Question A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed v (radians per second). The material of the bar has weight density y. (a) Derive a formula for the tensile stress σx in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress σmax?    

Given thatLength of the round bar, 2 LAngular speed, \omegaWeight density of the bar material, yConsider segment C B :Let F_{x} be the axial force in the bar at section DLet d_{m} be the mass of the element which is at a distance t from ' C^{\prime} Thickness of the element d tLet d F be the inertia force of the elementd F=d m\left(t \omega^{2}\right)d F=\left(\frac{\gamma}{g} A d t\right) t \omega^{2}Where A-cross sectional area of the bard F=\left(\frac{\gamma}{g} A \omega^{2}\right) t d t\int_{0}^{F_{x}} d F=\int_{x}^{L}\left(\frac{y}{g} A \omega^{2}\right) t d tF_{x}=\frac{\gamma}{g} A \omega^{2}\left(\frac{t^{2}}{2}\right)_{x}^{L}F_{x}=\frac{\gamma}{g} A \omega^{2}\left(\frac{L^{2}-x^{2}}{2}\right)F_{x}=\frac{y}{2 g} A \omega^{2}\left(L^{2}-x^{2}\right)(a) Tensile stress in the bar at distance x from the mid point ' C^{\prime}\begin{array}{l}\sigma_{x}=\frac{F_{x}}{A} \\\sigma_{x}=\frac{\gamma}{2 g} \omega^{2}\left(L^{2}-x^{2}\right)\end{array}(b) Maximum tensile stress \left(\sigma_{\max }\right) is obtained at x=0\begin{array}{c}\sigma_{\max }=\frac{\gamma \omega^{2}}{2 g}\left(L^{2}-0\right) \\\sigma_{\max }=\frac{\gamma \omega^{2} L^{2}}{2 g}\end{array} ...