In the derivation, F(x) doesn't change with cross section? Only cross-sectional area changes with length. Also, in the problem, the unit of density is KN/mm2. You meant specific gravity?
@@MechanicsofMaterialsLibre apologies sir. It's my mistake. Since you have directly written (L^2)/2 during the integration , i saw it like a y is missing. Sorry again.
thank you doc
Thank you
Are there any examples where the area is variable?
In the derivation, F(x) doesn't change with cross section? Only cross-sectional area changes with length.
Also, in the problem, the unit of density is KN/mm2. You meant specific gravity?
The force F(x) is constant on the cross section area. The unit of density is kN/m3 in the presented example.
I've tried the last exercise,,,,
I've gotten the internal force as (P-wy)... My final answer for displacement is
3.353×10^-(3) mm.
Is it correct.?
The internal force is (wy-P) but the answer you got is a bit off.
Sir i think in the last example one y is missing during the integration. Should not it be y^2?
Do you mean the hanging rod under its own weight? If so, which equation?
@@MechanicsofMaterialsLibre apologies sir. It's my mistake. Since you have directly written (L^2)/2 during the integration , i saw it like a y is missing. Sorry again.
what happened to the gravitation constant? it should be in the equation
If you are refrerring to Example 3-1-2, based on the unit of density, it already includes the g. Hope that helps.
@@MechanicsofMaterialsLibre ok it's already included, good, i really like your videos you do not only have good examples but also good explanations
is the answer 0.003685 mm ?
thank you doc