Torsion of Thin Walled Sections Questions and Answers

Answer: C

Correct answer: 3.04e6 N m^2

\[ GJ = 27 \times 10^9 \times 1.125 \times 10^{-4} = 3.04 \times 10^6\ \mathrm{N\,m^2}. \] Torsional rigidity is simply the product of the material property and the geometric one, so a change of material scales it directly - substituting steel at 80 GPa would treble it, at three times the density.

Answer: D

The rate is \( T/GJ = 90\,000/3.04 \times 10^6 = 0.0296 \) radians per metre, which is \( 0.0296 \times 57.3 = 1.70 \) degrees per metre. Over a ten-metre semi-span that would accumulate to seventeen degrees, which shows why wing torsional rigidity is watched so closely.

Answer: A

In the closed-section expression the thickness appears only in the line integral in the denominator, so \( J \) is proportional to it. Contrast the open-section case, where the thickness is cubed and the same change would give a factor of eight.

Answer: B

Using \( J = \pi(D^4 - d^4)/32 \), the bracket is \( 1.296 \times 10^7 - 2.56 \times 10^6 = 1.04 \times 10^7 \), and dividing by \( 32/\pi \) gives \( 1.02 \times 10^6 \) mm\( ^4 \). Removing the inner material has cost only about twenty per cent of the solid value while saving forty-four per cent of the weight.

Answer: C

Correct answer: 2880 mm^4

For an open section the torsion constant sums the contributions of the individual walls: \[ J = \tfrac{1}{3}\sum bt^3 = \tfrac{1}{3} \times 4 \times 80 \times 3^3 \] \[ = \tfrac{1}{3} \times 8640 = 2880\ \mathrm{mm^4}. \] The cube on thickness is what makes open sections so poor in torsion - the same four walls closed into a box would give a torsion constant thousands of times larger.

Answer: D

Correct answer: 4 degrees

Twist is proportional to length at constant torque and section: \[ \theta = \frac{TL}{GJ} \quad\Rightarrow\quad \theta \propto L, \] so doubling the length doubles the twist, from 2 degrees to 4. The relation is linear, not cubic - the cube belongs to deflection under a transverse load, which is a different problem.

Answer: A

For a thin circular tube \( J = 2\pi r^3 t \), so the cube on the radius gives a factor of \( 2^3 = 8 \). Enlarging the tube is therefore far more effective than thickening its wall, which would give only a factor of two.

Answer: B

Correct answer: 1000 N m

Rearranging the twist relation, \[ T = \frac{GJ\,\theta}{L} = \frac{4 \times 10^4 \times 0.05}{2} = 1000\ \mathrm{N\,m}. \] The angle must be in radians for this to work; using degrees directly would give an answer about 57 times too large, which is the commonest error in torsion arithmetic.

Answer: C

Correct answer: 2.0 degrees

Shafts in series carry the same torque and their twists add, exactly as extensions add for bars in series under the same load: \[ 1.2 + 0.8 = 2.0\ \text{degrees}. \] Shafts in parallel would be the other case, sharing the torque in proportion to their rigidities while twisting through the same angle.

Answer: D

Weight fixes the material area, and the tube places all of it at a large radius where it contributes most to the polar second moment. The stiffness gain can be several fold, which is why drive shafts and control rods are almost always tubular.