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: Axial force N* Bending moment M* Shear 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 <
f yw 250
(they only increase the buckling capacity)
Vertically – stiffened webs If a web has vertical stiffeners,, its strength g is increased markedlyy because: • •
the elastic local buckling coefficient is increased a benign tension field action develops similar to a truss action
Of course this only applies when
stiffener
Vu < Vw
flange
web
dw
s
stiffener
Verticallyy – stiffened webs Vu = α vα d V w ≤ V w
The nominal capacity is:
αv
⎧ ⎪ 82 ⎪ = ⎨ fy ⎪ dw ⎪ t 250 ⎩ w
αv
2
⎫ ⎪ ⎪ 2 ( ) 0 . 75 d / s +1 ⎬ w ⎪ ⎪ when ⎭
[
⎧ ⎪ 82 ⎪ = ⎨ fy ⎪ dw ⎪ t 250 ⎩ w
]
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
s ≥ dw
2
[(d
/ s ) + 0 . 75 2
w
]
when
s < dw
Verticallyy – stiffened webs αd =
1−αv
1 .15α v 1 + (s / d w )
2
+1
αv reflects the local buckling contribution.
Under shear, the local buckling coefficient is approximately:
k = 5 .35 + 4 (d w s )
s ≥ dw
k = 5 .35 (d w s ) + 4
s < dw
2
2
Verticallyy – stiffened webs αd reflects the TENSION FIELD contribution. M
stiffener
stiffener web
flange
Verticallyy – stiffened webs The stiffened web p panel acts like a truss,, but the compression p region g of the thin web plate is very slender and unable to resist much compressive stress.
Check the equations:
(d w / s )
2
→0
so
Let s >> 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<
(1 − N Equation
⎛ M oc ⎜⎜ ⎝ Mo
⎛ ⎞ N ⎜ ⎟⎟ = 1 − oc ⎜ Ny ⎠ ⎝ 2
N z ) > (1 − N N y )(1 − N N x )
⎞⎛ ⎟ ⎜ 1 − N oc ⎟⎜ Nz ⎠⎝
⎞ ⎟⎟ can then be approximated by: ⎠
N oc M oc 1 + =1 (1 − N c N x ) M o Ny This equation is the basis of the AS4100 design rule. rule
Out--o Out of--p of plane a e be behavior a o Since
M oc 1 − N oc N x corresponds to the amplified moment (that is, the second order M* from a stiffness package), this equation becomes:
or
M ≤ φM ox * x
out of plane
N* M* + ≤1 φN c φM b
φM ox = φM b (1 − N φN c ) *
Example a pe7 Check the out out-of-plane of plane member capacity of the 200UC52.2 200UC52 2 beambeam column of Grade 300 steel considered previously with end moments of 121.5 kNm and 124.7 kNm that bend it into reverse curvature if the effective ff ti lengths l th for f column l and d beam b lateral l t l buckling b kli are both b th Le = 7000 mm.
A = 6660mm 2 I x = 52.8 × 10 6 mm 4 I y = 17.7 ×106 mm 4 ry = 51.5mm J = 325 × 103 mm 4 I w = 166 × 109 mm 6
204 12.5
12.5
206
8.0
Example a pe7
Example a pe7
Example a pe7
Biaxial a a be bending d g More o e generally ge e y a beam-column be co u iss bent be about bou BOTH O major jo axes es by M*x and d M*y , as well as being subjected to a compressive force N*. Thi situation This it ti is i complex l to t model d l accurately. t l AS4100 requires i a CROSS CROSSSECTION CHECK to be made according to: or for compact doubly-symmetric I-sections
*
* x
M
* y
M N + + ≤1 φ N S φ M Sx φ M Sy
according to the less conservative:
⎛ M ⎜ ⎜ φM prx ⎝ * x
⎞ ⎟ ⎟ ⎠
γs
⎛ M +⎜ ⎜ φM pry ⎝ * y
⎞ ⎟ ⎟ ⎠
γs
≤1
M
prx
where: *
N ≤ 2..0 γ s = 1. 4 + φN s
M pry
⎛ N* = 1 . 18 M sx ⎜⎜ 1 − φN S ⎝
⎞ ⎟⎟ ≤ M sx ⎠
⎛ ⎛ N* = 1.19 M sy ⎜1 − ⎜⎜ ⎜ ⎝ φN S ⎝
⎞ ⎟⎟ ⎠
2
⎞ ⎟≤M sy ⎟ ⎠
Biaxial a a be bending d g AS4100 also requires a MEMBER CHECK to be made according to:
⎛ M ⎜⎜ ⎝ φ M brx * x
⎞ ⎟⎟ ⎠
γm
⎛ M +⎜ ⎜ φM bry ⎝
M brx = min[M ix , M ox ]
* y
⎞ ⎟ ⎟ ⎠
γm
M bry = M iy
≤1
γ m = 1 .4
Example a pe8 Determine the maximum design value M* of a 200UC52 200UC52.22 beam-column beam column of Grade 300 steel which has a design axial compressive force of N*= 143.9kN, major axis (x-axis) end moments of M* and 0.974M* causing reverse curvature t b bending, di and d minor i axis i (y-axis) ( i ) end d moments t off 0.4 04 M* and 0.4M* causing single curvature bending. Le = 7m for column and beam buckling. 204 12.5
Mx x
206
8.0
I x = 52.8 × 10 6 mm 4
ry = 51.5mm
I y = 17.7 ×106 mm 4
J = 325 ×10 mm 3
12.5
y
A = 6660mm 2
My
4
I w = 166 × 109 mm 6
Example a pe8
Example a pe8
Frames
Lecture ectu e Out Outline e 9 9
C l Column: 9
elastic buckling load
9
effective length factor
Frames: with simple joints 9 with semi-rigid joints 9 with rigid joints 9
Types:
Analysis Method:
9
First order (linear, no P-δ effects)
9
Second order (nonlinear, P-δ effects)
9
Elastic Buckling Analysis (braced and sway frames)
Effective length g factor – isolated column For column design or checking we generally use the effective length Le
Le = ke L L is the column length; ke is i the h EFFECTIVE LENGTH FACTOR Some standard cases are given below (Trahair & Bradford, 1998):
Frame Idealisation STRUCTURAL FRAMES are composed p of one dimensional members connected together in skeletal arrangements which transfer the applied loads to the supports.
