Impact Failure of Beams Using Damage Mechanics - Part I

International Journal of Impact Engineering 27 (2002) 837861

Impact failure of beams using damage mechanics:Part IAnalytical model

Marc!lio Alvesa, Norman Jonesb,*aDepartment of Mechatronics and Mechanical Systems Engineering, University of S *ao Paulo, S *ao Paulo,

05508-900, BrazilbDepartment of Engineering (Mechanical), Impact Research Centre, University of Liverpool, Brownlow Hill,

Liverpool L69 3GH, UK

Received 7 August 1999; received in revised form 24 October 2001; accepted 5 March 2002

Abstract

A simple theoretical method, which is based on ductile damage mechanics and which retains strain rateeffects, is presented for predicting the failure of beams made from a perfectly plastic material and subjectedto impact loads. For this class of materials, the strains can be estimated by dening a hinge length. Thedenition adopted here leads to reasonable predictions for the plastic strains and the strain rate, as shownby comparing the results with numerical calculations and experimental data. The equivalent strain and thestrain rate can be used in the damage model to predict the failure of beams, as shown in a companion paper(Alves, Jones, Int J Impact Eng 2002;27(8):86390). r 2002 Elsevier Science Ltd. All rights reserved.

1. Introduction

The analysis of metallic structures, subjected to static and dynamic loads producing largedisplacements and large plastic strains, is nowadays a routine task when using nite-element andother numerical techniques as well as the rigid-plastic methods of analysis [1]. However, regardlessof the method used, the prediction of failure due to material rupture is fraught with difculty [2].One approach to predict failure is based on the methods of micromechanics. However, in view

of the wide variety of void shapes and different mechanical properties for materials, it is difcultto establish a procedure which is suitable for the prediction of failure in engineering practice.Sophisticated micromechanic models for predicting ductile failure rely on several materialconstants, as in Curran et al. [3], and intense computational effort is required [47].

*Corresponding author. Tel.: +44-151-794-4858; fax: +44-151-794-4848.

E-mail address: [email protected] (N. Jones).

0734-743X/02/$ - see front matter r 2002 Elsevier Science Ltd. All rights reserved.

PII: S 0 7 3 4 - 7 4 3 X ( 0 2 ) 0 0 0 1 7 - 9

Nomenclature

B beam breadthC; q; %m; %n coefcients in the CowperSymonds equationD damageDd dynamic damage including strain hardening and strain rate effectsDd0 dynamic contribution to the total damageDd0 dynamic damage based on the ow stress s0; without strain hardeningDs0 static damage based on the ow stress s0; without strain hardeningDs static damage based on the ow stress s0; with strain hardeningDs0 static damage based on the initial ow stress s00; without strain hardeningDE0 damage due to hardeningE elastic modulusE0 hardening modulusf yield functionG impact massh H=L1H beam thickness%H Heaviside function

k dimensionless constant2L beam lengthL1; L2 distances dened in Fig. 2lM1 ; lM2 bending hinge lengthslN1 ; lN2 membrane hinge lengthslQ1 ; lQ2 shear hinge lengthsm mass per unit length of a beamM bending momentM0 collapse momentMy yielding momentp accumulated plastic strainpD accumulated threshold plastic strainQ transverse shear forceQ0 collapse transverse shear forceQy yielding transverse shear forceRn dened by Eq. (6)sf shape factor%S damage strength parametert timeV0 impact velocityw W=H%wf Wf=L1W beam displacement at the loading pointWf nal value of W

M. Alves, N. Jones / International Journal of Impact Engineering 27 (2002) 837861838

A global failure criterion is an alternative method which disregards any local detail. Theclassical concept of a yield surface has been expanded to embrace fracture surfaces, giving rise totechniques which predict failure based mainly on stress and strain parameters, as in Theocaris [8]

Ws indentationWsf nal indentationx; y and z rectangular coordinates

Greek symbols

g shear straine true straineD threshold straineeq true equivalent straineM bending straineN membrane strainy anglek curvaturel plastic multipliern Poissons ration1; n2 L1=H; L2=Hx L1=L2s true stresss0 ow stresss00 initial ow stressseq equivalent stresss0ij deviatoric stress componentsh hydrostatic stresssy yield stresst0 shear ow stress

Subscripts

0 initialcr criticald dynamiceq equivalents static

Superscript

: time derivative:: second time derivativeB effective variablep plastic component of the strain tensort true stress or strain

