2.4 Effects of dynamic loading
2.4.2 Inertia and its effects on energy absorption
1984; Reid and Gui, 1987; Hou et al., 1995) have found that the dynamic behaviour of an elastic-plastic cantilever under impact is dis- tinctly different from that predicted by a rigid-plastic analysis (Parkes, 1955).A thorough investigation made by Yu et al. (1997) has confirmed that these differences can be attributed to the interaction between reflected elastic flexural waves from the distal clamped end and the travelling plastic ‘hinge’. Thus, it is evident that, although the elastic flexural wave does not directly dissipate energy, it does affect the energy dissipation in the cantilever beam by altering the deformation pattern and influencing the evolution of plastic regions.
Now consider a case of dynamic loading and the applied force F(t) is supposed to be a rectangular pulse (Fig. 2.16(b)), that is
[2.59]
where Tdenotes the period of application of force F.
In 0 £ t £ T, the mass block undergoes an acceleration phase in the dynamic response of the model and its equation of motion is
[2.60]
where udenotes the displacement along the direction of the applied force.
Hence, at t =T, the velocity of the block is vT=[(F -Fy)/m]T.
In the subsequent deceleration phase (t >T), the equation of motion of the block is
[2.61]
so that the velocity of the block is
[2.62]
Thus, at tf=FT/Fy,v =0 and the motion of the block ceases. The variation of the velocity of the block with time is depicted in Fig. 2.16(c).
The total displacement of the block is found from the area of the trian- gle shown in Fig. 2.16(c), which gives
[2.63]
so that the energy dissipation of the system during the dynamic response to the applied force pulse is
[2.64]
where p∫FTand po∫(F-Fy)Trepresent the total impulse and the ‘over- loading’ impulse, respectively.
It is evident that here D is inversely proportional to the mass m, clearly indicating the effect of inertia on the energy dissipation in the system. When applying this analysis to a structural component, of course, the important but difficult issue is to determine the ‘equivalent mass’ m for the compo- nent concerned.
D F F F F
m T p p
y f m
y o
= = ( - )
= ¥
D 2 2
2
Df T f
y y
v t F F F mF T
= = ( - )
2 2
2
u˙ v v F
m t T F mT F
mt
T
y y
∫ = - ( - )= -
mu˙˙= -Fy mu˙˙=F-Fy
F t
t
F t T
t T ( )=
£
£ £
>
Ï ÌÔ ÓÔ
0 0
0 0
The role of inertia in dynamic performance of structural components
As a simple example for a structural component, examine a free-free beam of length 2L, as shown in Fig. 2.17(a). Since the beam has no support at all, it cannot sustain any quasi-static load in its transverse direction.
However, if a transverse step force,F, as shown in Fig. 2.17(b), is applied to the mid-point of the free-free beam, the force will make the beam move transversely with an acceleration ao=F/2rL, where rdenotes the mass per unit length of the beam. According to the d’Alambert principle, the inertia force rao = F/2L is uniformly distributed along the beam, as shown in Fig. 2.17(c). Together with the applied step force F, this inertia force pro- duces a bending moment diagram as shown in Fig. 2.17(d), in which the maximum bending moment appears at the mid-span of the beam and is equal to Mmax=(F/2)(L/2) =FL/4. Therefore, a plastic hinge will appear at the mid-span of the beam, if the magnitude of force Freaches a ‘dynamic’
collapse force
F(t)
F(t)
F
t 2L
(a)
F(t)
F(t) rao
(c)
(e) C
(b) 0
M
Mmax
L 2L
(d) 0
a a
F(t) (f) C
D D¢
a a
a¢ a¢
2.17 (a) A free-free beam loaded by a dynamic force at its mid-span;
(b) a step force; (c) the inertia forces developed in the beam;
(d) bending moment diagram; (e) deformation mechanism with one plastic hinge; (f) deformation mechanism with three plastic hinges.
[2.65]
Here it is worthwhile noting that there is no ‘static’ collapse force for the free-free beam concerned.