Reduction of 3-D framework to plane frames
Two-dimensional flexural frames TWO DIMENSIONAL FRAMES – a number of independent two-dimensional frames with in-plane loading.
THE STRUCTURAL BEHAVIOUR of a flexural frame is influenced by the behaviour of the member joints, which are usually considered to be either SIMPLE, SEMI SEMI-RIGID RIGID or RIGID, according to their ability to transmit moment.
Member joints Force & Moment connection
Force connection
Moment connection
Frames a es While most FRAMES are three-dimensional, they may often be considered as a series of parallel two-dimensional two dimensional frames, frames or as two perpendicular series of two-dimensional frames.
THE BEHAVIOUR of a structural frame depends on its arrangement and loading, and on the type of connections used.
Frames The members usually have substantial BENDING ACTIONS The members usually have substantial BENDING ACTIONS, and if they also have significant axial forces, then they must be designed as beam‐ties or beam‐columns. I f In frames with SIMPLE CONNECTIONS, the moments ith SIMPLE CONNECTIONS th t transmitted t itt d by the connections are small, and often can be neglected, and the members can be treated as isolated beams, or eccentrically loaded beam‐ties or beam‐columns. However, when the connections are SEMI‐RIGID or RIGID, there are important moment interactions between the members.
Frames with simple joints No moment is transmitted through a SIMPLE JOINT,, the members connected to the jjoint mayy rotate. If there are sufficient number of pin-joints to make the structure statically determinate then each member will act independently of the others, determinate, others and may be designed as: an isolated tension member, compression member, beam
or
beam-column. (if the th pin-jointed i j i t d structure t t i indeterminate, is i d t i t then th some partt off it may actt as a rigid-jointed frame)
Frames with simple joints One of the most common methods of designing frames with
SIMPLE JOINTS is often used for rectangular frames with vertical (column) and horizontal (beam) members under the action of vertical loads only. Th columns The l i suchh a frame in f are assumedd to t actt as if eccentrically t i ll loaded. (it should be noted that such a pin-jointed frame is usually incapable of resisting transverse forces, and must therefore be provided with an independent bracing or shear h wall ll system) t )
Frames with semi semi-rigid rigid joints SEMI-RIGID JOINTS are those which have d dependable d bl momentt capacities iti andd which hi h partially ti ll restrain the relative rotations of the members at the joints.
The action of these joints in rectangular frames is to reduce the maximum moments in the beams, beams and so the semi-rigid design method offers potential economies over the simple design method.
Frames with rigid joints Behaviour: RIGID JOINT – a joint j i which hi h has h sufficient ffi i rigidity i idi to virtually i ll prevent relative rotation between the members connected. Properly arranged welded and high-strength friction grip bolted joints are usually assumed to be rigid. There are important interactions between the members of frames with rigid joints, which are generally stiffer and stronger than frames with simple or semirigid joints. Therefore rigid frames offer significant economies.
Frames with rigid joints Behaviour: Although a RIGID
FRAME may behave in an approximately linear fashion
while its service loads are not exceeded, especially when the axial forces are small, it becomes non-linear near its in-plane ultimate load because of yielding and buckling effects.
When the axial compression forces are small, failure occurs when a sufficient number of plastic hinges have developed to cause the frame to form a collapse mechanism, in which case the load capacity of the frame can be determined by
plastic analysis of the collapse mechanism.
Frames with rigid joints g j Methods of Analysis: 9 First First‐order elastic analysis order elastic analysis 9 Second order elastic analysis 9 Advanced analysis
Stress resultants (M, N & V) ( , ) and deflections Equilibrium Stress‐strain Stress strain Compatibility
9 First First‐order plastic analysis order plastic analysis Collapse Load (Equilibrium, Mechanism)
Stability Buckling loads Effective lengths
9 Elastic buckling of braced frames 9 Elastic buckling of unbraced (sway) frames
First-order First order elastic analysis Assumptions: 9
Material behaves linearly (yielding effects are ignored);
9
The members behave linearly (no P-δ
9
The frame behaves linearly (no frame instability effects such as those
effects are taken into account)
caused by the moments of the vertical forces and the horizontal frame deflections P-Δ deflections,
effects, are taken into account)
i.e. Flexibility y method of analysis, y , Stiffness method of analysis y (commercial computer packages)
Second-order Second order elastic analysis Second order effects in elastic frames include additional moments moments. The second-order second order moments arising from the member deflections from the straight line joining the member ends are often called the P- δ
effects.
