We'll email you at these times to remind you to study
You can set up to 7 reminders per week
We'll email you at these times to remind you to study
In last lecture we have discussed the design philosophy of ordinary black bolt and we haveseen that how to calculate the design shear strength of black bolt, the bearing strength, tensilestrength and tensile strength of plate and combination of shear and tension. So how tocalculate the strength due to different force occurring that has been discussed and we havegone through a small example in the last lecture.Now in this lecture I will go through two examples of ordinary black bolt and we will seehow to calculate different type of strength due to shear, due to bearing, due to tension. Wewill also go through another example where a plate is connected with different number ofbolts and we will calculate the efficiency of that bolt that means the strength of bolt and thestrength of the solid plate. In first example we will try to calculate the strength of lap jointand then we will go for butt joint. So this is what we will cover in today’s lecture.
Example 1: Design the following joints using ordinary black bolts between two plates ofwidth 200 mm and thicknesses 10 mm and 18 mm respectively to transmit a factored load of150 kN. Use plates made of Fe 410 grade steel and 16 mm diameter bolt of grade 4.6.
Nominal diameter of bolt, d = 16 mmDiameter of hole, d0 = 18 mm (Ref. Table 19 of IS 800: 2007)For grade 4.6 bolts; fub = 400 MPaFor Fe 410 grade of steel; fu = 410 MPaPartial safety factor for bolt, γmb = 1.25So we will design the connection for the following three cases:a) Lap jointb) Single cover butt joint with cover plate of 8 mm.c) Double cover butt joint with 8 mm covers plates.a) Lap joint:In this case we can assume that shear plane is including in the threads so nn is 1 and ns as 0.For 16 mm diameter bolt; Net shear area of the bolt at threads is, Anb=157 mm2Design shear strength per bolt, (excluding reduction factor)
form shear point of view the number of bolts required is
So the bolt value will be minimum of bolt value will be minimum of two that is 29 and 74.78kilonewton.Therefore, bolt value = 29 kNNumber of bolts required ¿ 15029=5.2
So we can provide 6 nos of bolts and arrange them as shown in the above figure.b) Single cover butt joint with cover plate of 8 mm:
In this case two plates are joint with a cover right. So we have to provide certain as shown inthe above figure. So for this case also the design shear strength of bolt Vdsb will be 29 kNbecause this is also in single shear.Design bearing strength per bolt,
Therefore, the bolt value is 29 kN.Number of bolts required for this connection 15029=5.1Provide 6 bolts and arranging it in three lines as shown in the figure.
So bolt we are providing in single line in each case here and plates are connected, right sothis is how we can arrange the bolts in three lines and we can find out the arrangement of thebolts in this way.
So what will be the difference from earlier one that in this case as double cover butt joint isconsidered, we can assume that it is actually in a double shear.Therefore, the bolts will be in double shear.Assuming threads in the shear planes. Therefore nn=2;ns=0Since the two plates of thicknesses 18 mm and 10 mm are to be jointed, packing plate ofthickness (18-10) = 8 mm will be required and as the packing plate thickness is greater than 6mm we have to provide the reduction factor (β pkg) for the packing plate.As per clause 10.3.3.3 of IS:800 2007, design shear capacity of bolts will be decreased byfactor β pkg
So the design shear strength of bolt in double shear, If packing plates are not given than the shearing strength of the bolt due to double shear willbecome 2× 29=58kN but as we are providing packing plate which is more than 6 mm,we have to reduce certain amount which was given as a reduction factor of β pkg and thatvalue is coming 0.9, that is why in place of 58 this is coming 52.21 kilonewton. Now design bearing strength per bolt, As double cover butt joint has been provided that is why the cover plate thickness willbecome 8×2 that is 16 mm and thickness of other plates are 10 mm and 18 mm. So thicknessof the thinner plate will be 10 mm so that is why we are providing 10
Therefore, the bolt value is 52.21 kN.Number of bolts required for this connection
Now atleast we can provide we have to provide a single line of bolt here and we have toprovide a single line of bolt here right. So if we see these we can see that in plan it will looklike this is the two plates are joining here and bolts are connected here four number of boltsare there. So the detailing can be done in this way either we can consider 31 millimeter or wecan provide 35 may be 130 and 35 to cover this total 200 mm thickness, right and this we canprovide 65 mm or more whatever means suitably we have to provide, so this will be the thiswill be the arrangement of the bolts, right.Using 4 bolts to arrange the joint as single bolted double cover butt joint can be arranged asshown in the figure.
Example 2: Two plates 10 mm thick are joined by 16mm diameter bolts in a triple staggeredlap joint as shown in the figure. Find efficiency of the joint.
So if we draw the figure we will see this is a triple staggered lap joint. So bolt we can providein this way. The bolt lines are marked as 1-1, 2-2, 3-3.Now the failure may occur in any direction as marked in the above figure. So if we makesection 1-1, 2-2, 3-3 then we have to consider first failure at 1-1 that we will try to find outthen it will be going to fail in section 2-2. So when we are going to find out the strength at 2-2 that means we have to find out strength at bolt at 1-1 failure strength plus 2-2. So it will beclear when we will be going through this example.
So first let us consider section 1-1 means along 1-1 if it fails how it looks. So there first wehave to find out the this is in single shear so the P single shear if I write then due to singleshear Vdsb will be same as earlier and this will be 29 kN which we have already calculated inearlier case of 16 mm diameter because this is a single shear and diameter is 16 mm diameterso 29 kN will be the single shear strength shear strength due to single shear.Now design bearing strength per bolt
So bolt value due to shear it is coming 29 and due to bearing it is coming 60. So smaller ofthis two will be the bolt value, so we can consider the bolt value as 29 kN, right.So the strength of joint based on bolt value will become how much strength of joint because 7number of bolts are there so this will be 7×29=203kNAgain now we will see if it fails along section 1-1 then what will be the strength of that joint.Strength of joint along 1-1
So the strength of the joint will be minimum of 203 kN, 277.49 kN, 282.35 kN, 422.49 kN.Therefore, Strength of the joint = 203 KN
Now we have to find out the efficiency of the joint, so efficiency of the joint means thestrength of joint divide by the strength of the solid plate. So strength of solid plate we have tocalculate first.Strength of the solid plate
So in this lecture what we have seen we have gone through two examples, one was based onthe lap joint and butt joint and we found what will be the strength of the joint for differenttype of lap joint and butt joint that we have calculated in first example and in second examplewe have calculated the efficiency of the joint and how the joint fails mean joint can fail as ahole due to shear of the bolt or it can fail along a particular section.
Log in to save your progress and obtain a certificate in Alison’s free Advanced Diploma in Design of Steel Structures online course
Sign up to save your progress and obtain a certificate in Alison’s free Advanced Diploma in Design of Steel Structures online course
Please enter you email address and we will mail you a link to reset your password.