Mechanical Engineering - Strength of Materials - Discussion

When the load applied gradually then stress induce = W/A.

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

By comparing.

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

Can anyone please explain how it can be twice?

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.

No specific solution is available for this question.

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

Stress do not depend on rate of application of force I prefer equal in both the cases.

@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.

Because new load is P=P1+P2, where P1=P2 so the stress is twice.

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,
Stress = load/area.

So let p1 is stress when suddenly load and p2 is stress when load is gradually.

So p1 = w/area.

And p2 = (w/2)/area.

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

So p1/p2 = 2.

p1 = 2p2.

Thats proves that suddenly loaded stress is double to the gradually loaded stress.

Ya. As per the formula. Suddenly applied load doubled than gradully applied load.