Truss By Stiffness Matrix
∆em =T. ∆e
∆eG=T -1 ∆em
∆eG=TT ∆em
T -1 = T T
Hence T matrix is orthogonal.
Similar relationship is valid for forces.
(fe)m=TfeG
fes=TTfem
=TT Kem T ∆em
=TTKemT ∆eG
feG =KG ∆eG
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Structural Analysis -Truss By Stiffness Matrix, Study notes of Structural Analysis
Truss By Stiffness Matrix ,Solution by Gauss-elimination Method.
Typology: Study notes
2010/2011
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Truss By Stiffness Matrix
∆e
m
=T. ∆e
∆e
G
=T
∆e
m
∆e
G
=T
T
∆e
m
T
= T
T
Hence T matrix is orthogonal.
Similar relationship is valid for forces.
(fe)
m
=Tfe
G
fe
s
=T
T
fe
m
=T
T
Ke
m
T ∆e
m
=T
T
Ke
m
T ∆e
G
fe
G
=K
G
∆e
G
The Fig. shows a plane truss with 3 members. All members are of length 1000 mm and sectional area 600 mm^2 .Young’s Modulus is 150 kN/mm^2. Analyse by Stiffness Method. 80kN 10kN Y X
⁰
A
B
C
500√3mm 1000 mm
Force-displacement relations for all 3 members are written using equatio C^2 α CαCβ -C^2 α -CαCβ CαCβ C 2 β -CαCβ -C 2 β -C^2 α -CαCβ C^2 α CαCβ -CαCβ -C 2 β CαCβ C 2 β
FXAB
FYAB
FXAB
FYAB
EA/L
UA
VA
UB
VB
FXBA
FYBA
FXAB
FYAB
UB
VB
UA
VA
UB VB UA
VA
Member BA:
Member AC: 22.50 -38.97 -22.
-38.97 67.50 38. -67. -22.50 38.97 22. -38. 38.97 -67.50 -38.
FXAC
FYAC
FXCA
FYCA
EA/L
UA
VA
Uc Vc
UA
VA
UC
UA VA UC
Equilibrium equations are obtained by assembly procedure:
UA
VA
UC
Solution by Gauss-elimination Method: D = (values in mm)
FXBA
FYBA
FXAB
FYAB
Member end forces referred to global axes are initially calculated using eqn 1 along with the known displacements:
Fes
For Member BA:
FXAC
FYAC
FXCA
FYCA
For Member AC: (All values in kN)
Fes =
Axial forces in member are now calculated using equation:
Force = FXBA cos α +
FYBA cos β
BA: F= (-18.09)(0.5)-(31.34)(0.866)=
-36.18 kN. BC: F=(28.09)(1)+ =28.09 kN. AC: F=(-28.09)(0.5)-(48.66)(0.866) = -56.18 kN.