## 2 Materials and Methods
### Timoshenko Beam Theory  
Now, we try to find out how the moments and forces shown in Figure 2a,b would change due to change in the beam theory. The Timoshenko beam theory takes into account shear deformation and rotational inertia effects, making it suitable for describing the behaviour of short beams. For a homogenous beam of constant cross-section, the Timoshenko beam governing equations are as follows:  
\[\frac{d^{2}}{dx^{2}}\bigg{(}E_{s}I\frac{d\varphi}{dx}\bigg{)}=q(x,t)\frac{dw} {dx}=\varphi-\frac{1}{\kappa AG_{s}}\frac{d}{dx}\bigg{(}E_{s}I\frac{d\varphi}{ dx}\bigg{)} \tag{8}\]  
where \(\varphi\) is the angle of rotation of the normal to the mid-surface of the beam and \(\kappa\) is the shear coefficient factor. The coefficient, \(\kappa\), is a dimensionless quantity, dependent on the shape of the cross-section, which is introduced to account for the fact that the shear stress and strain are distributed not-uniformly over the cross-section [38]. In a linear elastic Timoshenko beam, the bending moment, \(M_{xx}\), and the shear force, \(Q_{x}\), are related to the angle of rotation, \(\varphi\), and the deflection, \(w\), by  
\[M_{xx}=-E_{s}I\frac{\partial\varphi}{\partial x}Q_{x}=\kappa AG_{s}\bigg{(}- \varphi+\frac{\partial w}{\partial x}\bigg{)} \tag{9}\]  
(a) A cantilever beam with lateral displacement but with no rotation in the end (Figure 3a): Since the distributed load (force per length), \(q(x)\), is zero, the solution to the first differential equation of Equation (8) can be expressed as:  
\[E_{s}I\frac{d^{3}\varphi}{dx^{3}}=q(x)=0\ \to E_{s}I\varphi=\frac{C_{0}}{2}x^{2}+C_{1}x+C_{2} \tag{10}\]  
Figure 2: Forces and moments required to cause (**a**) lateral displacement, \(\delta\), with no rotation at the free end of the beam, (**b**) rotation, \(\theta\), with no lateral displacement at the free end of the beam, (**c**) pure axial extension, and (**d**) pure twist in the free end of an Euler–Bernoulli beam [32; 36]. Figure 3: Forces and moments required to cause (**a**) lateral displacement, \(\delta\), with no rotation at the free end of the beam, (**b**) rotation, \(\theta\), with no lateral displacement at the free end of the beam, (**c**) pure axial extension, and (**d**) pure twist in the free end of a Timoshenko beam. By applying the boundary condition of no rotation at the root of the cantilever beam, Equation (10) gives \(C_{2}=0\). Similarly, applying the boundary condition of no rotation (\(\varphi=0\)) at the end of the beam (\(x=l\)) to Equation (10) gives \(C_{1}=-\frac{C_{0}l}{2}\). Substituting \(C_{1}\) and \(C_{2}\) in the second line of Timoshenko beam theory (Equation (8)) gives the deflection function of the beam as  
\[w=\frac{C_{0}}{2E_{s}I}\left(\frac{x^{3}}{3}-l\frac{x^{2}}{2}\right)-\frac{E_{ s}I}{\kappa AG_{s}}\frac{C_{0}}{2E_{s}I}(2x-l)+C_{3} \tag{11}\]  
The two other constants (\(C_{0},\ C_{3}\)) can be found from the boundary conditions in \(x=0\) and \(x=l\). At the root (\(x=0\)), the deflection equals zero, hence:  
\[C_{3}=-\frac{C_{0}l}{2\kappa AG_{s}} \tag{12}\]  
Moreover, the deflection at the end of the beam equals \(\delta\), which gives:  
\[C_{0}=\frac{-\delta}{\frac{\beta}{\Gamma E_{s}I}+\frac{l}{\kappa AG_{s}}} \tag{13}\]  
By substituting \(C_{0},C_{1},C_{2},\ C_{3}\) in the first line of Equation (8) and in Equation (11), the Timoshenko beam theory for a cantilever beam with lateral displacement but with no rotation could be obtained as follows:  
\[\varphi=\frac{-\delta}{\frac{\beta}{\Gamma E}+\frac{2E_{s}II}{\kappa AG_{s}}} \left(x^{2}-lx\right)w=\frac{-\delta}{\frac{l^{3}}{\Gamma E_{s}I}+\frac{l}{ \kappa AG_{s}}}\left(\frac{1}{2E_{s}I}\left(\frac{x^{3}}{3}-l\frac{x^{2}}{2} \right)-\frac{1}{\kappa AG_{s}}x\right) \tag{14}\]  
Now, it is possible to find the moments and forces shown in Figure 3a. As mentioned above, the bending moment, \(M_{xx}\), and the shear force, \(Q_{x}\), are related to the angle of rotation, \(\varphi\), and the deflection, \(w\), by Equation (9).