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CENG 6507 - Steel Steel Structures Structures
Course Instructor: Dr.-techn. Dr.-techn.
Medhanye B.
STRUCTURAL ENGINEERING CHAIR ETHIOPIAN INSTITUTE OF TECHNOLOGY
CENG 6507Steel Steel Struct Structure ures s
5. Ultimate bucklin buckling: g:
limit
MEKELLE UNIVERSITY
state
of
beambeam-colu columns mns, ,
built-up built-up members
Introduction • Limi Limitt state tate des design ign requir quire es the the str struct ucture ure to sat satisf isfy two two prin princi cipa pall crit criter eria ia:: the ul ulti tima mate te li limi mitt stat state e (ULS (ULS)) and and the the servic serviceabilit eability y limit state state (SLS). (SLS). • A limit state is a set of performance criteria (e.g. vibration levels, deflection, strength, stability, buck buckli ling ng,, twis twisti ting ng,, coll collap apse se)) that that must must be met met when when the the stru struct ctur ure e is subj subjec ectt to loads loads.. • All engi engine neer erin ing g desi design gn crit criter eria ia have have a comm common on goal goal:: that of ensuring a safe structure and ensuring the functi functiona onalit lity y of the struct structure ure..
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Ultimate Limit State
Serviceability Limit State
• To satisfy satisfy the ultimate ultimate limit state, the structur structure e must not collapse when subjected to the peak design load for which it was designed.
• To satisfy the servicea serviceability bility limit state criteria, criteria, a structure must remain functional for its intended use subject to routine (everyday) loading, and as such such the the stru struct ctur ure e must must not not caus cause e occupant discomfort under routine conditions.
• A structure structure is deemed to to satisfy the ultimate limit state criteria if all factored bending bending,, shear and tensile or compressive stresses are below the factor factored ed resist resistan ance ce calcul calculate ated d for the sectio section n under consideration. • Where Whereas as Magni Magnific ficati ation on Facto Factorr is used used for the loads, and Reduction Factor for the resistance of members.
Examples: Deflection, Vibration, Crack widths in concrete
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Ultimate Limit State
Serviceability Limit State
• To satisfy satisfy the ultimate ultimate limit state, the structur structure e must not collapse when subjected to the peak design load for which it was designed.
• To satisfy the servicea serviceability bility limit state criteria, criteria, a structure must remain functional for its intended use subject to routine (everyday) loading, and as such such the the stru struct ctur ure e must must not not caus cause e occupant discomfort under routine conditions.
• A structure structure is deemed to to satisfy the ultimate limit state criteria if all factored bending bending,, shear and tensile or compressive stresses are below the factor factored ed resist resistan ance ce calcul calculate ated d for the sectio section n under consideration. • Where Whereas as Magni Magnific ficati ation on Facto Factorr is used used for the loads, and Reduction Factor for the resistance of members.
Examples: Deflection, Vibration, Crack widths in concrete
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ULS: TYPES OF STEEL FAILURE
Failure can happen because of a) yielding of members b) failure of connections - most common cause of steel buildings collapse c) buckling of members - buckling of columns rarely has resulted in structural failure
Design consideration of tension members
Tension Members • Tension members are linear members in which axial forces act so as to elongate (stretch) the member. • Tension members carry loads mo mos st efficiently, since the entire cross section is subjected to uniform stress. • Unlike Unlike compression compression members, they do not fail by buckling
Tension members are efficient structural elements but its efficiency may be affected by: – End connection connection (e.g. Bolt Bolt holes reduce member member section) section) – Reversal of load (buckling) – Bending Bending moments moments
Design of Axially loaded Tension members ( EBCS-3, 1995 ) – The design design value value of of the the axial axial tensio tension n force force is N N t,Sd
≤
N N t,Rd
Where Nt,Rd is the design tension resistance capacity of the x-section taken as the smaller of:
1. The design plastic plastic resistance of the gross section section is
N pl , Rd =
Af y
γ M 1
2. The design ultimate ultimate resistance of the net section section at the bolt hole is 0.9 Aeff f u
N u, Rd =
γ M 2
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BUCKLING OF STEEL MEMBERS
Buckling of steel may happen in a) members under compression b) members under bending c) members under combined loading: compression + bending
Local Buckling and Section Classification Internal or outstand elements internal – are the webs of open beams or the flanges of boxes outstand – are the flanges of open sections and the legs of angles Outstand Internal Outstand
Web Web
Flange (a) Rolled I-section
Flange (b) Hollow section
• As the plate elements in structural sections are relatively thin compared with their width, when loaded in compression they may buckle locally
Internal
Internal Web
Local buckling
Internal
Flange (c) Welded box section
• Local buckling within the cross-section may limit the load carrying capacity of the section by preventing the attainment of yield strength.
