Hey youtubers welcome back to the engineering tool box channel where I give you the tools you need to solve real-world engineering problems
I'm really excited to share this video with you
I put a ton of work into it
And I think it turned out pretty well not to mention you're gonna be learning some really cool things
So let's check it out in this video
We're gonna be talking about
Fastened joints and how to create a really cool excel document that can help you find the strength of various fast and joint configurations
You'll learn the basics of stress and the different types of stresses that these joints are under and the failure modes that those stresses can
cause then
I'll show you how to use excel to assist in the calculations to figure all this out and
Finally how to use Excel solver function to optimize your joint configuration and get the maximum strength from the joint
Wow, that sounds like a lot, but don't worry. I'm gonna break it down for you before we get into building the spreadsheet
Let's first understand a little bit about stress and more
Specifically what types of stresses these joints are under so first of all what is stress?
Stress is defined as a force on an area so if we have a certain force applied to a body
That force would be dispersed throughout each cross sectional area that is resisting force in other words stress equals a force divided by an area
the equation being Sigma the symbol for stress equals F or force over a the area
Fastened joints like these undergo three types of stress shear stress tensile stress and compressive stress first. Let's understand shear stress
Here's a cross-section view looking perpendicular to the bolt or fastener
Let's add some forces so as these forces are applied to the base materials
There is a stress within the fastener as the base materials are forcing
It knobs of directions along what's known as the shear plane the cross-sectional area of this plan would look like this
where a
Equals pi times the radius to the fasteners squared if the stress in this area is greater than the ultimate shear strength of the fastener
Then the fastener will fail just like in this real example
I'll place a line on the center axis of the bolt for reference as the plates pull on the bolt the forces cause a shear
That eventually causes the bolt to deform and eventually fail
next let's look at the tensile or tearing stress that happens within the base materials as you is looking down from the top of the
plates that the fastener is holding together as
Forces are added to the plates the fastener transfers the opposing forces to each plate and creates a tension along this plane within the material
The cross-section of this plane would look like this
Where a the area equals view the width of the plate mine is D the diameter of fastener times the thickness of the plate?
Now remember tensile stress means the material is being pulled apart
So if the tensile stress exceeds the tensile strength of the material?
It will cause tearing that could lead the joint to fail in this way as well
Finally if we look down at the top of the joint again
we can see there is also a
Compressive or bearing stress that happens within the base material here in the area behind the hole as the fastener is forced into the material
At the aerial contact here is cross-section of this area where a equals the diameter of the fastener times the thickness of the material
Now if this stress exceeds the compressive or bearing strength of the material the fastener will actually begin to pull through the base materials
Let's take a look at a real example of this
Look at the holes notice
They are actually deformed or elongated from the force the bolt placed on the area behind it now the ultimate shear strength of the bolt
Was exceeded before the plate failed?
But I think this demonstrates this type of stress pretty well
So now that we understand three ways these joints can fail we can dive into the fun stuff and do some calculations to figure out
How strong our joint actually is given some parameters, I'll admit the spreadsheet does have some limitations, and it makes some assumptions
But it should cover many of the typical fast and drink connection
We will see in the field
For example the spread sheet as soon as the plate thicknesses are the same so this distance here
And this distance here are the same this distance in this distance and these types of fastened joints
It also assumes that the bolt diameters are the same if we have multiple bolts and our joint
And then we're also using the ultimate strengths of the materials to solve for the failure strength of the joints so
If we look here, we're using the ultimate shear ultimate tensile altima compressive
although all those easily could be
substituted for yield strength values in order to figure out the point or the strength of the joint at which it would actually start to
Deform where the materials would deform
Hopefully once I go through this and we understand how the spreadsheet works
Then you can modify it to your actual needs so with that in mind. Let's start going through this we have
It all pretty much set up as far as the formatting and labeling goes here
We can see our inputs in this column, so we're looking at the number of fasteners the fit the factor of safety
the number of shear planes so
That would be this plane here in this plane here if we have two plates on one side or multiple plates involved
We're looking at the ultimate shear strengths of the fastener
We'll be using the diameter of the fastener and for the plate
We'll be used using the ultimate tensile strengths the thickness the width and then the cross-section will be calculated
We'll be using the ultimate compressive strength to find the strengths of the whole remember again that area behind here that can potentially fail
We'll be using ultimate compressive strength to solve for that and then once we have the rest of those inputs
We will find the cross-section
Of that hole there
So over here, we have our outputs basically the things that we're going to be calculating
given our inputs over here
So what we're gonna be solving for is the amount of force that our joint can handle under shear tension and compression
So this is going to tell us how much force we can apply to the joint on each side here before the joint actually fails
So let's start filling in some values
just for starters, let's go with the single shear or
two plate one bolt design here
So the number of fasters in this case will just keep it up one the factor safety
is going to depend on your application obviously if
you're
Holding a person ten thousand feet in the air with this or something like that
You know you're going to want the factor of safety to be higher
But if you're just holding up a picture or something like that
You don't you don't need such a high factor of safety?
