Discussion :: Theory of Structures - Section 1 (Q.No.35)
| 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. |
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| 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. |
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| 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. |
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| 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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