Influence of strain hardening on bending moment–axial force interaction

International Journal of Mechanical Sciences 55 (2012) 65–77

Contents lists available at SciVerse ScienceDirect

International Journal of Mechanical Sciences journal homepage: www.elsevier.com/locate/ijmecsci

Influence of strain hardening on bending moment–axial force interaction H. Abbas a,n,1, N. Jones b a b

Specialty Units for Safety and Preservation of Structures, Department of Civil Engineering, King Saud University, Riyadh 11421, Saudi Arabia Department of Engineering (Mechanical), Impact Research Centre, University of Liverpool, Brownlow Hill, Liverpool L69 3GH, UK

a r t i c l e i n f o

abstract

Article history: Received 6 October 2010 Accepted 19 December 2011 Available online 26 December 2011

This article derives bending moment–axial force (M–P) interaction curves for mild steel by considering elastic–plastic and strain hardening idealisations with linear and parabolic strain hardening characteristics. The interaction relations can predict strains, which is not possible in a rigid, perfectly plastic idealisation. The parameters required are obtained from a standard uniaxial tension test and the procedure can be used for many other materials. & 2011 Elsevier Ltd. All rights reserved.

Keywords: Strain-hardening Rectangular section Bending Axial force Interaction Mild steel

1. Introduction Structural designers are often required to estimate the failure load of structural members for which they employ numerical techniques, such as the finite element method, but the analysis up to failure with large displacements and strains is usually difficult. Continuum damage mechanics has been used recently [1,2] for predicting the static and dynamic failure of beams, but the method requires the values for several parameters, some of which are difficult to obtain. An elastic–plastic model has been used, for example, to study the theoretical anomalous dynamic response of beams [3,4] and plates [5] for a short pulse loading causing small deflections. Another simpler and more attractive option for some problems is to carry out a rigid perfectly plastic analysis [6], the accuracy of which has been compared with the predictions of an elastic– plastic material [7,8]. However, a rigid, perfectly plastic analysis does not predict strains so that it is difficult to study failure unless some assumptions are made to overcome this difficulty. Zyczkowski [9] presented a comprehensive review of theoretical and experimental papers concerning combined plastic loading of sections having different shapes. Many investigators have developed interaction curves or carried out interaction studies pertaining to the combined action of different stress resultants but the studies were mostly for elastic or elastic perfectly plastic

cases [10–14]. Whereas, the interaction studies up to the ultimate capacity are either experimental or numerical [15,16]. In the present paper, kinematically admissible interaction curves for the simultaneous action of bending moment and an axial force on a rectangular section have been developed for elastic–plastic and strain-hardening material idealisations. These curves may be used for the failure analysis of structural elements.

2. Stress–strain diagram The stress–strain diagram for mild steel is idealised as bilinear for small strains, whereas, two models – linear and parabolic – are used for strain-hardening (Fig. 1). Direct tensile test results for a mild steel specimen t036 [2] are shown in Fig. 1. Thus, there are three zones in the idealised diagram: elastic zone from k¼ 0 to 1; yield zone without any strain-hardening from k¼1 to k1; and the strain-hardening zone from k¼k1 to k2, where, key is the strain. The stress in the strain-hardening range, sd, at any strain, e ¼ key (k1 r krk2), can be obtained from the following relations: ðsd syd Þ ¼ ðsud syd Þm

for linear-hardening

ðsd syd Þ ¼ ðsud syd Þmð2mÞ where     eeh kk1 m¼ ¼ eu eh k2 k1

for parabolic-hardening

ð1Þ ð2Þ

ð3Þ

n

Corresponding author. Tel.: þ966 1 4861058. E-mail address: [email protected] (H. Abbas). 1 On leave from Department of Civil Engineering, Aligarh Muslim University, Aligarh 202 002, India. 0020-7403/$ - see front matter & 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijmecsci.2011.12.004

The suffix d in the above expressions has been used to indicate dynamic values. The stress–strain curve can be used for high strength steel by substituting k1 ¼1 and many other materials can

66

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

Nomenclature

Greek symbols

a, b, c, C parameters A, q Cowper–Symonds’ parameters B width of rectangular section E modulus of elasticity F parameter for exponential model G shear modulus h H1 =H H depth of rectangular section H1 distance of neutral axis (NA) from extreme compression fibre L half span of beam m, m0 parameters M bending moment at the section M ðM=Myd Þ ¼shape factor when yield stress is syd 0 M ðM=M0yd Þ ¼shape factor when yield stress is s0ud Mu ðM u =M yd Þ ¼ultimate shape factor Myd dynamic yield moment¼ 16syd BH2 2 2 1 0 M0yd 3 M ud ¼ 6sud BH Mud dynamic ultimate moment n parameter for exponential model P axial force on the section Pyd yield force¼ sydBH Pud ultimate force¼ sudBH P P=Pyd Pu P=Pud R radius of curvature s ððsud =syd Þ1Þ V striking velocity y distance

a e ey em ¼key eh ¼k1ey

be easily represented by these equations for different values of the parameters. The exponential model [17,18] is another model, which is used for idealising the stress–strain relation:

s ¼ Ee þF e1=n

ð4Þ

where, E is the modulus of elasticity and the other parameters, F and n, can be found by fitting a known stress–strain curve, as shown in Fig. 1 for specimen t036 [2]. In the model shown in the figure, the linear term in Eq. (4) has been ignored. This model has

eu ¼k2ey e_ s sys syd sus sud s0yd

parameter normal strain yield strain strain in extreme fibre strain corresponding to end of yielding and beginning of strain-hardening ultimate strain strain rate normal stress static yield stress dynamic yield stress static ultimate stress dynamic ultimate stress modified ultimate dynamic stress for plastic bending ¼ð4M ud =BH2 Þ

Subscripts a c d e h s u y

axial collapse dynamic elastic beginning of hardening static ultimate yield

been used only to study the influence of pure bending because of its simplicity for this case. The bending moment–axial force interaction is more involved, so it is not considered in this article for the exponential model. The model parameters of the direct tensile test results for specimen t036 [2], whose stress–strain curve is shown in Fig. 1, have been used in the subsequent analysis. The strain-softening portion of the curve has been ignored in the present analysis.

