Question

A3. A direct tensile stress σ,-160 MN/m2 and a direct compressive stress σ,--120 MN/m2 are maximum principal stress in the material is limited to 200 MN/m2 (in tension). Use Mohr's circle to find the allowable value of shear stress at the point. [10 Marks] Final Answer: Max shear stress-radius of circle-180 MN/m

Answer 1

We know that the principal stress is given by
$\sigma _{1} = \left ( \frac{1}{2} \right )\left [\left ( \sigma _{X}+\sigma _{Y} \right ) + \sqrt{(\sigma _{X} - \sigma _{Y})^{2}+4\tau _{XY}^{2}} \right ]$
On putting the value
$200 = \left ( \frac{1}{2} \right )\left [\left (160-120\right ) + \sqrt{(160 - (-120))^{2}+4\tau _{XY}^{2}} \right ]$
On solving we get
$\tau _{xy} = 113.14$ MN/m²
Now drawing the point A and B on the mohr circle
Diameter of mohr circle
$= \sqrt{\left [ 160-(-120) \right ]^{2}+\left [ 113-(-113) \right ]^{2}}$
= 360
Now the maximum shear stress = Radius of mohr circle = 360/2 = 180 MN/m²

(160, 113) (-120,-113)