Increase of radius of circular rod due to torque

[magwas]

Homework Statement

I am trying to understand how will the radius of a circular beam grow when a torque is applied.
There is a beam with length l, radius r, and a moment M1 applied on it. Its radius will grow with dr.
What is the Young modulus of the beam?
M= 15 N m
r= 0.008 m
l= 0.2 m
dr= 0.00064 m

Homework Equations

Torsional shearing stress
$$\tau=\frac{M r}{J}$$
Polar moment of inertia
$$J=\frac{1}{2} \pi r^{4}$$
Young modulus definition
$$G=\frac{l \tau}{dx}$$
Contraction due to shear
$$dl=l - \sqrt{l^{2} - dx^{2}}$$

The Attempt at a Solution

expressing $$dx$$ with $$G$$
$$dx=\frac{l \tau}{G}$$
substituing $$dx$$ to $$dl$$
$$dl=l - \sqrt{l^{2} - \frac{l^{2} \tau^{2}}{G^{2}}}$$
substituing $$\tau$$ to $$dl$$
$$dl=l - \sqrt{l^{2} - \frac{M^{2} l^{2} r^{2}}{G^{2} J^{2}}}$$
substituing $$J$$ to $$dl$$
$$dl=l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}$$
dr is square root of dl (multiplying with $$\sqrt{m}$$ to keep units)
$$dr=- \sqrt{dl} \sqrt{m}$$
substituing $$dl$$ to $$dr$$
$$dr=- \sqrt{m} \sqrt{l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}}$$
solving for G
$$G=- 2 \mathbf{\imath} m \sqrt{M^{2}} \sqrt{l^{2}} \sqrt{\frac{1}{\pi^{2} dr^{4} r^{6} - 2 l m \pi^{2} dr^{2} r^{6}}}$$
$$G = 9215558692.57753 \frac{N}{m^{2}}$$

 

[KataKoniK]

Thank you for your question. To answer your question, we need to consider the equilibrium of the circular beam under the applied torque. We can use the torsional shearing stress equation to relate the torque (M) and the radius (r) to the applied stress (\tau), which is given by \tau=\frac{M r}{J}. Here, J is the polar moment of inertia, which is calculated using the beam's radius as J=\frac{1}{2} \pi r^{4}.

Next, we can use the definition of Young's modulus (G=\frac{l \tau}{dx}) to relate the applied stress (\tau) to the strain (dx) in the beam. The strain in the beam is the change in length (dl) divided by the original length (l).

We can also use the equation for contraction due to shear (dl=l - \sqrt{l^{2} - dx^{2}}) to calculate the change in length (dl) of the beam due to the applied torque.

Substituting the expression for dx into the equation for dl and solving for dl, we get dl=l - \sqrt{l^{2} - \frac{l^{2} \tau^{2}}{G^{2}}}.

Now, we can substitute the given values for M, r, and l into the equation for dl to calculate the change in length (dl) of the beam. We can then use the given value for dr to calculate the change in radius (dr) of the beam.

Finally, we can use the equation for Young's modulus (G=\frac{l \tau}{dx}) to relate the change in length (dl) to the change in radius (dr) and solve for G.

Using the values given in the question, we get G = 9215558692.57753 \frac{N}{m^{2}}.

I hope this helps clarify the concept for you. Let me know if you have any further questions.