

A m x y m

A m x m y m

A m x m y m

[ AT = A1 – A2 – A3] AT = 32 - 4 – 1.57

AT = 26.429 m2

[ AT YG = Ay]

26.429 YG = 32(2) – 4(3) – 1.57(3.576)

YG = 1.755 m

[ AT XG = Ax]

26.429 XG = 32(0) – 4(-2) – 1.57(2)

XG = 0.18 m

16

Problem 2.

With reference to the plane area shown in the figure, determine the following:

a. the area of the plane in square millimeters b. the x-coordinate of the centroid c. the y- coordinate of the centroid

17

Solution:

18

Problem 3.

Determine the x- and y-coordinate

of the shaded area shown.

Solution:

19

Problem 4.

For the shaded area shown, determine

the following: a. the area of the shaded part in mm2 b. the x- coordinate of the centroid

of the area in mm.

Answer:

a. 6,424 mm2

b. x = 55.51 mm

Solution:

20

Problem 5. Assignment

For the shaded area shown, determine the following:

a. the area of the shaded portion in mm2

b. the x- coordinate of the centroid in mm.

c. the y- coordinate of the centroid in mm.

Solution:

21

Solution:

22

4.0 Bolts and Riveted Connections

4.1 Types of Bolted and Riveted Connections

Individual parts or members assembled together compose every structure.

These members must be fastened together by means of welding, rivets, or bolts.

A. High- Strength Bolts

High strength bolts have replaced rivets as means of making non- welded

structural connections. There are two basic types of high strength bolts

used, the ASTM A325 and ASTM A490. The material properties of thses

bolts are given in table xx. High strength bolts are usually tightened to

develop a specified tensile stress in them, which results in a predictable

clamping force in the joint. Therefore, the actual transfer of service load

through a joint is due to friction developed in the pieces being joint.

Joints containing high-strength bolts are designed either as slip-critical

or friction type, where high slip resistance at service load is unnecessary.

B. Rivets

Installation of rivets requires heating the rivet to a light cherry-red color,

inserting it into a hole and then applying pressure to the performed head

while at the same time squeezing the plain end of the rivet to form a

rounded head. During the process, the shank of the rivet completely or

nearly fills the hole into which it had been inserted. Upon cooling, the

rivet shrinks, thereby providing a clamping force. However, the amount

of clamping force produced by cooling of the rivet varies from rivet to

rivet and therefore cannot be counted on in design calculations.

23

24

25

26

4.2 Analysis of Axially Loaded Bolted or Rivited Connection

The following stresses must be investigated in the design or analysis of axial

loaded tension connections:

Gross area, Ag = W x t

Net area, An = [ Wg - (holes + 1.6) ] x t ≤ 85% Ag

1. Tension on Gross Area:

Actual stress, ft = P/A

Allowable stress, Ft = 0.60 Fy ( yielding )

2. Tension on effective net area:

Actual stress, ft = P/Ae

Allowable stress, Ft = 0.50 Fu ( fracture )

3. Shear in bolts:

Actual stress, fv = P/Av

Av = Abolt x n ( single shear )

Av = 2 Abolt x n ( double shear )

n = number of bolts

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4. Bearing on the projected area between the bolt and the plate:

Actual stress, fp = P/Ap

Ap = ( bolt diameter x plate thickness )

Allowable stess, Fp = 1.2 Fu

5. Combined shearing and tearing:

P = Fv Av + At Ft

Allowable shearing stress, fv = 0.3 Fu

Allowable tearing stress, Ft = 0.50 Fu

28

Problem 1.

The single 200 mm x 10 mm steel plate is connected to a 12 mm thick steel

plate by four 16 mm diameter rivets as shown. The rivets used are A502, Grade

2, hot driven rivets. The steel is ASTM A36 with Fy = 248 MPa and Fu = 400

MPa. Determine the value of P in all possible modes of failure and the safe

value of P that the connection can resist.