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Classes of cross-sections
• Premature failure (arising from the effects of local buckling) may be avoided by limiting the width to thickness ratio - or slenderness - of individual elements within the cross section. • This is the basis of the section classification approach. • EBCS-3, 1995 defines four classes of crosssections. • The class into which a particular cross-section falls depends upon the slenderness of each element and the compressive stress distribution .
Four classes of cross-section are:1. Class 1 or plastic cross-sections:- – Design of Steel Structures for plastic elements. – A plastic hinge can be developed with sufficient rotation capacity to allow redistribution of moments within the structure. Only Class 1 section may be used for plastic design. 2. Class 2 or compact cross-sections: – The full plastic moments capacity can be developed but local buckling may prevent development of a plastic hinge with sufficient rotation capacity to permit plastic design. Class 2 sections can be used without restriction except for plastic design. 3. Class 3 or semi-compact sections:- – The stress at the extreme fibers can reach the design strength but local buckling may prevent the development of the full plastic moment. Class 3 sections are subjected to limitations on their capacity. 4. Class 4 or thin-walled sections:- – Local buckling may prevent the stress in a thin-walled section from reaching the design strength. Design of Class 4 sections requires special attention.
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Centroidal axis of effective cross-section
Centroidal axis of gross cross-section Centroidal axis of gross cross-section
Stress distribution (compression positive)
1 > ψ ≥ 0:
beff
eN
Effective width b eff
σ 1
σ 2
beff = ρ c c
Non-effective zones bc
bt
Effective crosssections for class 4 in compression and bending
Gross cross-section (a) Class 4 cross-sections - axial force
eM
beff = ρbc = ρc /1(− ψ ) σ 2 beff
ψ = σ 2 / σ 1
1
0
-1
1 ≥ ψ ≥ − 1
Buckling factor k σ
0,43
0,57
0,85
0,5 7− 0,2 1ψ + 0,07ψ 2
(class 4 x-section )
Non-effective zone
Centroidal axis
beff
σ 2
beff = ρ c
c
Non-effective zone beff
eM
ψ < 0:
σ 1
Centroidal axis
Centroidal axis of effective section bc
Gross cross-section (b) Class 4 cross-sections - bending moment
Stress distribution (compression positive)
σ 1
Effective width b eff
ψ = 1:
σ 2
be1
b = b - 3t beff = ρ b be1 = 0,5 beff
be2
be2 = 0,5 beff
b _ 0: 1 > ψ >
σ 1
σ 2
be1
b e1 =
be2
2beff
5 - ψ b e2 = beff - be1
b
bc
Effective widths of internal compression elements
b = b - 3t b eff = ρ b
bt
ψ < 0:
σ 1 σ 2
beff
b = b - 3t = ρbc = ρb / (1 - ψ ) be1 = 0,4b eff
be1
be2 = 0,6b eff
be2 b
ψ = σ 2 / σ 1
1
Buckling factor k σ
4,0
Alternatively, for
1 > ψ > 0 8,2 1,05+ ψ
_ ψ > _ - 1: 1>
Illustrated as rhs. For other sections
0 > ψ > - 1
0 7,81
-1
7,81- 6, 92ψ + 9,78ψ 2 23,9
k σ =
- 1 > ψ > - 2
598 , (1 -ψ )2
16 [(1 + ψ )2 + 0,112(1- ψ ) 2 ]0,5 + ( 1+ ψ )
b = d for webs b = b for internal flange elements (except rhs)
(EBCS-3, Tab. 4.3)
beff = ρbc = ρc /1(− ψ )
σ 2 bt
ψ = σ 2 / σ 1
1
1 > ψ > 0
0
Buckling factor k σ
0,43
0578 , ψ + 0, 34
1,70
0 > ψ > − 1
1,7 − 5ψ + 17,1ψ 2
Effective widths of outstand compression elements (EBCS-3, Tab. 4.4)
1 > ψ ≥ 0 :
σ 1
Centroidal axis of effective section
ψ < 0 :
σ 1
-1 23,8
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Compression Members • The strength of steel compression members is usually limited by their tendency to buckle. • If a 6mm diameter steel rod 1m long is placed in a testing machine subjected to a pull, it will be found to carry a load of about 7KN before failure occurs. If on the other hand this same rod had been subjected to compression, then the maximum load, which would have been carried, would be about 0.035KN , a very big difference • Failure in the first test occurs by the fracture of the member, in the second it is due to bending out of the line of action of the load. • The load at which a compression member becomes unstable is the buckling load .
