Journal of Constructional Steel Research 48 (1998) 79–87
A hysteresis model for plane steel members with damage cumulation effects1 Shen Zuyan*, Dong Bao, Cao Wenxian College of Structural Engineering, Tongji University, Shanghai 200092, PR China
Abstract In order to consider the effects of damage cumulation on the structural responses, a damage model for steel structural members based on plastic strains of the material has been proposed. Adopting the hypotheses of elasto-plastic damage hinges and the damage concentrated at the ends of a steel member, a hysteresis model for plane steel members with effects of damage cumulation has been built. The proposed model can consider the following effects: strain hardening effect and strength degeneration effect caused by damage cumulation. Finally, by means of this model, an elasto-plastic stiffness matrix is built for plane steel members with damage cumulation effects. 1998 Published by Elsevier Science Ltd. All rights reserved. Keywords: Damage cumulation; Damage index; Damage model; Elasto-plastic damage hinge; Hysteresis model; Steel member
1. Introduction In order to meet the needs of elasto-plastic analysis for earthquake-resistant design of steel structures, many hysteresis models for steel members have been established, the simplest one among which is the ideally elasto-plastic model [1,2]. Other models describing the strain hardening effect, such as the two-line and three-line models [3] were also proposed. In addition, Chen and Ausuta [4] suggested a constitution model, * Corresponding author. Tel: + 86 021 6598 2926; Fax: + 86 021 6501 5345 1 National Key Projects on Basic Research and Applied Research: Applied Research on Safety and Durability of Major Construction Projects. 0143-974X/98/$—see front matter 1998 Published by Elsevier Science Ltd. All rights reserved. PII: S 0 1 4 3 - 9 7 4 X ( 9 8 ) 0 0 1 9 4 - 1
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in which both the strain effect and Bauschinger effect were taken into account. However, none of the above-mentioned models considered the effect of damage cumulation on material. In fact, the hysteresis characteristic reflects the effects of damage cumulation on material properties in cycling loading. Test data have indicated that the damage and its cumulation will decrease the elastic modulus and yield strength of steel in a repeated loading and unloading process. Usually, the difference becomes more obvious as damage gets more serious. Therefore, the assumption that the unloading route in –⑀ curves is considered to be parallel with its initial loading route, which is often adopted in elasto-plastic mechanics, will no longer be suitable [5]. The effects of damage on elastic modulus and yield strength can be expressed as follows [13] ED = (1 − 1D)E, Ds = (1 − 2D)s
(1)
where D is the damage index, D = 0 means no damage, and D = 1 means complete failure of the material, E and ED are the elastic modulus in respect of D = 0 and D, respectively, s is the initial yield stress when D = 0 and Ds is the yield stress in respect of D. 1 and 2 are two material parameters. Since the complication of factors related with damage, different expressions for the damage index have been proposed. Kachanov [6] was the first one who described the damage using the alteration of the cross-section area; Hearn and Testa [7] pointed out that the damage could be expressed as the decrement of a certain property of the cross-section; Lardner [8] adopted strength degeneration to describe the damage; Park and Ang [9,10] used a linear combination of deformation and energy to represent damage; Iemura [11] introduced hysteresis energy to define the damage index. Considering the characteristic of damage in structural steel, in this paper plastic strain is adopted to measure the degree of damage. The damage index can be expressed as
冘
⑀pm ⑀pi +  ⑀pu ⑀p i=1 u N
D = (1 − )
(2)
where N is the number of half-cycles which cause plastic strain,  is the weight value, ⑀pi is the plastic strain during the ith half-cycle, ⑀pu is the ultimate plastic strain and ⑀pm is the largest plastic strain during all half-cycles.