So just keep that in mind the number of shear planes again is just one
So under that the ultimate shear strength in psi for a steel bolt. Maybe it is 50,000 psi
the diameter of our bolt, let's say is a quarter inch and
That rounded Adhamiya, so let's bump that up one the ultimate tensile strength
I'm gonna cheat and look at my
Finished spreadsheet will use sixty thousand psi our minimum thickness. Let's say is a quarter-inch
again our width will be
two inches
It's not the cross-section of the plate so again if you remember our
Formula that's going to be the width of the plate
- the
diameter the hole times the thickness of the plate
So that'll give us the cross sectional area of our plate
So right here - right here - the hole the ultimate compressive strength of the material. Let's go back to our
Cheater document and we'll use thirty six thousand psi
And in the cross-section of the hole again going back to our equations from before this is going to be the diameter of the hole
Here from times the thickness of the plate
Now I have some alternate values over here for the metric folks out there
Pretty cool function in Excel is the convert function whoops if I could spell convert convert?
Okay convert this number from our psi units to our
Megapascals units
So what that did is it looked at our value here, and it said we want to convert it from psi to megapascals?
So that's kind of a cool little feature we can then just take that and fill it down
I'll go without formatting and that just looked at every unit value over here and converted it to the unit
That I have over here
It's a nice little way to kind of give us some alternate units
Maybe you're working with both types of Units where you need to convert easily. That's a great way to do it now
I realized I was actually forgetting some things when I did these formulas
I was only accounting for this scenario where we have one bolt and one shear plane
So in order to calculate the correct Crocs cross section for our plate
We're gonna have to consider the number of fasteners
so we actually want to take the width of the plate minus the diameter of the fastener or
fasteners times
the number of fasteners
Times the thickness and then that'll give us the correct
cross sections
So if I bump up the decimal place, and then enter in two fasteners we can see that our cross section actually goes down
Also we will have to look at the hole here the cross section there will bump up the decimal place and
We'll have to take the diameter times the thickness times
The number of fasteners then now on the calculating our outputs here so in order to calculate the outputs
We have to remember our stress equation
Where stress is equal to the force over the area?
So since we're gonna be actually solving for the force we need to rearrange this formula
so we'll actually takes the stress times the area and
The ultimate shear strength and the ultimate tensile strength the ultimate compressive strength
those are the values for the amount of stress that the
material can handle in each of these types of stress, so
ultimate shear stress is the maximum amount that the material can handle in shear so
To solve for the maximum force then we'll have to use that value times the area
So let's do that before shear stress. We need to take the ultimate shear stress
Time's the area. That's under shear so that'll be
pi times
The radius, so it'll be
diameter divided by two and then will be squared and
then we're gonna take that times the number of fasteners times the
number of shear planes
And close that off
oops
Enter there, and then remember we need to consider our factor of safety
So we'll take all of that divided by oops not two we wanted to go by this itself
So that is the amount of force that our joint can handle in share?