M. Alves, N. Jones / International Journal of Impact Engineering 27 (2002) 837861 839

and Ukadgaonker and Awasare [9]. Failure theories have been developed as an extension ofplasticity theory [10], which use generalised stresses [11] and even fractal concepts [12]. Thesemethods are more arbitrary than micromechanics models, but are simpler to use and moresuitable for engineering design purposes.A rigid-plastic analysis was used in Refs. [13,14] to predict the dynamic behaviour of a beam

which fails either by tensile tearing or by transverse shearing. The beam is subjected to auniformly distributed velocity over its entire span, as an idealisation of a blast loading, and thethreshold velocities, which cause failures known as mode II and III, have been determined. Thetheoretical results are in reasonable agreement with experimental data available for beams whichare made from a non-strain rate sensitive material. For tensile tearing, a hinge length was denedand the uniaxial rupture strain was used in the calculations to predict failure.A similar procedure was used for a free-free beam [15], in order to determine a critical dynamic

pressure pulse which causes a tearing failure, and for a beam struck transversely by a mass [16].It was assumed in Ref. [17] that rupture occurred in a rigid-plastic structure when the

absorption of plastic work per unit volume reached a critical value. To calculate the actual plasticwork in beams, a hinge length was estimated from experimental data obtained by Menkes andOpat [18] on impulsively loaded aluminium beams.This methodology was reviewed in Ref. [11] and has been used to predict the failure of clamped

beams [17] and plates [19] under impulsive loading and clamped beams struck by a mass [20]. Themethod predicts reasonable values for the threshold impulses that produce tensile tearing andtransverse shearing failures in actual beams. Nevertheless, the failure of circular plates is not wellpredicted. However, some slippage at the clamped boundaries of the plates during the experimentsmight be responsible for the deviation and further study is required.The rigid, perfectly plastic models for failure are, essentially, global approaches relying on

generalised stress elds so that any details related to void coalescence and interaction, triaxialityand softening are disregarded.This paper explores the value of Continuum Damage Mechanics, CDM, to predict analytically

material failure in simple structures. CDM idealises a material as a continuum using the usualconcepts of stress and strain. The presence of voids, small cracks or second phase particles arecatered for by a single damage variable so that the constitutive laws might be formulatedusing continuous variables for materials exhibiting voids or small cracks. A simple ductiledamage model is explored for perfectly plastic and linear strain hardening materials, strain rateeffects being considered. The model requires strain and strain rate values, which are obtainedby dening a plastic hinge length. The procedure is employed in a companion paper [21] topredict the static and dynamic failure of beams and comparisons are made with experimentalresults.

2. CDM ductile model

Continuum Damage Mechanics offers a theoretical framework for the prediction of damageevolution in a structure. An upper bound to the growth of damage is the initiation of a crack,which occurs when a critical damage is attained in the continuum. The threshold between criticaldamage and crack growth is taken as a failure criterion.


A basic CDM model is proposed by Lemaitre [2224], with the damage evolving according to

D lqFDqY

; 1

or

D *s2eqRn2E %S

p %H/p pDS; 2

since

l=1 D p; 3

FD Y 2=2 %S1 D; 4

and

Y *s2eqRn=2E; 5

with

Rn 21 n=3 31 2nsh=seq2: 6

In Eqs. (2)(6), D is the damage, E is the original elastic modulus of a virgin material, n isPoissons ratio, *seq is the effective stress, dened by *seq seq=1 D [25], seq and sh are theequivalent and hydrostatic stresses, respectively. The Heaviside function %H indicates that D > 0when the accumulated plastic strain, p > pD; where pD is a function of the material and stress state.The parameter %S in Eq. (2), the damage strength material parameter, is regarded as a materialconstant [25,26], but can be seen as a material dependent, adjustable parameter, since its mainfunction is to bring the prediction of Eq. (2) closer to experimental data. Eq. (3) comes from the

denition of the accumulated plastic strain, p 23epij e

pij

qwhen using

epij l qf =qsij l3

2

s0ijseq

1

1 D;

where s0ij sij skkdij=3 [25].Strain rate effects on the damage evolution are taken into account in the present work. By

assuming that the elastic modulus, Poissons ratio, %S and the triaxiality, sh=seq; are strain rateindependent parameters, the strain rate only causes an increase in the ow stress, s0; as estimatedusing the widely known CowperSymonds equation [1]

s0d s0s %me%n; 7

where

%m s0sC1=q and %n q1; 8

with the subscripts s and d standing for static and dynamic, respectively, and C and q beingmaterial constants.Under these assumptions, it follows, when retaining strain rate effects according to Eq. (7), that

the dynamic damage evolution, Dd; is obtained by writing Eq. (2) in the form:

Dd s0s %me%n2Rn p %H/p pDdS=2E %S: 9


3. Particular integration of the damage equation

For some special cases, the dynamic damage equation may be integrated, as reported here forperfectly plastic and linear hardening materials. In the following integration, the triaxiality ratio,sh=seq; is assumed to remain constant for proportional loading, so that the factor Rn is invariantwith stress. The role of the triaxiality on the failure of beams will be explored in a companionpaper [21].Consider rst an elastic, perfectly plastic material with a static ow stress *seq s0: According

to Lemaitre [25], the damage law, Eq. (2), may be integrated to give a static perfectly plasticversion of the damage,

D s20Rnp pD=2E %S for pXpD: 10

In the case of proportional loading, the accumulated plastic strain, p; equals the equivalent strain,eeq; and pD eD; giving

1

D s20Rneeq eD=2E %S for eeqXeD; 11

which is due to Lemaitre [25].The material model can be improved by considering the linear hardening law

