SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

CIVL3340 L7: Introduction to Direct

Stiffness Method

BY JOE GATTAS, FARIS ALBERMANI

Introduction

Deformation methods of structural analysis are those that formulate equilibrium equations to cal-culate unknown joint displacements. The slope-deflection and moment-distribution displacementmethods are widely used historically, however modern computational structural analysis softwarealmost exclusively uses the direct stiffness method.

The direct stiffness method is a matrix analysis method which means equilibrium equations areformulated into a single matrix relationship. Free joint displacement equations can be auto-matically selected from the full system matrix and solved, so is very well suited to computer-automated analysis of statically indeterminate structures. The method is similar for statically de-terminate and indeterminate structures as it exploits kinematic redundancy (the number of jointdegrees-of-freedom DOF) rather than static redundancy. There is therefore always the samenumber of equilibrium equations as unknown displacements.

Truss Beam FrameNumber of unknown forces/member 1 2 3Number of displacements/node 2 2 3Number of equilibrium eqs/node 2 2 3

page 1 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

Example 7.1: Static and Kinematic Redundancy Review

Calculate the static and kinematic redundancy of the following structures.

Truss Beam Beam

m=4,n=5,R=8SR=1m+R-2n=2KR=2

m=1,n=2,R=2SR=2m+R-2n=0KR=2

m=2,n=3,R=2SR=2m+R-2n=0KR=4

The broad procedure for the direct stiffness method is as follows:

1. Define problem including structure nodal coordinates, element locations, element type,support restraints, and applied loads.

2. Calculate basic member properties including length L, element inclination angle ✓, andbasic stiffness k.

3. Transform element basic stiffness to local k0 and global K systems and assemble globalstiffness matrix K

S

.

4. Formulate applied load vector R and, if needed, fixed end action vector P . Stiffness solutionis then given by a single matrix equation {R} = [K

S

]{r} + {P} which can be partitioned

into free and supported displacement components⇢R

f

R

s

�=

K

ff

K

fs

K

sf

K

ss

�⇢r

f

r

s

�+

P

f

P

s

�.

5. Solve free nodal displacements r

f

, support reactions R

s

, and basic member forces S.

Variations to this specifics of this procedure are encountered between truss, beam, or framestructures and 2D or 3D systems. CIVL3340 lectures on the direct stiffness method are asfollows.

L7: Direct stiffness method: introduction, problem definition, basic element stiffness.

page 2 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

L8: Direct stiffness method: truss element transformation, global stiffness matrix assembly, andsystem solution.

L9: Direct stiffness method: truss element lack-of-fit, support settlement, and MATLAB solver.Introduction to commercial structural analysis softwares.

L10: Direct stiffness method: Beam element formulation.

L11: Direct stiffness method: Frame element formulation.

L12: Structural symmetry and introduction to advanced structural analysis.

Problem Definition

An automated matrix method of analysis requires a formal definition of the structural systembeing analysed. This includes geometry (node numbers and coordinates), topology/connectivity(element numbers, start and end nodes), material properties (E), section properties (I, A), sup-ports, and load conditions.

Example 7.2: Structural System Definition

Tabulate a definition of the following structural system.

5m 5m

8.66m

400kN

E=200GPaA=104 mm2

1.5A,

E

A,E

1.5A,

E

1.5A

,E

A,E

45o

Assign coordinate system, node numbers, element numbers, and element directions.

page 3 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

2

1 2

3 4

1

3 45

List element properties including element number (EID), start node i, and end node j.

EID i j E A

1 1 2 2⇥ 10

51⇥ 10

4

2 3 4 2⇥ 10

51⇥ 10

4

3 1 3 2⇥ 10

51.5⇥ 10

4

4 2 4 2⇥ 10

51.5⇥ 10

4

5 2 3 2⇥ 10

51.5⇥ 10

4

List node properties including node number (NID), coordinates, reactions R, and applied loadsP .

NID x y R

x

R

y

P

x

P

y

1 0 0 F

x,1 F

y,1 0 02 5 0 0 F

y,2 0 03 5 8.66 0 0 0 04 10 8.66 0 0 2.82⇥ 10

5 -2.82⇥ 10

5

Certain system properties can be calculated from the above problem definition. Element lengthL and element inclination angle ✓ are calculated from element node i and j coordinates:

L =

q(y

j

� y

i

)

2+ (x

j

� x

i

)

2

↵ = arctan

y

j

� y

i

x

j

� x

i

Basic element stiffness can also be calculated from these inputs as follows.

page 4 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

Basic Element Stiffness

Stiffness coefficient knm

is equal numerically to the force required at a point in a specified di-rection, n, to produce a unit displacement at another point in a specified direction, m. All otherdisplacements are assumed to be prevented (locked) at that point. For example, the below figureshows load P

n

that generates displacement �m

, with all other displacements locked at m. Thestiffness relationship between the two is therefore P

n

= k

nm

�

m

. This is the inverse of the flexibil-ity relationship �

n

= f

nm

P

m

so the stiffness coefficient is the inverse of the flexibility coefficientk

nm

= 1/f

nm

.

