Shock deflection of a beam

[jimothy]

I am trying to calculate the maximum deflection of a cylindrical beam under a dynamic transverse shock load. I can calculate the deflection assuming a static load but am not sure how to extend this to the dynamic case.

The beam is effectively a tube simply supported at each end. The load would be a dynamically applied distributed load of a given G-level over the entire length. The impulse would be represented by applying this load over a half sinusoid although it could be simplified to a square pulse for an initial analysis.

Thanks in advance for your help.

 

[Twoballcane]

Do you know the shock pulse (ie 20g@.008sec) and natural frequency of he beam? If so you will have to calcuate the Shock Amplification. You can find this in any good Vibration book.

Tobalcane
"If you avoid failure, you also avoid success."

 

[Cockroach]

You can do this analysis in a number of ways. This is my preferred method.

An object of known mass falls a distance down to the beam. At the moment of impact, it stops completely and transfers that energy to the cantilever beam. The change in momentum equals the impulse imparted to the beam. Forget about efficiency in elastic/inelastic collision, you can essentially find the force of impact given the change in momentum over a very small time period.

No put that load statically at the point of impact on the beam. Solve the problem like it was a static case for deflection.

Typically shock loads are three to four times the static weight of the object. This gives you an estimate on how reasonable your energy transfer from the object coming to rest is with respect to the beam.

And now you know.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[desertfox]

Hi jimothy

Go to this site and scroll down till you come to beams and the formula's are there.
Two things though one your distributed load would need to be converted to a single point load which would be conservative, secondly is this a dynamic load involving impact with another object?
The reason I asked the last question was that you can have a shock load that doesn't involve impact ie a short circuit on a three phase set of busbars is a shock load without impact due to the magnetic forces.

http://www.roymech.co.uk/Useful_Tables/Fatigue/Mechanics_Impact.html

 

[zekeman]

Timohenko and DenHartog (Mechanical Vibrations,, 4th edition,. Mc Graw Hilll, 1956 ,pp 148-151) do problems similar, so the following is my analysis mostly based on DenHartog’s method:

The problem is that a g loading on a simply supported uniform weight beam with a square time duration involves first getting the natural modes which expresses the Fourier components of the dynamic loading of
f(x)*h(t)
f(x) = g*rho for an uniform weight of rho on the beam length of L
and
h(t)= time pulse of unity over a time duration of to

f(x) and h(t) can be obtained from the Fourier series product
f(x)=summation
An*sin(nPIx/L)- n=1,3,5...
where
An=4*g*rho/nPI
g= number of g's
satisfying the boundaries of the simply supported beam
and
h(t)=summation (cos(wn(t-to)-cos(wnt))( taken from Den Hartog )

where wn and nPIx/L are connected by
wn=n^2PIi^2/L^2*sqrt(EI/rho)
which can be derived from the free vibration equation
EI*y''''=rho*w^2*y
So
f(x)*f(t)=
summation of 4*g*rho/(nPI)*sin(nPI*x/L))*coswn(t-to)-cos(wnt). The response of this dynamic is the sum of each term in n
The deflection is obtained by integrating the sine term 4 times and dividing by EI, so the response for each n is
Summation
4*g*rhol^4/(PI*n)^5)/EI*sin(nPI*x/L))*coswn(t-to)-cos(wnt)… n=1,3,5…
So by superposition you add each term.
As a practical matter only the first term is significant owing to the n^5 falloff, so you need only expand the first term to get a solution in time and space.The peak is obviously at the center at a time corresponding the maximum of cosw1(t-to)-cos(w1t)

 

[jimothy]

Thanks for the advice everyone. The shock loading I am considering is just the weight of the tube itself under a high G shock event. In practice imagine dropping the simply supported tube from a given height onto the supports and imagine the resulting deflection at the center.

The existing assumption here that I have been battling is that as the event is short (2ms for the half sine) the tube does not have time to reach the full static deflection before the load is removed and that therefore we can cut down on safety margin. Ho hum.

 

[zekeman]