*seq s00 E0eeq; 12

where, E0 is the hardening modulus and s00 is a initial ow stress, as shown in Fig. 1.By inserting Eq. (12) into Eq. (2) and performing the integration for proportional loading, the

static damage evolves with the plastic strain according to

D s200Rneeq eD=2E %S RnE0eeq eDeDE0eD 3s00

eeqE0eeq 3s00 E0eDeeq=6E %S for eeqXeD: 13

Consider now the integration of Eq. (2) when not only linear hardening but also strain ratesensitivity (Eq. (7)) is taken into account according to the equation2

*seqd s00 E0eeq %me %n; 14

which gives

Dd s200Rneeq eD=2E %S RnE0eeq eDeDE0eD 3s00 eeqE0eeq 3s00

E0eDeeq=6E %S Rn %me %neeq eD2s00 E0eD eeq %me %n=2E %S for eeqXeD: 15

Thus, the total damage, Dd; consists of three parts Ds0 (static), DE 0 (hardening) and Dd0 (dynamic),which correspond to the three terms on the right-hand side of Eq. (15), respectively.

1The effective equivalent stress was set equal to the ow stress, so that damage is coupled to the plasticity criterion.2A static value, eD; was used for the threshold dynamic plastic strain, eDd ; as suggested by the experimental data on a

mild steel in a companion paper [21].


For a perfectly plastic material with an initial static ow stress s00 s0; DE0 0; so thatEq. (15) becomes

Dd0 s0 %me%n2Rneeq eD=2E %S; 16

when strain rate effects are retained.Eqs. (11), (13), (15) and (16) can be used for the prediction of damage in a structure provided

the strains, and other parameters, are known. However, the stresses are not related uniquely to thestrains in a perfectly plastic material so that an approximate method for predicting the strains isexplored in the next section.

4. Strains in a perfectly plastic material

Strains might be evaluated approximately at a hinge in a perfectly plastic structure by choosinga hinge length to distribute an otherwise innite curvature.Nonaka [27] proposed a deection dependent hinge length for beams by using a theoretical

rigid-plastic analysis and a plane strain approximation. A similar approximation has been used inRef. [16] for beams, while Ref. [17] has analysed experimental data from Menkes and Opat [18]and obtained a hinge length related to the beam thickness and to the plastic work dissipated in ahinge. Wen et al. [28] have used a constant hinge length, equal to the beam thickness, forevaluating the strains associated with bending deformations.Another denition of hinge length is suggested here which allows the bending, membrane and

shear strains to be determined. The various hinge length denitions are tested against theexperimental results and numerical data which are available in the literature. Simple expressionsare then obtained for the equivalent strain and its rate.

eq

eq

oo

o

E'1

Fig. 1. Denition of the ow stress.


4.1. Bending hinge length

Consider the elastic, perfectly plastic beam3 in Fig. 2 which is subjected to a concentrated loadat any position on the span, and which produces innitesimal displacements. The beam material isfully plastic underneath the concentrated load where the moment achieves a peak value. Awayfrom this point, the inner elastic core of the beam cross-section increases in size until the entirecross-section is elastic. This decrease of the plastic zone size is related directly to the decrease ofthe bending moment. Thus, for small displacements of the beam, the bending hinge length may bedened as

Bending hinge length, lM, is the distance along an elastic perfectly plastic beam between the two

points A and B, where jMAj M0 and jMBj My: My and M0 are the initial yield and fullyplastic collapse moments for the cross-section.

As an example, consider that portion of the simply supported beam in Fig. 2 which is on theright-hand side of the concentrated load. From the relation4 M0=L1 M0 My=lM1 ; thebending hinge length is

lM1 L1sf 1=sf L1=3; 17

where the shape factor sf M0=My 32 for a beam with a rectangular cross-section and a uniformthickness H: A similar expression can be obtained for the left-hand side of the beam and resultsfor other beam congurations may be obtained, as summarised in Table 1.It is assumed that the change of curvature, k1; of the centroidal axis on the right-hand side of

the load in Fig. 2 is

k1 y1=lM1 ; 18

taken to be constant along the hinge length. Eq. (18) gives an axial strain e1 k1z; where z is thethrough-thickness co-ordinate. For moderate displacements, the angle y1 may be approximated asy1 W=L1; inducing a bending strain at the outermost surface, z H=2;

eM1 3HW=2L21: 19

A similar procedure can be developed for the left-hand side of the beam.According to the above denition for the bending hinge length, it is evident that there is a

discontinuity of curvature and strain underneath an off-centre concentrated load. The nearer aconcentrated load is to a support, the stronger is the discontinuity in the curvature at the loadingpoint. In fact, a very sharp change of the curvature occurs in this region and data to be presentedlater will conrm this observation. It should be emphasised that such a discontinuity violates thegeneral relationship between curvature and bending moment for elasto-plastic beams. This is anintrinsic difculty of the present approach which sets out to determine large plastic strains inbeams made of perfectly-plastic materials.

