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GATE_2018

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GATE_2018

14) The maximum and the minimum principle stresses are denoted by $\sigma_{1}$ and $\sigma_{3}$ respectively. The differential stress can have an absolute value grater than $\sigma_{1}$ when

A) $\sigma_{1}$ and $\sigma_{3}$ are both Compressive

B) $\sigma_{1}$ is Compressive and $\sigma_{3}$ is Tensile

C) $\sigma_{1}$ and $\sigma_{3}$ are equal 
 
D) $\sigma_{1}$ and $\sigma_{3}$ are both tensile

(Special Thanks to Dr.Rajeev Kumar Yadav, ISR)
(Thanks to Chandrasekhar, ANU)

Solution:

Differential stress is the difference between the greatest and the least compressive stress experienced by an object 
 
$\sigma_{D}$ = $\sigma_{1}$ - $\sigma_{3}$ .

Basically,
  Compressive means - (-Ve)

Tensile means + (+ve).

The condition we needed is $\sigma_{D}$ should be more than $\sigma_{1}$


Case(1):

$\sigma_{1}$ and $\sigma_{3}$ both are Compressive( i.e both are negative)

$\triangle\sigma = \mid-\sigma_{1}-(-\sigma_{3})\mid$

=$\mid-\sigma_{1}+\sigma_{3}\mid$



$\triangle\sigma<\sigma_{1} $hence this wrong

Case(2):

If$ \sigma_{1}$ is Compressive (-ve) and $\sigma_{3}$ is tensile (+ve)
$\triangle\sigma = \mid-\sigma_{1}-\sigma_{3}\mid$

              =$\mid-(\sigma_{1}+\sigma_{3})\mid$

             =$\sigma_{1}+\sigma_{3}$

$\triangle\sigma>\sigma_{1}$

This is the condition needed.


Reference:https://en.wikipedia.org/wiki/Differential_stress

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