## Transcript

INFORMATION PAPER
Determination of Elastic Critical Load Factor for Steel Structures
Structural Engineering Branch Architectural Services Department November 2012 Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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TABLE OF CONTENTS
1.
Introduction .............................. ..................................................... ............................................. ............................................ .................................. ............ 1
2.
Determination of Elastic Load Factor ............................................................. ..................................................................... ........ 1
3.
Effects of Horizonal Load .......................................... ................................................................ ............................................ ........................ 3
4.
Conclusions ............................................. .................................................................... ............................................. ......................................... ................... 4
5.
References ........................ ............................................... .............................................. ............................................. ......................................... ................... 4
Annex A
– Calculation of λ λcr by Deflection Method Example 1 – Calculation
Annex B
– Calculation of λ λcr by Eigenvalue Analysis Example 2 – Calculation
Annex C
Example 3 – Considerations – Considerations in the Determination of λ λcr by Eigenvalue Analysis
Annex D
– Effects of Horizontal Load on λ Example 4 – Effects on λcr
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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1.
Introduction
1.1
Code of Practice for Structural Use of Steel 2011 (the “Code” Code”) published by Buildings Department clause 6.3.2 on the elastic critical load factor, λ factor, λcr , has been revised to clarify the determination of λ of λcr by the deflection method. However, there had not been any elaboration on the use of the eigenvalue analysis method though the correct λcr is a critical step in the design and analysis of steel structures. determination of λ
1.2
The purposes of this paper are: λcr ; a) to summarize the available methods in the Code for the calculation of λ b) to give examples to calculate λcr and to provide guidance on the correct interpretation of λ λcr found by different methods; and λcr . c) to study the effect of horizontal load on the calculation of λ
2.
Determination of Elastic Critical Load Factor
2.1
In the Code, λcr of a steel frame is defined as the ratio by which the factored loads would have to be increased to cause elastic instability. λcr is an important parameter that “non-sway”, sway”, “sway” and “ultra-sensitive “ultra-sensitive sway” is required to classify a steel frame into “nonsway”. Clause 6.3.3 of the Code classifies frames with λcr ≥10 as non-sway, non-sway, and specifies that the P- effect is insignificant insignificant for such frames. For sway frames frames with λ with λcr <10, they are classified as either sway with 5≤ 5≤ λcr <10 or ultra-sensitive sway with λ with λcr <5, and clause 6.6.1 of the Code specifies that both the P- and P-δ P-δ effects should be considered in their design. In accordance accordance with clause 6.3.3 6.3.3 to 6.3.5 of the Code, λ Code, λcr can be calculated either by the eigenvalue analysis for general structures or the deflection method for geometrically regular and rectangular frame.
2.2
Deflection method In the deflection method, a frame is classified into either sway (or ultra-sensitive sway) or non-sway by considering the magnitude of the horizontal deflection of each storey due to the application of notional horizontal loads at each storey which are taken typically as 0.5% of the factored dead plus live loads on and above the floor considered. For clad structure provided that the stiffening effect of masonry infill wall panels or diaphragms of profiled steel sheeting is not explicitly taken into account, the frame is considered as non-sway if the deflection of every storey is less than 1/2000 of the storey height under under the action of the notional notional horizontal loads. With such limit on deflection, deflection, equation (6.1) of the Code therefore gives the following equation to calculate λcr of a steel frame, except for frames with sloping members having moment-resisting connections in which the calculation of λ λcr should refer to clause 8.11 of the Code: λcr
FN
h
FV δ N
where FV is the factored dead plus live loads on and above the floor considered; FN is the notional horizontal force taken typically as 0.5% of F V for building frames;
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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and
h is the storey height; δN is the notional horizontal deflection of the upper storey relative to the lower storey due to the notional horizontal force F N.
An example is included in Annex A to show how to determine the λcr for a typical geometrically regular and rectangular frame by deflection method. Further example on the calculation of λ of λcr can also be found in NCCI: Calculation of alpha-cr (King 2005) (available: http://www.access-steel.com/ , accessed: 18 September 2012). 2.3
Eigenvalue analysis
2.3.1 The determination of λ of λcr by eigenvalue analysis is usually calculated by commercially available computer software (e.g. SAP2000, NIDA). In eigenvalue analysis, the numbers of modes to be found can be specified in the computer program and each eigenvalue found is associated with a buckled mode shape. λcr should be taken as the eigenvalue corresponding corresponding to the first “sway” sway” buckling mode. An example is included in Annex B to demonstrate the determination of λ λcr by eigenvalue analysis with NIDA and SAP2000. It should be noted that the eigenvalue determined by the computer programme may either be a “sway buckling mode” mode ” or “non-sway buckling mode” mode ” (or “local buckling mode” mode”). Figure 1 has therefore been added in the Code in its 2011 version in order to distinguish sway buckling mode from local column buckling mode. Both the P- and P-δ P-δ effects effects are only required for sway frame, and hence only λcr for sway buckling mode is required.