From the solution I gave, assuming the 2ms square pulse I would get, (not much different from the 1/2 sine)
at the center
4*g*rho*L^4/(3.14^5*E*I)*maximum of cos(w1(t-to)-cosw1t
From
wn=n^2PIi^2/L^2*sqrt(386EI/rho)
n=1 first and dominant mode
w1=2*3.14^2*sqrt(386*EI/rho)/L^2
rho= lb/inch of beam
L length beam , inches
For example, take a simply supported steel beam WF 8x5.25, 6ft long
rho=1.41 lb/in
I=56.4 in^4
Putting in the numbers
w1=2*3.14^2*sqrt(386*30*10^6/1.41)/(6*12)^2=344 rad/sec
The maximum of the time function cos(w1(t-to))-cos(w1t) is
where
t=PI/2w1+to/2
and the maximum is
2sin(w1(to/2))=2*sin 344*.003/2)=0.11 and finally
assuming a 200g shock
defl=(0.11*4*200*1.41*(12*6)^4/(3.14^5*30*10^6*56)=.059"
The small deflection is due to the time duration of the pulse is too small at the frequency w1. The next harmonic at n=3 would yield a frequency 9*w1 or 3096 and a w2 *to=9.
so the maximum will peak at a time t<to
Looking for the maximum of
cos(w2(t-to)-cos w2t)is nearly 1, but the deflection function is attenuated by n^5, or 3^5=243, so the max contribution of this harmonic is only 1/.11/243=.05 of the 1st harmonic.
Looks like there is virtually no deflection of this beam and probably yours.

 

[rb1957]

seems odd to me that the thing designing the stiffness of a beam is it falling a prescribed distance ...

 

[flash3780]

Dropping a beam, eh? I'd look at it as a conservation of energy problem. The beam could be thought of as a spring.

Solve for the spring constant from a standard load case of beam deflections for a distributed load (p 716, Case 6, Beer/Johnston Mech of Materials 1992) and relate that to a spring constant (wL/? = P/? = k).

Then, simply relate the kinetic energy of the beam before deflection (.5 × m × v^2) to the potential energy of the spring... erm... beam (.5 × k × x^2) at maximum deflection.

Solve for "x" and you have your answer.

 

[flash3780]

Forgot to mention: Your answer should be a bit conservative vs real life. Materials tend to act stiffer at high strain rates. Usually a good thing to be a bit conservative though.

 

[rb1957]

would a standard distributed load apply here ? ... i don't have beer and johnson but i imagine the referred to case is a simply supported beam (ie supported at both ends). but dropping the beam would make it impact at an end, or along it's length.

faciously, don't drop the beam.

i'm still surprised that this is the critical design case, and that it is so important to control the deflection of a dropped beam ? ... unless this is a school problem ...

 

[zekeman]

New Postrb1957 (Aerospace)
12 Dec 09 8:32
would a standard distributed load apply here ? ... i don't have beer and johnson but i imagine the referred to case is a simply supported beam (ie supported at both ends). but dropping the beam would make it impact at an end, or along it's length."

No, under the vibration that follows the impact, the loading on the uniform beam would have to be be proportional to the deflection at each point, which would make that assumption invalid..
Only a combination of the normal modes can be used for the deflection, y(x), i.e,
y(x)= ay1(x)+ y2(x) +......
where y1,y2,.. are the normal modes.
Methods like Stodola or Raleigh can be used iteratively to get reasonable answers.

 

[flash3780]

You're right, the load isn't going to be the "fully distributed load case" or the "point load in the center of the beam case".

However, you could bracket the problem by assuming the loading is somewhere inbetween. Perhaps the best description of the loading in the deflected condition would be a half-sine distribution (which isn't a standard case in my text, so you may need to solve for it).

I don't think that the vibration of the beam after the initial deflection is relevant based on jimothy's description of the problem. It sounds like he's looking for the maximum deflection of a pipe dropping onto supports. I think that an energy based solution is most appropriate.

 

[zekeman]

Revisiting.......
There were a few errors that make a substantial difference in the example shown in my post of 12/10.
It read:

"For example, take a simply supported steel beam WF 8x5.25, 6ft long
rho=1.41 lb/in
I=56.4 in^4
Putting in the numbers
w1=2*3.14^2*sqrt(386*30*10^6/1.41)/(6*12)^2=344 rad/sec
The maximum of the time function cos(w1(t-to))-cos(w1t) is
where
t=PI/2w1+to/2
and the maximum is
2sin(w1(to/2))=2*sin 344*.003/2)=0.11 and finally
assuming a 200g shock
defl=(0.11*4*200*1.41*(12*6)^4/(3.14^5*30*10^6*56)=.059" "


It is corrected as follows:

For example, take a simply supported steel beam WF 8x5.25, 6ft long
rho=1.41 lb/in
I=56.4 in^4
Putting in the numbers
w1=3.14^2*sqrt(386*30*10^6*56.4/1.41)/(6*12)^2=1291 rad/sec
The maximum of the time function cos(w1(t-to))-cos(w1t) is
the max of cos(w1t-1291*.002) or maximum of cosw1t=cos(w1t-2.58)-cos(w1t)


2sin(w1(to/2))=2*sin( 2.58 rad/2)=1.91 and finally
assuming a 200g shock
defl=(1.91*4*200*1.41*(12*6)^4/(3.14^5*30*10^6*56)=1.02"