The dynamic deformation mechanism is as shown in Fig. 2.17(e), from which the angular acceleration of a half of the beam is found to be
[2.66]
where J =rL3/12 is the moment of inertia of a half of the beam about its own centre. Consequently, the energy dissipation rate is
[2.67]
whilst the input energy rate is
[2.68]
where aodenotes the transverse acceleration of the mass centre of the beam and vCdenotes the velocity of the mid-span C of the beam. By combining Eq. [2.67] with Eq. [2.68], the ratio of the plastic dissipation to the total input energy is found to be
[2.69]
where f ∫FL/Mpdenotes non-dimensional force applied, and this equation is valid only when f ≥4 (i.e.F ≥ Fd). It is known from Eq. [2.69] that the maximum value of ratio Rpoccurs when f ∫FL/Mp=6, at which Rp=1/3.
When fincreases further, this ratio reduces, e.g.Rp=2/9 when f=12.
A further analysis (refer to Lee and Symonds, 1952) indicates that when the applied force further increases to f ∫FL/Mp≥22.9, then two more plastic hinges will appear on the sides of the mid-span and the deformation mech- anism will contain three hinges, as shown in Fig. 2.17(f).
Figure 2.18 depicts the variation of the energy dissipation ratio Rp∫D/Ein with the non-dimensional force f ∫FL/Mp. Clearly, the dependence is rather complex because of the alteration of the deformation mechanisms. More details of the analysis can be found in Yang et al. (1998) and Yu (2002).
It is evident from this illustrative example that when a structural com- ponent is accelerated by dynamic loads, its inertia will create shear forces and bending moments on top of those produced by the applied loads;
consequently, its dynamic deformation mechanism and energy-absorption behaviour may be significantly altered, for instance:
• the dynamic load-carrying capacity of a structural component could be significantly different from its static counterpart (in fact, the free-free
R D E M FL FL M FL M f f f
p∫ in = ( p )( - p) ( - p)
= ( - ) ( - )
3 4 3
3 4 3
dEin dt=FvC =F a( o+aL 2)t=2Ft FL( -3Mp)
(
rL2)
dD dt=2Mp(dq dt)=2Mpat=6Mp(FL-4M tp)
(
rL3)
a =d2q dt2=(FL4-Mp) J=3(FL-4Mp)
(
rL3)
Fd=4Mp L
beam analysed has no static load-carrying capacity, but it can sustain dynamic loading);
• the ratio of plastic energy dissipation to the input energy may non- monotonically vary with the magnitude of the applied force;
• the dynamic deformation mechanism may be different from the quasi- static collapse mechanism and it may vary with the magnitude of the applied force.
Energy loss during collision
If a projectile collides with a structural component, no matter whether the projectile is rigid or elastic or elastic-plastic, a portion of the initial kinetic energy of the projectile will be lost during the collision, whilst the momen- tum of the system remains conservative.
In the classical analysis of central collision between two rigid spheres, as shown in Fig. 2.19, it is assumed that before collision the sphere of mass m1 was moving with velocity vo, and the sphere of mass m2was stationary. If the central collision is assumed to be completely inelastic, then the conser- vation of linear momentum requires
0 0.5 0.4 0.3 0.2 0.1
4 6 10 20 22.9 30 f = FL/Mp
Rp = D/Ein
2.18 The ratio of the plastic dissipation Dto the input energy varies with the magnitude of the applied step force F. (Yang et al., 1998)
m1 vo
m2
2.19 Central collision of two spheres.
[2.70]
where v¢denotes the common velocity of two spheres after the inelastic col- lision. Hence
[2.71]
Obviously, when the two spheres are regarded as a system, there is a loss in the kinetic energy of the system during the collision
[2.72]
Clearly, 0 <Kloss<Ko. It is also seen that the larger the mass ratio m2/m1, the larger the relative energy loss Kloss/Ko.
In the case where a structural component is struck by a projectile,m2 would represent an equivalent mass of that component. Obviously, a loss in the kinetic energy will take place during impact, especially if the structure’s mass is much greater than that of the projectile.
Thus, it is concluded that the structure’s inertia will notably alter the input energy during a collision/impact. This important issue will be further ad- dressed in Chapter 7, in which a typical inertia-sensitive energy-absorbing structure will be comprehensively discussed.