3. Strain rate effect The material strain rate effect for the dynamic yield and ultimate strength for a known value of strain rate, e_ , has been incorporated by using the Cowper–Symonds’ equation: " "  1=qy #  1=qu # e_ e_ , sud ¼ sus 1 þ , ð5a; bÞ syd ¼ sys 1 þ Ay Au

Fig. 1. Experimental stress–strain curve [2] and different models for mild steel.

where Ay, qy and Au, qu are the Cowper–Symonds’ coefficients for the yield and ultimate stresses, respectively. The values of these parameters are obtained from the results of uniaxial tensile tests for mild steel and are taken as: Ay ¼ 1300/s, qy ¼5, Au ¼6340/s, and qu ¼5 which are reported in Ref. [19] for ey ¼0.05 and eu ¼0.25, respectively. The strain rate effect on the rupture strain has been ignored [2,20]. The compressive and tensile behaviour are assumed to be the same. An estimate of the equivalent strain rate for a clamped beam struck by a mass at mid-span can be taken as [1] sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 V 9H2 8ks ð6Þ e_ ¼ þ L 2L2 3

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

4. Bending moment–axial force (M–P) interaction A rectangular section of beam of width B and depth H has been considered for studying the interaction of a bending moment, M, and an axial force, P (Fig. 2). The bending moment is assumed to cause tension at the bottom face. The axial force considered in the present analysis is tensile and the same relations can be used for a compressive axial force because the material behaviour in compression has been assumed to be the same as in tension. The interaction curves for different states of stresses have been obtained in the subsequent subsections. The extreme fibre strain in tension (i.e. at the bottom fibre) is taken as key. The extreme fibre strain at the top fibre is taken as k0 ey which is compressive when the neutral axis is inside the section, whereas, it is tensile when the neutral axis is outside the section. The value of k0 when the neutral axis lies inside the section is given by kh , k ¼ 1h 0

one for it lying outside the section, as shown in Fig. 3. The relations obtained for these cases are given below. The proof of Case I is given in Appendix A. Other expressions can be derived in a similar way. In the first case, the neutral axis is inside the section (Fig. 3a) and there is an elastic–plastic state in tension as well as one in compression, i.e. 1rk0 rk1. The M–P interaction relation obtained for this case is     1 P 1 2 ð11Þ M þ 1:5 1 þ 2 P þ 2 ¼ 1:5 1 2 : 3k k 3k The distance of the neutral axis, h, required for determining the value of k0 from Eq. (7) is given by h ¼ 0:5ð1PÞ

The proof of Eqs. (11) and (12) is given in Appendix A. It can be seen from Eq. (11) that for large values of the extreme fibre strain (i.e. k), the expression converts to Eq. (10). The curve represented by Eq. (11) is convex and its normality is checked in Appendix C. When there is no axial force (i.e. P ¼ 0), the equation gives the maximum value of the shape factor for the elastic–plastic case, B

ð7Þ

whereas, when the neutral axis lies outside the section, the value of k0 is given by kh 1þh

ð8Þ

where h¼H1/H; H1 is the distance of the neutral axis from the extreme compression fibre.

εy εy

There are well-established interaction curves for an elastic and a rigid, perfectly plastic section having a rectangular cross-section [6]: for Elastic Model

σyd Stress Diagram

B σ

k'εy

H1

ð9Þ

for Rigid, Perfectly Plastic Model

kεy Strain Diagram

C

εy

Neutral Axis

ð10Þ H

2 2 MþP ¼ 1 3

T=P+C

Beam Section

4.1. Elastic and rigid perfectly plastic models

MþP ¼ 1

C

H

Neutral Axis

σyd

k'εy

0.5H

0

k ¼

ð12Þ

H1 = h H

where H is the depth of the beam having a rectangular section; L is half span of the beam; V is the striking velocity of the mass; ks is a dimensionless shear strain parameter whose value for mild steel is 0.26 [2]. This expression does not capture the variation of the strain rate throughout the motion, but it does provide a reasonable estimate for an average strain rate during the initial response.

67

T=P+C

0.5H

The cross-section can only become fully plastic when the extreme fibre strain is infinite, which practically, is not possible. The advantage of Eq. (10) lies in its simplicity but its main disadvantage is that it cannot predict strains. To overcome this difficulty the elastic–plastic case is considered in the next subsection, which almost reduces to the rigid, perfectly plastic model for a large yield zone.

Beam Section

kεy

σyd

Strain Diagram

Stress Diagram

H1

Neutral Axis k'εy

σ

εy

H

There are three cases for this perfectly plastic model, which depend upon the position of the neutral axis. Two cases are associated with the neutral axis lying inside the cross-section and

h'H

B

4.2. Elastic–plastic model

M

P

T=P

0.5H

M

P Beam Section

Fig. 2. An element subjected to external pull and bending.

kεy Strain Diagram

σyd Stress Diagram

Fig. 3. Stress and strain variation in the section of a beam for different positions of neutral axis in the elastic–plastic model. (a) Case I (1r k0 rk1); (b) Case II (0rk0 r1); (c) Case III (1r k0 r1).

68

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

given by ! 3k 1 2

M¼

ð13Þ

2

2k

For a large value of k, the value of the shape factor given by the above equation, approaches 1.5, which is equal to the plastic shape factor of a rectangular section. In the second case, the neutral axis is also inside the section (Fig. 3b), but the compression remains elastic, i.e. 0 rk0 r1, whereas it is elastic–plastic in tension. The M–P interaction relation obtained for this case is 3

M þ 3Pð12hÞ

2

2kh ð3k 1Þ  ð1hÞ2 ¼ 0, 2 ð1hÞ k

ð14Þ

where h¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 þkðP2Þ þ k P 2kðP1Þ1 ðk1Þ2

ð15Þ

In the third case, the neutral axis is outside the section (Fig. 3c) so that the tension zone is in the elastic–plastic state, i.e. 0 rk0 r1 and becomes fully plastic when k0 ¼1. The M–P interaction relation obtained for this case is M þ 3ðP1Þ þ

ð1þ hkhÞ3 2

k ð1 þ hÞ

¼ 0,

ð16Þ

h¼

ðk1Þ

2

4.3.1. Linear-hardening For the linear hardening model, the six cases, which depend upon the position of the neutral axis, are presented below: Case I: In the first case, the neutral axis is inside the section (Fig. 5a) and there is strain-hardening in tension as well as in compression, i.e. k1 rk0 rk2. The M–P interaction relation obtained for this case is h i h2 02 0 0 M ¼ ð3k 1Þ þsm0 ðk k1 Þð2k þk1 Þ 2 k’ h i ð1hÞ2 2 þ ð3k 1Þ þ smðkk1 Þð2k þ k1 Þ 3Pð12hÞ, 2 k

ð18Þ

where   sud 1 , s¼

ð19Þ

syd

and m0 is given by Eq. (3) with k replaced by k0 . The value of h can be obtained from the quadratic equation: 2

ah þ bh þ c ¼ 0,

where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kð2PÞ1k ðkPÞ2 ðk1Þ2

the subsequent sections. The proof of the most difficult case, which is Case I for parabolic-hardening, is given in Appendix A. The remaining expressions can be derived in a similar way.

