Idealisation of materials’ behaviour

2

Methodology of analysing energy-absorption capacities

To provide basic models and tools for analysing the energy-absorp- tion capacities of materials and structures, this chapter illustrates the idealised models of materials’ behaviour, as well as fundamental concepts, principles and methods; the effects of large deformation and dynamic loading are also discussed.

The plastic deformation stage will end when the material coupon is even- tually broken under tension. Before fracture occurs the stress will reach a maximum, denoted by the ultimate stresssu. When fracture occurs, the cor- responding strain is termed the fracture strain denoted as ef. These two quantities (refer to Fig. 2.1(a)) represent the strength and ductility, respec- tively, of the material under tension.

When subjected to simple compression or pure shear, most engineering materials display behaviour similar to their behaviour under tension, although the relevant material constants may be different.

2.1.2 Idealised material models

In order to establish reasonably simple theoretical models to analyse the energy-absorption capacities of materials and structures, first of all the materials’ mechanical behaviour should be idealised so that their stress–strain relationship can be expressed by simple analytical functions.

When the deformation (strain) is small, a linear elastic material model can be adopted, while Young’s modulus Eand Poisson ratio nserve as two specified material parameters. After the material’s yielding, if the strain- hardening is insignificant, then an elastic,perfectly plasticmaterial model can be adopted, as shown in Fig. 2.2(a). Here the term ‘perfect plastic’

implies that strain-hardening is negligible; in other words, the material will continue to deform plastically from the initial yielding till fracture whilst the stress remains Y.

If the material’s strain-hardening after the initial yielding is significant, then either an elastic,linear hardeningmodel (Fig. 2.2(b)), or an elastic, power hardeningmodel (Fig. 2.2(c)) can be considered, depending on the material’s actual behaviour measured from its simple tensile test.

ssu

s_f

ey ef e

0

(a) Y

ssu

ey ef e

0

(b) Y

ssu

ef e 0

(c) 2.1 Stress–strain curves of materials under tension: (a) mild steel;

(b) aluminium alloy; and (c) knitted textile composite.

Corresponding to the models shown in Figs 2.2(a), (b) and (c), the idealised relationship between stress s and strain e can be analytically expressed as

[2.1]

[2.2]

and

[2.3]

respectively, where eyis the yield strain,E_pdenotes the hardening modulus, Kand q(hardening exponent) are material constants determined experi- mentally. It is obvious that if q=1 and K =Ep, the power hardening model is identical to the linear hardening model.

When used for an energy-absorption purpose, the materials, structural components and devices will usually undergo a large plastic deformation, as illustrated in Chapter 1. In these cases, the plastic strain will be much larger than the elastic strain, so the latter can be neglected in the analysis.

Effectively, Young’s modulus can be taken as infinite, so that the material exhibits rigid behaviour before initial yielding. Thus, the idealised material model is called a rigid-plasticmodel. Figures 2.3(a), (b) and (c) represent a rigid,perfectly plasticmodel, a rigid,linear hardeningmodel and a rigid, power hardeningmodel, respectively.

s e e e

e e e e e

= £ =

+ ( - ) £ <

ÏÌ Ó

E Y E

Y K

y y

q

y f

for for

s e e e

e e e e e

= £ =

+ ( - ) £ <

ÏÌ Ó

E Y E

Y E

y

p y y f

for for

s e e e

e e e

= £ =

£ <

ÏÌ Ó

E Y E

Y

y

y f

for for

(a) 0

(b) (c)

s

ey e Y

E 1

0 s

ey e Y

E

E_p

1

1

0 s

ey ef e

Y

2.2 Idealised stress–strain curves of materials under tension:

(a) elastic, perfectly plastic; (b) elastic, linear hardening; and (c) elastic, power hardening.

The idealised stress–strain relationship for rigid, perfectly plastic model (Fig. 2.3(a)) and rigid-hardening models (Fig. 2.3(b) and (c)) can be expressed as

[2.4]

[2.5]

and

[2.6]

respectively.

