Civil Engineering - Steel Structure Design - Discussion
Discussion Forum : 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
Discussion:
23 comments Page 1 of 3.
Arnab said:
1 decade ago
n=p/Rv so from the formula 35000/3425=10.22. So answer should be between a&b.
Shashank said:
1 decade ago
n = load applied/least strength of rivet = 35000/3425 = 10.22 = 11 rivet.
Rajesh said:
10 years ago
But the answer shows 13.
Pranjal said:
10 years ago
Shearing strength = 3425/1.25.
Answer 13.
Answer 13.
Lal bir said:
9 years ago
From where 1.25 comes?
Rabbi Dhillon said:
9 years ago
1.25P/least strength of rivet.
1.25 * 35000/3425 = 12.773.
1.25 * 35000/3425 = 12.773.
Atif Saeed said:
9 years ago
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.
So, 13 Rivets are required.
Ankit said:
8 years ago
Why 1.25 is multiplied in 35000?
M. Farooq said:
8 years ago
1.25 is Factor of safety. FOS= 1.25.
Prateek said:
8 years ago
Always even num of rivets are provided so 12 num of rivets is required.
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