Buckling Load Pcr =
π 2 EI 2 e
L
=
π 2 EI ( KL)
2
• The buckling load depends on the length, crosssection, and end conditions of the column and the stiffness of the material.
EFFECTIVE LENGTH OF COLUMNS
• The effective length, L , of a e
member hinged at its ends is the distance between the axes of the hinges. • K, is the ratio of the length (L ) of the equivalent column to the actual length (L); and the length of the equivalent column is the distance between two consecutive points of contraflexure (points of zero moment) in the actual column. e
• Pcr is the load at which the compression member becomes unstable • E is modulus of elasticity of steel • I is moment of inertia of the cross section • L is the length of the compression member • K is the effective length factor
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Effective Length Factors Buckling of a column in a non – sway frame
Buckling of a column in a sway frame
COLUMNS OF NON-SWAY FRAMES
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Design steps for Axially Compressed Members 1. 2. 3. 4.
5. 6. 7. 8. 9.
Determine the axial load, N sd. Determine the buckling length, Select a trial section (take into consideration economy, i.e. least weight per unit length). Determine the Class of the section according to Section 4.3.2 and Table 4.1. If the cross-section is classified as Class 4, determine A eff according to Section 4.3.4 and Table 4.4 (Sec 4.5.4.3) Determine the non-dimensional slenderness ratio (Section 4.5.4.3) Determine the appropriate buckling curve, Table 4.11 Determine the value of χ . Interpolation must be used to determine more exact values, Table.4.9 Calculate the design buckling resistance N b,Rd of the member. Buckling about both principal axes must be checked. Check the computed buckling resistance against the applied load. If the calculated value is inadequate or is too high, select another section and go back to Step 4.
Example-1: x
The column B – E on the Figure shown below is under the action of N Sd = 2800 kN. Both sides are pinned. Check the resistance of the column. Steel grade F e 430 is used.
x
P
P z
y
z
y
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Solution: Step 1: Axial load NSd = 2800 kN. Step 2:Buckling length L = 4000 mm (pinned end both sides. Frame non-sway mode). Step 3: The section is given. Step 4:Determine the class of the cross-section and check for a local buckling. The section is subjected to uniform compression. For the section to be classified as at least class 3, in order to avoid any modification to the full cross sectional area due to local buckling, the limiting width to thickness ratio for
• Outstand element of compression flange: c/tf ≤15 ε. • Web subject to compression only: d / tw ≤ 42 ε. • For Fe 430 steel grade fy = 275 N / mm2. Thus
ε = 235
275
= 0.92
This gives the following limiting values: • Outstand element of compression flange: c / tf = (254/2) / 16.3 = 7.78 < 15 x 0.92 = 13.8 OK.
• Web subject to compression only: d/tw = (310-2(33))/9.1 = 26.8 < 42 x 0.92 = 38.64
OK.
class 3 section are (See Table 4.1 EBCS-3).
• Therefore, the section belongs to at least Class 3. Thus, βA = 1.0
Step-5:
Step-6: Determine the appropriate column curves (Table 4.11 EBCS-3). • Use curve a for buckling about y-axis and curve b for buckling about z-axis.
Determine the non-dimensional slenderness ratio.
• For Fe 430 steel grade, λ1 = 93.9 ε = 93.9 x 0.92 = 86.39 • Slenderness ratio about y-axis: λy = L / i y = 4000/135 = 29.63 • Slenderness ratio about z-axis: λz = L / i z = 4000/63.6 = 62.89 • Hence, the non-dimensional slenderness ratio is determined as: λ 1 = 0.34 λ y = y β A = 29.63 86.39 λ 1
( )
λ λ z = z λ β A = 62.89 86.39 ( 1) = 0.73 1
h
b
= 310
254
= 1.22
and t f = 16.3 mm < 40
Step 7: Determine value of χ. Using Table 4.9 and interpolating: • For y-axis: curve a for λ y = 0.34 ⇒ χ y = 0.97 • For z-axis: curve b for λ z = 0.73 ⇒ χ z = 0.77 Therefore, buckling about the z-axis becomes critical.
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Step-8: Calculate the design buckling resistance. N b, Rd =
χβ A Af y
=
γ M 1
0.77 x1 x11000 x 275 1.1
(Only to show the method!!) Add an additional hinged support at mid-height to increase the resistance about the minor axis.
= 2117500 N = 2117.5 kN
Go to Step 5. Slenderness ratio about y-axis = 29.63 (doesn’t vary) Slenderness ratio about z-axis = 2000/63.6 = 31.45 Non dimensional slenderness ratio
Step 9: Because 2800 kN > 2117.5 kN, the column do not resist.
