Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 2 (Q.No. 13)
13.
Two steel plates each of width 150 mm and thickness 10 mm are connected with three 20 mm diameter rivets placed in a zig-zag pattern. The pitch of the rivets is 75 mm and gauge is 60 mm. If the allowable tensile stress is 150 MPa, the maximum tensile force that the joint can withstand is
Discussion:
8 comments Page 1 of 1.
ARUNPANDIYAN said:
1 decade ago
What is the formula for this questions?
(1)
Dishant shah said:
10 years ago
(b-dh)t*Ts.
b = Width of plate.
dh = Diameter of hole = 21.5.
t = Thickness of plate.
Ts = Tensile stress = 150.
b = Width of plate.
dh = Diameter of hole = 21.5.
t = Thickness of plate.
Ts = Tensile stress = 150.
HEMANT KESHARWANI said:
9 years ago
(b - d) * t * Pt, Nominal diameter 20, then gross diameter 20 + 1.5 = 21.5.
= (150 - 21.5) * 10 * 150.
= 192750 N.
Tearing strength of steel plate - 192.75 KN.
= (150 - 21.5) * 10 * 150.
= 192750 N.
Tearing strength of steel plate - 192.75 KN.
(3)
Manoranjan jena said:
7 years ago
Thanks for the answer.
Dhanu said:
7 years ago
How come d=21.5? Please explain anyone.
Krishna said:
7 years ago
For rivets of diameter 16 to 25mm, the rivet hole diameter is equal to 1.5mm + rivet diameter.
Vishal said:
6 years ago
The rivet pattern is given zig, zag so we need to consider pitch and gauge distance.
Rahul said:
6 years ago
We have to take all the possibilities and look for the minimum area, when we are taking section along a single bolt the area comes out as 1285mm2, similarly along the section containing both the bolts the area comes out as 1070mm2, and while taking the section along zig zag arrangement, area comes out as 1323.75mm2, accordingly we have to take critical area ie minimum which is 1070 and multiply it by 150, but the sol is not in the options, don't understand why, in the sol the second lowest value ie 1285 has been taken and multiplied by 150, could someone elaborate. ?
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