ð20aÞ

with ð17Þ

The M–P interaction curve considering all three cases is plotted in Fig. 4 for different values of k. It can be seen from this figure that the curve approaches the rigid, perfectly plastic case for large values of k (k Z10), as required by the limit theorems of plasticity [6]. 4.3. Strain-hardening models For linear as well as parabolic strain hardening, there are six cases depending upon the position of the neutral axis, three each for the neutral axis inside and outside the section as shown in Figs. 5 and 6, respectively. The extreme fibre strain at the top is compressive when the neutral axis lies inside the section, whereas, it is tensile when the neutral axis is outside the section. The relations obtained for these cases for the two types of hardening are given in

h¼

b7

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 b 4ac , 2a

ð20bÞ

where a ¼ 4kðk2 k1 Þ4skk1 ;

b ¼ 2kðk2 k1 ÞðP3Þ þ 2skð3k1 kÞ

c ¼ 2kðk2 k1 Þð1PÞ þskðk2k1 Þ:

ð21a  cÞ

The sign of discriminant in Eq. (20b) for admissible values of h for this particular case is negative. When there is no axial force (i.e. P ¼ 0) then Eq. (18) gives the maximum value of shape factor for linear-hardening, given by ! 2 3k 1 sm M¼ ð22Þ þ 2 ðkk1 Þð2k þ k1 Þ 2 2k 2k which reduces to Eq. (13) when s ¼0. When the extreme fibre strain in tension equals the ultimate strain (i.e. k¼k2) then the above equation gives the maximum value of the ultimate shape factor for linear-hardening. Case II: In the second case also, the neutral axis is inside the section (Fig. 5b) but there is an elastic–plastic state in the compression zone, i.e. 1rk0 rk1, whereas, there is a strain-hardening state in tension. The M–P interaction relation for this case is 02

M ¼ ð3k 1Þ

2 h i ð1hÞ2 h 2 þ ð3k 1Þ þ smðkk1 Þð2k þ k1 Þ 3Pð12hÞ, 02 2 k k ð23Þ

where h¼

2kþ smðkk1 Þ2kP : smðkk1 Þ þ4k

ð24Þ

Case III: In the third case also, the neutral axis is inside the section (Fig. 5c) but there is an elastic state in the compression zone, i.e. 0 rk0 r1, whereas, there is a strain-hardening state in tension. The M–P interaction relation for this case is Fig. 4. M–P interaction curves for elastic, plastic and elastic–plastic models. Note: For k¼ 12, Case I is valid for P/Py up to 0.85 and thus the extrapolation of this curve beyond its limits of validity almost matches with curves for Cases II and III and meets the P/Py axis at 0.995 as given by Eq. (56).

M¼

3 h i ð1hÞ2 2kh 2 þ ð3k 1Þ þ smðkk1 Þð2kþ k1 Þ 3Pð12hÞ, 2 ð1hÞ k

ð25Þ

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

69

B

εy εy εh

0.5H

εh

T=P+C

T=P+C k εy Beam Section

C

C

H

Neutral Axis

H1= h H

k'ε y

Linear Hardening

Parabolic Hardening

Stress Diagram

Strain Diagram

B σyd

k'ε y H1= h H

σyd

C

C

εy εh

0.5H

H

Neutral Axis

εy

T=P+C

T=P+C

k εy Beam Section

Strain Diagram

Linear Hardening Parabolic Hardening Stress Diagram

B H1

k'ε y

C

C

εy εh

0.5H

H

Neutral Axis

T=P+C

T=P+C k εy Beam Section

Linear Hardening

Parabolic Hardening

Stress Diagram

Strain Diagram

Fig. 5. Stress and strain variation in the section of a beam for positions of neutral axis inside the section for strain-hardening model. (a) Case I (k1 rk0 r k2); (b) Case II (1r k0 r k1); (c) Case III (0 r k0 r 1).

where h¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2ðkPaÞ 4ðkPaÞ2 4ðak Þða2kPÞ 2

2ðak Þ

, and

ð26Þ

a ¼ ð2k1Þ þ smðkk1 Þ:

ð27Þ

Case IV: In the fourth case, the neutral axis is outside the section (Fig. 6a) such that there are: elastic, plastic and strainhardening states in tension, i.e. 0rk0 r1. The M–P interaction relation for this case is M ¼ 3ð1PÞ

ð1 þhkhÞ3 2

k ð1 þ hÞ

þ

sm 2

k

ðkk1 Þð1 þ hÞðk1 hkhþ k1 þ 2kÞ: ð28Þ

in tension, i.e. 1rk0 rk1. The M–P interaction relation for this case is   4 4k1 ð1PÞ þ 1 ð1PÞ M¼ ð30Þ sm k The value of h, required for determining k0 from Eq. (8), is given by h¼

2kðP1Þ 1 smðkk1 Þ

ð31Þ

Case VI: In the sixth case also, the neutral axis is outside the section (Fig. 6c), leaving only a strain-hardening state in tension, i.e. k1 rk0 rk2. The M–P interaction relation for this case is M þ P ¼ 1 þsm

ð32Þ

The value of h can be obtained from the quadratic equation given by Eq. (20) taking discriminant with negative sign but the parameters of the quadratic equation are now:

The value of h, required for determining k from Eq. (8), is given by

a ¼ ðk1Þ2 smðkk1 Þ;

h¼

b ¼ 2kP4k2smðkk1 Þ þ2;

c ¼ 2kðP1Þsmðkk1 Þ þ 1:

ð29a2cÞ

Case V: In the fifth case also, the neutral axis is outside the section (Fig. 6b), but there are plastic and strain-hardening states

0

sðk2k1 Þ2ðP1Þðk2 k1 Þ 2ðP1Þðk2 k1 Þ2sðkk1 Þ

ð33Þ

Eq. (32) is a straight line in the PM plane. The M–P interaction curves of the linear hardening model for the different values of k are plotted in Fig. 7.

70

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

B k'ε y ε y εh

0.5H

H

H1

Neutral Axis

T=P k εy Beam Section

T=P

Linear Hardening

Strain Diagram

Parabolic Hardening

Stress Diagram

B H1

Neutral Axis k'ε y

σyd

σyd

0.5H

H

εh

T=P

T=P

k εy

H1

Parabolic Hardening

Stress Diagram

k'ε y

H

Neutral Axis B

Linear Hardening

Strain Diagram

Beam Section

T=P

0.5H

T=P

k εy Beam Section

Linear Hardening

Strain Diagram

Parabolic Hardening

Stress Diagram

Fig. 6. Stress and strain variation in the section of a beam for positions of neutral axis outside the section for strain-hardening model. (a) Case IV (0 r k0 r1); (b) Case V (1r k0 r k1); (c) Case VI (k1 r k0 r k2).

4.3.2. Parabolic-hardening Six cases which depend upon the position of the neutral axis are required for the parabolic hardening model as discussed below: Case I: In the first case, the neutral axis is inside the section (Fig. 5a) and there is strain-hardening in tension as well as in compression, i.e. k1 rk0 rk2. The M–P interaction relation for this case is  2 2 h ð1hÞ2 s m0 h 02 2 M ¼ ð3k 1Þ 02 þ ð3k 1Þ þ 0 2 2 k k k 0

02

2

0

ð8k2 k 3k 3k1 6k1 k þ4k1 k2 Þ þ

s m2 ð1hÞ2 2 2 ð8k2 k3k 3k1 6k1 k þ 4k1 k2 Þ3Pð12hÞ 2 2 k

ð34Þ

where m0 is given by Eq. (3) with k replaced by k0 ; the value of k0 is given by Eq. (7) and h can be obtained from: 3

að1hÞ3 þ bð1hÞ2 þ cð1hÞ þk ¼ 0,

ð35Þ

where Fig. 7. M–P interaction curves for linear hardening model. (Note: k2 in legend is k2 in text.)