3The moment distribution is obtained for a perfectly plastic material. However, the hinge length concept adopted

here invokes the elastic core of the beam. Moreover, the damage is measured from the change of the elastic modulus.

Hence, the use of the term elastic.4 It is assumed that L1pL2:


Following a similar procedure, the change of curvature and bending strain can be calculated fordifferent loads and boundary conditions.

4.2. Dynamic bending hinge length

The previous denition of a bending hinge length was based on a static moment distribution,which is strictly valid only for small displacements of a rigid, perfectly plastic beam. It is nowenquired whether or not the dynamic bending moment distribution occurring during the variousphases of a beam when loaded dynamically, would differ substantially from the static bendingdistribution.Consider a clamped beam struck at the mid-span by a mass, G; travelling with an initial

velocity, V0; as shown in Fig. 3. For a beam to tup mass ratio, 2mL=G; smaller than one,approximately, the second phase of motion dominates the beam response [1,29] and the rsttransient phase of motion might be ignored. Moreover, a tensile failure would be more likely tooccur during the second phase when transverse shear effects are not considered.After the rst phase of motion, when the travelling plastic hinges have propagated from the

centre towards the supports, the bending moment distribution along the beam is [1]

M M01 2b23%a b=%a23 2%a 6b=%a3 2%a; 0pbp%a; 20

Fig. 2. (a) A beam loaded off-centre showing the bending hinge length denition. (b) Associated bending moment

distribution. The beam displacement is exaggerated.


Table 1

Bending hinge lengths for various beam congurations. sf is the shape factor used in plasticity

Beam conguration Bending hinge length

Rectangular cross-section General cross-section

lM L

3lM 1 s1f L

lM1 L1

3lM1 1 s

1f L1

lM2 L2

3lM2 1 s

1f L2

lM L3

p lM L1 s1f

q

lM L

6lM

sf 12sf

L

lM1 L1

6lM1

sf 12sf

L1

lM2 L2

6lM2

sf 12sf

L2

lM L6

p lM L21 s1f

q


where b mx=G; %a mL=G and m is the mass per unit beam length. Eq. (20) is plotted in Fig. 4for two different ratios of 2mL=G: It is clear that the bending moment distribution changes withthe tup mass; the smaller tup mass having a longer hinge length. Nevertheless, from anengineering perspective, the dynamic moment distribution with 2mL=G51; is similar to thecorresponding static one [29].Consider now a clamped beam loaded impulsively throughout its entire span with an initial

velocity V0; Fig. 5. The bending moment distribution during the second phase of motion withstationary plastic hinges is [1]

M=M0 1 x=L3 3x=L2: 21

A difference of only 5.3% is obtained between the values of x=L for which My=M0 23 in Eq. (21)and the static moment distribution for a uniform pressure loading on a beam with a rectangularshaped cross-section.

Fig. 3. Clamped beam struck at the mid-span by a falling mass.

-1

-0.5

0

0.5

1

x/L

M/M

0

2mL/G=12mL/G=5

static

0.20 10.80.4

Fig. 4. Static and dynamic bending moment distributions during the second phase of motion, along the clamped beam

shown in Fig. 3 when struck by different masses. The horizontal thin line denes My=M0 23:


Further calculations for other beam congurations show that the difference between the staticand dynamic bending moment distributions is not signicant during the stationary hinge or modalphase of motion. Accordingly, any possible increase in accuracy in the bending hinge lengthaccording to the present denition when using a dynamic bending moment distribution isovershadowed by the simplicity of the static moment expressions. Thus quasi-static methods ofanalysis should be adequate for the present investigation with low impact velocities [29].

4.3. Shear hinge length

A transverse shear hinge length may be dened as

A transverse shear hinge length, lQ, is the distance along the length of a beam between two points Aand B, where jQAj Q0 and jQBj Qy: Qy and Q0 are the initial yield and the plastic collapsetransverse shear forces of the cross-section, respectively.

The evaluation of the shear hinge length is more involved than the hinge length associated withbending, partly because transverse shear effects are potentially more signicant for dynamic loads,particularly during the rst, or transient, phase of motion [1]. Thus, lQ must be obtained from ananalysis of the rst phase of motion for a dynamically loaded beam.Consider the simply supported rigid, perfectly plastic beam in Fig. 6, which is subjected to a

uniformly distributed impulsive velocity of magnitude V0: During the rst phase of motion, thetransverse shear force is [1]

Q Q0x x0L x0

2; x0pxpL; 22

when LX3H=2 and, from symmetry, only the right hand side of the beam in Fig. 6 is considered.Here, Q0Es0BH=2 is the transverse shear force necessary to fully deform, plastically andindependently, a beam having a rectangular cross-section with dimensions B and H made from amaterial with a ow stress s0: x0 L 3H=2 denes the position of the stationary bending hingeduring the rst phase of motion.Now, for beams with rectangular cross-sections, the transverse shear force which just causes the

material to yield at the beam centre is Qy BH23t0; whereas Q0EBHt0 is the transverse shearforce which causes the entire cross section to become fully plastic. Hence, when using Qy=Q0 23;

