Torsion of Thin Walled Sections Questions and Answers

Answer: A

The twists of successive elements add, and each element contributes its torque times its length divided by the local rigidity. Integrating that quantity along the member gives the total twist, exactly as flexibilities add in series.

Answer: B

Correct answer: Zero, if warping is unrestrained

Pure torsion produces shear stress only. Direct stresses arise when the section is prevented from warping out of plane, because the axial displacements it wants to make are resisted - and a closed section left free to warp makes them without hindrance. Restrain it, at a root fitting or a rib, and a self-equilibrating system of direct stresses appears.

Answer: C

The magnitude of the restraint stress depends on how much warping is being prevented. An open section wants to warp a great deal, so restraining it generates large direct stresses; a closed section warps very little and the effect is minor.

Answer: D

The shear flow varies inversely with the enclosed area, so enlarging the cell reduces both flow and stress directly. A higher shear modulus reduces the twist but not the stress, and thinning the wall raises the stress rather than lowering it.

Answer: A

With the enclosed area changing along the span the shear flow changes with it at constant torque, so both the stress and the local rate of twist vary. A wing box, tapering in both chord and depth, is exactly this case.

Answer: B

Correct answer: 6.14e5 mm^4

For a solid circular section \[ J = \frac{\pi d^4}{32} = \frac{\pi (50)^4}{32} = \frac{\pi \times 6.25 \times 10^6}{32} \] \[ = 6.14 \times 10^5\ \mathrm{mm^4}. \] Using \( \pi d^4/64 \), which is the second moment of area for bending, gives half this and is the usual slip - the polar value is twice the diametral one.

Answer: C

Correct answer: 81 MPa

\[ \tau_{max} = \frac{Tr}{J} = \frac{2 \times 10^6 \times 25}{6.14 \times 10^5} = 81\ \mathrm{MPa}. \] Note the units: the torque must be in newton-millimetres to match a \( J \) in millimetres to the fourth power. The radius used is the outer one, since shear stress in a circular shaft grows linearly from zero at the axis to its maximum at the surface.

Answer: D

From \( \theta = TL/GJ \), the numerator is \( 2 \times 10^6 \times 1000 = 2 \times 10^9 \) and the denominator \( 80\,000 \times 6.14 \times 10^5 = 4.91 \times 10^{10} \), giving \( 0.0407 \) radians. Multiplying by \( 57.3 \) converts this to 2.33 degrees.

Answer: A

For a thin closed circular tube \( J = 2\pi r^3 t = 2\pi \times 512\,000 \times 3 = 9.65 \times 10^6 \) mm\( ^4 \). The general closed-section formula gives the same answer, since \( 4A^2/(\oint ds/t) \) reduces to this for a circle.

Answer: B

Correct answer: 1.13e-4 m^4

Bredt’s formula for a single-cell closed section gives \[ J = \frac{4A^2t}{s} = \frac{4(0.15)^2(0.002)}{1.6} \] \[ = \frac{1.8 \times 10^{-4}}{1.6} = 1.13 \times 10^{-4}\ \mathrm{m^4}, \] with \( A = 0.5 \times 0.3 \) the enclosed area and \( s = 2(0.5 + 0.3) \) the perimeter. The enclosed area is squared, which is why enlarging the box is far more effective than thickening its wall.