1D elastic impact stress wave in a rod

[Hannes]

Hello all,

 

the problem: A rod of length L, cross sectional area A, Young's modulus E and density rho moves with velocity v, hitting with its end on a rigid plane.

 

Questions:

1. What is the maximum impact stress sigma?

2. How propagates the elastic stress wave in the rod (time and coordinate dependence)?

 

An answer to question 1 could be sketched in this way:

The impact stress is sigma = rho c v, where the wave velocity c is the square root of E/rho. This follows from the impuls change of the rod dp in a time interval dt, (i) dp = sigma A dt, substituting (ii) dp = rho A v c dt.

Some explanations: Equation (i) invokes the stress sigma at the impacting cross section which is equal to the impuls change dp per time and area of the rod. Equation (ii) calculates the impuls change dp = rho dV v, where dV = A c dt is the increase of the compressed stress wave volume with nearly zero velocity at the impacting side of the rod and v is the velocity of the decreasing uncompressed region outside the stress wave.

 

I am looking for the solution procedure for question 2. Any hints, a book or internet links are appreciated.

[studiot]

The problem (why is this not in homework?) should say whether the rod is brought to rest at the point of impact or if it bounces off.

[Hannes]

The impact should be elastic, so the rod may bounce off. However, I am only interested in a solution of the initial stress wave before it reaches the end of the rod.

[Enthalpy]

Hi Hannes, willkommen here!

 

OK for the maximum stress and propagation speed - more subtleties later.

 

The stress propagates backwards in the rod at sound speed; before the wave front passes (hence further from the impact) the material isn't stressed but keeps its speed; after the front (nearer to the impact) it is immobile, compressed and stays compressed.

 

Once the front reaches the far end of the rod (at this time the whole rod is immobile and compressed), the compression converts into speed, and a new front propagates forward: before the second front the matter is immobile and compressed, after the front the matter has backward speed and no stress.

 

One the second front reaches the impact point, all matter has backward speed and no stress; the rod just detaches from the target and continues bouncing back.

 

Subtleties:

 

A target much stiffer than the rod is uncommon. More usually in solids, as in acoustics, the constrast between materials is small, so the impact spreads its effect into both objects. The ratio of acoustic impedances determines much IF both waves are (were) plane.

 

A flat shock is quite difficult to achieve. But a spherical contact, computed with Hertz's (static) formula, and the metal's static yield strength, gave me good results.

 

Young's modulus E isn't always enough - as in acoustics. If the objects are wide and the shock has fast time components, then matter has no time to move sideways. Though, E supposes a free side move. When this side move isn't possible at all, one should replace E with E/(1-2µ²) where µ is Poisson's ratio, as in acoustics [or maybe E/(1-2µ²/(1+µ)), please check]. This means also that short compression shocks and sounds propagate a bit faster than long ones, depending on the rod's diameter.

 

Most metals and ceramics have E constant over a very wide frequency range, at least many 100MHz. Other materials vital to shock engineering, especially elastomers, depend fundamentally (E varies by 100 or *1000) on the frequency (coupled strongy with the temperature). This is a consequence of the desired damping. Check viscoelasticity, glass transition temperature. Data for elastomers is scarce and difficult to use (big deformations make E meaningless), so most designs are experimental.

 

A rod is the simple case. Parts use to have complicated shapes, and in metals and ceramics, all echoes, impedance mismatches and the like apply. For instance, I had an aluminium part hit frontally at 30m/s a massive steel target. A small ring was milled (single part) at the rear. The front kept intact as expected, but the concentrated wave at the narrower rear broke the ring and ejected the bits backwards.

 

Shocks and acoustics aren't very developed theories. Since waves and their propagation in shapes, materials, impedances... are essentially the same as an electric signal in a stripline, and theory is much more advanced there, radiocomm engineering is a better formation for shock and acoustics engineering; just pick their books.

[studiot]

 

the problem: A rod of length L, cross sectional area A, Young's modulus E and density rho moves with velocity v, hitting with its end on a rigid plane.

 

The impact should be elastic, so the rod may bounce off. However, I am only interested in a solution of the initial stress wave before it reaches the end of the rod

 

Since the (flat?) end of the rod is much smaller than the rigid plane I would start by modelling the wave as a plane wave.

