Shear Flow Example – calculate connector spacing – Mechanics of Materials

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Shear Flow Example – calculate connector spacing – Mechanics of Materials

all right what’s up everyone welcome back to structure fruit learners and headed this video we’re gonna talk about shear flow again for built-up shapes and do an example problem to determine nail spacing of a wooden cross section but before we do that we talked a little bit about shear flow the thing you want to do when we talk about shear flow is remember the shear flow formula first which involve this little Q is equal to the internal shear on the beam so this would be the shear that we get from internal shear diagram of a beam structure this first moment of area which might be the most difficult thing to calc is the first moment of area associated with the location on the cross section that you want to compute and we’ll talk about that that’s why the most challenging part and this moment this area moment of inertia which is for the entire cross section now the units of this Q this little Q shear flow is force per length which you know something like Newton’s per millimeter or pounds per inch depending on where you are or what country or things you’re operating with in any case this shear flows really useful in built-up shapes especially when you want want to connect a bunch of different cross-sectional elements together and and when you want to determine how many nails you need to make sure that this whole thing acts together this entire beam cross section or or let’s say the glue the strength of the glue you need to bond the different layers together so that your cross section again behaves like a beam together so you might be wondering like what do these connections look like so hey I got some pictures right here I’ve got here on the top are pictures of wood cross-sections used in timber construction I don’t know if you can see the different layers here for this these have different layers of let’s say like two by fours or or one by ones or something just pieces of material connected together and in this case these things are all glued and I know they’re all layered and they take all the pieces of wood glue them together so that the moment of inertia is a lot stronger and you have this deeper beam without needing a you know a piece of wood that’s all single cross-section right you don’t need to find a giant 2 by 12 you could just make it right and here this is an i-beam and and here in this eye beam it’s glued over here and here and you take essentially what might be like a 2 by 1 or 2 by 4 and then have a groove on each and then add this oriented strand board this OSB which is really just a bunch of wood fibers glued together and it’s able to make a a strong beam out of cheap material it could be more cost effective the shear flow is useful just taking a bunch of different materials putting it together so that you have a beam that is either stronger or stiffer than what what it is that you could get out of a solid piece of wood which is typically much more expensive and then here what I have down here are different cross sections for here this is a carbon fiber beam this is out of carbon fiber CFRP and it’s got different layers you know this got all it’s got layers to because it’s basically layers of carbon fabric that are glued together using some sort of epoxy or another polymer and in here I’ve got this aluminum beam that’s that is made by with an aluminum plate in the middle for the web and then for angle shapes or for L shapes all riveted or bolted together and that’s pretty cool because you know that new f-150 is made of aluminum as well and all those bolts are riveted because you can’t weld aluminum in any case you can see the applications for designing the spacing of these connectors or the strength of the glue to hold pieces of material together so that you have a more cost-effective cross section and something that may be stronger and stiffer than a solid piece right all right so one of the most basic problems associated with shear flow is that you’re given let’s say the beam geometry in terms of length do you have the loading of the beam and you’re even given the strength of the bolt or the nail or the glue a lot of times which are asked to do is find the spacing of the connectors that you need or in the case of the glue it would be maybe like the strength of the glue that you need or you know how much glue ah with the glue that might be a little bit trickier but the spacing of the connector so that you know how many bolts or nails you’re going to need to make sure this beam acts like a beam the other way that this problem is set up is that you’re given the strength of the connectors the beam geometry the strength and sometimes they ask you to

