# Increase of radius of circular rod due to torque

1. Feb 15, 2010

### magwas

1. The problem statement, all variables and given/known data

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

2. Relevant 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
$$dl=l - \sqrt{l^{2} - dx^{2}}$$

3. 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
$$G=- 2 \mathbf{\imath} m \sqrt{M^{2}} \sqrt{l^{2}} \sqrt{\frac{1}{\pi^{2} dr^{4} r^{6} - 2 l m \pi^{2} dr^{2} r^{6}}}$$
$$G = 9215558692.57753 \frac{N}{m^{2}}$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution