Question 1.26: Derive strain-displacement relations in polar coordinates.
Derive strain-displacement relations in polar coordinates.
An element ABCD shown in Fig. 1.35, deforms to \acute{A} \acute{B} \acute{C} \acute{D} . The radial and tangential components of displacements are represented by u and v respectively.
Considering side AB, the radial strain \varepsilon_r is given by,
\varepsilon_r=\frac{\left(u+\frac{\partial u}{\partial r} d r-u\right)}{d r}=\frac{\partial u}{\partial r} (i)
For tangential strain \varepsilon_\theta , one part due to radial displacement is given by,
\frac{(r+u) d \theta-r d \theta}{r d \theta}=\frac{u}{r}Another part of tangential strain \varepsilon_\theta , due to radial displacement of D to \acute{D} is given by,
\frac{\nu +\frac{\partial \nu }{\partial \theta} d \theta-\nu }{r d \theta}=\frac{1}{r} \frac{\partial \nu }{\partial \theta}Therefore the total tangential strain,
\varepsilon_\theta=\frac{u}{r}+\frac{1}{r} \frac{\partial \nu }{\partial \theta} (ii)
For shearing, angular displacement of side AD,
\frac{\frac{\partial u}{\partial \theta} d \theta}{r d \theta}=\frac{1}{r} \frac{\partial u}{\partial \theta}Similarly angular displacement of side AB,
\frac{\frac{\partial v}{\partial r} d r}{d r}=\frac{\partial v}{\partial r}However side \acute{A} \acute{B} has an initial rigid body displacement ν/ r, which does not contribute to deformation. Thus the total shearing deformation is given by,
\gamma_{r \theta}=\frac{1}{r} \frac{\partial u}{\partial \theta}+\frac{\partial \nu }{\partial r}-\frac{\nu }{r} (iii)
