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Video 1
In this lecture the design procedure of bolts will be discussed. Bolt may be two types whichare commonly used, one is ordinary black bolt and another is high strength friction grip bolt.Now at first I will discuss about the design procedure of ordinary black bolt. As we know that the design procedure of bearing type of bolt and friction type bolt are givenin clause 10.3 and clause 10.4 of IS: 800-2007 respectively. So while we will see the designprocedure I would suggest the participants to follow the codal provisions and open the clause10.3 while designing the ordinary black bolt. When we are going to design the ordinary blackbolt, we will see the two type of joints will come across openly, one is lap joint where the twoplates are overlapped together at a certain length and another is butt joint where two platesare in same plane and joint with some cover may be single cover or may be double cover. Sothese two types of joints will be covered in this lecture. Now for designing of ordinary black bolt we will see that what are the failure criteria andfrom that failure criteria what are the strength like it may fail due to shear. We know thatblack bolt may fail at shear plane when two plates are connected through bolt/bolts. So whilewe will go for design of that bolt first we have to find out what is the strength of the bolt inshear then we will go for bearing strength of bolt and so on. Bolt connection may fail alsodue to bearing of bolt or due to tension in the bolt or due to tension in plate or due thecombined shear and tension. So we have to design the connection against the lowest strengthof bolt/plate. In case of plate failure in tension, we have to find the net strength of the platewhich is reduced due to the presence of hole. So that also we will check and we will seewhich one is the least strength, which will be the bolt strength. So we will try to find out thebolt strength for different failure criteria as discussed above. Next let us come to design of ordinary black bolt. Now as per clause 10.3 of IS 800: 2007, nominal capacity of bolt is given by Where, Vnsb = nominal capacity of bolts in shearfub = ultimate tensile strength of boltsnn = no. of shear planes with threads intercepting the plane.
ns = no. of shear planes without threads intercepting the planeβlj = reduction factor which allows for overloading of end bolts that occur in longconnectionsβlg = reduction factor for large grip lengthsβpkg = reduction factor for packing plates thicker than 6mmAnb = net tensile area of bolt to be considered at the root of the threads = 0.78×π×d2/4Asb = nominal plain shank area of the boltSo for different type of bolts the ultimate tensile strength of bolt will be different that we canfind out from the code. Now if you see the above picture, where entire bolt grip is shown, you can identify nn and ns,means the shear plane with threads and without threads intercepting the plate. So now I am going to discuss about the design of ordinary black bolts, this is given in clause10.3 of IS: 800. Now as I discussed earlier that the design of ordinary black bolts has to bedesigned under shear, tension, bearing of bolt and tension of the plate. So we will go throughone by one and we will see what are the codal provisions made and accordingly we will try tofind out the design strength of the bolt under shear, under bearing and under tension. So using the above formula given in clause 10.3, we can find the nominal shear capacity ofbolt (Vnsb) where fub is the ultimate tensile strength of bolt which depends on the materialproperty of the bolt. So what type of material property we are going to use depending on thatwe can find out the value of fub. As we discussed earlier about the number of shearing plane with and without threads, herefrom the above figure we can found the shear planes as ns= 2 and nn=1. Now we have to findout the Anb and Ans. Ans is the cross sectional area of the plane shank of the bolt and Anb is thecross sectional area of the threaded portion of the bolt. Now we know the cross sectional areaof the shank portion will be πd2/4, where d is the nominal diameter of the bolt but when weare going to calculate the net area of the threaded portion we will reduce to a certain extentwhich is suggested by the code as 0.78 times the cross sectional area of the shank area, thatmeans this will be reduced to 0.78× πd2/4. Now we will discuss about the reduction factors which will be used to determine the nominalshear capacity of bolt. βlj is the reduction factor which allows for overloading of end boltsthat occur in long connections. If a connection is quite long, then there will be a factor whichwe have to multiply with the capacity whatever is coming this factor is called βlj. Similarly,βlg is the reduction factor due to large grip length, if the grip length is large that means platethickness is high or several plates are given then grip length increases, say suppose I amincreasing the number of plates and joining with a bolt. So if grip length is more then wehave to multiply a factor which is called reduction factor for large grip lengths. Now βpkg isthe reduction factor for packing plates. Suppose we have a butt joint where two plates havedifferent thickness and we have to join with certain number of bolts. So for this case we mayhave to provide packing plates. These packing plates when we are going to