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For uniaxial load in volumetric strain ev={ [sigma(p/A)/young Modulus (E)]× (1-2m)} From Hooke's laws stress directly proportional to strain Sigma=young Modulus E× strain Strain(e)=sigma÷ Young's modulus E So volumetric strain(ev) in uniaxial load={strain(e)×1-2m}
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Your answer for dv is 4500 but its wrong. lnew=6000+3=6003 tnew=20-0.0025=19.9975 bnew=150-0.01875=149.98125 Vnew=lnew X bnew X tnew =18004498.03 dV=Vnew-Vold=18004498.03-18000000=4498.031391
Much appreciated sir,i have fully understood it with in a minute 🙏💫
Thank you very much
Thank you sir ✨~
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E=200X10^-3 N/mm^2
sir there is a mistake in volumetric strain formula ev=3e(1-2u)
Ashish Adiley are you sure about that?
Ashish Adiley you are wrong
bhai we have applied load in one direction only so its correct
Ashish bhai ap galat Bata rehe ho
For uniaxial load in volumetric strain ev={ [sigma(p/A)/young Modulus (E)]× (1-2m)}
From Hooke's laws stress directly proportional to strain
Sigma=young Modulus E× strain
Strain(e)=sigma÷ Young's modulus E
So volumetric strain(ev) in uniaxial load={strain(e)×1-2m}
Formula to find change in volume of circular section
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Your answer for dv is 4500 but its wrong.
lnew=6000+3=6003
tnew=20-0.0025=19.9975
bnew=150-0.01875=149.98125
Vnew=lnew X bnew X tnew =18004498.03
dV=Vnew-Vold=18004498.03-18000000=4498.031391
you should have solve the solution before in a sheet and u just need to explain because lot of time is being wasted in writing