Chapter 3: Problem 7
A uniform bar of rectangular section is elastically bent into a circular arc by applying a pair of equal and opposite couples \(M_{0}\) to its two ends. By using the fact that each "fiber" under tension (or compression) shrinks (or expands) sideways an amount given by Poisson's ratio \(\sigma\) times the fiber elongation, show that the neutral surface does not remain plane but has a lateral radius of curvature \(R^{\prime}\) given by $$ R^{\prime}=\frac{R}{\sigma} $$ where \(R\) is the longitudinal radius of curvature given as usual by \(R=Y I / M_{0}\). If the bar is of thickness \(t\), then on the top surface the lateral and longitudinal curvatures are \(\rho^{\prime}=R^{\prime}+\frac{1}{2} t\), \(\rho=R-\frac{1}{2} t\), and $$ \sigma=\frac{R}{R^{\prime}}=\frac{\rho+\frac{1}{2} t}{\rho^{\prime}-\frac{1}{2} t} $$ (By polishing the surface of the undistorted bar to optical flatness, it is possible to use interference fringes, made with the help of an optically flat glass plate, to measure \(\rho\) and \(\rho^{\prime}\), from which an accurate value of \(\sigma\) can be computed.)
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.