a ¼ ðkk1 Þ2 ð3k2 2k1 kÞðk þ k1 Þ2 ð3k2 2k1 þ kÞ þ

6k ðk2 k1 Þ2 ; s

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

3k ðP þ1Þðk2 k1 Þ2 ; s 2 2 c ¼ 2k ðk þ k1 Þk ð3k2 2k1 þkÞ: b ¼ 3kðkþ k1 Þðkk1 þ2k2 Þ

h can be found from the roots of Eq. (35), thus giving qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b 2cos b 2  b 3ac, h ¼ 1þ 3a 3a

ð36a2cÞ

3

cos 3b ¼

ð37Þ

3

2

2ðb 3acÞ3=2

:

The value of h can be obtained from the quadratic equation given earlier by Eq. (20) taking discriminant with positive sign but with 2 2 2 sm ð3k2 2k1 kÞk þ 2k1; 3 4 b ¼ sm2 ð3k2 2k1 kÞ2kP þ 4k2; 3 2 2 c ¼ sm ð3k2 2k1 kÞ2kP þ2k1: 3 a¼

where b can be obtained from 9abc2b 27a2 k

71

ð38Þ

The proof of Eqs. (34)–(38) is given in Appendix A. When there is no axial force (i.e. P ¼ 0) then this case gives the maximum value of the shape factor for parabolic-hardening, thus giving: ! 2 3k 1 sm2 2 2 þ M¼ ð8k2 k3k 3k1 6k1 kþ 4k1 k2 Þ ð39Þ 2 2 2k 4k and when k¼k2 then it gives the shape factor for the ultimate moment: ! 2 3k2 1 s 2 2 Mu ¼ ð40Þ þ 2 ð5k2 3k1 2k1 k2 Þ 2 2k2 4k2 Case II: In the second case also, the neutral axis is inside the section (Fig. 5b), but there is an elastic–plastic state in the compression zone, i.e. 1rk0 rk1, whereas, there is strain-hardening state in tension. The M–P interaction relation for this case is  2 h sm2 02 2 2 ð8k2 k3k M ¼ ð3k 1Þ 02 þ ð3k 1Þ þ 2 k i ð1hÞ2 2 3k1 6k1 k þ 4k1 k2 Þ 3Pð12hÞ, ð41Þ 2 k

ð46a2cÞ

Case V: In the fifth case also, the neutral axis is outside the section (Fig. 6b), but there are plastic and strain-hardening states in tension, i.e. 1rk0 rk1. The M–P interaction relation for this case is M¼

3 b ð1PÞð2Pa2Þ þ 2 ðP1Þ2 , a a

ð47Þ

where a¼

sm2 ð3k2 2k1 kÞ; 3k

b¼

sm2 2

2k

2

ð48a; bÞ 0

The value of h, required for determining k from Eq. (8), is given by h¼

ðP1Þ 1 a

ð49Þ

Case VI: In the sixth case also, the neutral axis is outside the section (Fig. 6c), leaving only a strain-hardening state in tension, i.e. k1 rk0 rk2. The M–P interaction relation obtained for this case is M ¼ 3ð1PÞ þ

s

2

2ðk2 k1 Þ2

2

0

0

ð50Þ 2

3k þ sm ð3k2 2k1 kÞ3kP sm2 ð3k2 2k1 kÞ þ 6k

0

ð42Þ

and k0 is given by Eq. (7). Case III: In the third case, the neutral axis is inside the section (Fig. 5c), but there is an elastic state in the compression zone, i.e. 0 rk0 r1, whereas, there is strain-hardening in tension. The M–P interaction relation for this case is   3 2kh sm2 2 2 2 þ ð3k 1Þ þ ð8k2 k3k 3k1 6k1 kþ 4k1 k2 Þ M¼ ð1hÞ 2 2



02

ð8k2 k3k þ 6k1 k 12k1 k2 2kk þ 4k2 k Þ

where h¼

2

ð8k2 k3k 3k1 6k1 k þ4k1 k2 Þ

ð1hÞ 2

k

3Pð12hÞ

a¼

smð2k2 k1 kÞ P þ 1; ðk2 k1 Þ 2

2

b¼

sð9k2 k3k 12k1 k2 þ 6k1 Þ

c¼

sð3k2 kk 6k1 k2 þ 3k1 Þ

3ðk2 k1 Þ2 2

2ðP1Þ;

2

2

3ðk2 k1 Þ

ð43Þ

The value of h can be obtained from the quadratic equation given earlier by Eq. (20) taking discriminant with negative sign but with 2 2 sm ð3k2 2k1 kÞ1; 3 4 b ¼ 2kP4k sm2 ð3k2 2k1 kÞ þ2; 3 2 2 c ¼ sm ð3k2 2k1 kÞ2kP þ2k1: 3

where k is given by Eq. (8), which is a function of h; and h can be determined from the quadratic equation given earlier by Eq. (20) taking discriminant with positive sign but with the parameters of the quadratic equation given by

ðP1Þ:

ð51a  cÞ

The M–P interaction curves of parabolic hardening model for different values of k is plotted in Fig. 8. The range of P and M for different cases of the three models is given in Appendix B. 4.4. Discussion

2

a ¼ k þ 2k þ

ð44a2cÞ

Case IV: In the fourth case, the neutral axis is outside the section (Fig. 6a) such that there are all three: elastic, plastic and strain-hardening states in tension, i.e. 0rk0 r1. The M–P interaction relation obtained for this case is   sm2 2 2 2 ð8k2 k3k 3k1 6k1 kþ 4k1 k2 Þ M ¼ ð3k 1Þ þ 2 

ð1 þ hÞ2 2

k

3



2kh 3Pð1þ 2hÞ ð1þ hÞ

ð45Þ

When there is no axial force, i.e. bending alone, the moment– curvature relations obtained from Eqs. (22) and (39) for linear and parabolic hardening have been plotted in Fig. 9 for the material characteristics from the tension test results of specimen t036 [2]. The variable in the abscissa is the parameter, k, which is proportional to the curvature and is equal to the ratio of extreme fibre strain to the yield strain: k¼

H , 2Rey

ð52Þ

where R is the radius of curvature. It is observed from this figure that there is little difference (  6%) between the predictions of the linear and parabolic hardening models for the particular material considered in Fig. 1. The moment–curvature relation for the linear

72

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

Fig. 8. M–P interaction curves for parabolic hardening model. (Note: k2 in legend is k2 in text.)

Fig. 10. M–P interaction curves for different models (k¼ k1 ¼ 12 for elastic–plastic and k¼ k2 ¼127 for linear and parabolic hardening curves).

