Mechanical Engineering - Strength of Materials - Discussion

In case of impact loading:

σ or δ(impact)=(I.F.)*σ or δ(static) I.F.=Impact Factor.

I.F.=1+(1+2h/δ).

h= height of load drop, δ= static change in dimension.

Here in this case h=0.

So, I.F.= 2.

So, stress and strain will be double.

In stress it is two times.

In strain it is four times.

Please, define "gradually" and "suddenly".

Only if the load is related to weight loading, (dropping a weight vs supporting a weight) , only then the rate of the loading is actually a factor that affect the stress induced to the body. If the load is related to a spring (releasing a compressed spring that is already in contact to a body) then the rate of the loading does not play any role.

At gradually (f=w only or m.g) but in sudden there will inertia in addition to it so ( f=w+m.a) then will (f=2w).

Yes, it is experimentaly proved.

Here, σ impact load = (F/A)[1 + (1 + 2hAE/FL)^0.5].

As in gradually applied load, load varies from 0 to a.

The net load applied is w1 = w/2.

So, the stress is p = w/(2a).

For suddenly applied load stress p1 = w/a.

p1/p = 2, p1 = 2p.

So, stress due to suddenly applied is twice of gradually applied load.

Strain energy stored in the body = Work done by the load in deforming the body.
Strain energy stored in the body = Area of the load-extension curve.
Strain energy stored in the body = P. x.
U = P. x.

As we know that maximum strain energy stored in the body U will be provided by the following expression as mentioned here.
U=(σ^2/2E)*V.
U=(σ^2/2E)*A*L.
P*x=(σ^2/2E)*A*L.

Now we will secure the value of extension x in terms of Stress, Length of the body and Young's modulus of the body by using the concept of Hook's Law.
According to Hook's Law.

Within elastic limit, stress applied over an elastic material will be directionally proportional to the strain produced due to external loading and mathematically we can write above the law as mentioned here.

Stress = E. Strain
Where E is Young's Modulus of elasticity of the material.

σ = E. ε.
σ = E. (x/L).
x = σ. L/ E.

Let use the value of the extension or deformation "x" in the above equation and we will have
P*(σ/E)*L=[(σ^2)/(2*E)]*A*L.
P=σ*A/2.
σ=2(P/A).

Therefore, we can say here that maximum stress induced in the body due to suddenly applied load will be twice the stress induced in the body with the same value of load applied gradually.

@Amol.

Then it should be one-half is the answer.

L is length of the plate.