5.2 Elastic Strain Energy

Impact and Dynamic Loading
Section 5.2
Consider the case of a weight P dropped instantaneously onto the end of an elastic bar. If
the weight P had been applied gradually from zero, the strain energy stored at the final
1 force P and final displacement Δ 0 would be 2 P Δ 0 . However, the instantaneously
applied load is constant throughout the deformation and work done up to a displacement
Δ 0 is P Δ 0 , Fig. 5.2.14. The difference between the two implies that the bar acquires a
kinetic energy (see Eqn. 5.1.19); the material particles accelerate from their equilibrium
positions during the compression.
As deformation proceeds beyond Δ 0 , it is clear from Fig. 5.2.14 that the strain energy is
increasing faster than the work being done by the weight and so there must be a drop in
kinetic energy; the particles begin to decelerate. Eventually, at Δ max = 2Δ 0 , the work
done by the weight exactly equals the strain energy stored and the material is at rest.
However, the material is not in equilibrium – the equilibrium position for a load P is Δ 0 –
and so the material begins to accelerate back to where Δ 0 .
Figure 5.2.14: non-equilibrium loading
The bar and weight will continue to oscillate between 0 and Δ max indefinitely. In a real
material, internal friction will cause the vibration to decay.
Thus the maximum compression of a bar under impact loading is twice that of a bar
subjected to the same load gradually.
Example
P
Δ Δ
0
max
Consider a weight w dropped from a height h. If one is interested in the final, maximum,
displacement of the bar, Δ max , one does not need to know about the detailed and complex
transfer of energies during the impact; the energy lost by the weight equals the strain
energy stored in the bar:
1
w ( h + Δ max ) = PΔ
max
(5.2.37)
2
where P is the force acting on the bar at its maximum compression. For an elastic bar,
P Δ EA/
L
P = kΔ
,
= max , or, introducing the stiffness k so that max
Solid Mechanics Part I 191
Kelly