Edupanda » Strength of Materials » Castigliano's theorem - calculating deflections and rotations in beams

# Castigliano's theorem

## Calculating deflections and rotations in beams

**4 examples**from this section

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**In this text you will learn**what Castigliano's theorem is and how it works. You will see

**the algorithm for solving**problems with this method.

**You will find example solutions**

### Castigliano's method

The value of the accumulated elastic energy in a bent rod (due to the bending moment alone).\begin{aligned} U^M=\int_{0}^{l}{\frac{M_g^2}{2EI}}dx \end{aligned}

In the case of a sudden change in the cross-section of the rod or a system composed of several rods - we determine the elastic energy as the sum of the energy accumulated on the length \(l_i\) of the rod.

\begin{aligned} U^M=\sum \int_{0}^{l_i}{\frac{M_g^2}{2EI}}dx \end{aligned}

and after differentiation we get

\begin{aligned} \delta_i=\sum \frac{1}{EI} \int_{0}^{l_i} (M_g \cdot \frac{\partial M_g}{\partial P_i} dx) \end{aligned}

The above equation is of a general nature.

- when determining __deflection__, we differentiate the bending moment with respect to the force acting at the location and in the direction of the desired displacement, and

- when determining __rotation angle__, we differentiate the bending moment with respect to the applied moment as described above.

#### Example 1

#### Content

Calculate the deflection and rotation angle at the point where the concentrated force is applied. Load on the beam is in [kN].

#### Solution

##### First, we calculate the deflection at point B.

When calculating__deflection__, we differentiate the function of the

**bending moment**with respect to the

**force**acting at the location and in the direction of the desired displacement.

To be able to differentiate with respect to force, we need to assign it some notation, let's take

**P**=30 [kN].

Therefore (this is

**very important**) - in the task there IS already a force acting at the location and in the direction of the desired displacement

*(the 30 kN force is a concentrated vertical force - it corresponds to the desired vertical displacement).*

Hence, we don't need to do anything extra, we calculate the task as it is given.

If we were looking for

**rotation angle at point B**, then we would notice that there is NO force corresponding to the desired displacement at point B (the

**rotation angle**corresponds to the

**concentrated moment**).

If there is no concentrated moment applied where we need to calculate the rotation angle, we need to add a fictitious moment \(M^*=0\), calculate the reactions considering this moment, perform the differentiation with respect to \(M^* \) . In the final step, we substitute \(M^*=0\).

##### We calculate support reactions

\begin{align} & \sum{Y}=0\\ & R_{CY}=P^*\\ & \sum{M_C}=0\\ & M_{A}-P^*\cdot 2=0\\ & M_{A}=2P^* \\ \end{align} We write down the moment function in the AB range and BC range from the left side.The function can also be written from the right side, one interval from left A-B, and the other from right C-B.

We encourage you to check different variants for practice, calculate the desired displacement and compare the results.

Moment functions in the A-B range => x∈<0;2)m and BC range => x∈<3;5)m

\begin{aligned} &M_{g1}^{AB}=M_A=2P^*\\ &M_{g2}^{BC}=2P^*-P^*\cdot (x-3)\\ \end{aligned} Derivatives of moment functions with respect to \(P^*\) in each range, i.e. \(\frac{\partial M g_i(x)}{\partial P^*}\)

\begin{aligned} &\frac{\partial M_{g1}}{\partial P^*}=2 \\ &\frac{\partial M_{g2}}{\partial P^*}=2-(x-3)=5-x \\ \end{aligned} We calculate the deflection at point B.