For tension we need to take the ultimate tensile strength times
the cross section and
Then divide that by the factor of safety no since our cross section here already considered all of the geometric things like
Our diameter and thickness and all that stuff. That's all we have to do to calculate that
For a compression. It's also pretty simple. We just need to take the ultimate compressive strength
times the cross-sectional area
divided by the factor of safety then and
Then to find our overall joint strength we're gonna take the minimum of those three values so basically we're finding the weakest link
Between those three failure modes so we can see that our joint is actually weakest in compression and then one last thing here
We'll use the convert
Function again that time I spelled it right at least so we'll convert that
value from pounds force to units
And then again, we'll just right-click drag fill without formatting and then that has our converted values as well alright
So now we pretty much have a completed spreadsheet here. We can play around with the different inputs. We have we can
Make this so now it's a two fastener two shear plane
Jointed connection and we can see what that does to the joint strength we can see it actually went up in this case
We can lessen the width of the plate and see what that does didn't do anything
You know so we can just sit here and play around with different combinations of joints and see how strong the joint actually is
Which is pretty cool? The next step and probably the coolest part of all this is
We're gonna use the solver function now to actually maximize this joint strength given our
Parameters, or we're gonna set some limits on these parameters
So it's actually gonna solve for the strongest joint within the ranges that we let it adjust our parameters
so I have it activated here under this analyse tab if you don't have that activated you go to options and
Then add-ins, and then you're gonna have to go to the analysis toolpak oops. Sorry solver add-in and then hit
Go and then
We're gonna add this analysis toolpak and hit ok so now if you go to the data tab you should have this solver and data
analysis tab over here
Or group over here. I should say so if we click solver it opens up this window and what we're gonna
Do is select an object that's asking for so we want to set our joint strength
hit enter there
We want to solve for a max
We could actually solve for a value up so let's say we just need to know we need to
Attain a certain strength to hold up a certain mass or a certain
Force or something we can actually set it to a specific value and then optimize
the inputs in order to reduce the let's say number of fasteners or something like that to achieve the
value that we want but in our case let's just play out with this and let's do the maximum so given the ranges that we
Are going to use for our inputs? We're gonna solve for the maximum joint strength
So we want to add our input cells
So we're gonna say solve or set this value our joint strength to the maximum by changing
Variable cells, and we want to say the number of fasteners don't want to affect the factor of safety
Will add the shear planes and then will allow it to change the diameter
Will allow it to change the thickness and will allow it to change the width so
I guess we're assuming right now that we have our plate
material selected and our bolt material selected where we're not going to allow it to change the properties or the
ultimate shear strength or the tensile strength or anything like that of our materials that would chose them, but we're going to
allow it to change the dimensions of the
materials that we're using
So let's look here now. We have subject to the constraints
We're gonna add some constraints to our inputs, so we're gonna say the number of fasteners has to be less than or equal to two
we'll add that we'll say the number of shear planes should be less than or equal to two add that and
Let's say the diameter the bolt has to be less than or equal to
0.625 add that and then the thickness has to be less than or equal to 0.375
Add that
The width has to be less than or equal to two inches, and then we'll hit OK and see what we got
They also have some cool things. We can add
an integer
constraint to the number of fasteners because you can't have one and a half fasteners or anything like that so we have to add an
Integer constraint there, and then the same thing with the shear planes. We'll add an integer constraint to that and that
Should be pretty much it so
no, all we have to do is just hit the solve button and
It cranks out a value and then
It's gonna ask us. We want to keep the results first or the original values, and I'm just gonna hit OK and
So now it we can see that it adjusted some some of our values over here to give us the maximum
Strength and really this isn't that impressive
I mean all it did was it bumped up the number of fasteners a number of sheer planes
I mean it pretty much used the maximum of
Everything that we allowed it to change
So if we think about this even though this example isn't any interesting there are some things here that are kind of at odds
So there's gonna be an optimal point where
The let's say the number of fasteners is
Going to increase the shear strength always, but as you add fasteners you're poking holes into the plate
So there's gonna be an optimal area where you have a certain number of fasteners that is gonna
Give it the most shear strength
But also not so much that it starts to decline on the tensile strength so that's where this can come into handy
maybe if we would have bumped the
Number of fasteners up to let's say four it might have found the optimal number of fasteners to be two or three
but most likely not four because
Every time you add a diameter or fastener with a diameter of you know 5/8 of an inch
That's a lot of a lot of cross section that you're taking out of the plate, so that's where this comes in handy definitely oh
That was a lot there was a lot in that video it went pretty long
I hope you guys made it to the end, and I definitely. Hope you found it valuable
Definitely let me know if you have any questions I can always post follow-up video is to clarify anything
Let me know what you guys want to see next time
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Thank you guys so much. We'll see you next time