Maxwell-Mohr Method

Introduction to energy methods

The course “Energy methods - Displacements in statically determinate systems” focuses on energy methods that utilize the principles of energy conservation in elastic systems to predict the behavior of materials under load.

Maxwell-Mohr method - simplified approach

(considering only the effect of bending moment)

The Maxwell-Mohr method is another important energy technique used for analyzing displacements in structures. It is a more general method, which includes both mechanical and thermal loads, geometric and elastic supports. The Maxwell-Mohr formula takes into account the influence of all possible interactions on the structure. In this method, displacements due to mechanical loads are calculated by integrating the products of bending moment functions, shear forces, and normal forces for initial and unit states.

In this basic course we will only address the influence of moments on displacement due to mechanical loads. Furthermore, we want to demonstrate the basic approach, in which we will perform analytical integration of moment functions.

Basic formula of the method

According to the Maxwell-Mohr method, determining the displacement u reduces to calculating the integral, under which the bending moment caused by the actual external load (Mg) appears, as well as the bending moment that would be caused by a unit fictitious force P=1 odpowiadaj corresponding to the sought displacement (\(M_1\)).

\[ u=\int_{0}^{l} \frac{M_{g} \cdot M_{1}}{E I} dx \]

Integration can be performed analytically or graphically. In the following example, we will show both approaches.

See graphical integration - Wereszczagin method

From this course, you will learn:

Maxwell-Mohr method (analytical integration of moment functions):

how to determine bending moment functions in intervals,
how to define the unit state and determine moment functions,
how to analytically integrate products of moment functions over intervals,
how to calculate deflection and angle of rotation of a beam at any point.

Wereszczagin method:

how to determine moment diagrams in the initial and unit states,
how to graphically integrate moment diagrams,
how to calculate deflection and angle of rotation of a beam at any point.

Example 1

Content

Calculate the vertical displacement of point D.

Consider the influence of bending moments.

Przykład 1 - metoda Maxwella-Mohra — *Fig. 1. Beam diagram - example 1*

Solution

We break the hinged beam into straight beams, calculate the support reactions, and draw the internal force diagrams. Note - reactions can be calculated without breaking into straight beams. Checking the extremum is not necessary to calculate the displacement.

Rozbicie belki — *Fig. 2. Breaking the hinged beam into straight beams*

Support reactions

\[ \begin{aligned} & \Sigma M_{B}=0 \\ & 10+2 V_{C}=0 \\ & V_{C}=-5 \text{ kN} \\ & \Sigma Y=0\\ & V_{B}+V_{C}=0 \\ & V_{B}=5 \text{ kN} \end{aligned} \]

II

\[ \begin{aligned} & \Sigma M_{A}=0 \\ & -M_{A}-6 \cdot 4 \cdot 2-4 V_{B}=0 \\ & M_{A}=-68 \text{ kNm} \\ & \Sigma Y=0 \\ & V_{A}-6 \cdot 4-V_{B}=0 \\ & V_{A}=29 \text{ kN} \end{aligned} \]

Bending moment functions - initial state

\[ \begin{aligned} & \text{AB} \quad 0 < x < 4 \\ & M_1(x) = -68 + 29 \cdot x - 6 \cdot \frac{x^2}{2} \\ & \quad M_1(0) = -68 \text{ kNm} \\ & \quad M_1(4) = 0 \text{ kNm} \\ \\ & \text{BC} \quad 4 < x < 6 \\ & M_2(x) = -68 + 29 \cdot x - 6 \cdot 4 \cdot (x - 2) \\ & \quad M_2(4) = 0 \text{ kNm} \\ & \quad M_2(6) = 10 \text{ kNm} \\ \\ & \text{DC} \quad 0 < x < 2 \\ & M_3(x) = 10 \\ & \quad M_3(0) = 10 \text{ kNm} \\ & \quad M_3(2) = 10 \text{ kNm} \end{aligned} \]

Stan jednostkowy

W celu obliczenia pionowego przemieszczenia w punkcie D przykładamy pionową siłę jednostkową i rysujemy wykres momentów.

