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Bending of an irregular shape

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dave319
#1
Mar27-12, 12:31 PM
P: 13
I am looking for an equation (or method to develop the equation) for bending of an irregular shape. I have only been able to find bending equations for beams. The attached picture shows the shape of the gusset being bent and the location of the force. If this were a beam it would be a cantilever but since it is a thin plate I'm not sure what equation to use.

Thanks!
Attached Thumbnails
Bending of gusset.png  
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pongo38
#2
Mar28-12, 05:33 PM
P: 699
This is sometimes modelled as a diagonal strut and a horizontal tie, but additional assumptions need to be made to bring the model to completion.
dave319
#3
Apr2-12, 12:46 PM
P: 13
I was thinking about cutting it horizontally and treating it as multiple beams but I don't know if that would be a good method to use or how that would work since, after the first beam, the length wouldn't extend to the location of the point force. Any suggestions?

pongo38
#4
Apr2-12, 05:04 PM
P: 699
Bending of an irregular shape

It's too short in span to be treated by normal beam theory, which has an implicit assumption that the beam is long in relation to its depth. A span/depth ratio of at least 10 is considered reasonable to apply stress = My/I formula. So that's no good then. Hence a truss analogy with strut and tie model is favoured.
dave319
#5
Apr2-12, 06:55 PM
P: 13
So the truss would look like the following picture then? The force in each member would need to be solved for?
Attached Thumbnails
Bending of gusset 2.png  
pongo38
#6
Apr4-12, 03:28 AM
P: 699
It's simpler than that. Just a triangular truss. Forget the hatching on the right. The top member is horizontal and connected to the support with a hinge. The diagonal member goes from the load to the lower support where there is another hinge. Now you have a 3-hinged arch which you can analyse for forces. The next step is bit harder, requiring judgement (or code of practice guidance) of how wide to take the members concerned in order to determine stresses.
dave319
#7
Apr16-12, 03:43 PM
P: 13
This is what I've got so far. I know I must be forgetting something because I don't have enough info to solve for By and Cy. Can you point me in the right direction?

