Calculate deflection in a shelf made from sheet metal

[keaAstrac]

How can I calculate the maximum load a shelf made from sheet metal can take at a certain deflection limit? Have a number of different sizes and gauges and need to calculate without testing.
Any help is much appreciated.

The shelf is supported in the 4 corners similar to photo below
https://www.physicsforums.com/attachments/upload_2015-6-1_11-43-56-png.84337/

 

[SteamKing]

Your image didn't get attached to your post for some reason.

 

[keaAstrac]

The shelf is supported in the 4 corners by simple pins. Not sure what's up with the image as i can see it when I'm on this forum.

 

[jack action]

Assuming a uniformly loaded shelf:

The deflection at the center is:

Rewriting the equation, the total load supported [itex]F[/itex] ([itex]=qL[/itex] in N or lb) is:

[tex]F = \frac{384EI\delta_C}{5L^3}[/tex]​

Where:

[itex]I[/itex] is the area moment of inertia about the horizontal plane (m⁴ or in⁴);
[itex]E[/itex] is the Young's modulus of the shelf material (Pa or psi);
[itex]\delta_C[/itex] is the maximum deflection (m or in);
[itex]L[/itex] is the beam length (m or in).

The area moment of inertia depends on the shelf side view geometry. For example, if the shelf is a rectangular board with height h and depth b (in m or in):

The area moment of inertia about the horizontal axis (X) is:

 

[keaAstrac]

Thanks for your reply jack action. I have used the method above to calculate the maximum load but the results are incorrect.

Example: Shelf size 1000mm x 300mm.
Taking a cut through the shelf and using inventor to calculate the Second Moment of Area, (I = 18654.4 mm^4) - this value is correct.
The edge of the shelf have extra folds - take as simply supported.
Maximum Deflection = 1000/100 = 10mm at centre of shelf
E=205 GPa
L=1000 mm

Result: Max shelf load from calculation = 299.38 KG
Tested result = 160KG

Not sure why the calculated result is so far out??

 

[jack action]

When tested, was it uniformly loaded or only one force at the center of the shelf? If so the equation becomes:

[tex]F = \frac{48EI\delta_C}{L^3}[/tex]​

And in that case, the load is 187 kg.

 

[keaAstrac]

Sorry it was tested using a UDL, similar to below.

 

[jack action]

All I can say is that usually there is a safety factor involved, although it is usually in term of failure rather than deformation. Below is my personal reference sheet to help me define the proper safety factor. I took it from a class book when I was an undergrad.

If I used that for your problem, I would of probably taken a safety factor of 3, nothing below 2 for sure. You have a safety factor less than 1.9 (= 299.38/160).

--------------------------------------------------------------------------------------------------
For ductile material (usually based on yield strength), the safety factor should be max(N1, N2, N3).
For a brittle material (usually based on ultimate strength), the safety factor should be 2 * max(N1, N2, N3).

Where:

Information.........Quality of information......Factor
.........................N1
Material property available from tests:
............The actual material used was tested........1.3
..........Representative material test data are available...2
...........Fairly representative material test data are available...3
..........Poorly representative material test data are available...5+

.........................N2
Environmental conditions in which it will be used:
...........Are identical to material test conditions.....1.3
...........Essentially room-ambient environment......2
............Moderately challenging environment........3
............Extremely challenging environment......5+

.........................N3
Analytical models for loading and stress:
..........Models have been tested against experiments.....1.3
............Models accurately represents system.....2
............Models approximately represents system......3
.............Models are crude approximations......5+

 

[keaAstrac]

We have measured a deflection of 10mm (at centre of shelf) at a UDL load of 160 KG. The calculation should therefore provide the same result, it does not.
Think it may be something due to the thickness less than 1mm and the folds on the outer edges.
Their must be a different formula for this type of structure?

The link I found below has different formula but they don't give good results either. ( The load using that formula is 10N which is completely wrong)
http://www.agriculture.purdue.edu/fnr/faculty/eckelman/documents/198406a.pdf

Any help is appreciated thanks..

 

[SteamKing]

We have measured a deflection of 10mm (at centre of shelf) at a UDL load of 160 KG. The calculation should therefore provide the same result, it does not.
Think it may be something due to the thickness less than 1mm and the folds on the outer edges.
Their must be a different formula for this type of structure?

The link I found below has different formula but they don't give good results either. ( The load using that formula is 10N which is completely wrong)
http://www.agriculture.purdue.edu/fnr/faculty/eckelman/documents/198406a.pdf

Any help is appreciated thanks..

You haven't really provided any clear pictures/drawings/descriptions of this shelf. The images attached are lacking in detail. I'm not even sure what the image in Post #3 is supposed to show.

You've calculated a value for I, but does the shelf act like a beam, or something else? Is the I value constant throughout the length of this shelf? Are there any discontinuities in the edges of the shelf? When loaded, does part of the shelf start to flex some, which suggests that the stresses may not be distributed as in a prismatic beam?

 

[keaAstrac]

The image in post 3 shows the pin type supports. The section is constant throughout as in the image below, only their is a fold over the two ends as seen in post 3. Shelf size 1000mm x 300mm.

"does the shelf act like a beam, or something else?" - am trying to determine this. Think the shelf bends down in the centre due to it being light their, this is probably where most stress is. But I am most interested in finding the deflection along the edge?

 

[jack action]

From the link you gave, you can see that shelves are assumed to act like a beam ... in all directions.

How is your design depth-wise? It acts like a beam in this direction as well. Assuming your ends are not closed, the second moment of area in the other direction is 42.7 mm⁴ (= 1000 * 0.8³ / 12). With the equation above - assuming you have a shelf 300 mm wide X 1000 mm depth - the mass needed to get a 10 mm deflection would be only 25 kg. Even if the ends are closed, with the length, you can see the margin you have to play with.

If your shelf changes shape depth-wise under loading, it means the area for the second moment of area (length-wise) will change shape also (curved line towards the centroid instead of a straight line away from the centroid), thus affecting the second moment of area. Can it explain a factor of 1.9? I don't know, but it has some effect for sure (hence the use of a safety factor when «models approximately represents system»).

 

[keaAstrac]

The deflection at the middle of the shelf is not a problem, it also must get some support from the edges.

If your shelf changes shape depth-wise under loading, it means the area for the second moment of area (length-wise) will change shape also (curved line towards the centroid instead of a straight line away from the centroid), thus affecting the second moment of area. Can it explain a factor of 1.9? I don't know, but it has some effect for sure (hence the use of a safety factor when «models approximately represents system»).

Think you are correct the I value will change due to the other axis deflecting. I have checked this and it will change as below.

I = 18654.4 mm^4 at start ---- now; I = 11300 mm^4 with centre deflecting worst case - which gives a load of 181 KG
The load above is closer but not exact...

Any other thoughts??