The second-order moments arising from the joint displacements Δ are often called ll d the h P-Δ PΔ 9
effects ff t .
I braced frames, the In th joint j i t displacement di l t Δ are small, ll
and only the P-δ are important. 9
I unbraced In b d
fframes, the th PP Δ effects ff t are iimportant, t t
and often much more so than the P- δ effects.
Second-order elastic analysis y P P
Δ
ISOLATED COLUMN
δ
P-δ
P
P
P
P
δ
δ
P-Δ Δ
Δ
δ
δ
P
FRAME braced
sway
Second-order Second order elastic analysis
Independent Behaviour
Interactive Behaviour
First order plastic analysis PLASTIC ANALYSIS tends to be used less commonly, y, even if a steel structure satisfies the more stringent conditions that must be imposed for rational plastic analysis. All instability effects are ignored. The collapse p strength g of the frame is determined by y using g the rigid-plastic g p assumption and finding the plastic hinge locations which first convert the frame to a collapse mechanism. All members must be ductile so that the plastic moment capacity can be maintained at each hinge over a range of hinge rotations sufficient to allow the plastic collapse mechanism to develop develop.
Plastic analysis and design will be considered in subsequent lectures, but it is worth noting that the plastic analysis is dependent on the elastic buckling load factor (λc) of the frame.
Advanced analysis Ideally, the member stress resultants (M, Ideally (M V, V N) should be determined by a method of frame analysis which accounts for both second second-order order effects (P (P-δδ and P P-Δ) Δ), inelastic behaviour, residual stresses and geometrical imperfections, and any y local or out-of-plane p buckling g effects.
Such a method has been described as an advanced analysis.
Advanced analysis is still basically a research tool.
Elastic buckling of braced frames Analysis: The results of an elastic buckling analysis may be used to approximate any second-order effects.
The set of member forces Nom which causes buckling depends on the distribution of th axial the i l forces f in i the th frame, f and d iis often ft expressedd in i terms t off a load l d factor f t λc by b which the initial set of axial forces Nim must be multiplied to obtain the member forces Nom at the frame buckling, buckling so that Nom = λc Nim. Alternatively, it may be expressed by a set of effective length factors km which define the member forces at frame buckling by Nom = π2EIm/(kmLm)2 The determination of the frame buckling load factor λc may be carried out using a suitable computer program (i.e. buckling analysis option in Microstan)
Elastic buckling of braced frames A l i Analysis: Alternatively, the
effective length factor km of each compression member
may be obtained by using estimates of the
member relative end stifnesses
γ1, γ2 in a braced member chart (AS4100).
The direct application of this chart is limited to the vertical columns of regular rectangular frames with regular loading patterns in which each horizontal beam has zero axial force, and all the columns buckle simultaneously in the same mode.
Effective length g factor – frame column More generally a column may be in a braced frame or a sway frame. Elastic spring restraints against rotation and translation
6
5
B
4
2
A
3
1
B
Elastic spring p g restraints against rotation
A
A is restrained by Column A-1 and beams A-2 and A-3 B is restrained by column B-6 and beams B-5 and B-4
Relative end stiffnesses – braced frame If the frame is BRACED AGAINST SWAY, and is of regular rectangular geometry with negligible axial force in the beams, ke is the solution of (Trahair & Bradford, 1998)
γ Aγ B ⎛ π ⎞
2
γ A + γ B ⎞⎛ π π ⎞ tan π ke ⎛ ⎜⎜ ⎟⎟ + ⎜ =1 ⎟⎜⎜1 − cot ⎟⎟ + 4 ⎝ ke ⎠ ⎝ 2 ⎠⎝ ke ke ⎠ π ke where at end A
γA
(I L ) ∑ = ∑ β (I L )
columns
e
and similarly at end B
γB
modifying factor
beams
(I L ) ∑ = ∑ β (I L )
columns
e
beams
ke is then obtained from Fig. 4.6.3.3 of AS4100. The values of βe are given below.
Stiffness factor,, βe – far end restraint conditions Sway Frame
Non Sway (Braced) Frame M
M
β e = 1.0 far end rigid g
β e = 1.5 far end pinned
β e = 2.0
M
far end fixed 4EI/L
M 3EI/L
β e = 0.5
3EI/L
β e = 0.67
M 6EI/L
β e = 1.0
2EI/L
M
M
M 4EI/L
Relative end stiffnesses FIXED BASE
PINNED BASE column
column
B
B
beam
beam
A
γ A = 0.6
A
γ A = 10
Buckling g load factor – braced frame For a BRACED FRAME ((i.e. one that is not free to sway) y) the procedure to calculate the effective length is: STEP 1. 1 Calculate C l l t γA and d γB for f each h off the th columns. l STEP 2. 2 Use U th the chart h t to t determine d t i ke for f each h off th the columns. l
STEP 3. Calculate
N om =
π 2 EI
(k e L )
2
for each of the columns.