Fig. 5. A clamped beam loaded impulsively with a velocity V0:


the denition of a transverse shear hinge length gives

lQ 36

pH=2 0:275H: 23

A similar procedure for clamped supports at both ends yields

lQ 36

pH 0:551H: 24

Ref. [30] has studied the case of a mass, G; impacting with an initial velocity, V0; at any positionon the span of the clamped beam in Fig. 7, which has a rectangular cross-section. It turns out thatthe analysis consists of three cases: case i, 3on1pn2; case ii, 1on1p3 and 3on2; case iii,0on1p1 and 3on2; where n1 L1=H and n2 L2=H:For case i, the shear force is symmetrical about the struck point and using the relations from

Ref. [30], it might be shown that

lQ 36

pH 0:551H; 25

where x0 3H [30]. Eq. (25) is identical to the case of a fully clamped beam loaded impulsivelythroughout the span.For case ii, with the further restriction5 1:5pn1o3; it transpires that there are two different

hinge lengths because the shear force distribution on the left-hand side of the tup is different fromthat on the right-hand side. In the rst phase of motion, it might be shown that

lQ1 L13H L1 7L21 16HL1 9H

2

q=3H L1 26

and

lQ2 36

pH; 27

at the right- and left-hand sides of the tup, respectively.For case iii, the impact load acts at a distance from the support which is smaller than the beam

thickness and gives the shear hinge lengths

lQ1 L1 28

Fig. 6. A simply supported beam loaded impulsively with a velocity V0:

5For 1on1o1:5; the time at the end of the rst phase of motion depends on the positions of the plastic hinges and isgiven by Eqs. (32) or (35) in Ref. [30]. By substituting these equations into the transverse shear force expression, it turns

out that it is difcult to obtain a closed form expression for the shear hinge length, according to the present denition.


and

lQ2 36

pH; 29

at the right- and left-hand sides of the tup, respectively.It is observed from the above equations that the shear hinge length depends on the external

loading position, type of support and beam thickness, but not on the yield stress of a perfectlyplastic material. Also, it is rather small when compared with the bending hinge length.

4.4. Membrane hinge length

Large displacements in a broad class of engineering structures produce membrane forces,which, in the case of beams, are constant along the length when disregarding axial inertia. Thissuggests that the membrane strain spreads evenly along the beam length. In this work, membranehinge lengths

lN1 L1 and lN2 L2; 30

are adopted.

4.5. Equivalent strain

The following equations are deduced for the particular case of a clamped beam subjected toeither a static or a dynamic concentrated load acting at any position on the span.The equivalent strain, based on the von Mises yield criterion, can be dened in cartesian

co-ordinates as

eeq 2

p exx eyy

2 eyy ezz2 ezz exx

2 6e2xy e2xz e2yzq

=3: 31

Now for a beam in the xz plane, it is assumed that exy eyz 0 and eyy ezz exx=2; so thatEq. (31) reduces to

eeq e2xx g2xz=3

q; 32

L2

x

L1

V0

G

Fig. 7. Clamped beam struck transversely by a mass G:


where gxz 2exz is an engineering shear strain. The two strains in Eq. (32) are calculated in thefollowing two sections.

4.5.1. Membrane strainThe membrane strain, eN ; on the right-hand side of a fully-clamped beam, due to a normal force

N; becomes

eN1 DL1=lN1 1 W=L1

2

q 1: 33

Eq. (33) can be averaged with a similar expression for the left-hand side of the beam, leading to

eN 1 x2hw2=4; 34

for W=L1251 when dening

x L1=L2; h H=L1 and w W=H: 35

This strain is assumed to be distributed evenly along the beam span and throughout the cross-section.

4.5.2. Shear strain

The shear strain, which is taken as constant along the shear hinge length, is evaluatedapproximately as

gxz Ws=lQ1=2; 36

where Ws measures the indentation of the load device on the beam. The factor 2 in Eq. (36) wasused in Refs. [28, 31], based on experimental results. The latter authors suggested that the factor 2should be used for shear failure taking place at the proximal side of a beam, i.e., on the surface hitby the drop mass. In the experimental programme reported in the companion paper [21], some ofthe impacted specimens failed by shear, but it was not possible to detect the region where failurestarted.The results in Section 4.3 suggest an approximate value for the shear hinge length of

lQ1 lQ2 H=2 37

for a clamped beam, so that Eq. (36) gives

gxz 4Ws=H: 38

A shear hinge length of about one-half of the beam thickness was found experimentally bymeasuring the angles of initially square grids on impacted beams [21]. Moreover, Ref. [32] arrivedat a value of lQ 0:433H based on the static equilibrium equations for a beam.As far as the shear displacement, Ws; is concerned, experimental data in Ref. [21], for a fully

clamped beam impacted by a mass, suggests the relation

Wsf kHWf=L1; 39

where Wsf is the nal indentation (transverse shear displacement), H is the beam thickness, Wf isthe nal transverse displacement of a beam (including Wsf ) and k is a constant. Also, numericalinvestigations in Ref. [33], suggest that it is reasonable to assume

