# End loaded beam with spring supports

### Derivation

#### Force equilibrium at A

\begin{align} F_{A} + S &= R_{F1} \\ S &= R_{F1} - F_{A} \label{eq1}\tag{Eq.1} \end{align}

#### Moment equilibrium at A

\begin{align} R_{M1} + M_{1} &= M_{A} \\ M_{1} &= M_{A} - R_{M1} \label{eq2}\tag{Eq.2}\\ &= M_{A} - k_{M1} \theta_{A} \label{eq3}\tag{Eq.3} \end{align}

#### Force equilibrium at B

$F_{B} = S + R_{F2}$

Substitute $$S = R_{F1} - F_{A}$$ from $$\ref{eq1}$$.

\begin{align} F_{B} &= R_{F1} - F_{A} + R_{F2} \\ R_{F1} + R_F{2} &= F_{A} + F_{B} \\ k_{F1} \delta_{A} + k_{F2} \delta_{B} &= F_{A} + F_{B} \label{eqA}\tag{A} \end{align}

#### Moment equilibrium at B

\begin{align} M_{2} + M_{B} &= R_{M2} \\ M_{2} &= R_{M2} - M_{B} \\ &= k_{M2} \theta_{B} - M_{B} \label{eq4}\tag{Eq.4} \end{align}

#### Moment equilibrium on beam

$M_{1} + SL = M_{2}$

Substitute $$M_{1} = M_{A} - R_{M1}$$ from $$\ref{eq2}$$ and $$S = R_{F1} - F_{A}$$ from $$\ref{eq1}$$.

\begin{align} M_{A} - R_{M1} + (R_{F1} - F_{A})L &= R_{M2} - M_{B} \\ k_{M1} \theta_{A} + k_{M2} \theta_{B} - k_{F1} \delta_{A} L &= M_{A} + M_{B} - F_{A}L \label{eqB}\tag{B} \end{align}

#### Compatibility of beam

For derivation of compatibility equations, see #euler-beam-end-loaded-compatibility.

##### Compatibility equation 1

\begin{align} \theta_{A} - \theta_{B} &= \frac{L}{2EI}(M_{1} + M_{2}) \\ \end{align}

Substitute $$M_{1} = M_{A} - k_{M1} \theta_{A}$$ from $$\ref{eq3}$$ and $$M_{2} = k_{M2} \theta_{B} - M_{B}$$ from $$\ref{eq4}$$.

\begin{align} \theta_{A} - \theta_{B} &= \frac{L}{2EI}(k_{M2}\theta_{B} + M_{A} - M_{B} - k_{M1}\theta_{A}) \\ \theta_{A} - \theta_{B} + \frac{k_{M1}L}{2EI}\theta_{A} - \frac{k_{M2}L}{2EI}\theta_{B} &= \frac{L}{2EI}(M_{A} - M_{B}) \\ \bigg(\frac{k_{M1}L}{2EI} + 1 \bigg) \theta_{A} - \bigg(\frac{k_{M2}L}{2EI} + 1 \bigg) \theta_{B} &= \frac{L}{2EI}M_{A} - \frac{L}{2EI}M_{B} \label{eqC}\tag{C} \end{align}

##### Compatibility equation 2

\begin{align} \delta_{A} - \delta_{B} + \theta_{A}L &= \frac{L^{2}}{6EI}(2M_{1} + M_{2}) \\ \end{align}

Substitute $$M_{1} = M_{A} - k_{M1} \theta_{A}$$ from $$\ref{eq3}$$ and $$M_{2} = k_{M2} \theta_{B} - M_{B}$$ from $$\ref{eq4}$$.

\begin{align} \delta_{A} - \delta_{B} + \theta_{A}L &= \frac{L^{2}}{6EI}\big(2(M_{A} - k_{M1}\theta_{A}) + k_{M2}\theta_{B} - M_{B}\big) \\ \delta_{A} - \delta_{B} + \theta_{A}L + \frac{k_{M1}L^{2}}{3EI}\theta_{A} - \frac{k_{M2}L^{2}}{6EI}\theta_{B} &= \frac{L^{2}}{6EI}(2M_{A}-M_{B}) \\ \delta_{A} - \delta_{B} + \bigg(L + \frac{k_{M1}L^{2}}{3EI}\bigg)\theta_{A} - \frac{k_{M2}L^{2}}{6EI}\theta_{B} &= \frac{L^{2}}{3EI}M_{A} - \frac{L^{2}}{6EI}M_{B} \label{eqD}\tag{D} \end{align}

#### Summary

Equations $$\ref{eqA}$$, $$\ref{eqB}$$, $$\ref{eqC}$$ and $$\ref{eqD}$$ in matrix form:

$\begin{bmatrix} k_{F1} & k_{F2} & 0 & 0 \\ -k_{F1}L & 0 & k_{M1} & k_{M2} \\ 0 & 0 & \frac{k_{M1}L}{2EI} + 1 & -\frac{k_{M2}L}{2EI} - 1 \\ 1 & -1 & L + \frac{k_{M1}L^{2}}{3EI} & -\frac{k_{M2}L^{2}}{6EI} \\ \end{bmatrix} \begin{bmatrix} \delta_{A} \\ \delta_{B} \\ \theta_{A} \\ \theta_{B} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 & 0 \\ -L & 0 & 1 & 1 \\ 0 & 0 & \frac{L}{2EI} & -\frac{L}{2EI} \\ 0 & 0 & \frac{L^{2}}{3EI} & -\frac{L^{2}}{6EI} \end{bmatrix} \begin{bmatrix} F_{A} \\ F_{B} \\ M_{A} \\ M_{B} \end{bmatrix}$