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In this lecture I will discuss about the eccentric connections where load is lying in the planeof joint and using weld connection, here basically fillet weld will be used for connections.Now in case of load lying in the plane of joint two types of stresses will come into picture inthe weld. One is due to direct force which would be the shear, and another will come due toeccentricity and the stress will come due moment.So we have to make calculation of the stresses due to direct force and due to moment andthen we have to find out its equivalent forces that means resultant forces, and that resultantforces should be less than the weld strength. If the resultant stress is less than the weldstrength then our design factor is ok means design whatever we have designed is okotherwise we have to increase the weld size or length of the weld size or orientation meansthe weld length and width has to be changed and then again we have redesign it.
So coming to this picture if we see here that basically if there is a force P then moment willbecome P×e. Now because of that the stresses will be generated on the wild. Now if weld isconnected in this way, so in the bracket throughout the bracket if it is connected then we haveto find out the cg of the weld group.Once we find out the cg of weld group then we can find out what will be the stresses actingon each portion of the weld say for example in this point stress will be acting one stress willbe acting in this direction vertically downward due to direct force and another will be 90degrees to the radial distance from cg, and then we have to find out the resultant right.Now similar as bolt connection we have to find the resultant stress. But we know the stresseswill develop much more at the extreme point building stresses, so the radial distance of thispoint and radial distance of this point if we compare it will be more here that means thecritical point will be this point. Similarly critical point will be this one and this point and thispoint will be same because this distance and this distance are same, it is symmetric along thisdirection.Therefore the radial distance between this two will be same, the angle between two forceswill be same, therefore the stresses the resultant stresses at this point and this point will besame. So at any point any of these two points if we calculate then we can find out the criticalstresses of the section right. So in place of calculating entire weld length, in place ofcalculating in different position of the weld if we calculate only one point then we can findout the critical strength.
So critical strength will be developed only at this point because of its position. Now directshear stress in the weld can be found out from the following formulae
Where P is the vertical load, L is the total length of the weld and t is the throat thickness.Say for example if we draw this figure whatever we have seen and this is a column which isconnected by a weld. The bracket is taking a load of P right and this distance is suppose somedistance say e1 and this we have this distance we have so we can find out the cg of thisdistance and then we can find out the x and cg e.So we do not know what is the M value right now unless we know the cg of the weld groupright. So weld if it is made through this periphery then total length here this L, L is the lengthof weld, which can be calculated as
Where M is the moment which is calculated as P×e and r is the radial distance of the weldingpoint from the cg of the weld. Now unless we know the distribution of the weld we cannotfind out the value of e, e is the distance between load and the cg of the weld group
So before calculating the stress we have to provide certain distribution of the weld, then wecan find out the cg of the weld, then we can find out the stress how much it is coming due tomoment. So here Ip is basically polar moment of inertia of the weld and this can be calculatedas Ixx + Iyy.Now at any point the resultant stress is given by
Where θ is the angle between ps and pb.For critical condition developed stress should be less than the permissible stress in the weld.
Example:A bracket is subjected to a load of 50 kN and is connected to a stanchion by welding asshown in the figure. Find the size of the weld so that the load can be carried safely.
Solution:Firstly, we have to find out the cg of the weld group. So along vertical directions say in ydirection weld group is symmetric but along horizontal direction say x-direction we have tofind out the cg.Distance of the CG of the welded area from weld line BC,
Hence, provide 5 mm size of the weld.Now from minimum and maximum criteria means what should be the minimum size of weldthat from the codal permission we have to check. And then we have to see whether it isexceeding or not if it is less than the minimum then we have to provide the minimum. So forthis case the minimum thickness will be 3 mm so we can provide the size of the weld as 5mm.Now say for example, if the size of the weld is given and the distribution is given then alsowe can find out what will be the load carrying capacity of the weld joint. The reverse wayfirst we have to find out the strength of the weld then according to the size of the weld we canfind out what is the Ps and Pb then from that we can find out the load acting at a particulardistance P.So either of the way we can find out right. So the weld joint when load is lying in the plane ofjoint then how to design the joint that has been discussed in this lecture.
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