Ws Wsf W=Wf ; 40


or

Ws kHW=L1; 41

where W is the beam displacement.From Eqs. (38) and (41), it follows that the transverse shear strain at the impact point of a

clamped beam is

gxz 4kW=L1 4khw: 42

4.5.3. Bending strain

The bending strain, at the point of the maximum beam displacement, is

eM 1

2

W

L1lM1z

W

L2lM2z

; 43

when taking an average value of the strains at the left- and right-hand sides of the load point. Fora clamped beam subjected to a concentrated load, the bending hinge lengths are

lM1 L1=6 and lM2 L2=6 44

and the interest here is concentrated at the point of maximum bending strain, i.e. z H=2:Accordingly, the bending strain in Eq. (43) reduces to

eM 3h21 x2w=2: 45

For the calculation of exx is should be noted that the bending strains dominate the responseup to wE1: For larger displacements, the axial strain is mainly due to membrane effects. It followsthat

exx eM 3=21 x2h2w for wp1 46

and

exx 1 x2h2w2 5=4 for w > 1; 47

where the last equation was obtained by imposing the requirement eM eN at w 1:Eqs. (42), (46) and (47) can now be substituted into Eq. (32) to give an estimate for the

maximum equivalent strain in a clamped beam,

eeq 3=2h21 x2w2 4khw=

3

p2

qfor wp1; 48

and

eeq h21 x2w2 5=42 4khw=

3

p2

qfor w > 1; 49

which occurs underneath the concentrated load.A further simplication is possible by noting that, for beams hit close to a support, say x50:1;

it is expected that shear effects would dominate bending and membrane strains. Thus, theengineering equivalent strain for impacts close to a support is

eeq 4khw=3

pforw51: 50


5. Strain rate

A rough estimate for the average strain rate, em; in a clamped beam is [1]

em WfV0=32

pL2; 51

where Wf is the nal transverse displacement and V0 is the initial impact velocity of a mass at themid-span. This equation was obtained by Perrone and Bhadra [34], who studied a simple model ofa mass supported by two strain rate sensitive wires of equal length L: Wen et al. [28] estimated thestrain rate in beams loaded at any position, L1; from the supports, using Eq. (51) with L L1:Now, consider the strain rate variation underneath an aluminium alloy beam impacted on the

top surface at the one-quarter span position (H 7:62mm, L1 25:4mm, V0 5:34m/s,Wf 6:8mm). The strain rate behaviour according to a numerical simulation of this problem inRef. [33], is shown in Fig. 8. The prediction according to Perrone and Bhadra [34], Eq. (51), is alsoplotted in the same gure. The predicted strain rate underestimates the peak strain rate, in partbecause the moment is disregarded in the model which is used to obtain Eq. (51), as well as theneglect of material strain rate sensitivity.The poor result given by Perrone and Bhadras equation (51), motivates the search for a more

accurate description of the strain rate in a clamped beam.The equivalent strain rate is dened as eeq

2eij eij=3

p; which reduces to

eeq e2xx g2xz=3

q; 52

in the present analysis. Now, differentiating Eqs. (42) and (46) and substituting in Eq. (52), givesan expression for the equivalent strain rate which is a function of the velocity of the beam and,hence, changes continuously with time. A further simplication is possible using the observation

-40

0

40

80

120

160

t (ms)

(1/s).

Perrone&Bhadra [34]

Yu&Jones [33]

0 21 3 4

eq. (53)

Fig. 8. Strain rate vs. time underneath an aluminium beam impacted on the top surface at the one-quarter span

position. G 5 kg, V0 5:34m/s, 2L 101:6; B 10:16 and H 7:62mm. The thick line is the tensile strain rate onthe beam surface, according to the numerical simulation in Ref. [33]. The horizontal continuous lines are the predictions

according to Perrone and Bhadra [34], Eq. (51), and to Eq. (53).


of Perrone and Bhadra [34] that the maximum strain rate occurs when W V0=2

p;

approximately, yielding a constant equivalent strain rate

eeq V0

L1

9=81 x22h2 8k2=3

qforwp1: 53

Moreover, for beams hit close to a support, only shear strains are assumed to be important (seeEq. (50)), leading to a strain rate expression

eeq 8=3

p kV0L1

forw51: 54

Observe that Eq. (46), valid for wo1; was used instead of Eq. (47), valid for w > 1; to obtainEq. (53). If Eq. (47) were to be used, it would be necessary to estimate the nal beamdisplacement. Also, calculations show that the difference in the strain rate values using eitherequation are not large. Finally, the comparison in Fig. 8 gives a good prediction for the strain ratecalculated using Eq. (53). The gure shows the strain rate underneath the clamped beam in Ref.[33], which was struck at the one-quarter span position, x 1

3: The estimated strain rate of

eeq 116:2 s1 using Eq. (53) compares reasonably well with the nite-element results in Ref. [33].

