5. Imperfections. Josef Machacek. Czech Technical University in Prague

Transcription

1 5. Imperfections Josef Machacek Czech Technical University in Prague

2 Objectives of the lecture This lecture describes various forms of of structures. It is shown, how these, whether at all, should be introduced into. Finally some basic examples are presented. 2

3 Outline of the lecture global 4. Imperfections of structures for of Imperfections versus

4 1. Geometrical : variance of dimensions of a structure or a member, e.g.: b ± Δ lack of verticality of a structure and straightness or flatness of a member, e.g.: φ h φ h e 0 4

5 1. Material : variance of material properties, e.g.: σ ± Δ ε residual stresses (distribution in a cross section usually considered uniform along the member): σ r f y σ r e. g. in I sections (both hot-rolled and welded) 5

6 1. Structural : variance of boundary conditions, eccentricities in joints, e.g.: e 0 6

7 2. : In strut (frame) all types of are usually introduced as equivalent geometrical (with increased value of amplitude e ). 0d In plate (plated structure) t geometrical ti and residual stresses are introduced to derive buckling factors. 7

8 2. In frame the following shall be introduced: of frames or bracing systems (cover lack of verticality for frames or straightness of structure restrained by ) Local (member) of individual members (cover lack of straightness or flatness of a member and residual stresses of the member) 8

9 2. Other mentioned are covered by partial factors in the Limit State Design procedure. In introduction of the equivalent geometrical (i.e. deflections) there is necessary to determine: shape of the deflection (buckling mode); amplitude of the deflection. 9

10 3. global 1. In general, the first critical buckling mode (η cr ) of the structure may be investigated and applied as imperfection shape for GNIA. The amplitude of the shape (e 0d ) shall be determined d from Eurocode 3, eq , securing required reliability in the most axially stressed cross section. 2. Approximately, the global imperfection in sway mode (φ) and local geometrical (e 0d ) of individual members may be introduced, see next page: 10

11 3. global sway φ: V 1 φ V V 1 1 V V 2 2 φ V 2 φ For value of φ see Eurocode 3, eq In general, the sway are introduced into as corresponding horizontal loadings H i = φ V i. Sway may be disregarded if the rate of horizontal/vertical loading is high ( 0,15), so that their contribution to internal forces is negligible. 11

12 3. global Local geometrical are given as values e 0d /L in Eurocode 3 (Tab. 5.1), which may be replaced by corresponding transverse uniform loadings giving the same bending moments. N Ed N Ed 4 NEde0d L e 0d Ed 2 8 N L e 0d N Ed N Ed 4 N Ed L e 0d 12

13 3. global Usually these local are ignored in global and covered by reduction factors χ and χ LT in member checks, unless the frame is sensitive to 2nd order effects, i.e.: -a member has at least one moment resistant end joint; - and has simultaneously high slenderness given in Eurocode 3, eq

14 4. Imperfections of structures for of Bracing systems may provide lateral stability of a strut in compression or a beam in bending. The strut/beam should be considered with a geometrical imperfection (initial bow) of amplitude e = 0 L/500 or less, taking number of strut/beams into account according to Eurocode 3, eq The bow with amplitude e 0 may be replaced by transverse uniform loading q d. 14

15 4. Imperfections of structures for of The loading q d corresponds to impact of sum of the amplitude e 0 and the in-plane deflection of the bracing system δ q. Such requires 2 nd order calculation or iterative procedure, see Eurocode 3, eq. 5.13: member in compression (or compression flange force of a beam) bracing system N Ed e 0 N Ed q N d = Ed8 e 0 + δ L 2 q L 15

16 4. Imperfections of structures for of s supporting a splice of compression members have to be verified for additional force N Ed /100. N Ed N Ed 100 N Ed 16

17 Formative Assessment Question 1 Describe types of. How the are introduced into design of a steel structure? Describe form of global and their design model. Explain form of imperfection for (e.g. rafter bracing in a roof of an industrial building) and how to encompass it for design. 17

18 5. Imperfections In general, the influence of member (local) is covered by reduction factors (in columns and beams by χ, χ χ LT, in plates by ρ, χ w, χ F ). Instead, when using GNIA (geometrically nonlinear with or approximate second order ), the of critical shape are taken with amplitudes as follows: 18