2. A non-linear hysteresis model with damage cumulation effects In the forthcoming study the following assumptions will be adopted: (1) planes before deformation remain planes after deformation; (2) shear strains are neglected; (3) plasticity concentrates at both ends of a member; (4) damage concentrates at both ends of a member. The combination of (3) and (4) can be termed as the hypothesis of elasto-plastic damage hinge. Since damage in a cross-section is usually not uniformly distributed, for sake of simplicity in numerical analysis, a virtual uniformly distributed damage ˜ is substituted for the actual damage. D ˜ is called equivalent damage varivariable D
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able in the same cross-section, which can be calculated as follows. A cross-section can be divided into many subsections (Fig. 1). For the ith subsection, Ai denotes its area, yi denotes the distance from the centre of the subsection to the neutral axis of the whole cross-section and Di denotes its damage variable. The following formula can be written:
冕
DidAi
˜ = D
S
(3)
A
Let the yield function for a plane member without axial force be
=
| | M MDp
(4)
where M is the bending moment on the cross-section, and MDp is the ultimate yield ˜. bending moment when the cross-section possesses damage D Let the hysteresis parameter for a plane member be R = Et/E
(5)
where Et denotes the tangent modulus of elasticity at a certain point of the –⑀ curve. Then, as for the nth loading,
冦
˜ 1 − 1D
˜) R = (1 − 1D ˜) q(1 − 1D
冋
1 + (q − 1)
− s,n p,n − s,n
册
(for ⬍ s,n) (for s,n ⱕ ⱕ p,n)
(6)
(for > p,n)
where s,n denotes the value of the initial yield during the nth loading, p,n denotes
Fig. 1. Subsections on a cross-section.
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the value of the perfect yield during the nth loading and q is the strain hardening coefficient.
3. The elasto-plastic stiffness matrix with damage cumulation effects Supposing a member is shown in Fig. 2, the applied force increment vector and its corresponding deformation increment vector can be expressed as follows 兵dF其 = [dQi dMi dQj dMj ]T,
(7)
兵d␦其 = [dvi di dvj dj ]T
(8)
In an arbitrary state, the deformation increment vector {d␦} at the ends of the member can be divided into two parts: the elastic deformation {d␦e} without damage, and the plastic deformation {d␦pD} with damage D, i.e. 兵d␦其 = 兵d␦e其 + 兵d␦pD其
(9)
兵d␦其 = [兵d␦ 其 ,兵d␦ 其 ] , 兵d␦ 其 = [dv ,d ] ,兵d␦ 其 = [dv ,d ] i T
j T T
i
i
i T
j
j
j T
(10)
According to assumptions (3) and (4), the damage and plasticity cause the elastoplastic damage hinge only at the ends of the member, therefore 兵d␦ ipD其 = [0 dipD]T = [g]兵1其dipD, 兵d␦ jpD其 = [0 djpD]T = [g]兵1其djpD where 兵1其 = [1 1]T, [g] =
冋 册 0 0 0 1
(11)
(12)
The force increment vector at the ends of the member can be decomposed into two orthogonal components {dFn} and {dFt}, i.e. 兵dF其 = 兵dFt其 + 兵dFn其
Fig. 2.
Forces and deformations at both ends of a member.