This simple model does not capture the variation of the strain rate throughout the motion, butit does provide a reasonable estimate for an average strain rate during the initial response.Unfortunately, due to a lack of numerical and experimental data, it was not possible to furthercompare the strain rate equations for beams loaded at other positions. However, it is well knownthat although material strain rate effects are important for steel and other materials, their highlynonlinear behaviour means that approximate estimates of the strain rate can give good estimatesfor the ow stress.

6. Validation of the hinge length denitions

It is important to assess whether the approximate hinge lengths dened in Section 4 and theassociated strains have any experimental support.

6.1. Qualitative validation

Menkes and Opat [18] present experimental data for aluminium alloy beams loaded impulsivelyalong the entire span. For moderate impulses, when transverse shear effects are not dominant, onecan plot the deformed prole of the beam and mark the boundaries of the bending hinge lengthdenition which, for a clamped beam and a uniformly distributed load, is L=

6

pfrom Table 1. It

is observed in Fig. 9 that the hinge length, as here dened, predicts that most of the beam span hasa non-zero curvature, as observed in the test specimens.Fig. 10 shows the permanent deformed proles of two mild steel beams impacted by a mass and

with the bending and shear hinge lengths marked in the gure. The bending hinge length for theseclamped beams is L=6 from Table 1, while the shear length varies according to the load positionbut, as discussed in Section 4.5.2, is assumed to be one-half of the beam thickness, regardless ofthe tup position.


Fig. 9. Displacement proles of fully clamped aluminium beams loaded impulsively throughout the span [18]. The

smaller distances between the vertical lines dene the regions where the curvature is zero, according to the bending

hinge length based on a static moment distribution.

Fig. 10. (a) Beam struck by a round tup at its mid-span by G 6:5 kg travelling at V0 14:9m/s. (b) Beam struck by asharp tup at L1=2L 15 by G 19:5 kg with V0 6:14m/s. The regions between the set of vertical white lines mark the(a) bending and (b) transverse shear hinge zones.


It is evident that the boundaries of the bending hinge length more or less coincide with theregions where the curvature becomes zero. Moreover, the grids marked on the beam in Fig. 10(b)show that the regions where transverse shear is present more or less coincide with the shear hingedimensions here dened.

6.2. Quantitative validation

A nite-element simulation was conducted in Ref. [33] for the impacted beam experimentsreported in Ref. [35]. These results are especially interesting in the present context because theyprovide a detailed description of some eld variables.Consider a fully clamped aluminium alloy beam with a length of 101.2mm struck by a 5 kg

mass at the one-quarter span position and at a velocity of 5.34m/s. The curvature prole given bythe numerical results is shown in Fig. 11 for the instant when the maximum deection of the beamis reached. This gure also contains the theoretical predictions for the change in curvature (seeEq. (18))

k1 W=L1lM1 6W=L21 and k2 W=L2lM2 6W=L

22; 55

where the subscripts 1 and 2 refer to the right- and left-hand sides of the tup, respectively. It isassumed that these are the maximum values which are constant up to the boundaries of thehinges.The good correlation in Fig. 11 for the change in curvature does not necessarily imply accurate

values for the strains. However, the theoretical predictions from Eq. (48) for x 13when

transverse shear is neglected,

eeq 5h2w=3; 56

-0.08

-0.04

0

0.04

0.08

x/L1

(1/mm)

predictionnumerical

tup

0 0.5 11.5

Fig. 11. Curvature prole across part of the span of a fully clamped aluminium beam (H 7:62mm) struck by a fallingmass at the one-quarter span position (x=L1 0 is at the clamped support closer to the impact point and x=L1 1 is atthe impact point). Thin and thick lines represent the nite-element results [33] for W 7:4mm at t 2:572ms andEq. (55), respectively. The tup width is 5.08mm.


can be compared with the maximum axial strain (no shear) on the bottom surface underneath theimpact region reported in Ref. [33]. Fig. 12 shows a reasonable correlation between the numericalvalues and the theoretical prediction of Eq. (56) for small displacements.Consider now the transverse shear strains in an aluminium beam struck at the one-quarter span

position. Yu and Jones [33] calculated the transverse shear strains across the beam cross-sectionimmediately to the right- and left-hand sides of the tup, when the beam reaches the maximumdeection. An average value is about exz 0:05; or gxz 0:10; while from Eq. (36), the averageshear strain is

exz Ws=lQ 0:225=H=2 0:06 or gxz 0:12; 57

with H 7:62mm and Ws 0:225mm given in Ref. [33]. Thus, the method in Section 4 canpredict reasonable values for the strains in a clamped beam, at least for the above case.The nite-element results in Ref. [36] have been compared with experimental data from tests on

fully clamped mild steel beams, struck at the mid-span by a sharp tup. The equivalent strain onthe lower surface of the beam, underneath the striker, evolves according to Ref. [36] with thedimensionless displacement, as shown in Fig. 13. It is also shown in the same gure with a thickline that Eqs. (48) and (49) with x 1 predicts the strain

eeq 3h2w2 4khw=

3

p2

qfor wp1 58

and

eeq h2w2 5=22 4khw=

3

p2

qfor w > 1; 59

which are in reasonable agreement with the numerical data for displacements up to around twicethe beam thickness.