According to the formula presented earlier:

\begin{aligned} \delta_i=\sum \frac{1}{EI} \int_{0}^{l_i} (M_{gi} \cdot \frac{\partial M_{gi}}{\partial P^*} )dx \end{aligned}

where:

\(\delta_i\) - the desired displacement at point B \(\Rightarrow \Delta_{y_B}\)

So, finally:

\begin{align} & \Delta_{y_B}=\frac{1}{EI}\left[ \int_{0}^{3} (M_{g1} \cdot \frac{\partial M_{g1}}{\partial P^*} )dx + \int_{3}^{5} (M_{g2} \cdot \frac{\partial M_{g2}}{\partial P^*} )dx \right] \\ & \Delta_{y_B}=\frac{1}{EI}\left[ \int_{0}^{3} (2\cdot P^* \cdot 2 )dx + \int_{3}^{5} (2P^*-P^*\cdot (x-3)) \cdot (5-x)dx \right] \\ \end{align} In Castigliano's method tasks,

**at this stage we remind ourselves of the value of the force after which we performed the differentiation.**In this case, \(P^*=30\ [kN]\) and we substitute that value. \(\begin{aligned} & \Delta_{\mathrm{yB}}=\frac{1}{\mathrm{EI}}\left[\int_0^3 2 \cdot 30 \cdot 2 \mathrm{dx}+\int_3^5[2 \cdot 30-30 \cdot(\mathrm{x}-3)](5-\mathrm{x}) \mathrm{dx}\right] \\ & \Delta_{\mathrm{yB}}=\frac{1}{\mathrm{EI}}\left[\int_0^3 120 \mathrm{dx}+\int_3^5\left(30 \cdot \mathrm{x}^2-300 \cdot \mathrm{x}+750\right) \mathrm{dx}\right] \\ & \Delta_{\mathrm{yB}}=\frac{1}{\mathrm{EI}}\left[120 \cdot 3+30 \cdot \frac{\left(5^3-3^3\right)}{3}-300 \cdot \frac{\left(5^2-3^2\right)}{2}+750(5-3)\right]=\frac{440}{\mathrm{EI}} \end{aligned}\)

#### Second part of the solution - calculation of rotation angle

**We notice that there is NO**force corresponding to the desired displacement at point B (the

**rotation angle**corresponds to the

**concentrated moment**).

If there is no concentrated moment applied where we need to calculate the rotation angle, we need to add a fictitious moment \(M^*=0\), calculate the reactions considering this moment, perform the differentiation with respect to \(M^* \) . In the final step, we substitute \(M^*=0\).

##### We calculate new support reactions

\begin{align} & \sum{Y}=0\\ & R_{CY}=30\\ & \sum{M_C}=0\\ & M_{A}-30\cdot 2+M^*=0\\ & M_{A}=60-M^* \\ \end{align} Moment functions in the A-B range from the left side => x∈<0;3)m and in the C-B range from the right side => x∈<0;2)m\begin{aligned} &M_{g1}^{AB}=M_A=60-M^*\\ &M_{g2}^{CB}=R_{Cy}\cdot x=30\cdot x\\ \end{aligned} Derivatives of the moment function with respect to \(M^*\) in each interval, i.e. \(\frac{\partial M g_i(x)}{\partial M^*}\)

\begin{aligned} &\frac{\partial M_{g1}}{\partial M^*}=-1 \\ &\frac{\partial M_{g2}}{\partial M^*}=0 \\ \end{aligned} We calculate the rotation angle at point B.

\begin{align} & \varphi_{B}=\frac{1}{EI}\left[ \int_{0}^{3} (M_{g1} \cdot \frac{\partial M_{g1}}{\partial M^*} )dx + \int_{0}^{2} (M_{g2} \cdot \frac{\partial M_{g2}}{\partial M^*} )dx \right] \\ & \varphi_{B}=\frac{1}{EI}\left[ \int_{0}^{3} (60-M^*) \cdot (-1) )dx + \int_{0}^{2} (30\cdot x \cdot 0)dx \right] \\ \end{align} In Castigliano's method problems, we always

**remind ourselves at this stage what is the value of the force after which we differentiated.**In this case, \(M^*=0 \) and we substitute this value. \(\begin{aligned} & \varphi_{B}=\frac{1}{EI}\left[ \int_{0}^{3} (60-0) \cdot (-1) )dx + \int_{0}^{2} (0)dx \right] \\ & \varphi_{B}=\frac{1}{EI}\left[ \int_{0}^{3} (-60) dx \right] = \frac{1}{EI}\left[ -60\cdot 3 \right] =\frac{-180}{EI} \\ \end{aligned}\)

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