(13)
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83
where 兵dFt其 = [兵dFit其T,兵dFjt 其T]T, 兵dFit其 = [dQi,0]T, 兵dFjt 其 = [dQj ,0]T
(14)
兵dFn其 = [兵dFin其T,兵dFjn其T]T, 兵dFit其 = [0,dMi]T, 兵dFjn其 = [0,dMj ]T
(15)
Apparently, there is a constant relationship between {dF} and {d␦e}, i.e. 兵dF其 = [Ke]兵d␦e其
(16)
where [Ke] is the elastic stiffness matrix of the member element. Rearranging the above equation in matrix forms, we have
冋 册 冋 兵dFi其
兵dFj 其
=
册冋 册
[K iie] [K ije ] 兵d␦ ie其
(17)
[K jie] [K jje ] 兵d␦ je其
Supposing {dFn} is only interrelated with {d␦pD}, {dFn} can be written as 兵dFin其 = [K ih]兵d␦ ipD其, 兵dFjn其 = [K jh]兵d␦ jpD其 where [K sh] = Bsr =
冋 册 Bs1 0
0 Bs2
(18)
(s = i,j)
(19)
Rs kss (r = 1,2;s = i,j) 1 − Rs err
(20)
in which kiierr and kjjerr are the rth row and the rth column element of matrix [K iie] and [K jje ], respectively. Ri and Rj are the parameters for hysteresis force at end i and j, respectively. From Eqs. (9), (13), (17) and (18), one gets 兵dFit其 = [K iie]兵d␦ i其 + [K ije ]兵d␦ j 其 − ([K iie] + [K ih])兵d␦ ipD其 − [K ije ]兵d␦ jpD其 兵dFjt 其 = [K jje ]兵d␦ j 其 + [K jie]兵d␦ i其 − ([K jje ] + [K jh])兵d␦ jpD其 − [K jie]兵d␦ ipD其
冎
(21)
From Eqs. (11) and (18) it can be inferred that vectors 兵d␦ ipD其 and 兵d␦ jpD其 are parallel to vectors 兵dFin其 and 兵dFjn其, respectively. Therefore, vectors 兵d␦ ipD其 and 兵d␦ jpD其 are orthogonal with vectors 兵dFit其 and 兵dFjt 其, respectively [12], i.e. 兵d␦ ipD其T兵dFit其 = ([g]兵1其dipD)T兵dFit其 = dipD兵1其T[g]T兵dFit其 = 0 兵d␦ jpD其T兵dFjt 其 = ([g]兵1其djpD)T兵dFjt 其 = djpD兵1其T[g]T兵dFjt 其 = 0
冎
(22)
For the sake of clarity, several different conditions, which possibly occur in a member, will be discussed separately. Damage and plastic yield occur at both ends i and j of a member. Such being the case, dipD⫽0 and djpD⫽0. From Eq. (22), one can have
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兵1其T[g]T兵dFit其 = 0, 兵1其T[g]T兵dFjt 其 = 0
(23)
Introducing Eq. (21) into the above equation, we get 兵dpD其 = [L][E]T[G]T[Ke]兵d␦其 [G] =
冋
册 冋
[g] [0] [0] [g]
[E] =
册 冋 册
兵1其 兵0其
兵0其 兵1其
[L] =
kii kij
(24)
−1
kji kjj
kii = 兵1其T[g]T([K iie] + [K ih])[g]兵1其, kij = 兵1其T[g]T[K ije ][g]兵1其 kji = 兵1其T[g]T[K jie][g]兵1其, kjj = 兵1其T[g]T([K jje ] + [K jh])[g]兵1其
(25)
冎
(26)
Combining Eqs. (24) and (11), the following formula can be obtained 兵d␦pD其 = [G][E][L][E]T[G]T[Ke]兵d␦其
(27)
From Eqs. (9), (27) and (16) we have 兵dF其 = ([Ke] − [Ke][G][E][L][E]T[G]T[Ke])兵d␦其
(28)
Therefore, when damage and plastic yield occur at both ends of a member, the elasto-plastic matrix can be expressed as follows: [KpD] = [Ke] − [Ke][G][E][L][E]T[G]T[Ke]
(29)
Damage and plastic yield occur only at end i of a member. Under this condition, dipD⫽0, dipD = 0. Following the similar deduction mentioned above, Eq. (29) is also obtained, but in which [L] =
冋 册 1/kii 0 0
0
(30)
Damage and plastic yield occur only at end j of a member. Since dipD = 0 and d ⫽0, we have Eq. (29), and j pD
[L] =
冋 册 0
0
0 1/kjj
(31)
4. Test and verification For steel material (Q235), it can be ascertained from the test results [13] that  in Eq. (2) is 0.081, 1 and 2 in Eq. (1) are 0.227 and 0.119, respectively. A computer programme has been developed according to the analysis presented in this paper.
