A body is subjected to a bi-axial state of stress as shown in figure below: The magnitude of normal and shear stresses acting on the 45 degree plane in MPa

Correct option is C

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$$\textbf{Given:} \\\sigma_x = 60\ \text{MPa}, \\\sigma_y = -40\ \text{MPa (compressive)}, \\\tau_{xy} = 30\ \text{MPa}, \\\theta = 45^\circ \Rightarrow 2\theta = 90^\circ. \\\textbf{Stress-transformation equations:}\\\sigma_n = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos(2\theta)+ \tau_{xy}\sin(2\theta)\\\tau = -\frac{\sigma_x - \sigma_y}{2}\sin(2\theta)+ \tau_{xy}\cos(2\theta)\\\text{For } \theta = 45^\circ,\ \sin(90^\circ)=1,\ \cos(90^\circ)=0.\\\textbf{Now calculate:}\\\sigma_n = \frac{60 - 40}{2}+ \frac{60 + 40}{2}\cdot 0+ 30 \cdot 1= 10 + 0 + 30 = 40\ \text{MPa}.\\\tau= -\frac{60 + 40}{2}\cdot 1+ 30 \cdot 0= -50 + 0= -50\ \text{MPa}.$$​​