STEP 4. Estimate the BUCKLING LOAD FACTOR, λm f each for h off the h columns, l where h N* is i the h axial i l compression in the column.
λm
N om = N*
Effective length factor – braced frame STEP 5. 5 Estimate the FRAME BUCKLING LOAD FACTOR from λcr = min of the λm values STEP 6. Recalculate Nom = λcrN* for each of the columns
STEP 7. By using
N om =
π 2 EI
(L e )2
= λ cr N *
recalculate the effective length for each of the columns from
Le =
π EI 2
N om
Example a pe1 16 kN
40 kN
E 24 kN 60 kN 56 kN 100 kN D
A
5m
A
100UC14
1890
3.18x106
B, E
250UB25
3270
35.4x106
C, D
150UC23
2980
12.6x106
F
250UB37
4750
55.7x106
F
B C 6m
Section A (mm2) I (mm4)
Member
5m Find the effective length for each of the columns.
8m
A first order matrix stiffness analysis produces the following axial forces in this braced frame: Member N* (kN)
A -39.5
B
C
D
-0.8
-197.4
-31.9
E -7.6
F +4.7
Example a pe1 A
C
Example a pe1 Elastic s c buc bucklingg loads: o ds:
D
Buckling load factor: A C D
Example a pe1 Recalculate ec cu e buc bucklingg loads o ds in each e c member: e be : A C D
Effective lengths A
C
C
Elastic buckling of sway (unbraced) frames Analysis: The determination of the frame bucking load factor λc of a rigid-jointed rigid jointed unbraced frame may also be carried out using a suitable computer program.
For isolated unbraced members of very simple unbraced frames, a buckling analysis l i may also l be b made d from f first fi principles. i i l
Alternatively the effective length factor, km of each compression member may be obtained by using estimates of the relative end stiffnesses γ1, γ2 in a chart such as that of
AS3600.
Relative end stiffnesses – sway frame If the frame is FREE TO SWAY and is of regular rectangular geometry with negligible axial force in the beams, ke is the solution of:
γ Aγ B (π ke ) − 36 ⎛ π ⎞ ⎛ π ⎞ = ⎜⎜ ⎟⎟ cot⎜⎜ ⎟⎟ 6(γ A + γ B ) ⎝ ke ⎠ ⎝ ke ⎠ 2
2
where (as for a braced frame)
γA
(I L ) ∑ = ∑ β (I L )
columns
e
beams
and d
γB
(I L ) ∑ = ∑ β (I L )
columns
e
beams
ke is obtained from Fig. 4.6.3.3 of AS4100
Effective length factor – sway frame The procedure for determining Le in a SWAY FRAME is: STEP 1. Calculate γA, γB and use the chart to determine ke for each column STEP 2. Calculate
N om =
π 2 EI
(k e L )
2
for each of the columns
STEP 3. For each STOREY calculate its buckling load factor from
λ ms =
∑ (N ∑ (N
STEP 4. 4 Estimate frame bucking load factor from λcr = min of the λms values STEP 5. Recalculate Nom = λcrN* for all the columns STEP 6. Recalculate
Le =
π 2 EI N
om
for all the columns
om *
L) L
)
Example a pe2 20 10
40
20 kN
Member A (mm2) I (mm4)
C B
50
50
80
D
5m
A, E
6660
52.8x106
B, D
3860
17.6x106
C
3270
35.4x106
F
5210
86.4x106
30 A
E 6m
5m Find the effective length for each of the columns.
6m
A first order matrix stiffness analysis produces the following axial forces in this braced frame: Member N* (kN)
A -116.2
B
C
D
-37.5
-22.5
-42.5
E -143.9
F -8.9
Example a pe2 Elastic s c buc bucklingg loads: o ds: Upper storey columns
Example a pe2 Elastic s c buc bucklingg loads: o ds: Lower storey columns
Example a pe2 Buckling uc g load o d factor: co : Upper storey columns
Lower storey columns
Example a pe2 Recalculate ec cu e buc bucklingg loads: o ds: Upper storey columns
Lower storey columns
Design of connections I
Lecture Outline Bolts bolt group Bolts, Welds,, weld ggroup p butt weld, fillet weld
Force connections splices in tension and compression, truss joints, shear splices and connections in beams
Moment connections b beam momentt splice li
Force and moment connections seat for the beam-to-column connection, semi-rigid beam to-column connection, the full strength beam splice
Connections Connections are used to transfer the forces supported pp by y a structural member to other parts of the structure or to the support. They are also used to connect braces and other members which provide restraints to the structural member. Connections join members using CONNECTORS CONNECTORS, such as bolts, bolts pins, pins rivets, or welds, and may include additional plates or cleats. The arrangement off the Th h components iis usually ll chosen h to suit i the h type off action (force or moment) being transferred and the types of member ((tension or compression p member,, beam,, or beam-column)) A connection is designed by analysing the method of force transfer from the member through the connection and its components to the other parts of the structure, and by proportioning each component so that it has sufficient capacity for the force that it is required to transmit.
