Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 22)
22.
A triangular section having base b, height h, is placed with its base horizontal. If the shear stress at a depth y from top is q, the maximum shear stress is
Discussion:
21 comments Page 2 of 3.
Tailam mahesh said:
8 years ago
Well said @Krishna.
Prabhu said:
8 years ago
The formula is 3/2(F/A).
F=S & A = 1/2(bh).
F=S & A = 1/2(bh).
Harish said:
9 years ago
Yes, the right answer is 1.5 s/area for triangle.
Yps said:
9 years ago
The right answer should be 1.5s/bh.
Krishna said:
9 years ago
For a triangle max shear stress = 3/2 times the average shear stress.
T(avg) = S/(area).
area = bh/2.
T(mx) = 3s/(bh).
T(avg) = S/(area).
area = bh/2.
T(mx) = 3s/(bh).
(1)
Kiran said:
10 years ago
Area only 0.5*bh. Not 0.5*bh^2.
Dhrubajyoti said:
10 years ago
Please show me the whole calculation.
Tathagata Paul said:
1 decade ago
T = 12y(h-y)/bh^3.
Vyshu said:
1 decade ago
Here S is force.
Logu said:
1 decade ago
What is S?
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