 Hello friends, I am Mr. Sanjeev B. Naik, working as assistant professor in Mechanical Department, Washington Institute of Technology, Singapore. In this video, I am explaining the design of welded joints subjected to an eccentric load. At the end of this session, students will be able to design welded joints subjected to an eccentric load acting in the plane of the weld. So, this is a welded connection of two plates having center of gravity of weld section over here as a G, whereas the load is acting at certain distance from this center of gravity. And this load is called as an eccentric load. So, when a welded connection is subjected to this type of eccentric load, according to the principle of applied mechanics, this eccentric load is replaced by an equal and similarly directed force P acting through CG as well as a couple is considered about the center of gravity and that is P into E. So, this eccentric load is replaced by a direct load P and a couple M acting about the CG. And that is why two types of stresses are induced in the weld section. One is because of the direct load P and one is because of the moment M. So, just pause the video and just recall the procedure of finding CG of different cross section because we consider the effect of eccentric load about CG. It is very important to locate the CG of different weld cross sections. So, now we have seen that the eccentric load is replaced by a direct load P which introduces a direct shear stress because the failure because of this load is a shear failure of the weld section. And that is why direct shear stress is introduced because of the force P and it is calculated as tau 1 is equal to P by A. P is the load applied and A is the area of the weld section. So, resisting area of the weld section is considered. For example, area of this weld plus area of this weld is a total area of weld section is considered as A. This is the way we can calculate the direct shear stress. However, the another shear stress is introduced because of the couple as a torsional moment. It is calculated over here as a secondary shear stress and that is calculated by considering the torsion equation. Tau 2 is the torsional shear stress is equal to torque into R by G where torque is a moment P into E into R by polar moment of inertia GG. So, this is the way we can calculate the secondary shear stress tau 2 in the weld section as shown over here by considering the effect of moment or a couple. And that is why the total resultant shear stress is tau and that is been considered by having tau 1 and tau 2 and that is why the equation becomes resultant shear stress tau is equal to square root of tau 1 square plus tau 2 square plus 2 times tau 1 tau 2 cos of theta. So, finally considering for safe design the value of tau as a permissible limit or permissible shear stress whatever unknown of design can be calculated by considering this equation. Now, calculating the polar moment of inertia for different sections of the weld is a tedious job and that is why we can refer the standard values from design data book. For example, if there is a single weld line about CG horizontal weld line then having dimension A as a length we can calculate its polar moment of inertia by this equation. Or if the C section as shown over here then the polar moment of inertia can be calculated by using this equation. So, this is the time of design. So, we can refer the standard data from design data book shown over here. Now, let us consider an example of welded connection between two plates it is connected by a C section like this one two horizontal welds having 100 mm line and one vertical weld having 200 mm line whereas it is subjected to an eccentric load of 50 kN at the end as shown over here. So, CG of this section will be somewhere here and the load is acting away. So, it is a case of eccentric load acting in the plane of the weld. So, important point is to locate the CG of this section. So, this is considered over here as a C section having three elements top then this is second element and bottom third element. So, now this is a symmetrical about x axis naturally the CG will be lying along x axis having y distance or y bar distance is 100 and 100. Whereas we have to locate x bar distance that is distance of x x from reference point g3. And the procedure is that we consider the area moment here as x bar is a1 x1 plus a2 x2 plus a3 x3 upon a1 plus a2 plus a3. So, this is a1 area first element. So, the area is 100 into h whereas its distance from the CG g1 to g3 the reference is 50. So, this is the way we get a1 x1 similarly a2 x2 a2 is area of vertical element. So, it is 200 into h whereas its CG and the reference CG as a g3 point they are one and same. So, distance is 0 considered over here. So, likewise we consider for third element and then we divide by total area of element 1 is 100 h, 200 h and 300 h. And we can find out x bar as a distance 25 mm. That means the location of g from g3 as a reference will be 25 mm. So, this is way along x axis with the reference g3 the g is located as 25 mm and whereas a y bar is already we have seen that is 100 mm half of this. So, this is the way we can locate CG of this c-section. Now, we can also calculate the theta angle of joining g with the extreme position of well section as a distance r. We can get tan of theta as 100 by 75. That means that this y bar is a vertical height 100 whereas the horizontal distance is calculated as 75 because total width is 100 out of that x bar is 25 mm minus and that is why remaining horizontal line will be 75 mm length. And that is why it is 100 by 75 is a tan of theta. And that gives a calculation of theta is around 53.13 degrees. So, this theta is 53.13 degrees. Similarly, we can calculate r that is a distance of CG to the extreme position of well section. And that has been vertical distance 100. So, 100 square plus 75 square its square root is a value of r and that r is 125 mm. So, this way we get theta r we can calculate. Now, we come to a problem and we know that the eccentric load acting in the plane is replaced by a direct force p and a couple m. The direct force p induces the direct shear stress tau 1 throughout the well section as uniform distribution we say and that is why tau 1 is equal to p by a. p is applied load which is 50 kilo Newton and total area of well section c is 100 into t throat area which is a failure area and 200 into 2 is a failure area that is a resisting the load and 100 t. So, that way total area is this much. So, if we take the value tau 1 now it is 125 by t Newton per mm square. This is the value of tau 1 we obtained by these equations. Similarly, we can calculate tau 2 which is induces because of the torsional shear stress and the equation for tau 2 is moment into r upon jj. r is already calculated the load is known e is eccentric distance which is shown over here and jj is a polar moment of inertia. So, jj of this polar moment of inertia of this section we can get directly from standard table as I said it by using this equation where a is 100 in our case and b is 200. So, substituting a is 100 and b is 200 by geometrical dimensions of well section we can calculate value of polar moment of inertia as something this value in terms of throat t side. So, from that we get tau 2 as I said is a load p applied 50 kilo Newton eccentricity is 275 because this well section is 200 here and then 100. So, total distance of load up to the back of this end is 300. However, x bar we have to make it minus because cg lies over here and that is why this is 25. So, out of 300 we minus 25. So, that remaining distance is eccentricity e and that is 275 and r distance already we have calculated is at 125. So, that way we will get a tau 2. So, getting tau 1 and tau 2 we require to know tau that is resultant shear stress. And already we have seen that the equation for tau is root of tau 1 square plus tau 2 square plus 2 tau 1 tau 2 cos of theta where tau is replaced as a permissible shear stress of weld material. So, tau 1 already we have seen 125 by t Newton per mm square and tau 2 is this much 457.43. So, substituting that we get tau is 640.287 by t is a throat size. So, substituting permissible stress given in the problem of 70 Newton per mm square the value of this throat size 9.40 mm. However, the size of the weld is normally specified as a leg size or size of the weld in terms of h where h is t upon 0.707 and that is why the size of the weld design value h is 12.93 mm. So, thus the size of the weld is calculated for a problem of eccentric load as we have seen it acting in the same plane and that is something 12.93 mm. So, this is the way by using the equations we can solve the problem and find out the required size of the weld. My references are design opportunities by Bandari and PhD design data book. Thank you.