Question

Consider the plane strain field given by \(\varepsilon_{xx} = 10\,x y^{2}, \; \varepsilon_{yy} = -5\,x^{2}y \text{ and } \gamma_{xy} = A \, x y (2x - y)\) where \(A\) is a constant and \(\gamma_{xy}\) is the engineering shear strain. The value of \(A\) for the strain field to be compatible is .................. (rounded off to 1 decimal place).

Hint:

For 2D small-strain fields, use \(\displaystyle \frac{\partial^{2}\varepsilon_{xx}}{\partial y^{2}}+\frac{\partial^{2}\varepsilon_{yy}}{\partial x^{2}} =\frac{\partial^{2}\gamma_{xy}}{\partial x\,\partial y}\). Remember that the engineering shear strain is \(\gamma_{xy}=2\varepsilon_{xy}\).

Answer 1

Step 1: Write the 2D strain-compatibility equation.
For small strains in plane strain, the compatibility condition is \[ \frac{\partial^{2}\varepsilon_{xx}}{\partial y^{2}} \;+\; \frac{\partial^{2}\varepsilon_{yy}}{\partial x^{2}} \;=\; \frac{\partial^{2}\gamma_{xy}}{\partial x\,\partial y}, \] since \(\gamma_{xy}=2\varepsilon_{xy}\Rightarrow 2\,\partial^{2}\varepsilon_{xy}/\partial x\partial y = \partial^{2}\gamma_{xy}/\partial x\partial y\). Step 2: Compute the left-hand side (LHS).
\[ \varepsilon_{xx}=10xy^{2} \;\Rightarrow\; \frac{\partial\varepsilon_{xx}}{\partial y}=20xy,\quad \frac{\partial^{2}\varepsilon_{xx}}{\partial y^{2}}=20x. \] \[ \varepsilon_{yy}=-5x^{2}y \;\Rightarrow\; \frac{\partial\varepsilon_{yy}}{\partial x}=-10xy,\quad \frac{\partial^{2}\varepsilon_{yy}}{\partial x^{2}}=-10y. \] Thus, \[ \text{LHS}=20x-10y. \] Step 3: Compute the right-hand side (RHS).
\[ \gamma_{xy}=Axy(2x-y)=A(2x^{2}y-xy^{2}). \] \[ \frac{\partial \gamma_{xy}}{\partial y}=A(2x^{2}-2xy),\qquad \frac{\partial^{2}\gamma_{xy}}{\partial x\,\partial y}=A(4x-2y). \] Hence, \[ \text{RHS}=A(4x-2y). \] Step 4: Enforce compatibility for all \(x,y\).
\[ 20x-10y \;=\; A(4x-2y). \] Matching coefficients of \(x\) and \(y\): \[ 20=4A \Rightarrow A=5, \qquad -10=-2A \Rightarrow A=5 \;(\text{consistent}). \] \[ \boxed{A=5.0} \]