10.3. Determining displacements and stresses under the impact
Consider
the case
of the longitudinal impact
of the load on an immobile body. Let us put the load of the weight G
fall from the height h
on the immobile bar (Fig. 10.3 a). The body velocity in the impact
moment is determined by the known formula of freely falling bodies
This velocity in a very short interval of the impact time, calculated
by thousandth or hundredth parts of the second, falls to zero. In
consequence of the large acceleration (moderation) a considerable
force of inertia arises which determines the impact action.
However, it is difficult to derive the theoretical law of the velocity change and consequently the force of inertia. Here another way is applied, based on the approximate utilization of the energy preservation law and the following assumptions:
1) the stresses under the impact do not surpass the proportional limit so that Hooke’s law keeps its action under the impact;
2) bodies do not fall off after the impact;
3) the mass of the bar under the impact is considered small compared with the mass of the body which makes the impact and therefore it is not taken into consideration;
4) we ignore the loss of the energy part passed in the heat and the energy of the hitting bodies vibration motion. Equate the falling load work to the internal strain energy of the bar.
a)
b)
Fig. 10.3.
The work accomplished by the weight of the falling load is as follows
(10.5)
where
is the displacement at the impact point which is equal to the bar
shortening. The internal strain energy in compression has the form
(10.6)
From these two equations we get
(10.7)
or
(10.8)
Dividing all terms of this equation by EA we get
(10.9)
But
is the bar shortening from the action of the statically applied load
G.
Then
(10.10)
Having
solved this quadratic equation about
we obtain
(10.11)
Leaving the
plus sign (because
)
we get finally
(10.13)
where
is the dynamic coefficient.
Having divided two parts of the last equation by the bar length and multiplied by the elasticity modulus E, we will pass on deformations to the stresses:
(10.14)
It can be seen from these formulas that the dynamic stresses and displacements depend on the static deformation of the body subject to the impact. The more a static deformation is, the less the dynamic stress is.
That is why the gaskets (rubber, spring) are applied for the impact softening as they give large deformations.
The dynamic
stresses must not surpass the critical stresses under the compressive
impact to avoid buckling. The analogical forms have the formulas and
for the case of the lateral bending impact but in this case exept
we have to consider the static bar deflection at the impact place 
and instead of
the dynamic deflection
(Fig. 10.3 b).
Special cases.
1. If h=0,
i.e. a
suddenly
applied load takes place,
we get
from the formulas (10.12) and (10.13). Under a suddenly applied load
the deformations and the stresses are twice as much than under the
statically action of the same load.
2. If the
height h
of the fall is much more the statically deformation
to determine the dynamic coefficient we get the following approximate
formula:
(10.15)
Example.
The load G=1kN
from the height h=10
cm=0,1
m falls on the middle of the steel double - T profile №27 a by the
span 3 m. The second moment of the section is
the section modulus is
(from the tables of the profiles),
Determine the maximum beam deflection and the maximum stresses for its cross section.
Solving. Calculate the statical beam deflection under the load
(10.16)
The dynamic coefficient is
(10.17)
In the given case the dynamic effect of the falling load surpasses its statical effect in 64 times.
Calculate the statically stress from load G. The maximum bending moment will be at the middle beam section:
The maximum statically stress is
The maximum dynamic stress is
It can be seen from this example how the dynamic loads are dangerous by their action. Therefore the allowable working stresses under the impact are accepted less than under the action of the statically loads.