To clarify any confusion, when you are trying to find out how much energy exactly is spent to grow crack from length a to a+da is given as follows: 1. constant load: Increase the load gradually to P1, at which crack increases from a to a+da and corresponding displacement changes from v1 to v2. Hence, external work done at this point is 0.5*P1v1 + P1(v2-v1). Part of this work is stored in elastic energy, the other part was spent in advancing crack. Unloading this configuration releases the elastic energy stored, which equals to 0.5*P1v2. Hence, work done to advance crack from length a to a+da is 0.5*P1v1 + P1(v2-v1) - 0.5*P1v2 = 0.5*P1(v2-v1). This work is stored as inelastic strain energy = DeltaU = 0.5*P1(v2-v1). In differential form, dU = 0.5*P*dv 2. Constant displacement: Similar to the first case, pull the double cantilever beam until displacement v1 and load P1. crack advances while keeping the displacement constant and dropping the load to P2. After this unload the double cantilever beam by reducing the displacement to zero. The net work done to advance crack under constant load will be 0.5(P2-P1)v1. This work done is stored in inelastic strain energy in a differential form: dU = 0.5*v*dP. Now, in the first case, the term dv (v2>v1) is positive which implies that strain energy increases, while in the second case, dP (P2
Look at it this way: We know that the change in strain energy is represented by that small obtuse angled triangle shown shaded. The area of this triangle is: (Area of the rectangle formed by P1, V1 and V2) + (Area of right triangle formed by P1 and V1) - (Area of th right triangle formed by P1 and V2) This is = P1(V2 - V1) + 0.5P1V1 - 0.5P1V2 = 0.5P1(V2 - V1)
At 33:31, can anyone explain to me how 0.5P1(v2 - v1) corresponds to the shaded portion of the graph, indicating change in strain energy? I'm not clear on this. Thanks in advance!
To clarify any confusion, when you are trying to find out how much energy exactly is spent to grow crack from length a to a+da is given as follows:
1. constant load:
Increase the load gradually to P1, at which crack increases from a to a+da and corresponding displacement changes from v1 to v2. Hence, external work done at this point is 0.5*P1v1 + P1(v2-v1). Part of this work is stored in elastic energy, the other part was spent in advancing crack. Unloading this configuration releases the elastic energy stored, which equals to 0.5*P1v2.
Hence, work done to advance crack from length a to a+da is 0.5*P1v1 + P1(v2-v1) - 0.5*P1v2 = 0.5*P1(v2-v1).
This work is stored as inelastic strain energy = DeltaU = 0.5*P1(v2-v1).
In differential form, dU = 0.5*P*dv
2. Constant displacement:
Similar to the first case, pull the double cantilever beam until displacement v1 and load P1. crack advances while keeping the displacement constant and dropping the load to P2. After this unload the double cantilever beam by reducing the displacement to zero. The net work done to advance crack under constant load will be 0.5(P2-P1)v1.
This work done is stored in inelastic strain energy in a differential form: dU = 0.5*v*dP.
Now, in the first case, the term dv (v2>v1) is positive which implies that strain energy increases, while in the second case, dP (P2
Look at it this way:
We know that the change in strain energy is represented by that small obtuse angled triangle shown shaded. The area of this triangle is:
(Area of the rectangle formed by P1, V1 and V2) + (Area of right triangle formed by P1 and V1) - (Area of th right triangle formed by P1 and V2)
This is = P1(V2 - V1) + 0.5P1V1 - 0.5P1V2
= 0.5P1(V2 - V1)
Wonderful class!
At 33:31, can anyone explain to me how 0.5P1(v2 - v1) corresponds to the shaded portion of the graph, indicating change in strain energy?
I'm not clear on this. Thanks in advance!
energy consumed as crack length increase, under same loading over a period, say months
GD