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CVEN3302 Structural Design Semester 2,, 2009

School of Civil and Environmental Engineering, UNSW

Introduction by

Zora Vrcelj

Lecture Outline Introduction – ‘Housekeeping rules’ Structural steel p p properties, fabrication ,

Limit states design to AS4100 g 4 Strength, serviceability

Loads Dead, live, wind, earthquake , , , q

SStructurall Steel - Properties

This is what you end up with  y p if you don’t build with steel!

Bad choice of material can  cause premature failure

Steel is by far the most useful  material for building structures with  g of approximately ten times  pp y strength that of concrete, steel is the ideal  material for modern construction. 

Steel is also a very eco‐friendly  material and steel structures can  b   il  di be easily dismantled and sold as  tl d  d  ld    scrap. 

Due to its large strength to weight  ratio, steel structures tend to be more  economical than concrete structures  for tall buildings and large span  buildings and bridges. 

To get the most benefit out of steel, steel  structures should be designed and  t t   h ld b  d i d  d  protected to resist corrosion and fire.  Steel structures are ductile and robust  and can withstand severe loadings such  as earthquakes.  The effects of temperature should be  considered in design. Special steels and  protective measures for corrosion and fire  i    f   i   d fi   are available and the designer should be  familiar with the options available. To prevent development of cracks under  fatigue and earthquake loads the  p connections and in particular the welds  should be designed and detailed properly.  They should be designed and detailed for  easy fabrication and erection. Good  quality control is essential to ensure  lit   t l i   ti l t     proper fitting of the various structural  elements.

Thus the lifecycle cost of steel structures,  which includes the cost of construction,  maintenance, repair and dismantling, can be  less than that for concrete structures. Since  steel is produced in the factory under better  quality control, steel structures have higher  li l l h hi h reliability and safety.

Steel structures can be constructed very fast  and this enables the structure to be used early  y g y thereby leading to overall economy. Steel  structures can be easily repaired and  retrofitted to carry higher loads.

Structural Steel – Mild Steel E Economic, Ductile i  D til Hot‐rolled into  standard shapes

Box sections

Easily fabricated by welding

Standard sections

Cell Form Beams

Typical Steel Fabrication Shop

f bi i   fabrication workshop k h

h   lli hot rolling

Ductility  The most important material characteristic of mild steel is its  ductility.  Ductility allows very large strains to develop with little increase in  stress, prior to failure.  The advantages of ductility are: It can give prior warning of impending failure It allows energy absorption in dynamic loading  or in resisting brittle fracture It allows for redistribution of actions, which is usually benign , y g p , in achieving g a ductile stress-strain curve it requires q y N.B At present, the yield stress fy to be less than 450 MPa. The yield stress is also called the Grade of the mild steel, i.e. Grade 350 steel has fy = 350 MPa.

Properties of mild steel p St Stress

Not to scale

Upper yield stress

fu

Strain hardening Est

fy

Plastic

Tensile rapture

El ti E Elastic

εy

εst

Strain

Idealised stress‐strain relationship for structural steel N.B. N B The same stress‐strain curve is assumed in compression, but we shall see that  Th     t t i    i   d i   i  b t    h ll   th t  buckling of members and elements in compression usually prevents high strains from  being realised

Idealised Stress‐Strain (σ−ε) Diagram Mild Strength Steel  σ ≤ 450 MPa Mild Strength Steel, σ σ

ε

High Strength Steel  σ ≥ 450 MPa High Strength Steel, σ σ

ε

Yielding under biaxial stresses Mises Yield Criterion Mohr Circle Construction

Uniaxial tension

-1.0 P i i l stress Principal

Uniaxial compression

f1, f2 – normal stresses τ – shear stress

Princcipal stress

ratio f2/fy

Pure shear

1.0

1.0 ratio i f1/fy

-1.0

Maximum distortion‐energy  criterion: f12-f1f2+f2 2+ 3τ2= fy3

Introduction to  Structural Steel Design

Design Process Problem!

Includes: setting criteria, constraints functional and structural requirements

Definition of problem (Design brief)

Information search Includes: design data data, information from other consultants, loads

Preliminary designs and selection Structural systems (C (Conceptual t l design) d i ) Includes: type of system, spacing of major members, fixity of connections construction techniques

Detailed design Drawings and specifications Advice on construction

Solution (completed job)

When the need for a new structure arises, an individual or agency  has to arrange the funds required for its construction.  The individual or agency henceforth referred to as the owner then  approaches an architect. 

The architect plans the layout so as to satisfy the functional requirements and also  ensures that the structure is aesthetically pleasing and economically feasible.  In this process, the architect often decides the material and type of construction as well. 

The plan is then given to a structural engineer who is expected to locate the structural elements  so as to cause least interference to the function and aesthetics of the structure.  He then makes the strength calculations to ensure  safety and serviceability of the structure. 

This process is known as structural design. 

Finally, the structural elements are fabricated and erected by the  contractor  If all the people work as a team then a safe  useful   contractor. If all the people work as a team then a safe, useful,  aesthetic and economical structure is conceived. 

However in practice, many structures fulfil the requirements  only partially because of inadequate coordination between the  l   ti ll  b   f i d t   di ti  b t  th   people involved and their lack of knowledge of the capabilities  and limitations of their own and that of others.  • Since a structural engineer is central to this team, it is necessary for him  to have adequate knowledge of the architects and contractors work   to have adequate knowledge of the architects and contractors work.  • It is his responsibility to advise both the architect and the contractor  about the possibilities of achieving good structures with economy. 

Ever since steel began to be used in the construction of  structures, it has made possible some of the grandest  t t  it h   d   ibl     f th   d t  structures both in the past and also in the present day.

Successful Structures Functional requirements – set by client SAFETY– building life including construction period  SAFETY b ildi  lif  i l di   t ti   i d  STRUCTURAL ENGINEERS Aesthetic satisfaction – set by architects Economy – Capital cost is not just the structural component  but also financing and construction speed maintenance costs can effect long term life cycle costing

Steel Structures Code Section 9 Section 5 to 8 Section 5 4 Section 6,7243& 8 Section Section AS 4100 Connections Engineered g Capacity p1 y Design Member Section Properties Method ofof Timber of subjected Connections to- Design Capacity Introduction Timber Structural •nails, screws Products - •Bolting bending g pProperties scope, of Members •eng. eng. Australian Standard Analysis •bolts, •b coach h screws b lt•plywood l d •Welding definitions, strength (f ’) •tension Steel Structures •shear connectors, •Elastic •poles stiffness (E) •Section capacity notation, •compression split rings •Plastic •Modif’n •Modif n factors •Member units itd (N, (N •Member buckling •combined bicapacity d di •bending b•glulam k•Frame modifies strength •LVL buckling mm, MPa) actions

Standards Australia

Limits States Design to AS4100 g 4 Forrest Centre, Perth WA Firstt M Fi Melbourne lb building b ildi to use concrete filled tubular steel columns Central C t l concrete t core with steel beams and metal formwork

Steel skeleton connected to the central core

Casseldon Place, Melbourne

Columns were concrete filled composite steel box columns

Design approach of AS4100 g pp 4 ‹ Based on limit state design ‹ Principal limit states •

Strength (ultimate limit state), concerned with  ‘collapse’: – – –

•

yyielding g buckling  overturning

Serviceability limit state, concerned with  i bili li i d h ‘function’: – –

deflection d fl i vibration

Limit States Design Aim ‐ Satisfactory performance under a variety of different uses  or load scenarios

Strength ‐ Rare scenarios: want safety for occupants  no failure

Serviceability y ‐ Common scenarios: want satisfactory performance in service under common loadings no cracking, no bouncing, satisfactory appearance and function

Strength limit state design principles ‹ Define relevant limit states ‹ Determine appropriate actions ‹ Analyse using appropriate methods and 

g y accounting for variability to determine: • Design effects { S* }, and • Design resistance { φ R } ‹ Ensure no limit state is exceeded

*

S ≤ φR

Strength limit state design principles Effect of Factored Loads ≤ Factored Resistance

*

S ≤ φR For load combinations, the effect of factored loads (S*) is the structural effect due to the specified loads multiplied by load factors.

Variability of actions a ab ty o act o s ‹ Precision of modelling actions varies: •

dead loads related to material density and  thickness • imposed loads based on type of occupancy • wind and snow loads based on meteorological  data ‹ Probability of load combinations varies y

ACTIONS DEAD LOADS Weights of the various structural members and the weights of any  objects that are permanently attached to the structure (i e  self‐weight  objects that are permanently attached to the structure (i.e. self‐weight  of the structure + superimposed dead load)

LIVE LOADS • Building loads • Bridge loads B id  l d • Wind loads • Snow loads • Earthquake loads • Hydrostatic and Soil Pressure • Other Natural Loads (the effect of blast, temperature changes, different  settlement of the foundation)

DESIGN LOADS ‐ GENERAL For the design of structural steelwork the following loads and influences shall be  considered: G  ‐ Dead loads, including the weight of steelwork and all permanent materials of  construction  partitions  stationary equipment  and additional weight of  construction, partitions, stationary equipment, and additional weight of  concrete and finishes resulting from deflections of supporting members, and the  forces due to prestressing; Q – Live loads, including load due to intended use and occupancy of structures;  movable equipment, snow, rain, soil, or hydrostatic pressure; impact; and any  ot e e oad st pu ated by t e egu ato y aut o ty; other live load stipulated by the regulatory authority; T – Influences resulting from temperature changes, shrinkage, or creep of  p , ; component materials, or from different settlements; W – Live load due to wind; E – Live load due to earthquake

G – Dead Load Loads act in vertical direction. The specified dead load for a structural member consists of: ¾

the weight of the member itself,

¾

the  eight of all materials of constr ction incorporated into the weight of all materials of construction incorporated into

¾

the building to be supported permanently by the member,

¾

the weight of partitions,

¾

the weight of permanent equipment, and

¾

the vertical load due to earth, plants and trees. ,p

Dead Load, G , Services (ventilation, electricity ducts, etc.)

Superimposed dead load

Load path?!

Variability of material and section  y properties ‹ Resistance {R } is related principally to material 

and section geometry ‹ Yield strength of steel is guaranteed e d st e gt o stee s gua a teed ‹ Other properties, notably Young’s modulus (E), 

are much less variable

Strength (Ultimate) Limit State ‹ The following conditions should be considered: • •

Stability: overturning (equilibrium) Strength: including local and overall buckling effects where  appropriate

‹ It may also be necessary to consider: • •

Second order effects Rupture (due to fatigue)

Strength Limit State i e  GRAVITY LOADS:     1.2G i.e.  GRAVITY LOADS:     1 2G + 1.5Q + 1 5Q ≤ fR or

S ≤ φR *

R = resistance capacity (at failure) Left hand side is factored strength load effect, S* φ = capacity reduction factor

Capacity Factor φ Gives consistent reliability to whole structure φ = 0.9    0 9  (for steel members, M (f   t l  b  Mu, V  Vu, N  Nu) φ = 0.8  (connectors and connections)

Strength Limit State g Strength ‐ avoid failure in the lifetime of the structure

i.e. live load combination   

1.2G 1 2G + 1 1.5Q 5Q 0.8G + 1.25Q 1.2G 1 2G + Wu

i.e. wind load combination

0.8G 0 8G + Wu

(1) and (3) combinations are used if the loads act in the same direction (2) and (4) are used if the loads act in opposite directions.

Strength Limit State g Loads are generally: UDL’s (or pressures = Force/Area)  g y p and Point Loads Load effects are: ƒAxial force N* ƒBending moment M* ƒShear force V* When these load effects are determined using factored loads (*) they are called design loads When the load effects are determined without using factored loads they are called nominal loads

G – nominal deal load (given in loading case AS1170.1) Q – nominal live load (given in loading code AS1170.1) (g g 7 ) Wu – nominal ultimate wind load (given in loading code AS1170.2)

φMu, φNu, φVu – the strengths are determined from the steel code AS4100

S*

φR

Strength Limit State g An example of the design equation may be establishing that: f

M ≤ φM b *

where M* is the factored bending moment in a beam (determined from  structural analysis) Mb is the bending strength of the beam (that accounts for lateral  buckling) and φ = 0.9

Strength Limit State g Another example might be establishing that

N ≤ φN c *

where N* is the factored axial compression in a column (the load effect or  design axial compression) Nc is its strength that accounts for the effects of column buckling if  the column is slender (the nominal compressive strength) and  φ = 0.9

Serviceability Limit State y ‹ The following conditions may need to be considered: •

excessive deflections

•

excessive vibrations

‹ Both conditions are associated with stiffness rather 

than strength g ‹ For most buildings, controlling deflections will also limit 

vibrations b

Serviceability of beams y Deflection limits for beams:  appearance (sagging) fitness for purpose (machinery, pipe grades) str ct ral (a oid  nintended load paths) structural (avoid unintended load paths)

Serviceability Limits y Code gives guidance only g g y (i.e. Δ/L = 1/250, 1/500, etc.)

Main message is ‐ THINK and discuss with client

Design for Serviceability g y how to size a member? 1. Agree on deflection limits   A    d fl i  li i  δ lim with client i h  li

2. Evaluate serviceability load combinations that have limit δ lim • •

split combination into constituent loads wi estimate duration of each constituent load

Design for Serviceability g y 3. 

⎛ 5 L4 ⎞⎛ w ⎟⎟ ⎜⎜ I ≥ ⎜⎜ ⎝ 384 E ⎠ ⎝ δ lim

⎞ ⎟⎟ (udl, ss) ⎠

Note: Design load factors used for Strength Limit State do not apply to  Serviceability Limit State y ((i.e. we use w not w*)

4. Se Select ect c cross-section oss sect o to g give e

I

5. Check Ch k bending, b di shear, h axial i l strength t th

φMu

φVu

φNu

Reference Material AS4100  1998 Steel Structures:  AS4100 – 1998 Steel Structures  Standards Australia, Sydney. S d d  A li  S d AS1170.1 & AS1170.2 Loading Codes: Standards Association of Australia,  Sydney.

NS Trahair & MA Bradford: The Behavior and Design of Steel  g Structures to AS4100, 3rd Australia edition, E&FN Spon, London, 1998. MA Bradford, RQ Bridge & NS Trahair: Worked Examples for  MA Bradford  RQ Bridge & NS Trahair  Worked Examples for  Steel Structures, 3rd edition, AISC, Sydney, 1997. ST Woolock, S Kitipornchai & MA Bradford: Design of Portal  Frame Buildings, 3rd edition, AISC, Sydney, 1999.

DESIGN  OF LATERALLY  RESTRAINED BEAMS

Lecture Outline B Beam Design:  D i Name governing Limit States? g g _______________ _______________ _______________ _______________

Lecture Outline Steel beam Modes of failure Local Buckling and Section Classification g Compact Non‐compact Slender

Section capacity in bending

Section Capacity  or 

Design Capacity of Fully Laterally Restrained Beams or

In Plane Capacity In‐Plane Capacity or 

Design Capacity of very Short Laterally Unrestrained Beams

Steel beam Beams are members of structures which carry loads transverse to their length. These members resist flexure (bending) and shear,  and sometimes torsion, introduced by transverse  loads. Purlins, rafters, joists, spandrels, lintels, floor beams, stringers and  other similar structural parts are all beams. other similar structural parts are all beams Members subjected to bending and axial compression simultaneously are beam‐columns.

Steel beam Beam (UDL -

major axis loading)

Beam

(couple)

Beam

(torsion) ( )

Beam (UDL -

minor axis loading)

Beam-column

(axial compression + transverse loading)

Steel Beam – where do we use it?

Strength limit state bending  moment capacity Design equation for bending strength

(φ M ) ≥ M * Design capacity > factored strength limit state  D i   it    f t d  t th li it  t t   moment

In plane bending In‐plane bending X (u)

Y (v)

Modes of failure The usual strength modes of failure for  structural steel beams are: • Plastification • Flange local buckling • Lateral buckling • Web crippling • Web local buckling under shear We shall consider each of these strength limit states in turn.

Plastification A steel beam may fail when the PLASTIC MOMENT Mp develops, or when sufficient plastic hinges develop to form a  mechanism.

Mp = S fy

S – plastic section modulus

fy – yield stress

2) ( (MPa = N/mm /

S is tabulated for most rolled sections in handbooks (mm3).

Ductile stress‐strain curve Recall Mp is reliant on a DUCTILE STRESS is reliant on a DUCTILE STRESS‐STRAIN CURVE STRAIN CURVE.

σ fy E 1

εy

Long plastic plateau means ductile

ε

εy = yield i ld strain t i = fy / E = fy / 2 2x10 105

MPa

Maximum moment Maximum moment that can be attained is the PLASTIC MOMENT Mp

equal areas C h T plastic neutral axis Nominal capacity Mmax = Ms = Mp Ms is called the SECTION CAPACITY.  It is the moment to cause failure of the cross‐section. Here, Mp = C x h = T x h

Example 1 Calculate Mp for the section shown below.

500

20

200

C1 C2

h2

C1

h1

C2

10

T2 20 200

fyf = 350 MPa    MP

T1

Example 2 Calculate the factored load W* to cause plastic collapse of the  beam shown below: W* 410UB60

Sx = 1190 x 103 mm3 fyf = 300 MPa MP 3000 mm

M = φ Sx f yf = 0.9 ×1190×10 × 300 Nmm *

3

= 321.3 kNm

M = W L = 3W *

*

*

* M 321.3 * W = ≤ 3 3

W * ≤ 107.1 kN

Design bending capacity g g p y M ≤ φM S *

φ = 0 .9

However, lateral However  lateral and local  buckling usually result in  l lowering M i  Mmax below M b l  Mp.

MS = MP Beams are usually unable to reach Mp because of  the occurrence of premature BUCKLING

Local Buckling  g and  S ti  Cl ifi ti Section Classification

Basis of section classification Rolled or welded sections may be considered as an assembly  of individual plate elements Some are outstand

Outstand Internal

‐ flanges of I beams fl f b ‐ legs of angles and T‐sections

Some are internal ‐ webs of open beams  ‐ flanges of boxes

Internal Web

Web

Flange Rolled I-section

Flange Hollow section

Basis of section classification Rolled or welded sections may be considered as an assembly  y y of individual plate elements Some are outstand Outstand

‐ flanges of I beams fl f b ‐ legs of angles and T‐sections Web

Internal

Internal te a

Some are internal ‐ webs of open beams  ‐ flanges of boxes

Flange

Welded box section

Basis of section classification • As the plate elements are relatively thin, when loaded in compression they  may buckle locally   • The tendency of any plate element within the cross section to buckle may  limit the axial load carrying capacity, or the bending resistance of the  section, by preventing the attainment of yield.   • Avoidance of premature failure arising from the effects of local buckling  may be achieved by limiting the width‐to‐thickness ratio for individual  may be achieved by limiting the width to thickness ratio for individual  elements within the cross section.  Outstand Internal

O t t d Outstand Internal

Internal Web

Web

Flange

Rolled I-section

Web

Flange

Hollow section

I t Internal l

Fl Flange

Welded box section

Flange local buckling g g Buckled flange

Compression flange Buckled web Web

Flange and top compressive region of  the web DISTORT, but the line  junction between the flanges and web  remains straight. Occurs in slender COMPRESSION FLANGES

Flange local buckling g g If the compression flange of a beam is slender, it may buckle LOCALLY and prevent the  beam from reaching its maximum bending strength Mp (PLASTIC MOMENT).

The stress to cause ELASTIC LOCAL BUCKLING σol is given (see Trahair & Bradford Chapter 4):

σ ol

π E ⎛⎜ t f =k 12 (1 − ν 2 ) ⎜⎝ b f 2

⎞ ⎟ ⎟ ⎠

2

Called FLANGE OUTSTANDS tf

where:

k = the local buckling coefficient that depends

on edge and loading conditions (= 0.425 here) E = Young Young’ss modulus = 200 x 103 MPa

ν = Poisson’s ratio (0.3 for steel)

bf

bf

Plate buckling • A thin flat rectangular plate subjected to compressive forces along its short edges has an elastic critical buckling stress ((σcr ) g given byy

σ cr

kσ π E ⎛ t ⎞ = ⎜ ⎟ 2 12(1 − ν )⎝ b ⎠ 2

2

kσ is the plate buckling parameter which accounts for edge support conditions, conditions stress distribution and aspect ratio of the plate

Plate buckling in g compression p Bounded plate in uniform compression

n = number of ½ sine waves

For bounded flanges kb = 4

Flange plate behaviour in compression Critical buckling coefficient kσ therefore depends on: Flange in Compression

• Boundary conditions L t

• Stress distribution • Aspect ratio (width/thickness)

(b)

(a)

b

Simply supported on all four edges

Buckling coefficient k 5 b

Simply supported

4

longitudinal edge

b

L

Free

Exact

3

k = 0.425 0 425 + (b/L) 2

NOTE; for a web in pure compression both longitudinal edges g are simply p y supported pp and kσ = 4.0.

