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Failure of a steel shelving unit

by Ralff
Tags: failure, shelving, steel, unit
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Ralff
#1
Oct31-12, 07:06 PM
P: 8
I am currently working on a project to determine how much weight a mild steel shelving units can support. This is a project for a friend and for my own interest.

I am having trouble deciding exactly what is necessary to determine the weight the shelving units will hold. Each shelf on the unit is supported by four pins. Do I analyze each shelf as a beam with a uniformly distributed load? Do I use the yield stress to determine the max load possible after applying a safety factor? For the sides, is it necessary to determine buckling?

I've taken a mechanics of materials class, so I understand, bending stress etc.. I am just not sure exactly how to use what I know to determine the maximum capacity for each shelf and the shelving unit as a whole.

I will appreciate any help. The shelving unit is similar to the one in the link below. Thanks.

http://www.menards.com/main/storage-.../p-1863678.htm
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Studiot
#2
Oct31-12, 07:18 PM
P: 5,462
Your first task will be to determine what you mean by "how much a shelf can support" ie to specify your failure criteria.

For mild steel the deflection will become unacceptable long before there is any failure to support load.
Chronos
#3
Oct31-12, 08:32 PM
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Static load capacity far from the only issue here, you have multiple possible failure modes [substrate load capacity, fasteners, etc.] and load conditions to consider. Safety factors are routinely applied to simplify matters. Finite element analysis is an option, but, rather impractical in most cases.

Vadar2012
#4
Nov1-12, 11:54 PM
P: 208
Failure of a steel shelving unit

It's not too hard, just quite a bit of work. You'll have to go through each failure mode like Chronos said. Start with the easy ones, like what is the maximum load each shelf will have to take and determine the maximum stress in the beams. Compares these values with the relevant standards. For example, deflection should be limited to span/250 for a beam with fixed ends. Then you might add up the total load from all the shelves and determine the maximum stress in the bottom sections. It's a pretty easy one, the hard part are the holes in the material...
Jupiter6
#5
Nov3-12, 04:22 AM
P: 128
Quote Quote by Ralff View Post
I am currently working on a project to determine how much weight a mild steel shelving units can support. This is a project for a friend and for my own interest.

I am having trouble deciding exactly what is necessary to determine the weight the shelving units will hold. Each shelf on the unit is supported by four pins. Do I analyze each shelf as a beam with a uniformly distributed load? Do I use the yield stress to determine the max load possible after applying a safety factor? For the sides, is it necessary to determine buckling?

I've taken a mechanics of materials class, so I understand, bending stress etc.. I am just not sure exactly how to use what I know to determine the maximum capacity for each shelf and the shelving unit as a whole.
I'd start with figuring the moment of inertia for the shelf cross section. From there you can figure the max point load you can put in the middle of the shelf and get an idea of your FOS based on crap-steel yield....36ksi. You can distribute the load along a shelf but the max moment will still be the same.

The design is limited to how easily the joints can pull apart and how far the frame can move. Do a force balance on the structure to figure geometry.

According to the ad each shelf can take 700 .lbs. If each Shelf is 16 Ga. over a 60" span, I'm guessing they're pushing an FOS of a mere 1 per shelf.

The next thing to figure out is how substantial the boltless interface is.

To answer your question, if your support system is solid, your shelf material and lengths can be easily figured.
Ralff
#6
Nov7-12, 01:33 AM
P: 8
So, first, I should analyze the shelves individually as simply supported beams by finding the cross section moment of inertia and calculating the deflection and max bending stress in each shelf (beam).

So, to determine the max weight capacity for a shelf, I need to make the max stress in the shelf much less than the yield stress and fracture stress, and I also need to make the max allowable deflection acceptable say span/250?

This is not taking into account that the fasteners could fail. Would this be correct so far?
Studiot
#7
Nov7-12, 03:05 PM
P: 5,462
Ralff, is this a school or college project.

If so at what level?
Ralff
#8
Nov26-12, 04:08 PM
P: 8
Studiot, this is a project I am working on only to gain experience. I am a college student. I have a degree in Mathematics, and I am working on my degree in engineering, and I have only taken the first semester of a mechanics of materials class. Sorry about the late reply, I was away on break.

I am doing this because I am curious what goes into deciding weight limits on shelving units, stands, etc.. I figured the best way to find out is do it myself and compare results with actual weight limits.
Studiot
#9
Nov26-12, 04:17 PM
P: 5,462
Well I'm sorry to tell you that this shelving project is too difficult for elementary mech of mat.

The shelves are edge stiffened plates, not beams and they are supported at the corners.

You can have some fun, however, calculating things like the tearout shear of the bolts and the buckling strength of the uprights and the many things others have said.

go well with your project and studies
Ralff
#10
Nov26-12, 04:30 PM
P: 8
Do you know where I can find more information about deflection in edge stiffened plates?
Studiot
#11
Nov26-12, 05:10 PM
P: 5,462
Well there was much research into this in the second half of the 20th century, but most of the published work is papers. Google will find some of these.

Classic theory as presented in tabulated books of formulae such as those by Roark or Pilkey only treat flat plates.

There is some stuff pertaining to support conditions in the classic treatment by Timoshenko

The Theory of Plates and Shells.

With your background you could cope with

Advanced Structural Mechanics by D Johnson

He has a chapter on plates and membrane theory (and much else) that would be useful.

I would approach this as follows

1) Consider the edges as stiff beams. This will generate support conditions.

2) Consider the main area of shelf as a flat plate or membrane with the support conditions provided by (1). You will have to experiment with moments transferred to the supports v simple support. Johnson's treatment comes in handy here.
the_emi_guy
#12
Nov27-12, 01:11 AM
P: 587
You have me curious, are there simulators available for this type of analysis?
Studiot
#13
Nov27-12, 02:04 AM
P: 5,462
You have me curious, are there simulators available for this type of analysis?
You can use various structural design (computer) packages for this. Some offer analysis of framed structures as opposed to FE analysis. These would be with rigid or semi rigid joints.

A sheet metal shelf such as this is really a thin diaphragm or membrane supported on a stiff rectangular edge. Thin plates (diaphragms or membranes) carry their load by in-plane forces, unlike beams which have provide the resistance moments by forces above and below the (neutral) plane. Thick plates or slabs can be considered as two dimensional or two way beams.

In bridge deck analysis one form is known as a grillage, this could be pressed into service here.
Jupiter6
#14
Dec1-12, 10:21 AM
P: 128
Quote Quote by the_emi_guy View Post
You have me curious, are there simulators available for this type of analysis?
Don't waste your time. It's not that complicated.

While I'm a Timoshenko fanboy, there are no plates or shells to be analyzed on this unit. Each shelf is comprised of formed channels with lips to support the filler material (particle board). While the particle board is cheap crap, in addition to holding stuff it supplies weight to drive the shelves into the post locks creating a stiffer joint.

Note there are no diagonal truss members which would be absolutely necessary if the shelves were a thin material e.g. a 22 ga home depot $10 unit.

Mc/I should do it.

The actual joints will need to be seen.


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