0

0.1

0.2

0.3

0 0.2 0.4 0.6 0.8 1W/H

x m

ax

Fig. 12. Maximum axial strain (no shear) on the bottom of the fully clamped aluminium beam underneath the load

point vs. the dimensionless displacement. The beam was struck by a falling mass at the one-quarter span position. Thin

and thick lines represent numerical results from Ref. [33] and the prediction of Eq. (56), respectively.


Eqs. (58) and (59) depend on the beam geometry and on the factor k: The factor k can beconsidered as a material parameter with the experimental results for mild steel beams from Ref.[21] giving k 0:26: It is not known whether k would change signicantly for other ductile metals.The shear strains for a mild steel beam with length 2L struck at the mid-span have also been

evaluated in Ref. [36]. The maximum value quoted throughout the entire response is gxz 0:66:This value is reduced to gxz 0:45 at the centre of the striker when the beam reaches its maximumdeection. For comparison, the present shear hinge denition yields a maximum shear strain

gxz Ws=lQ=2 Ws=H=2=2 2kWf=L=2 20:2621:81=50:5=2 0:45; 60

where Wf is the displacement at the impact point of beam SB09 in Ref. [36].

7. Final comments

The beams tested in the companion paper [21] were loaded by indenters with sharp and roundnoses at various positions across the span. Ductile failure occurred for transverse displacementsup to three times the beam thickness. Due to the reasonable predictions for the equivalent strainand its rate given by Eqs. (48)(50), (53) and (54), and bearing in mind their simplicity, they areused in the companion paper to predict strains, strain rate and failure, when using the presentCDM model, for steel beams struck by masses at several positions across the span.In the case of proportional loading and for perfectly plastic materials, the damage predicted by

Eq. (11), is proportional to the equivalent plastic strain which agrees with some theoreticalevidence. Ortiz and Molinari [37], for instance, studied the dynamic expansion of a spherical void

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4

W/H

eq

static(FEM)

dynamic(FEM)

Fig. 13. Equivalent strain on the lower surface of a mild steel beam (2L 101:6; H 6:29; B 10:10mm) immediatelyunderneath the striker vs. displacement. The beam is struck at the mid-span with a mass G 5 kg travelling with animpact velocity V0 10:5m/s. Thick line represents the engineering equivalent strain (Eqs. (58) and (59)). Thin lines arethe true equivalent strain calculated numerically in Ref. [36].


in an unbounded solid and concluded that a void expands linearly with time or, for proportionalloading, linearly with strain.Although the dynamic damage evolution equation was integrated for perfectly plastic and

bilinear materials, it is possible to adopt other stressstrain laws but the integration could bedifcult, as it is evident from the complexity of Eq. (13) for the linear hardening model.The strains for a perfectly plastic material can be estimated when assuming a nite length for

the regions where plastic ow takes place. Though some denitions for hinge length are availablein the literature, they have not always been compared directly with experimental results or nite-element calculations. New denitions for estimating the bending, membrane and shear hingelengths in the context of beam theory are developed, and explored, in order to obtain thecurvature, extensional, shear and equivalent strains for a perfectly plastic material. The proposedhinge lengths give strains and curvatures which are in reasonable agreement with thecorresponding experimental results and numerical data for aluminium and mild steel beamswhich were impacted at different positions along the span. However, it should be noted thatneither the theoretical model for the calculation of strains nor the numerical data in Ref. [33] takedamage into account.It is evident that the various generalized stresses interact, as explored numerically in Ref. [20],

for example, but this phenomenon is not taken into account here. Such an interaction is importantand changes with time, from the initial dominance of transverse shear forces to the developmentof a membrane state as motion progresses. However, to the authors knowledge, no analyticalsolutions are available which cater for a single hinge having a variable length to reect all thechanges in the generalised stresses.The equivalent strains along the span of a beam and across the thickness can now be used in

conjunction with the ductile strain rate-sensitive CDM model developed in this paper. Thus, thefailure of beams due to static and dynamic loads can be predicted and the results compared withsome new experimental data, as discussed in a companion paper [21].

Acknowledgements

The nancial support of CAPES, a Brazilian research funding agency, is greatly acknowledged,as well as the Impact Research Centre at The University of Liverpool.

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Impact failure of beams using damage mechanics: Part I-Analytical modelIntroductionCDM ductile modelParticular integration of the damage equationStrains in a perfectly plastic materialBending hinge lengthDynamic bending hinge lengthShear hinge lengthMembrane hinge lengthEquivalent strainMembrane strainShear strainBending strain

Strain rateValidation of the hinge length definitionsQualitative validationQuantitative validation

Final commentsAcknowledgementsReferences

Impact Failure of Beams Using Damage Mechanics - Part I

Documents

strain hardeningde0

impact failure of beams

damage model

equivalent strain

prediction of failure

ductile failure

ductile damage mechanics

total damagedd0 dynamic