Non-sway buckling mode Sway buckling mode Figure 1 Sway and non-sway buckling modes 2.3.2 In the determination of λ of λcr by eigenvalue analysis, it is therefore important to study the form of each buckling mode to see if it is a sway mode or a non-sway mode. King (2005) commented that when using eigenvalue analysis in finding the first sway- mode, “it is important to study the form of each buckling mode to see if it is a frame mode or a local column mode. In frames where sway stability is ensured by discrete bays of bracing (often referred to as “braced frames”), it is common to find that the eigenvalues of the column buckling modes are lower than the eigenvalue of the first sway mode of the frame. Local Loca l column co lumn modes may also als o appear a ppear in unbraced frames at columns hinged at both ends or at columns that are much more slender than the average slenderness of noted that “where the columns columns in the same storey.” Similarly, Rathbone (2002) noted are axial load predicated, many of the lower buckling modes will be [local] column buckling modes. It is the [sway] buckling mode of the whole structure that is important ” to include second-order effects. Therefore, λcr should be taken as the factor corresponding corresponding to the first sway buckling mode rather than the local buckling mode. Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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2.3.3 Therefore, if the first mode of the eignevalue analysis analysis is a local buckling mode, then the lowest eigenvalue found does not correspond to the first sway buckling mode. Instead, the eigenvalue analysis finds the eigenvalue for the local buckling mode. In such case, PSE should take care in using the eigenvalue analysis and not just use the lowest eigenvalue which may only be local buckling mode, and it is not the original intent to use it to define sway sensitivity. Should eigenvalue analysis be adopted, PSE should scan the output to see which eigenvalue corresponds to the first sway buckling mode. Alternatively, the deflection method can be used to calculate λcr for the first sway buckling mode. An example is included in Annex C to illustrate the correct determination of λ of λcr for a typical portal frame by eigenvalue analysis. Further real example can also be found in SEB Information Paper - A Case Study of Using Metal Scaffold System for Demountable Grandstand: The Opening Ceremony of Hong Kong 2009 East Asian Games (available: http://asdiis/sebiis/2k/resource_centre/ ).
3.
Effects of Horizontal Load
3.1
Clause 6.3.2.2 of the Code states that “… of a geometrically regular and rectangular frame subjected to gravitational loads or gravitational loads plus horizontal load (e.g. …”, and this wind), the elastic critical load factor for a sway frame may be calculated as …”, means that the deflection method is applicable to frames with or without horizontal load. However, it should be noted that equation (6.1) of the Code is only a function of vertical load and the storey height height and is irrelevant of the horizontal load. Horne (1975) remarked that “[i]t should be noted that the elastic critical load of a frame is affected to an entirely negligible effect by horizontal loading, and only the vertical loads need be considered.” In order to verify the accuracy of the deflection method and Horne’s remark, the values of λcr of a simple portal frame (at Annex D) under various combinations of horizontal and vertical loading by both deflection method and eigenvalue analysis have been analysed.
3.2
From the example at Annex D, the following can be observed that: a) The results from both deflection method and eigenvalue analysis are similar when λcr is less than 10. b) When λcr becomes larger, the differences between the deflection method and eigenvalue analysis also become larger but this is not a concern as it has already been a non-sway frame with λ with λcr 10. c) Under the same applied vertical load, λcr determined by deflection method remains unchanged while that determined by eignevalue analysis remains of the same order of magnitude under the action of different magnitude of horizontal loads. d) For the case where the frame is subject to horizontal load only (from horizontal load of small to large magnitude), the values of λ of λcr so obtained are always substantially larger than 10, i.e. non-sway frame. fr ame. Therefore, all of the above observations agree with Horne ’s remark that “the elastic critical load of a frame is affected to an entirely negligible effect by horizontal loading.”
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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4.
Conclusions
4.1
λcr , PSE should use the eigenvalue In using eigenvalue analysis for the determination of λ corresponding to the first sway buckling mode, not the local buckling mode, as λcr . Alternatively, the deflection method can be used to calculate λcr for geometrically regular and rectangular frame structure.
4.2
Horizontal loading has negligible effect on λcr of a frame, and hence only the vertical λcr . loads need to be considered in the t he determination of λ
5.
References BS EN 1993:2005 – Eurocode 3: Design of Steel British Standard Institution (2005), BS EN Structures (London: BSI).
