Problem 302:
What is the minimum diameter of a solid steel shaft that will not twist
through more than 3° in a 6-m length when subjected to a torque of 12
kN·m? What maximum shearing stress is developed? Use G = 83 GPa.
Solution:
Given,
Length, L = 6 m
Torque, T = 12 KN.m
Angle of twist, θ = 3°
G = 83 GPa.
Minimum diameter, d =?
Maximum shearing stress, τmax =?
We know that , Angle of twist,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdMZ37V1Y1-__lMfdlnKlXmKhMVW0K138Gj5puFyBkK0xFOaYcrX-CA2m_iR_IjgQzgGfeO1dwcVd6v5lWwxqK1IWm5us6_8SytRI3o5ZM5P33epo_xl-a7zc13tKUSWLZYKUG5x5EkJo/s1600/302-1.jpg)
So, Minimum diameter = 113.98 mm
We also know that, maximum shearing stress,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRM35NLl_PLlppafep5gbygWktR3V2OnsFQuDVDqdUEMLkETZ5pNqMKcWSRc0_f-Ih4PSEEQzZOpXmpc3I-NZAYTSE4CxXE8DcifnQYVEb3I68zGrE6pkFfmo1V9wQjPMF52hslat6RKQ/s1600/302-2.jpg)
So, Maximum shearing stress = 41.27 MPa
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