lecture 2 slides

LECTURE 2

1

May 26, 2014

09:00 am – 11:00 am

VCB 3022

Design of Steel Structures

Lecturer

Dr Zubair Imam Syed

Email: [email protected]

Ph: 05 368 7313

Room: 14.03.13

mailto:[email protected]

LEARNING OUTCOME:

2

Desig

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tructu

res• Basics related to Design of Steel Structures

• Classification

• Cross section properties used in design

CLO 1:

Distinguish the properties of steel and determine Section

classification

Design of steel and

prestressed concrete

structures

3

THE TENSION TEST

4

Desig

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Highlights from Last Lecture

The important characteristics of steel for

design purposes are:

• yield stress (Fy)

• ultimate stress (Fu)….tensile strength

• modulus of elasticity (E)

• percent elongation ()

• coefficient of thermal expansion ()

Grade of Steel:

four commonly used grades are S235,S275, S355, S450

OBJECT OF STRUCTURAL DESIGN

Safety (the structure doesn’t fall down during lifetime)

Serviceability (how well the structure performs in term of

appearance and deflection)

Fulfill requirements of client

Economy (an efficient use of materials and labor)

Alternatives

Several alternative designs should be prepared and their

costs compared

Engineering Design consists of Two stages

Feasibility Study/ Conceptual design

Involves comparison of the alternative forms of structure and selection of most suitable type

Detailed design

• involves detailed design of the chosen structure

• The detailed also requires these attributes but is usually more dependent upon a thorough understanding of the codes of practice for structural design namely EC2 and EC3

CONCEPTUAL DESIGN OF BUILDING

Design – process by which an optimum solution is obtained. In any design, certain criteria must be established to evaluate whether or not an optimum has been achieved.

Design: Determination of overall proportions and dimensions of the supporting framework and the selection of individual members.

Aim of Structural Design – To provide with due regard to economy a structure capable of fulfilling its intended function and sustaining the specified loads for its intended life. The design should facilitate safe fabrication, transport, handling and erection- account future maintenance, final demolition, recycling and reuse of materials.

Responsibility: The structural engineer, within the constraints imposed by the architect (number of stories, floor plan,..) is responsible for structural design.

Philosophies/ Theories used for design: Elastic design, Plastic design and Limit State Design

LIMIT STATE DESIGN

Also called LRFD (Load and Resistance Factor Design) in USA.

The structure is deemed to be satisfactory if its design load effect

does not exceed its design resistance

Design load effect ≤ Design resistance

(effect of specified loads x γg,Q) specified resistance / γM factor

Though limit state design method is presented in a deterministic

format, the partial factors are obtained using probabilistic models

based on statistical distributions of loads and structural capacity

Each load effect (DL, LL, ..) has a different load factor which its

value depends on the combination of loads under consideration.


LIMIT STATE CONCEPT IN DESIGN

Stated in cl 2.2 EN 1993-1-1 2005 :Eurocode 3

Design of Steel Structures Part 1-1: General rules

and rules for buildings, British Standards

The standard gives recommendations for the

design of structural steel work using hot rolled

sections, flats, plates, hot finished structural hollow

sections and cold formed structural hollow sections,

in buildings and allied structures

Structures should be designed by considering the

limit states beyond which they would become unfit

for their intended use.


Limit state design

Ultimate limit states (ULS)

Strength

Stability against overturning and sway stability

Fatigue

Brittle fracture

Serviceability limit states(SLS)

Deflection

Vibration

Wind induced oscillation

Durability

11

ACTIONS

BS EN 1990:2002 : ACTIONS ARE A SET OF FORCES

(LOADS) applied to a structure, or/and deformations

produced by temperature, settlement or earthquakes

Values of actions are obtained by determining

characteristic or representative values of loads or forces

Ideally, loads applied to a structure during its working life,

should be analysed statistically and a characteristic load

is determined.

Characteristic Load: is the representation of the real load,

which is defined as the load with 95% probability of not

being exceeded throughout its lifetime

Characteristic Load = Average Load +1.64 x Standard

deviation

CLASSIFICATION OF ACTIONS

PERMANENT ACTIONS (G)

are due to weight of the structure i.e. walls, permanent partitions,floors, roofs, finishes and services

The actual weights of materials (Gk) should be used in design calculations;but if not known use density in kN/m3 from EN 1991-1:2002.

Also included in this group are water and soil pressures, forces due tosettlement etc

VARIABLE ACTIONS (Q)

Imposed floor Loads (Qk) are variable actions; given for variousdwellings in EN 1991-1-1:2002.

