00:01
Hi, this is a question based on euler's buckling formula.
00:04
So, euler's buckling formula.
00:09
So, here we are given we are not given it the specified.
00:14
So, we are not specified with a particular end.
00:17
So, particular end.
00:19
So, particular end is not given.
00:23
So, particular end condition of the column is not given.
00:33
So, is not given.
00:35
So, we have to consider all possible loads.
00:38
So, consider all possible loads.
00:45
So, first one is.
00:46
So, first one is both end are hinged.
00:51
So, both ends are hinged.
00:55
So, hinge we have the length effective is equal to l.
00:59
So, that is the given length.
01:01
So, we have pcr for this case is given by pi square into e i divided by l square where e is the modulus of.
01:11
So, elasticity and i is the moment of inertia l is the length.
01:14
So, we have this is pi square into e is 200 kilo pascal that is 10 power 3 into 70 power 4 divided by 12 divided by length is 1000 square.
01:29
So, by this we have.
01:30
So, i is b power 4 by 12.
01:32
So, b is given as 70.
01:35
So, side b is 77 centimeter that is.
01:40
So, that is 0 .7 meters.
01:45
So, now we have or 70 millimeters.
01:49
So, 70 millimeters then this will be.
01:54
So, moment of inertia will be b power 4 by 12.
01:57
So, substituting here we have for the first case we have pcr for a is equal to 3 .949 mega newtons.
02:08
For the second parts we will let us consider that both the ends are fixed.
02:14
So, both ends are fixed...