# Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 10)

10.

The stress induced in a body, when suddenly loaded, is __________ the stress induced when the same load is applied gradually.

Discussion:

25 comments Page 1 of 3.
Afeesar said:
4 years ago

In case of impact loading:

Ïƒ or Î´(impact)=(I.F.)*Ïƒ or Î´(static) I.F.=Impact Factor.

I.F.=1+(1+2h/Î´).

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

Here in this case h=0.

So, I.F.= 2.

So, stress and strain will be double.

Ïƒ or Î´(impact)=(I.F.)*Ïƒ or Î´(static) I.F.=Impact Factor.

I.F.=1+(1+2h/Î´).

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

Here in this case h=0.

So, I.F.= 2.

So, stress and strain will be double.

(3)

Ofigileo said:
5 years ago

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=(Ïƒ^2/2E)*V.

U=(Ïƒ^2/2E)*A*L.

P*x=(Ïƒ^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.

Ïƒ = E. Îµ.

Ïƒ = E. (x/L).

x = Ïƒ. L/ E.

Let use the value of the extension or deformation "x" in the above equation and we will have

P*(Ïƒ/E)*L=[(Ïƒ^2)/(2*E)]*A*L.

P=Ïƒ*A/2.

Ïƒ=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.

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=(Ïƒ^2/2E)*V.

U=(Ïƒ^2/2E)*A*L.

P*x=(Ïƒ^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.

Ïƒ = E. Îµ.

Ïƒ = E. (x/L).

x = Ïƒ. L/ E.

Let use the value of the extension or deformation "x" in the above equation and we will have

P*(Ïƒ/E)*L=[(Ïƒ^2)/(2*E)]*A*L.

P=Ïƒ*A/2.

Ïƒ=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.

Rajkotha said:
5 years ago

In stress it is two times.

In strain it is four times.

In strain it is four times.

Sabari said:
5 years ago

@Amol.

Then it should be one-half is the answer.

Then it should be one-half is the answer.

Anomi said:
6 years ago

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.

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.

(1)

Syed mudassir said:
7 years ago

L is length of the plate.

Amar k m s said:
7 years ago

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).

(1)

Sam said:
7 years ago

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

What is L in this formula?

What is L in this formula?

Avipsa said:
7 years ago

Yes, it is experimentaly proved.

Akash said:
7 years ago

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

(1)

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