These loads include a small allowance for impact and otherdynamic effects that may occur in normal occupancy. Do notinclude forces resulting from the acceleration and braking ofvehicles or movement of crowds. The loads are usually given asdistributed loads or an alternative concentrated load

Wind Actions (Wk) : Are variable but for convenience are expressed asstatic pressures in EN 1991-1-4(2002).

Thermal effects need to be considered for chimneys, cooling towers, tanksand cold storage services. Classified as indirect variable actions.

Actions to be taken for adequate performance in fire

ACCIDENTAL ACTIONS(A)

Accidental actions during execution include scaffolding, props and

bracing (EN 1991-1-6:2002). These may involve consideration of

construction loads, instability and collapse prior to completion of the

project

Earthquake Loads (the effects of ground motion are simulated by

a system of horizontal forces):EN1998-8(2004)

Actions induced by cranes and machinery : EN 1991-3(2004)

Impact and Explosions covered in EN 1991-1-7(2004).

ACTIONS

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CHARACTERISTIC AND DESIGN LOAD

When checking the safety of a member, the designer cannot be

certain about the load the member must carry because (a) of the

variability of the occupancy or environmental loading, and (b)

because of unforeseen circumstances which may lead to an

increase in the general level of loading, errors in analysis, errors

during construction etc

To cover this uncertainty the characteristic value is used.

The characteristic load is the value above which the load lies in

only small percentage of cases.

Statistical principles cannot be used at present to determine

characteristic loads because sufficient data is not available.

Therefore the characteristic loads are normally taken to be the

design loads from other codes of practice.

Design Load is the value used in design calculations – product of

characteristic load and partial safety factors in order to increase

reliability.

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COMBINATIONS OF DESIGN ACTIONS

FOR THE ULTIMATE LIMIT STATE, three alternative combinations of actions, modified by appropriate partial safety factors (γ), must be investigated

(a) Fundamental: a combination of all permanent actions including self weight(Gk), the dominant variable action (Qk) and combination values of all other variable actions(ψ0Qk)

(b) A combination of the dominant variable actions (ψ0Qk). This combination assumes that accidents of short duration have a low probability of occurrence

(c) Seismic: reduces the permanent action partial safety factor(γG)with a reduction factor (ξ)between 0.85 and 1

FOR SERVICEABILITY LIMIT STATE : 3 alternative combination of actions must be investigated

The characteristic rare combination occurring in cases exceeding limit state causes permanent local damage or deformation

Load partial factors γF , γG ,γQ

Partial factor for variability of strength γM

Different types of load have different probabilities of

occurrence and different degrees of variability, and

that the probabilities associated with these loads

change in different ways as the degree of overload

considered increases. Because of this different load

factors should be used for the different load types.

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CHARACTERISTIC AND DESIGN MATERIAL STRENGTH

The material strength may be less than intended because (a) of its variable composition, and (b) because of the variability of the manufacturing conditions , and other effects such as corrosion.

The characteristic strength is the value below which the strength lies in only small percentage of cases.

The characteristic value is determined from test results using statistical principles, and is normally defined as the value below which not more than 5% of the test results fall.

The overall effect of items under (b) is allowed for using a partial safety factor : γm for strength.

PROPERTIES OF MATERIALS

13

PROPERTIES OF MATERIALS

Design Strength is obtained by dividing the characteristic strength by the partial safety factor for strength

The value of γm depends upon the properties of the actual construction materials being used.

BS EN 1993-1-1(2005) covers the design of structures fabricated from structural steels conforming to the grades and product standards specified. If other steels are used, due allowance should be made for variations in properties, including ductility and weldability.

The design strength should be taken as 1.0Ys but not greater than Us /1.2 where Ys and Us are respectively the minimum yield strength and the minimum tensile strength specified in the relevant product standard.

Design strength

16

19

For the more commonly used grades and thicknesses of steel the

value of design strength values can be obtained from Table 3.1.

Design of steel and


structures

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STANDARD CROSS-SECTIONAL SHAPES

Design of steel and


structures

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Design of steel and


structures

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COMPOUND SECTIONS

Compound sections are formed by:

• Strengthening a rolled section (say UB) by welding a cover

plate

• Combining 2 separate rolled sections like in crane girder

• Connecting two members to form a combined strong

member. Example: laced and braced members

Design of steel and


structures

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FABRICATED SECTIONS/ BUILT-UP SECTIONS

Fabricated sections can be welded or bolted

Cold formed Rectangular Hollow sections

Cold-formed steel has been widely used in building construction, from residential houses to industrial buildings.

Cold-formed steel offers versatility in building because of its lightweight and ease of handling and use.

Cold-formed steel represents over 45 percent of the steel construction market in US, and this share is increasing

The hot-rolled steel shapes are formed at elevated temperatures while the cold-formed steel shapes are formed at room temperature.

