Truss and Energy method
Truss and Energy method
(OP)
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
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






RE: Truss and Energy method
have to think why strain energy is 1/2 the work done ...
another day in paradise, or is paradise one day closer ?
RE: Truss and Energy method
RE: Truss and Energy method
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 ?
RE: Truss and Energy method
RE: Truss and Energy method