2.1.3 Moment–curvature relationship for plastic beams

For beams (or other 1-D structural components, such as rings and arches) made of elastic-plastic materials, the relationship between applied bending moment Mand the curvature kof its central axis will be linear if the applied moment is small (M < Me, with Me being the maximum elastic bending moment), or non-linear if M>M_e. The actual M–krelation can be derived from an integration of the appropriate s–e relation over the beam’s thickness.

Typically, for a rectangular cross-sectional beam (or another 1-D struc- tural component, such as a ring or an arch) made of elastic, perfectly plastic material, the stress profile across the beam’s thickness is as shown in Fig.

2.4(a), which contains an elastic ‘core’ sandwiched by two plastic deforma-

s e

s e e e

£ =

= + < <

ÏÌ Ó

Y

Y K ^q f

for for

0 0

s e

s e e e

£ =

= + < <

ÏÌ Ó

Y

Y Ep f

for for

0 0

s e

s e e

£ =

= < <

ÏÌ Ó

Y

Y _f

for for

0 0

s

e 0

(a) Y

s

e 0

1 E_p

(b) Y

s

e 0

(c) Y

2.3 Idealised stress–strain curves of materials under tension: (a) rigid, perfectly plastic; (b) rigid, linear hardening; and (c) rigid, power hardening.

tion zones. Consequently, the moment–curvature relation is expressed as (e.g. refer to Section 1.4, Yu and Zhang, 1996)

[2.7]

where Me=Ybh²/6 is the maximum elastic bending moment,ke=Me/EI= Me/(Ebh³/12) =2Y/Ehis the maximum elastic curvatureand Mp=Ybh²/4 denotes the fully plastic bending momentof the beam of rectangular cross- section, with b and h being the width and thickness of the cross-section, respectively.

Equation [2.7] can be recast into a non-dimensional form as

[2.8]

where m = M/Meand f = k/kedenote non-dimensional bending moment and non-dimensional curvature, respectively.This relation is sketched in Fig.

2.4(b).

An important special case is when a rigid, perfectly plastic relation between stress and strain (see Fig. 2.3(a) and Eq. [2.4]) is adopted, the stress profile across the beam’s thickness will contain plastic zones only, if the beam’s section has a non-zero curvature, as shown in Fig. 2.5(a). Conse- quently, the relation between bending moment and curvature is expressed as a step function, as shown in Fig. 2.5(b). That is

f=

£ £

- £ <

Ï ÌÔ ÓÔ

m m

m m

for for

0 1

1

3 2 1 3

2 M

M

M M M M M

e e

e

e

e p

=

£ £

- Ê Ë

ˆ

¯ < <

Ï ÌÔÔ Ó ÔÔ k k

k k

for for

0 3

2 1 2

2

h/2

h/2

–Y +Y

Plastic zone

Plastic zone

(a) (b)

Elastic core

0 1 2 3 4 5

1.0 1.5 m = M/M_e

f = k/ke

2.4 Bending of an elastic, perfectly plastic beam: (a) stress profile across the thickness; (b) non-dimensional moment–curvature relation.

[2.9]

2.1.4 Plastic hinge and hinge-line

When a beam (or another 1-D structural component, such as a ring or an arch) is idealised as being rigid, perfectly plastic, its plastic deformation will be concentrated at one or a few cross-sections, where the magnitude of the applied bending moment reaches the fully plastic bending moment of the cross-section,M_p(e.g.M_p=Ybh²/4 in the case of rectangular cross-sectional beams). These cross-sections are termed plastic hinges. Any bending moment whose magnitude is larger than Mp, is not statically admissiblefor an equilibrium configuration (see Section 2.2 for further illustrations). On the other hand, any bending moment whose magnitude is smaller than M_p, will produce no plastic deformation.