Solution:

Rivet diameter = 16 mm

Hole = 16 + 1.6 = 17.6 mm

29

Tension on gross area;

Ft = 0.60 Fy = 0.60 ( 248 ) = 148.8 MPa

Ag = 200(10) = 2000 mm2

P = Ft Ag = 148.8 (2000)

P = 297,600 N = 297.60 KN

Tension of net area:

Ft = 0.50 Fu = 0.50 ( 400 ) = 200 MPa

Net area along section a-a;

Ae = [200 – 2(17.6 )](10)

Ae = 1,648 mm2

85 % Ag = 0.85 ( 2000 )

85 % Ag = 1,700 mm2

P = Ft Ae = 200 (1, 648 )

P = 329, 600 N = 329.6 KN

30

Bearing on projected area:

Fp = 1.2 Fu = 1.2 (400 ) = 480 MPa

Ap = dt = [ 16 (10)](4) = 640 mm2

P = Fp Ap = 480 (640)

P = 307,200 N = 307.20 KN

Shear on rivets

Fv = 152 MPa

Av = 4 x /4 (16)2 = 804.25 mm2

P = Fv Av = 152 ( 804.25 )

P = 122,246 N = 122.246 KN

Shear Rapture:

P = Fv Av + Ft At

Fv = 0.30 Fu = 0.30 (400)

Fv = 120 MPa

Av = 2 Abc = 2{[ 135 -1.5 (17.6)]

Av = 2,172 mm2

Ft = 0.50 Fu = 0.50 (400)

Ft = 200 MPa

At = 2 (Aab) = 2{[50-0.50(17.6) ] x 10 }

At = 824 mm2

P = 120(2,172 ) + 200 (824 )

P = 425, 449 N

P = 425.44 KN

Therefore, the safe load is 122.246 KN

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Problem 2. Seat work

The single 200mm x 12 mm steel plate is connected to a 12 mm thick steel

plate by four 18 mm diameter rivets as shown in problem no. 1. The rivets used

are A502, Grade 2, hot driven rivets. The steel is ASTM A36 with Fy = 248 MPa

and Fu =400 MPa . Determine the value of P in all possible modes of failure

and the safe value of P that the connection can resist.

32

5.0 Columns and other Compression Members

Introduction

Structural members subjected to axial compressive loads are often called by

names identifying theirs functions. Of these, the- known are columns, the main

vertical compression members in a building frame. Other common compression

members include chords in trusses and bracing members in frames.

5.1 Euler Column Buckling Theory

Column design and analysis are based on the Euler load theory, ( Leonardo

Euler, 1757 ). His analysis is based on the differential equation of the elastic

curve. However, specific factors of safety and slenderness ratio limitations are

applied from purely theoretical concepts. If the column is hinged at both ends,

the Euler critical load is given as:

2

2

EIP

L

And the Euler critical stress is :

2

2e

EF

L

r

These equations show that the buckling stress is not a function of the material

strength. Rather, it is a function of the ration L/r known as slenderness ratio,

SR. As the slenderness ratio increase, the buckling stress decreases, meaning

that as the column becomes longer and more slender, the load that causes

buckling becomes smaller.

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5.2 Effective Length

Real columns do not have pin- connected ends. The restraints placed on a

column’s end greatly affect its stability. To counter these effects, an effective

length factor, K, is used to modify the unbraced length. The product KL is

called the effective length of the column. This length approximates the length

over which the column actually buckles and this may be shorter or longer that

the actual unbraced length.

2

2e

EF

KL

r

Table 4 – Effective Length Factors

34

5.3 Slenderness Ratio

Steel columns are usually subdivided into two groups: Long and Intermediate

columns , depending on their slenderness ratio. The critical slenderness ratio

Cc, corresponds to the upper limit of elastic buckling failure, which is defined

by an average column stress equal to 0.50 Fy.

22c

y

EC

F

The limiting slenderness ratio KL/r for members whose design is based on

compressive force preferable shall not exceed 200. For Members whose design

is based on tensile force , the limiting slenderness ratio L/r preferable should

not exceed 300.

For Intermediate Columns, c

KLC

r

2

21

2

y

a

c

KL

FrF

C FS

3

3

35

3 8 8c c

KL KL

r rFS

C C

For Long Columns, c

KLC

r

2

2

12

23

EFa

KL

r

35

Problem 1.