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85
Table 1 The material properties and dimensional size of I-shaped steel column specimen Young’s modulus E (MPa)
Yield strength fy (MPa)
2.01 × 105
310.5
Ultimate Yield strain strength fu ⑀y (MPa) 453.5
0.00154
Section height h (mm)
Section width b (mm)
240
180
Flange Web thickness tf thickness tw (mm) (mm) 8
6
Inputting any given load route, this programme will output the correspondent displacement route of the free end of an I-shaped steel cantilever column and output the damage value of the member. The following is a calculation example for a tested I-shaped steel cantilever column [14]. The material properties and dimensional size are listed in Table 1. The cantilever length of the column specimen is 1200 mm. The cross-section of the fixed end is discretized into 60 elements (Fig. 3). At the end of loading, the damage value Di of each element is listed in Table 2. The comparison between calculation and test can be seen in Fig. 4, where Py is the horizontal force applied on the top end of the tested column specimen, and Dy is the corresponding horizontal displacement. The calculated damage value for the column specimen ˜ = 0.103. is D 5. Conclusion A hysteresis model for plane steel members with damage cumulation effects has been proposed in this paper. Based on the proposed model, an elasto-plastic stiffness matrix for plane steel members with damage cumulation effects has been derived,
Fig. 3.
Element discretization of the cross-section.
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Table 2 The damage value of each element at the end of loading No. of Damage No. of Damage No. of Damage No. of Damage No. of Damage No. of Damage element value element value element value element value element value element value 1 2 3 4 5 6 7 8 9 10
0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136
11 12 13 14 15 16 17 18 19 20
0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123
Fig. 4.
21 22 23 24 25 26 27 28 29 30
0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123 0.123
31 32 33 34 35 36 37 38 39 40
0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136 0.136
41 42 43 44 45 46 47 48 49 50
0.107 0.083 0.042 0.000 0.000 0.000 0.000 0.042 0.083 0.107
51 52 53 54 55 56 57 58 59 60
0.107 0.083 0.042 0.000 0.000 0.000 0.000 0.042 0.083 0.107
Comparison of measured and calculated results.
which can be applied to the structure damage cumulation analysis under earthquake conditions.
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[3] Park R. Theorisation of structural behaviour with a view to defining resistance and ultimate deformablity. Bull. NZ Soc. Earthquake Engng 1973;2:52–70. [4] Chen WF, Ausuta T. Theory of beam-columns, Vol. 2. McGraw–Hill, 1976. [5] Yin SZ. Fracture and damage theories and their application (in Chinese). Beijing, China: Tsinghua University, 1992. [6] Kachanov LM. Introduction to continuum damage mechanics. Dordrecht: Martinus Nijhoff Publishers, 1986. [7] Hearn G, Testa RB. (1991) Model analysis for damage detection in structures, ASCE, ST10, 3042–3063. [8] Lardner RW. A theory of random fatigue. J. Mechanics Physics Solids 1967;15(3):205–21. [9] Park YJ, Ang AHS. (1985) Mechanistic seismic damage model for reinforced concrete, ASCE, ST4, 722–739. [10] Park YJ, Ang AHS, Wen YK. (1985) Seismic damage analysis of reinforced concrete buildings, ASCE, ST4, 740–757. [11] Iemura H. Hybrid experiments on earthquake failure criteria of reinforced concrete structures. In: Proceedings of the Eighth WCEE, Vol. 6, 1984:103–110. [12] Li GQ, Shen ZY. A nonlinear analysis model for elasto-plastic static and dynamic response of steel frames (in Chinese). J. Build. Struct. 1994;2:51–9. [13] Shen ZY, Dong B. An experiment-based cumulative damage mechanics model of steel under cyclic loading. Adv. Struct. Engng. 1997;1:39–46. [14] Liu WG. Study on hysteretic characteristics of steel members (in Chinese), Master Degree Thesis, Tongji University, Shanghai, China, 1990.