### Discussion :: Steel Structure Design - Section 2 (Q.No.41)

Thinn said: (Sep 19, 2013) | |

How to check? Can I get any reference? |

Annon said: (Oct 24, 2015) | |

Why strain have unit? |

Ganesh said: (Oct 26, 2015) | |

Explain? |

Michigan said: (Oct 4, 2016) | |

How to find the answer? Please explain it. |

Rahul said: (Nov 8, 2016) | |

Load = 2300 kg/cm^2 = 23 kg/mm^2. A = pi() x 20 x 20/4 = 314 sq.mm. Axial tensile stress = P/A = 23/314 = 0.073 kg/sq.mm = 0.73 N/sq.mm. Allowable tensile stress till yield point = 0.6 fy = 0.6 x 250 = 150 N/sq.mm. 0.73 < 150, Hence in elastic range. |

Atif Saeed said: (Feb 19, 2017) | |

There should be no units to strain. So t is assumed that the length is in meter to Strain= 0.004 / 100 = 0.00004 (m/m). E=Stress/Strain so Strain= Stress/E = 2300/(2.1x10^6)=0.001. Since Strain here is 00004< 0.001, the member is in Elastic Range. |

Nilraj said: (Oct 28, 2018) | |

Here Area A=πD^2/4=3.14*20*20/4=315 mm sq.=3.14 cm sq. Load P=2300 Kg. Stress=P/A=2300/3.14=732.48kg/cm sq. Now, The stress/strain= 732.48/0.004=1.83 * 10^5 Kg/cm sq.....which is less than E=2.1*10^6 kg per cm sq. So the bar is in elastic range. |

Gill said: (Feb 5, 2019) | |

@Nilraj you are wrong. 2300 is not load check units duffer. |

Kalyan said: (Oct 6, 2019) | |

Can any one give me different types of ranges with their suitable values. |

Shashwat Kapoor said: (Mar 24, 2020) | |

@All, If we consider stress-strain curve of mild steel, Young's Modulus (longitudinal stress/longitudinal strain) is slope of stress-strain curve (up-to proportionality point) and the region in left side of elastic point (nearly equal to proportionality point) represents elastic range and right side of curve represents plastic range. Given: Axial tension = 2300 kg/cm^2 (axial stress), Axial strain = 0.004, Therefore, Axial stress/Axial strain = 2300/0.004 = 5.75 X 10^6 kg/cm^2, which is lesser than Young's modulus of steel (2.1 x 10^6 kg/cm^2). So, the bar is in the elastic range. |

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