2

L

(c) 1 0.425

Free longitudinal edge

FREE FLANGE OUTSTAND

0

(d)

1

2

3

4

Plate aspect ratio L / b

5

Example 3 p 3 What must we restrict the width to thickness ratio bf/tf to in order to ensure yielding at fyf will occur before elastic local buckling?

If buckling is to be prevented, prevented then from:

σ ol

π E ⎛⎜ t f =k 12 (1 − ν 2 ) ⎝⎜ b f 2

f yf > σ ol

so that

bf tf

⎞ ⎟ ⎟ ⎠

2

0 .425 × π × 200 × 10 = 12 1 − 0 .3 2 2

(

f yf ≤ 277 (fy in MPa)

)

3

⎛ tf ⎜ ⎜b ⎝ f

⎞ ⎟ ⎟ ⎠

2

Section classification AS AS4100 defines three types of cross section:  d fi  th  t   f    ti

(a) COMPACT SECTION (b) NON ‐ COMPACT SECTION ( ) (c) SLENDER SECTION

Section classification Section classification depends on: Section classification depends on • slenderness of each element (defined by  a width‐to‐thickness ratio)    idth t thi k   ti )  • the compressive stress distribution 

Slenderness parameter

⎛ bf λ e = ⎜⎜ ⎝ tf

⎞ fy ⎟ ⎟ 250 ⎠

fy in MPa units

Variations in λey and λep due to residual stress effects

The section slenderness,  , λe From the previous example we saw that the section slenderness (bf/tf) is  important in enforcing yielding to occur before elastic buckling. It will be shown how this can be extended even further. t be s o o t s ca be e te ded e e u t e

N.B. The SLENDERNESS λe of a flange is defined by:

⎛ bf λ e = ⎜⎜ ⎝ tf

⎞ fy ⎟ ⎟ 250 ⎠

fy must be in units of MPa ( = N/mm2) the significance of the f y term is apparent from f Example l 3. The “normalising” with respect to 250 MPa is historical as most steels once had fy = 250 MPa. Yield stresses are now higher. (b f t f ) f y 250 obviously is more transparent than (b f t f ) f y

Section classification ((a) COMPACT SECTION ) These sections allow the FULL PLASTIC MOMENT Mp and for  the strain hardening region to be entered before ELASTIC  BUCKLING occurs. Sections must be COMPACT if plastic analysis/design is to be utilised.

The section slenderness is governed by:

λ e ≤ λ ep λe p is constant

The limits on λεp are much ‘tighter’ th when than h fol = fy in i Example E l 3 because higher strains at fy are needed to make local buckling occur in the strain hardening region.

Section classification

There are limits for flange and web classification

Th limits The li it are therefore: th f

λep = 10  [stress‐relieved flanges] = 9    [hot‐rolled]       [h t ll d] = 8    [welded] [ ] The difference is due to initial geometric out‐of‐ straightness and to different residual stresses.

Residual Stresses Hot rolled Hot‐rolled

Welded

Compact section p M - ρ of a PLASTIC SECTION

M Moment t

MP MY

Inelastic local buckling well into the strain-hardening range

curvature -ρ The design equation is then:

M

*

≤ φM

S

φ = 0 .9

M S = M P = fyS

Example 4 p 4 530UB92.4 Sx = 2370 x 103 mm3 fyf = 300 MPa

5333

209

10.2

209

N.B. Some UB’s have λe > 9. We have not considered the compactness of the web (this will be done latter).

Section classification (b) NON‐COMPACT SECTION These sections allow the FIRST YIELD MOMENT My to be reached,  but buckle locally before Mp can be attained. Their moment/curvature response is:

The design  equation is then:

Moment

M * ≤ φM S

M-ρ of a NON-COMPACT SECTION

φ = 0 .9

MP MY Inelastic local buckling before Mp is reached

and for a  NON‐COMPACT  SECTION

λe ≤ λey

curvature -ρ

M S = f y Ze

Non‐compact section For non-compact non compact sections:

λep ≤ λe ≤ λey

The λey limits are ‘looser’ than the λep limits and essentially correspond to the coincidence of yielding and elastic local buckling, but they are modified to include residual stresses and initial geometric imperfections in the strength.

The limits are therefore

λey = 16 [most flange outstands]

= 15 [welded flange outstands]

N.B. We saw in Example 3 that first yield [MY ] and elastic local buckling coincided when (b f t f ) f y = 277 , or

(b

f

t f ) f y 250 = λe = 277

250 = 17.5

This is close to the above limits.

Non‐compact section p Ze is the effective section modulus. of course, f 

Ze = S S if  MS = MP Ze = Z if  MS = MY

[Z = elastic section modulus,  MY = fy Z ]

Non‐compact section p R l behavior Real b h i

Moment

MP MS MY IInelastic l ti local l l buckling

ρy

Linear approximation

curvature -ρ

For non-compact sections we can interpolate linearly between MY and MP, based on the value of λe.

Non‐compact section p Linear approximation

Moment

MP MS MY

⎡ λey − λe ⎤ Ze = Z + ⎢ ⎥ (S − Z ) ⎢⎣ λey − λep ⎥⎦

λep Section strength of section with λe

λe

λey

λe

Check: λe = λep

Ze = Z + 1(S-Z) 1(S Z) = S

[plastic]

λe = λey

Ze = Z + 0(S-Z) = Z

[non-compact]

Section classification ((b) SLENDER SECTION ) These sections buckle locally even before the yield stress (and My) are reached.

The moment/curvature response is: h i

For slender  sections:

Moment M-ρ of a SLENDER SECTION

λ e > λ ey

φ = 0 .9

MP MY

and for a  SLENDER  SECTION:

Buckling failure prior to MY

curvature -ρ

M S = f y Ze

Slender section The effective section modulus may be calculated by two methods:

Method 1: An effective width approach omits from each flange the width in excess of that  which corresponds to λey. be

be

tf

compression flange (partially effective, 2be)

ineffective (ignore) (g )

tension flange (fully effective, b)

Slender section The effective width be is defined such that: be tf

fy 250

= λ ey

or

λ ey be = b λe

Although accurate, the method may be cumbersome for beam cross‐sections as the  effective section becomes MONOSYMMETRIC, i.e. Need to calculate new centroid and I: Centroid of  original section

Centroid of  d f ti   ti defective section

yC

yT

ZC = I

yC

Z e = min[Z C , Z T ]

ZT = I and since

Ze = I

yC

yT

yT < yC

Slender section Method 2:   Method 2 An easier and simpler method to use:

Ze

⎛ λ ey = ⎜⎜ ⎝ λe

⎞ ⎟⎟ Z ⎠

where Z is the elastic modulus calculated for the full section.

Section classification based on web  slenderness l d So far we have considered the compression flange which may  f f buckle locally under UNIFORM STRESS. The web is subjected to bending stress (compression along one edge,  tension along the other edge) and may also buckle locally. tension along the other edge) and may also buckle locally

C

C

web T

T

Under bending, the coefficient k in web is approximately 23.9.

Web local buckling g Stocky flange Slender web

Buckled web

Stocky flange

occurs in slender webs  with large bending and/or  shear stress h   t

Web crippling Occurs due to localised yielding of the  g web near concentrated loads. Behavior is dominated by gross yielding  over a small web region.

web b

Section classification based on web  slenderness Webs can be classified similarly to flanges as Webs can be classified similarly to flanges as: COMPACT:

λe < λep

NON‐COMPACT: NON COMPACT: SLENDER:

λep ≤ λe < λe y

λe ≥ λe y

The limits are:

fy

dw λe = tw

250

λep = 82

λey = 115

tw dw

Section classification based on web  slenderness l d For a SECTION to be COMPACT:  both the FLANGES and WEB must be compact. For a SECTION to be NON‐COMPACT:  EITHER the FLANGE or WEB or BOTH are non‐compact. For a SECTION TO BE SLENDER:  EITHER the FLANGE or WEB or BOTH are slender.

Compact flange Slender web

ii.e. this SECTION is  hi i classified as  SLENDER

Example 5 p 5 Calculate the design bending (section) capacity for  the cross‐section shown. 240 4

8 10 240

fy = 250 MPa throughout

Example 5 p 5 The cross section is therefore NON‐COMPACT. f

Example 5 p 5

Box‐cross‐sections For the compression flange, k p g , = 4.0 For the web in bending, k = 23.9

bf

compression i flange

b bf

tf

Welded box column in  bending

d

Rectangular (or square) hollow section  [RHS or SHS]

Box‐cross‐sections For the compression flange, k = 4.0 The classifications are the same as for flange outstands, but with:

λep = 30

λey = 45

COMPACT if again

= 40 = 35 3

λe = (b f t f ) f y 250 < λep

NON COMPACT if NON‐COMPACT SLENDER if

[ hot-rolled] hot rolled] [ lightly welded] [ hheavily il welded] ld d]

λep ≤ λe < λey

λe ≥ λey

The webs of box cross‐sections are clearly the same as those of I‐sections.

DESIGN OF LATERALLY UNRESTRAINED BEAMS

Lecture Outline Lateral torsional buckling Elastic lateral buckling Twisting moment warping

Moment gradient factor, αm Slenderness reduction factor, αs Idealised end conditions full, lateral, partial and unrestrained

In-plane bending X (u)

Y (v)

Out-of-plane buckling

φ

φ X (u)

or Lateral–Torsional Buckling

Y (v)

Lateral Torsional Buckling or

Flexural-Torsional Buckling or

Member Capacity or

Lateral-Torsional buckling or

Out-of-Plane buckling

Lateral buckling

flexural torsional buckling

or

φ u Buckled web

Buckled configuration Original configuration

u – lateral displacement φ – twist

Introduction Clamp at root • Slender structural elements loaded in a stiff plane tend to fail by buckling in a more flexible plane. • In the case of a beam bent about its major axis, failure may occur by a form of buckling which involves both lateral deflection and twisting.

