 In the previous video we have discussed the design steps for the design of tension member. Now we will see an example. Myself Kaji Syed Shujat Ali, currently working as assistant professor in civil engineering department, Valchen Institute of Technology, Sulapur. At the end of the session, students will be able to design a tension member. We have seen all these things. The strength in gross section yielding, in rupture and in block shear. In our previous video, just a recall of that. The formula for design strength in gross section yielding Tdg is equal to A g Fy divided by gamma m0. In the case of net section rupture, for the angle section, Tdgn is equal to this formula. If you want to see the detail formulas, you can refer my previous video. For the angle section, this is the formula. Beta is equal to this one. And ANC is the area of connected leg. AG0 is the area of outstanding leg. Fy is the yield strength. Fu is the ultimate strength. These are the partial safety factors. And beta is the correction factor. Beta is equal to this formula. W is given in this figure. T, thickness of member, Fy, Fu, Bsl. This value should be less than or equal to this one. And this value should be greater than or equal to 0.7. Then the strength in block shear failure is given by this formula. Tdb1 and Tdb2. The strength in block shear failure. These are the design steps. You can pause the video and just read these steps. We'll start the question. It is required to design a bridge trust diagonal to carry a factor tensile load of 250 kN. The length of member is 2.8 m. The tension member is connected to a gusset plate of 16 mm thickness with one line of 20 mm diameter bolts of grade 8.8. We have to design a single angle section. We have to design an angle section by using the grade of steel as 410 MPa. The given data is Fe equal to 410. For that one, Fe equal to 410 Newton per mm square. Fy equal to 250 Newton per mm square. The partial safety factor as per IS 800, 2007 equal to gamma M0 equal to 1.1. Gamma M1 equal to 1.25 and gamma Mb equal to 1.25. For grade 8.8, grade of bolt 8.8, ultimate strength equal to 830 Newton per mm square. For diameter greater than or equal to 16 mm. The net area of the bolt for 20 mm diameter is equal to 0.78 pi by 4 d square. So, d is equal to 20 mm, 0.78 pi by 4 d square which is equal to 245 mm square. We will take an assumption here. That is the thread will cut the shear plane. The thread will cut the shear plane. And the shear strength of bolt Vdsb is equal to Feb by root 3 gamma Mb into Anb equal to this 93.92 kilo Newton. Feb is equal to 830. This is root 3 gamma Mb 1.25. Anb is equal to 245 which is equal to 93.92 kilo Newton. Now, then we will determine the strength in bearing given by 2.5 kbd into t into Feb divided by gamma Mb into Anb. Now, to determine this kb, we have to determine pitch edge distance and end distance normally pitch equal to 2.5 into nominal diameter of bolt and end edge distance equal to machine flame cut edges. It is equal to 1.5 gross diameter of bolt and for hand flame cut edges it is equal to 1.7 into gross diameter of bolt. Vdpb equal to kbd into t into Feb divided by gamma Mb into Anb. We all know this except kb. kb is determined by using these formulas. kb equal to the minimum of e divided by 3 d0 p divided by 3 into d0 minus 0.25 then fub divided by fub or 1 e is the end distance p is the pitch e divided by 3 d0 equal to 0.682 this equal to this this equal to this the lesser is 0.682 we use that value then diameter then thickness here we assume the thickness of the member fub equal to 410 then area divided by gamma Mb we get the value as 89.48 then we will take the lesser of these two shear strength as well as bearing strength so the lesser is 89.48 number of bolt equal to the given factor first divided by 89.48 which is equal to 3 we will provide it in a single line all the three bolts then we will determine the area by using net section fracture An is equal to 0.9 An into Fe divided by gamma Mb we will rearrange the term we will determine An which is equal to 846.88 Mb then we will increase this area by 20% to get the gross area after increasing 20% we get this one we also determine the area by using the gross section yearlings t equal to Ag into Fe divided by gamma Mb rearranging the terms Ag is equal to t into gamma Mb divided by Fe which is equal to 110 Mb then we will select nth section depending on this area ISA 100 by 75 by 8 area is this, diameter of bolt hole is this connected area is this, outstanding area is this I hope you know how to calculate these areas then we will determine the design strength due to net section which is governed by the tearing strength at net section tdn equal to 0.9 into An into Fe divided by gamma Mb plus beta into Ag0 into Fe divided by gamma Mb we have to calculate this beta by using this formula and this value should be less than this value and this value should be greater than or equal to 0.7 first of all we will determine this value w equal to outstanding leg 75 Mb ws 50 Mb assume it, shear leg width equal to bs equal to w plus ws minus t equal to 117 mm lc equal to length of connection equal to 130 mm beta comes out to be 1.009 then we will determine this value 0.9 into Fe into gamma Mb divided by Fe into gamma Mb this comes out to be 1.3 it is greater than 0.7 we will compare this value with beta it is 1.3 and beta comes out to be 1.009 therefore beta is less than this value we take beta as 1.009 we put the value of beta here we will determine the strength tdn which comes out to be 305.01 kN greater than 250 or given force therefore safe then we will determine the strength in block shear first of all Avg is the this is the shear plane this is the tension plane for the gross shear we take this much length so 2 into 65 plus 45 into 8 and net area we have to subtract the diameter of these poles 2 into 65 plus 45 minus 2.5 into 22 how this 2.5 comes this one, this one and this half we will get the net area and shear for tension gross is our similar 50 into 50 is this length 50 into 8 and to determine the net area we have to subtract this half bolt for shear this for the shear plane it cuts this bolt completely it cut this bolt completely and it cut this bolt partially half bolt gets cut here and for tension similar this bolt gets cut only partially then by using the formula for shear yield and tension fracture TDB1 we have determined in the previous slides the Avg and Atn Atg and Avn we will put all these terms and we will determine TDB1 equal to 275.8 kN and TDB2 equal to 254.5 kN we will compare it we will compare it with this one whichever is less we will take that as an our strength in block shear failure we will determine the block shear failure by we have determined this here and we will take the lesser one and we compare it our given factor force it is greater than our given factor force therefore it is safe in block shear failure then we take a check on cylinderness ratio cylinderness ratio is given by effective length divided by least radius of variation it is equal to 176 now the permissible cylinderness ratio as per IS 800 2007 table number 3 is equal to 180 therefore 176 is less than 180 safe it is safe in cylinderness ratio these are some review questions you can pause the video and answer this question the design strength of the steel section due to gross section yielding TDG is given by so TDG is equal to whichever the which formula we use for the design of for the design of the section in gross section yielding the answer is option A and the another question is the design strength of plate due to rupture of critical section is given by the answer is option D these are my references thank you