(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]\tau=\frac{M r}{J}[/tex]

Polar moment of inertia

[tex]J=\frac{1}{2} \pi r^{4}[/tex]

Young modulus definition

[tex]G=\frac{l \tau}{dx}[/tex]

Contraction due to shear

[tex]dl=l - \sqrt{l^{2} - dx^{2}}[/tex]

3. The attempt at a solution

expressing [tex]dx[/tex] with [tex]G[/tex]

[tex]dx=\frac{l \tau}{G}[/tex]

substituing [tex]dx[/tex] to [tex]dl[/tex]

[tex]dl=l - \sqrt{l^{2} - \frac{l^{2} \tau^{2}}{G^{2}}}[/tex]

substituing [tex]\tau[/tex] to [tex]dl[/tex]

[tex]dl=l - \sqrt{l^{2} - \frac{M^{2} l^{2} r^{2}}{G^{2} J^{2}}}[/tex]

substituing [tex]J[/tex] to [tex]dl[/tex]

[tex]dl=l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}[/tex]

dr is square root of dl (multiplying with [tex]\sqrt{m}[/tex] to keep units)

[tex]dr=- \sqrt{dl} \sqrt{m}[/tex]

substituing [tex]dl[/tex] to [tex]dr[/tex]

[tex]dr=- \sqrt{m} \sqrt{l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}}[/tex]

solving for G

[tex]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}}}[/tex]

[tex]G = 9215558692.57753 \frac{N}{m^{2}}[/tex]

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

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Increase of radius of circular rod due to torque

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