Reakcje podporowe - stan jednostkowy

I

\[ \begin{aligned} & \Sigma M_{B}=0 \\ & -1 \cdot 4+2 V_{C}=0 \\ & V_{C}=2\\ & \Sigma Y=0 \\ & V_{B}+V_{C}-1=0 \\ & V_{B}=-1 \end{aligned} \]

II

\[ \begin{aligned} & \Sigma M_{A}=0 \\ & -M_{A}-4 V_{B}=0 \\ & M_{A}=4 \text{ m} \\ & \Sigma Y=0 \\ & V_{A}-V_{B}=0 \\ & V_{A}=-1 \end{aligned} \]

Bending moment functions - unit state

\[ \begin{aligned} & \textbf{AB:} \quad 0 < x < 4 \\ & m_1(x) = 4 - x \\ & m_1(0) = 4 \text{ m}, \quad m_1(4) = 0 \text{ m} \\[0.5em] & \textbf{BC:} \quad 4 < x < 6 \\ & m_2(x) = 4 - x \\ & m_2(4) = 0 \text{ m}, \quad m_2(6) = -2 \text{ m} \\[0.5em] & \textbf{DC:} \quad 0 < x < 2 \\ & m_3(x) = -x \\ & m_3(0) = 0 \text{ m}, \quad m_3(2) = -2 \text{ m} \end{aligned} \]

Obliczenie przemieszczenia - całkowanie analityczne

Wzór ogólny:

\[ \Delta_{D}=\int_0^4 \frac{M_1(x) \cdot m_1(x)}{EI} dx+\int_4^6 \frac{M_2(x) \cdot m_2(x)}{EI} dx+\int_0^2 \frac{M_3(x) \cdot m_3(x)}{EI} dx \]

Podstawienie funkcji:

\[ \begin{aligned} \Delta_{D}= & \frac{1}{EI} \cdot\left[\int_0^4\left(-68+29 \cdot x-3 \cdot x^2\right) \cdot(4-x) dx+ \right.\\ & \left. +\int_4^6[-68+29 \cdot x-24 \cdot(x-2)] \cdot(4-x) dx + \int_0^2 10 \cdot(-x) dx\right] \end{aligned} \]

Simplification of expressions under the integrals:

\[ \begin{aligned} \Delta_{D}=\frac{1}{EI} \cdot & \left[\int_0^4\left(3 \cdot x^3-41 \cdot x^2+184 \cdot x-272\right) dx+ \right.\\ & \left. +\int_4^6\left(40 \cdot x-5 \cdot x^2-80\right) dx+\int_0^2-10x dx\right] \end{aligned} \]

Step-by-step integration solution:

\[ \begin{aligned} & \int_0^4\left(3x^3-41x^2+184x-272\right) dx+\int_4^6\left(40x-5x^2-80\right) dx+\int_0^2-10x dx= \\ & {\left[\frac{3x^4}{4}-\frac{41x^3}{3}+92x^2-272x\right]_0^4+\left[20x^2-\frac{5x^3}{3}-80x\right]_4^6+\left[-5x^2\right]_0^2=} \\ & \left(\frac{3 \cdot 256}{4}-\frac{41 \cdot 64}{3}+92 \cdot 16-272 \cdot 4\right)+\\ & +\left(\left(20 \cdot 36-\frac{5 \cdot 216}{3}-80 \cdot 6\right)-\left(20 \cdot 16-\frac{5 \cdot 64}{3}-80 \cdot 4\right)\right)+(-5 \cdot 4)= \\ & \left(192-\frac{2624}{3}+1472-1088\right)+\left((720-360-480)-\left(320-\frac{320}{3}-320\right)\right)-20= \\ & \left(\frac{576}{1}-\frac{2624}{3}\right)+\left(-120+\frac{320}{3}\right)-20= \\ & -\frac{896}{3}-\frac{40}{3}-20=-\frac{936}{3}-20=-312-20=-332 \end{aligned} \] \[ \Delta_{D}=\frac{-332}{EI} \]

The fact that the displacement came out negative means that it will occur in the opposite direction to that assumed for the unit force.

Calculation of displacement - graphical integration

This example will clearly show the advantage of graphical integration over analytical when calculating displacements.

\[ \Delta_{D}=\sum \int \frac{M_{P} M_{1}}{EI} dx \]

Całkowanie graficzne — *Fig. 4. Moment diagrams - initial and unit states*

Distribution of the parabola into simple figures:

Rozkład paraboli — *Fig. 5. Distribution of the parabola into simple figures*

If both diagrams are on the same side of the beam - the result of integrating a given section is positive, if on opposite sides of the beam - the result is negative. It can be conventionally assumed that at the bottom we have a positive moment, at the top a negative one.

\[ \begin{aligned} \Delta_{D} &=\sum \int \frac{M_{P} M_{1}}{EI} dx\\ &= \frac{1}{EI}\left[-\frac{1}{3} \cdot 4 \cdot 68 \cdot 4+\frac{1}{3} \cdot 4 \cdot 4 \cdot \frac{6 \cdot 4^{2}}{8}-\frac{1}{3} \cdot 2 \cdot 2 \cdot 10-\frac{1}{2} \cdot 2 \cdot 2 \cdot 10\right]\\ &=-332 \frac{1}{EI} \end{aligned} \]

More information on graphical integration can be found here - Wereszczagin method

Strength of materials

Maxwell-Mohr Method

Introduction to energy methods