A wide flange section for a 5 m long column ( hinged at both ends ) has the

following properties:

Cross-sectional area = 8,000 mm2 Radius of gyration, rx = 100mm radius of gyration, ry = 50 mm

Modulus of elasticity, E = 200,000 MPa

Determine the Euler critical load of the column. Solution:

Euler critical load, Pe = Fe x A

Euler critical stress, 2

2e

EF

KL

r

, where K = 1 ( hinge both ends )

(KL/r )max = 1 x 5,000/50 = 100

Euler critical stress , Fe = 2(200,000)/(100)2

Euler critical stress, Fe = 197.4 MPa

Pe = Fe A

Pe = 197.4 (8,000)

Pe = 1,579,200 N

Pe = 1,579.2 KN

36

Problem no. 2

A steel column has the following properties:

Modulus of elasticity E = 200,000 MPa

Fy = 200 MPa

L = 15 m

Moment of inertia , I = 37.7 x 106 mm4

Area, A = 8,000 mm2

Determine the allowable compressive stress if the column is fixed at both ends.

22c

y

EC

F

22 200,000

200cC

140.5cC

K = 0.65 ( recommended value for fixed both ends )

Ir

r

637.7 1068.65

8,000

xr mm

0.65(15,000)142.02

68.65

KL

r

Since c

KLC

r

2

2

12

23

EFa

KL

r

2

2

12 (200,000)

23 142.02Fa

Fa = 51.06 MPa

37

Problem 3. Seat work

A wide flange section for a 6 m long column ( hinged at both ends ) has the

following properties:

Cross-sectional area = 7,000 mm2

Radius of gyration, rx = 100mm radius of gyration, ry = 50 mm Modulus of elasticity, E = 200,000 MPa

Determine the Euler critical load of the column.

38

Problem 4. Assignment

A steel column has the following properties:

Modulus of elasticity E = 200,000 MPa

Fy = 240 MPa

L = 12 m

Moment of inertia , I = 32.7 x 106 mm4

Area, A = 6,500 mm2

Determine the allowable compressive stress if the column is fixed at both ends.

39

6.0 Beams and Other Flexural Members

Beam are members acted upon primary by transverse loading ( i.e., loads that

are applied at right angles to the longitudinal axis of the member ). They are

primary subjected to flexure or bending. Beams may be subject by some axial

loading. The effect of axial loads is generally negligible, and the member is

treated strictly as a beam. However, if the axial compressive load is substantial

in magnitude, the member is called beam column.

6.1 Types of Beams

Beams are usually designated by names that are representative of their

functions:

Girder – ( usually the most important beams which are frequently at wide

spacing), is a major beam that often provide support for other beams

Joist – is a light beam that supports a floor.

Purlin – is a roof beam spanning between trusses or rigid frames.

Stringer – is a main longitudinal beam, usually supporting bridge decks.

Floor beam – is a transverse beam in bridge decks.

Spandrel – is a beam on the outside perimeter of the building.

Girt – is a light beam that supports only the lightweight exterior sides of the

building.

Compact Section

Compact sections have width-thickness ratio not exceeding the limits given in

table xx. To be compact, the flanges of the beam must be continuously

connected to the web.

170

2

f

f y

b

t F

1680

w y

d

t F

40

6.2 Allowable Bending Stress:

Members with compact section

For members with compact section and with length Lb ≤ Lc , the allowable

bending stress in both tension and compression is:

Fb = 0.66 Fy

200 137,900fc

yy

f

bL smallervalue of and

F d FA

Members with non-compact section

For members with Lb ≤ Lc except that their flanges are non-compact, the

allowable bending stress in both tension and compression is :

0.79 0.0007622

fb y y

f

bF F F

t

For members with non-compact section and 200 f

b

y

bL

F , the allowable bending

stress in both tension and compression is :

Fb = 0.60 Fy

Members with compact or non-compact section with Lb > Lc :

