Hello, I am going to discuss about design of Fillet welds. In last lecture I havediscussed about the fillet welds, means different parameters used for fillet welds like what isthe effective length of the weld, what is the total length of the weld, what is the size of thewelds, what is the effective thickness of the weld etc. We have also discussed to find out themaximum allowable throat thickness of the size and minimum size of the weld on the basis ofthe plate thickness and finally we have discussed about the design strength of fillet weld,right.
This is the same formula to calculate the design strength of the fillet weld,
Lw = length of weld in mmfu= ultimate stress of weld in MPate = effective throat thickness = 0.7Sγmw = partial safety factor= 1.25 for shop welding and= 1.5 for site weldingS = size of weld in mmWe generally use right angle and for that it is 0.707S and on the basis of this, we will gothrough one workout example. If some load is given then how to find out the length of theweld and how to distribute the length of the weld in different site that will see through thisworkout example.
Example:A tie member of a roof truss consists of ISA 100×75×8 of Fe410 grade, is welded to a 10 mmgusset plate. Design the welded connection to transmit a tensile load, T. Assume connectionare made in the workshop.
So here the thickness is given means thickness of the gusset plate is given 10 mm andthickness of the angle is 8 mm, so from these two we can find out the size of the weld right.So this is one thing second thing is that this is an angle section so its cg distance will not be atthe middle not at will be at the centre. So that means the weld length will not be distributedequally in upper side and lower side so, the design of weld has to be made in such a way theequivalent strength passes through the cg.
Solution: Minimum weld size = 3 mm [Table 21, IS 800]Maximum weld size = ¾ ×8 =6 mm [clause 10.5.8.2, IS 800]Therefore, Let us adopt 5 mm thick fillet weld.Throat thickness, te = 0.7 × 5 = 3.5 mmFor ISA 100×75×8, Gross area, Ag = 1336 mm2 , Cz = 31mmFull strength of the angle,
So whatever you are getting you have to represent in terms of drawing so that engineer canunderstand at the site. Here one thing we have to remember, we provide the effective lengthsuppose length whatever we are providing is effective length and engineer has to add to this that means the size of the weld it has to add and then it has to fabricate right. So this is oneexample.
Example Design a suitable fillet weld to connect web plate to flange plate and flange plate toflange cover plate of a built-up girder as shown in the figure, for the following data. Assumeshop welding.Web plate: 1200 mm × 12 mmFlange plate: 450 mm × 20 mmFlange cover plate: 350 mm × 16 mmMaximum Factored shear force: 1600 kN
This is how we can check the joint whether it is ok or not right. So in todays lecture what wehave seen that two type of problem we have come across and we have seen how to calculatethe design strength of the weld or how to design the weld joint. And in first case we have seenhow the distribution of the weld will be done for an angle section because in angle section cgdistance is not at the centre so we have to make the weld connection in such a way thatstrength of the weld connection coincide with the cg of the joint. So this is what we havedone, this is all for todays lecture, thank you.
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