λ y = 0.34
(doesn’t vary)
31.45 (1) = 0.36 λ z = Values of χ: 86.39 y-axis: χy = 0.97 doesn’t vary z-axis: Curve b for λ z = 0.36 ⇒ χ z = 0.94
N.B. Please note that even the plastic capacity of the section (2750 kN) is less than the applied load (2800 kN)
Hence buckling about the z-axis becomes critical N b, Rd =
Add two plates 200 x 10 mm to reinforce the weak axis.
Now: i z =
I z = I zW + 2 I z
=
10 x 2003
5.78 x107
12 = 62 mm
= 44.5 x106 + 13.3 x106 = 5.78 x107 mm 4 4000
64.52
; then λ z = 62 = 64.52 and λ z = 86.39 = 0.72 For column curve b; χz = 0.77 And N = 0.77 x1.0 x15000 x 275 = 2887.5 kN > 2800 kN OK A
1.5 x104
b , Rd
1.1
0.94 x11000 x 275 1.1
= 2585 kN < 2800 kN . doesn ' t resist
Example-2: Determine the design buckling resistance of a 457 x 152 x 52 UB used as a pin-ended column. The column is 3m long and its steel grade is Fe 360.
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Solution:
For the 457x152x52UB profile, the actual values are : Outstand element of compression flange: c / tf = (152.4 / 2) / 10.9 = 7 < 15 OK. Web subject to compression only: d / tw =(449.8 – 2x10.9 – 2x10.2) / 7.6 = 53.60 > 42
Step 1: buckling strength is required! Step 2: Buckling length = 3000 mm.
Step 3: The section is given. Step 4: Determine the class of the cross-section and check for local buckling. For Fe-360 steel grade f y = 235 N / mm2. Thus, ε = 235 =1 f y These the limiting values for an I-section are: Outstand element of compression flange: c/t f≤15ε=15 Web subject to compression only: d / t w ≤ 42 ε = 42
Therefore, the flange satisfies the Class 3 requirement, but the web is Class 4 section. Consequently, there must be a reduction in the strength of the section to allow for the local buckling which will take place in the web. Therefore, the effective area, Aeff must be determined for the web!
The reduction factor ρ is given by, a) ρ = 1; if λ p ≤ 0.673
(λ p − 0.22) 2 b) ρ = λ
if
λ p > 0.673
p
Where λ p = b t (28.4ε k σ )
The effective width is beff = reduction factor x b = The
ρ
x b.
method to calculate the effective area (A eff) is explained in section 4.3.4 of EBCS-3.
In which : t is the relevant thickness. k is the buckling factor corresponding to the stress ratio ψ from Table 4.3 or 4.4 as appropriate. b = d for webs. σ
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In this example, since the column is axially loaded the stress distribution is uniform, i.e. σ1 = σ2. Table 4.3 is used to calculate the effective width. Thus, σ1 / σ2 = 1 & k = 4.0 (see lower part of Table 4.3 ) σ
Therefore the area that should be ignored at the center of the web is: ∆ A = (407.7 − 331.2 ) x7.6 = 581.4 mm 2 And then Aeff (6650 − 581.4) β A =
b = d = 407.6 mm b
t w
= 407.6
7.6
= 53.6
λ p = 53.6
(28.4 x1 x 4 ) = 0.944 > 0.673 (λ − 0.22) = (0.944 − 0.22) ∴ ρ = p
2
0.944
λ p
2
= 0.812
And b eff = ρ b = 0.812 x 407.6 = 331.2 mm
For h/b = 449.8/152.4 = 2.95 > 1.2; & t f = 10.9 < 40 mm;
use curve b for buckling about z-axis . Step 7: Determine the value of χ. Using Table 4.9 and interpolating, z-axis: curve b for λ z = 0.98 ⇒ χ z = 0.6034 Step 8: Design buckling resistance: N b, Rd =
γ M 1
=
0.6034 x0.913 x6650 x 235 1.1
=
6650
= 0.913
Step 5: Determine the non-dimensional slenderness ratio (axis-z govern). 3000 λ z = = 96.5 31.1 λ 1 = 93.9ε = 93.9 Hence the non dimensional slenderness ratio becomes
λ (96.5) 0.913 = 0.98 λ z = z λ β A = 93.9
1
Bending/Flexural Members
Step 6: Appropriate column curve:
χβ A Af y
A
Strength of flexural members are limited by: – Local buckling of a cross section – Lateral–torsional buckling of the entire member – Development of a plastic hinge at a particular cross section General Stability (Lateral – torsional buckling) Stability
Due to shear on the web. Local Stability
= 782660 N
Answer: The design buckling resistance will then be
N b, Rd = 782.66 kN .