British Standard Institution (2000), BS 5950-1:2000: Structural Use of Steelwork Design of Steelwork in Building Buildin g (London: BSI). Buildings Department (2011), Code of Practice for the Structural Use of Steel 2011 (Hong Kong: Buildings Department) (available: http://www.bd.gov.hk/english/ documents/ , accessed: accessed: 18 September 2012). Buildings Department (2007), Explanatory Materials to Code of Practice for the Structural Use of Steel 2005 (Hong Kong: Buildings Department) (available: http://www.bd.gov.hk/english/documents/ , accessed: 18 September 2012). Horne, M R (1975), “An Approximate Method for Calculating the Elastic Critical Loads of Multi-storey Plane Frames,” Frames, ” The Structural Engineer , 6(53), pp. 242-248 King, C (2005), NCCI: Calculation of Alpha-cr (Ascot: http://www.access-steel.com/ , accessed: accessed: 18 September 2012).
SCI)
(available:
Rathbone, A J (2002), “Second-order “Second -order Effects – Who Needs Them?” The Structural Engineer , 80(21), pp. 19-21 (available: www.istructe.org www.istructe.org,, accessed: 18 September 2012).
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Annex A Example 1 Calculation of cr cr by Deflection Method
This example shows the procedures to determine the value of cr cr of a 3-storey portal frame at 5m c/c (Figure A1) by using the deflection method. Design Loadings: 2
Dead Load = 7.5 kN/m 2 Live Load = 5.0 kN/m Steel Members: Steel Grade: Beam Size: Column Size:
S355 356×171×51 356×171×51 kg/m UB 254×254×107 254×254×107 kg/m UC
Design Load Case
= 1.4
Factored Load
= (1.4
Load per floor
= 92.5 12
DL + 1.6
LL
7.5 + 1.6 5.0) 5 = 92.5 kN/m
Therefore, Notional Horizontal Force
= 1110 kN = 0.5% of vertical load = 0.5% 1110 = 5.55 kN
Figure A1
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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Annex A
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Annex A
The structrue was analsyed with QSE, and the displacement of each floor obtained is shown in Figure A2:
Figure A2 Deformed Shape 1
= 5.585 mm
h1 = 3.0m
2
= 4.230 mm
h2 = 3.0m
3
= 2.704 mm
h3 = 3.0m
Using equation (6.1) of the Code, cr,1 cr,1
=
cr,2 cr,2
=
cr,3 cr,3
=
5.55 1110
3000
5.585 4.230
5.55 2 1110 2 5.55 3 1110 3
3000
4.230 2.704 3000
2.704
= 11.07 10 = 9.83 10 = 5.55 10
Therefore, the value of cr cr should be taken as the smallest of cr,1 cr,1, cr,2 cr,2 and cr,3 cr,3, i.e. 5.55.
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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Annex A
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Annex B Example 2 Calculation of cr cr by Eigenvalue Analysis
The same example at Annex A is now analysed by the eigenvalue analysis with NIDA and SAP2000. Design Loadings: 2
Dead Load = 7.5 kN/m 2 Live Load = 5.0 kN/m Steel Members: Steel Grade: Beam Size: Column Size:
S355 356×171×51 356×171×51 kg/m UB 254×254×107 254×254×107 kg/m UC
The computer analysis by NIDA and SAP2000 is carried out in accordance with the t he following steps: (A) NIDA Step 1: The geometry of the structure to be analyzed is input in t he computer model with the factored design load case of 1.4 DL +1.6 LL.
Structural Engineering Branch, ArchSD Information Paper on Calculation of Elastic Critical Load Factor for Steel Structures Issue No./Revision No. : 1/-
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Annex B
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Annex B
Step 2: In the analysis option, select the “Eigen-Buckling “Eigen-Buckling Analysis” Analysis” as the analysis type.
Step 3: After the completion of analysis, the load factor corresponding to the first buckling mode for the load combination is found to be 5.36.
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Annex B
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Annex B
Step 4: Check the mode shape of the model to confirm if the structure deforms in a sway buckling mode.
Therefore, λ Therefore, λcr should be taken as 5.36 (the load factor corresponding to the first sway buckling mode) which is about the same value as that obtained from the deflection method at Annex A. (B) SAP2000 Step 1: The geometry of the structure to be analyzed is input in the computer model with the factored design load case of 1.4 DL +1.6 LL.
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Annex B
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Annex B
Step 2: In the definition of load cases, cases, select the “Buckling “Buckling”” as the load case type.
Step 3: After the completion of analysis, the load factor corresponding to the first buckling mode for the load combination is found to be 5.29. Check the mode shape of the model to confirm i f the structure deforms in a sway buckling mode.