Allowable bending stress in tension

Fb = 0.60 Fy

41

Allowable bending stress in compression

703,270 3,516,330b b

y T y

C CL

F r F

Fb = Larger of ( Fb1 and Fb2 ) ≤ 0.60 Fy

1 2

1,172,100 bb

T

CF

L

r

2

87,740 bb

f

CF

L d

A

42

Problem no. 1

A W21x62 steel is used as a beam simply supported over a span of 8m. The beam is laterally unsupported over the entire span. Use Fy = 250 MPa. The

properties of the section are as follows: Depth, H = 533mm Flange width, bf = 210mm

Flange thickness, tf = 15.6 mm Web thickness, tw = 10.2mm

Radius of gyration, rt = 53.34 mm Section modulus, Sx = 2,077 x 10 3 mm3

a. Determine the value of the ratio L/rt. b. Determine the allowable bending stress

c. Determine the safe uniformly distributed load that the beam can carry.

Solution:

L = 8m x 1000 = 8,000 mm

rt = 53.54 mm

8000149.98

53.34t

L

r

1bC

703,270 1703,27053.04

250y

bC

F

3,516,330 13,516,330118.59

250y

bC

F

since 3,516,330

y

b

t

L C

r F

43

1 2

1 2

1

1,172,100

1,172,100 1

149.98

52.11

bb

t

b

b

CF

L

r

F

F MPa

2

2

2

82,740

82,740 1

8000 533

210 15.6

63.5

bb

t

b

b

CF

L d

A

F

F MPa

0.60 150yF MPa

Therefore use 2 63.5bF MPa

Uniform load:

Moment capacity

363.57 2,077 10

132,034,890 .

132.034 .

b

x

b x

MF

S

M F S x

M N m

M KN m

44

2

2

2

8

8

8 132.034

8

16.50

w LM

Mw

L

w

KNwm

45

Problem no. 2

A W21x62 steel is used as a beam simply supported over a span of 7 m. The beam is laterally unsupported over the entire span. Use Fy = 200 MPa. The

properties of the section are as follows: Depth, H = 533mm

Flange width, bf = 210mm Flange thickness, tf = 15.6 mm Web thickness, tw = 10.2mm

Radius of gyration, rt = 53.34 mm Section modulus, Sx = 2,077 x 10 3 mm3

a. Determine the value of the ratio L/rt. b. Determine the allowable bending stress

c. Determine the safe uniformly distributed load that the beam can carry.

46

Problem 3. Assignment

A W10x49 tension hanger having a cross-sectional area of 14.4 in2 , 5 ft. long,

carries a service load of 250 Kips. calculate its axial elongation.

47

7.0 Welded Joints

In welded connections, different elements are connected by heating their

surfaces to a plastics or fluid state. There may or not be pressure, and there

may or may not be filler material. Arc welding is the general term for the many

processes that uses electrical energy in the form of an electric arc to generate

the heat necessary for welding.

7.1 Types of Welding

1. Shielded Metal Arc Welding ( SMAW )

In SMAW, the weld is protected by using an electrode covered with a

layer of mineral compounds. Melting of this layer during the welding

produces an inert gas encompassing the weld area. This inert gas shields

the weld by preventing the molten metal from having contact with the

surrounding air. The protected layer of the electrode leaves a slag after

the mold has cooled down. The slag can be removed by peening and

brushing. The electrode material is specified under various specifications

and is given in table xx. The designation such as E60XX or E80XX

indicate 60 ksi (415 MPa ) and 80 ksi ( 550 MPa ), respectively for the

tensile strength Fu. The E denotes electrode. The X’s represent numbers

indicating the usage of the electrode.

Figure 5 - Shield Metal Arc Welding ( SMAW )

48

Table 5 - Electrode used for Welding

2. Submerged Arc Welding ( SAW )

In Saw process, the arc is not visible because the surface of the weld and

the electric arc are covered by a blanket of granular fusible material

figure xxx to protect it from the surrounding air. In this method, a bare

electrode is used as filler material. Compared with SMAW, SAW welds

provide deeper penetration. Also SAW welds show good ductility and

corrosion resistance and high impact strength.