Connections The need for connections: gap
N
N
σ P
Axial action (longitudinal stresses ) stress/force t /f transfer t f required i d
M Shear action (shear stress ) surfaces slide past each other
R
Bending g action (longitudinal stresses)
M
σC σT Lack of compatibility
P Torsion action (shear stress) bracket rotates
P
CONNECTIONS (a) hold parts together (b) allow transfer of internal actions
Connections Real connections usuallyy have combinations of these actions:
V
Moment action due to V & H.
T Tension action due to H.
H S
e
M
Shear action due to V V.
V T
Torsion action due T = Ve. Shear action due to V.
S
Connections Splice: p
w
design for M* and V* BMD
V M*
M
M
SFD
V*
V
Bolts Several different types of bolts may be used in structural connections connections, including ordinary structural bolts (i.e. commercial or precision bolts and black bolts), and high strength bolts. Bolts may transfer loads by shear and bearing, by friction between plates clamped together, together or by tension. tension
i.e. 4.6 – ultimate tensile strength Æ fuf (f-fastener) “400 MPa” - structural or “block bolts” (industrial buildings) 8.8 – high strength bolts Æ fuf = 830 MPa (yield is 0.8 of 830 Æ 664 MPa)
Pins Pin connections Pi ti used d to t be b provided id d iin some triangulated t i l t d frames f where h it was thought to be important to try to realize the common design assumption p that these frames are p pin-jointed. j
The cost of making a pin connection is high because of the machining required for the pin and its holes, and also because of difficulties in assembly.
Pins are now rarely used, except in special applications where it is necessary to allow relative rotation to occur between the member being connected. t d
Rivets In the past, past hot hot-driven driven rivets were extensively used in structural connections. They were often used in the same way as ordinary structural bolts are used in shear and bearing and in tension connections. There is usually less slip in a riveted connection because of the tendency for the rivet holes to be filled by the rivets when being hot-driven. Shop riveting was cheaper than filed riveting, and for this reason shop riveting was often combined with field bolting. However,, the use of riveting g has declined considerably, y, and has been largely replaced by welding or bolting.
Bolts, rivets
Bolt dimensions 4 6 commercial bolts 4.6
8 8 high strength bolts 8.8
σ
σ fuf = 400 MPa MP
fy
fyf = 240 MPa
fy
ε nominal diameter D of fastener
fuf = 830 MPa proof stress fyf = 660 MPa
0.2% permanent strain
dc core dia.
ds
ε
diameter of tension stress area
Types of filed connections
Joints between beams Force & M Moment connection i
Force connection
M Moment connection i
Force connection
Force & Moment connection Force & Moment connection
Beam--to Beam to--column joints Force & Moment connection M Moment t connection ti
Force connection Moment connection
Axial shear
Force connection
Force connection
Force connection
Force connection
Eccentric shear / Tension Force & moment connection ti
Force connection
Flexible end connections
Any moment transfer can b neglected be l t d Force connections
Rigid and semisemi-rigid connection
Force & moment connection
Force & moment connection
Semi--rigid Semi g connection
Force & moment connection
Prying action
Connections to concrete cores
Connections to concrete cores
Connections to concrete cores
Web panel in shear (instability)
Axial compression and bending moment
A h Anchorages Base plate
Contact steel-concrete and concrete in compression
Bolt tightening g g SNUG TIGHTS (S) – achieved by full effort of a person using a standard spanner (friction between two surfaces is not too high) BERARING (B) – design action transferred by shear in bolts and bearing on connected parts at STRENGTH LIMIT STATE FRICTION (F) – bolts tightened to induce a minimum tension so that damping action transfers shear at SERVICEABILITY LIMIT STATE due to friction TENSION ((T)) – snugg tightened g and then tightened g further a certain number of fraction of turns
Bolts Bolts suffer only 2 actions, tension and shear.
bolts in tension T
C
P
bolts in shear P/2 P
P/2
P
Torsion
(Pe causes shear)
e
P
S P P/2
P/2
M
two shear planes
S
P one shear plane
Bolt categories g very flexible (semi (semi-rigid rigid connection) 4.6/S 8 8/S 8.8/S 88.8/TB 8/TB 8.8/TF
only snug tightened
can be snug tightened too but it is wasteful friction capacity capacit (serviceability (ser iceabilit limit state)
Commercial C i l packages k (i.e. (i Microstran, Mi t Spacegas, S etc.) t ) assume moment connection (hence connections are able to resist bending moment)
Modes of failure B lt in Bolts i TENSION Bolts in SHEAR B lt in Bolts i BEARING Desi n checks required: Design a. Bolt strength g b. Plate bearing c. Plate tearing: i Straight i. St i ht across ii. Staggered
Modes of force transfer
Force connection
Bolted plates in shear and tension ae df
Tearing out at end of a plate, due to tension
V*v ≤ φ Vv φ =0.7 Vv= ae t fup
fup - ultimate lti t ttensile il strength t th off plate material
