# 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.
Akshay patil said:
1 decade ago

Suddenly applied load, having higher velocity compared to gradually applied load.

Dara Singh said:
1 decade ago

Twice stress why & which formula used this it may be more than twice or little less than twice.

Abinash said:
1 decade ago

When the load is applied slowly due to viscous dissipation by the atmosphere only half of the energy is stored as compared to the case when the load is applied suddenly.

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.

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.

K.p said:
1 decade ago

It is experimentally proved that the stress developed by impact load is twice than developed by gradually applied load.

Ashu said:
1 decade ago

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

AMAN said:
1 decade ago

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.

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.

Sanjay said:
1 decade ago

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

Rishi said:
10 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.

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.

Krishna said:
9 years ago

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

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