find the loading what’s the maximum loading that can be applied to this structure and that’s a very common that’s an important question as well because a lot of times things fail at connections so let’s go ahead and just start the example problem so let’s say in my example problem the way that the example problem will work so for this example problem I’m going to look at a problem where I’m designing the st. the spacing of some nails I am going to be given a simply supported beam that has a uniformly distributed load so we’ll go with the simple loading in this case so that this is a first example and we know that the allowable shear force on a wood screw 650 Newtons what I would like to do is find the minimum required spacing of the nails or the screws and here is what my cross-section and my beam look like so in 3d my simply supported beam would look something like this let’s see if I can draw I will say we have like an eye shaped cross-section all right that’s as good is this going to get for me in this 3d looking eye shape cross-section I will say that it is simply supported at the end so if you can imagine we’ll say that there’s a pin support on this side oh I know it’s a large structure technically it should be supported right at the center but we will just put a pin support here you get the idea and a roller down here bail and this thing this I shaped cross-section is built up so it’s got a bunch of nails right so here in gray it’s got nails connecting these plates these three plates together to make an eye shape and we’ll say the nails go right on top right here or screws the wood screws go on top and there’s also wood screws the same on the bottom as well and what we’re trying to do is find the spacing of these screws and the cross-section looks like this so it’s 150 millimeters for the web 30 millimeter thick flanges and let’s see the width of this thing is 100 millimeters and the width of the web here is 25 millimeters and there’s a uniformly distributed load oh that’s going to get crazy here but it’s a uniformly distributed load which I will just draw it’s a line load that looks like this and hopefully this drawing is not getting too convoluted but here BAM this is a uniformly distributed load that means it’s uniform or constant all the way across and this thing is 2 kilonewtons per meter all right so let’s say I’m doing this problem I look at it it’s on the final and I’m like Oh Frank I have no idea how to do this problem I’ve never even heard the terms before and you know one of the things I would do is because this whole class has been or this whole topic of mechanics of materials about taking internal loading and converting into some sort of stress or transforming it into something else I would look at this thing and say hey you know when I got a beam and I’ve got some loading on it I should at least try to draw the shear and maybe the moment diagram because those are the internal loads that I need to convert into something else so I can design this beam or figure out something or solve the problem so the first thing I would do is you know do my statics and none of this is stuff that you learn you learn in mechanic’s this is all stuff you would have learned in the prerequisite for mechanics and that would be just calculate the reactions draw the shear moment diagram and I’m going to do this fast because this is a a simply supported beam and it’s not a big deal if you feel like you need practice on shared moment diagrams this is probably one of the first places you want to start it and you want to go ahead and practice that on your own and in this case I just need the shear diagram because the question only asks for the spacing of the of the screws and those screws are associated with shear flow and shear flow is associated only with internal shear loading so here it is here is what my beam looks like if I draw it with just stick figures and because the loading and the geometry of the beam are symmetrical I know that the rear and also the reactions or the external reactions here are also everything the loading the external reactions and the geometry of this beam are symmetrical that means I know that the rig that my reactions are going to be the same or equivalent so that means each one let’s see the resultant is 2 times 3 which is 6 kilonewtons we’ll have a 3 kilonewton reaction and then I can go ahead and draw my shear diagram vertical lines that the discontinuities ma’am boom right here here’s my shear kilonewtons and let’s see I go up 3

kilonewtons and across so this is a C up 3 linear because I have a uniformly distributed load and going left to right BAM there’s my shear diagram plus 3 here minus 3 here and my maximum shear is going to be at the support so those or that’s the region I’m interested in the length of the beam if my connectors fail or my beam fails in shear it’s likely to happen at or near the supports all right so there’s my max shear some V in this case for this problem V is equal to 3 kilonewtons alright good so I finished my statics and now I want to apply the basic design relationship for one of these nails or screws which are going through my board they are going through like this they’re coming in from the bottom and from the top and connecting those boards together so that I can have an eye shape and and in order for us to look at this you know we have to understand the basic design relationship that’s applied and so in order to do that I want to draw if you will the web or half of the web of this beam and focus on one nail okay i’m or one screw and so if i look i just called this 3 and i’ll call this the b dr and if I look at one nail of this cross-section so if I could uncover one of that top flange board and if I drew this one little space right here so if you can imagine I’ve taken off the top flange and what you see here is or one single screw like this into the top of the board and it continues here and it’s BAM sharp and it’s connecting that top board to the web aboard this screw is responsible for certain or connecting a certain region of the beam just like a zone defense you know you’re responsible for a certain area right and so this is responsible if you will half way to the next nail and half way to the other nail so if I go back up here real fast like this screw right here if that’s the one I drew this one is responsible for halfway and halfway or in between the other connector so it’s responsible for connecting that zone right that zone of the beam together or that flange to the web in that distance assuming everything is you know uniformly spaced and everything that distance for this problem would be halfway to each and in this case if everything is spaced the same distance each of these would be s over 2 s over 2 so the area or the length of responsibility is actually this full length s the spacing of the screw is that is that region of responsibility the force that gets applied right on one of the screws which I will just boom go like this right here is related to the shear flow between the layers and here this is this Q right here there’s that shear flow and that applied force is equal to the shear flow which was in units of Newton’s per meter or Newton’s per millimeter or force per length and this is a flow right so a force per length this Q and in order to turn this into a force I’ve got to multiply it by the spacing or really the length of responsibility of my connector it is another way of putting it and I’m going to add also right here the number of shear planes because if I have because sometimes that shear flow or that you know that zone is that zone or that area of responsibility of the connector has more than one shear plane like in this case here we’re looking at a problem that has only a single shear plane right here and then a single shear plane right here but it depends on the area that you select for that capital Q that first moment of area and we’ll talk about that in a second so here this is my force applied and my basic design relationship that says this is force applied on the on the connector or the screw here has to be less than or equal to whatever is allowed and if you can look at the mechanics model for this for a nail this is a single lap shear connection and we’re getting the way we’re going to look at this nail is like a single lap shear connection and we’re going to use an technically what we do is we cut it and we use an average shear stress across that area but nonetheless