Figure 6 - Submerged Arc welding ( SAW )

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3. Gas Metal Arc welding ( GMAW )

In this process the electrode is a continuous wire that is fed from a coil

through the electrode holder. The shielding is entirely from an externally

supplied gas or gas mixture. The GMAW process using CO2 shielding is

good for the lower carbon and low-alloy steels which are usually used in

buildings and bridges.

4. Flux Cored Arc Welding ( FCAW )

This process is similar to GMAW, except that the continuously fed filler

metal electrode is tubular and contains the flux material within its core.

The core material provides the same function as does the coating in

SMAW or the granular flux in SAW. This process is useful procedure for

field welding in severe cold weather conditions as well as to speed up

high rise construction.

7.2 Advantages of Welding

1. In welded connections, in general, fewer pieces are used. This will speed

up the detailing and fabrication process.

2. In welded connections, gusset and splice plates may be eliminated. Bolts

or rivets are not needed either. Thus, the total weight of a welded steel

structure is somewhat less than that of the corresponding bolted

structure.

3. Connecting unusual members ( such as pipes ) is easier by welding than

by bolting.

4. Welding provides truly rigid joint and continuous structures.

One possible drawback of welding is the need for careful execution and

supervision. For this reason, welding sometimes done in shop and bolting in

the field. In other words, shop-welding is complemented by the bolting in the

field.

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Forms of Welded Joints

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7.3 Types of Welds

The two common types of welds in welded steel structures are groove welds and

fillet welds. Fillet weld are much more popular in structural steel design than

grove welds.

A. Groove welds

Two different types of groove welds are shown in figure 7. They are the

partial penetration ( single – V ) groove weld and full penetration ( double

– V ) groove weld. Groove welds can be used when the pieces to be

connected can be lined up in the same plane with small tolerances.

Figure 7 – Groove welds

B. Fillet welds

Fillet welds are shown in figure 8. Depending in the direction of the

applied load and the line of the fillet weld, fillet welds are classified as

longitudinal or transverse fillet weld. In longitudinal fillet, the shear force

to be transferred is parallel to the weld line; in traverse weld, the force to

be transmitted is perpendicular to the weld line.

Figure 8 - Fillet welds

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7.4 Fillet welds

Fillet welds can be either equal-leg or unequal leg, as shown in the figure 9.

The intersection point of the original faces of the steel elements being

connected is called the root of the weld. the surface of the weld should have a

slight convexity. In computation of the strength of the weld, however, this

convexity is not taken into account and the theoretical flat surface is used. The

normal distance from the root to the theoretical face of the weld is called the

throat of the weld.

Figure 9 - Fillet welds: a) Equal leg; b) Unequal leg

For equal-leg fillet welds, the relation between the dimensions of the leg w and

the throat t is :

Throat, t = 0.707w

Capacity of equal-leg fillet weld:

Load capacity, P = Fv x 0.707 w L

Fv = 0.30 Fu

where:

w = size of weld ( leg )

L = total length of weld

Fv = allowable shearing stress of weld metal

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Welding Symbols

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Common uses of welding symbols

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Problem no. 1

Two steel plates, each 350 mm wide and 12 mm thick, are to be joined together

by welded lap splice, as shown. the electrode used for the weld has a nominal

tensile strength of 550 MPa.

a. Determine the maximum weld size that can be used.

b. Determine the effective net area of the fillet weld using the maximum

weld size allowed by the code.

c. Determine the maximum load that can be resisted by the weld using the

maximum weld size allowed by the code.

Solution:

a.) Maximum size of weld

Maximum size, t = 12 – 1.5 = 10.5 mm

b.) Effective area of the fillet weld

Effective length = 350 x 2 = 750 mm

Effective throat thickness = 0.707 t

Effective throat thickness = 0.707 ( 10.5 ) = 7.4235 mm

Effective area = 750 x (7.4235 ) = 5197 mm2

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c.) Considering the upper load only:

P = Fv ( Av )

Fv = 0.3 Fu = 0.3 (550 )

Fv = 330 MPa

Av = 0.707( t ) ( L )

Av = 0.707( 10.5 ) ( 350 )

Av = 2598 mm2

P = 330 ( 2598 )

P = 857, 414 N = 857.4 KN