In this text you will learn what the Wereszczagin Method is used for when integrating bending moment diagrams/functions while calculating displacements of different constructions using the Maxwell-Mohr method formula.

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Introduction to the Wereszczagin Method

step-by-step presentation

To practice, we leave you with the calculation of the rotation angle at point A, expect a result of (=-12.5/EI) :-)

So Wereszczagin noticed that

\(u=\int \frac{M_g \cdot M_1}{E I} \cdot d x=\frac{1}{E I} \cdot \Omega \cdot \eta\)
If we need to integrate together figures (meaning functions), where:
- one is or can be parabolic (quadratic function, under constant continuous load) - it will be the diagram in the Initial State (from given loads),
- the other is non-parabolic - it will be the Unit State, when calculating displacements this will be the diagram from the unit force applied at the point and in the direction of the desired displacement - See more on this topic - Maxwell-Mohr Method,
then the result of such integration can be reduced to the formula:
\(u=\Omega \cdot \eta\)
where:
\( \Omega \) - area of the figure on the moment diagram in the Initial State,
\( \eta \) - ordinate on the diagram in the Unit State at the position (at the height) of the center of gravity of the figure from the Initial State.
The following image shows this graphically.

A few practical notes

1) What does it actually mean if the displacement is negative?
This means that the direction of the displacement will be opposite to the direction of the assumed unit force.

In the case of vertical displacement at point B, a unit force with downward direction was assumed,
the displacement came out negative, which means that in reality it will be upward.
The direction of the unit force corresponding to the desired rotation angle at point A was assumed to be clockwise.

2) When integrating using the Wereszczagin method, we need to pay attention to whether the areas of the integrated figures are on the same side of the beam or on the opposite side.
If they are on the same side, the result of this single integration is positive, if they are on opposite sides, it is negative.

3) Another important thing is the division into simple figures. Both the diagram in the Initial State and in the Unit State, if they are more complicated than: rectangle, right triangle, parabola, we can and will mentally divide them into simple figures.

Examples of division into simple figures

Trapezoid 1

An example similar to the one we considered when deriving the Wereszczagin method.

Trapezoid 2

Now a figure that we actually identify at first glance as a trapezoid.

Sometimes it is useful to calculate the value at the midpoint of the diagram, how can this be done easily?
If the diagram changes linearly, then at the midpoint there is the arithmetic mean of the two extreme values, for example:
- on the first plot \( e=\frac{-a+b}{2} \),
- on the second plot \(e=\frac{-a-b}{2} \),

If the diagram changes parabolically, then to this mean value we need to add or subtract the value \( f=\frac{q\cdot l^2}{8} \) representing the bulge of the parabola that occurs if we have a continuous load on the element.

So \( e=\frac{a+b}{2}\pm \frac{q\cdot l^2}{8} \), where whether we give plus or minus depends on whether the action of the continuous load causes the parabola to be convex or concave, but see the examples below.

Triangle-parabola

The action of the continuous load causes the parabola to be convex, so \( e=\frac{0+a}{2}+\frac{q\cdot l^2}{8} \)

Parabolic trapezoid 1

Now something more interesting and in fact nothing more difficult than two further cases can happen.

\( e=\frac{a+(-b)}{2}-\frac{q\cdot l^2}{8} \)

Parabolic trapezoid 2

\( e=\frac{a+(-b)}{2}+\frac{q\cdot l^2}{8} \)

Parabolic trapezoid 3

\( e=\frac{-a+(-b)}{2}-\frac{q\cdot l^2}{8} \)

Parabolic trapezoid 4

\( e=\frac{-a+(-b)}{2}+\frac{q\cdot l^2}{8} \)

Parabolic trapezoid 5

I think the principle of operation is already clear.
Finally, one more, troublesome at the beginning case. We must remember that if the continuous load is not perpendicular to the rod, then when distributing into simple figures, when calculating the height of the parabola \( f \) as the length, we take the length of the continuous load, and not the length of the rod!.
See the example below

\( e=\frac{0+a}{2}+\frac{q\cdot c^2}{8} \)