# Fixed end moment due to support settlement Tanvesh
Masters in Structural Engineering | Research Interest - Artificial Intelligence and Machine learning in Civil Engineering | Youtuber | Teacher | Currently working as Research Scholar at NIT Goa

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## How to use calculator

1. Select the units from drop down (metric / Imperial)
2. Select fixed end moment due to support settlement or support rotation.
3. Enter the geometric details of the beam i.e. length, youngs modulus, area moment of inertia and flexural rigidity.
4. Select the end of the beam where the support settlement or support rotation takes place. The left end is named as ‘End A’ and the right end is named as ‘End B’.
5. The sign convention for settlement is upward settlement of a support is positive, downward settlement is negative. For rotation of support anticlockwise rotation of support is positive and clockwise rotation of support is negative.

## Fixed end moment due to support settlement derivation \begin{aligned} &M_{x}=R_{B} * x-M_{B}\\&M_{x}^{2}=R_{B}^{2} * x^{2}-2 * R_{B} * M_{B} * x+M_{B}^{2}\\&U=\int_{0}^{L}\frac{M_{x}^{2} d x}{2 E I} \\&U=\int_{0}^{L} \frac{\left(R_{B}^{2} * x^{2}-2 * R_{B} * M_{B} * x+M_{B}^{2}\right) d x}{2 E I} \\&U=\frac{1}{2 E I} *\left(\frac{R_{B}^{2} * L^{3}}{3}-R_{B} * M_{B} * L^{2}+M_{B}^{2} * L\right) \\&\frac{\partial U}{\partial M_{B}}=0 \\&-R_{B} * L^{2}+2 * M_{B} * L=0 \\&\frac{\partial E}{\partial R_{B}}=\frac{R_{B} L}{2} \\&\frac{\partial U}{2 E I} *\left(\frac{2 * R_{B} * L^{3}}{3}-M_{B} * L^{2}\right)=\delta \\&R_{B}=\frac{12 E I \delta}{L^{3}} \\&M_{B}=\frac{6 E I \delta}{L^{2}}\end{aligned}

## Fixed end moment due to support rotation derivation

\begin{aligned}&M_{x}=R_{B} * x-M_{B} \\&M_{x}^{2}=R_{B}^{2} * x^{2}-2 * R_{B} * M_{B} * x+M_{B}^{2} \\&U=\int_{0}^{L} \frac{M_{x}^{2} d x}{2 E I} \\&U=\int_{0}^{L} \frac{\left(R_{B}^{2} * x^{2}-2 * R_{B} * M_{B} * x+M_{B}^{2}\right) d x}{2 E I} \\&U=\frac{1}{2 E I} *\left(\frac{R_{B}^{2} * L^{3}}{3}-R_{B} * M_{B} * L^{2}+M_{B}^{2} * L\right) \\&\frac{\partial U}{\partial R_{B}}=0 \\&\left(\frac{2 * R_{B} * L^{3}}{3}-M_{B} * L^{2}\right)=0 \\&R_{B}=\frac{3 M_{B}}{2 L} \\&\frac{\partial U}{\partial M_{B}}=\theta \\&\frac{1}{2 E I}\left(-\frac{3 M_{B}}{2} * L+2 * M_{B} * L\right)=\theta \\&M_{B}=\frac{4 E I \theta}{L} \\&R_{B}=\frac{6 E I \theta}{L^{2}}\end{aligned}</p><p>

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