Due to compressive stress on the flange Beams can be • Laterally restrained (supported) • Laterally non-restrained (unsupported)
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Restrained &unrestrained Beams Restrained Beams: • Beams which are unable to move laterally are termed restrained . and are unaffected by out-ofplane buckling (lateral-torsional instability). • Beams may often be designed on basis of bending moment resistance. • Moment resistance is dependent on section classification. • Co-existent resistance.
shear
forces,
affect
moment
Resistance to bending moment • According to EBCS-3, for bending about one axis in the absence of shearing force, the design value of bending moment .
Restrained Beams (Cont’d) Beams may be considered restrained if • Full lateral restraint is provided by for example positive attachment of a floor system to the top flange of a simply supported beam • Adequate torsional restraint of the compression flange is provided • Closely spaced bracing elements are provided such that the minor axis slenderness is low
Moment resistance • In a simple single span beam, failure occurs when the design value of the bending moment (M sd) exceeds the design moment resistance of the cross-section (M C,Rd) , • The magnitude of which depends on the section shape, material strength and section classification
Restrained Beams
The design moment resistance (M c,Rd) may be taken as: For class 1 & 2 cross-sections, the design plastic resistance moment of the gross section M c.Rd = M pl.Rd = Wpl. f y γ M0
M Sd ≤ M c, Rd
Resistance to shear • The design value of the shear force VSd at each cross-section shall satisfy
V Sd ≤ V pl , Rd
For a class 3 cross-section, the design elastic resistance moment of the gross section M c.Rd = M el.Rd = W el. f y γ M 0 For a class 4 cross-section, the design local buckling resistance W f M c.Rd = M o.Rd = eff . y γ M 1 Shear resistance the plastic shear resistance, V pl.Rd of a shear area (Av) ( f y / 3 ) V pl . Rd = Av γ MO
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Resistance for bending and shear • The theoretical plastic resistance moment of a crosssection is reduced by the presence of the shear. • For small values of the shear force this reduction is not significant and may be neglected. However, when the shear force exceeds half of the plastic shear resistance, allowance shall be made for its effect on plastic resistance moment. • That is, if the value of the shear force V Sd does not exceed 50% of the design plastic shear resistance no reduction need be made in the resistance moments • When VSd exceeds 50% of Vpl the design resistance moment of the cross-section should be reduced to M v,Rd obtained as stated in EBCS Article 4.6.1.3 (3).
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Local Stability During bending, part of the web and at least one flange is under compressive stress, therefore can be subjected to loss of stability. 1) Shear buckling resistance: Near the support, where there is a considerable acting shear force, the web of the beam can lose its stability through the formation of web folds of buckles.
2. Flange induced buckling As we can see in the figure below, the upper flange is under the action of the compressive stress and may lose it local stability
This problem is prevented by providing transverse stiffeners as shown in the figure below.
Laterally Unrestrained Beams • Beams bent about the major axis may fail by buckling in a
Clamp at root
more flexible plane
– This form of buckling involves both lateral deflection and twisting - Lateral-torsional buckling
• When the beam has a higher bending stiffness in the vertical plane compared with the horizontal plane, the beam can tw ist sideways under the action of the load as s hown in the Figure below. • The applied moment at which a beam buckles by deflecting laterally and twisting reached is the elastic critical moment
slender cantilever beam loaded by a vertical end load.
Unloaded position Buckled position
Dead weight load applied vertically
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•
A reduction factor for lateral torsional buckling , χ LT is
used to determine the capacity
• The design buckling resistance moment (Mb.Rd) of a laterally unrestrained beam is thus taken as:
M b , Rd = χ LT β wW pl , y f y / γ M 1
The elastic critical moment (Mcr) for LTB is
M cr
π 2 EI z k I w (kL) GI t 2 = C1 + + (C 2 z g ) 2 π EI z (kL)2 k w I z 2
2
M cr =
π 2 EI z I w L2 GI t + L2 I z π 2 EI z
0.5
− C2zg
0, 5
EBCS-3, Section 4.6.3 P(78-84)
The values of the imperfection factor αLT for lateral torsional buckling should be taken as: αLT = 0,21 for rolled sections αLT = 0,49 for welded sections Refer to the same table like those of members under compression and with: • for rolled sections: curve a (α = 0,21) • for welded sections: curve c (α = 0,49)
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Resistance of web to transverse forces The resistance of an unstiffened web to transverse forces applied through a flange, is governed by one of the following modes of failure: 1. Crushing of the web close to the flange, accompanied by plastic deformation of the flange. See (a). 2. Crippling of the web in the form of localized buckling and crushing of the web close to the flange, accompanied by deformation of the flange. See (b). 3. Buckling of the web over most of the member. See (c).