Therefore, λ Therefore, λcr should be taken as 5.29 (the load factor corresponding to the first sway buckling mode) which is about the same value as that obtained from the deflection method in Annex A and NIDA.
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Annex B
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Annex C Example 3 Considerations in the Determination of cr cr by Eigenvalue Analysis
The value of λ of λcr of a single storey portal frame with a diagonal bracing ( Figure C1) is determined by both the deflection method and eigenvalue analysis. analysis. Design Loading: 1000kN at ¼th span of the first bay (factored (f actored load) Steel Members: Steel Grade: Beam Size: Column Size: Bracing Size:
S355 152 × 152 × 23 kg/m UC 152 × 152 × 23 kg/m UC 88.9 × 6.3mm CHS 1000 kN
3m
3m
3m
Figure C1
By Deflection Method 5 kN
Notional horizontal defelction defelction due to the notional notional horizontal force (0.5%×1000 = 5kN) is 0.149mm (using computer software NIDA) Therefore, λ Therefore, λcr = 3000/200/0.149 = 100.6
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Annex C
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Annex C
By Eigenvalue Analysis
Mode
Mode Shape
st
1
Sway Mode
The ratio by which the load has to be increased to cause elastic instability
Non-sway buckling mode/ Local column buckling mode
7.1
Non-sway buckling mode/ Local column buckling mode
25.1
Non-sway buckling mode/ Local column buckling mode
27.4
Sway buckling mode
99.0
Lateral deflection at top of frame = 0.72mm
nd
2
Lateral deflection at top of frame = 2.69mm
rd
3
Lateral deflection at top of frame = 2.3mm
4
th
Lateral deflection at top of frame = 51.5mm
Therefore, λ Therefore, λcr should be taken as 99.0 (the load factor corresponding to the first sway buckling mode) which is of about the same value as that obtained from the deflection method.
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Annex C
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Annex D Example 4 Effects of Horizontal Load on cr cr
The value of λ of λcr of a single storey portal frame ( Figure D1) is determined by both the deflection method and eigenvalue analysis under different combinations of horizontal and vertical loads. Steel Members: Steel Grade: S355 Beam Size: 254 × 146 × 373 kg/m UB Column Size: 203 × 203 × 46 kg/m UC Vertical Load, Fv Horizontal
B
C
Load, Fh 3m
A
D 4.5m
1.5m
Figure D1 Vertical Load, Fv (kN)
λ cr(Deflecton)
λ cr(Eigenvalue)
Ratio
Case
Horizontal Load, Fh (kN)
(Deflection Method)
(Eigenvalue Analysis)
( λ cr(Deflection) / λ cr(Eigenvalue))
1 2 3 4 5 6 7 8 9 10 11
50 100 200 400 50 100 200 400 10 20 30
400 400 400 400 50 50 50 50 10 10 10
8.09 8.09 8.09 8.09 64.7 64.7 64.7 64.7 332.93 332.93 332.93
7.5 7.5 7.4 7.2 59 56 48 34 295 278 258
1.08 1.08 1.09 1.12 1.12 1.17 1.36 1.90 1.13 1.20 1.29
12 13
40 50
10 10
332.93 332.93
239 219
1.40 1.52
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Annex D
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Annex D
From the analysis results, it can be observed that: a) b)
c)
The results of both deflection method and eigenvalue analysis are similar when λcr is less than 10. When λcr becomes larger, the differences obtained from the deflection method and eigenvalue analysis also become larger but this is not a concern as it has already been a non-sway frame for λ for λcr 10. Under the same applied vertical load, λcr as determined by deflection method remains unchanged unchanged while that as determined by eignevalue analysis remains of the th e same order of magnitude under the action of different magnitude of horizontal loads.
λcr of the same structure under the action of horizontal load only ( Figure D2) has Values of λ also been determined as follows:
Horizontal
B
C
Load, Fh 3m
A
D 6m
Figure D2
Case
Horizontal Load, Fh (kN)
Vertical Load, Fv (kN)
(Eigenvalue Analysis)
Lateral Deflection (mm)
Deflection/ Height
14 15 16
10 20 30
0 0 0
1783 891 594
9 18 27
1/333 1/167 1/111
17 18 19 20 21
40 50 100 200 400
0 0 0 0 0
446 356 178 89 45
36.1 45.1 90.2 180.3 360.6
1/83 1/67 1/33 1/17 1/8
λ cr(Eigenvalue)
In the above cases where the frame is subject to horizontal load only (from horizontal load of small to large magnitude), λcr so obtained are always substantially larger than 10, i.e. nonsway frame and therefore the horizontal load will not affect sway stability for frame classification.
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Annex D
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