Question

Two separate load cases act at a point. Case I: pure uniaxial tension of \(10\ \text{N/mm}^2\) along the \(x'\)-axis; Case II: pure uniaxial compression of \(10\ \text{N/mm}^2\) along the \(y'\)-axis. The axes \((x',y')\) are rotated by \(45^\circ\) with respect to the \((x,y)\) axes. Find the sum of the maximum and minimum principal stresses for the combined loading (rounded to 1 decimal place). \begin{center} \includegraphics[width=0.7\textwidth]{15.jpeg} \end{center}

Hint:

In plane stress, the sum of principal stresses equals \(\sigma_x+\sigma_y\) (the trace). Superposition plus this identity lets you avoid drawing Mohr's circle when only the sum is asked.

Answer 1

Step 1: Transform Case I (uniaxial \(x'\)-tension).
Given \(\sigma_{x'}=10,\ \sigma_{y'}=0,\ \tau_{x'y'}=0,\ \theta=45^\circ\). Using plane-stress transformation with \(\cos 2\theta=\cos 90^\circ=0,\ \sin 2\theta=1\): \[ \sigma_x^{(I)}=\tfrac{\sigma_{x'}+\sigma_{y'}}{2}+\tfrac{\sigma_{x'}-\sigma_{y'}}{2}\cos2\theta = \tfrac{10}{2}+ \tfrac{10}{2}\cdot 0=5, \] \[ \sigma_y^{(I)}=\tfrac{\sigma_{x'}+\sigma_{y'}}{2}-\tfrac{\sigma_{x'}-\sigma_{y'}}{2}\cos2\theta = 5, \] \[ \tau_{xy}^{(I)}=\tfrac{\sigma_{x'}-\sigma_{y'}}{2}\sin2\theta= \tfrac{10}{2}\cdot1=5. \] Step 2: Transform Case II (uniaxial \(y'\)-compression).
Here \(\sigma_{x'}=0,\ \sigma_{y'}=-10,\ \tau_{x'y'}=0\). \[ \sigma_x^{(II)}=\tfrac{-10}{2}+\tfrac{0-(-10)}{2}\cdot0=-5,\quad \sigma_y^{(II)}=-5,\quad \tau_{xy}^{(II)}=\tfrac{0-(-10)}{2}\cdot1=5. \] Step 3: Superpose the two cases.
\[ \sigma_x=\sigma_x^{(I)}+\sigma_x^{(II)}=5+(-5)=0,\quad \sigma_y=5+(-5)=0,\quad \tau_{xy}=5+5=10. \] The combined state is \emph{pure shear} with \(\tau_{xy}=10\ \text{N/mm}^2\). Step 4: Principal stresses and their sum.
For pure shear, principal stresses are \(\sigma_{1,2}=\pm \tau_{xy}=\pm 10\ \text{N/mm}^2\). Hence \[ \sigma_1+\sigma_2= (+10)+(-10)=0.0\ \text{N/mm}^2. \] \[ \boxed{0.0\ \text{N/mm}^2} \]