 Hello friends, I am Mr. Sanjeev B. Naik, Assistant Professor in Department of Mechanical Engineering, Waltham Institute of Technology, Solapur. In this video, I am explaining design of welded joints subjected to bending moment. At the end of the session, students will be able to design welded joint subjected to bending moment. So, this is a cantilever beam having rectangular section, cross section, which has been welded to the support by two horizontal fillet welds as shown over here. And an eccentric load P is acting at the end of the cantilever beam having eccentric distance E. And this causes the eccentric force causes the bending moment. So, according to the principle of applied mechanics, the eccentric force P can be replaced by an equal and similarly directed force P acting through CG of the weld section and also a couple that is a bending moment. And that is why two stresses are induced in the weld section. One is because of this force P which is acting in the plane of the weld section and that is why it causes the shear failure. So, it introduces the shear stress whereas this bending moment introduces the bending stress. So, this cantilever beam of welded joint is subjected to combined shear stress and bending stress which are calculated as, so this is a primary shear stress which is calculated as the one load upon area P by A where A is the total area of the weld section is considered. Whereas the moment M causes the bending stresses in the welds and we know that the bending stress is equal to M y by I. M is a moment y is a distance from neutral axis and I is a moment of inertia of all the welds. Also we know that I upon y can be substituted as the section modulus z and that is why bending stress can be also calculated by M by z. So, this is the way we calculate sigma B that is a bending stress and tau n as a primary shear stress and that is why the resultant shear stress can be obtained by principle shear stress theory that tau is equal to square root of sigma B by 2 whole square plus tau 1 square. So, this is the way we can calculate the resultant shear stress induced in the weld material and that is why for safe design this value of tau is equated to the permissible value and the size of the weld can be calculated. So, pause the video for a certain time and just recall how to calculate section modulus z of different cross section. Now, here is a table which is given as a standard table from design data book where z value that is section modulus value for different cross section of the welds or different sections of the welds are being given. For example, if it is a circular weld then we can calculate this value of z is pi t d square by 4 t is a throat size because we know that the failure occurs along the throat and that is why throat size is important and d is the diameter of the circular section. So, this is the way for different standard sections of the weld we can get the values from the table. Now, let us consider an example where a rectangular beam is being welded to a vertical support by two parallel two fillet welds vertical fillet welds as shown over here and subjected to an eccentric load at certain distance. So, as we have seen that this type of loading eccentric loading causes a direct force and the bending moment which causes the direct or primary shear stress to 1 which is calculated as a load by area and the area is the total area of the throat calculated as 2 into 15 into t because length of this weld one weld is 50 mm whereas the size cross section is a t throat size. So, two welds are there so 2 into 15 into t so total area. So, considering this and a load 3 kilo Newton we calculate the primary shear stress tau 1 as a 30 upon t Newton per mm square. The second effect is the bending moment which causes the bending stress. So, bending stress is m by z and z value for this section is obtained as a standard tb square by 3 t is a throat size b is a length of the weld 50 mm so that is what considered. So, we calculate the bending stress by this equation and it is obtained to be 540 upon t Newton per mm square. So, this is the way for given problem we are getting primary shear stress tau 1 and sigma b calculated then just we have to calculate maximum principle shear stress that is a tau by using sigma b and tau 1 in this equation we can calculate the value of tau and that is what resultant shear stress it is a 271.66 upon t Newton per mm square. Now, using that maximum shear stress value it is restricted to permissible shear stress in the weld for the safe design. The permissible shear stress given is 50 Newton per mm square so by equating this value we get the size of the throat t is equal to 5.43 mm and then actually the size of the weld which is been specified is a leg size and that is noted as h. So, h is equal to t upon 0.707 and that results into a size of 7.68 mm and this is what required to design the weld to find the leg size of the weld for a given problem. Now, let us consider one more example where the rectangular beam is subjected to eccentric load but here the weld is a box type section that means there are two horizontal welds and two vertical welds according to the rectangular section of the beam. So, this length is 150 mm this length is 100 mm. So, once again with the case of cantilever beam subjected to a bending load and that is why the weld section is been subjected to combined primary shear stress and bending stress. So, primary shear stress tau 1 is calculated as load upon area here the area is a total area of the weld. So, two horizontal weld area plus two vertical weld area is to be considered and that is 2 into 150 times t plus 100 t results into something 500 t substituting that we get tau 1 primary shear stress as 50 by t Newton per mm square whereas the bending stress induced is calculated by bending moment m upon z where z for this section can be obtained from standard table and that is been given some b a into b square by 3 bracket t where b is 150 this length and a is 100 substituting these values the value of z is obtained and then the value of bending stress is obtained as 550.55 upon t Newton per mm square. So, this is the way we get tau 1 and sigma b and then we can calculate maximum principle shear stress in the weld tau as square root of sigma b by 2 whole square plus tau 1 square and substituting the values we get tau equal to 282.24 upon t Newton per mm square based upon this we can decide the size of the weld by using this value of tau permissible value as a 75 Newton per mm square. So, that gives a safe design and that is why using this equation we calculate the t throat size and then the size of the weld is t upon 0.707. So, naturally it results into 5.32 mm or say 6 mm. So, this is the way we design this cantilever beam with the welded section or weld section is designed subjected to bending moment. One more example considering the round weld section where the round bar is welded to vertical column by a round weld and subjected to bending moment. So, here direct or primary shear stress is obtained as a tau 1 load upon area. Area is a circumferential area of this weld which is pi d into thickness throat thickness and p is applied load. So, that is why we find that tau 1 obtained is 63.66 upon t Newton per mm square. The bending stress is obtained by bending moment sigma b is m upon z. The z for this circular weld section obtained from table standard table is pi t d square by 4 d is the diameter of the circular weld. So, substituting these values we get the value of z in terms of t and then by using equation of sigma b we calculate bending stress something as 1018 upon t Newton per mm square. So, once we get bending stress and shear stress we calculate the resultant shear stress tau and that is been obtained by the equation sigma b by 2 whole square plus sigma tau 1 square and that results to be 513.21 upon t Newton per mm square. So, once we get the resultant shear stress value we equate it to the allowable limit 95 Newton per mm square and by using that equation we get throat size as 540 mm 5.40 mm is a throat size and that is why we calculate the size of the weld h as tf 0.707 and that results to be 7.64 mm. So, this is the way for this round bar welded to the vertical support subjected to bending moment the size of the weld is calculated to be h is equal to 7.64 mm. So, these are my references. Thank you.