00:01
Let us first draw the cross -sectional diagram.
00:04
So it is like this.
00:28
Let us this is the section 2 and this is section 1.
00:37
Now the length, this is given as 100 mm, this side.
00:42
This side is given as 50 mm and this is given as 10 m.
00:52
Now the given that the material of the column is 836 steel.
00:58
Material is 836 steel.
01:02
For 836 steel, we have the value, young's modulus of elasticity, e is equal to 200 multiplied by 10 to the power of 9 pascal.
01:16
The length of the column is given as l is equals to 5 meter.
01:21
Also the yield strength for 836 steel is equals to 250 megapascal.
01:31
Now given that the column is fixed at both end, column fixed at both ends, column fixed at both ends.
01:47
So now, let us now write the uler's critical load.
01:52
So it will be p .r, it is equal to pi square, i divided by l effective square, l effective square.
02:03
Here i will be equals to i1 minus i2...