Question

The state of stress at a point, for a body in plane stress, is shown in the figure below. If the minimum principal stress is 10 kPa, then the normal stress ${\sigma }_y$ (in kPa) is

• A. 9.45
• B. 18.88
• C. 37.78
• D. 75.50

Answer 1

The correct option is C 37.78

Given ${\sigma }_x=100kPa,{\tau }_{xy}=50kPa,{\sigma }_2=10kPa$ Minimum principal stress,
${\sigma }_2=\frac{{\sigma }_x+{\sigma }_y}{2}-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

$10=\frac{100+{\sigma }_y}{2}-\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+{50}^2}$
$\therefore \sqrt{{(50-\frac{{\sigma }_y}{2})}^2+{50}^2}=50+\frac{{\sigma }_y}{2}-10=40+\frac{{\sigma }_y}{2}$
By squaring
$2500+\frac{{\sigma }_y^2}{4}-50{\sigma }_y+2500=1600+\frac{{\sigma }_y^2}{4}+40{\sigma }_y$
$\therefore 90{\sigma }_y=3400$
${\sigma }_y=37.78kPa$