19 5. Imperfections For compression struts the equivalent initial bow may be used with form of flexural buckling and amplitude e 0d in accordance with Eurocode 3 (Tab. 5.1, e 0d /L given), e.g.: L is system length; e e 0d depends on buckling curve 0d and type of. For beams in bending only equivalent initial bow in the direction of weak axis of the beam may be used, with amplitude 0.5 e 0d (where e 0d is as above): e 0d 19

20 5. Imperfections For plates the geometric should have amplitude of 80 % of assembly and residual stresses according to fabrication (say - compression stresses from 0.10 f y to 0.25 f y ) or equivalent initial plate deflections with amplitude b/200 only and equivalent initial stiffener bow with amplitude L/400. For unstiffened compression plating e.g.: e 0 = b/200 welds f y σ res = f y b 20

21 6. Imperfections versus Tolerances of structures/members are given in product standards and Eurocode EN Eurocode distinguishes essential (required for due resistance) and functional (class 1 and more rigorous class 2 for fit up and appearance requirements). Essential have to be confirmed by inspection and testing to determine quality of the structure. 21

22 6. Imperfections versus Comparison of some e 0d for and essential : columns: e = 0d L/100 L/350 tolerance: L/750 girders: 0,5 e 0d, i.e. L/200 L/700 tolerance: L/750 22

23 6. Imperfections versus unstiffened plates and stiffeners: plates: stiffeners: e 0d = b/200 e 0d = b/400 tolerance: b/100 tolerance: b/400 b (obviously incorrect) frames, e.g. simple portal frame: Δ Δ d = h/200 h tolerance: h/500 23

24 Formative Assessment Question 2 How member are commonly introduced into design? How member are introduced into GNIA? Explain differences between and. 24

25 7. Example 1: Two-bay braced frame of a building Example 2: Portal frame Example 3: Rafter bracing 25

26 7.1 Example 1 Example 1: Two-bay braced frame of a building i f ti The frames spaced at distance of 6 m, bracing each 12 m. B HE 160 B HE 160 B HE Geometry and cross sections: composite floor beams: A= 9345 mm 2, I = mm 4 26

27 7.1 Example 1 Loading [kn] and reactions imp imp imp 3 H Ed,1 H Ed,2 V Ed,1 V Ed,2 V Ed,3 H Ed,3 Note: Wind (horizontal) loading is due to this bracing. 27

28 7.1 Example 1 For formulas see Eurocode 3: Σ H Ed = kn i.e. < 0.15 ΣV Ed = kn ( φ need to be considered) Ed Ed Sway imperfection for global : 2 2 α h but α h,min = 3 α 2 = = h = = 1 + = m 3 m φ = φ0 αh αm = =

29 7.1 Example 1 ( ) 1 6 kn imp 1= φ V = =. imp 2 = imp 3 = = φ V = =. ( ) 1 5 kn These belongs to each cross frame. For of the bracing frame appropriate total values (as for wind loading) need to considered. Here they are doubled (belonging g to two cross frames): 29

30 7.1 Example 1 Internal forces due to loading + doubled : M Ed [knm] N Ed [kn] 30

31 7.1 Example 1 Local for global only if simultaneously (bottom central column): - exists moment resistant end joint: no (M Ed 0) - slenderness Af y λ > 0, 5 = 0, 5 = NEd true, because: λ λy = λ 1 = 4200 / = Therefore, the local could be ignored for global in this example. 31

32 7.1 Example 1 local In LA (linear ) the local are covered by reduction factors (χ and χ LT ). 32

33 7.1 Example 1 In GNIA (see lecture 6 for details) The following should be used: 1. Generally either together with sway also approximate local (sinusoidal) bows with amplitudes in accordance with Eurocode 3, Tab. 5.1: Buckling curve Elastic Plastic e 0 /L a 0 1/350 1/300 a 1/300 1/250 b 1/250 1/200 c 1/200 1/150 d 1/150 1/100 33