 

The wave will be the same whether the rod is moving and strikes the plane or the something like a large hammer with a flat face strikes a stationary rod.

Solutions to the latter are well developed in the literature. Such devices are used in oceanography and seismology. Avoid hammers striking drums that is a different situation.

 

If the rod is long you will need to consider Rayleigh waves developing at the surfaces.

[Hannes]

Hi Hannes, willkommen here!

 

OK for the maximum stress and propagation speed - more subtleties later.

 

The stress propagates backwards in the rod at sound speed; before the wave front passes (hence further from the impact) the material isn't stressed but keeps its speed; after the front (nearer to the impact) it is immobile, compressed and stays compressed.

 

Once the front reaches the far end of the rod (at this time the whole rod is immobile and compressed), the compression converts into speed, and a new front propagates forward: before the second front the matter is immobile and compressed, after the front the matter has backward speed and no stress.

 

One the second front reaches the impact point, all matter has backward speed and no stress; the rod just detaches from the target and continues bouncing back.

 

Subtleties:

 

A target much stiffer than the rod is uncommon. More usually in solids, as in acoustics, the constrast between materials is small, so the impact spreads its effect into both objects. The ratio of acoustic impedances determines much IF both waves are (were) plane.

 

A flat shock is quite difficult to achieve. But a spherical contact, computed with Hertz's (static) formula, and the metal's static yield strength, gave me good results.

 

Young's modulus E isn't always enough - as in acoustics. If the objects are wide and the shock has fast time components, then matter has no time to move sideways. Though, E supposes a free side move. When this side move isn't possible at all, one should replace E with E/(1-2µ²) where µ is Poisson's ratio, as in acoustics [or maybe E/(1-2µ²/(1+µ)), please check]. This means also that short compression shocks and sounds propagate a bit faster than long ones, depending on the rod's diameter.

 

Most metals and ceramics have E constant over a very wide frequency range, at least many 100MHz. Other materials vital to shock engineering, especially elastomers, depend fundamentally (E varies by 100 or *1000) on the frequency (coupled strongy with the temperature). This is a consequence of the desired damping. Check viscoelasticity, glass transition temperature. Data for elastomers is scarce and difficult to use (big deformations make E meaningless), so most designs are experimental.

 

A rod is the simple case. Parts use to have complicated shapes, and in metals and ceramics, all echoes, impedance mismatches and the like apply. For instance, I had an aluminium part hit frontally at 30m/s a massive steel target. A small ring was milled (single part) at the rear. The front kept intact as expected, but the concentrated wave at the narrower rear broke the ring and ejected the bits backwards.

 

Shocks and acoustics aren't very developed theories. Since waves and their propagation in shapes, materials, impedances... are essentially the same as an electric signal in a stripline, and theory is much more advanced there, radiocomm engineering is a better formation for shock and acoustics engineering; just pick their books.

Hello, thank you for the welcome and for providing the description of the stress wave and the subtleties (danke).

Coming back to the most simple case, could you or anybody indicate the systematic mathematical way to derive the impact solution to the wave equation of the displacements? This is now my main question.

The relations below, being in agreement with your description, at the same time satisfy the wave equation for the displacements and thus appear to be a successful guess of the correct solution. The impact stress wave is described by Heavisides function, theta(x>0) = 1 and theta(x<0) = 0:

sigma = - rho c v theta(ct - x) after the impact (0 < t < L/c) and

sigma = - rho c v theta(2L - ct - x) after the first reflection (L/c < t < 2L/c).

This implies for the displacements relative to the unstressed rod (u = 0 for zero stress):

u = rho c v/E (ct - x) theta(ct - x) after the impact (0 < t < L/c) and

u = rho c v/E (2L - ct - x) theta(2L - ct - x) after the first reflection (L/c < t < 2L/c).

Best regards,

Hannes

[studiot]

 

Coming back to the most simple case, could you or anybody indicate the systematic mathematical way to derive the impact solution to the wave equation of the displacements? This is now my main question.

 

When you derive the wave equation acting in a system you do not use the external input force.

You use the restoring force, which is due to the mechanics of the system.

The wave action is due to the presence of this restoring force, If the restoring force is proportional to the dispalcement you obtain the linear solutions you indicated.

The input force is used to provide an initial displacement and does not appear directly.