Calculation of true thickness of the bed

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