### Discussion :: Strength of Materials - Section 5 (Q.No.4)

Shraddha said: (Sep 22, 2015) | |

Answer: 8 times. Deflection at the free end due to point load = Wl*L*L/3EI. Put 2L in place of L, it becomes 8 times. |

Subhra Nandi said: (Sep 24, 2015) | |

No answer is wrong its will be ANS-A (8). |

Ashish Gusain said: (Jul 9, 2016) | |

Option B is wrong. Deflection = Wl.l.l/3EI, Given that L = 2l, Then deflection = W.2l.2l.2l/EI, => 8x W.l.l.l/3EI. So option A is right. |

Shweta Puri said: (Aug 28, 2016) | |

Read the question carefully, they are not telling about length they said if the load at the free end is double it mean W = 2W. The deflection of the free end is WL^3/3EI when we put W = 2w it become1/2 which is not given in the answer but the right answer is 1/2. |

Anni said: (Sep 23, 2016) | |

@Shweta. I think, your answer is wrong. |

Baloch said: (Nov 6, 2016) | |

It's not doubling the load, it's doubling the length. L^3, 2L^3 = 8, And is C. |

Surjya said: (Jan 15, 2017) | |

It will be option A i.e. 8. |

Don Dai said: (Feb 9, 2017) | |

More W & L, more deflection. More B & D, less Deflection. That's why 8 times. |

Minesh Rathore said: (Jun 6, 2017) | |

I think the Answer should be A. |

Aaqib said: (Apr 20, 2018) | |

Wl^3÷3Ei÷w (2l) ^3÷3Ei, =1/8 Answer. I think option B, please correct me if I m wrong. |

Sushil Thakur said: (Sep 30, 2018) | |

Right @Aaqib. But it is not ask about ratio they ask about how much increase in deflection. So, x1/x2 =1/8, i.e x2=8x1 (answer). I hope you understand. |

Vishal Amit Patil said: (Jul 1, 2019) | |

@All. Here, it is mentioned to double the load not length. |

Jinish said: (Aug 14, 2019) | |

Deflection 1= wl^3/ 3EI and, If l = 2 then deflection 2 = 8wl^3/3 EI. So deflection = deflection 1/ deflection 2 = 1/8. Correction answer is 1/8. Here, mentioned length is to doubled l = 2l. |

Dhru said: (Apr 21, 2020) | |

Wl^3/3EI upon W(2l^3) /3EI. Solve then the answer will be 1/8. |

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