Impact load
Impact load
(OP)
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.






RE: Impact load
Testing is often the best way, I have heard that .3 0r .2 seconds is often used.
RE: Impact load
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Impact load
RE: Impact load
A short column shape beam with a high strength/stiffness ratio would be best (unless there are other factors).
RE: Impact load
RE: Impact load
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.
RE: Impact load
RE: Impact load
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.
RE: Impact load
RE: Impact load
RE: Impact load
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Impact load
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!
RE: Impact load
I wasn't saying otherwise. I used the term deceleration to clear up the concept for Tomfh, precisely because the terms are interchangeable.
RE: Impact load
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.
RE: Impact load
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.
RE: Impact load
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.
RE: Impact load
RE: Impact load
RE: Impact load
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Impact load
so what is k, is that the equivalent spring constant?
RE: Impact load
Thanks for checking that for me. It turns out I was mistaken in mmy math, however. I think I accidentally squared the constant 'k' when I used my calculator to get the number, so I divided by k^2 at one point rather than 'k'.
While waiting for your response, I looked it up in a few books. Since after impact the problem becomes a 'free vibration' problem, if we neglect damping it may be assumed the beam oscillates as a sine function wrt time.
u(t) = A x sin (w x t) + B x cos (w x t) + C
I believe the term that captures the kinetic energy in the falling weight is the first term, A x sin (w x t). The constant 'A' is the value of the deflection needed to absorb the K.E.
The other two terms capture the deflection resulting from the system going from one equilibrium state to a new state, and this is the consequence of loading the beam in an instantaneous manner. The values of 'B' and 'C' are the magnitude of the static deflection, although they have opposite signs.
CSD,
'k' is the spring constant of the SDOF system.
RE: Impact load
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Impact load
?
RE: Impact load
"the longer the period of time a given force is applied the greater the acceleration "
Surely you mean velocity, not acceleration. If you have a constant force acting on a constant mass the acceleration will be constant.
"In statics the period of time does not matter but in dynamics it is essential."
That's generally true of course, but I was responding to Tomfh who seemed to imply that the rate of deceleration doesn't affect the beam's capacity because the same amount of work is done.
RE: Impact load
I think you may have forgotten to convert pounds (force) to slugs (mass). The calculated deflection should be around 7" assuming a DF#2 2x4 with E = 1,600,000 psi
The equivalent static load is 1668lb, with a resulting bending stress of 16,338psi.
Also, calculating the kinetic energy is an unnecessary step. Just equate the potential energy before the mass falls to the potential energy after the mass falls.
Note that, for more accuracy, the deflection of the beam could be added to the fall height and the calculation iterated until the before and after potential energy numbers converge. Generally, I would consider the extra distance added by beam deflection a second-order effect and neglect it, but 7" of deflection after a 60" fall is not negligible.
RE: Impact load
moment = sqrt (6 P E I h / L)
moment = sqrt (6 * 100 * 1.6e6 * 5.359 * 60 / 120)
= 50718 in lb.
Bending stress = 50718/3.0625 = 16561 psi
I'm curious about the 1% difference in our answers, what input values did you use?
Note that if you include the extra 7 inches drop height the moment is about 5% higher. This will be well under the energy lost as sound etc.
RE: Impact load
we used the same values; I suspect rounding differences. I didn't take the time to come up with a streamlined formula, so there were several intermediate steps with the answer rounded on each.
I went ahead and did the analysis with the deflection included in the energy calculation, and iterated until the "before" and "after" energy values converged. I got a 6% increase in bending stress.
RE: Impact load
Ho do I calculate the assumed deflection (delta)?