34 7.1 Example 1 - i.e. at columns (HE profile): e 0 = L/250 = 4 200/250 = 16.8 mm (buckling curve b) - at composite beam approx.: e 0 = L/200 = 4 200/200 = 21.0 mm (buckling curve c) - at bracing diagonals (L profile): e 0 = L/250 = 3 662/250 = 15.0 mm (buckling curve b) Note: For this example however, the local can be ignored as shown above. 34

35 7.1 Example 1 2. Or unique global and local imperfection in the shape of the critical buckling mode (received from LBA) corresponding to buckling of respective member with amplitude e 0. The first buckling mode in the present frame corresponds to buckling of the central column (non-sway mode): column (non sway mode): e 0 The first critical buckling mode: α c1 = 5.51 c,1 (Note: The first sway mode is the 15 th, where α cr,15 = ) 35

36 7.1 Example 1 Example: Warning: In some cases the first critical mode corresponds to other (less important) members, e.g. hinged diagonals. The critical mode corresponding to required member (e.g. column) may be of higher level. This higher mode shall be taken for column (otherwise the solution is conservative). If in our frame the bottom diagonals are 2L 70x6 and overhead diagonals 2L 60x6, the 4 th critical mode should be taken for column design: 1 st mode 2 nd mode 3 rd mode 4 th mode α cr,1 = 1.66 α cr,2 = 2.16 α cr,3 = 4.80 α cr,4 =

37 7.1 Example 1 According to Eurocode 3, eq. 5.10: e 0 = α χ λ 1 γ MRk M1 ( λ 0. 2) 2 N Rk 1 χ λ 2 - where for central bottom column: Nc, Rk = Afy = = α N 3 c,rk ult, k = = = 1. 3 NEd N 37

38 7.1 Example 1 - for diagonal 2 x L110/10 (other members not relevant): 3 N c,rk = Af y = = N 3 Nc,Rk αult, k = = = N Ed Ed Lower, i.e. column decides. λ = α α ult,k cr = = For buckling curve b (α = 0.34): χ =

39 7.1 Example 1 M pl, Rk 3 = W f = = pl y 6 knm Resulting amplitude of in the shape of the first critical buckling mode: e = 0 = α ( λ 0. 2) M N Rk Rk χ λ 1 γ 1 χ λ = M1 2 = ( ) mm

40 7.2 Example 2 Example 2: Portal frame global linear HE 340 B IPE imp 1 H Ed,1 12 kn/m' 40 kn 40 kn H Ed,2 V V Ed 2 loading and reactions V Ed,1 Ed,2 geometry and cross sections 40

41 7.2 Example 2 For formulas see Eurocode 3: Σ H Ed = 0 i.e. < 0.15 V Ed (consider φ) Sway imperfection for global (imp 1): α h = b t α h,min = 3 but = h αm = = = 0.87 m φ = φ 0 α h α m = 0.87 = ( ) kn imp 1= φ V = =

42 7.2 Example 2 Internal forces (loading + ): -374,6 387,1 144,5-38,7 143,5-483,2-183,5-184,5-37,5 38,7 M Ed [knm] N Ed [kn] V Ed [kn] M Ed [knm] N Ed [kn] V Ed [kn] 42

43 7.2 Example 2 Local for global only if simultaneously (column concerned): -exists moment resistant t end joint: OK - slenderness λ > Afy , 5 = 0, 5 = 2, 33 N Ed not true, because λy λ = = λ / = 0.73 There, the local can be ignored in global. 43

44 7.2 Example 2 local In LA (linear ) the local are covered by reduction factors (χ and χ LT ). 44

45 7.2 Example 2 Alternative GNIA In GNIA the following should be used: 1. Generally either together with sway also approximate local (sinusoidal) bows with amplitudes in accordance with Eurocode 3, Tab. 5.1: at columns: e 0 = L/250 = 10000/250 = 40 mm (buckling curve b) at beam: e 0 = L/300 = 24000/300 = 80 mm (buckling curve a) Note: For this example however, the local can be ignored as shown above. 45

46 7.2 Example 2 2. Or unique global and local imperfection in the shape of the first critical buckling mode received from LBA, with amplitude e 0 : p 0 e 0 The first critical buckling mode: α cr,1 =

