Shaft on bearings, in two planes

The shaft is solved in two planes, the vertical and the horizontal, one at a time. In each, the two bearings are simple supports and the loads are that plane’s components: of every force, of every applied moment and, in the vertical plane, the shaft’s own weight where it is counted. Two simple supports leave nothing indeterminate: the two reactions follow from the two equations of equilibrium, and the shear and the bending moment at a section from the forces on one side of it. The diameters do not enter any of these, except through the weight of the shaft.

Slope and deflection are where the diameters count. The shaft is cut at every shoulder, bearing, force and applied moment, and on each piece a plane’s bending moment is one polynomial — a straight line, or a parabola under the shaft’s own weight — over one flexural stiffness EI, with I = π(D⁴ − d⁴)/64 for an outer diameter D and a bore d. Integrating M/EI once gives the slope, a cubic at most, and once more the deflection, a quartic at most: closed forms, written out term by term and evaluated where they are asked for, with no numerical integration and no interpolation. Slope and deflection carry over unbroken from one piece into the next, across a shoulder as across an applied moment, and the two constants of integration are fixed by the deflection being zero at the two bearings.

A section feels the two planes together. The resultant — the square root of the sum of the squares of the two components — is taken of the shear, the bending moment, the slope and the deflection alike. For a round section that is exact: its moment of inertia is the same about every diameter, so the moments of the two planes are one moment about one diameter. The resultant deflection is how far the centre of the section has moved, and the resultant slope the angle its axis has turned through, whatever the direction of either; a bearing’s radial load is the resultant of its two reactions.

The bending stress is the resultant moment over the section modulus of the segment the section is in, Z = π(D⁴ − d⁴)/(32D). At a shoulder with no applied moment on it, the moment is one value and the section is two, so the stress there is two values, the higher on the side of the smaller modulus. Each segment is given the largest moment within its own length over its own modulus, and the largest stress on the shaft is the largest of those: it may be at a shoulder, far from the largest moment, as in the example the page opens with.

The largest resultant is not, in general, where either plane’s own diagram peaks. On each piece the sum of the squares of the two planes’ values is a polynomial, and the largest moment, slope and deflection are looked for at the sign changes of its derivative, and at the ends of the pieces: each is the value at the place it is printed with, not the largest of the points plotted.

To check the page by hand, four shafts with a closed form. L is the distance between the two bearings, and a shaft of one solid diameter D has I = πD⁴/64.

One diameter, a force P midway between the bearings: each reaction P/2; deflection under the force PL³/(48EI); slope at each bearing PL²/(16EI).

One diameter, a force P on an overhang, at a distance a beyond one bearing: that bearing carries P(L + a)/L, the other Pa/L the opposite way; moment Pa at the near bearing; deflection under the force Pa²(L + a)/(3EI).

The first case with P at 45° between the two planes: each plane carries P/√2 and gives 1/√2 of that deflection and slope; the resultants are the first case’s, whole.

The first case on a stepped shaft, its middle half of stiffness nEI and its two end quarters of EI: deflection under the force PL³(1 + 7/n)/(384EI); slope at each bearing PL²(1 + 3/n)/(64EI).