Amazing sir but doubt in last question . If we consider the equation. L(Alfa)∆t =.2mm then we will get ∆t required to change in length by .2 mm. Then we subtract the ∆t into next equation. E(alfa)(∆T-∆t) to calculate the thermal stress. ..... Is this approach is right for this because I got the same answer by this method also , please let me know if you have seen.
Genuine Doubt !!!! Sir At time 1:17:36 you have considered length = 250, Why not 250+0.2 ??? Mechanical stress will only develop after elongation touches the support. That's why I think lengthy should be taken as 250+0.2 Am I right or wrong ? Kindly Reply to me sir.
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thanks sir
Amazing sir but doubt in last question . If we consider the equation. L(Alfa)∆t =.2mm then we will get ∆t required to change in length by .2 mm. Then we subtract the ∆t into next equation. E(alfa)(∆T-∆t) to calculate the thermal stress. ..... Is this approach is right for this because I got the same answer by this method also , please let me know if you have seen.
Great
Que 1 - time 01:02:45: will there be any self weight related stress?
💯
Sir, please release CAE series
at 10:51, why he is not considering effect of Poisson ratio in Ex ??
Genuine Doubt !!!!
Sir At time 1:17:36 you have considered length = 250,
Why not 250+0.2 ???
Mechanical stress will only develop after elongation touches the support.
That's why I think lengthy should be taken as 250+0.2
Am I right or wrong ?
Kindly Reply to me sir.
It is a change length ∆l. We have to consider Change in length
Sir ,fusion 360
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