Bending of Thin Walled Beams Questions and Answers

Answer: A

For a section symmetric about the plane of bending the stress varies linearly with distance from the neutral axis, giving the familiar expression. It rests on plane sections remaining plane, a linearly elastic material and bending about a principal axis, and every one of those assumptions has to be revisited for an unsymmetric section.

Answer: B

With no axial force applied, the resultant of the direct stresses must vanish, and that condition places the neutral axis through the centroid. The shear centre is a different point entirely and governs torsion rather than bending.

Answer: C

Symmetry about either reference axis makes the integral of the product of coordinates cancel in pairs, so the product of inertia is zero and the reference axes are principal axes. An unsymmetric section such as a Z or an angle has a non-zero value and needs the general bending theory.

Answer: D

Rotating the reference axes changes the second moments and the product of inertia, and at one particular orientation the product vanishes. Those are the principal axes, and about them the second moments take their maximum and minimum values and simple bending theory applies again.

Answer: A

Only when bending occurs about a principal axis does the neutral axis lie perpendicular to the moment. With a non-zero product of inertia the section prefers to bend about a different direction, so the neutral axis rotates away and a moment applied in one plane produces deflection in another.

Answer: B

The coupling introduced by the product of inertia means a purely vertical moment produces curvature about an inclined axis. The beam therefore moves sideways as well as downward, which is exactly why an unsymmetric section cannot be analysed with the simple formula.

Answer: C

Shifting the reference axis away from the centroid adds a transfer term equal to the area times the square of the distance moved. For a thin-walled section built from booms far from the centroid, that transfer term dominates and the members' own centroidal inertias are often neglected entirely.

Answer: D

Concentrating the direct-stress-carrying area into discrete booms and leaving the skin to carry shear reduces the section to a handful of areas at known positions. The bending calculation then becomes a summation rather than an integration, and the accuracy is good because the stringers really do carry most of the direct load.

Answer: A

The standard boom formula shares the panel's area between its two ends in proportion to a term containing the stress ratio, so that the idealised section carries the same direct load as the real one. For a panel in uniform stress the split is simply half to each end.

Answer: B

Treating the wall as a thin ring and integrating the square of the distance from the diameter gives pi r cubed t. The cube on the radius is the important part: doubling the radius of a tube at constant wall thickness multiplies its bending stiffness eightfold.