1. Trang chủ
  2. » Kỹ Thuật - Công Nghệ

Marine_Structural_Design Episode 5 potx

40 89 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

138 PART I Strucntral Design Principles Minimize degradation at system level so that when local fatigue degradation occurs, there are no significant effects on the system’s ability to perform satisfactorily. Here good fatigue design requires system robustness (redundancy, ductility, capacity) and system QA. Inspections and monitoring to disclose global system degradation are another strategy to minimize potential fatigue effects. Cyclic strains, material characteristics, engineering design, specifications, and life-cycle QA (inspections, monitoring) are all parts of the fatigue equation. This is the engineering equation of “fail safe design” fatigue may occur, but the structure can continue to hction until the fatigue symptoms are detected and repairs are made. The alternative is “safe life design” no significant degradation will occur and no repairs will be necessary. Safe life designs are difficult to realize in many long-life marine structures or elements of these structures. This is because of the very large uncertainties that pervade in fatigue design and analysis. Safe life design has been the traditional approach used in fatigue design for most ocean systems. The problems that have been experienced with fatigue cracking in marine structures and the extreme difficulties associated with inspections of all types of marine structures, ensure that large factors of safety are needed to truly accomplish safe life designs. For this reason, fail-safe design must be used whenever possible. Because of the extreme difficulties associated with inspections of marine structures and the high likelihood of undetected fatigue damages, it is not normally reasonable to expect that inspections will provide the backup or defenses needed to assure fatigue durability. 7.4 1. 2. 3. 4. 5. 6. References NTS (1998), “NORSOK N-004, Design of Steel Structures”, Norwegian Technology Standards Institution, (available from: www.nts.no/norsok). API (2001), “API FV 2A WSD, Recommended Practice for Planning, Designing and Constructing Fixed Offshore Platforms - Working Stress Design”, American Petroleum Institute, Latest Edition. API (1993), “API Rp 2A LRFD - Recommended Practice for Planning, Designing and Constructing Fixed Offshore Platforms - Load and Resistance Factor design, First Edition. 1993. API (2001), “MI FW 2FPS, Recommended Practice for Planning, Designing and Constructing Floating Production Systems”, First Edition. API (1997), “API FV 2T - Recommended Practice for Planning, Designing and Constructing Tension Leg Platforms”, Second Edition. IS0 Codes for Design of Offshore Structures (being drafted). Part 11: Ultimate Strength Part I1 Ultimate Strength Chapter 8 Buckling/Collapse of Columns and Beam-Columns 8.1 This Chapter does not intend to repeat the equations and concept that may be found from exiting books on buckling and ultimate strength, e.g. (Timoshenko, 1961 and Galambos, 2000). Instead, some unique formulation and practical engineering applications will be addressed. 8.1.1 Buckling Behavior For a column subjected to an axial force, the deflection produced by the axial force will be substantially amplified by the initial imperfections. Buckling Behavior and Ultimate Strength of Columns P II P Y Figure 8.1 Coordinate System and Displacements of a Beam-Column with Sinusoidal Imperfections Let's consider the case in which the initial shape of the axis of the column is given by the following equation, see Figure 8.1 : .m w, = w,,, wn- 1 Initially, the axis of the beam-column has the form of a sine curve with a maximum value of worn in the middle. If this column is under the action of a longitudinal compressive force P, an additional deflection w, will be produced and the final form of the deflection curve is: w=w,+w, (8.2) The bending moment at any point along the column axis is: M = P(wo + w,) (8.3) 142 Part XI Ultimate Strength Then the deflection w, due to the initial deformation is determined from the differential equation: EI- - Substituting Eq. (8.1) into Eq. (8.4), we may obtain: d2wl m -+ k2w, = -k2w,,, sin- dX2 I where, The general solution of Eq. (8.5) is: 1 ?r2 k212 .m w,,, sin - wI = Asinkx+Bcoskx+- 1 I (8.4) To satisfy the boundary condition (w, = 0 for x = 0 and B = 0. Also, by using the notationa for the ratio of the longitudinal force to its critical value: x = 1)for any value of k, A = where, Z’EI PE =- I2 we obtain the following: a m w, =- wonax sin - 1-a I The final form of the deflection curve is: (8.9) This equation shows that the initial deflection w,,, at the middle of the column is magnified at the ratio - by the action of the longitudinal compressive force. When the compressive force P approaches its critical value, a approaches 8.0, the deflection w increases infinitely. Substituting Eq. (8.9) into Eq. (8.3), we obtain: w=wo+wl =w,,,sin-++w,,,sin-=~sin- ma .m m I 1-a I 1-a I a 1-a (8.10) Chapter 8 BucklingKollapse of Columns and Beam-columns 143 I 2 From Eq. @.lo), the maximum bending moment at x = - may be obtained as follows: I 2 The maximum stress in the cross section where x = - is: ', MMAX OMM: =- AW Eq. (8.12) may be rewritten as follows: where, W = Section modulus A C = Area of the cross section = Distance from the neutral axis to the extreme fiber r W S = Radius of the core: s = - A = Radius of gyration of the cross section By taking the first term of the Fourier expansion pE -I+-+ - + (3 P-P, PE we obtain: P PE -1+- P-PE PE Combining Eq. (8.14) and Eq. (8.13), the maximum stress is given by: (8.1 1) (8.12) (8.13) (8.14) (8.15) 8.1.2 Perry-Robertson Formula A simple method to derive the ultimate strength of a column is to equate oMAX in Eq. (8.12) to yield stress cry : (8.16) 144 The above equation may be written as: OiLT -[ +(l++)OE]Ou" +UEQY =o where, Its solution is called the Perry-Robertson Formula and may be expressed as: QULT J+rl+Y-JGnT CY 2Y where, Part II Utimate Strength (8.17) (8.18) In Perry-Robertson formula, the effect of initial deflection is explicitly included. Comparison with more precise solutions such as finite element results demonstrates that the formula is accurate when the initial deflection is at the range of fabrication tolerance. When the initial deflection is due to in-service damages, which may be up to 1 per cent of the column length, the formula may under-estimate ultimate strength. The formula may be extended to account for the effect of residual stress explicitly. Perry-Robertson Formula has been frequently used in European steel structure codes. 8.1.3 Johnson-Ostenfeld Formula The effect of plasticity may be accounted for by correcting the Euler buckling stress using the Johnson-Ostenfeld approach (see Galambos, 2000), Figure 8.2: QULT = QE for Q, /cy S 0.5 (8.19) (8.20) ,A 4 1.0 g PT * inelastic elastic Figure 8.2 Johnson-Ostenfeld Approach Curve Chapter 8 Buckling/Collapse of Columns and Beam-columns 145 The Johnson-Ostenfeld approach was recommended in the first edition of the book "Guide to Stability Design Criteria for Metal Structures" in 1960 and has been adopted in many North American structural design codes in which a moderate amount of imperfection has been implicitly accounted for. The Johnson-Ostenfeld formula was actually an empirical equation derived from column tests in the 1950s. It has since then been applied to many kinds of structural components and loads, see Part 2 Chapters 10 and 11 of this book. 8.2 8.2.1 Beam-Column with Eccentric Load Buckling Behavior and Ultimate Strength of Beam-Columns Figure 8.3 Beam-Column Applied Eccentrical Load Consider a beam-column with an eccentricitye, at each end, see Figure 8.3. The equilibrium equation may be written as: d2W ak2 EI- + P(w+ e, ) = 0 The general solution of Eq. (8.21) is: w= Asinkx+bcoskx-e, Using boundary conditions I w =o ut x=f- 2 d2w I EI- =-Pel ut x=+- dr2 2 the integral constant may be obtained and the solution of Eq. (8.21) is: w=e, see-coskx-1 c 1 The maximum deflection at the middle of the beam-column is given by: (8.21) (8.22) (8.23) (8.24) kl wMM = e, sec- 2 The maximum moment and stress at the middle of the beam-column are expressed as follows: 146 Part II Ultimate Strength 1 M, = Pel - kl 2 cos - (8.25) (8.26) Eq. (8.26) is called the secant formula. Taking the first two terms of the formula expansion: and substituting Eq. (8.27) into Eq. (8.28), we obtain: n2 e, P (8.27) (8.28) 8.2.2 The deflection for a beam-column in Figure 8.4 may be obtained easily by superposition