# 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.
equal to
one-half
twice
four times
Explanation:
No answer description is available. Let's discuss.
Discussion:
25 comments Page 1 of 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.

When we loaded suddenly load 'w' then we consider 'w' load in stress while we loaded gradually load "w' then we consider average value of 'w' which will be w/2 in the stress calculation.

So we know that,

So p1 = w/area.

And p2 = (w/2)/area.

So p1/p2 = [w/aria]/[ (w/2) /aria].

So p1/p2 = 2.

p1 = 2p2.

Siva bala said:   1 decade ago
In case of impact loading the dynamic force experienced by the body is given by P(dyn.) = W*[1+sqrt(1+2h/D)].

Here P = dynamic load experienced by the body.
W = static load which is impacting.
D = stact deflection due to above load.
h = height from which load is acting.

In this case h=0, So dynamic load experienced by the body is twice the load applied. As load to area is stress so stresses induced will be twice.

Anomi said:   6 years ago

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)

Mani deepak said:   8 years ago

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.

Afeesar said:   4 years ago

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.
(4)

Rishi said:   9 years ago
@Aman.

How can you say that the average load is w/2?

In case w is split into 4 loads (w1+w2+w3+w4) applied gradually then average would come w/4.

In this case answer would come 4 times. Your method is completely wrong.

AMOL MORE said:   8 years ago

When the load applied suddenly then stress induce = 2 * W/A.

By comparing.

(stress) suddenly = 2 * (stress) gradually.