STAVEBNÍ MECHANIKA 3 - SM 3 - Jenin

také v maticovém tvaru. Definujme
[Cp] = L
⎡
1
⎢ 0
⎢ 0
EA ⎢ 0
⎣ 0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1.414
0
0
0
0
0
0 0 0 0 0 1.414
{X1} = {0, 0, −1, 0, 0, 1.414} T
{X2} = {−0.707, −0.707, −0.707, −0.707, 1, 1} T
{F} = {0, 0, −1, 0, 0, 0} T
[X] = [{X1}, {X2}]
Užitím těchto vzorců dostaneme
[CK] =
δ11
δ12
δ21 = δ12 δ22
= [X] T [Cp] [X]
∆ 0 1 = {X1} T [Cp] {F} · F = 1, 0 · F srovnej s rov. (154)
∆ 0 2 = {X2} T [Cp] {F} · F = 0, 71 · F
⎤
⎥
⎦