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# Clebsch Method (Macaulay's Method/Double Integration Method)

## Calculating Deflections and Rotations in Beams

**In this text, you will learn** what the **Clebsch Method** (also known as **Macaulay's Method** or the **Double Integration Method**) is and how it works. You will see **how to write boundary conditions** (kinematic) depending on the method of beam support. **You will find examples of solutions**.

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### Clebsch Method

The differential equation of the deformed axis of the beam that we start with in the**Clebsch Method**has the following form:

- E - Young's modulus,

- J - moment of inertia of the beam cross-section relative to the horizontal axis.

The "-" sign on the right side of the equation is due to the adopted coordinate system and the convention specifying the sign of the bending moment.

To determine the deflections of the beam, we integrate the above equation twice and obtain the

**first derivative**- the

**rotation function**of the beam

and the

**second derivative - deflection function**

C and D are constants of integration.

We determine the integration constants from the kinematic boundary conditions, i.e. conditions for zero deflections and rotation angles in a specific type of support. See what types of supports are used for beam on a plane

As we noticed, the

**Clebsch method**, while maintaining certain conditions of notation, allows us to obtain an equation of the deflection line containing only two unknowns (integration constants) independently of the number of intervals for a simple beam.

The principles governing the use of the Clebsch method can be summed up in 4 points:

- **the ordinates in all intervals must be measured from the same point**

we assume one coordinate system for a simple beam, we can't write parts of the function from one side and parts from the other side of the beam

- **in the case of continuous loading, it cannot be interrupted**

if such a case occurs, the continuous load should be extended to the end of the beam, while adding an equal load with the opposite sign (counter-load)

- **all newly occurring terms in the expression for the bending moment must contain the factor \( (x- l_{i-1}) \),**

where:

\(l_{i-1}\) indicates the coordinate of the beginning of the i-th beam interval

- in the case of a concentrated moment M - **we multiply the moment by the moment arm raised to the power of 0**

- **integration should be performed without expanding expressions in parentheses**

the integration constants apply to the entire beam (for all intervals)

If the coordinates \(l_{\mathrm{i}}\) specify the position of the concentrated forces \(P_{\mathrm{i}}\)
or the beginnings of continuous loads \(q_{\mathrm{i}}\),
the expressions like \(P_i\left(x-l_i\right)\) or \(q_i \frac{\left(x-l_i\right)^2}{2}\) are integrated according to the scheme

And now let's look at the example below and see the solution in practice.

**See an Example**

##### A beam with three intervals and different types of loads.

Calculate the deflection and rotation at a given point on the beam.

#### Solution to the example from the video course

#### Text

#### Solution

We calculate the support reactions

\begin{aligned} &\sum{M_B}=0 -20\cdot 3+30\cdot 1.5+10+15\cdot 6-R_C\cdot 3=0 R_A=28.33kN\\ &\sum{M_C}=0 -20\cdot 6+R_B\cdot 3-30\cdot 1.5+10+15\cdot 3=0 R_B=36.67kN\\ &\sum{P_iY}=0 -20+R_B-30+R_C-15=0 L=P\\ \end{aligned}We write the moment function on the left side.

You can also write the function on the right side. We encourage you to check this variant, calculate the desired displacement, and compare the results.

\begin{aligned} &M_g(x)=-20x+R_B(x-3)-\frac{1}{2}q(x-3)^2+R_C(x-6)+10(x-6)^0+\frac{1}{2}q(x-6)^2\\ &EJ\cdot w"=-M_g(x)=20x-R_B(x-3)+5(x-3)^2- 28.33(x-6)-10(x-6)^0-5(x-6)^2\\ &EJ\cdot w'=20\frac{x^2}{2}-36.67\frac{(x-3)^2}{2}+5\frac{(x-3)^3}{3}-28.33\frac{(x-6)^2}{2}-10(x-6)-5\frac{(x-6)^3}{3}+C\\ &EJ\cdot w=20\frac{x^3}{6}-36.67\frac{(x-3)^3}{6}+5\frac{(x-3)^4}{12}-28.33\frac{(x-6)^3}{6}-10\frac{(x-6)^2}{2}-5\frac{(x-6)^4}{12}+Cx+D\\ \end{aligned}Boundary conditions

\begin{aligned} &w(x=3)=0 \Rightarrow 90+3C+D=0\\ &w(x=6)=0 \Rightarrow 588.735+6C+D=0\\ &C=-166.245\\ &D=408.735\\ \end{aligned}We calculate the deflection at point A.

If we have chosen the coordinate system at the left end of the beam, then point A has the coordinate x=0.

Therefore \begin{aligned} &w_A(x=0)=\frac{1}{EI}\cdot (D)\\ \end{aligned} \begin{aligned} &w_A=\frac{1}{EI}\cdot (408.735)\\ \end{aligned}

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