Beam Analysis by the Direct Stiffness MethodStevenVukazich

SanJoseStateUniversity

ConsideraninclinedbeammemberwithamomentofinertiaI andmodulusofelasticityEsubjectedtoshearforceandbendingmomentatitsends.

x’axis(local1axisinSAP2000)i =initialendofelementj=terminalendelement

φ"

x’

y’ y

x

i

j

L

Δj

Δi

θi

θj

Vi

Vj Mj

EI

Mi

Note the sign convention

Beam Element Stiffness Matrix in Local Coordinates

Each Column of the Frame Element Matrix in Local Coordinates is derived from indeterminate fixed end moment solutions

φ

x

yy

x

i

j

L Δ

i =1

θi = 0

θj = 0

Δi = 0

Vi =12EIL3

M j =6EIL2

Mi =6EIL2

Vj = −12EIL3

set Δi = 1 and θi = θi = Δj = 0

First Column of the Frame ElementMatrix in Local Coordinates

ViMi

Vj

M j

!

"

##

$

##

%

&

##

'

##

=

12EIL3

6EIL2

−12EIL3

6EIL2

6EIL2

4EIL

−6EIL2

2EIL

−12EIL3

−6EIL2

12EIL3

−6EIL2

6EIL2

2EIL

−6EIL2

4EIL

)

*

++++++++++

,

-

.

.

.

.

.

.

.

.

.

.

Δi

θiΔ j

θ j

!

"

##

$

##

%

&

##

'

##

Frame Element Matrix in Local Coordinates

ConsiderthebeamcomprisedoftwoelementsNotethatsincethex’andxaxesarethesamethelocalandglobalDOFareinthesamedirectionsandso [k’] = [k] Eachbeamjointcanmoveintwo

directions2DegreesofFreedom(DOF)perjoint• Numbereachjoint;• NumbereachstructureDOF.

BeamStructureConnectivityTabletoAssemble

2.JointandDOFnumbering

1.Numbereachelement

Beam Structure System of Equations from Element Contributions

Labeltherowsandcolumnsofeach4x4elementstiffnessmatrixwithcorrespondingstructureDOFfromconnectivitytable;

Assemble6x6Structurestiffnessmatrixfromelementcontributions

Assembly of Beam Structure Stiffness Matrix from Element Contributions

DOF3,4and6arefreeDOF;

DOF1,2,and5arerestrained(support)DOF

AtrestrainedDOF(yellow), weknowthedisplacementsbuttheforces(supportreactions)areunknown

AtfreeDOF(grey),weknowtheforces(appliedjointloads)butthedisplacementsareunknown

FirstSolveFreeDOFEquationsettofindunknowndisplacementsK33 K34

K43

K63

K44

K64

K36

K46

K66

!

"

####

$

%

&&&&

Δ3θ4θ6

(

)*

+*

,

-*

.*=

−P00

(

)*

+*

,

-*

.*

solve free DOF equation set to find Δ3, θ4, and θ6

Structure System of Equations: Restrained and Free DOF

DOF3,4and6arefreeDOF;

DOF1,2,and5arerestrained(support)DOF

After displacements are found, multiply to find unknown forces (support reactions) V1

M2

V5

!

"#

$##

%

&#

'##

=

K13 K14 0K23 K24 0K53 K54 K56

(

)

****

+

,

----

Δ3θ4θ6

!

"#

$#

%

&#

'#

Δ3, θ4, and θ6 found from previous step

Find Support Reactions

Workinunitsofkipsandinches

For the beam shown the properties of the elements are: Member Section I E 1 W8x10 30.8 in4 29000 ksi 2 W8x10 30.8 in4 29000 ksi !

y, y’

x, x’

!

12 k 3

4

2 2 !1 !3 !

!!

12 ft 6 ft

1 ! !

1

2 !!

6 !!

5

!!!

(144in) (72in)

Lab Problem

k[ ] = !k[ ] =

12EIL3

6EIL2

−12EIL3

6EIL2

6EIL2

4EIL

−6EIL2

2EIL

−12EIL3

−6EIL2

12EIL3

−6EIL2

6EIL2

2EIL

−6EIL2

4EIL

#

$

%%%%%%%%%%

&

'

((((((((((

Construct the 4x4 Element Stiffness Matrix for Each Element

Assemble the 6x6 Structure Stiffness Matrix from the 4x4 Element Stiffness Matrices and Connectivity Table

K33 K34

K43

K63

K44

K64

K36

K46

K66

!

"

####

$

%

&&&&

Δ3θ4θ6

(

)*

+*

,

-*

.*=

−P00

(

)*

+*

,

-*

.*

Multiplytoverifythatthesolutionis:

∆!= !−1.238155!!" !! = !0.0051590!!"# !! = !0.0232154!!"#

Solve the free DOF Equation Set to find the Unknown Displacements

–P = –12

1. Find Support Reactions (V1, M2, V5) from the joint displacement results and matrix multiplication;

2. Verify your results for V1, M2, and V5 satisfyequilibrium using statics.

V1M2

V5

!

"#

$##

%

&#

'##

=

K13 K14 0K23 K24 0K53 K54 K56

(

)

****

+

,

----

Δ3θ4θ6

!

"#

$#

%

&#

'#

∆!= !−1.238155!!" !! = !0.0051590!!"# !! = !0.0232154!!"#

Find the Support Reactions and Verify Equilibrium