An aluminum (E = 10 x 106 lb/in2, ν = 0.333) bar of length (L = 49 inch) has a circular cross-section of radius (r = 0.7 inch). The beam is fixed at one end. A torque (T = 1,666 lb-in) is applied to the free end. Calculate the twist of the bar free-end. (Units: degree)

Respuesta :

Answer:

1.24 degrees

Explanation:

The twist angle can be calculated using the following formula for torsion:

[tex]T = \frac{E I \theta}{L}[/tex]

Where [tex]I = \frac{\pi r^4}{2} = \frac{\pi 0.7^4}{2} = 0.377 in^4[/tex] is the polar inertia of the circular cross section.

[tex]1666 = \frac{10^7 * 0.377 \theta}{49} = 76969 \theta [/tex]

[tex]\theta = 1666 / 76969 = 0.0216 rad = 0.0216 * 180 / \pi = 1.24^o[/tex]