Mechanical Engineering - Machine Design - Discussion

When a load is suddenly applied, as in an impact, the force and consequently the stress are increased according to a prescribed formula.

In the case of weight, suddenly applied from 0 distance the force applied is 2X.

As this distance and the velocity increases so does the applied force. Therefore, it suffices to say that stress does not increase only as a function of the weight.

Case1: (if external load increases).

Stress-induced in the bar will be increased if the external load increases because the applied load is increasing but the cross-sectional area is constant (stress=F/A).

Case2: (if self-weight increases).

To increase the self-weight, the density of bar should increase or area of bar should increase but the bar is made of same material, therefore, density is constant everywhere, therefore, the only parameter is the area.

So here selfweight increases with the increase of area. That means both are increasing simultaneously (stress=F/A), change in self-weight equal to change in the area therefore induced stress is constant.

W1 is the weight of bar hence incorrect is right.

I think stress depends upon the applied load not it's own weight and also the stress depend on uoon the cross-sectional area.

For suddenly applied loads impact stress will be acted (heavily). Think this as a case of Izod and Charpy test then you can understand after suddenly applied load the bonds will break.

Given the case, as increasing load increases stress but after that increasing the load is no use.
There won't be any change in stress.

Explain?