# Civil Engineering - Steel Structure Design - Discussion

### Discussion :: Steel Structure Design - Section 2 (Q.No.11)

11.

Strengths of a rivet in single shearing, in bearing and in tearing are 3425 kg, 4575 kg and 5025 kg respectively. If the load in the member is 35 tonnes, the number of rivets required, is

 [A]. 10 [B]. 11 [C]. 12 [D]. 13 [E]. 15

Explanation:

No answer description available for this question.

 Arnab said: (Aug 6, 2014) n=p/Rv so from the formula 35000/3425=10.22. So answer should be between a&b.

 Shashank said: (Dec 1, 2014) n = load applied/least strength of rivet = 35000/3425 = 10.22 = 11 rivet.

 Rajesh said: (Sep 5, 2015) But the answer shows 13.

 Pranjal said: (Jan 26, 2016) Shearing strength = 3425/1.25. Answer 13.

 Lal Bir said: (Sep 2, 2016) From where 1.25 comes?

 Rabbi Dhillon said: (Nov 10, 2016) 1.25P/least strength of rivet. 1.25 * 35000/3425 = 12.773.

 Atif Saeed said: (Feb 19, 2017) Load Factor of 1.2 gives total load as 35 * 1.2 * 1000 = 42000 kg. Hence the Number of Rivets = 42,000/3425 = 12.26. So, 13 Rivets are required.

 Ankit said: (Mar 17, 2017) Why 1.25 is multiplied in 35000?

 M. Farooq said: (May 3, 2017) 1.25 is Factor of safety. FOS= 1.25.

 Prateek said: (May 28, 2017) Always even num of rivets are provided so 12 num of rivets is required.

 Ravi Kumar Vishwakarma said: (Jun 5, 2017) The answer should be 11 because; n = load/rivet value. n = 35000/3425, n = 10.2 = 11.

 Selvi said: (Jul 7, 2017) No of rivet = load/rivet vale. Rivet value is the least value of bearing and shearing. SO, n = 35000/3425 = 10.22 say 11.

 Pavan said: (Jul 19, 2017) Rivets are used in WSM method. So why we multiply the load by load factor? Then answer is 11.

 Priya said: (Nov 10, 2017) We know number of rivet = force /rivet value. But here shearing, tearing many value given how can solve explain it.

 Dipak said: (Nov 26, 2017) 11 is the right answer.

 Rakesh said: (Dec 5, 2017) Bearing, shearing, and tearing strength is called Rivet value. In bearing tearing and shearing value we take always Lower value whichever is less Here my Rivet value is= 3428kg. Load is = 35tonnes =35x1000=35000. Then, No of Rivet= load/ rivet value. Then, 35000/3425= 10.22. We can say 11 rivet.

 Viru Kapoor said: (Mar 7, 2018) Guys it's 13. Take fOS also.

 Deep said: (Mar 15, 2018) It is 11. And no need to add FOS for matching with answer.

 Amit Kumar said: (Apr 1, 2018) Agree @Deep. It should be 11.

 Manu said: (Apr 5, 2018) Yes, I also agree @Deep. It is no need of FOS.

 Parth said: (Oct 24, 2018) While calculating the shear and bearing capacity of rivet the factor of safety had been calculated. Is it necessary to consider the factor of safety twice? Please, anyone, tell me.

 Veer said: (Feb 20, 2020) No need to use FOS because of strength is already underestimated.

 Rohan Mukherjee said: (Aug 26, 2020) The answer is wrong! Rivet value is taken as minimum value of Shearing, Bearing and tearing! No of rivets = Force/Rivet value. Then the answer is 10.21.