Twist of an open versus closed cylinder

[Robin91]

Homework Statement

a) Determine what the thickness should be in a closed tube versus an open tube to have the same twist angle
b) Determine what the thickness should be in a closed tube versus an open tube to have the same max shear stress

G=20GPa
T=50Nm
tr=1mm (for the open tube)

Also see the attachment

Homework Equations

a)
theta=(TL)/(GJ)

For closed:
J=pi/2(Ro^4-Ri^4)
Ro=Outer radius
Ri=inner radius

For open:
J=st^3/3 => J=1/3(D+tr)=1/3*pi*(D+1)
where:
s= circumference of circle (2pi*rm), rm is the radius up to the middle of the bar (between inner and outer, so 1/2D+1/2t)
t=thickness

b)

For closed:
tau=T*Ro/J

For open:
tau=T*t/J

The Attempt at a Solution

I calculated the J for the open and closed tube. However, I get a fourth degree equation, because I tried to substitute Ro=tl+D/2 into the equation to calculate J. After expansion I had terms containing tl^4, tl^3, tl^2 and tl, which resulted in a really long derivation of tl, however, I don't think that is necessary for this assignment. For the shear it was even worse, the resulting equation for tl didn't even fit on my paper.

Do I miss something here?

Thanks in advance,
Robin

P.s. Sorry for the equations written in this way, Latex didn't seem to work, it gave errors (while I was sure I typed it correctly)

 

[Robin91]

Is there anyone who can enlighten me?

Latex does work now, I'll restate the relevant equations (I can't edit my post)..

2. Homework Equations
a)
\theta=\frac{TL}{GJ}

For closed:
J=\frac{\pi}{2(R_0^4-R_0^4)}
Ro=Outer radius
Ri=inner radius

For open:
J=\frac{st^3}{3} => J=\frac{1}{3}(D+tr)=\frac{1}{3}\pi(D+1)
where:
s= circumference of circle (2pi*rm), rm is the radius up to the middle of the bar (between inner and outer, so 1/2D+1/2t)
t=thickness

b)

For closed:
\tau=\frac{TR_0}{J}

For open:
\tau=\frac{Tt}{J}