As a result, the deformed configuration of a rigid, perfectly plastic beam (or a ring or an arch) will contain one or more plastic hinges only. As indi- cated by Eq. [2.9] and Fig. 2.5(b), the curvature at a plastic hinge can take an arbitrary value or infinity. Therefore, a finite relative rotation qoccurs at a plastic hinge as the result of the application of Mp.

Away from those plastic hinges all the segments in the beam (or the ring or arch) will remain rigid and their curvatures will remain unchanged (i.e.

k=0). However, those rigid segments are allowed to have translation and/or rotation, provided these motions are kinematically admissible in a pro- posed deformation mechanism (see Section 2.2 for further illustrations).

From the viewpoint of energy dissipation, when a rigid, perfectly plastic beam deforms, plastic dissipation takes place only at the discrete plastic

M M M M

p p

£ =

= >

ÏÌ Ó

for for

k k

0 0

h/2

h/2

–Y 0 k

M

M_p +Y

(a) (b)

2.5 Bending of a rigid, perfectly plastic beam: (a) stress profile across the thickness; (b) moment–curvature relation.

hinge(s). The total energy dissipation in the beam, therefore, can be calcu- lated by

[2.10]

where nis the total number of the hinges in the beam, and qidenotes the relative rotational angle at the i-th plastic hinge.

In plates or shells, the rigid, perfectly plastic idealisation of materials will result in their plastic deformation being concentrated at discrete plastic hinge-lines. Along those hinge-lines the magnitude of the bending moment per unit length must be equal to M_o, which denotes the fully plastic bending moment per unit length,Mo=Yh²/4. Note that the unit of Mois N, while the unit of Mpis Nm.

In a similar way to the behaviour of plastic hinges in a beam, relative rota- tions are allowed along plastic hinge-lines in a plate or a shell. Consequently, the total energy dissipation in the plate or shell can be calculated by

[2.11]

where nis the total number of the hinge-lines in the plate or shell,qidenotes the relative rotational angle at the i-th plastic hinge-line, and Liis the length of the i-th plastic hinge-line.

It is observed that by adopting the rigid, perfectly plastic idealisation of material, the plastic deformation in a structural component of dimension N (N=1 for beams, rings, arches, etc and N=2 for plates, shells, etc.) will be concentrated in discrete regions of (N-1) dimension. This will greatly sim- plify the plastic analysis of structures and will be very useful for the theo- retical modelling of energy-absorption components, as will be seen in the following sections.

2.1.5 Mechanical models for materials’ idealisation

The idealised materials’ behaviour can be demonstrated by mechanical models such as those shown in Fig. 2.6. The mechanical model shown in Fig.

2.6(a) contains a linear elastic spring and a friction pair, so that when a force Fapplies along the axis of the spring, the relationship between force Fand axial displacement Dwill be

[2.12]

where k is the spring constant, and F_yis the critical friction when relative motion begins. The clear similarity between Eq. [2.12] and Eq. [2.1] indi-

F k F k

F

y y

y y f

= £ =

£ <

ÏÌ Ó

D D D

D D D

for for D M_{o i}L

i n

= i

Â

= ^q 1

D M_{p i}

i n

=

Â

= ^q 1

cates that the mechanical model shown in Fig. 2.6(a) appropriately repre- sents the behaviour of an elastic, perfectly plastic material. If the elastic spring is taken away from the model shown in Fig. 2.6(a), which means that the elastic deformation of the material is negligible, then the mechanical model (Fig. 2.6(b)) demonstrates a rigid, perfectly plastic behaviour.

When the bending behaviour of beams is considered, the rigid, perfectly plastic relation between bending moment Mand relatively rotational angle q can be similarly demonstrated by a mechanical model as shown in Fig.

2.6(c). In this model, the two rigid segments are connected by a mechani- cal hinge that can rotate freely. The angular friction pair in the model requires a critical moment Mp (applied in either the clockwise or the counterclockwise direction) to motivate its non-zero relative rotation.

