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)
$\triangle\sigma
= \mid-\sigma_{1}-\sigma_{3}\mid$
=$\mid-(\sigma_{1}+\sigma_{3})\mid$
=$\sigma_{1}+\sigma_{3}$
This is the condition needed.
Reference:https://en.wikipedia.org/wiki/Differential_stress
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