Answer:
The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2
Explanation:
Given that
[tex]\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?[/tex]
From equation of normal strain in x direction:
[tex]\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta[/tex]
Substituting the values:
[tex]\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}[/tex]
