12 ton frame lowered on plastic --> What Happens?

[OlPhyz]

Summary:: 12 ton rectangular frame is lowered by a 4 point crane attached to each corner of the frame.
Frame will be lowered onto plastic supports, these supports have guiding ramps of 60 degrees to help the operator lower the frame in the correct spot.
The frame has 4 feet, one on each corner, as the frame is lowered onto the guiding-plastic support what happens to the forces? Especially on the guiding ramp part.

[Mentor Note -- thread moved from the ME forum to the schoolwork forums. This is for a schoolwork project.]

Hello Engineers,
I am coming to you for help with my problem.

Context :
We have a 12 ton frame with 4 aluminium feet. This frame is lifted then lowered onto plastic support spacers.
Lifting is done with 4 cranes attached to the 4 corners of the frame, they lower the frame parallel to the ground. (so the feet are parallel to the ground as they approach the plastic supports, one on each corner where the feet will go)

The frame's lowering speed is not known precisely, but i think we can go with 0.0254 m/s it can probably go slower though.
That will be determined later on, according to the results.

The 4 plastic supports have a 60 degree guiding ramp to help the operator lower the frame on the correct spot on the plastic supports.
We need to know if the plastic will hold lateral forces exerted by the frame as it is touching the guiding ramps.
Knowing what proportion of the 12t is actually being transferred to the plastic spacers would help.
And then a case of looking at the lateral resultant of the applied force.

Problem :
What's the approximate value of the lateral forces transmitted to the plastic supports? (guiding ramp)

Plastic properties :
Yield strength -- 65 MPa
Tensile modulus -- 3000 MPa
Flexural strength -- 115 MPa
Flexural modulus -- 2900 MPa
Friction coefficient -- 0.32

Data Recap :
Frame mass -- 12 tons
Frame lowering speed -- 0.0254 m/s
Point of contact -- Plastic guiding ramp angled at 60 degrees

Oh and I just noticed in the image, you might think the two feet are hitting the guiding ramps, this is not the case, only one guiding ramp is used by the operator, the others have enough play so that the feet are not all hitting their respective ramps.

If I missed out on any important details, just tell me.
Thanks for helping out!

 

[Lnewqban]

Is the frame free to rotate about its vertical axis?

 

[OlPhyz]

Is the frame free to rotate about its vertical axis?

Ah yes I did not point that out. Thank you, no it cannot.
It also cannot pivot, and does not have any give in the lifting system.

 

[Lnewqban]

That greatly reduces any inertial force on the guides.
I believe that it is close to impossible to synchronize four cranes working simultaneously.
If that is true, some oscillations will still happen.
The magnitude of any lateral impact force on the guides is directly related to the velocity of impact, the moment of inertia of the frame, and how rigid those guides are.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi

 

[OlPhyz]

That greatly reduces any inertial force on the guides.
I believe that it is close to impossible to synchronize four cranes working simultaneously.
If that is true, some oscillations will still happen.
The magnitude of any lateral impact force on the guides is directly related to the velocity of impact, the moment of inertia of the frame, and how rigid those guides are.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi

Yes, indeed it simplifies things and takes us one more step away from reality.
But I have not been informed to take this into account, therefore I'm guessing it is minimal movement.
(though minimal movement with 12 tons will always create big results haha)

Lets start simple, and add variables later if need be for a more accurate depiction.

How does one know how much of the 12 tons are actually resting on the guiding ramps?
Is this weight split with the two of the four guiding ramps? (the two on the same side as the operator)
I'm guessing so, since the frame would be hitting them at the "same" time.
Is the weight transfer dependent on the friction coefficient and the angle on which it resting upon?

 

[jack action]

This is the way I would approach the problem (if I understood it correctly):

The only lateral force $F_L$ on the ramp comes from the friction force $F_f$ sliding on the ramp, so:
$$F_L = F_f \cos \theta$$
Where $\theta$ is ramp angle.

The friction force depends on the force normal to the surface $F_N$, so :
$$F_f = \mu F_N$$
Where $\mu$ is the friction coefficient on the ramp.

The normal force can be found by imagining the weight hitting the ramp at velocity $v\cos\theta$ and causing a deflection $x_N$ to the ramp, perpendicular to the inclined surface. The reaction force $F_s$ of the material (modeled as a spring) will be:
$$F_N = F_s = kx_N$$
Where $k$ is the equivalent spring constant of the material.

To determine $x_N$, we equate the kinetic energy transferred with the elastic energy absorbed by the spring:
$$\frac12 m(v\cos\theta)^2 = \frac12 kx^2_N$$
Or:
$$x_N = \sqrt{\frac{m}{k}}v\cos\theta$$
Putting it all together, we get the maximum lateral force possible:
$$F_L = \mu\sqrt{km}\ v\cos^2\theta$$
Personally. I would assume the whole mass hit a single ramp, which is the worst-case scenario.

 

[berkeman]

Welcome to PF.

Hello Engineers,
I am coming to you for help with my problem.

What is your level of education in Engineering, and what professional certifications do you currently hold? And why are you turning to the Internet to work out this problem? Does your work insurance company know that you are having to turn to the Internet to do these calculations?

 

[OlPhyz]

Welcome to PF.
What is your level of education in Engineering, and what professional certifications do you currently hold? And why are you turning to the Internet to work out this problem? Does your work insurance company know that you are having to turn to the Internet to do these calculations?

I'm still at school, working on being a technician. No certifications yet, I'm turning to the internet for this problem because the teacher told us to use forums for help understanding certain issues, (there are a lot of students, so he can't be there for all)
This is for a school project where we have to make guiding systems from different materials. :)

 

[berkeman]

Oh my goodness, that's different! Okay, the thread is fine, but I'll move it to the schoolwork forums where it should be. We thought that somehow you were an apprentice at an actual company, designing 12-ton crane apparatus for real. Yikes!

 

[OlPhyz]

This is the way I would approach the problem (if I understood it correctly):

The only lateral force $F_L$ on the ramp comes from the friction force $F_f$ sliding on the ramp, so:
$$F_L = F_f \cos \theta$$
Where $\theta$ is ramp angle.

The friction force depends on the force normal to the surface $F_N$, so :
$$F_f = \mu F_N$$
Where $\mu$ is the friction coefficient on the ramp.

The normal force can be found by imagining the weight hitting the ramp at velocity $v\cos\theta$ and causing a deflection $x_N$ to the ramp, perpendicular to the inclined surface. The reaction force $F_s$ of the material (modeled as a spring) will be:
$$F_N = F_s = kx_N$$
Where $k$ is the equivalent spring constant of the material.

To determine $x_N$, we equate the kinetic energy transferred with the elastic energy absorbed by the spring:
$$\frac12 m(v\cos\theta)^2 = \frac12 kx^2_N$$
Or:
$$x_N = \sqrt{\frac{m}{k}}v\cos\theta$$
Putting it all together, we get the maximum lateral force possible:
$$F_L = \mu\sqrt{km}\ v\cos^2\theta$$
Personally. I would assume the whole mass hit a single ramp, which is the worst-case scenario.

Thank you for your insight Jack, very well explained and straight to the point.
Indeed this is very helpful for getting my head around how it is possible to determine lateral forces on tapered ramps.

 

[OlPhyz]

Oh my goodness, that's different! Okay, the thread is fine, but I'll move it to the schoolwork forums where it should be. We thought that somehow you were an apprentice at an actual company, designing 12-ton crane apparatus for real. Yikes!

Oops, sorry. Yes indeed it should be in there, thank you.