47 7.2 Example 2 According to Eurocode 3, eq. 5.10: e 0 = α χ λ 1 γ MRk M1 ( λ 0. 2) 2 N Rk - where for columns: 1 χ λ Nc, Rk = Afy = = N α 3 c,rk ult, k = = 3 Ed = N N 21,8 47

48 7.2 Example 2 - for beam: Nc, Rk = Af y = = N α 3 c,rk ult, k = = 3 Ed = N N 81,6 Lower, i.e. column decides. λ = α α ult,k cr 218, = = For buckling curve b (α = 0.34): χ =

49 7.2 Example 2 M pl, Rk = W f pl y = = knm Resulting amplitude of in the shape i f ti of the first critical buckling mode: e ( λ 0.2) M N χ λ 1 γ 0 Rk M1 = α 2 = Rk 1 χ λ = ( ) = 75.2 mm 2 49

50 7.3 Example 3 Example 3: Rafter bracing purlin rafter IPE x 6 = 60 m L = 8 x 3 = 24 m q d = 1 knm 6 m δ q = 4.5 mm 24 m Plan of the roof Deflection of bracing 50

51 7.3 Example 3 Initial deflections with amplitude e 0 of the bracing system will be replaced by equivalent stabilizing force q d : q d e 0 = α m L/500 Data from former calculations: l - max. moment in the rafter: M Ed = knm - max. force in the compression flange of the rafter: N Ed = M Ed /h = 362/ = kn - external loading per one bracing system: q d,ext = 3.70 kn/m Number of braced flanges per one bracing system: m = 11/3 =

52 7.3 Example 3 Amplitude e 0 : α = = = m e0 = αm L / 500 = / 500 = 38 4 mm 0. Equivalent stabilizing loading q d requires iterative procedure. To avoid the iteration, suitable guess of the total deflection δ q,(0) from stabilizing loading q d and all external loading q dext d,ext is necessary. Say: δ L = 500 ( 0) 48 0 mm q. 52

53 7.3 Example 3 and therefore the equivalent stabilizing loading : e0 + δ q(0) 3 N 8 = ( ) q d = Ed 8 8 = N/mm 2 L Check of the guess of δ q(0) : δ ( q + q ) δ = ( ) mm q ( 1) = d d,ext q(q= 1) =. The guess was OK, because conservative: δ q(0) = 48.0 mm > δ q(1) = 30.1 mm 53

54 8. 1) Imperfections significantly influence strength of structures. 2) In frame structures equivalent geometrical (initial deflections) may substitute all kinds of. 3) In plated structures preferably both initial deflections and residual stresses should be introduced into design. 4) Generally, global and local have to be considered. 54

55 8. 5) Shape of the initial deflections is generally given by the first critical mode, approximately in the form of initial sway imperfection and individual bow of members. 6) Amplitude of the initial shall correspond to values given in Eurocode 3 (chapter 5.3.2) to secure required reliability of design. 7) In common design, the influence of is usually covered by global geometrical and reduction factors for members. 8) Compression residual stresses shall correspond to expected mean values. 55

56 to Users of the Lecture This session is for of structures t and requires about 60 minutes lecturing and 60 minutes for tutorial session. Within the lecturing, three types of necessary to account for in of a structure are described. In particular, introduction of global, for bracing systems and of individual members in compression and bending are shown. Attention is also paid to required by Eurocode EN 1090 for execution. Further readings on the relevant documents from website of and relevant standards of national standard institutions are strongly recommended. Formative questions should be well answered before the summative questions completed within the tutorial session. Keywords for the lecture: initial deflections, residual stresses, global, for bracing systems, member, buckling mode, equivalent horizontal force,. 56

57 for lecturers Subject: Imperfections of structures. t Lecture duration: 60 minutes plus 60 minutes tutorial. Keywords: initial deflections, residual stresses, global, for bracing systems, member, buckling mode, equivalent horizontal force,. Aspects to be discussed: types of, necessity of their introduction into. Within the lecturing, the introduction of global and member imperfection should be practised and for bracing system in a roof of an industrial building as well. Further reading: relevant documents and relevant standards of national standard institutions are strongly recommended. Preparation for tutorial t exercise: see examples within the lecture. 57