Checking the Soviets: Determining the Bending Moment and Stress, and the Parallel Axis Theorem

In this section we’ll discuss the bending moment of the section which connects the two rotors. It’s worth noting that, from a vibratory operational standpoint, the two sets of rotors and eccentrics and self-synchonising, which means that, as they rotate, the amplitude of the vibrations synchronise the rotations. This was applied to vibratory equipment by Tünkers. In any case it is necessary to join the two rotating assemblies, and the cross-section shown above does just that.

The cross-sectional area is not “regular” but is an assembly of rectangles whose moment of inertia around their own centroids is easy to compute. To determine their moment of inertia about the central (neutral) axis it is necessary to use the parallel axis theorem. With pile driving equipment this is usually associated with leaders, and in that application it is discussed in the post Section Modulus of Pile Hammer Leaders. For each rectangle the moment of inertia about the neutral axis is given by the equation

{\it I}=1/12\,b{h}^{3}+bh{a}^{2} (1)


  • I = moment of inertia, cm4
  • b = base of rectangle, cm
  • h = height of rectangle, cm
  • a = distance from the centroid of the rectangle to the centroidal (neutral) axis of the entire section

There are two sizes of rectangles here: the large rectangles which extend from one side to the other, and the small “ribs” which are placed on top and bottom. The dimensions for each of these (and some of the results) are as follows:

ParameterLarge RectanglesSmall Rectangles
Total Number214
Base b, cm421
Height h, cm1.53.5
Distance from Neutral Axis a, cm12.214.5
Cross-Sectional Area of Each Rectangle, cm2633.5
Moment of Inertia of Each Rectangle at neutral axis using Equation (1), cm49389739

Adding up all of these produces a total moment of inertia of (9389)(2) + (739)(14) = 29,130 cm4. This distance from the neutral axis to the extreme fibres is 16.5 cm, thus the section modulus S is 29130/16.5 = 1765 cm3.

The maximum moment is estimated as a point load in the centre of the beam and is given by the equation

M=1/4\,g_{{{\it impact}}}WL (2)


  • M = maximum moment, kgf-m
  • gimpact = peak impact deceleration = 150 g’s (see the previous post)
  • W = weight of each rotor-eccentric assembly = 600 kgf
  • L = length of beam = 34 cm

Substituting yields a moment of 765,000 kgf-cm. Since the maximum fibre stress is M/S, it can be computed as 765,000/1765 = 433 kgf/cm2. All of these calculations correlate well with the originals except that they are more precise.

An interesting question arises: the ribs look insignificant, are they really helpful. Lets consider that case. The following changes take place if we remove the ribs:

  • The moment of inertia decreases to 18,777 cm4
  • The distance from the neutral axis to the extreme fibre decreases to 12.95 cm
  • The section modulus decreases to 1450 cm3
  • The maximum stress increases to 765,000/1450 = 527 kgf/cm2

The maximum stress thus increases 22%, certainly a significant increase, even with the rather modest ribs. This is why an important objective of any moment-resisting beam structure is to increase the section modulus by increasing the moment of inertia, the distance to the extreme fibre, or both.


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