# Nonlinear Finite Element Analysis Formulation for Shear in Reinforced Concrete Beams
## 5 Force-Based Element Formulation of the Timoshenko Beam
### Direct Integration of the Equations of Equilibrium  
Define the following functions as integrals of the applied transverse and axial loads [16]:  
\[Q(\xi)\equiv\int_{0}^{\xi}q(\eta)d\eta \tag{15a}\] \[M(\xi)\equiv\int_{0}^{\xi}\eta q(\eta)d\eta\] (15b) \[P(\xi)\equiv\int_{0}^{\xi}p(\eta)d\eta \tag{15c}\]  
The integration of Equation (1c) gives:  
\[m^{\prime}(\xi)=Q(\xi)+c_{1}=\Big{(}1-\frac{\xi}{L}\Big{)}Q(\xi)+\frac{\xi}{L} Q(\xi)+c_{1} \tag{16}\]  
Applying the integration by parts, and integrating the equation above, leads to:  
\[\begin{array}{ll}m(\xi)&=\Big{(}\xi-\frac{\xi^{2}}{2L}\Big{)}Q(\xi)-\int_{0} ^{\xi}\Big{(}\eta-\frac{\eta^{2}}{2L}\Big{)}\eta(\eta)d\eta+\frac{\xi^{2}}{2L }Q(\xi)-\int_{0}^{\xi}\frac{\eta^{2}}{2L}q(\eta)d\eta+c_{1}\xi+c_{2}\\ &=\xi Q(\xi)-\int_{0}^{\xi}\eta q(\eta)d\eta+c_{1}\xi+c_{2}\\ &=\xi Q(\xi)-M(\xi)+c_{1}\xi+c_{2}\end{array} \tag{17}\]  
By solving for the constants and rearranging the equation above, we obtain:  
\[m(\xi)=m_{0}b_{1}(\xi)+m_{L}b_{2}(\xi)+\frac{\xi}{L}\overline{M}-M(\xi)+\Big{[} Q(\xi)-\overline{Q}\Big{]}\xi \tag{18}\]  
where, \(b_{1}(\xi)\equiv 1-\xi/L;b_{2}(\xi)\equiv\xi/L;m_{0}\equiv m(0);m_{L}\equiv m(L); \overline{M}\equiv M(L)\); and \(\overline{Q}\equiv Q(L)\). The integration of Equation (1e) and rearrangement give:  
\[\
u(\xi)=-Q(\xi)+\
u_{0} \tag{19}\]  
where, \(\
u_{0}=\
u(0)\). From Equation (19) and knowing that \(m^{\prime}=-\
u\), we can obtain:  
\[\begin{array}{ll}m(\xi)&=\int_{0}^{\xi}Q(\eta)d\eta-\
u_{0}\xi+m_{0}\\ &=\xi Q(\xi)-\int_{0}^{\xi}\eta q(\eta)d\eta-\
u_{0}\xi+m_{0}\end{array} \tag{20}\]  
Along with Equation (15), substituting \(L\) into Equation (20) and solving for \(\
u_{0}\) give:  
\[\
u_{0}=-\frac{m_{L}}{L}+\overline{Q}-\frac{\overline{M}}{L}+\frac{m_{0}}{L} \tag{21}\]  
Rearranging Equation (19) leads to:  
\[\
u(\xi)=-Q(\xi)-\frac{m_{L}}{L}+\overline{Q}-\frac{\overline{M}}{L}+\frac{m _{0}}{L} \tag{22}\]  
The integration of Equation (1a) and rearrangement give:  
\[n(\xi)=n_{0}+P(\xi) \tag{23}\]  
where, \(n_{0}=n(0)\).  
Let us denote the force field and the discrete element force parameters as:  
\[\sigma(\xi)\equiv[n(\xi),\ m(\xi),\ \
u(\xi)]^{T} \tag{24}\]  
\[\mathbf{s}\equiv[n_{0},\ m_{0},\ m_{L}]^{T} \tag{25}\]  
The force field can be written as:  
\[\sigma(\xi)=\mathbf{b}^{T}(\xi)\mathbf{s}+\sigma^{p}(\xi) \tag{26}\]  
where, \(\mathbf{b}^{T}(\xi)\) and the particular force field \(\sigma^{p}(\xi)\) are defined as:  
\[\mathbf{b}^{T}(\xi)\equiv\left[\begin{array}{ccc}1&0&0\\ 0&b_{1}(\xi)&b_{2}(\xi)\\ 0&1/L&-1/L\end{array}\right] \tag{27}\]  
\[\sigma^{p}(\xi)\equiv\left[\begin{array}{c}P(\xi)\\ \frac{\xi}{L}\overline{M}-M(\xi)+\left[Q(\xi)-\overline{Q}\right]\xi\\ -\frac{\overline{M}}{L}+\overline{Q}-Q(\xi)\end{array}\right] \tag{28}\]  
The end forces can be defined as:  
\[\hat{\mathbf{s}}\equiv\left[-n_{0},\ \
u_{0},\ -m_{0},\ n_{L},\ -\
u_{L},\ m_{L} \right]^{T} \tag{29}\]  
as shown in Figure 4.  
This can be expressed in terms of the essential force parameters and the applied loads as [16]:  
\[\hat{\mathbf{s}}=\mathbf{Ls}+\mathbf{P} \tag{30}\]  
where, the equivalent nodal load vector is denoted:  
\[\mathbf{P}=\left[\begin{array}{cccc}0&\overline{M}/L-Q&0&\overline{P}&- \overline{M}/L&0\end{array}\right] \tag{31}\]  
and the matrix \(\mathbf{L}\) is given by:  
\[\mathbf{L}^{T}=\left[\begin{array}{cccc}-1&0&0&1&0&0\\ 0&1/L&-1&0&-1/L&0\\ 1&1/L&0&1&-1/L&1\end{array}\right] \tag{32}\]