Question

An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsional spring with spring constant k. If a concentrated torque of magnitude Ta is applied in the center of the bar, what is the rotation at the end of the bar, φ(L)? Assume a constant shear modulus G and polar moment of inertia J.

Answer 1

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.