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
∆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
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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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

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.