Truss and Energy method

[Integrator62]

I got this doubt suddenly. I have shown a two member truss having same E and A and Pin joint end conditions.

Let,

W= work done by external force, which is, Fdy

E1 is the Work done by Member 1, which is, F(1)^2*L/EA

E2 is the work done by Member 2, which is same as E1, F(2)^2*L/EA

(I may found F(1) and F(2) using coordinates or any other ways)
-------------
Is,

W=E1+E2
or, Fdy= F(1)^2*L/EA + F(2)^2*L/EA?

Or, I have to Half of every value to console Strain Energy method (although it makes the same value in this case)? Thanks

 

[rb1957]

I think you're ok with internal work done = external work done

have to think why strain energy is 1/2 the work done ...

another day in paradise, or is paradise one day closer ?

 

[CANPRO]

I believe the 1/2 comes from the assumption that the load is gradually applied to the system in order to produce the deflection. So the average load applied to the system during the displacement is 0.5*F. If the load was applied instantaneously, the 1/2 would not be there.

 

[rb1957]

it seems odd that the internal strain energy is 1/2 the internal work done; off the cuff i'd've said they should be the same ?

internal work done is force*distance = F*(strain*L) = F*F/AE*L
internal strain energy is 1/2*stress*strain*volume (integration for the stress/strain curve) = 1/2*F/A*(F/AE)*AL ... wouldn't've expected that ...

another day in paradise, or is paradise one day closer ?

 

[CANPRO]

rb1957, I think there should be a 1/2 in your equation for internal work done. Assuming the load is applied gradually, when the strain is zero, the load is zero and when the strain is maximum, the load is F. So the entire load F, does not go along for the entire ride (strain). If you were to integrate the force/distance curve you would have 1/2*F*D

 

[Integrator62]

Thanks everyone. I was of internet for days. Its clear to me right now