Bearing failures top p plate p
V* B
beam reaction
other p plate
V* B
per bolt
local crushing under washer
Force connection
Strength g limit state Shear
V
* f
≤ φV f
Vf* = design shear force Vf = nominal shear capacity of bolt Vf = 0.62 fufkr (nnAc + nxAo)
φ = 0 .8 (i)
Connections are less ductile than beams (ii) We do not want connection to be the ‘weakest link’ in the structure
nn= number of shear planes with treads IN the shear plane whose bolt area is Ac nx= number of shear planes without treads EXCLUDED from shear plane whose bolt area (shank area) is Ao
Shear distribution in a force connection
Strength limit state kr= reduction factor for a lapped pp connection of length g Lj If:
Lj < 300 mm Æ kr = 1.0 300 < Lj < 1300 mm Æ kr = 1.075 – Lj /4000 nx = 1 nn = 3
Lj > 1300 mm Æ kr = 0.75
Ao
T*
T* Lj Not exactly in equilibrium Æ moment imbalance
nx Shear planes
Ac
nn nn nn
Tension
N
* tf
≤ φ N tf
φ = 0 .8
Ntf* = design tension force φNtf = design tension capacity of bolt Ntf = As fuf As = tensile stress area of bolt
Shear and tension Bolts o s aree only o y subjected subjec ed too two wo actions: c o s: TENSION NS ON and dS SHEAR,, caused by axial forces, shear forces, bending moments and torsion, when tension and shear occur together, the bolt strength has to satisfy the following interaction equation: 2
⎞ ⎛ N ⎟ +⎜ ⎟ ⎜ φN ⎠ ⎝ tf * tf
2
⎞ ⎟ ≤ 1 .0 ⎟ ⎠
φ = 0 .8
INTERACTION CURVE pure shear
SHE EAR
⎛ V ⎜ ⎜ φV ⎝ f * f
safe
pure tension AXIAL
Vf , Ntf = nominal i capacities i i determined i earlier i (snug tighten bolts Æ there is no tensile force in bolts)
Shear and tension
Shear and tension – typical examples NO BENDING
BENDING
PV
P
P
PV CG of bolt group
PH
CG of bolt group
e PV /n
Line of P through CG
S
Shear force from loading
e
Pv
M Moment from frame action
BENDING
M = Pv e M causes extra t tension per bolt
PH Shear per Sh bolt V*vf
PH /n Tension per b lt V*tf bolt
Beam--to Beam to--column joints Force & Moment connection Moment connection
Force connection
Moment connection
Shear and tension – typical examples BENDING
Tension in these bolts T1 T
M S
h
h1 h 2
h3 dn/3
NA
y3
y
y 1
T23
2
C
d dn
b
dn/3
Plates compressed mp together g Could calculate position of NA by trial and error until ΣT but usually dn
≈ h/6.
≈ ΣC,
Shear and tension – typical examples (i) Take moment about line of force C.
M = n1T1h1 + n2T2h2 + etc.
n1= No. of bolts on level 1
(ii) From linear force diagram
T2 = T1 y2/y1 , T3 = T1 y3/y1 (iii) Substitute into M equation and rearrange.
T1 = M y1 /(Σnihiyi) per bolt at level 1. (iv) Shear per bolt = S / total number of bolts. (v) Now apply shear and tension interaction equation. However if there are horizontal forces, such as Phe , eccentricities etc, include their effects and then apply interaction equation.
Ply in bearing Bearing force on ply Vb* due to a bolt ply
ae
V ≤ φ Vb * b
bolt
φ = 0 .9
Vb = nominal bearing capacity of ply
Vb = 3.2d f t p fup > aet p fup df = bolt diameter tp = thickness of ply fup = ultimate tensile strength of ply
Example 1 Checking the capacity of a connection – lap splice connection 35 70 70 35 40 70
Splice l plates, l 2x10mm 10 thick h k fy = 260 MPa steel, fu = 410 MPa 35 70
N*
70 35
N* N* *
10
N*
20 70
10
40
9M24,8.8/S
ELEVATION
PLAN
Spliced plate, 20mm thick fy = 250 MPa steel, fu = 410 MPa
a. Bolt strength M24 8.8 /S in dholes= 26mm
Threads intercept one shear plane plane, plain shank the other shear plane. plane
Design capacity of bolts in shear = φVfn + φVfx = 133 + 186 = 319 kN [TA2.2]
b Plate bearing b.
(spliced plate)
Vb = 3.2 x 24 x 20 x 410 = 630 kN ≤ 35x 20 x410 = 287 kN Since 287 < 319, plate bearing capacity governs.
Example 1
Serviceability limit state for bolt For 8.8/TF where slip at service load is to be limited Although a serviceability limit state still uses a capacity reduction factor φ
Shear:
V ≤ φ V sf φ * sf
= 0 .9
V*sf = nominal bolt shear capacity for friction-type connection Vsf = μneiNtikn
μ = slip factor (coefficient of friction) taken usually as 0.35 nei = number of effective interfaces Nt = minimum i i bolt b lt pretension t i in i generating a friction-type connection
kn = 1.0 for standard holes = 0.85 for short slotted or oversize holes = 0.7 for long slotted holes
Shear + Tension ⎛ V ⎜ ⎜ φV ⎝ sff * sf
2
⎞ ⎛ N ⎟ +⎜ ⎟ ⎜ φN ⎠ ⎝ ti * tf
2
⎞ ⎟ ≤ 1 .0 ⎟ ⎠
φ = 0 .7
Vsf* = design serviceability shear force Ntf* = design strength shear force (bolt pre-tension) Nti = bolt p pretension Minimum Nti VALUES (kN) M16 – 95 M20 – 145 M24 – 210 M30 – 335 M36 - 490
Bolt group subjected to inin-plane loading doubler plate weld
bolt
Momentt M connection column stiffener
stiffener
Consider n bolts of equal area subjected to force P eccentric e to the y-axis which passes through the bolt centroid. y
e
C
Pe
P
≡ x
P
end plate
Bolt group subjected to inin-plane loading Separate shear caused by P and Pe are hard to calculate and sum vectorially. Instead we calculate position of INSTANTANEOUS CENTRE OF ROTATION. Force Vfi* on i-th bolt and radium ri from centre of rotation.