Thanks
Attached Thumbnails
001.jpg  
pongo38
#8
Apr16-12, 05:10 PM
P: 699
So far so good. Assuming that BC is rigid, you could also use the principle that the Bending moment at A in AB is zero. Also, using the Bending moment at A in AC is zero acts as a numerical check. Another way of looking at it is the recognise that your reaction components are actually two single forces at B and at C. So what are their obvious directions? A principle of equilibrium in two dimensions (which never lets you down) is that if a body in static equilibrium is acted on by just three forces, then they must be concurrent. Why? Because if they weren't, there would be an out of equilibrium moment.
dave319
#9
Apr17-12, 12:36 PM
P: 13
I performed joint calculations and determined that Cy is actually pointing down. Is this correct? I read your comments but I'm still not sure how to calculate By and Cy. You mentioned setting the bending moment at A in AB and AC to be zero but the bending of this truss (really a solid gusset) is what I really want to calculate isn't it? In reality I want to know how thick the gusset needs to be so that it won't deform when experiencing the 33000lbs.
pongo38
#10
Apr17-12, 03:44 PM
P: 699
The analysis we have been discussing (and which remains incomplete) is just one small part of the design process. It is a mathematical model you can use to estimate stresses if you have a trial thickness, and make other assumptions. If you recognise that the edge AC is in compression then a maximum slenderness ratio of 15, say, for that edge should avoid any buckling problems. How do you get Cy is pointing up? Do you agree that Cx is pointing left? If you do, you will see that Cx has a clockwise moment effect about A. So, for the bending moment in AC to be zero at A, what is the appropriate direction for Cy? The fact is that this is a short beam for which simple bending theory does not apply. Hence the truss analogy. You are right to concern yourself with deformation, but that is an effect distinct from stress, or from buckling; design needs to consider all those things (strength, stiffness, and stability).
dave319
#11
Apr17-12, 04:53 PM
P: 13
After doing the joint analysis I realized that Cy was initially pointing the wrong direction. This better?
Attached Thumbnails
Force Calculations 2.jpg  
pongo38
#12
Apr19-12, 12:37 PM
P: 699
Not any better I am afraid. What you haven't realized is that because AB is horizontal, there is theoretically a zero value for BY. The "member" BC is rigid. So it doesn't deform. So it has a zero force in it. Your original direction for Cy upwards was correct. By asking you to tell me how you got this, your answer revealed a weak understanding of the assumptions in the model. Have you heard of the triangle of forces? In this case there is a triangle of forces at A, which, as happens in this case is identical to the triangle of forces for the whole frame, and whose geometry is geometrically similar to the geometry of the frame itself. I think a principle that has passed you by is that the force in AC must be equal and opposite to the reaction at C. I think if you had started with this triangle, and then worried about components afterwards (if you needed to, which you don't), you might have arrived at the answer sooner. If you can now succeed, it may help you to summarise back to me what you have learned from this exercise.
dave319
#13
Apr19-12, 03:57 PM
P: 13
Statics is definitely not my strong suite and I appreciate your patience. My understanding of a triangle of forces is that the resultant of 3 vector forces is equal to 0. When I draw these vectors (attached picture) and see that the B vector only has a horizontal component it follows that By = 0. Since Cy is the only vertical force opposing the 33000 lbs, I conclude that Cy = 33000 lbs in the opposite direction. Am I back on track?
Attached Thumbnails
Triangle of forces.jpg  
pongo38
#14
Apr19-12, 04:42 PM
P: 699
yes. well done. A better model would depend on how the bracket was fixed to the wall, but the forces you have uncovered are pretty inescapable, whatever the method of fixing. You have a moment resisted by the horizontal reaction components, and a shear resisted by Cy. So now you almost have the force in AC and you can design that edge for that force.
dave319
#15
Apr20-12, 11:13 AM
P: 13
The force in AC = Cy(13.9/11) = 41700 lb. So now that I want to make sure this member is thick enough to withstand this compression force without buckling do I use the equations for a column? You mentioned that a slenderness ratio of 15 should suffice. Does it mean that the radius of gyration (Rg) = 13.9/15 = .93?
pongo38
#16
Apr23-12, 10:51 AM
P: 699
Yes. Engineers in practice would do trial and error. for example, try 1 inch thick (btw, always state your units). What would you say was a credible dimension in the plane of the web? (some fraction of 13.9" ). Work out the safe load, and compare it with your applied load. Adjust if necessary.
dave319
#17
Apr23-12, 06:03 PM
P: 13
I made an initial guess for the thickness to be 1/2" and used the equations for a short column located in Marks' Standard Handbook for Mechanical Engineers. So using this method I have determined that if my b dimension (thickness) is 1/2" than my other dimension must be 6.68". How do I decide if this is a reasonable thickness or not since this second dimension varies?
Attached Thumbnails
Marks' 1.jpg   Marks' 2.jpg   Dimension Calc.jpg  
pongo38
#18
Apr24-12, 02:17 AM
P: 699
There is no formulaic answer to a question such as yours which requires judgement and the taking of risk. You have to bear in mind that your bracket is unaware of the assumptions you are making about its behaviour (elastic, initially straight, without any residual stresses from its manufacture, departure of the mathematical model from the reality etc). Another (rough) way of looking at slenderness ratio is the ratio 13.9/0.5 = 27.8 Personally I am wary of calling anything over 28 as 'short'. In my mind it has a buckling tendency that a ratio of less than 20 doesn't. Another consideration is this: At the mid length of the strut, where bending due to buckling is most severe, do you have 6.68" available? Well....yes....mmmmm...but isn't that asking quite a lot. btw 6,68 is a bit precise given the variables here. That is 7", practically. I would be happier with half that as a credible strut. There may be others reading this who could contribute another perspective. You have definitely strayed into 'design' from 'analysis'.


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