Impact load

[dcredskins]

If I drop a mass of 100 lbs from a height of 5 ft to a 10 ft simply supported beam (at the middle), how do I find out the actual moment/effect due to impact? I would appreciate help. I know it has something to do with impact loading, but I am not sure how to analyze.

 

[MiketheEngineer]

There are a bunch of threads about this. Search the forum. Biggest problem is determining the de-acceleration either in time or deflection. If you have either of these you can find what you need.

Testing is often the best way, I have heard that .3 0r .2 seconds is often used.

 

[GregLocock]

You actually have a classic single degree of freedom system there so the answer is straightforward. Work out the spring stiffness, then you can just match PEmass_initial=KEmass_at_impact= ElasticEnergy in beam to get a max deflection.

m*g*h=1/2*k*y^2

Then drop the deflection y back into your beam equation.

Obviously this ignores the local stresses due to impact.


Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Tomfh]

Greg's way is best since it elegantly sidesteps all the deceleration period nonsense.

 

[csd72]

It is worth noting that greater flexibility is your friend in this case as the more the beam deflects the lower the acceleration.

A short column shape beam with a high strength/stiffness ratio would be best (unless there are other factors).

 

[Tomfh]

Why does acceleration matter?

 

[frv]

What csd72 is referring to is deceleration, and it matters because F=ma.

If you know the deceleration of the object, you can calculate the force it imparts on the beam.

The energy method, as stated before is an alternate method. I haven't thought through the method thoroughly, but I see no reason why it wouldn't be fine.

 

[Tomfh]

In reality the rate of decceleration does not affect the beams capacity, so why pretend it does?

 

[csd72]

Tom,

You have obviously forgotten your physic fundamentals:

As F=ma, if you reduce the acceleration and you reduce the force.

frv,

deceleration is just acceleration in the opposite direction, as far as physics is concerned there is no difference other than the direction.

 

[Tomfh]

Reducing the acceleration does reduce the force, but it also increases the duration of the force so that the same amount of work gets done.

 

[apsix]

But the beam doesn't care if the load duration is 5 or 10 or whatever milliseconds, it is the magnitude of the load that dictates if the beam is adequate.

 

[GregLocock]

Can't help wondering if 10 minutes of maths might save a lifetime of embarrassment.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Lion06]

Tomfh-
Think of it this way........ damage to your car aside, would you rather stop by hitting a 2' thick concrete wall or by hitting your brakes as you normally would? The deceleration matters.

Greg-
Nice post!

 

[frv]

csd72-

I wasn't saying otherwise. I used the term deceleration to clear up the concept for Tomfh, precisely because the terms are interchangeable.

 

[csd72]

apsix,

A common trap when you are typically dealing with statics is that you forget that the force needs to be applied for a certain peiod of time get an acceleration, the longer the period of time a given force is applied the greater the acceleration (google impulse - which is the physics term for a force applied over a period of time).

In statics the period of time does not matter but in dynamics it is essential.

frv,

Sorry, just clarifying my statement.

 

[frv]

csd72-

Precisely. This is exactly why you use the assumption of "slow loading" when deriving almost any equation in structural analysis. Otherwise you may create a dynamic response.

 

[Dinosaur]

I have a question, and the answer may clear up the rest of the discussion.

I ran some numbers becuase I always do this as a habit to get a reality check on a formula. I believe GregL's formula is correct for capturing the kinetic energy from the fall. However, I am wondering if the static weight needs to be added to the value from GregLs formula to obtain the total deflection.

When I did a quick check, I computed a moment of 3,286 in-lb developed in the beam as it stops the weight from falling, but the static moment for the load was 3,000 in-lb. I recall that dropping a weight from a height of nearly zero, doubled the max moment from the static moment in order to account for the real dynamic behavior. Basically, either I am a bit too rusty, or something is still missing in the solution.

If I am correct that there is still a missing term needed to compute the maximum deflection, then this may explain some of the earlier confusion.

If you would like to check my arithmatic, here is what I did ...

Drop a 100 lb weight five feet (60 inches) onto a 2x4 (which is 1.5 x 3.5) with a ten feet simple span and a modulas of elasticity of 1.1 x 10^6 psi. The kinetic energy I computed as 6,000 in-lb. The MOI of the beam was 5.36 in^4. The stiffness was 48xExI/L^3. I computed a max deflection of 0.67 inches. The moment from this deflection would be 3,286 in-lb. The static moment would be 3,000 in-lb. And the static deflection would be about 0.60 inches. It seems to me the static deflection should be less than half the total deflection in this problem, unless I am forgetting something.

 

[Tomfh]

Several people here have made the point that acceleration matters. If that's the case, then why can you solve the problem without taking acceleration as a parameter?

 

[civilperson]

Kinetic Energy: 1/2 M V<2

 

[GregLocock]

Yes, there is an additional deflection term due to the change in PE of the block between the initial impact and the point of maximum deflection, if you want to continue in energy terms.

So the equation then becomes

m*g*(h+y)=1/2*k*y^2

which of course is still easy to solve. Note that one of the answers to that quadratic is a bit odd!




Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.