Lateral-torsional Lateralbuckling

Buckled position

Dead weight load applied vertically

Unloaded position

Consider an I-beam ….. •

Perfectly elastic, initially straight, loaded by equal and opposite end moments about its major axis.

M

M L Section Elevation

•

Unrestrained along its length.

Plan

•

End Supports …

z

– Twisting (φ) and lateral deflection (u) prevented. – Free to rotate both in the plane of the web and on plan.

x u

y

φ

Strength limit state bending moment capacity

Design equation for bending strength

Design capacity ≥ factored strength limit state moment

Lateral buckling Lateral buckling is the most influential strength limit state in the design of steel beams. Lateral buckling is also called flexural-torsional buckling (or member buckling) and involves the lateral or sideways instability of long slender beams. Beams with FULL LATERAL RESTRAINT do not buckle laterally and their strength is the CROSS-SECTION STRENGTH defined by:

M * ≤ φM S

M S = Ze f y

φ = 0 .9

Lateral buckling Most commonly beams do not have full lateral restraint and the nominal strength Ms must be reduced to the MEMBER BENDING STRENGTH Mb.

Lateral buckling is catastrophic and so represents a STRENGTH LIMIT STATE.

The design equation is:

*

M ≤ φM b

φ = 0 .9

Lateral buckling

or

flexural torsional buckling

Occurs in slender, laterally unrestrained beams. Beam deflects laterally ( = sideways) by u and twist

φ.

Like columns, the beam reaches an energy configuration at which it prefers to snap into an OUT-OF-PLANE buckled position rather than continuing to bend IN-PLANE.

M

{unstable if u, φ = 0 after buckling moment} buckling moment or “point of bifurcation”

stable buckled position

u, φ

Lateral buckling

We must therefore calculate Mb based on lateral buckling of the beam.

This is done by undertaking firstly an ELASTIC BUCKLING ANALYSIS.

Elastic lateral buckling The elastic buckling resistance depends on the following cross-section properties: Minor axis bending stiffness EIy Torsion resistance GJ Warping resistance EIw Iy = minor axis second moment of area (mm4) J = torsional constant (mm4) Iw = warping constant (mm6)

E = Young’s modulus 200,000 MPa G = Shear modulus 80,000 MPa ν = 0.3

E for G = 2(1 + ν ) steel

Elastic lateral buckling Iy, J, Iw are all tabulated in the One Steel Handbook. Alternatively for the doubly-symmetric I-section: 3

bf tw tf

Iy = dw

bf t f

(N.B. web ignored)

6

1 n 1 3 3 J = ∑ bi ti = 2t f b f + t w3 d w 3 i =1 3

(

)

I w = I y h 2 4 (doubly symmetric )

Twisting moment - torque Consider a built-in cantilever subjected to a twisting moment ( ≡ torque)

Mt move “in”

ELEVATION

move “in”

move “out”

PLAN

Mt

Warping An interpretation of “warping”:

A situation where plane sections do not remain plane during torsion. This occurs with lateral buckling and its effect is reflected in the warping constant Iw.

Torsion The equation of torsion is: 3

dφ dφ M t = GJ − EI w 3 dz dz uniform torsion resistance

warping torsion resistance

φ = angle of twist

Lateral buckling Basic model for lateral buckling is a “simply supported I-beam” subjected to a uniform bending moment M.

“Simply supported” in the lateral buckling sense means lateral deflection and twist are prevented at the beam ends (u = 0, φ = 0), but the flanges are free to rotate in their planes when the beam buckles laterally.

Simply supported I-beam - model M

M

free to rotate in plane during buckling, but u = φ = 0

L ELEVATION

M

BMD

PLAN M

buckled top flange

Elastic buckling moment The ELASTIC BUCKLING MOMENT is:

2

Mo =

π EI y 2

L

2

⋅ GJ +

π EI w 2

L

(N.B. This is stated without proof. See Chapter 6 of Trahair & Bradford.)

Elastic buckling Recall for the elastic buckling of a pin-ended column:

2

N oc =

π EI y 2

L

pin-ended column deformed shape

Euler buckling load (elastic critical buckling load – pin-ended column)

Example 1 Calculate the elastic lateral buckling moment for a simply supported 460UB82.1 beam of length L = 3m subjected to uniform bending.

460

191

From OneSteel Tables section properties handbook: 16

I y = 18.6 × 106 mm 4 J = 701× 103 mm 4

9.9

I w = 919 × 109 mm6 16

191

460UB82.1

N.B. this is the ELASTIC BUCKLING MOMENT and not the actual BUCKLING STRENGTH Mb which also depends on the yield stress fy. Mb will be determined later.

Moment gradient factor, αm Of course beams in reality are rarely loaded in uniform bending, nor are they pin-ended (or “simply supported”) so the formula for Mo needs some modification.

So far we have considered uniform bending only. This is very conservative as the elastic buckling moment (Mo) is increased by unequal moments, transverse loads etc.

This effect is reflected in the αm values given in Table 5.6.1 of AS4100.

Moment gradient factor, αm M

single curvature

M

M

L M

L + = compression

M

+

M

double curvature

M BMD

- = tension +

BUCKLED SHAPE

moment is highest in this region

M moment is very small in this region

Moment gradient factor, αm The effect of the moment gradient is reflected in the αm values given in Table 5.6.1 of AS4100 (see attachment). Alternatively, the BMD is often given using analysis software (Microstran, Spacegas, Multiframe etc.). Therefore

M*m = maximum αm =

* 1.7M m ≤ 2.5  *2 *2 * 2  M 2 + M3 + M 4 

( ) ( ) ( )  M*m x

M*3 M*4

M*2 S/4

S/4

S/4

S/4

design moment within a segment where

M*2, M*4 = design moments at quarter points of a segment

M*3 = design moment at the mid-length of a segment

i.e. BMD segment

Moment modification factors

Slenderness reduction factor, αs The SLENDERNESS REDUCTION FACTOR αs converts the ELASTIC REFERENCE BUCKLING MOMENT Mo into a DESIGN STRENGTH.

1/ 2   2  where M  M s    + 3 − s  αS = 0.6  Mo    Mo    

Mo - the elastic reference

buckling moment determined from Le (effective length) Mo =

αs represents a transition between Ms) and elastic buckling (at Mo)

full yielding (at

π 2 EI y 2 e

L

⋅ GJ +

π 2 EI w L2e

Ms – the section bending

capacity depending on whether the cross-section is compact, non-compact or slender =

Ze fy

Slenderness reduction factor, αs αs

Section strength at Ms

1.0 Elastic buckling at Mo

αs 0

Le Short beam does not buckle laterally and αs = 1

Long beam is not influenced by yielding as its bucking moment is very small

Idealised end conditions There are four types:

• FULL RESTRAINT (F) • PARTIAL RESTRAINT (P) • LATERAL RESTRAINT (L) • UNRESTRAINED (U)

Idealised end conditions • FULL RESTRAINT (F) Lateral deflection and twist are effectively prevented, i.e.

brace stiffeners

compression flange

seat support

shear connector

concrete slab

Idealised end conditions •PARTIAL RESTRAINT (P) Lateral deflection prevented at some point other than at the compression flange, and partial twist thus occurs during bucking.

P

compression flange

buckled configuration

tension flange Seat support restrains tension (T) flange fully at ends.

φ

Idealised end conditions •LATERAL RESTRAINT (L) Compression flange is restrained against translation during buckling, but the cross-section is free to twist during buckling.

screws

brace thin roof sheeting – quite stiff in-plane but flexible in bending

thin sheeting “bends” during buckling

Idealised end conditions UNRESTRAINED (U) Free to both displace and twist during buckling, i.e. cantilever tip. W*

cantilever tip

Idealised end conditions For these idealised end conditions, AS4100 specifies a

TWIST RESTRAINT FACTOR kt RESTRAINTS FF end 1

FP

PP

FL end 2

PL

kt

LL FU 1

PU

2

1.0  d w  t f 1+     L  2t w

  

 d w  t f 1 + 2    L  2t w

3

  

3

Effects of load height Load applied above the shear centre ( ≡ centroid for doubly symmetric I-section) causes an increased “destabilising torque” that lowers the buckling load. W W

W

W

δ

load at shear centre

load at top flange level

Effects of load height To account for the height of application of the load, AS4100 specifies a

LOAD HEIGHT FACTOR kl For load WITHIN THE BEAM SEGMENT RESTRAINTS FF PP

FP PL

FU

PU

FL LL

AT SHEAR CENTRE

AT TOP FLANGE

1.0

1.4

1.0

2.0

For load AT THE END OF THE BEAM SEGMENT RESTRAINTS FF PP

FP PL

FU

PU

FL LL

AT SHEAR CENTRE

AT TOP FLANGE

1.0

1.0

1.0

2.0

Load and rotation factors: AS4100

conservatively kr = 1

Lateral restrain classifications

Lateral restrain classifications

End restraints: examples

End restraints: examples

End restraints: examples

Twist factor: AS4100

Effective length Le The reference buckling moment Mo is written in terms of the

EFFECTIVE LENGTH Le similarly as previously as: 2

Mo =

π EI y 2 e

L

⋅ GJ +

π 2 EI w L2e

L = segment length or length of a sub-segment between full and/or partial restraints

kt = twist restraint factor kl = load height factor

Le = kt kl L N.B. AS4100 also has a “rotational restraint factor” kr that is difficult to quantify and which we shall take equal to unity (conservatively):

Le = kt kl kr L = kt xkl x 1.0 x L L e = kt kl L

Bending capacity, Mb Finally, the design equation for bending within a segment is:

*

M ≤ φM b M b = α mα s M s ≤ M s φ = 0 . 9 αm reflects the effect of the distribution of the bending moment along the beam.

αs reflects the elastic lateral buckling (via Le

Mo) and yielding (Ms). It accounts for load height and restraint (via Le).

Bending capacity, Mb Clearly if αmαs < 1.0, the full SECTION STRENGTH in bending is not attained (Ms), and the beam will buckle laterally at Mb. This is very often the case.

Mb is the MEMBER STRENGTH in deference to Ms which is the SECTION STRENGTH.

αm reflects the bending moment effects (moment gradient) and is reasonably difficult to control as the loading is fixed.

αs can be increased by using a bigger section (larger Iy, Iw, J) or by bracing the beam to decrease Le.