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Example: Bending members A simply supported beam 7.0 m span is laterally supported at the third points and carries unfactored uniform loads of 18.5 kN/m imposed load and 9.4 kN/m permanent load. In addition, the beam carries unfactored concentrated loads of 50 kN permanent load and 50 kN imposed load at mid span . Find a universal beam of grade Fe 430.
Solution: Geometry, materials and loads Factored loads: Imposed loads: Permanent loads:
q = 1.6 x 18.50 = 29.60 kN/m Q = 1.6 x 50.00 = 80.00 kN. g = 1.3 x 9.40 = 12.20 kN/m
G = 1.3 x 50.00 = 65.00 kN. Fe 430; fy = 275 N/mm2 (assume t ≤ 40 mm)
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Step 1: Maximum bending moment and shear force: 2 max M = max V =
(29.6 + 12.2) x7
+
8
(80 + 65) x7 4
(29.6 + 12.2) x7 (80 + 65) +
2
2
Step 3: Selection of the profile: The relevant section properties are:
= 509.8 kN − m
The relevant section properties are:
= 218.8 kN
Step 2: Required plastic modulus: W el =
M f y γ M 1
=
509.8 x10 2 (kN − cm) 27.5 (kN / cm2 ) 1.1
= 2039 cm
3
h = 533.1 mm
w = 0.92 kN/m
It = 76.2 cm4
d = 476.5 mm
A = 118 cm2
Iw = 1.6 x 10 6 cm6
tf = 15.6 mm
Iy = 55400 cm4
tw = 10.2 mm
Iz = 2390 cm4
b = 209.3 mm
Wel,y = 2080 cm3 Wpl,y = 2370 cm3
Try 533 x 210 x 92 UB. Class of section
ε =
235 275
209.3 2
= 6.70 < 11 x 0.92 OK 15.6 476.5 = 46.7 < 83 x0.92 OK 10.2
= 0.92
The section is at least Class 2.
Step 4: moment Resistance . (for class 2 section). M pl , Rd =
W pl f y
2370 x 275 x10
384 EI y
= 592.5 kN − m 1 .1 Check of self-weight of the beam. (w = 0.92 kN/m) Factored weight: 1.3 x 0.92 = 1.2 kN/m Additional moment: (1.2 x 7 2)/8 = 7.35 kN-m. Total moment: 509.8 + 7.35 = 517 kN-m < 592.5 kN-m OK.
γ Mo
=
Step 6: Check for deflection (SLS). 4 5wL For uniformly distributed load: δ =
−3
3
For concentrated load: δ =
Dead load deflection: δ DL =
Step 5: Check for shear. Shear resistance of section:
V pl , Rd =
and
t w
=
476.5 10.2
(
3
γ Mo V Sd V pl , Rd
=
)
=
233 816.4
(
1.04 x533.1 x10.2 275 1.1 = 0.27 < 0.5
14
385 x1.1634 x10
+
5 x18.5 x70004 384 x1.1634 x1014
50 x103 x 70003 48 x1.1634 x1014 +
= 5.59 mm.
50 x103 x 70003 48 x1.1634 x1014
= 8.04 mm
= 46.7 < 69 x 0.92 = 63.5 OK . L
Allowable deflection for imposed load: δ = 350 =
Shear buckling resistance must not be verified.
AV f y
& EI y = 2.1 x105 x55400 x1014 = 1.1634 x1014 N − mm2
5 x9.4 x 70004
Imposed load deflection: δ LL =
Maximum shear force, VSd = 218.8 + (1.2 x 7)/2 = 223 kN. d
PL
48 EI y
3
) x10−3 = 816.4 kN > 233 kN
Therefore, no reduction of design moment resistance is required.
OK .
7000 350
= 20 mm. > 8.04 mm OK
Total deflection: δ max = 5.59 + 8.04 = 13.63 mm Allowable total deflection:δ =
L
250
=
7000 250
= 28 mm > 13.63 mm OK
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Step 7: Check for lateral – torsional buckling. χ LT β wW ol , y f y . c = 209.3 = 6.70 < 10 x0.92 = 9.2 M b, Rd = 15.6 t f γ
λ LT =
M 1
The section is Class 1 and β w = 1 Determination of M cr. Lateral support to the beam is provided at the ends at the third points. Therefore the effective buckling length is: L = span/3 = 7000/3 = 2333 mm. The critical moment for lateral-torsional buckling is: π 2 EI z I w L2GI t Take C1 = 1.132; G = 80 Gpa M cr = C 1
and
G
π 2 E M cr =
2
I z
L
=
+
π 2 EI z
80000
π 2 x210000 2
2333
M cr
=
π 2 x 2.1 x105 x 2370 x103
12
1.6 x10
2390 x10
4
2
+
4
0.39 x2333 x76.2 x10 4
2390 x10
9 = 2.67 x10 N − mm
2.67 x109
λ 1 = 93.9ε = 93.9 x 0.92 = 86.8 and λ LT =
42.9 86.8
= 42.9
= 0.4942 > 0.4
For rolled section curve a is used. Therefore χ LT = 0.9250 M b , Rd =
0.9250 x1 x 2370 x103 x 275 1.1
Therefore, resistance adequate in bending.