if you can understand this part this basic design relationship you’re you’re pretty much home you know I’m saying it’s pretty crazy alright so here for the shear flow I’m just going to substitute for that shear field that’s V Q over I times s over the number of shear planes and this number of shear planes again is related to this capital Q that we choose for this problem this F allow in this case would be 600 50 Newtons so let’s see what do we know what do we not know we know capital V yes we’re trying to figure out this spacing all right and this number of shear planes is dependent on this first moment of area and then we can calculate the moment of inertia it’s just of an i shaped cross-section so what we’re really looking at our geometric properties next so 1 2 3 4 geometric props and in particular I’m talking about Q and in particular I’m talking about the moment of inertia and this first moment of area capital Q all right so to do that let’s talk AB let’s go ahead and calculate the moment of inertia of this cross-section which is this thing right here make a copy of this and bring it down oh yeah there it is and I you know this is a symmetrical cross section so I know that the centroid the geometric centroid is right in the middle bull is to calculate the moment of inertia about the horizontal axis here or the neutral axis in this case I can take let’s see the I can do a shortcut and take the big box if you will right here which is 1/12 the base 100 millimeters times the height which is 150 plus 30 plus 30 so 210 millimeters cubed minus the voids which is this orange area and this orange area and I can subtract the voids in this problem because the centroid of the voids is also on the centroid of my larger or from my entire cross section so this let’s see there’s two of these voids and I’m going to have 1/12 one of these rectangles so this is 100 minus 25 that’s 75 so this this the total this distance right here is thirty seven and a half millimeters so I have 30 7.5 millimeters times the height which is 150 millimeters cubed and now if I go ahead and I calculate this thing the moment of inertia is fifty six point oh eight one times ten to the six millimeters to the fourth so there is my moment of inertia about the horizontal axis now for my first moment of area ah this is the challenge usually with shear flow so here when I calculate capital Q I have to make sure or in essence I’m wanting to calculate the shear flow or shear stress in a way where my connectors are okay so that’s one thing that’s one thing to start I have to look at the location where my connectors are or where the elements of my cross section are being connected so that would be here and here okay but I’m not going to put a red line here now one choice I could just choose this point right here and I have to when I calculate my capital Q that equation for capital Q is the sum of a prime times y bar Prime and a prime or this area for my first moment of area is all the area above or below where I want to calculate my shear flow in this case so if I chose this as my only shear plane then I would have this as my area this would be a prime and the NY bar prime is the distance I’ll do this in purple is the distance from the centroid of this area to the neutral axis or I should say it’s a distance from the neutral axis to the centroid of a prime this is this Y bar prime and so if I chose this area right here as I know that I’m interested in the connection I choose all the area above then this in this case this a prime is equal to 100 times 30 millimeters which is 3,000 millimeters squared and y bar

