Civil Engineering - Steel Structure Design - Discussion

Discussion Forum : Steel Structure Design - Section 4 (Q.No. 24)
24.
In a truss girder of a bridge, a diagonal consists of mild steel flat 4001.S.F. and carries a pull of 80 tonnes. If the grossdiameter of the rivet is 26 mm, the number of rivets required in the splice, is
6
7
8
9
12
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Avdhesh said:   6 years ago
The load carried by splice = total load÷2.
Rivet value = 0.785*26*26*90 = 47.78KN,
Number of rivet = 400÷47.78 = 8.37.
= 9 rivets.
(1)

Aditya anand said:   7 years ago
In this case, flat will be in double shear and assume power driven shop rivets.

Thickness of 400 ISF=12mm(assume)
Shearing strength=2*3.14*26^2*100/4/1000=106.22kn,
Bearing strength=300*26*12/1000=93.6kn,
Rivet value=min(106.22,93.6)=93.6kn,
No of rivets=800/93.6=8.54=9,
Hence correct answer is D.
(1)

VIKASH said:   7 years ago
80000/dt,
80000/ 400*26.

ASHISH M SHAJI said:   8 years ago
Can anyone give the correct solution, please?

Akshay said:   8 years ago
Number of rivets= sf/area=40001/(3.14*26^2/4)=8.
(1)

Anku said:   8 years ago
No. of rivets = load/rivet value.
load=800kn.
The rivet value=shear strength of rivet= (3.14(26)^2x(100/1000))/4.

Raj said:   9 years ago
How to solve this?

Arch said:   9 years ago
Anyone explain it.

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