### Part E : Stress Analysis

Purpose of this section is to derive the equations needed to determine the stresses inside the long arm (length b) of the beam in a cross-section a distance f away from the tip of the beam. Here we have the perpendicular and normal component of the force the pumpkin is exerting via the rope, Pp and Np, respectively.. At the center of mass of this beam section we have the gravitational force, m g. At the cross-section itself we have a shear force, Ts, a normal force Ns, and the moment Ms, all of which we need to determine the maximum stresses at this location as function of time.

The location of the center of mass of this section is given by the distance from the pivot as $$r = b - f/2$$ and the angle α. We also know the angular velocity $$\dot{\alpha}$$ and the angular acceleration $$\ddot{\alpha}$$ at each and every time step during the integration procedure. With that the tangential and centripedal acceleration, $$a_t = r \dot{\alpha}^2$$ and $$a_c = r \ddot{\alpha}$$, respectively, of the center of mass of the beam section are known as well.

F = m a for the center of mass :

\begin{align*} m g \cos\alpha + N_f - N_p &= m (b{-}f/2) \dot{\alpha}^2 \\ m g \sin\alpha - T_f + P_p &= m (b{-}f/2) \ddot{\alpha}\\ T_f f/2 + P_p f/2 + M_f &= I_f \ddot{\alpha} \tag{E.1} \end{align*}

In above equation m is the mass and $$I_f$$ is the mass moment of inertia of the beam section :

\begin{equation*} I_f = \frac{1}{12} m f^2 \tag{E.2} \end{equation*}

We solve Eq.(E.1) for $$N_f , T_f , M_f$$ and get from that the stresses with Af as the area and $$S_f = w_f h_f^2/6$$ as the section modulus of the cross-section of the beam :

\begin{align*} \text{Normal stress } \sigma_N &= N_f/A_f \\ \text{Maximum bending stress } \sigma_B &= M_f / S_f \\ \text{Maximum shear stress } \tau &= 1.5 T_f / A_f \tag{E.3} \end{align*}

Zig Herzog; hgnherzog@yahoo.com
Last revised: 03/18/20