### Discussion :: Strength of Materials - Section 1 (Q.No.28)

Nitin Ingole said: (Sep 26, 2013) | |

Answer should be 0.25 Poisson's ration = ( E-2G )/2G = (200-2x80)/2*80 = 0.25. |

Rathi said: (Sep 28, 2013) | |

Relation between young's modulus(E), Rigidity modulus(G) and Poisson's ratio(u). E = 2G(1+u). u = E/2G - 1. = 200/2*80 - 1. u = 0.25. |

Manasa Sundaram said: (Mar 23, 2014) | |

c = mE/2(m+1). 1+(1/m) = (2*10^5)/(2*(8*10^4)). 1+u = 1.25. u = 0.25. |

Vikas Kumar said: (May 11, 2014) | |

Formula of poisson ratio = (E-2G)/2G. Where E = Young's modulus. G = Modulus of Rigidity. Now U = (200-2*80)/2*80 = 0.25. |

Mayuri said: (Dec 31, 2014) | |

E = 2G (1-1/m). 200 = 2*80(1-1/m). 200 = 160-160/m. 40 = 160/m. 1/m = 40/160. 1/m = 0.25. |

Piyush Singh said: (Feb 28, 2015) | |

Young modulus = E. Modulus of Rigidity = G. Poisson's Ratio = U. Formula => E = 2G (1-u). Therefore: A) 200 = 2*80 (1-u). B) 1-u = 1.25. C) u = 1.25-1. D) u = 0.25. |

Susmita said: (Jul 30, 2015) | |

We know that C = mE/2 (m+1). Where C = 80GN/m2. E = 200GN/m2. 1/m = Poisson's ratio = 0.25. |

Abhay said: (Feb 2, 2016) | |

Young's modulus E = 2G(1 + μ) where, G = Modulus of rigidity. => μ = E/2G - 1. => 200/(2*80) - 1. = 1.25 - 1. μ = 0.25. |

Krishna said: (May 12, 2016) | |

You are correct @Rathi. E = 2G(1+mu) then u = 0.25. |

Mujru said: (Aug 11, 2016) | |

The equation related to bulk modulus and modulus of rigidity is; E = 2n(1+u). E = bulk modulus, n= modulus of rigidity, u = possion raito. So, u = E/2n -1. = 200/(80 * 2) - 1. = 0.25. |

Ammulu said: (Dec 28, 2016) | |

E = 2g(1 + u), 200 = 2 * 80(1 + u), 200 = 160 + 160u, U = -40 ÷ 160, U = 1 ÷ 4, U = 0.25. |

Abigail said: (Feb 23, 2017) | |

I tried d formulae for the relationship between bulk modulus and modulus of rigidity but it's not giving me the right answer. Can anyone help me? |

Anusha Reddy said: (May 9, 2017) | |

Good explanation @Ammulu. |

Bikash Kabiraj said: (May 17, 2017) | |

Relation between e[young], k[bulk], c[mod of rig] and 1/m[poiss]. e = 3k(1-2/m)=2c(1+1/m)=9kc/3k+c, e = 200 and c=80. e = 2c(1+1/m) so 200=2*80(1+1/m) so 1/m=40 /160 =. 25ans. |

Gondelamahesh said: (Jun 9, 2017) | |

If the modulus of rigidity is 80 kn/mm square and bulk modulus 140 kn/mm square position ratio is what anyone explain it? |

Shruti said: (Jul 26, 2017) | |

E = 2G( 1+U), 200= 2 * 80( 1+U), 200/160 = (1+U), 5/4-= (1+U), 5 = 4(1+U)5=4+4U, 4U = 5-4, u= 1/4 = 0.25. |

Chakradhar Padhan said: (Aug 9, 2017) | |

Relation between e[young], k[bulk], c[mod of rig] and 1/m[poiss].e = 3k(1-2/m)=2c(1+1/m) =9kc/3k+c,e = 200 and c=80.e = 2c(1+1/m). So 200=2*80(1+1/m) so 1/m=40 /160 =. 25. |

Pradip Behera said: (Feb 5, 2018) | |

Here, E=2G(1+u)=3K(1-2u)=9KG/(3K+G). |

Akshata said: (Jul 11, 2018) | |

1st find Youngs Modulus (E) using E= (9KG)/(G+3K) where K=bulk modulus & G= modulus of rigidity then E=2G (1+u) where u= Poisson's ratio. |

Viru Kapoor said: (Jul 23, 2018) | |

E=2G(1+U). Given E=200. G=80. 200=2 * 80(1+u). 200/2 * 80=1+u, 1.25=1+u, u=1.25 - 1=0.25. |

Kk Bhal said: (Apr 16, 2019) | |

The answer is supposed to be 0.4. C = (mE)/2(m+1). Relationship between Young's modulus of elasticity and Rigid modules. Ref: RS Khurmi. |

Maxwell said: (May 12, 2019) | |

Please anyone solve this, If the modulus of rigidity is given as 42MN/mm square and the modulus of elasticity is 200 KN/mm square, what will be an acceptable value for bulk modulus if the material has a cuboid configuration? |

Kowsalya said: (Jun 24, 2021) | |

E = 200*10^9N/m^2. G = 80*10^9N/m^2. U=? E=2G(1+U). (200 * 10^9)/(2 * 80 * 10^9) = (1+U), 100/80 = 1+U, (10/8)-1 = U, (5/4)-1 = U, (5-4)/4 = U. 1/4 = U==>U = 0.25. |

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