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 2 of 3.
Priya said:
8 years ago
We know number of rivet = force /rivet value.
But here shearing, tearing many value given how can solve explain it.
But here shearing, tearing many value given how can solve explain it.
Pavan said:
8 years ago
Rivets are used in WSM method. So why we multiply the load by load factor?
Then answer is 11.
Then answer is 11.
Arnab said:
1 decade ago
n=p/Rv so from the formula 35000/3425=10.22. So answer should be between a&b.
Ravi kumar vishwakarma said:
8 years ago
The answer should be 11 because;
n = load/rivet value.
n = 35000/3425,
n = 10.2 = 11.
n = load/rivet value.
n = 35000/3425,
n = 10.2 = 11.
Prateek said:
8 years ago
Always even num of rivets are provided so 12 num of rivets is required.
M. Farooq said:
8 years ago
1.25 is Factor of safety. FOS= 1.25.
Ankit said:
8 years ago
Why 1.25 is multiplied in 35000?
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.
Rabbi Dhillon said:
9 years ago
1.25P/least strength of rivet.
1.25 * 35000/3425 = 12.773.
1.25 * 35000/3425 = 12.773.
Lal bir said:
9 years ago
From where 1.25 comes?
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