prime so my Y bar prime is the distance from the neutral axis to the center of this which is 105 millimeters minus 15 millimeters which takes me to 90 millimeters and so for this case my capital Q would be three thousand millimeters squared times 90 millimeters which is equal to two hundred and seventy thousand millimeters cubed which I could also write as point two seven zero times 10 to the 6 millimeters cubed all right so that’s one way and in this case right here this is the first moment of area associated this with this one shear plane so based on my selection here I have only gone with one shear plane now check this out I could alternatively I could look at this cross section and say hey I want to solve it considering both shear planes both of these connectives connections and let me do this again so if that were the case I could do again everything above or below but I’ve got to have some boundaries it’s like I can only go from shear plane to share plane if you will right here so I could choose as one possibility and this drawing is going to get messy but I will choose this in orange here I have I could choose this area as potentially my a prime so let’s say I did okay so I have this a prime so I’m choosing Q this alternative Q with this a prime and this a prime times y bar prime and I’ve got a single element this time it’s associated with two shear planes and in this case here I’ve got let’s say boom – boom right there and this area is 25 millimeters by 150 millimeters but the centroid of this bad boy is right here it’s right on the neutral axis the centroid of this orange area is right on the neutral axis so oh snap right that’s that’s just that’s not going to work for us because look when I look for y-bar Prime the way that this is defined is the distance from the neutral axis to the center of the element and in this case that’s stinking zero which is equal to zero and that would say capital Q is zero which means the shear flow is zero which means hey I don’t need any nails to make this thing work together which makes absolutely no sense right right right if you have a boat three boards and have no nails you don’t have a beam son you just got three boards all right so that’s why that’s just that’s just not right and that’s not going to work for us so this anytime you get the first moment of area equal to zero you know there’s something wrong right so the other choice that I could have is everything if you will beyond beyond the the shear planes and that would mean that I would choose this area here this blue area again so I’ll shade again even darker and I would choose the bottom here and when I calculate this capital Q the sum of a prime times y bar prime would I do this I would have aa this area here we already know is 3000 millimeters squared so I’ll just write it it’s 100 millimetres times 30 millimeters and the distance from the neutral axis to the center of this element which is right there that we found was 90 millimeters and if I add the other term this area down here this area is the same it’s also 100 millimetres times 30 millimeters and the distance from the neutral axis to the center here is also 90 millimeters which would just give me double this thing right here which would be 540 thousand millimeters cubed and that would be my first motive area and remember this thing is associated with two shear planes it was our choice all right so now I definitely feel like I’m balling cuz dad you know I count all the geometric properties I got the internal shear force I even have an understanding of the number of shear planes that are associated with the way I’m going to solve it and I’m going to go ahead and calculate for the spacing here this s so let’s see let’s say I take the first way that I calculated this first moment of area and the one associated with one shear plane so for the one shear plane area you know when I go ahead and I plug in chuck numbers let’s see that shear force here and this relationship was three kilonewtons that first moment of area was 0.27 zero times ten to the

sixth millimeters cubed the moment of inertia was fifty six point zero eight one times ten to the six millimeters to the fourth and for this first moment of area I had one shear plane less than or equal to 650 Newtons which I could convert into kilonewtons and this thing is just point six five kilonewtons and when I you know look at how convenient this is the 10 to the six cancels out millimeters cube per millimeter you know I know that this kilonewton that kilonewton is going to cancel out I’m going to be left here boom that millimeters cute I’m just gonna be left with millimeters which makes sense for the spacing and when I solve this I should get a spacing that’s required is less than or equal to 45 millimeters Wow look at how nice that number turned out so this is with one shear plane and look and you probably wanna know hey does it work with that two shear plane thing and you know and I know it’s going to work because I had this five hundred forty thousand and this bottom number is going to become two shear plane so here if I do this again for the two shear plane case again the same internal shear force three kilonewtons that first moment of area this time was 0.5 four zero times 10 to the 6 millimeters cubed divided by the moment of inertia 56.0 eight one times 10 to the 6 millimeters to the fourth times s divided by this time two shear planes less than or equal to point six five the Newtons again and here you can see already when I divide this point five four by two I’m going to get the exact same numbers as that first case and I would get that the spacing required has to be less than or equal to forty five millimeters and if I were designing this and trying to stay within my labels I made sure something like forty millimeter spacing just to make sure I have a little bit of safety right all right I hope this was a useful introduction to shear flow for built-up shape so to free