y (xc, 0)
P
e ri
Vfi* x
Bolt group subjected to inin-plane loading V = kri A * fi
n
(
I z = A∑ x + y 2
k = constant A = area of bolt 2
)
about centroids
Pe k = Iz P xc = − kn A
N.B. (1) force in x x-direction direction as well its effect can be included in the same way; and V
* fi
= {V fi* x 2 + V
* fi
y 2 }1 / 2
(2) if welds are used replace the bolt area with the weld size and summations by integrals along the weld
Example p 2 A typical web side plate connection is shown in the figure on the next slide in which a single 10mm thick side plate is bolted t th to the web b off a b beam and d iis welded ld d tto th the fl flange off th the column. l In designing the welds, the beam reaction is assumed to act through the centroid of the bolt group at a distance of 90mm from the face of the column. In designing the bolts, the beam reaction is assumed to act at the line of the weld at a distance of 90mm from the centroid of the bolt group. Problem: For a design beam reaction of 250 kN, determine the maximum shear force in a bolt of the bolt group
Example p 2
Example p 2 Solution: The calculations are based on the instantaneous centre of rotation approach of Chapter C9 of the Commentary (AS4100). By inspection, the centroid of the bolt group is at its geometric centre.
Example p 3 Determine the maximum shear in the bolt group in the beam splice shown. 25 55 140 30 65 35 70 70 35
75
Member design actions at bolt group centroid Shear = +160kN Moment = +20kNm
8-M20 8.8/S bolts. Threads in shear plane. 2x280mm E48XX fillet welds. Single web plate. This problem demonstrates the in-plane elastic analysis of a bolt group.
Example p 3
Example p 4 Check an M20 88.8/S 8/S bolt whose treads intercept a single shear plane for a design shear of Vf* = 43.1kN 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 plane. 2x280mm E48XX fillet welds. Single web plate.
This problem illustrates the determination of the shear capacity of a bolt.
Example p 4
Example p 5 Determine the maximum design tension force that can be transmitted in conjunction with a design shear force of Vf* = 43.1kN byy an M20 8.8/TF bolt whose threads intercept p a single g shear plane. Member design 25 55 140 30 75 65 35 70 70 35
actions at bolt group centroid Shear = +160kN Moment = +20kNm 8-M20 8.8/S bolts. Threads in shear plane. plane 2x280mm E48XX fillet welds. Single web plate.
This problem illustrates the checking of the strength of a bolt under combined shear and tension.
Example p 5
Example p 6 Determine the maximum serviceability tension force that can be transmitted in conjunction with a serviceability shear force of Vsf* = 30.0kN byy an M20 8.8/TF bolt in a standard hole. This problem illustrates the checking of the serviceability of a bolt in a frictiongrip connection with a single interface.
Bolt groups in torsion e
PV
Pys = P/n
y Bolt shear due to torsion
r
y CG
T = PVe
PTx
x
PTy
CG of bolt group
bolt shear due to torsion
From mechanics of solids Æ PT
= torsional constant x r
Torque per bolt Æ Ti = PT x r = C x r2 = C x (xi2 For n bolts, total torque Æ T = Σ Torsional constant Æ C =
x PT
+ yi2)
Ti = C x r2 = C Σ (xi2 + yi2)
T /Σ (xi2 + yi2)
Bolt g groups p in torsion In bending Æ σ = My/I In torsion Æ PT = Tr/IP or torsional components p Æ PTx = Ty/I y P, where Æ IP =
PTy = Tx/IP
Σ (xi2 + yi2)
CRITICAL BOLT – vectorial effect of components is maximum PTy
PTy
PTy
Pmax = P
PTy PTy
PTy
+ (PTy + PyS )
2
Tx
VECTORIAL SUMMATION
PTy PTy PTy
2
PTy PTy
PTy
PTy
Design of connections II
Lecture Outline Welds, weld group butt weld weld, fillet weld
M Moment t connections ti beam moment splice
Force and moment connections seat for the beam-to-column connection, semi-rigid beam to-column connection, the full strength beam splice
Welds Structural connections between steel members are often made by arc arc-welding welding techniques, in which molten weld metal is fused with the parent metal of the members or joint plates being connected. Welding is often cheaper than bolting because of the great reduction in the preparation p p required, q , while ggreater strength g can be achieved,, the members or plates no longer being weakened by bolt holes, and the strength of the weld metal being superior to that of the material connected. In addition, welds are more rigid than other types of load-transferring connectors. On the other hand, welding often produces distortion and high local residual stresses, t and d results lt iin reduced d dd ductility, tilit while hil fi field ld welding ldi may b be diffi difficult lt and costly.