= 0.039
2 5 4 1.132π x 2.1 x10 x 2390 x10
π 2 EW pl , y
− x10 = 548 kN − m > 517 kN − m 6
of
the
member
is
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Beam-Column members
Beam-columns are defined as members subject to combined bending and compression. In principle, all members in frame structures are actually beamcolumns
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Buckling Resistance of members with moments.
compression
The total stress due by the combined action of axial force and bending moment is:
σ N + σ M , y + σ M , z ≤ , then we can write
f y
γ M 1
σ M , y σ N σ Mz + + ≤1 f y γ M 1 f y γ M 1 f y γ M 1
and finally: N A ⋅ f y γ M 1
+
M y W y ⋅ f y γ M 1
+
M z W z ⋅ f y γ M 1
≤1
Now, taking into account the problem of the loss of stability, the design according with EBCS-3 is as follows:
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Example: A 4.00 m pin-ended column supports a beam with a reaction of 100 kN permanent load and 150 imposed load. Assuming the beam reaction to be applied 75 mm from the face of the flange. Check the adequacy of a 203 x 203 x 46 UC grade 430 steel profile.
Table 4.1
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BM-Coln
Sec 4.5.4.3
Fig 4.13
Table 4.11
Tab.4.9
BenMem bding
Table 4.12
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Built-up compression members • Build-up members, made out by coupling two or more simple profiles for obtaining stronger and stiffer section are very common in steel structures, usually for realizing members which are usually under compression. • Two of the most common arrangements for built-up members are represented in the figure to the right.
•Although the discrete nature of the connections between the two members connected by lacings and/or battenings, the models that will be described in the following for analyzing and checking built-up members are based on the assumption that the member is regular and smeared mechanical properties (such as, flexural stiffness) can be assumed throughout the member axis and utilized in calculations. •Consequently, some regularity requirements are usually imposed in designing these members and can be listed below as a matter of principles: - the lacings or battenings consist of equal modules with parallel chords; - the minimum numbers of modules in a member is three.
•The key models which can be utilized for the stability check of this kind of members will be discussed in this section.
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Shear Flexibility of members and critical load: While shear flexibility can usually be neglected in members with solid sections, built-up members are hugely affected by these parameters as a result of the axial deformation of lacings and out-of-plane flexural flexibility of the chord members.
Since imperfections play an even important role in both strength and stability checks of built-up members, a conventional eccentricity e 0 is usually introduced for simulating their effect in amplifying the axial action N Sd. For instance, EBCS-3, 1995 provides the following value of eccentricity as a function of the member span length L:
Consequently, the critical load of built-up members have to be evaluated taking into account the role of shear stiffness Sv. By definition, the shear stiffness, S v, is the force resulting in a unit value of the shear deformation γ .
Nevertheless, it is worth emphasizing the role of shear flexibility on the value of eccentricity to be adopted in verifications; indeed, since second order effects are usually of concern, the following magnified value of eccentricity has to be considered to take into account its total value:
The expression of the magnification factor considered in the above equation is based on consideration of the role of shear flexibility.
As a matter of principle, the above eccentricity of the axial force results in an external moment M which can be defined as follows:
Then the compression in one of the two connected members can be estimated as follows:
where h0 is the distance between the centroids of the two chord members.
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Laced compression members: Laced members are made out of two chords connected by a bracing system with inclined lacings, in which each segment of longitudinal profile between two braced nodes can be regarded as an isolated beam-column, whose lateral slenderness is considerably reduced at least throughout the plane of lacings. The stability check, along with all the structural verification dealing with members and connections, have to be carried out by assuming an accidental eccentricity due to imperfections. Consequently the design action in the single chord of a laced member with total axial force is N Sd can be derived according to the following equation:
The value of the critical load N cr has to be determined by neglecting the shear flexibility influence. Consequently, the usual expression can be considered: where the effective moment of inertia I eff is defined for one of the two axes which does not cross all the connected chord sections: Af being the area of the single chord section and h0 the distance between their centroids.
