Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 35)
35.
The maximum magnitude of shear stress due to shear force F on a rectangular section of area A at the neutral axis, is
Discussion:
8 comments Page 1 of 1.
Engineer andualem asnake said:
6 years ago
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.
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.
Parvat kumar said:
6 years ago
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.
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.
Ajay kumar said:
7 years ago
Thanks @Amulya Rajbhar.
Aniruddh said:
8 years ago
At NA share stress is max.
So.3F/2A is equalled to 1.5 x F/A.
So.3F/2A is equalled to 1.5 x F/A.
Priyanka said:
8 years ago
Anybody explain this.
Amulya Rajbhar said:
9 years ago
(max Share stress) / (avg shear stress) = 1.5.
Max shear stress at N. A.
Avg Shear Stress = F/A.
Max shear stress at N. A.
Avg Shear Stress = F/A.
Sandeep said:
9 years ago
How is it?
Kalu charan parida said:
1 decade ago
Why it is the right answer?
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