Example 2 Determine the maximum design moment M* of a 200UC52.2 . The effective length Le = 3.5m and the end moments are as shown. M*

206

204

12.5

8.0

S x = 570 × 103 mm3 12.5

204

200UC52.2 fyf = 300 MPa

I y = 17.7 × 106 mm4 J = 325 × 103 mm4 I w = 166 × 109 mm6

0.4M*

Example 2 1. Determine SECTION CAPACITY Ms

Example 2 2. Determine MEMBER CAPACITY Mb

Example 2

Example 2 In this case the SECTION STRENGTH than the MEMBER STRENGTH lateral buckling.

Ms governed rather

Mb that is determined by

This is because of: • the low value of

Le

am that produced a very high elastic buckling moment Mo . αm.

• the high moment modification factor

This is NOT always the case and commonly:

Mb < Ms.

Example 3 The loads are applied on the top flange. Determine the maximum design value of W*. 4W* B

A

C

D

W*

200UC52.2 Grade 300 P

L 3m

3m

206

204

3m

12.5

U

S x = 570 × 103 mm3 I y = 17.7 × 106 mm4 J = 325 × 103 mm4

8.0

I w = 166 × 109 mm6 12.5

204

Example 3 In-plane analysis

(

)

* * * M = 0 = 6 R − 3 4 W + 3 W ∑ C A

R A* = 1.5W * M B* = 1.5W * × 3 = 4.5W * * C

*

M = −W × 3 = −3W

BMD

3W*

*

B

A

+ 4.5W*

C

D

Example 3 Segment ABC – moment gradient

F

3β m FL 16

Example 3 Segment ABC – effective length

Example 3 Segment ABC – elastic buckling capacity

Segment ABC – section capacity

Example 3 Segment ABC – slenderness reduction factor, αs

Example 3 Segment CD – elastic buckling capacity

Example 3 Segment CD – slenderness reduction factor, αs

Example 4 A simply supported beam with a span of 15m has a nominal central concentrated live load of 100 kN acting on the top flange. The beam is restrained against lateral displacement and twist only at the ends, and is free to rotate in plan. Design a suitable WB in accordance with AS4100 of Grade 300 steel.

150kN

15m

Example 4 Assume fyf = 300 MPa, compact section

α m = 1.35 Guess α s = 0.25 M sx ≥ 562.5 × 10 6 / (0.9 × 1.35 × 0.25) = 1851.9 kNm Sx ≥ 1851.9 × 10 6 / 300 = 6172.8 × 103 kNm Try a 800WB192 b f = 300 mm

A g = 24400 mm 2

t f = 28 mm

S x = 8060 × 10 3 mm 3

d = 816 mm

I y = 126 × 10 6 mm 4

t w = 10 mm

J = 4420 × 10 3 mm 4 9

I w = 19600 × 10 mm

6

fyf = 280 MPa

Example 4

Example 4

DESIGN SHEAR CAPACITY

Lecture Outline Strength g limit state Local buckling Intermediate transverse stiffeners YIELD limit li it state t t BUCKLING limit state

Combined shear and bending L d bearing Load b i stiffeners tiff YIELD limit state BUCKLING limit state

Strength limit state shear capacity Design equation for shear strength t th

(φ Vv ) ≥ V * D i Design capacity it ≥ factored f t d strength t th limit li it state shear

Local buckling in shear The WEB off a steel Th t l member b resists i t th the SHEAR STRESSES. Local buckling in shear may restrict the SHEAR CAPACITY of a BEAM. The design equation is:

V ≤ φVv *

φ = 0.9

Vv = nominal

shear h capacity it

Local buckling in shear Consider firstly y when the shear stress in the web is APPROXIMATELY UNIFORM.

shear stress

Parabolic P b li but b approximately uniform

d dw tw

*

τ

V = (d w t w )

For this case DEFINE Vv = Vu

(u – uniform)

Local buckling in shear Equation τ ol

π E = k 12 (1 − ν 2

2

)

⎛ tw ⎜⎜ ⎝ dw

⎞ ⎟⎟ ⎠

2

can also be used for local buckling with shear stresses.

Example p 1 Unsitffened web yielding in shear before buckling locally. If the web is to yield before buckling locally then:

fy

τy =

3

≤ τ ol

For a long web, k = 5.35 and using E = 200 x 103 MPa, ν = 0.3 produces 2

f yw 3

≤ 5.35

π × 200 × 10 ⎛ t w ⎞

dw which rearranges to tw

2

3

(

12 1 − 0.3

f yw 250

2

)

≤ 81.9

⎜⎜ ⎟⎟ ⎝ dw ⎠

Example p 1 The yield capacity is then

Hence in AS4100 if

Aw

f yw

dw < tw

3

82 f yw 250

The web yields before buckling and where

Vw = 0.6 Aw f yw

For universal sections Aw = dtw For welded sections Aw = dwtw

≈ 0.58 Aw f yw

Vu = Vw

Example p 1 when

d w t w > 82 /

local buckling capacity

dw and so when ≥ tw

f yw 250 the capacity equals the

τ ol × Aw 82 f yw / 250

th webb buckles the b kl before b f it yields i ld andd Vu = α vVw where Vw = 0.6 Aw f yyw and

α

v

⎧ ⎪ 82 ⎪ = ⎨ f yw ⎪ dw ⎪ t 250 ⎩ w

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

2

Example p 1 N.B. g , it is common to add ¾ When bucklingg governs, vertical stiffeners as Vu drops off rapidly as 1 (d w t w )2 ¾The provision of these vertical stiffeners will not i increase the th capacity it when h yielding i ldi governs, i.e. i when 82

d w tw
> d (no stiffeners)

82 αv = d w f yw t w 250

⎧ Hence ⎪ 82 ⎪ Vu = ⎨ ⎪ d w f yw ⎪t ⎩ w 250

1 αv 1− αd → +1 →1 1.15αv ⋅ ∞

2

⎫ ⎪ ⎪ ⎬ × 1 × Vw ⎪ ⎪ ⎭

which is the same as for an unstiffened web.

Example p 2 Calculate the design shear capacity for a 200x2 web plate if stiffeners are placed at 400 mm intervals.

Example p 2

Example p 3 What stiffener spacing in the 200x2 web plate is needed to resist a design shear V* = 56kN?

Example p 3

This design Thi d i equation ti agrees with ith the th previous i design d i check h k and d illustrates ill t t the bilinear interpolation in the (αv αd) table.

Non-uniform shear stress distribution The shear may be nonuniform in a monosymmetric beam:

parabola

τ

2Vu Vu = ⎛ f vm 0.9 + ⎜⎜ ⎝ f va

⎞ ⎟⎟ ⎠

The h strength h Vu is defined d f d as earlier l for f uniform f shear h stress

fvm = maximum shear stress fva = average shear stress

For design:

V ≤ φVv φ = 0.9 *

Non-uniform shear stress distribution τvm = fvm = maximum shear stress

*

V Qmax = It w

area A at this location Q = A ⋅ y

y

τva = fva = average shear stress τ va = f va

V* = d wt w

f vm Qmax For this case: = ⋅ dw I f va

Proportioning intermediate transverse stiffeners Intermediate transverse stiffeners are used to PREVENT BUCKLING OF THE WEB IN SHEAR when subjected to a design shear force V*. Recall the design check is:

Vu = α vα d V w ≤ V w

αv and αd are calculated based on a stiffener spacing s.

V w = 0 .6 Aw f yw

The stiffener must be STIFF ENOUGH to prevent local buckling, as well as STRONG ENOUGH to carry the tension field component without failing. They must therefore be designed for STRENGTH and STIFFNESS.

Proportioning intermediate transverse stiffeners MININUM AREA: A minimum area is required for the stiffeners to carry the vertical force in the tension field action utilised in the use of αv.

The minimum area rule is: 2 ⎧⎪ s ( s / dw ) As ≥ 0 .5γAw (1 − α v )⎨ − 2 d w ( ) 1+ s / dw ⎪⎩

⎫⎪ ⎬ ⎪⎭

γ = 1.0 for a pair of stiffeners

=1 1.8 8 for a single stiffener comprising of an angle section = 2.4 for a single plate stiffener

αv = factor determined earlier = f ((d w / t w ) f yw / 250 , s / d w )

Proportioning intermediate transverse stiffeners MININUM STIFFNESS: Stiffeners must be stiff enough to enforce a node at the web/stiffener junction at local buckling

I s ≥ 0 .75 d w t w3 3 3 w w 2

1 .5 d t ≥ s

s / dw < s / dw ≥

web

2

2 stiffener

Iw is the second moment of area of the stiffener about the web centreline.

Buckling g capacity p y When intermediate stiffeners are used, used the capacity of the web is Vb = αvαdVw. Hence:

V ≤ φ ( Rsb + Vb ) *

φ = 0 .9

Rsb = buckling capacity determined in the same way as for a load bearing stiffener

Vb = αdαvVw as earlier, where Vw = 0.6Awfy as earlier

⎛ dw αv = f ⎜ ⎜ tw ⎝

s ⎞⎟ , 250 d w ⎟⎠ f yw

End p panel At the end of the beam the tension field must be “anchored” anchored .

sep

end panel – tension field action cannot be mobilised This can be achieved by choosing s in the design equation:

V * ≤ φ (α vα d )V w

with

α d = 1 .0

Connection of intermediate stiffener to web < 4tw

< 4tw

Stiffener/web connection must be capable of resisting a design shear force of:

t w2 0.008 f ys [kN / mm] bs tw = web thickness in mm bs = stiffener outstand in mm

fys in MPa (N/mm2)

Stiffener outstand As for a load bearing stiffener – local buckling of stiffener must not occur prior to attainment of the yield stress, i.e.

⎛ bs ⎞ ⎜⎜ ⎟⎟ ⎝ ts ⎠

f ys 250

≤ 15

Vertically – stiffened webs local b buckling ckling of the web eb in shear yes

YIELD LIMIT STATE

V * ≤ φVu

Vw = 0.6fyAw

αv ≤ 1

Vu = α vV w

yes

yes

END

no

proportion INTERMEDIATE (TRANSVERSE) stiffeners

V b = α vα d V w

(spacing)

V ≤ φVb

no

yes

decrease s

*

no

N.B αvαd includes s

BUCKLING LIMIT STATE

no

V * ≤ φVw

END

increase Aw

END

Combined shear and bending Sometimes Vv has to be reduced where high shear forces V* and high moments M* coincide. (for example, in a continuous beam where V* and M* are high at an internal support)

-

BMD

SFD

-

Combined shear and bending g Design capacity:

Vvm = Vv

V ≤ φVv m *

φ = 0.9

M ≤ 0.75(φM s ) *

when

⎧ 1.6M ⎫ Vvm = Vv ⎨2.2 − ⎬ φM s ⎭ ⎩

*

*

when

M 0.75 ≤ ≤ 1.0 φM s

MS is BENDING SECTION CAPACITY.

