Problem 301:
A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque of 15 Kip.ft. Determine the maximum shearing stress and the angle of twist. Use G = 12 × 106 psi.
Solution:
Given,
Length, L= 3 ft = 3 x 12 in
Diameter, D = 4 in
Torque, T = 15 kip.ft = 15000 x 12 = 180000 psi.in
G = 12 × 106 psi.
Maximum Shearing stress, τmax = ?
Angle of twist, θ = ?
We know that, maximum shearing stress,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfUn8YmkUA6w8oA4ScftOvtzdXRg4TyOy6Mm03TYEROVJHt6hKbx6jx83YERyMaq-bkSZN70KDMJqzXSWAOsAjqzbmeqT0ZlhcI9_nJ6PiEV0RVlgK4CnvqZwKGd4af7GH0zZSl8alOG8/s1600/301-1.jpg)
So, the maximum shearing stress = 14.32 ksi
We also know that, angle of twist,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjeRX6zICS-PbCQ7DB2KsbWvaroiYGdf2bbWv1VzZaCQfqE5rhMAan4y-a9UKFsFuaWAiVNYbi4apXP5U_Zxhplx1WOVQXDbBfXauXDtZU4N-O2jouHVDd1kT2xCafb3H-6gM1lHMoFpk0/s1600/301-2.jpg)
So, the angle of twist = 1.23o
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