Table 1 Extent of different zones on M–P curve represented by different cases covering different positions of neutral axis. Case

I II III IV V VI

Elastic–plastic

Linear-hardening

Parabolic-hardening

Range of P

% age

Range of P

% age

Range of P

% age

0–0.848 0.848–0.959 0.959–1.000 – – –

84.8 11.1 4.1 – – –

0–1.036 1.036–1.213 1.213–1.226 1.226–1.232 1.232–1.255 1.255–1.509

68.7 11.7 0.9 0.4 1.5 16.9

0–1.106 1.106–1.289 1.289–1.303 1.303–1.310 1.310–1.339 1.339–1.509

73.3 12.1 0.9 0.4 2.0 11.2

Fig. 9. Moment–curvature relation for different models.

hardening model is nearly linear, whereas, for parabolic hardening model there is a small curvature. One quadrant of the M–P interaction curves for the elastic– plastic, linear and parabolic hardening models have been plotted already in Figs. 4, 7 and 8, respectively. The M–P interaction curves for all the models are plotted in Fig. 10. The ranges of P and the corresponding percentages of the maximum range for the different zones of elastic–plastic, linear and parabolic hardening models for the mild steel results considered earlier and illustrated in Figs. 4, 7 and 8, are given in Table 1. The observations made from these figures and the table are: (a) The elastic–plastic interaction curve for k¼ k1 is close to the rigid, perfectly plastic curve for large values of the yield strains (k1 Z10). There are three parts in the elastic–plastic curve arising due to the three positions of the neutral axis in Fig. 3. The third case, i.e. with the neutral axis outside the section in Fig. 3, represents a small zone of the interaction curve with P 40:95 for large values of k1 (say, k1 Z10). (b) Cases III–V of the strain hardening models in Figs. 5 and 6 are required for a continuous curve, but from a practical perspective, the portion of the interaction curve represented by these cases is very small and could be ignored for linear as well as parabolic hardening materials. Therefore, only three cases: I, II

and VI (first two for the neutral axis inside the section and the last for the neutral axis outside the section) are sufficient for practical purposes for defining the interaction curve of mild steel rectangular sections. All the three possible cases are required to define the interaction curve for an elastic–plastic material. (c) The strains and hence the strain rates due to bending and an axial force can be separated only for the linear-elastic case because the principle of superposition is not valid for the nonlinear case. (d) The difference between the interaction curves for linear and parabolic hardening for the particular material illustrated in Fig. 1, is small (6% when P is zero) which was evident from the small gap between the moment–curvature plots for the two cases. (e) The M–P interaction curves for the strain hardening models, are non-convex for some of the cases. The curvature of M–P interaction curves for elastic–plastic and strain hardening models for the material parameters considered in the study are reported in Table 2. It is observed from the table that the concavity of interaction curve for both of the strain hardening models starts with Case III when there is an elastic state in the compression zone, whereas, there is strain-hardening in tension zone. For linear strain hardening, the curve remains non-convex for Cases IV and V but becomes straight line for

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

Table 2 Curvature at the extremities of different zones of M–P curve represented by different cases covering different positions of neutral axis.a Case Elastic–plastic

Linear-hardening

Parabolic-hardening

73

strain hardening models may be taken as     0 1 P 1 0 02 M þ1:5 1þ 2 P u þ 2u ¼ 1:5 1 2 , 3k 3k k

ð57Þ

where Value of P Curvature

Value of P Curvature Value of P Curvature

I

0.000 0.848

 3.007  3.007

0.000 1.036

 0.937  2.469

0.000 1.106

 2.124  2.279

II

0.848 0.959

 1.300  2.796

1.036 1.213

 2.580  2.580

1.106 1.289

 2.490  2.490

III

0.959 1.000

 2.796 1.213  N(  5e7) 1.226

5.068 33.345

1.289 1.303

4.850 30.804

IV

– –

– –

1.226 1.232

35.123 2656

1.303 1.310

31.237 2480

V

– –

– –

1.232 1.255

15.714 15.714

1.310 1.339

 13.259  13.259

VI

– –

– –

1.255 1.509

0.0 0.0

1.339 1.509

0.0  0.005

2

2

a Curvature is given by d M=dP ; its negative sign indicates convexity of M–P curve.

Case VI. Whereas, for parabolic hardening the curve is nonconvex only for Cases III and IV.

4.5. Simplified M–P interaction curve It is evident that the M–P interaction curve, for the strainhardening models, is non-convex for large values of the axial force except for the rigid perfectly plastic case. A convex curve can be obtained by adopting a simpler modified elastic–plastic model. It is seen from the elastic–plastic model that the M–P interaction curve represented by Case I is convex but it is valid for   k1 2ð3k1Þ and M Z Pr ð53a; bÞ kþ1 ðkþ 1Þ2 and beyond this zone, a tangent to this curve may be taken, which is quite close to the curves represented by Cases II and III. The equation of this tangent is 2

Mþ

3

ð3k 3k þ2Þ ð3k þ3k2Þ P¼ kðkþ 1Þ kðkþ 1Þ2

ð54Þ

This straight line meets the P axis at 3

P¼

ð3k þ 3k2Þ 2

ðk þ 1Þð3k 3k þ2Þ

2

ð3k 1Þ 2

ð3k þ 1Þ

0

ð58a; bÞ 0

M ¼

M ; M 0yd

Pu ¼

P P ud

2 1 M ¼ s0 BH2 3 ud 6 yd " # ! 2 3k2 1 s 2 2 2 s0yd ¼ ð5k 3k 2k k Þ s , þ 1 2 2 1 2 2 3 yd 2k2 4k2

M0yd ¼

ð59a; bÞ

ð60Þ

ð61Þ

where s0yd is the equivalent yield stress giving a plastic moment equal to the ultimate moment obtained from the parabolic model. Thus, s0yd is the yield stress of an equivalent elastic–plastic model such that the plastic moment of this equivalent model is equal to the ultimate moment of the parabolic hardening model. Eqs. (59)–(61) ensure that Eq. (57) passes through P ¼ 0 point on the parabolic hardening curve and Eq. (58) ensures that the curve passes through M ¼ 0 point. The lower and upper bound curves given by Eq. (57) are plotted in Fig. 8. It is observed from Fig. 8 that for small axial forces, the lower bound curve is slightly outside the parabolic hardening curve but the difference is small, so it is an approximate lower bound. The proposed expression of the lower bound has been adopted due to its simplicity.

5. Conclusions The M–P (bending moment–axial force) interaction curves have been developed for elastic–plastic and for linear and parabolic strain hardening idealisations of mild steel. The M–P interaction relations are expressed in terms of the extreme fibre strains, which is not possible for a rigid, perfectly plastic model. Simplified equivalent, yet accurate, elastic–plastic M–P interaction curves are suggested as upper and lower bounds for the purposes of failure analysis of beams of rectangular section. A uniaxial tension test provides all the parameters, which are required for the idealisation of mild steel. However, the formulation could be used to develop the interaction curves of many other ductile materials.

ð55Þ

which is slightly greater than unity. Solving Eq. (55) for k for the purpose of analysis may be difficult. Therefore, an even simpler M–P curve is Case I of the elastic–plastic model (Eq. (11)) for the whole range, i.e. extended beyond its range of validity. This curve is convex for the whole range and is close to the curves corresponding to the other two cases (Fig. 4). This curve meets the P axis at P¼

0

Pu ¼ ðs þ1Þ0:8 Pu for lower bound; and P u ¼ Pu for upper bound

ð56Þ

which is slightly smaller than unity for k Z4, approximately. The advantages of using this curve is that there is only one equation representing the whole curve and that Eq. (11) can be solved easily for k. Also, for large values of k, Eq. (11) reduces to the rigid, perfectly plastic case. This curve for an elastic–plastic model can also be adopted with some modification for the strain-hardening model. Upper and approximate lower bounds to the parabolic

Acknowledgements The first author is grateful to the Commonwealth Scholarship and Fellowship Commission, UK for awarding the fellowship (Ref. INCF-2003-93) and the Aligarh Muslim University, Aligarh, India for granting leave.