Combined shear and bending

1.0

Vvm/Vv

0.75

1.0

M*/φMs

Proportioning method – an alternative procedure If the flanges g have enough g capacity p y to resist the bending g moment,, the web may resist all of the shear force. Thus,

M ≤ φM f φ = 0.9 *

Mf = Cdf = Tdf = Af fyf df Af = area of flange df = distance di t b between t flange centroids DESIGN THE WEB ONLY FOR SHEAR

V ≤ φVv *

then

C = Af fyf

df

≡

T = Af fyf

Mf

Web crippling limit state Occurs due to localised yielding of the web near concentrated loads. Behavior is dominated by gross yielding over a small web region.

web b

Web crippling limit state “Crippling” Crippling is caused by YIELDING of the web due to concentrated loads (point loads and reactions). CRIPPLING IS PREVENTED BY LOAD BEARING STIFFENERS PLACED IN PAIRS ON EACH SIDE OF THE WEB AND WELDED TO THE FLANGE.

We have to design g for two limit states:

Stiffener p plates welded to web and flanges

• YIELD and • BUCKLING

web

Web crippling limit state “CONCENTRATED CONCENTRATED LOADS” LOADS are only an abstraction (structural idealisation) used in structural analysis

In reality, y, the loads are applied pp to the flange g byy a STIFF BEARING PLATE

P*

Stiff bearing plate

flange ELEVATION

web

Web crippling pp g limit state We will consider firstlyy the case when NO STIFFENERS are p present. If either the YIELD or BUCKLING limit state fails for the unstiffened web then stiffeners must be design for for. The bearing stress is assumed to disperse through the flange at 1:2.5 bearing plate

1

R* bbf

flange

R*

1 2.5

2.5

ELEVATION

web b END ELEVATION

Web crippling pp g – YIELD limit state For the YIELD LIMIT STATE:

R ≤ φRby *

φ = 0.9

Rby = 1.25bbf t w f y yielding bearing

The 1.25 factor accounts for the benign g triaxial stress state at the flange/web interface where h the th restraining t i i actions allow large stresses to be resisted so that the strength is taken as ((1.25fy)

Web crippling – BUCKLING limit state For the BUCKLING LIMIT STATE, S the h stress is dispersed as below to obtain bb:

bb

dw/2 1

SUPPORT REACTION CASE

1

2.5

R* POINT LOAD CASE

dw/2

1

The web is then considered as an “effective column” as below: effective column

R*

N tw

1 2.5

1

1

1 1

1

bb

2.5

dw/2

Le

bb

dw/2 N

Web crippling – BUCKLING limit state The BUCKLING LIMIT STATE is then:

R ≤ φRbb *

φ = 0.9

Rbb is the buckling capacity of a column bb wide by tw thick with a slenderness ratio Le/r = 3.5 (dw/tw) buckling bearing

r

≡

radius di off gyration ti

≡ I/A

Web crippling – BUCKLING limit state The BUCKLING capacity Rbb of the “column” column can be obtained from the following recipe: 1. Calculate the MODIFIED SLENDERNESS: λn =

Le r

fy

dw = 3.5 250 tw

fy 250

2. Use the column tables with λn and αb = 0.5 to determine αc (T6.3.3(3)) 3. The column capacity is then:

Rbb = α c ⋅ (t wbb ) ⋅ f y

If Rb = min [Rby, Rbb] the check is then:

R ≤ φRb φ = 0.9 *

Web crippling If either the YIELD or BUCKILNG limit states fail for an unstiffened web then PAIRS OF LOAD BEARING STIFFENERS will be required.

tw

bs bs

ts A-A

A

A

web

If the stiffeners are to be adequate they must again satisfy YIELD and BUCKLING limit states.

Web crippling N.B. • load l d bearing b i stiffeners iff resist i ““concentrated” d” lloads d • intermediate transverse stiffeners (which were considered earlier) resist local buckling of the web in shear.

The YIELD LIMIT STATE is:

R ≤ φRsy *

Rsy = Rby + As f ys

yield stiffened

Rby b - capacity of unstiffened web (as earlier) As - area of stiffener (2bsts) fys - yield stress of stiffener

φ = 0.9

Example 4 Check the adequacy of a pair of load bearing stiffeners 100 x 16 plates with fys = 350 MPa for the girder shown with a design load of 1600 kN applied through a stiff bearing of length 300 mm. Check YIELD only.

1500

300

25

10

Rby b = 1.25bbf t w f y 25

300

Rsy = Rby + As f ys

Web crippling pp g – BUCKLING limit state The BUCKLING LIMIT STATE is:

φ = 0.9

R ≤ φRsb *

stiffened

buckling

Rsb is the column buckling capacity of the following effective cross-section: cross section:

tw

web

ts

ls

bs ls

Eff ti cross-section Effective ti

bs 17.5t w or s / 2 f y / 250

s = spacing between stiffeners

(whichever is less)

Web crippling pp g – BUCKLING limit state The column buckling capacity Rsb can be obtained from the following recipe:

1. Determine the slenderness ratio:

Le r

Le = 0.7dw if the flanges are restrained by other structural members against twist rotation, i.e.

Le = 1.0 dw otherwise

members prevent twist

Web crippling pp g – BUCKLING limit state r ≡

I/A

For the effective section:

I ≈ (2 b s + t w ) × t s / 12 3

(

A = 2 b s t s + 2 × 17 . 5 t w /

)

f y / 250 t w

or

A = 2 b s t s + 2 × (s / 2 )t w

whichever is less

Web crippling pp g – BUCKLING limit state 2. Calculate the modified slenderness ratio:

Le λn = r

fy 250

3. Determine αc from tables using αb = 0.5 4. The BUCKLING STRENGTH is then:

Rsb = α c Af y

N.B. The load bearing stiffener must YIELD before it buckles locally it lf Th itself. Therefore, f

outstand of stiffener

bs ts

f

ys

250

≤ 15

yield stress of stiffener

Example p 5 Check the BUCKLING LIMIT STATE in Example 4. 4

1500

300

25

10

16 100

10

100 25 300

ls

ls EFFECTIVE STIFFENER/WEB CROSS SECTION CROSS-SECTION

Example 5

(Flanges unrestrained i against i twist) i

Example p 5

Clearly load bearing stiffeners design is a TRIAL & ERROR PROCESS requiring design checks.

Web crippling load bearing stiffeners resist “concentrated” loads INCLUDE STIFFENERS

WITHOUT STIFFENERS YIELD LIMIT STATE

BUCKLING LIMIT STATE

YIELD LIMIT STATE

BUCKLING LIMIT STATE

Rsy = Rby + As f ys

Rsb = α c Af y

Rbb = α c ⋅ (t wbb ) ⋅ f y

Rby = 1 .25bbf t w f y

R* ≤ φRb

Rb = min[Rbb, Rby] yes

END

no

R ≤ φRs *

Rs = min[Rsy, Rsb] yes

END

no

increase As

Steel Beam-Columns

Strength Limit State

Reference e e e ce Material ate a 9AS4100 – 1998 Steel Structures: Standards Australia, Sydney 9AS1170 1 & AS1170 9AS1170.1 AS1170.2 2L Loading di C Codes: d Standards Association of Australia, Sydney, 1998.

9NS Trahair & MA Bradford: The Behavior and Design of Steel Structures to AS4100, 3rd Australia edition, E&FN Spon, London, 1998. 9MA Bradford, RQ Bridge & NS Trahair: Worked Examples for Steel Structures, 3rd edition, AISC, Sydney, 1997. 9ST Woolock, S Kitipornchai & MA Bradford: Design of Portal Frame Buildings Buildings, 3rd edition, edition AISC, AISC Sydney, Sydney 1999. 1999

Lecture ectu e Out Outline e 9

Steel Beam-column 9

9

Strength limit state

Types of failure: in-plane failure 9 lateral-torsional l t l t i l buckling b kli

9

9

biaxial failure

BMD, Mx

Frames a es

x - direction

2-D frame BMD, My

y

x w

3 D frame 3-D y x

y - direction

w, UDL applied in both  x and y direction

In-plane p a e be behavior a o STEEL BEAM COLUMNS are subjected to COMBINED BENDING and COMPRESSION N M1

This is typical of a

B

column in a rigid frame: A

M2 N

Steel beam Beam (UDL -

Beam

major axis loading)

(couple)

Beam (torsion)

Beam (UDL minor axis loading)

Column (axial compression)

Beam-column (axial compression + transverse loading)

In--plane In p a e be behavior a o There are three STRENGTH LIMIT STATES that may have to be considered:

(i) IN IN-PLANE PLANE FAILURE when the member is loaded about its major axis with full lateral support so that lateral beam buckling cannot occur (or column buckling cannot occur around the minor axis) or when the column is bent about its minor axis

(ii) FLEXURAL-TORSIONAL BUCKLING when the member is bent about its major j axis and there is insufficient lateral support pp

(iii) BIAXIAL FAILURE when the member is loaded about both principal axes

In--plane In p a e be behavior a o

X (u)

Y (v)

Out--ofOut of-plane p buckling g

φ

φ X (u)

or Lateral–Torsional Buckling or Flexural–Torsional Buckling

Y (v)

In--plane In p a e be behavior a o β mM

M N

N Lateral restraints

−1 ≤ βm ≤ 1

For a column with an imperfection δo the moment under the application of an axial force N at mid-height mid height is:

N ×δ o 1− N NE [P-δ effect (second order effects)]

N E = π EI / L 2

2

Euler buckling load

In--plane behavior In R ll Recall, e0

N

P-δ effect

N M M

z

δ

Nδ

δ L M [1-(1+β) z/L]

βM N

eL

M

eccentrically applied axial load results in double curvature bending

FIRST ORDER EFFECTS N SECOND ORDER EFFECTS

In--plane In p a e be behavior a o N

N

δ ο

The first-order moment Nδo is therefore amplified by:

In a beam-column, the imperfection may be thought of as being caused by the moments M and βmM at the ends

