Two strain gauges are fixed such that they make on angle of 40° t

SOLUTION

Concept:

For rotation by θ in anticlockwise sense

\( = \frac{{{_x} + {_y}}}{2} + \frac{{{_x} - {_y}}}{2}\cos 2\theta + \frac{{{\gamma _{xy}}}}{2}\sin 2\theta \)

Sum of strains on any two mutually 1 planes is same

\({\epsilon_1} + {\epsilon_2} = \epsilon_x' + \epsilon_y'\)

Principal stresses are given by

\({{\sigma }_{1}}=\frac{E}{1-{{\mu }^{2}}}\left( {{\epsilon }_{1}}+\mu {{\epsilon }_{2}} \right)\)

\({{\sigma }_{2}}=\frac{E}{1-{{\mu }^{2}}}\left( {{\epsilon }_{2}}+\mu {{\epsilon }_{1}} \right)\)

Also,

E = 2G (1+ μ) = 3 k (1 – 2μ)

\(\frac{G}{k}=\frac{3}{2}\left( \frac{1-2\mu }{1+\mu } \right)\Rightarrow ~G=0.103~k\)

Calculation:

 E = 210 × 10³ MPa

μ = 0.45

\(\begin{align} & \epsilon _{x}^{'}=445\times {{10}^{-6}},~\epsilon _{y}^{'}=-40\times {{10}^{-6}} \\ & {{\epsilon }_{1}}+{{\epsilon }_{2}}=\left( 445-40 \right)\times {{10}^{-6}}=405\times {{10}^{-6}} \\ \end{align}\)

At θ = 40°

\(\begin{align} & =445\times {{10}^{-6}}=\frac{{{\epsilon }_{1}}+{{\epsilon }_{2}}}{2}+\frac{{{\epsilon }_{1}}-{{\epsilon }_{2}}}{2}\cos 80{}^\circ \\ & 445\times {{10}^{-6}}={{\epsilon }_{1}}\left[ \frac{1}{2}+\frac{1}{2}\cos 80{}^\circ \right]+{{\epsilon }_{2}}\left[ \frac{1}{2}-\frac{1}{2}\cos 80{}^\circ \right] \\ \end{align}\)

⇒ 0.5868 ϵ₁ + 0.4131 ϵ₂ = 445 × 10^-6

ϵ₁ + ϵ₂ = 405 × 10^-6

⇒ ϵ₁ = 1598.70 × 10^-6

ϵ₂ = -1193.70 × 10^-6

∴ σ₁ = 279.52 MPa