Appendix A. Mathematical proofs A.1. Elastic–plastic (Case I) The total compressive force on the section can be obtained by integrating the stresses in Fig. 3(a) over the section in compression, thus giving,  Py  1 ðHH1 Þ C ¼ syd BH1  syd B ¼ ð2k þ1Þh1 2 k 2k

ðA:1Þ

74

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

The above equation can be written in the following form:

Similarly, the total tensile force in Fig. 3(a) is 1 ðHH1 Þ Py ¼ ð1hÞð2k1Þ T ¼ syd BðHH1 Þ syd B 2 k 2k

ðA:2Þ

The distance of the line action of the resultant compressive and tensile forces from the neutral axis can be found by taking the moment of the different force components, thus giving 2 2

y1 ¼

½3k h ð1hÞ2 H 3k½ð2kþ 1Þh1

ðA:3Þ

3

3

ap þbp2 þ cp þk ¼ 0 where p ¼ 1h;

ðA:16Þ

a ¼ ðkk1 Þ2 ð3k2 2k1 kÞðk þ k1 Þ2 ð3k2 2k1 þ kÞ þ b ¼ 3kðk þk1 Þðkk1 þ 2k2 Þ

and

2

2

y2 ¼

ð3k 1Þð1hÞH : 3kð2k1Þ

ðA:4Þ

The position of the neutral axis can be found by considering the equilibrium of forces P ¼ TC,

ðA:5Þ

6k ðk2 k1 Þ2 ; s

3k ðP þ 1Þðk2 k1 Þ2 ; s

2

c ¼ 2k ðk þk1 Þk ð3k2 2k1 þ kÞ:

ðA:17a2cÞ

Eq. (A.17a–c) are same as Eq. (36a–c) of the paper. The roots of the cubic equation, given by Eq. (A.15), can be determined by eliminating quadratic term and making it identical to Eq. (A.22), given latter. So, putting p ¼t  u in Eq. (A.15), we get, 3

at3 þ ðb3auÞt 2 þ ð3au2 2bu þ cÞtau3 þ bu2 cu þk ¼ 0

which using Eqs. (A.1) and (A.2) gives P ¼ syd BðH2H1 Þ ¼ Py ð12hÞ

ðA:6Þ

from which the position of neutral axis is h ¼ 12ð1PÞ

ðA:7Þ

which is same as Eq. (12) in the paper. The moment of bending resistance of the section is   H H1 M ¼ Cy1 þ Ty2 P 2

ðA:8Þ

or

ðA:18Þ

2

For eliminating t term from the above equation by replacing u by b/3a, thus giving, ! ! 2 3 b 2b bc 3 þ k ¼0 ðA:19Þ tþ  at3 þ c 3a 27a2 3a Putting t ¼ rcos b in Eq. (A.19), ! ! 2 3 3 b 3ac 9abc2b 27a2 k b  cos3 b cos ¼0 3a2 r2 27a3 r3

ðA:20Þ

Considering the identity,

Cy1 Ty2 M¼ þ 3Pð12hÞ My My

ðA:9Þ

which substituting Eqs. (A.1)–(A.4), and eliminating h with Eq. (A.7), we get Eq. (11), i.e.     1 P 1 2 M þ 1:5 1 þ 2 P þ 2 ¼ 1:5 1 2 ðA:10Þ 3k k 3k

cos 3b ¼ 4 cos3 b3 cos b

cos3 b34 cos b14cos 3b ¼ 0 Eq. (A.20) is identical to Eq. (A.22), provided, ! qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b 3ac 3 2 2 , or b 3ac ¼ r ¼ 4 3a 3a2 r2

The total compressive force on the section can be obtained by integrating the stress over the section in compression in Fig. 5(a), thus giving,

cos 3b ¼ 4

ðA:11Þ

3

2

 Py s½ðkþ k1 Þhk1    Py  ð2kþ 1Þh1 þ ð3k2 2k1 Þð1hÞkh 2k 3kðk2 k1 Þ2 ð1hÞ2 ðA:12Þ

Similarly, the total tensile force can be evaluated   P y sðkk1 Þ2 1 ð1hÞ þ ð3k2 2k1 kÞð1hÞ T ¼ P y 1 2k 3kðk2 k1 Þ2

3

9abc2b 27a2 k 27a3 r3

!

3

cos 3b ¼

Eq. (A.5) obtained from the equilibrium of forces is also valid for this case. Putting the values of C and T from Eqs. (A.12) and (A.13), respectively, into Eq. (A.5), we get " # ðkk1 Þ2 ð3k2 2k1 kÞð1hÞ3 s P ¼ 12h þ 2 3kð1hÞðk2 k1 Þ2 fðk þk1 Þhk1 g fð3k2 2k1 Þð1hÞkhg ðA:14Þ

ðA:24Þ

3

9abc2b 27a2 k 2

3=2

2ðb 3acÞ

ðA:25Þ

which is same as Eq. (38) of the paper. The value of p can be determined from: p ¼ tu

ðA:26Þ

which is the substitution made in Eq. (A.15). Using Eqs. (A.16) and (A.26), the value of h can be determined, thus giving, h ¼ 1t þ u

ðA:13Þ

ðA:23Þ

Putting the value of r from Eq. (A.23),

0

Putting the value of k from Eq. (7) and using Eq. (19), this equation converts to

ðA:22Þ

2

and,

ðsud syd ÞBHh 0 BHh 0 0 C ¼ syd ðk k1 Þ2 ð3k2 2k1 k Þ 0 ð2k 1Þ þ 0 2k 3k ðk2 k1 Þ2

ðA:21Þ

which can be written as

A.2. Parabolic hardening (Case I)

C¼

ðA:15Þ

Substituting the values of t and u in Eq. (A.27), qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b 2cos b 2  b 3ac h ¼ 1þ 3a 3a

ðA:27Þ

ðA:28Þ

which is same as Eq. (37) of the paper. For this case as well, the moment of resistance of the section is given by Eq. (A.9). The values of non-dimensional moment of the resultant compressive and tensile forces about the neutral axis can be obtained by taking the moment of force components,

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

thus giving,   0 2 Cy1 1 sðk k1 Þ2 h 2 0 02 2 0 ¼ 3 02 h þ 02 ð8k2 k 3k 3k1 6k1 k þ 4k1 k2 Þ My k 2k ðk2 k1 Þ2 ðA:29Þ   Ty2 1 sðkk1 Þ2 ð1hÞ2 ¼ 3 2 ð1hÞ2 þ 2 My k 2k ðk2 k1 Þ2 2