N

M

β mM

N

In--plane In p a e be behavior a o Thee deflected de ected sshape ape of o the t e beam-column bea co u iss ggiven ve by ((Trahair andd Bradford, d o d, 1998 998 – Chapter 7):

v=

M [cos μz − (β m cos ecμL + cot μL )sin μz − 1 + (1 + β m ) z L] N where

when

N π2 N μ = = 2 EI L N E 2

N β < − cos π N E the maximum moment is:

2 ⎧⎪ ⎛ ⎫⎪ ⎞ N N ⎟ ⎬ + cot π M m = M ⎨1 + ⎜⎜ β m cos ecπ NE N E ⎟⎠ ⎪ ⎪⎩ ⎝ ⎭

and when

β ≥ − cos π

N NE

the maximum moment is (i.e. at the end of the member)

In--plane In p a e be behavior a o These equations are clearly cumbersome cumbersome, and the maximum moment is approximated by:

M max

cm =M ≥M 1− N NE

where h cm = 0.6-0.4 0 6 0 4βm

The maximum stress in the beam-column is:

σ max

N M max = + A Z

Axial stresses

Bending stresses

In--plane In p a e be behavior a o If the beam beam-column column reaches its maximum strength when σmax = fy then:

σ max

N M max = fy = + A Z

cm N M ⎛ ⎜⎜ + 1= NY M Y ⎝ 1 − N N E

/fy ⎞ ⎟⎟ ⎠

from

N M max 1= + Af y Zf y M max

or

cm =M ≥M 1− N NE

where In AS4100: NY is replaced by the strength Ns (=kf Ag fy)

NY = Afy is the squash load MY = Zfy is the first yield moment

MY is replaced by the bending strength Ms = Ze fy Effective modulus

so

In--plane In p a e be behavior a o D i equation: Design ti

The DESIGN MOMENT M* is usually obtained from a secondorder elastic frame program (i.e. Microstran or Spacegass)

cm which has the amplification p factor build into it. 1 − N NE Alt Alternatively, ti l the th maximum i momentt M*max can be b obtained bt i d ffrom a fi firstt order (linear) analysis and then amplified to produce M*.

In--plane In p a e be behavior a o AS4100 thus requires the design actions N* and M* to satisfy:

The strength φMs(1-N*/φNs) is called φ Mr .

* ⎛ ⎞ N * ⎟⎟ M ≤ φM S ⎜⎜1 − ⎝ φNS ⎠

Reduced from Ms due to axial force.

φN S

[N.B. for a R-C cross-section]:

N*

φNo N * φM S

strength p envelope

M*

strength envelope

M*

φM o

Example a pe1 Determine D t i the th d design i major j axis i section ti momentt capacity it φMrx of a 200UC52.2 of Grade 300 steel which has a design axial compressive force of N* = 143.9 kN 204

f y = 300 MPa

12.5 2206

8.0

A = 6660mm2 S x = 570 ×103 mm3

12.5

Plastic section modulus

Example a pe1 For compression:

F bending: For b di

Example a pe1 For combined actions:

Plastic ast c capac capacity ty The equation:

(

M ≤ φM S 1− N φN S *

*

is based on first yield at

)

If the cross cross-section section is COMPACT (i.e. local buckling is not a problem and λe < λep) then the stress distribution at failure is:

f il failure, i.e. i PLASTIC BEHAVIOUR

fy Stress reaches first yyield fy

M*

N* ELASTIC BEHAVIOUR

M*

N*

fy

Example a pe2 Pl ti analysis Plastic l i off a rectangular t l cross-section ti f y = 300MPa

b

M*

N* N

Geometric centroid

fy C d

dn > d / 2

A

T fy Plastic neutral axis

C = bxdnxfy T = bx((d-dn)xfy

Example a pe2 N = C − T = bd n f y − bdf y + bd n f y = − N S + 2d nbf y d ⎛ d dn ⎞ M = C ×⎜ − ⎟ +T × 2 ⎝2 2 ⎠

These equations may be solved for dn, and using

N S = bdf y bd 2 MS = fy 4 produces at plastic failure ⎛ N ⎜⎜ ⎝ NS

2

⎞ M ⎟⎟ + =1 ⎠ MS

In--plane In p a e be behavior a o ⎛ N M + ⎜⎜ φM S ⎝ φN S *

1.0

*

N* φN S

2

⎞ ⎟⎟ = 1 ⎠

M* N* + =1 φM S φN S M * φM S

1.0

The linear interaction may therefore often be too conservative. In AS4100, if the crosssection is COMPACT and EFFECTIVE (kf = 1) and doubly symmetric then

M

*

≤ φM

rx

= φM

S

⎛ N* × 1 . 18 ⎜⎜ 1 − φN S ⎝

⎞ ⎟⎟ ≤ φ M ⎠

S

φ = 0 .9

Example a pe3 The 200UC52 Th 200UC52.2 2 considered id d earlier li iis d doubly-symmetric bl t i compact (λe < λep for bending) and effective (λe < λey for compression). Hence φM rx = 1.18 × 153.9(1 − 143.9 1798) = 167.1kNm

204

* ⎛ N M * ≤ φM rx = φM S × 1.18⎜⎜1 − ⎝ φN S

12.5

12.5

2206

8.0

⎞ ⎟⎟ ≤ φM S ⎠

but φM S = 153.9kNm < 167.1kNm The section is thus unaffected by axial compression and fails plastically at

φM S = 153.9kNm

In--plane In p a e be behavior a o The equation:

M ≤ φM rx *

is a check on the strength of the cross-section under bending (M*) and d compression i (N*) If th the b beam-column l is i too t slender l d (i.e. (i Le/r / is i too t great) t) it may fail f il by column buckling at NC rather than yield at NS. N.B. We have considered it to be laterally restrained, so it cannot buckle laterally at Mb < MS. We will check this later.

In--plane In p a e be behavior a o M x ≤ φM ix *

Hence in AS4100:

⎛ N* M ix = φM sx ⎜⎜1 − ⎝ φN C In-plane p strength

⎞ ⎟⎟ ⎠

bent about x-axis

φ N C = φ (α C k f Af y ) is the column strength

N.B. We have assumed the b beam-column l fails f il in-plane i l by b column buckling about the xaxis. It cannot fail by buckling about the y-axis because it is laterally restrained. NC must be determined therefore from λnx

Effective length g factor For column design or checking we generally use the effective length Le (although the above example has shown how λn can be determined from the elastic bucklingg load determined byy a computer p program p g or from charts in text books).

Le = ke L L is the column length; ke is i the h EFFECTIVE LENGTH FACTOR Some standard cases for isolated columns are given below (Trahair & Bradford, 1998):

Effective length g factor l =L

l = L/ 2

l = 0.7L

NE

NE

l = L/ 2

l = 2L

NE

NE

NE

l L

L

l

L

l

L L

l

l

l NE =

Theoretical

ke AS4100

ke

π 2EI

2π 2EI NE = 2 L

L2

4π 2EI NE = 2 L

1.0

0.5

0.7

0.5

2.0

1.0

0.5

0.85

0.7

2.2

4π 2EI NE = 2 L

NE =

π 2EI 4L2

Example a pe4 Check Ch k th the 200UC52 200UC52.2 2 considered id d earlier li if it iis subjected bj t d tto design moment of 121.5 kNm and 124.7 kNm as well as an axial compression of 143.9 kN. The effective length Le about the x-axis is 7.0 m. f y = 300MPa 204

rx = 89.1mm

12 5 12.5

12.5

206

8.0

124.7 kNm 143.9 kN

121.5 kNm 143.9 kN

Example a pe4

⎛ N* M ix = φM sx ⎜⎜1 − ⎝ φN C

⎞ ⎟⎟ ⎠

In--plane In p a e be behavior a o The linear interaction equation is essentially based on first yield at fy and may in some cases be too conservative. Thus for compact doubly symmetric sections (with kf = 1), AS4100 uses: 3 * ⎡ ⎧ ⎫ ⎛ 1 β + N ⎪ ⎛ * m ⎞ ⎪ M ≤ φM ix = φM sx ⎢⎨1 − ⎜ ⎟ ⎬⎜⎜1 − ⎢⎣⎪⎩ ⎝ 2 ⎠ ⎪⎭⎝ φN C

⎛1+ βm ⎞ + 1.18⎜ ⎟ ⎝ 2 ⎠ ≤ φM rx

3

N* 1− φN C

⎞⎤ ⎟⎟⎥ ⎠⎥⎦

Example a pe5 For the 200UC52.2 in the previous example 204

f y = 300MPa

12.5

rx = 89.1mm 206

80 8.0

124.7 kNm

121.5 kNm

12 5 12.5

143.9 kN

143.9 kN

Example a pe5

In--plane In p a e be behavior a o If the cross-section is bent

about the y-axis then in AS4100:

M ≤ φM ryy φ = 0.9 * y

* ⎡ N ⎤ M ry = M sy ⎢1 − ⎥ ⎣ φN S ⎦ Section capacity about y-axis

and out-of-plane buckling CANNOT occur.

In--plane In p a e be behavior a o Further if the cross-section is doubly symmetric and compact, the less conservative equation below may be used:

M ≤ φM ry * y

φ = 0.9

⎡ ⎛ N M ryy = 1.19M syy ⎢1 − ⎜⎜ ⎢⎣ ⎝ φN S *

⎞ ⎟⎟ ⎠

2

⎤ ⎥ ≤ M syy ⎥⎦

Section capacity about y-axis

Example a pe6 Determine D t i the th d design i minor i axis i section ti momentt capacity it φMry of a 200UC52.2 of Grade 300 steel which has a design axial compressive force of N* = 143.9 kN S y = 264 ×103 mm3

204

Z y = 174 ×103 mm3

12.5

b 206

80 8.0

N.B. for a rectangular section

S = bd 4 Z = bd / 6 2

12 5 12.5

d

2

So S = 1.5Z and the code requires that

Z e ≤ 1.5Z

Example a pe6

Out--o Out of--p of plane a e be behavior a o A beam beam-column column bent about its principal (x) axis may buckle laterally and twist, if there is insufficient lateral support, at a load which is significantly less than the maximum load predicted by an in-plane analysis. This flexural-torsional buckling may occur while the member is still elastic, or after some yielding due to in-plane in plane bending and compression has occurred occurred.

