00:01
Hi, in this question it is given that modulus of elasticity for steel is 200 gigapascal and modulus of elasticity for aluminium is 70 gigapascal and the moment of inertia for aluminium that is ia is equal to 1 divided by 12 then 50 times 30 cube minus 46 times 26 whole cube and similarly is will be equal to 1 divided by 12 multiplied by 46 times 26 whole cube.
00:50
Now hs that is the ratio of modulus of elasticity of es and ea so this will be 200 divided by 70 and this is approximately equal to 2 .857.
01:07
Similarly ha we will consider this as approximately equal to 1.
01:11
Now the moment of inertia of the transformed section i will be equal to ha multiplied by ia plus hs multiplied by i.
01:26
So now let's put all this value here so this is i equal to 1 multiplied by 1 divided by 12 multiplied by 50 times 30 cube minus 46 times 26 cube and then plus 2 .857 multiplied by 1 divided by 12 and then 46 times 26 cube and all this will be in millimeter so the unit will be millimeter raised to the power 4.
02:00
So on solving this we will get i is equal to 237 .614 multiplied by 10 raised to the power minus 9 meter.
02:08
Now after calculating all this in the first part of the question we need to find maximum stress in the steel that is sigma s max so we can write sigma s max value will be equal to hs times m times ys divided by i.
02:32
So this will be equal to 2 .857 multiplied by 300 multiplied by 26 divided by 2 and we will convert this into meter so we will have to multiply 10 raised to the power minus 3.
02:46
Then whole divided by 237 .614 multiplied by 10 raised to the power minus 9 and this will be in power square...