Cold-formed steel structural members are shapes commonly manufactured from steel plate, sheet or strip material.

The manufacturing process involves forming the material by either press-braking or cold roll-forming to achieve the desired shape. Examples of the cold-formed steel are corrugated steel roof and floor decks, steel wall panels, storage racks and steel wall studs.

Design of steel and


structures

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DIFFERENCES BETWEEN COLD FORMED AND HOT ROLLED SECTIONS

Press-braking is often used for production of small quantity of

simple shapes.

Cold roll-forming is the most widely used method for production

of roof, floor and wall panels. It is also used for the production

of structural components such as Cees, Zees, and hat sections.

Sections can usually be made from sheet up to 1.5m wide and

from coils more than 1,000m long.

During cold roll-forming, sheet stock is fed longitudinally

through a series of rolls, each of which works the sheet

progressively until it reaches the desired shape. A simple

section may require as few as six pairs of roll, but a complex

shape can require as many as 24 to 30. The thickness of

material that can be formed generally ranges between 0.10mm

up to 7.7mm, although heavy duty cold forming mills can handle

steel up to 19mm thick.

Design of steel and


structures

25

thickness

shapes.

Since cold-formed steel members are formed at room temperature, the material becomes harder and stronger.

Its lightweight makes it easier and more economical to mass-produce, transport and install.

One of the main differences between designing with cold-formed steel shapes and with hot-rolled structural shapes is that with the hot-rolled, one is primarily concerned about two types of instability: column buckling and lateral buckling of unbraced beams. The dimensions of hot-rolled shapes are such that local buckling of individual constituent elements generally will not occur before yielding.

This is not the case with cold-formed members. Here local buckling must also be considered because, in most cases, the material used is thin relative to its width. This means that the individual flat, or plate, elements of the section often have width to thickness ratios that will permit buckling at stresses well below the yield point.

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DIFFERENCES BETWEEN COLD FORMED AND HOT ROLLED STEEL

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Cold rolled shapes

CROSS SECTION PROPERTIES

Elastic section properties

Plastic section properties

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ELASTIC SECTION PROPERTIES

The exact section dimensions

The location of the centroid if the section is

asymmetric about one or both axes

Area of cross section

Moments of inertia about various axes

Radii of gyration about various axes

Moduli of section for various axes

PLASTIC SECTION PROPERTIES

Plastic moduli of section

29

OTHER IMPORTANT PROPERTIES OF UNIVERSAL BEAMS, JOISTS AND

CHANNELS USED FOR DETERMINING THE BUCKLING RESISTANCE

MOMENTS

Buckling parameter, u

Torsional index, x

Warping constant, H

Torsional constant , J

These properties are given in standard tables or

can be calculated using formulae given in the code.

ELASTIC SECTION PROPERTIES

1212

33 dtB

BDYYinertiaofMoment axis

1212

2 33 dtTBZZinertiaofmoment axis

A

IgyrationofRadius

y

y

A

IgyrationofRadius z

z

30

Area = 2BT+dt

Symmetrical I section

2/,sec

D

IZtionofModulus z

zz

2/,sec

B

IZtionofModulus

y

yy

Modulus of section is the elastic modulus of section

EXAMPLE

Determine the properties of a plated UB section “610 UB 125”

strengthened by welding a 300 mm x 20 mm plate to each flange.

Determine the section properties Ix and Zx

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Desig

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229 mm

612

Wel,y = 986 x 106 mm4

Wpl,y=3680 x 10 3 mm3

19.6 mm

11.9 mm

Solution:

The properties of the UB are available in tables and are shown above.

Because of the symmetry of the section the centroid of the plated UB

Is at the web centre.

2

sec fromdistance2 UBplateplateUBtionwelded CGtoCGplateofAreaIII

4626 1021842/2061220300210986 mmIwelded

33

6

106700

2

202612

102184

2/mm

D

IZ x

x

32

The properties Ix and Zx are elastic properties i.e. the whole section is effective

Moment of inertia of plate about its own centroid is small compared to other values , so

omitted.

