### Discussion :: Theory of Structures - Section 3 (Q.No.23)

George Zac said: (Mar 4, 2014) | |

The correct answer is option D. 6 * (4+1)/2 * 20,000 = 300,000 N 300,000 N / 4 *1 = 75,000 N/m2 |

Vishu said: (Aug 9, 2016) | |

@George Zac. Please explain why you divided it by 4 * 1? |

Songeta said: (Nov 27, 2016) | |

It was first multiplied by 4 * 1 to get the value in N then again divide by the width of base 4 to get the value in N/mm2. |

Mughal Saltanat said: (Dec 3, 2016) | |

Firstly, you have to take an assumption that length of dam is 1m, Then the answer is correct only for empty dam condition. If full reservoir condition has to be taken into account then due to moment factor value will be lesser. In that case, the answer will be 75000-135000 = - 60KN i.e tension will be at heel. Hope it is useful. |

Tanooja said: (Jun 3, 2017) | |

Weight of the dam per unit length. W = (a+b)/2 * H * 1 * density. a = top width of dam, b = bottom width of dam, W = 3 * 10^5N. Minimum stress=W/b(1-(6 * e)/b). e = X-b/2. X = moment/load, e = 0.6. Substitute all values. Minimum stress = 7500N/m^2. |

Priya said: (Jun 5, 2019) | |

How to find X? i.e Moment load not given. How e=0.6? |

Usthad said: (Jan 27, 2020) | |

For stability e= (b/6). |

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