Calculation Troubles! 1

[Stanch]

I am trying to work out induced compressive stress in a member from a dynamic load (a dropped member) using a formula taken from strength of materials, Case and Chilver second edition - 22.6 (rod struck by a moving mass). Has anyone ever worked through the example? Its part of an overall problem I have - controlling a potentially dropped (max 75mm high) steel girder onto timber cushions on top of a rigid steel frame. My theory is that the timber will compress enough to slow the girder to a stop and allow me to calculate and equivalent static load.

 

[Once20036]

I don't have a copy of that book, but I've never had luck with this type of excercise in the past.
F=ma
The mass is easy, but what's the deceleration? How long it takes to stop the falling load is going to depend of the stiffness of the wood. How much will it deflect from the energy of the falling load?
Good luck.

 

[Archie264]

Possibly some mechanical guy is going to correct me here but I think it can be done iteratively in the following manner (or, at least close enough for your purposes). Assume a deflection and set energy (.5mv^2) equal to work (Fd) and solve for the force. Use the force (Luke ) to solve for the deflection and see how close it was to your assumption. Refine. Lather, rinse, repeat. Iterate…but don’t go blind. And watch your units, whether you’re using slugs or Newtons or whatever…

If you’re wondering about your results post them here and we might be able to double-check them.

 

[Stanch]

Thanks Once/Archie.

I am essentially following your advice Archie by estimating a deflection within the timber but I am trying to justify that by showing induced compressive stress is in fact beyond yield in compression but also not ridiculously beyond that the timber cushions are ineffective. I have taken from the text an equation for induced compressive stress at impact but I am struggling with the units and not sure that my answer is correct:-

velocity,v = √(2*a*h) = √(2*9.81*0.075) = 1.21m/s
density,ρ = 350 kg/m3
Elastic modulus,E = 4600 N/mm2

induced compressive stress, v*√(ρ*E)

I work out the above to be 1.21*√(350*4600*10^-3) = 48.6N/mm2 which seems like a very big number.

Do I have the calculation correct?

 

[XR250]

Are you assuming your rigid frame will not deflect?
I would look at it as an energy equation 1/2mv^2 = 1/2kx^2
Model your frame with a 100n point load or so and find the deflection, you should be able to back K of your frame out of that.
Then solve for x from the above equation.
Once you have X, you can get the static load by F=kx.
Probably get you in the ballpark.

 

[Stanch]

ExcEng thanks. Yes assuming a rigid frame. you will need to excuse my ignorance but I don't follow 1/2kx^2 what are k and x?

Can you take a look at my induced compressive stress equation above, have I got my units correct for the result?

 

[boyze]

ExcelE is equating the kinetic energy of the beam at impact to the compression energy of the planks after impact. k is the linear stiffness and x the total compressed deflection from the impact. The compression energy is the area under the load deflection curve. You can also get there by using wh = 1/2kx^2 where w is the beam weight and the h the drop height. The force you are after will be F=kx. Your result will be very sensitive to what you use for the stiffness.

Here's a 1 dof online shock calculator that may help.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[Archie264]

It looks like Excel and Boyze are putting you on a good path. All I I had in mind was ignoring the affect of the timber and just running the operation as a point load on a steel beam. Assuming the beam is simply-supported and the load is applied at the center the relevant formulas are:

E=.5mv^2

W=Fd

d=F(l^3)/(48EI)

 

[XR250]

@Archie;

I was ignoring the wood as well.
Seems like there are many ways to skin a cat.