PLASTIC MOMENT OF A SECTION

33

yf

yf yf

M=σ X ZxxMe=Py x Zxx

Mp=Py x Sxx

Derivation:

Plastic Moment = Py x area in compression x d/2

= Py x area in tension x d/2

= Py ( area in compression x d/4 +area in tension x d/4)

= Py x algebraic sum of first moments of area about equal area axis

= Py x Sxx where Sxx= plastic section modulus

Neutral axis

PLASTIC MODULUS AND SHAPE FACTOR

Shape factor of a section is defined as ν, where

2

sec

D

tionofMomentPlasticModulusPlastic

xx

xx

Z

S

uluselastic

ulusplastic

mod

mod

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Problem: Determine the elastic moment, plastic moment,

elastic section modulus, plastic modulus and shape factor for the

rectangular section 10 mm

500m

m

Elastic properties

433

7.10416666612

50010

12mm

bdInertiaOfMoment

367.416666

2

500

7.104166666

2

mmD

IModulusSectionElastic xx

Desig

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yyxx ffZtionofMomentElastic 7.416666sec

Plastic properties

axis area equalabout area of momentsfirst of sum algebraicmodsec ulustionPlastic

Equal area axis coincides with the centroid of section

42sec

disqualAreaAxareaAboveEstionModuluPlastic

36250004

500

2

500102 mmulusPlasticMod

yy ffulusPlasticmomentPlastic 625000mod

5.167.416666

625000

xx

xx

Z

SFactorShape

DETERMINE THE SHAPE FACTOR FOR 610X229X125 UB S275

From the table of properties of Universal

Beams, the properties can be obtained:

Zxx = 3220 cm3

Sxx=3680 cm3

143.13220

3680

xx

xx

Z

S

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The value of shape factor for most I-sections is about 1.15

Tributary Areas

TRIBUTARY AREAS/ LOADING

A. ONE – WAY SYSTEMS

L2/L

1≥ 2.0 where L

2> L

1

http://classes.mst.edu/civeng217/concept/02/01c/002.jpg



A. ONE – WAY SYSTEMS

L2/L

1≥ 2.0 where L

2> L

1








B. TWO – WAY SYSTEMS

L2/L

1< 2.0 where L

2≥ L

1

EXAMPLE 1:

THE FLOOR PLAN SHOWN IS SUBJECTED TO A UNIFORMLY DISTRIBUTED LOAD OF 5 KN/M2.

IGNORE THE SELF WEIGHT OF THE BEAM:

A) HOW THE LOAD WILL BE DISTRIBUTED TO THE BEAMS ?

B) CALCULATE THE LOAD ON EACH BEAM

L/B RATIO = 8/3 = 2.66 >2, THEN THE LOAD WILL BE

DISTRIBUTED IN ONE DIRECTIONS AS SHOWN BELOW

Total load on all beams = 60 x 2 + 120 x 3 = 480 kN

46

B1

B2

ly

B1

B2

ly

B2

B2

B2 B2

B1 B1 B1

B1 lx

lx

EXAMPLE 2:

THE FLOOR PLAN SHOWN IS SUBJECTED TO A UNIFORMLY DISTRIBUTED LOAD.

DRAW THE IDEALIZED BEAMS LOADING(AFFECTED BY THE SHADED AREA).

47

lx

R1 R1

2(w.lx/2) kN/m

+

Self weight of the beam will act

as a uniformly distributed load

R1 R1

lx

Reaction of B1 will

be transferred to

supporting columns

Load on B1

48

R2

+

Self weight of the beam will act

as a uniformly distributed load

R2 R2

ly

R2

2(w.lx/2) kN/m

ly

Load on B2

49

B1

B2

lx

B2

B1

lx

B2

B1

EXAMPLE 3:


DRAW THE IDEALIZED BEAMS LOADING( AFFECTED BY THE SHADED AREA).

50

R2

2(w.lx/2) kN/m

ly

R2 R2

2(w.lx/2) kN/m

ly

R2

Load on B2

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2R2

R1 R1

2(w.lx/2) kN/m 2(w.lx/2) kN/m

lx lx

Load on B1

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EXAMPLE 4:


DRAW THE IDEALIZED BEAMS LOADING( AFFECTED BY THE SHADED AREA).

53

R2

2(w.lx/2) kN/m

ly

R2 R2

2(w.lx/2) kN/m

ly

R2

2R2

R1 R1

lx lx

Load on B1

54

A B

C D

E F

6m

6m

8m Figure 1

Example 5:

The floor plan shown in Fig.1 is subjected to a uniformly distributed load

of 5 kN/m2. Ignore the self weight of the beam:

a) How the load will be distributed to the beams ?

b) Calculate the load on each beam

c) Calculate the load on each column

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SOLUTION

AS RATIO IS = 8/6= 1.3 < 2, THE LOAD WILL BE DISTRIBUTED IN

TWO DIRECTIONS AS SHOWN BELOW

58

• Total load on all beams = 45 x 4 + 75 x 2 +150 = 480 kN

To check: Total Area = 2 x 6 m x 8 m = 96 m2

Total Load = 5 kN/m2

x 96 m2

= 480 kN

lecture 2 slides

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