2

ðA:30Þ

Putting the values of Cy1 =My , and Ty2 =M y from Eqs. (A.29) and (A.30), respectively, in Eq. (A.9), we get Eq. (34) of the paper, i.e.,  2 2 h ð1hÞ2 s m0 h 02 2 M ¼ ð3k 1Þ 02 þð3k 1Þ þ 0 2 2 k k k 2

MLH3 ¼

MLH4 ¼

0

ð8k2 k 3k02 3k1 6k1 k þ 4k1 k2 Þ

MLH5 ¼

s m2 ð1hÞ2 2 2 ð8k2 k3k 3k1 6k1 k þ4k1 k2 Þ3Pð12hÞ þ 2 2 k ðA:31Þ

Appendix B. Range of P and M for different cases of various models

smðkk1 Þ , 2ðk1Þ

P LH6 ¼ 1 þ

sm , 2

PLH7 ¼ 1 þsm

ðB:12e2gÞ

2

MLH1 ¼

MLH2 ¼

ð8k2 k3k 3k1 6k1 k þ4k1 k2 Þ

0

PLH5 ¼ 1 þ

75

MLH6 ¼

ð3k 1Þ þ smðkk1 Þð2k þk1 Þ

ðB:13aÞ

2

2k

4ð3k1 k1Þ þ smðkk1 Þðk þ5k1 Þ 2ðk þk1 Þ2 4ð3k1Þ þ smðkk1 Þðkþ 2k1 þ 3Þ 2ðkþ 1Þ2 ð3k2Þ þsmðkk1 Þðk þ 2k1 Þ

ðB:13cÞ

ðB:13dÞ

2

2k

smðkk1 Þðk þ 2k1 3Þ

ðB:13eÞ

2ðk1Þ2 sm , 2

ðB:13bÞ

M LH7 ¼ 0

ðB:13f; gÞ

The above Eqs. (B.12) and (B.13) for k¼k2 convert to:   k2 k1 s

ðB:14a; bÞ PLH1 ¼ 0, P LH2 ¼ 1þ 2 k2 þk1 2ðk2 1Þ þ sðk2 k1 Þ , 2ðk2 þ 1Þ

The range of P and M for different cases of the three models are given below:

PLH3 ¼

B.1. Elastic–plastic model

PLH5 ¼ 1 þ

sðk2 k1 Þ , 2ðk2 1Þ

P LH4 ¼

P LH6 ¼ 1 þ

s , 2

ð2k2 1Þ þsðk2 k1 Þ 2k2 PLH7 ¼ 1þ s,

ðB:14c; dÞ

ðB:14e2gÞ

2

Case I : P EP1 r P rP EP2 and M EP1 ZM ZM EP2

ðB:1a; bÞ

Case II : PEP2 r P rP EP3 and M EP2 Z M Z MEP3

ðB:2a; bÞ

Case III : PEP3 r P r 1 and M EP3 Z M Z 0

MLH1 ¼

MLH2 ¼

ð3k2 1Þ þ smðk2 k1 Þð2k2 þ k1 Þ 2

2k2 4ð3k1 k2 1Þ þsðk2 k1 Þðk2 þ5k1 Þ 2ðk2 þ k1 Þ2

MLH3 ¼

   k1 1 , and PEP3 ¼ 1 , P EP1 ¼ 0, P EP2 ¼ k þ1 2k

4ð3k2 1Þ þ sðk2 k1 Þðk2 þ 2k1 þ 3Þ 2ðk2 þ 1Þ2



M EP1 ¼

ð3k 1Þ 2

2k

, MEP2 ¼

2ð3k1Þ ðkþ 1Þ2

and M EP3 ¼

ðB:15bÞ

ðB:3a; bÞ

where

2

ðB:15aÞ

ðB:4a2cÞ MLH4 ¼ ð23kÞ 2

2k

ðB:5a2cÞ MLH5 ¼

The maximum value of parameter k in the above Eqs. (B.4) and (B.5) is k1.