consider we areusing a reduction factor for packing plates however if this packing plate is thicker than 6 mmthen we have to multiply a reduction factor of beta βpkg.Now the design shear force, V V dsb nsb mb = g where gmb is the partial safety factor, this gmb isgiven in IS code in table 5, in table 5 you will get a different type of partial factor for boltedconnection, for welded connection and for different cases the value of partial safety factor ofthe bolt or rivet bolt or welded according to the material it has been given. Now this gmb weconsider here as for bolt we use 1.25, right. Now let us come to the reduction factor how do we calculate βlj which is the reduction factordue to long joint. So long joint means if the length of joint become more than 15 times ofnominal diameter of the bolt is consider as long joint and the reduction factor βlj can becalculated as follows: Next we will calculate the value of βlg which is reduction factor due to large grip lengthwhich is consider if the grip length, lg is more than 5d, where d is the nominal diameter of thebolt. Another reduction factor is for packing plates that is βpkg which can be calculate as follows: If thickness of packing plates tpkg > 6mm, then
Video 2
We have seen first is that bolt in shear now for bearing as I told that bolt may fail due toshearing effect and due to bearing effect and bolt may fail due to tension also the joint mayfail due to tension of the plate tension failure of the plate that has also have to be consider.Now the nominal bearing strength of bolt (Vnpb) can be calculated from the following formula: Where,Vnpb = nominal bearing strength of boltfu = ultimate tensile stress of plated = nominal diameter of boltt = summation of thickness of connected plates experiencing bearing stress in samedirectionRemember, earlier we have calculated the ultimate tensile stress ultimate stress of bolt inearlier formula but here it is ultimate tensile strength of plate because it is bearing on plate.Now another factor is a constant Kb which can be calculated as follows: Where, fub = ultimate tensile stress of boltsd0 = diameter of bolt holep = pitch of fastener along bearing directione = edge distanceNow we can find out the design shear force as Where gmb is the partial safety factor of bolt and this value is 1.25 which we can find outfrom table 5 of IS: 800-2007. Now considering the failure of bolt due to tension we can calculate the nominal capacity ofbolt in tension as follows: Where, Tnb = nominal capacity of a bolt in tensionfub = ultimate tensile stress of boltsAn = net tensile stress areaAsb = shank area of boltfyb = yield stress of boltγm0 = partial safety factor = 1.1 (table 5 of IS: 800)γm1 = partial safety factor = 1.25 (table 5 of IS: 800)Now the design tensile force, T T db nb mb = g and we know γmb is basically 1.25.So what we have seen that the strength of bolt due to shearing, due to bearing and due totension we have calculated.Now another aspect is that the joint may fail due to tensile failure of the plate. So if plate failsthen the joint is going to fail. So the tension capacity of the plate also has to be calculatedwhile calculating the bolt strength of the joint. So we will calculate now the tension capacityof plate. The tension capacity of plate is given by Where, Tnd = tension capacity of platefu = ultimate tensile stress of plateAn = net effective area of plateγml = partial safety factor = 1.25The net effective area of plate is calculated from the following formula:
Where, b = width of platen = number of holes along width perpendicular to the direction of loadd0 = hole diameter = nominal diameter of bolt + clearance of the holet = thickness of plate
Now another failure may come which is 5th one that is bolt with combined shear and tension.Sometimes bolt are exerted to combined shear and tension. So when we are calculatingindividually the shear strength and tensile strength of the bolt we have to also calculate that ifboth the shear and tension acts then what will be the combined strength of the bolt and that isfound from this interaction formula that it has to fulfill these criteria and it has to be less than1.0.
Where, V = applied shear forceVsd = design shear capacityTe = externally applied tensionTnd = design tension capacityHere V should be less than Vsd and Te should be less than Tnd but also this check has to beconducted and the summation of these two will be less than 1.
Now whatever we have discussed we will go through one example and we will try tounderstand that how to calculate the bolt strength.Example: Calculate the shear strength of 16 mm diameter bolt of grade 4.6. The bolt is undertriple shear as shown in the figure below.
nn = no. of shear planes with threads intercepting the plane =1ns = no. of shear planes without threads intercepting the plane =2Nominal diameter of bolt, d =16 mm
So this is a small example we have shown where only shear strength has been calculated dueto multiple shear that means we have tried to understand here that what will be the value ofnn, how we will calculate the value of nn and ns and what will be the Anb and Ans andaccordingly what will be the Vdsb the design shear strength due to shear in the bolt.So due to multiple shear in this case the triple shear the example has been worked out and hasbeen shown I hope you have understood this example, thank you very much.
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