 Hello friends, I am Sanju Vinay, Assistant Professor in Department of Mechanical Engineering, Walshchand Institute of Technology, Solapur. In this video, I am going to solve some problems on design of welded joints. At the end of this session, students will be able to solve problems on design of welded joints. Pause the video and just recall the design procedure of various welded joints you have learned. This is an example in which two plates are joined together by means of flat weld. So these are the two plates which are joined by parallel flat weld as shown over here. And the leg dimension of the weld is 10 mm that is the size of the weld is 10 mm and the permissible shear stress at the throat section is 75 Newton per mm square. Determine the length of each weld L as shown over here if 15 mm weld length is required for starting and stopping of the weld run. So as far as the solution is concerned these are the two plates and these are the parallel flat welds which are subjected to load of 35 kilo Newton so P value is 35 kilo Newton and the size of the weld given in the problem is 10 mm and that is H and allowable shear stress allowable shear stress is given. So now to design this weld joint it is subjected to shearing action. So this weld and this weld they are subjected to shearing stress and that is why the strength of this welded joint is obtained as shearing area of one weld which is obtained as 0.707 into H into L multiplied by allowable shearing stress this is the strength of one fillet whereas there are two fillets so total strength of the joint that is load carrying capacity joint will be P is equal to 2 times 0.707 HL into tau substituting the value is given we obtain this and then from this we get length required is a 33 mm so the length required here for carrying the load effectively is a 33 mm. Here we have to add 15 mm length to this for starting and stopping of weld run because when welder starts welding initial sum of the length a proper weld is not been obtained and the same case happens when you want to stop the weld run and that is why some allowance is been kept at the start and the end and that is been given here as a 15 mm so total length required is 48 mm that is given as a 40 so this length L is required to be 48 mm this is another example where two plates are joined by fillet welds as shown over here which are transverse so transverse fillet welds are perpendicular to the load so these two welds are transverse and the plate width is 120 mm and thickness is 12.5 mm and these are joined by double transverse weld as I shown here the welded joint is subjected to tensile force of 150 kilo Newton so total force applied here is 150 kilo Newton tensile determine the length of the weld if the allowable tensile stress for the weld material is 120 Newton per mm square so we are given a transverse fillet weld two transverse fillet weld we have to design the length for this so to obtain the solution the data given is here load is given width of the plate and size of the weld is equal to size of the thickness of the plate so that is why h is also 12.5 mm now this particular transverse fillet weld is subjected to tensile failure so we have to consider the tensile strength of the joint welded joint so once again the failure area is a throat area and that has been taken as 0.707 h into L now there are two such welds are there so two times the area is total area of failure multiplied by allowable tensile stress because the failure stress is a tensile so we consider the safe limit that is allowable to decide the capacity of this welded joint to sustain the load and that we call a strength so strength of this weld required is this equation so we know that it has to be subjected to 150 kilo Newton load so we take this as a p-value and we solve it for L so L obtained here is 70.72 mm so this is the way we can design the required parameter of the joint that is the length and solution is obtained this is another example of angle so asymmetric section is considered here and this is an angle subjected to parallel welds having some length L1 and L2 as shown in the figure and the size of the angle is 200 mm height 100 mm width and 10 mm thickness as shown in this figure and the total force acting is 150 kilo Newton and it is passing through CG of the weld section and permissible shear stress is 70 Newton per mm square so what we have to do is determine the length L1 and L2 that is at the top and bottom so here we find that the total length if it is L that is L1 and L2 the load applied causes the shearing failure so parallel welds are subjected to shearing failure when such load is applied and that's why total shearing strength can be considered here as a p and that's why total shearing strength of the weld p is a throat area multiplied by tau that is allowable shear stress so that total load applied is 150 kilo Newton and that's why the total length L comprising L1 plus L2 is calculated first as a L value 303.09 mm now however this L has to be divided into L1 and L2 now to divide this one we take one more static law into account that is a summation of moment about CG0 here so we take this load into distance is a moment p1 into y1 and p2 into y2 however the strength p1 is proportional to the length L1 because throat size is same and material property is same so just load is proportional to L1 so the moment if I want to get to obtain the length L1 and L2 I get the moment as L1 y1 is equal to L2 y2 where L1 is a distance y1 is a distance shown over here which is nothing but 71.8 and L2 say y1 is a distance which is a 200 minus 71.8 and y2 y1 is a distance y2 is a distance 71.8 so substituting this y1 and y2 we can get the relationship between L1 and L2 and also we know that the total length L is L1 plus L2 solving this 1 and 2 we can come to conclusion that is what is L1 and what is L2 so here L1 obtained is 108 mm 0.81 mm and 194.22 28 mm is the value of L2 so this is the way required length is calculated now consider another example where 100 mm wide plate and 10 mm thick plate is welded to another steel plate by two parallel fillet welds so by means of double parallel fillet weld and a single transverse fillet weld so one single transverse and double parallel fillet weld so this way two plates are connected and the permissible tensile and shear stresses for the weld material and the plate are 17 Newton and 15 Newton per mm square respectively find the length of each parallel plate weld that means we have to decide this length L assuming the tensile force acting on the plate as static that is the load acting is a static so in such case being the strength of the plate and strength of the joint is same first we have to calculate the solution for strength of the plate is considered so as we know that the plate is also subjected to tensile failure the load is equal to failure area of the plate cross sectional area that is width into thickness and allowable tensile stress and that gives us p is equal to 100 is the width of the plate 10 is the thickness and 70 is allowable tensile stress as a given data and that's why load is 70 000 Newton so the strength of the plate is 70 000 and that's why the strength of the weld is also required to be 70 000 Newton total strength now that strength is been sustained by parallel plates as well as transverse plate so now if i assume that p1 is the strength of this parallel plate weld parallel plate is subjected to shear failure so considering shearing failure i can calculate the strength of this parallel plate as a p1 and that is two times the area of single weld because it is a double parallel plate weld so already we know that is 2 into 0.707 hL into tau and substituting this value we can obtain value of p1 in terms of L as 707 into L now we take p2 is the strength which has been for transverse where it is subjected to tensile failure so tensile area failure area into tensile stress is the value of p2 and that has been obtained as 49,490 Newton so this is the way we have calculated the strength of transverse plate and strength of parallel as well as the total strength of the weld and that's why we know that total strength of the weld is p p1 and p2 are obtained and if we equate this one we can calculate the required length L to be 29.01 mm so that's why the length required area is 29.01 mm as a solution we can obtain so these are my references thank you