Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 20 (Q.No. 39)
39.
What is the effective net width of plate shown in the given sketch, for carrying tension ?
Discussion:
7 comments Page 1 of 1.
Tanvir sayyed said:
1 year ago
Suppose upper rivet 1 , middle 3, and bottom 2.
The section can fail along either 1--2 or 1--3--2 we give a minimum net area for the strength of the member.
Let, sectn 1-2, the net area = (b-nd+n's²/4g)t.
For width, t=1 let, (300-2×25) for pitch zero:
Width 250mm
Now, for sect 1-3-2.
Net Width= (300-3×25+2×50²/4×100) = 237.5
I.e minimum among 237.5mm✓
The section can fail along either 1--2 or 1--3--2 we give a minimum net area for the strength of the member.
Let, sectn 1-2, the net area = (b-nd+n's²/4g)t.
For width, t=1 let, (300-2×25) for pitch zero:
Width 250mm
Now, for sect 1-3-2.
Net Width= (300-3×25+2×50²/4×100) = 237.5
I.e minimum among 237.5mm✓
(1)
Priyanka said:
4 years ago
B is correct.
(B -3d)+S * 2 ÷ 4g +S *2 ÷4g.
300-3 * 75) + 50*2÷4 * 100 +(50 * 2÷4 * 100),
= 237.5.
(B -3d)+S * 2 ÷ 4g +S *2 ÷4g.
300-3 * 75) + 50*2÷4 * 100 +(50 * 2÷4 * 100),
= 237.5.
(3)
Santu naskar said:
8 years ago
Answer is b.
300 - 3 * 26.5 + (2 * 50^2/4 * 100) = 233 mm.
300 - 3 * 26.5 + (2 * 50^2/4 * 100) = 233 mm.
Rahul said:
3 years ago
B - nD + n'P^2/4G.
Here;
P = pitch.
G = gauge distance.
Here;
P = pitch.
G = gauge distance.
(1)
Praveen Kumar said:
3 years ago
300-3 * 25 + {(50^2)/4*100}*2 = 237.5 mm.
(2)
Arun bamniya said:
2 years ago
237.5 is the correct answer.
Dolly said:
7 years ago
It should be option B.
(1)
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