- #1
c0der
- 54
- 0
Hi,
I have been stuck on a problem for a while now (3.24 part c).
My attempt is as follows:
Internal virtual work = external virtual work
T/2 ∫0->L (∂u/dx)(∂v/dx)dx + ∫0->L (∂^2u/∂t^2)vdx = ∫0->L (Pv)dx
Stationarity is already invoked on this functional as it's the principle of virtual work dv/dx = δv
Using the basis functions:
u = { 3w1L/x for 0<x<L/3,
(2 - 3/L*x)w1 + (-1 +3/L*x)w2 for L/3<x<2L/3,
3w2(1 - x/L) for 2L/3<x<L }
Extracting the functions for v from the basis functions:
v = { 3L/x for 0<x<L/3,
(2 - 3/L*x) + (-1 +3/L*x) for L/3<x<2L/3,
3(1 - x/L) for 2L/3<x<L }
Integrating over each interval does not get me the answer
Help please?
I have been stuck on a problem for a while now (3.24 part c).
My attempt is as follows:
Internal virtual work = external virtual work
T/2 ∫0->L (∂u/dx)(∂v/dx)dx + ∫0->L (∂^2u/∂t^2)vdx = ∫0->L (Pv)dx
Stationarity is already invoked on this functional as it's the principle of virtual work dv/dx = δv
Using the basis functions:
u = { 3w1L/x for 0<x<L/3,
(2 - 3/L*x)w1 + (-1 +3/L*x)w2 for L/3<x<2L/3,
3w2(1 - x/L) for 2L/3<x<L }
Extracting the functions for v from the basis functions:
v = { 3L/x for 0<x<L/3,
(2 - 3/L*x) + (-1 +3/L*x) for L/3<x<2L/3,
3(1 - x/L) for 2L/3<x<L }
Integrating over each interval does not get me the answer
Help please?