00:01
Hello students, let's solve this question.
00:01
As we know, for a thin cylinder, there are two directional stress acting.
00:05
First of all, longitudinal stress, longitudinal stress, which is represented by sigma l is equals to pr divided by t.
00:16
And second is hoop stress, that is sigma h is equals to pr divided by 2t.
00:23
And according to vaughan mass, according to vaughan, mrs.
00:30
We get sigma t square is equal to sigma l squared plus sigma h square plus sigma h sigma l.
00:38
And by putting values we will get sigma t squared is equal to pr divided by t square plus pr divided by 2t squared plus p r multiplied by p r divided by t multiplied by p r divided by t multiplied by 2t.
00:53
Therefore this will be equals to p r divided by t whole square multiplied by 1 .75, which implies t is equal to p r divided by sigma t multiplied by square root of 1 .75.
01:11
Now by putting the values, we will get t equals to 35 multiplied by 30 divided by 700 multiplied by square root of 1 .75...