Shear stress in a column.

[interface222]

Hi,

I'm trying to determine the shear stresses in a polymer plate acted upon by a circular cross section rod.

The average shear stress (Tau avg) at the circumference of the rod interface is not difficult to determine. But as the material is a polymer I'm trying to determine the shear stresses near the face of interaction between the rod and the plate as I am worried about plastic deformation at the bottom face of the plate.

My questions are:

In modeling the problem, I think that I should take a "rectangular" cross section of the "column" of polymer and approximate the stress for this cross section at each end based on the total circumferential shear stress.

I can then use this rectangular model to determine the shear stresses at the face using equations for distribution of shear stresses in a rectangular section.

- So.. Does this sound like the correct approach?

I haven't been able to find approximations for the shear stresses in columns.

- Can anyone point to models for shear stresses in columns?

 

[Dinosaur]

I don't understand the geometry of the problem from your description. You mention column behavior and mention the bottom of a plate. Is the rod a polymer? Is the rod loaded as a column? How do you take a rectangular cross section of a circular rod? How is the plate attached to the rod? In what orientation? You say you are concerned that the plate may experience plastic deformations on the bottom. What about at the point where the rod meets the plate?

 

[interface222]

Thanks for taking interest.

The plate is a polymer being acted upon from below in the center by a vertically fixed circular metal rod. In this case it would be glued to the bottom without a change in cross-section.

I'm trying to find the shear force at the bottom face of the fixed polymer plate.

 

[vonlueke]

interface222: I think the peak shear stress, prior to local deformation of the plate, will occur at the plate midplane, not the interface surface, and will be tau = 1.5*P/(pi*D*t). So you'd probably want to ensure FS*tau < 0.577*Sty, where FS = factor of safety, Sty = polymer tensile yield strength, D = rod diameter, t = plate thickness, and P = rod axial load. Secondly, you’d probably want to take a look at flatwise compressive stress on the plate surface on a small annulus near the rod perimeter, sigma = P/[0.25*pi*(D^2 - d^2)], and ensure FS*sigma < 1.5*Sty.

 

[vonlueke]

Third, calculate the plate bending stress, sigma_b, above the rod centerpoint, and ensure FS*sigma_b < Sty. This bending stress is the case of load P applied over a small circular area at the center of the plate, and can be found in, e.g., Roark.

 

[vonlueke]

My above posts assume load P is applied perfectly concentric with the rod centerline. If P is instead eccentric, having eccentricity distance e (i.e., if there's also an applied moment M, where e = M/P), then multiply tau in my first post by [1 + (4e/D)], and multiply sigma in my first post by {1 + [8eD(D^2-d^2)/(D^4-d^4)]}. Notice you can also compute the stress on the tension side of the rod by changing the plus sign in the above factors to a minus sign. These factors assume the rod is bonded to the plate. (They would have a different derivation if the plate can separate on the tension side when sigma changes sign.) By the way, P is applied load plus tributary weight of the plate.

 

[Dinosaur]

I think vonlueke covered it fairly well. The maximum theoretical shear stress away from the local effects occurs at mid-plane and I believe is corectly given in his post.

My concern in your problem is the combined stresses in the transition zone around the point of application. I think the stress-strain behavior (e.g. plasticity) of your polymer will have a lot of influence on your final design choices.

This is an example where I would initiate a testing program to verify your finite element model. There are many F.E. programs that will handle bi-linear stress-strain diagrams. Get one of these and a good book on advanced mechanics of materials such as the one of the same name written by Boresi & Sidebottom and published by Wiley.

Good Luck

 

[interface222]

Vonlueke and Dino thanks for the responses.

I believe I was concerned with local deformation but this study won't take the time to model the problem using FEA software.

I do think that compressive strength is really the telling factor because until there is local deformation, there shouldn't be a change in mid-plane location of the max shear stress as Vonlueke correctly stated. The material needs to stay in the linearly elastic region.


Vonluke:
A couple of questions,

1. Could you elaborate where the constants (ie .577 and 1.5) in your equations came from?

2. Why would you take the full compressive load over a small annulus at the rod diameter? And again why the 1.5 X Str Yield?

3. Out of curiosity the eccentric load would just be applied to one edge of the rod?

 

[vonlueke]

interface222: (1) 0.577*Sty and 1.5*Sty are rough estimates of shear yield strength and bearing yield strength for isotropic, ductile materials. But of course use the published material data for these strength properties if available for your material.

(2) There would tend to be higher contact pressure near the rod perimeter, depending on the relative stiffness of the plate, etc. You can concentrate the compressive stress as much, or as little, as you see fit depending on the value of lowercase d you decide to use. Using bearing yield strength (1.5*Sty) here is debatable. Some might prefer to play it safer and use flatwise compressive yield strength, or tensile yield strength, instead of bearing yield strength.

(3) If you also have an applied moment M, it can be alternately expressed as e*P, where e = M/P. Eccentricity e is the perpendicular distance of load P from the rod centerline such that P would also simulate the resultant applied moment M (if any). If M = 0, then e = M/P = 0.

 

[vonlueke]

interface222: I decided to check my assumption that the peak shear stress would occur at the plate midplane, and found it is incorrect. The peak shear stress occurs at the plate interface surface, near the rod perimeter. The good news is, the same formula I gave for tau seems to be a good estimate of the peak shear stress, except it occurs at the plate interface surface, not the midplane.

No change regarding what I wrote about sigma nor sigma_b, which seemed to prove correct. In most problems (depending on the proportions), the bending stress sigma_b above the rod centerpoint is almost identical to the bending stress at the rod perimeter. Therefore, as Dinosaur correctly stated, the peak stress is the combined stress at the plate interface surface near the rod perimeter, and for an isotropic, ductile material, can be estimated as sigma_e = (sigma_b^2 + 3*tau^2)^0.5, where sigma_b and tau are mentioned in my previous posts. So you'd probably want to ensure FS*sigma_e < Sty, and FS*sigma < 1.5*Sty (or Sty), where sigma is mentioned in my previous posts.

 

[interface222]

Looks right. My intuition was telling me that there'd be more force at the interface between rod and plate. Though, I was initially agreeing with you regarding peak shear stress at the midplane when you mentioned it. I think that's bending shear only I suppose.

The rough estimates (.577 and 1.5) were ringing bells but it's been several years. Thanks for the clarification.

I need to purchase Roark's Formulas, but I don't recognize
sigma_e = (sigma_b^2 + 3*tau^2)^0.5. I would have expected to use mohr's circle.

Sorry for the extra questions.

 

[vonlueke]

interface222: The peak shear stress away from the transition zone occurs at the plate midplane, but it rapidly migrates to the plate interface surface in the transition zone near the rod perimeter due to local effects, as pointed out by Dinosaur. Yes, the peak shear stress at the midplane away from the transition zone is due to bending.

The sigma_e is actually a form of von Mises stress, assuming you have ductile material. If you use Mohr's circle in your usual fashion, then plug the results into von Mises, you'll get the same answer.