Out-of-plane behaviour In-plane behaviour Flexural-torsional bucking of beam-columns

Out--ofOut of-plane p buckling g

φ

φ X (u)

or Lateral–Torsional Buckling or Flexural–Torsional Buckling

Y (v)

Out--o Out of--p of plane a e be behavior a o Consider an elastic beam-column in uniform bending:

Moc

Moc Noc

Noc L SIMPLY SUPPORTED

Out--o Out of--p of plane a e be behavior a o The combination of moment Moc and compression Noc to cause lateral buckling (i.e. buckling out of the plane of loading) is derived in Chapter 7 of Trahair and Bradford (1998) as

⎛ M oc ⎜⎜ ⎝ Mo

⎛ ⎞ N oc ⎜ ⎟⎟ = 1 − ⎜ Ny ⎠ ⎝ 2

⎞⎛ ⎟ ⎜ 1 − N oc ⎟⎜ Nz ⎠⎝

⎛ π EI y ⎞ ⎛ ⎜ ⎟ ⎜ GJ + π EI w 2 ⎜ L2 ⎟ ⎜ L ⎝ ⎠⎝ 2

where

GJ Nz = 2 ro

Mo =

⎞ ⎟⎟ ⎠

⎛ π 2 EI w ⎜⎜ 1 + 2 GJL ⎝

⎞ ⎟⎟ ⎠

2

is the Euler buckling load

⎞ ⎟⎟ ⎠

Ny =

π 2 EI y L2

is the torsional buckling load (i.e. the load to buckle the member by twisting about its longitudinal axis, axis and ro2 =

Ix + Iy A

, ro = polar radius of gyration)

Out--o Out of--p of plane a e be behavior a o Equation

⎛ M oc ⎜⎜ ⎝ Mo

⎛ ⎞ N oc ⎜ ⎟⎟ = 1 − ⎜ Ny ⎠ ⎝

⎞⎛ ⎟ ⎜ 1 − N oc ⎟⎜ Nz ⎠⎝

2

⎞ ⎟⎟ ⎠

does not account for the amplification of Mc due to Ncxδ, where δ is the mid1 span deflection. We have seen that the amplification is 1− N N X where

N x = π 2 EI x L2

Hence equation

⎛ M oc ⎜⎜ ⎝ Mo

⎛ M oc ⎜⎜ ⎝ Mo

2

as the member is bent about its x-axis

⎛ ⎞ N ⎟⎟ = ⎜ 1 − oc ⎜ Ny ⎠ ⎝ 2

⎞ ⎛ N ⎟⎟ = ⎜⎜ 1 − oc Nx ⎠ ⎝

⎞ ⎟⎟ ⎠

2

⎞⎛ ⎟ ⎜ 1 − N oc ⎟⎜ Nz ⎠⎝

⎞ ⎟⎟ can be replaced by: ⎠

⎛ ⎜ 1 − N oc ⎜ Ny ⎝

⎞⎛ ⎟ ⎜ 1 − N oc ⎟⎜ Nz ⎠⎝

⎞ ⎟⎟ ⎠

Out--o Out of--p of plane a e be behavior a o For most hot-rolled sections Noc 8.0) tt = throat th t thi thickness k

φ = 0 .8 SP

=t/ 2

fuw = ultimate tensile strength g of weld = 480 MPa for E48XX electrodes (most common) = 410 MPa for E41XX electrodes

(high degree of inspection) GP – general purpose (low degree of inspection)

Weld stress trajectories j

Weld symbols

Weld symbols

Example 1 Determine the maximum shear per unit length in the uniform thickens weld group caused by a design shear force of 160kN through g the centroid of the bolt ggroup p and a moment of 20kNm N about the centroid of the bolt group. 25 55 140 30 75 65 35 70 70 35

Member design actions at bolt group centroid Shear = +160kN Moment = +20kNm

8-M20 8.8/S bolts. Th d iin shear Threads h plane. l 2x280mm E48XX fillet welds. Single web plate. This problem demonstrates the in-plane elastic analysis of a fillet weld group under combined shear and bending.

Example 1 The centroid of the weld group is (70 + 30 + 25 + 37.5) = 162.5 mm from the centre of the bolt group.

V * = −160 kN, M * = 20 + 160 × 162.5 / 1000 = 46.0 kNm

{

( I x + I y ) / t = 2 × 280 3 / 12 + 2 × 280 × (75 / 2 )

2

}

= 4 . 446 × 10 6 mm 3 A / t = 2 × 280 = 560 mm

(

x c = −V * (I x + I y ) / M * A

)

(

= 160 × 10 3 × 4 .446 × 10 6 × t / 46 .0 × 10 6 × 560 × t = 27 .6 mm

rmax =

(75 / 2 + 27 .6 )

= 154 .4 mm

2

+ 140

2

)

yc = 0

vw* = M *rmax t / (I x + I y ) = 46.0 × 10 6 × 154.4 / 4.446 × 10 6 = 1.597 kN/mm

Example 2 Determine the weld leg size required for the equal leg fillet weld group, if the weld category is SP and the electrode is E48XX. 25 55 140 30 75 65 35 70 70 35

Member design actions at bolt group centroid Shear = +160kN Moment = +20kNm 8-M20 8.8/S bolts. Threads in shear p plane. 2x280mm E48XX fillet welds. Single web plate.

This problem illustrates the design of a fillet weld group

Example 2 f uw = 480 MPa

k r = 1.0

φ = 0.8 vw* = 1.597 kN / mm 1.597 × 103 ≤ 0.8 × 0.6 × 480 × tt × 1.0

tt ≥ 6.93mm t ≥ 6.93 × 2 = 9.8mm

Use 10x10 SP E48XX weld.

N.B. A smaller weld could be used if the weld group dimensions were increased increased. This would require the reanalysis of the weld group.

Example 3 An 8mmx8mm SP fillet weld from E48XX electrodes has a longitudinal design shear per unit length of vwL* = 1.0kN/mm and transverse design shears per unit length of vwx* = 0.6kN/mm 0 6kN/mm and vwy* = 0.4 kN/mm. Check the adequacy of the weld. This problem illustrates the checking of a fillet weld under combined loadings. loadings

vw* =

(1.0

2

)

+ 0.6 2 + 0.4 2 = 1.233kN / mm

f uw = 480 MPa tt = 8 / 2 = 5.66mm k r = 1.0 φ = 0.8 φvw = 0.8 × 0.6 × 480 × 5.66 × 1.0 = 1.303kN / mm > 1.233kN / mm = vw* Therefore OK.

Example 4 Box section fillet welded to end plate (Fillet loaded out-of plane) y 1

2 3

8

305

450 kN

weld group cetroid

x

90 kNm 4

7 5

203

6

Design actions: Px* = 0 , Py* = − 450 kN , Pz* = 0 M

* x

= 90 kNm kN , M

* y

= 0, M

* z

=0

Weld g group p properties: p p

L w = 2 (305 + 203 ) = 1016 mm

Example 4 If it is i assumedd that th t the th vertical ti l shear h is i primarily i il taken t k by b the th webs b of the box section, then this vertical shear must be assumed to be transferred through the vertical fillet weld only. Hence, L w = 2 × 305 = 610 mm d = 305 mm b = 203 mm

I wx = d

3

6 + bd / 2 = 14 . 2 × 10 mm 2

6

at ppoints 1,, 2,, 3,, 8

y = 152.5 mm

4, 5, 6, 7

y = − 152 . 5 mm

3

Example 4 Global set of design actions per unit length v*x = 0

v*z = (90000) × ( ± 152.5)/14.2×106 v*y = - 450/610 = + 0.967 at points1, 2, 3, 8 ( y = + 152.5) = - 0.738 at points 3, 4, 7, 8 =0 at points1, 2, 5, 6 = - 0.967 at points4, 5, 6, 7 (y = -152.5)

R l Resultant force f per unit i length: l h v w* =

(− 0 . 738 )2

+ (± 0 . 967

)2

= 1 . 22 kN/mm

Weld capacity:

8 mm E48XX fillet weld

φ v w = φ 0.6 f uw t t k r = 1 .30 kN/mm > v w*

φ = 0.8 fuw = 480MPa t w = 8 tt = 8 2 = 5.66 kr = 1

Example 5 Fillet welded bracket loaded In-Plane y 175

275

180 kN

d = 300 mm b = 275 mm

3 2

1 Weld group centroid

300

x 4 5

weld centroid x

b2 x = = 89 . 0 mm 2b + d

6

design actions Px* = 0 Py* = − 180 kN Pz* = 0

critical points

M

* x

=0 M

M

* z

= − 180 × (275 + 175 − 89 . 0 )

* y

=0

= − 64980 kNmm

Example 5 Weld group properties: Lw = 2× 275+ 300= 850mm

assume Lwx = Lwy = Lwz = Lw = 850mm I wp = I wx + I wy 3002 (6× 275+ 300) 2753 (275+ 2× 300) I wp = + 12 3(2× 275+ 300) = 21.8×106 mm3 at points 1, 6 :

x = 275− 89.0 = +186 y = ± 300/ 2 = ±150

at points 2, 3, 4, 5 :

x = − 89.0 y = ±150

Example 5 Global design actions per unit length: * − M − 64980×150 zy v*x = =− I wp 21.8×106

= + 0.447 at points 1, 2, 3 ( y = +150) = − 0.447 44 at points i 4, 5, 6 ( y = −150 1 0) * M x −180 − 64980×186 v*y = + z = + Lwy I wp 850 21.8×106

Fy*

= −0.767 at points 1, 6 (critical) −180 − 64980× (− 89.0) = + 850 21.8×106 = +0.054 at points 2, 3, 4, 5 (not critical)

Resultant R lt t fforce per unit it length: l th points 1, 6

vw* =

(± 0.447)2 + (− 0.767)2

= 0.888 kN/mm 6 mm E48XX fillet weld φ v w = φ 0.6 f uw t t k r = 0 . 978 kN/mm > v w* ∴ OK

φ = 0.8 fuw = 480MPa t w = 6 t t = 6 2 = 4.24

kr = 1

Reference material NS Trahair & MA Bradford: The Behaviour and Design of Steel Structures to AS4100, 3rd Australian edition, E&FN Spon, London, 1998. ST Woolcock, S Kitipornchai & MA Bradford: Design of Portal F Frame Buildings, B ildi 3rd edition, diti AISC, AISC S Sydney, d 1999 1999. TJ Hogan & IR Thomas: Design of structural connections, 4th edition, AISC, Sydney, 1994.