Civil Engineering - Theory of Structures - Discussion

35. 

The maximum magnitude of shear stress due to shear force F on a rectangular section of area A at the neutral axis, is

[A].
[B].
[C].
[D].

Answer: Option C

Explanation:

No answer description available for this question.

Kalu Charan Parida said: (Jan 16, 2015)  
Why it is the right answer?

Sandeep said: (Sep 1, 2016)  
How is it?

Amulya Rajbhar said: (Dec 3, 2016)  
(max Share stress) / (avg shear stress) = 1.5.

Max shear stress at N. A.

Avg Shear Stress = F/A.

Priyanka said: (Jan 21, 2018)  
Anybody explain this.

Aniruddh said: (Jan 27, 2018)  
At NA share stress is max.

So.3F/2A is equalled to 1.5 x F/A.

Ajay Kumar said: (Jan 24, 2019)  
Thanks @Amulya Rajbhar.

Parvat Kumar said: (Aug 6, 2019)  
J max=FA Y/IZ.
A = b(d/2-y).
Y BAR = {1/2(d/2+y)+y}
Z = b
I = be^3/12.

Jmax-6F{(d/2)^2-y^2}/bd^3.

Here y = 0,
Then;
Jmax = 6Fd^2/4bd^3.
Jmax = 3F/2A.

Engineer Andualem Asnake said: (Aug 18, 2019)  
Maximum shear stress divided by average shear stress = 3/2.
Hence maximum shear stress = 3/2 average shear stress.
Average shear stress = F/A.
Insert this equation into the 1st equation then you will get, maximum shear stress = 3F/2A.

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