MLH6 ¼

ð3k2 2Þ þ sðk2 k1 Þðk2 þ 2k1 Þ 2

2k2 sðk2 k1 Þðk2 þ2k1 3Þ 2ðk2 1Þ2 s , 2

MLH7 ¼ 0

ðB:15cÞ

ðB:15dÞ

ðB:15eÞ

ðB:15f; gÞ

B.2. Linear hardening model Case I : P LH1 r P rP LH2 and M LH1 Z M Z M LH2

ðB:6a; bÞ

Case II : P LH2 r P r PLH3 and M LH2 Z M Z M LH3

ðB:7a; bÞ

Case III : P LH3 r P r PLH4 and M LH3 Z M Z M LH4

ðB:8a; bÞ

Case IV : PLH4 rP rP LH5 and M LH4 Z M Z MLH5

ðB:9a; bÞ

Case V : PLH5 rP rP LH6 and MLH5 Z M Z MLH6

ðB:10a; bÞ

Case VI : PLH6 rP rP LH7 and MLH6 Z M Z MLH7

ðB:11a; bÞ

B3. Parabolic hardening model

where P LH1 ¼ 0,

P LH3 ¼

P LH2 ¼



 kk1 sm

1þ 2 k þk1

2ðk1Þ þsmðkk1 Þ , 2ðk þ1Þ

P LH4 ¼

ðB:12a; bÞ ð2k1Þ þ smðkk1 Þ 2k

ðB:12c; dÞ

Case I : P PH1 rP r PPH2 and M PH1 ZM ZM PH2

ðB:16a; bÞ

Case II : P PH2 r P rP PH3 and M PH2 Z M Z M PH3

ðB:17a; bÞ

Case III : P PH3 r P rP PH4 and MPH3 Z M Z M PH4

ðB:18a; bÞ

Case IV : P PH4 rP r P PH5 and M PH4 Z M Z MPH5

ðB:19a; bÞ

Case V : PPH5 rP r P PH6 and M PH5 ZM ZM PH6

ðB:20a; bÞ

Case IV : P PH6 rP r P PH7 and M PH6 Z M Z MPH7

ðB:21a; bÞ

where PPH1 ¼ 0,

PPH2 ¼

3ðkk1 Þ þ sm2 ð3k2 k2k1 Þ 3ðk þk1 Þ

ðB:22a; bÞ

76

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

P PH3 ¼

3ðk1Þ þ sm2 ð3k2 k2k1 Þ 3ðk þ1Þ

ðB:22cÞ

eA ¼

2

P PH4 ¼

Fig. 3(a) is

3ð2k1Þ þ 2sm ð3k2 k2k1 Þ 6k

ðB:22dÞ

sm ð3k2 k2k1 Þ 3ðk1Þ

P PH6 ¼ 1þ

sm2 ð3k2 k2k1 Þ , 3ðkk1 Þ

ðB:22eÞ

P PH7 ¼ 1 þsmð2mÞ

2

M PH1 ¼

2

2

2

ðB:23bÞ

2ðkþ k1 Þ2 2

4ð3k1Þ þsm2 ð2k2 kk 3k1 þ 4k1 k2 2k1 kþ 6k2 2k4k1 Þ 2ðk þ 1Þ2 ðB:23cÞ 2

M PH4 ¼

2

2

sm2 ð2kk2 k 3k1 þ 4k1 k2 2kk1 6k2 þ4k1 þ2kÞ

ðB:23eÞ

2

2ðk1Þ 2

M PH6 ¼

ðB:23dÞ

2k

2ðkk1 Þ2

P PH3 ¼

P PH2 ¼

, and M PH7 ¼ 0

ð3þ 2sÞðk2 k1 Þ 3ðk2 þk1 Þ

3ðk2 1Þ þ2sðk2 k1 Þ , 3ðk2 þ 1Þ

P PH5 ¼ 1þ

2sðk2 k1 Þ , 3ðk2 1Þ

ðB:23f; gÞ

ðB:24a; bÞ

P PH4 ¼

P PH6 ¼ 1 þ

3ð2k2 1Þ þ 4sðk2 k1 Þ 6k2 ðB:24c; dÞ

2s , 3

P PH7 ¼ 1 þs

ðB:24e2gÞ

2

M PH1 ¼

M PH2 ¼

M PH3 ¼

M PH4 ¼

M PH5 ¼

M PH6 ¼

ð3k2 1Þ þsmðk2 k1 Þð2k2 þ k1 Þ 2

2k2 4ð3k1 k2 1Þ þ sðk2 k1 Þðk2 þ 7k1 Þ 2ðk2 þ k1 Þ2 4ð3k2 1Þ þ sðk2 k1 Þðk2 þ 3k1 þ 4Þ 2ðk2 þ1Þ2 ð3k2 2Þ þsðk2 k1 Þðk2 þ3k1 Þ 2

2k2 sðk2 k1 Þðk2 þ 3k1 4Þ 2ðk2 1Þ2 s , and M PH7 ¼ 0 2

P yd k 2 2 ¼ ¼ ¼ eA Hð12hÞ HP 3Myd P

ðC:3Þ

k_ Myd 1 ¼ e_ A Pyd 3P

ðC:4Þ

where e_ A P yd and k_ M yd are the generalised strain rates along P and M axes, respectively. From Eq. (11), @M ð1 þP þ3Pk Þ ¼ 2 @P k

ðB:25aÞ

ðB:25bÞ

ðB:25cÞ

ðB:25dÞ

ðB:25eÞ

ðB:25f; gÞ

Appendix C. Normality for Case I of elastic–plastic model The axial strain in a rectangular section corresponding to the axial force, is the strain at the centroidal axis which from

ðC:5Þ

and,

@P

The above Eqs. (B.22) and (B.23) for k¼k2 convert to P PH1 ¼ 0,

Using Eqs. (C.1), (C.2) and (12)

@2 M

2

sm2 ð2kk2 k þ k1 2k1 k2 Þ

ðC:2Þ

2

2

ð3k2Þ þ sm2 ð2k2 kk 3k1 þ 4k1 k2 2k1 kÞ 2

M PH5 ¼

key key ¼ ðHH1 Þ Hð1hÞ

or,

2

4ð3kk1 1Þ þ sm2 ð2kk2 k 4kk1 þ10k1 k2 7k1 Þ

2

M PH3 ¼

ðB:23aÞ

4k

2

M PH2 ¼

k¼

ðB:22f; gÞ

2ð3k 1Þ þ sm2 ð8k2 k3k 3k1 6k1 k þ4k1 k2 Þ

ðC:1Þ

and the curvature is approximately given by

2

P PH5 ¼ 1þ

key ðH=2H1 Þ key ð12hÞ ¼ ðHH1 Þ 2ð1hÞ

2

2

¼

ð1 þ 3k Þ

ðC:6Þ

2

k

2

Since the value of @2 M=@P given by Eq. (C.6) is negative for all values of k, therefore, it is always convex. Eq. (C.4) is not exactly normal to the tangent to the non-dimensional failure curve given by Eq. (C.5) and it approaches for large values of k. The difference is due to the approximate value of curvature taken in Eq. (C.2). References [1] Alves M, Jones N. Impact failure of beams using damage mechanics: Part I—Analytical model. Int J Impact Eng 2002;27:837–61. [2] Alves M, Jones N. Impact failure of beams using damage mechanics: Part II—Application. Int J Impact Eng 2002;27:863–90. [3] Symonds PS, Genna F, Ciullini A. Special cases in study of anomalous dynamic elastic–plastic response of beams by a simple model. Int J Solids Struct 1991;27:299–314. [4] Qian Y, Symonds PS. Anomalous dynamic elastic–plastic response of a Galerkin beam model. Int J Mech Sci 1996;38:687–708. [5] Bassi A, Genna F, Symonds PS. Anomalous elastic–plastic responses to short pulse loading of circular plates. Int J Impact Eng 2003;28:65–91. [6] Jones N. Structural impact. Paperback edition 1997 (2nd Ed., 2012), Cambridge: Cambridge University Press; 1989. [7] Symonds PS, Frye CWG. On the relation between rigid-plastic and elastic– plastic predictions of response to pulse loading. Int J Impact Eng 1988;7: 139–49. [8] Yu TX. Elastic effects in the dynamic plastic response of structures. In: Jones N, Wierzbicki T, editors. Structural crashworthiness and failure. London, New York: Elsevier Applied Science; 1993. p. 341–84. [9] Zyczkowski M. Combined loadings in theory of plasticity. Int J Non-Linear Mech 1967;2:173–205. [10] Ma´s RI, Ma´s FI. Elastic interaction graphs for steel H-sections subjected to bending, shear and axial forces. Int J Solids Struct 2000;37:1327–37. [11] Ma´s RI, Ma´s FI. Shear–bending–torsion elastic interaction diagrams in circular steel sections. Int J Solids Struct 2001;38:435–43. [12] Shen WQ. Interaction yield hypersurfaces for the plastic behaviour of beams—I. Combining bending, tension and shear. Int J Mech Sci 1995;37: 221–38. [13] Aiello MA, Tegola AL, Ombres L. Coupled instability of thin-walled members under combined bending moment, axial and shear force. Thin-Walled Struct 1994;19:285–97. [14] Ma´s RI, Ma´s FI. Biaxial bending–axial force elastic interaction diagrams in hollow steel sections. Int J Solids Struct 2001;38:423–33. [15] White GJ, Grzebieta RH, Murray NW. Maximum strength of square thinwalled sections subjected to combined loading of torsion and bending. Int J Impact Eng 1993;13:203–14.

H. Abbas, N. Jones / International Journal of Mechanical Sciences 55 (2012) 65–77

[16] Kim K, Yoo CH. Ultimate strengths of steel rectangular box beams subjected to combined action of bending and torsion. Eng Struct 2008;30: 1677–87. [17] Johnson W, Mellor PB. Engineering plasticity. London: Van Nostrand Reinhold Co.; 1973. [18] Phillips A. Introduction to plasticity. New York: The Ronald-Press Co; 1956.

77

[19] Jones N. Some comments on the modelling of material properties for dynamic structural plasticity. In: Harding J. (Eds.) Institute of Physics, International conference on mechanical properties of materials at high rates of strain, Series No. 102; 1989. p. 435–45. [20] Jones N. Dynamic material properties and inelastic failure in structural crashworthiness. IJCrash 2001;6:7–18.