Beam Analysis by the Direct Stiffness Method

Steven VukazichSan Jose State University

Stiffness Based Approach

! = # ∆

Can think of a structure acting as a multi-dimensional spring

Δ& ! ! = &Δ

Δ' () Δ*(+

Δ, (-

!'

!*

!,

!) !+ !-

Consider an inclined beam member with a moment of inertia I and modulus of elasticity E subjected to shear force and bending moment at its ends.

x axis (local 1 axis in SAP 2000)i = initial end of elementj = terminal end element

Note the sign convention

Beam Element Stiffness Matrix in Local Coordinates

!"

#"

#$ !$

%Δ" Δ$

'

()

*"*$

+ ,

-

. = 0 1

We want to find this 4x4 matrix

12#$%& Δ

Δ

%

#$

6#$%) Δ

12#$%& Δ

6#$%) Δ

4#$% +

%

#$

6#$%) + +

2#$% +

6#$%) +

Each Column of the Frame Element Matrix in Local Coordinates is derived from Indeterminate Fixed End Moment Solutions

set Δi =1and θi =θi =Δj =0

Find the First Column of the Frame ElementStiffness Matrix in Local Coordinates

)* =12,-./

01 =6,-.3

4Δ* = 1 Δ1 = 0

.

,-

5* = 0

6 7

8

51 = 0

0* =6,-.3

)1 = −12,-./

Frame Element Stiffness Matrix

! = # $

%&'&%('(

=∆&*&∆(*(

12-./0

6-./2 −12-./0

6-./2

6-./2

4-./ −6-./2

2-./

−12-./0 −6-./212-./0 −6-./2

6-./2

2-./ −6-./2

4-./

Consider a beam comprised of two elements

Structure Stiffness Matrix

y

x

34

12 6

5

L2

EI1 EI2

L1

! = # ∆

The 6x6 structure stiffness matrix can be assembled from the element stiffness matrices

Each beam joint can move in two directions: 2 Degrees of Freedom (DOF) per joint

2. Joint and DOF numbering

Assemble Structure Stiffness Matrix

y

x

34

1

L1

12 6

5

23

1 2

L2

EI1

EI2

2n - 1

2nn

1. Number each element

Element DOF

1 2 3 4

Associated global DOF for element 1 1 2 3 4

Associated global DOF for element 2 3 4 5 6

3. Connectivity table to assemble structure stiffness matrix

34

Li

12 EIi

Label the rows and columns of each 4x4 element stiffness matrix with corresponding structure DOF from connectivity table

Assembly of Structure Stiffness Matrix from Element Contributions

Form 4x4 element stiffness matrix for element 1 from EI1 and L1

Form 4x4 element stiffness matrix for element 2 from EI2 and L2

Assemble 6x6 structure stiffness matrix

DOF 3 are 4 are free DOF;

DOF 1, 2, 5,and 6 are restrained (support) DOF

At restrained DOF, we know the displacements but the forces (support reactions) are unknown

At free DOF (blue), we know the forces (applied joint loads) but the displacements are unknown

Structure System of Equations: Free DOF

!"" !"# !"$ !"% !"& !"'!#" !## !#$ !#% !#& !#'!$" !$# !$$ !$% !$& !$'!%" !%# !%$ !%% !%& !%'!&" !&# !&$ !&% !&& !&'!'" !'# !'$ !'% !'& !''

(")#($)%(&)'

Δ"+#Δ$+%Δ&+'

=

x

34

L1

12 6

5EI1

EI2

DOF 3 are 4 are free DOF;

DOF 1, 2, 5,and 6 are restrained(support) DOF

At restrained DOF, we know the displacements (all equal to zero) but the forces (support reactions) are unknown

At free DOF (blue), we know the forces (applied joint loads) but the displacements are unknown

Free DOF System of Equations

!"" !"# !"$ !"% !"& !"'!#" !## !#$ !#% !#& !#'!$" !$# !$$ !$% !$& !$'!%" !%# !%$ !%% !%& !%'!&" !&# !&$ !&% !&& !&'!'" !'# !'$ !'% !'& !''

(")#($)%(&)'

00Δ$,%00

=

x

34

12 6

5V3

M4

Suppose we have loads applied to joint 2

Free DOF Equation Set

Solve for Displacements at Free DOF

x

34

12 6

5V3

!"" !"#!#" !##

∆"%# = '"

(#

Solvefor∆" and%#

M4

DOF 1, 2, 5 and 6 are restrained (support) DOF

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

Δ3 and θ4, found from previous step

Find Support Reactions

!"" !"# !"$ !"% !"& !"'!#" !## !#$ !#% !#& !#'!$" !$# !$$ !$% !$& !$'!%" !%# !%$ !%% !%& !%'!&" !&# !&$ !&% !&& !&'!'" !'# !'$ !'% !'& !''

(")#($)%(&)'

00Δ$,%00

=

!"$ !"%!#$ !#%!&$ !&%!'$ !'%

(")#(&)'

Δ$,%=