A virtual shear force V S, for the strength check of the connections can be determined as a function of the above N sd ⋅ e0 π ⋅ M s π eccentricity e0: V s =
L
=
L
⋅
1−
N sd N cr
−
N sd Sv
Moreover, the diagonal members have to be checked considering the following value for axial force N d:
Example:
Laced compression member
Let us consider the laced member shown below stressed in compression under an axial force whose design value is NSd=3500 kN. The member is 10 m high and simply hinged at its ends. Chord members are IPE 450 profiles while diagonals consists of steel plates with 60x12 mm 2 rectangular section both made out of grade Fe-360 steel.
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Cross-Sectional Properties:
The following values of the length-to-thickness ratios can be evaluated for flange and web:
Finally, the profile IPE 450 made out of steel FE-360 is in class 3 if loaded in compression.
Solution: Stability check according to EBCS-3 provisions: Step #1: classifying the transverse section: Since the thicknesses of the elements are less than 40mm, the yield strength of the steel used is f y=235 Mpa and the value ε=1.
The shear stiffness for the current configuration will be given by
The elastic critical load N cr can be then easily evaluated as:
in which the overall effective length L=10000 mm has been considered since the calculation is aimed at deriving the total effect of the eccentricity e 0 on the beam-column as a whole. Indeed, the moment M S induced by the eccentricity e 0 can be evaluated as follows:
taking into account the magnification effect due to second order displacements.
Step #2: evaluating the design actions: The eccentricity e0 which has to be considered for taking into account the possible imperfection effects is: The effective value of the moment of inertia of the built-up section Ieff can be also calculated as:
Finally, the actions on the various members can be easily derived as follows:
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Step #3: performing the stability check of the chord members: The reduction factor χ due to the slenderness of the member has to be calculated by looking after the possible instability in both the principal direction y and z.
The non-dimensional slenderness can be derived as a function of the elastic critical load N cr,y as follows:
Step #3.1: calculation of χy for instability in the z direction: As far as the possible instability in the plane orthogonal to the yaxis (namely, buckling in z direction) is considered, the value of the effective length coincides with the overall span length of the member, since no lacings restraints buckling in the considered direction, lying the diagonal members in a plane parallel to the y-axis. Consequently L o,y=10000 mm and the moment of inertia of the single longitudinal chord member is I y:
The profile follows the curve a and, consequently, the following value of the reduction factor χy can be evaluated:
Step #3.2: calculation of χz for instability in the y direction: On the contrary, as far as the possible instability in the plane orthogonal to the z-axis (namely, buckling in y direction) is considered, the value of the effective length coincides with the diagonal spacing, since buckling in y direction is forced by lacings to develop only between two adjacent nodes.
The profile follows the curve b and, consequently, the following value of the reduction factor χy can be evaluated as:
Consequently L0,z=1000 mm and the moment of inertia of the single longitudinal chord member is I z, have to be determined according to:
Step #3.3: calculating the axial bearing capacity : Since χy<χz, the strong axis of the built-up section is the z-axis, which is the weak one for the single member. This observation points out the huge effect of lacings in changing the behavior of the single member to obtain a built-up one. Finally, the axial load bearing capacity can be evaluated as follows:
The non-dimensional slenderness can be derived as a function of the elastic critical load N cr,z as follows:
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Battened members: Battened members are widely utilized as a technological solution for realizing beam-columns in industrial buildings. Due to the significant flexural stiffness of battenings and the related connections with the chord members, the nodes between the longitudinal profiles and the horizontal battenings is usually assumed as completely fixed, rather than hinges as usually considered for laced members. Consequently, the lateral shear flexibility of this kind of members can be determined considering the following three contributions: - bending in longitudinal members; - bending in battening; - shear strains in battenings.
Since battenings are usually assumed infinitely stiff with respect to the chord sections, EBCS-3 provided a lower bound for their moment of inertia I b with respect to one of the chord member and other geometrical parameters:
…..‘ a’ being the battening spacing.
The compressive force N f,Sd acting on the single chord member can be determined as follows for taking into account the effect of eccentricity e 0 due to imperfections:
where the moment M s is defined as follows:
The elastic critical load N cr can be evaluated according to the following expression:
and the effective moment of inertia can be estimated as follows:
In the general case, the shear stiffness S v has to be evaluated as follows:
The parameter slenderness λ:
µ
is basically defined as a function of
where the mentioned slenderness is defined as: and with I1 equal to I eff assuming µ=1
The shear force to be considered in local verifications according to the equilibrium conditions can be evaluated as already described for laced members.
