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Module 1: Eccentric Connections

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Eccentric Loading in Plane of Bolted Joint – Concepts

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Video 1
In this lecture we are going to discuss about eccentric connections. In last few lectures, wehave discussed about the different types of connections but mainly those are concentric. Nowin case of eccentric connections c.g. of the connection and c.g. of the load will be in differentposition so if it does not coincide then eccentricity will develop and because of eccentricityextra moment will come into picture and because of the moment extra stress will develop.So therefore we have to design the joint taking consideration of the direct load as well as dueto eccentricity. In case of eccentric joint we have different type of joint like when load islying in the plane of joint it will be one type of eccentricity again that can be designed byweld connection as well as by bolt connection and similarly when load is lying in theperpendicular to the plane of joint then another type of eccentricity come into picture meansanother type of load reaction will come and that also can be designed using bolt joint andusing weld joint. So basically four type of joints will be considered for design.
Now I will discuss today mainly on the bolted type connection where load is lying in theplane of joint. Say for example if we see above, there is a typical example of load lying in theplane of bolted joint, the cg of the bolted joint and the load are acting in different positions.Suppose the load is P with the eccentricity of e then the additional moment will come intopicture that is P×e. At the joint we have to calculate what are the load coming at each bolt,accordingly we have to calculate the bolt strength and we have to see whether this is safe ornot.
If we draw the above figure we can see that a column is carrying some eccentric load whichis coming from beam or say gusset plate. Now P is the load so the eccentricity will be e. Sothe additional moment due to eccentricity will be P×e. If n number of bolts are there, thenload in each joint will be P/n assuming that the loads are distributed equally to each joint.So if we consider that every bolt is sharing equal load then load at each bolt, F =P/n and thisis the load which is coming due to direct load. Now another load will come because ofmoment.We can say r is radial distance as shown in the above figure. So for each bolt there are twotypes of forces will be acting, one is called Fa due to direct load and another is called Fm dueto moment. Now we have to find out what is the resultant of this loads and these resultantload will be the load acting on the bolt. The resultant force can be calculated as follows
So we have to find out the resultant force for each bolt, where the extracted force for eachbolt will be different and we have to see which one is the most critical one.
So if we start with the load lying in the plane of bolted joint then we can see that this case canbe made equivalent to M (M = Pe) + = direct load as shown in the above figure. So we aredividing into two parts, one is moment due to eccentricity another is direct centroid load.
So this two will be added and we have to find out the equivalent load. So this is what I havediscussed that if we can find out the direct force that will be Fa and this Fa will be simply P/nand similarly we can find out moment, which will be P×e where P is the axial load and e iseccentricity. So if we add this direct force and moment we can find out the combination ofthis.
Similarly, here resultant will be the simply adding and here we can see that this will be againadding so out of this six bolts in this particular case will see the extreme case will come eitherthis one or this one because this bolt and this bolt have similar magnitude because the thetawill be same here so resultant force will be same because distance radial distance from hereto this bolt and to this bolt is same.That is why resultant force will become same for this case however in this case it will besimply adding but in this case we will see it will be reduced, this Fr value will be less andhere it will be Fa minus Fm so it will be much more less than this and similarly this is equalto this one so it will be also less so what we can see that we have to check only two boltsmeans if we can check only these two bolts then we will be able to understand which one isthe critical bolt because this bolt and this bolt are same so we do not need to check one meansif we check the extreme one that is sufficient and also we have to check this one then we haveto find out the critical one
So from this we can find out shear due to moment that will be Fm that will be varying with rbecause in this case if we see that the moment will be more if the distance is more. But Fmwill be less in this direction in this bolt, so it varies with the distance. So we can write as Where, r is the radial distance of bolts from the cg bolt.We can write
So this is how the force on a particular bolt can be calculated from this formula where e is theeccentricity and r is the radial distance of bolt to the cg group of the bolt, so Fm the force on aparticular bolt due to moment can be calculated. And Fa for each bolt can be calculated asfollows
Where n is the number of bolt
Example: Calculate the safe load F that can be carried by the connection as shown in thefigure below. HSFG bolts of grade 8.8 with 20 mm diameter are used. Assume the thicknessof the bracket plate as 12 mm and column used is ISWB 350. Assume no slip is permitted andslip factor (μf) as 0.5. All dimensions in the figure below are in mm.
 
Video 2
Now let us go through one example right, so this example let us go through. Say calculate thesafe load F that can be carried by the connection as shown in the figure below I will show thefigure. HSFG bolts of 8.8 grade with a 20 mm diameter are used. Assume the thickness ofbracket as 12 mm and column used is ISWB 350. Assume no slip is permitted and slip factorMu has been given that is Mu f as 0.5. So all dimensions in figure are given in millimetre.
So if we try to solve this problem then it will be clear to us say in this problem the figure isgiven like this say it is a ISWB 350, is connected with a bracket and it has a force F this is Fand it is situated at a distance of 180 millimetre. Now bolts are placed in this way six boltsare placed with a pitch distance of 60 mm and edge distance of 40 mm right. Say this is boltnumber 1, this is bolt number 2 and this bracket thickness is 12 mm and this is ISWB 350section.
So now the force due to direct load in each bolt will be Fa is equal to P by 6 right.
So 0.599P is equal to 55 we can equate and then we can find out P as 55 by 0.599 and thiswill become 92 kilonewton that means to withstand the load on this arrangement of bolt weneed means we can apply maximum value of P as 92 kilonewton at a distance of 170 mm. Sothis is how if we know the distribution of bolt and the loading distribution then either we canfind out the load considering the maximum capacity of bolt due to shearing and bearing or wecan apply certain load and we can check whether the existing bolt in what way it isdistributed are safe or not means if we provide certain load then we can check what is theload coming on these bolt or in the critical bolt what is the load coming and that load if it ismore than the bolt strength then it is unsafe otherwise the bolt is safe.So this is how in other way we can do either of the way we can do, either we can find out thecritical load or we can find out whether the section is mentioned the bolt connection group issafe or not. So in this way the calculation can be made when the eccentric connection in asense that load is lying in the plane of connection if it is then I can find out the sectionwhether it is adequate or not.