Welds Force connection
Force connection
Intermediate web stiffeners
Butt welds Butt welds Fillet welds Butt welds are frequently used to splice tension members members. A full penetration weld enables the full strength of the member to be developed, while hil the th butting b tti together t th off the th members b avoids id any joint j i t eccentricity. t i it Butt welds often require q some machining g of the elements to be jjoined. Special welding procedures are usually needed for full strength welds between thi k members thick b to t control t l the th weld ld quality lit and d ductility, d tilit while hil special i l inspection i ti procedures may be required for critical welds to ensure their integrity.
Butt weld
Fillet welds We will w only o y co consider s de equal equ leg eg fillet e we welds ds here: e e: throat t
t
2
throat
t weld size
Design g actions are calculated/unit length g of weld on p plane of throat: Longitudinal shear, transverse shear, normal force all act on throat and are summed vectorially to produce:
vw* = design force per unit length of weld
Fillet weld joint
Forces on weld elements The forc force p perr un unitt length ngth of fillet weld in the x,, y and z directions may be determined using the familiar expressions: general fillet weld group
* * P M * x zy − vx = Lw I wp
centroid of fillet weld group
Py*
* M * zx vy = − Lw I wp
* z
* x
P* y M* y
* y
M y M x P v = + − Lw I wx I wy * z
y
P*x M*x
M*
z
z
P*
z
Weld in x-y plane, z = 0
General fillet weld group
x
Design equation v ≤ φv w * w
Strength S e g des design: g :
v
* w
=
(v ) + (v ) + (v ) * x
2
* y
2
* 2 z
D i off any generall fillet Design fill t weld ld group subject bj t to t a generall design d i action ti sett
(P *x, P *y, P *z, M *x, M *y, M*z ) may be obtained by evaluating the property set
Lwx, Lwy, Lwz, Iwx, Iwy, Iwp (see Table on next slide) and substituting into the governing equation (fillet weld group loaded ‘in-plane’ in-plane and ‘out-of-plane’) out-of-plane ), checking that the governing inequality is satisfied, at each of the critical points.
Practical fillet weld g groups p Many y fillet weld groups g p comprise p lines of welds parallel p to the x and y axes. For such relatively regular fillet weld groups, the identification of possible critical points is correspondingly more straightforward. The possible critical points for a fillet weld g group p consisting g of lines of weld parallel to the x and y axes only are numbered 1 to 8. 8
y 2
1
3
8
x 7
4 5
6
Possible critical p points in particular weld group
Forces on weld elements where:
Lw Iwx, Iwy
- the total length of the weld;
Iwp
- the p polar moment of area of the weld elements about the centroid of the weld group (treated as a line element) = Iwx + Iwy
- the second moments of area of the weld elements (treated as a line element) about the x and y axes respectively;
The previous expressions can be sightly modified in order to allow them to reflect realistic distributions of the design force set ((P*x, P*y, P*z) between components of the total length of the weld group, as follows:
Px* M z* y v = − Lwx I wp * x
v = * y
Py* Lwy
M z* x − I wp
* Pz* M x* y M y x + − v = Lwz I wx I wy * z
where:
Lwx, Lwy, Lwz the lengths of the weld assumed to receive the component forces along the individual x, y and z axes respectively
Fillet Weld Group Loaded ‘i -plane’ ‘in‘in l ’ Fillet weld group loaded ‘in-plane’ by a common design action set of forces (P*x, P*y) and design moment (M*x): * * P M * x zy vx = − Lwx I wp
v = * y
v*z = 0
Py* L wy
Py* Pz*
* z
M x + I wp
M z*
Fillet Weld Group Loaded ‘out--of ‘out of--plane’ Fillet weld group loaded ‘out-of-plane’ by a common design g action set of forces (F*y, F*z) and design g moment (M*x)):
v*x = 0
v *y =
Py* L wy
* * M P * xy z vz = + Lwz I wx
Py* Pz* Mx*
Fillet Weld Group Loaded ‘out out--of out of--plane plane’ P
z
Line weldsÆ unit thickness
e Mx Py
y
Lw1
x
x
yt
1 Lw2
yc Centroid of weld group
y
Lw1
A
y
v*y = P*y/Lw = force per unit length acing in y-direction Lw = total weld length = 2Lw1 + 2Lw2 v *z = M *xyc/Iwx = normal force per unit weld in x-direction at point A Iwx = second moment of area of unit weld about centroid (mm3)
Weld group subjected to outout-of of-plane loading vn vz
throat
vy
(perpendicular)
vt force per unit length
vz = p produces normal component p p vz / 2 and transverse component on throat.
{ (
vw = v + vz 2 y
Therefore at A (say)
=
) + (v
z
/
2
)}
2 1/ 2
v y2 + v z2 n
For bolts calculate Ix as
2
2
A∑ y 2
Py V = y and n
etc.
vz / 2
Design g equation q Strength S e g des design: g :
v ≤ φv w * w
vw = nominal i l capacity it off fillet fill t weld ld per unit it llength th vw = 0.6fuwttkr
φ = 0 .6 GP SP – special p ppurpose p
kr = reduction factor for length of weld Lw(m) = 1.0 (Lw < 1.7) = 1.10 1 10 –0.06L –0 06Lw (1.7 (1 7 < Lw < 8.0) 8 0) = 0.62 (Lw > 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.
