### 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. |

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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