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- - **A solid steel shaft is subjected to a torque of 45 KN/m.**
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A solid steel shaft is subjected to a torque of 45 KN/m.A solid steel shaft is subjected to a torque of 45 KN/m. If the angle of twist is 0.5 degree per meter length of the shaft and the shear stress is not exceed 90 MN/m^2 (I) find the suitable diameter of the shaft . Take C = 80 GN/m^2 (II) Maximum shear strain we can solve the question by torsion equation T/Ip = e/R = CQ/l T maximum twisting torque in Nm Ip polar moment of inertia = pid^4/32 in m^4 e shear stress in N/m^2 R radius of the shaft Q the angle of the shaft in rad l length if the shaft in m [color=#FF0000]my question know why in solving fined two diameter ? [/color] |

Re: A solid steel shaft is subjected to a torque of 45 KN/m.Do you mean why do you need to find the value of twice the radius? What is the relationship between the radius and the diameter of a circle? |

Re: A solid steel shaft is subjected to a torque of 45 KN/m.yes why ? and here I have the solution the fist answer is Diameter on the basic of twist = 0.16 m and then diameter on the basis of shear stress : D = 0.1365 m Now my question why in answer give us two diameter ? and which one is correct ? |

Re: A solid steel shaft is subjected to a torque of 45 KN/m.and my other question why in question written the unit of torque kN/m not KNm |

Re: A solid steel shaft is subjected to a torque of 45 KN/m.see here the solution may will help http://store2.up-00.com/June12/yq190516.jpg http://store2.up-00.com/June12/zj390516.jpg http://store2.up-00.com/June12/dUP90516.jpg http://store2.up-00.com/June12/K5m90516.jpg My questions! 1 - Why in the solution there is two diameter 2 - I second diameter how they got the answer 0.13 m when I do the calculation I got 0.14 not 0.13 m 3 - In last from where he got the formula of maximum shear strain = RQ/l ? |

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