When a body is subjected to a direct tensile stress $$\left( {{\sigma _{\text{x}}}} \right)$$ in one plane accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ the minimum normal stress is
A. $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
B. $$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
C. $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$
D. $$\frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
Answer: Option B
This Question Belongs to Mechanical Engineering >> Strength Of Materials In ME
The minimum normal stress (often called the minimum principal stress) is calculated using the formula:(sigma_{min} = rac{sigma}{2} - sqrt{( rac{sigma}{2})^2 + au_{xy}^2})Why it happens:When a material is pulled (direct tensile stress, (sigma )) and twisted or slid (shear stress, ( au _{xy})), these forces combine to create an "angled" stress inside the material.The Plus/Minus Rule: Mohr's circle (a graphical method for stress analysis) proves that this combination of stresses results in both a maximum and minimum normal stress on the element.The Radius Concept: The value ( rac{sigma }{2}) represents the center point of the combined stresses. The square root portion (sqrt{( rac{sigma }{2})^{2}+ au _{xy}^{2}}) represents the radius of the combined stress circle.The Minimum Point: By taking the center point (( rac{sigma }{2})) and subtracting the radius, you find the lowest normal stress acting on the material, which frequently flips to become a compressive stress (indicated by a negative result).
The answer is option A