of Eq. (8.9) and Eq. (8.23), the total deflection is: Beam-Column with Initial Deflection and Eccentric Load sin-+A cos kx -cos- [ (: ) :] w=- Woma e kl 2 cos- 1 -a The maximum deflection occurs at the center of the beam-column: The bending moment at any section x of the beam-column is: M = P(e, + w) r 1 kl (: )I 2 me =p w,,, sin-+Lcos kr 1 cos - (8.29) (8.30) (8.31) and the maximum moment at the center of the beam-column is: (8.32) From Eq. (8.15) and Eq. (8.28), the maximum stress at the center of the beam-column is: (8.33) Chapter 8 BucklingKoIlapse of Columns and Beam-columns 147 Figure 8.4 An Initially Curved Beam-Column Carrying Eccentrically Applied Loads 8.2.3 Ultimate Strength of Beam-Columns For practical design, a linear interaction for the ultimate strength of a beam-column under combined axial force and bending is often expressed as: - +- Muq1 (8.34) where PuLT and MuLT are the ultimate strength of the beam-column under a single load respectively. Based on Eq. (8.34), the maximum moment in a beam-column under combined axial forces and symmetric bending moments M, , is given by: P ’ULT *ULT (8.35) Then, the ultimate strength interaction equation may be expressed as: (8.36) Determining the exact location of the maximum bending moment for beam-columns under non-symmetric bending moments is not straightforward. Instead, M, is substituted by an equivalent moment, MEQ = CUM,. where (Galambos, 2000) Cu =0.6-0.4-20.4 MB M* (8.37) (8.38) where MA and MB are end moments. [...]... Wall- (-) (-1 Area A(m2) 00.2 p 02 (kpf/mm2) ( k g f ) 00.2 po.2 00 .5 4 .5 (kgWm2) (kgfl-’) (kgf) A 1.0 97.0 301 .59 45. 00 13 751 .55 35. 25 10631. 05 40.00 12063.60 B 1.2 81.2 362.67 58 .00 21034.86 37 .50 13600.13 44 .50 16138.82 C 1.6 61.4 4 85. 56 54 .23 26341.63 37.00 179 65. 72 42. 75 20 757 .69 D 2 .5 40.0 7 65. 76 46. 75 357 99.28 33.00 252 70.08 38. 25 29290.32 Part II Ultimate Strength 160 P 40 - - 20 8 I tension... (o t n t no ) 19,6 45 35. 25 7 .51 DENT % 19,6 45 35. 25 5. 75 DENT 0.13 1116 19,6 45 35. 25 9.78 DENT 0. 25 1/16 19,6 45 35. 25 1,6 35 0.10 1/32 19,616 37 .50 9.90 DENT 1.430 0.61 1/16 19,616 37 .50 9.10 DENT I 2 1,430 I 02 I /8 19,616 37 .50 7. 95 DENT 98.2 I 6 1,430 0.44 1/32 19,160 3 7m 13.76 DENT 98.2 1.6 1,430 0.64 I116 19,160 37.00 11.90 98.2 1.6 1,430 1.40 118 19.160 37.00 9.99 HDI 100.0 2 .5 1.430 0.73 1/32... 100.0 2 .5 1,430 0.63 1116 18,809 33.00 17. 95 DENT HD3 100.0 2 .5 1,430 0.87 1/8 18,809 33.00 14. 95 DENT HD4 100.0 2 .5 1,6 35 I 44 114 18,809 33.00 13.46 COS ms 100.0 2 .5 8 95 0. 35 1132 18,809 33.00 26. 85 HD6 100.0 2 .5 5 75 0. 35 1116 18,809 33.00 30 .55 cos cos - HA0 97.0 1.O 1.430 HA2 97.0 1.o 1,6 35 0.43 HA3 97.0 1 o 8 95 HA4 97.0 1 o 6 05 HBI 97.4 1.2 HB2 97.4 1.2 HB3 97.4 HCl HC2 HC3 BAl 97.0 1 o 650 BB 1... BB 1 97.4 I 2 650 892 91.4 1.2 650 BC 1 98.2 1.6 650 BC2 98.2 1.6 650 BDI 100.0 2 .5 650 10.08 DENT DENT cos _- 19,6 45 35. 25 2. 75 DENT 19,616 37 .50 3.09 DENT BENDING 19,616 37 .50 3. 05 DENT BENDING 19,610 37.00 4.68 DENT BENDING _- BENDlNG BENDING 19,610 37.00 4.66 DENT BENDING 18,809 33.00 7.84 DENT Test specimens of alternative sizes are also tested The inner diameter is kept as 95 mm, and the... eccentric axial compression 167 Chapter 9 Buckling and Local Buckling o Tubular Members f P (ton) 16 12 97.0 U d I5 35 HA0 HA2 HA3 HA4 8 0 14 1 l 55 3 8 95 6 05 1116 1116 4 0 P (ton) 12 8 81.2 u-I) mi HE2 HB3 4 0 4 8 12 16 20 24 D I im 1,6 35 1,6 35 1116 1,6 35 118 28 u ftnd u-I) HC1 HC? HC3 D I Im 1,6 35 1,6 35 1116 118 1,631, I 0 Figure 9.9 4 8 12 16 20 24 28 32 36 Load-End Shortening Curves for Small Scale Test... z=-br+y&, A bh2 2 bh2 W=- (8.49) (8 .50 ) 6 Z =i7=1 .5 (8 .51 ) Tubular Cross-Section (t . Member 1800 14 .5 124 15. 5 631.3 Test Specimen 8000 50 8 6.4 78 15. 7 177.4 Chapter 9 Buckling and Local Buckling of Tubular Members ”*_ _-_ b I C 750 4 750 - 8000 157 Taking E and. bh2 2 bh2 W=- 6 Z =i7=1 .5 (8.48) (8.49) (8 .50 ) (8 .51 ) Tubular Cross-Section (t<<d) (8 .52 ) z I = -d’t 8 W = -d2t 4 (8 .53 ) 7t (8 .54 ) z W f =-=1.27 I-Profile. Pinned-Pinned: & 1 .o I Fixed-Guided: 5 9-NX 4.0 0 .51 Fixed-Pinned: & 2. 05 0.71 Fixed-Free % #- & 0. 25 21 Figure 8 .5 End-Fixity Coefficients and Effective Length

Ngày đăng: 13/08/2014, 09:21

Xem thêm: Marine_Structural_Design Episode 5 potx