2.1.6 Validity of rigid-plastic idealisation

Physically, the rigid-plastic idealisation of materials’ behaviour is based on the fact that the elastic strain (typically limited by eyª0.002 for structural metals) is much smaller than the plastic strain occurring in structural com- ponents used for energy-absorption purposes. However, the elastic defor- mation is always a precursor of subsequent plastic deformation. No matter whether the structure is subjected to a quasi-static load or a dynamic load, the first phase of its deformation is always elastic and after the plastic defor- mation is completed, the structure must undergo an elastic springback to achieve its final deformed configuration. Owing to these reasons, the valid- ity of rigid-plastic idealisation should be thoroughly examined.

A simple way to conduct this examination is to employ the one- dimensional mechanical models shown in Fig. 2.6, and to compare the response of the elastic-plastic model (Fig. 2.6(a)) with that of the rigid- plastic model (Fig. 2.6(b)).

First, assume that a force Fis quasi-statically applied to the left end of the elastic-plastic model shown in Fig. 2.6(a). Consider a deformation process of the model in which the displacement at the left end gradually increases from zero to a total displacement Dtlep

>Dy=F_y/k, where the sub- script tlpertains to the total displacement produced, and the superscript ep pertains to the elastic-plastic model. Correspondingly, force Ffirst increases

(a)

F k F_y

(b) (c)

F

M M

F_y

2.6 Mechanical models for materials’ idealisation: (a) an elastic, perfectly plastic model; (b) a rigid, perfectly plastic model; and (c) a plastic hinge.

from zero to Fy(as 0 <D<Dy) and subsequently remains Fy(as D>Dy). In this elastic-plastic deformation process, the total input energy (i.e. the work done) is

[2.13]

where E^e_max=F_yDy/2 is the maximum elastic energy which can be stored in the model, and Dpep

=Dtlep

-Dyrepresents the plastic (permanent) compo- nentof the displacement. The total input energy Eingiven in Eq. [2.13] can be represented by the shadowed area shown in Fig. 2.7(a).

Next, consider that a force F is quasi-statically applied on the left end of the rigid-plastic model shown in Fig. 2.6(b). In this case, the deformation becomes possible only when force Ftakes the value of Fy. Assume that the final displacement at the left end of the model is Dtlrp, where the superscript rppertains to the rigid-plastic model, and then the total input energy (i.e.

the work done) in the process is

[2.14]

Here the plastic component of the displacement,Dprp, is equal to the total displacement Dtlrp because the elastic component is neglected. The total input energy Eingiven in Eq. [2.14] can be represented by the shadowed area shown in Fig. 2.7(b).

Suppose the same amount of energy is input into these two models (i.e.

the values of E_inin Eqs. [2.13] and [2.14] are identical) and define an energy ratio

[2.15]

R E

er E

in

∫ e max

E_in=F_yD^rp_tl =F_yD^rp_p

Ein Fy y Fy tl E F

ep

y e

y epp

= ¹₂ ^D +

(

^D -^D

)

^{= max+ D}

(a) 0

F

F_y

Dy D

D_p D^ep

D_tl D^ep

(b) 0

F

F_y

D D_p

D^rp

D_tl D^rp

2.7 Force–displacement relationship of perfectly plastic models:

(a) elastic-plastic model; (b) rigid-plastic model.

then Eqs [2.13] and [2.14] result in

[2.16]

Divided by Ein, Eq. [2.16] leads to

[2.17]

or

[2.18]

Therefore, the relative ‘error’ of employing a rigid-plastic model in pre- dicting the plastic displacement is found to be

[2.19]

which indicates that the plastic deformation predicted by the rigid-plastic model is always slightly larger than that obtained from the corresponding elastic-plastic model, whilst the ‘error’ caused by the rigid-plastic idealisa- tion reduces with increasing energy ratio R_er∫E_in/E^e_max. For instance, if R_er

=10 in a particular structural problem, then the ‘error’ of the rigid-plastic model in predicting the plastic deformation is about 11 %, which is accept- able for most engineering applications.

2.2 Limit analysis and bound theorems