Dynamic pin shear force 2

[LeonhardEuler]

How do I calculate the strength of a pin that will have a dynamic force on it from an object falling at several feet. Is using the simple formula of f=mv^2 adequate, or do I need to consider kinetic energy and impulse? Thanks

 

[BAretired]

Dynamics is not my strong suit, but I think you have to consider the distance traveled after initial contact.

BA

 

[WARose]

I think one of the better methods of figuring impact force is the method/formulas in Blodgett's 'Design of Welded Structures'......it takes into account stiffness. (See Section 2.8.)

 

[LeonhardEuler]

Blodgett book was not very descriptive about how to apply the ultimate energy resistance formula. It is given in units of psi, or lb-in/in^3. Where is the volume coming from? Is it supposed to be the total volume of the member? The failure plane will likely be shear across the C.S. so it seems like only a square area should be considered. Not very clear.

 

[WARose]

Blodgett book was not very descriptive about how to apply the ultimate energy resistance formula. It is given in units of psi, or lb-in/in^3. Where is the volume coming from? Is it supposed to be the total volume of the member? The failure plane will likely be shear across the C.S. so it seems like only a square area should be considered. Not very clear.

I'm not sure what you are talking about. You have a object falling on a pin from a certain height yes? Well with Blodgett's book you can figure the impact load from that collision and then compare that to your allowable forces/stresses. On page 2.8-3 of the book, they give a very easy to use formula for figuring a impact force. Volume is not a variable.

 

[LeonhardEuler]

I am ok with how to find the force from page 2.8-3, which considers stiffness and deflection of the pin as well as kinetic energy of the falling object. What I have difficulty with is finding the allowable ultimate "strength" of the member subjected to impact force. This seems to be given in page 2.8-6 as the ultimate energy resistance, but it is in units of energy/volume. What I assumed is that I need to multiply this ultimate energy resistance value by the impact section properties of I/c^2 found on the same page; however, this guidance does not seem to be given anywhere and is essentially a guess.

 

[BAretired]

One of the difficulties with this problem is in defining the spring constant of the landing area. It can't be just the spring constant of the pin in shear. It must also be dependent on the member to which the pin is attached.

It seems to me that the impact force must be partially dependent on the falling body. The center of gravity of the falling body will move down following initial impact.

Not an easy problem to solve.

BA

 

[WARose]

What I have difficulty with is finding the allowable ultimate "strength" of the member subjected to impact force.

Depends on the material. But if we are talking a pin made of steel, with the load not being repeated numerous times (i.e. no fatigue), and a case where there will not be significant localized damages from contact stresses......you would use the same allowable you would use for a static load. Most books on blast loading give you a bit more (for shear).....but they take into account strain rate effects and so on.

 

[LeonhardEuler]

I thought the spring constant was simply 48*E*I/L^3
The force estimated by Blodgett's method on 2.8-2 does not seem accurate. It shows that a force of 200 lb falling at 80 inches creates a force of nearly 200 kips. This doesn't seem reasonable. Does it seem like a reasonable number to anyone else?

 

[WARose]

The force estimated by Blodgett's method on 2.8-2 does not seem accurate. It shows that a force of 200 lb falling at 80 inches creates a force of nearly 200 kips. This doesn't seem reasonable. Does it seem like a reasonable number to anyone else?

Nope. Most likely you have misestimated something (perhaps the stiffness at the point of impact).

 

[LeonhardEuler]

K=48*E*I/L^3. L=4 in, I=0.049 in^4 for a 1 in pin, E:29000 ksi, K=1068kip/in, F=W.b+(sqrt(w.b^2+2*K*Wb*h)) where wb is 200 lbs, F=160.27 kip

 

[BAretired]

That formula for K makes no sense to me. Where did it come from?

BA

 

[WARose]

K=48*E*I/L^3. L=4 in, I=0.049 in^4 for a 1 in pin, E:29000 ksi, K=1068kip/in, F=W.b+(sqrt(w.b^2+2*K*Wb*h)) where wb is 200 lbs, F=160.27 kip

I still think the issue is with the assumed stiffness.

 

[LeonhardEuler]

61 kips would be great. I could work with 61 kips. Which of your numbers were different than mine in the equation?

BAretired it is the inverse of the deflection equation for a simple beam less the external force. An example of using this as a spring constant is also given in Roarks.

 

[LeonhardEuler]

k=F/d where d is the deflection the rest of the formula comes from beam deflection table for simply supported beam

 

[WARose]

61 kips would be great. I could work with 61 kips. Which of your numbers were different than mine in the equation?

I think I made a error in the calculation.

As far as the stiffness goes......are you sure the supports will have that kind of stiffness as well? (A picture would help.)

 

[WARose]

By the way.....what is impacting the pin?

 

[BAretired]

LeonhardEuler,

By the title of the thread, I thought we were talking about a pin in shear, not one acting as a beam.

BA

 

[rb1957]

I think your approach, of using the stiffness of a SS beam, is what BA suggested initially, ie Fd = 1/2kd^2 = mgh ... if a SS beam is a good representation of the structure.

another day in paradise, or is paradise one day closer ?

 

[LeonhardEuler]

BAretired it is a pin in shear, but in order to find what force it takes to stop the falling object I need to set work done by the pin equal to the initial potential energy. Work done by the pin is obviously force*distance of deflection, distance of deflection is characterized by the pin in"bending" with the corresponding spring constant.

WARose a 1/2" plate

After playing around with these equations a bit it has become evident that this calculation might be better as a FEA simulation due to such a large variability in results from small changes in inputs.