Pn

Δm

Δm

Pn

knm

Pm

Δn

f nm

In the direct stiffness method, basic member stiffness k is the stiffness coefficient that relatesbasic forces and displacements on truss, beam, and frame members. Basic member force S

and basic member deformations v are therefore related with {S} = [k]{v}. It is numericallyequivalent to the member force required to generate a unit member displacement.

Example 7.3: Basic Element Stiffness

Formulate the basic element member force, member deformations, and stiffness for truss, beam,and frame elements.

Truss elements have a single axialtranslational displacement and singlecorresponding axial force. Basic stiff-ness can be derived from basic linearelastic material elongation: " = �/L =

�/E = P/(EA).

{v} = �

{S} = P

[k] =

AE

L

Δ

P

Beam elements have a rotational dis-placement and corresponding momentat each node. Basic stiffness canbe obtained from relations derived inL2 Slope Deflection Method: M

i

=

2EI

L

(2✓

i

+ ✓

j

) and M

j

=

2EI

L

(✓

i

+ 2✓

j

).

{v} =

⇢�

i

�

j

�

{S} =

⇢M

i

M

j

�

[k] =

2EI

L

2 1

1 2

� φjφi

MjMi

page 5 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

Frame elements have combined rota-tional and axial translational displace-ments.

{v} =

8<

:

�

i

�

j

�

9=

;

{S} =

8<

:

M

i

M

j

P

9=

;

[k] =

2

44EI

L

2EI

L

0

2EI

L

4EI

L

0

0 0

AE

L

3

5

MjMi

Δ

P

φjφi

Formulas

Nomenclature

Name Symbol

Element length L

Element inclination angle ↵

Local transformation matrix [T ]

Inclination component matrix [L

B

]

Global transformation matrix [L

D

]

Basic member force {S}Local member force {F 0}Global member force {F}Basic member deformations {v}Local member deformations {r0}Global member deformations {r}Basic member stiffness [k]

Local member stiffness [k

0]

Global member matrix [K]

Assembled global stiffness matrix [K

S

]

Applied load {R}Equivalent applied fixed end actions {P}Equivalent distributed node loads {v⇤}

page 6 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

General Formulas

c = cos↵

s = sin↵

{v} = [T ]{r0}{r0} = [L

D

]{r}{F 0} = [T ]

T{S}{F} = [L

D

]

T{F 0}[k

0] = [T ]

T

[k][T ]

[K] = [L

D

]

T

[k

0][L

D

]

[L

D

] =

2

4LB

0

0 L

B

3

5

{S} = [k]{v}{R} = [K

S

]{r}+ {P}8<

:R

f

R

s

9=

; =

2

4Kff

K

fs

K

sf

K

ss

3

5

8<

:r

f

r

s

9=

;+

2

4Pf

P

s

3

5

�

P

=

F

p

L

AE

�

T

= ↵L�T

{v⇤} =

wL

3

24EI

8<

:1

1

9=

;

Truss Member

[T ] = [ �1 0 1 0

]

[L

B

] =

2

4 c s

�s c

3

5

k =

AE

L

[k

0] =

AE

L

2

6666664

1 0 �1 0

0 0 0

1 0

Sym. 0

3

7777775

[K] =

AE

L

2

6666664

c

2cs �c

2 �cs

s

2 �cs �s

2

c

2cs

Sym. s

2

3

7777775

[K] =

AE

L

2

6666664

c

2cs �cc

0 �cs

0

s

2 �c

0s �ss

0

c

02c

0s

0

Sym. s

02

3

7777775

Beam Member

[T ] =

2

4 1/L 1 �1/L 0

1/L 0 �1/L 1

3

5

[k] =

2EI

L

2

4 2 1

1 2

3

5

[K] = [k

0] =

2

6666664

12EI

L

36EI

L

2�12EI

L

36EI

L

2

4EI

L

�6EI

L

22EI

L

12EI

L

3�6EI

L

2

Sym. 4EI

L

3

7777775

page 7 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L7

Introduction to Direct Stiffness Method

Frame Member

[T ] =

2

6664

0 1/L 1 0 �1/L 0

0 1/L 0 0 �1/L 1

�1 0 0 1 0 0

3

7775

[L

B

] =

2

6664

c s 0

�s c 0

0 0 1

3

7775

[k] =

2

6664

4EI

L

2EI

L

0

2EI

L

4EI

L

0

0 0

AE

L

3

7775

[k

0] =

2

6666666666664

AE

L

0 0

�AE

L

0 0

12EI

L

36EI

L

2 0

�12EI

L

36EI

L

2

4EI

L

0

�6EI

L

22EI

L

AE

L

0 0

12EI

L

3�6EI

L

2

Sym. 4EI

L

3

7777777777775

[K] =

2

6666666666664

AE

L

c

2+

12EI

L

3 s

2 AE

L

cs� 12EI

L

3 cs

�6EI

L

2 s

�AE

L

c

2 � 12EI

L

3 s

2 �AE

L

cs+

12EI

L

3 cs

�6EI

L

2 s

AE

L

s

2+

12EI

L

3 c

2 6EI

L

2 c�AE

L

cs+

12EI

L

3 cs

�AE

L

s

2 � 12EI

L

3 c

2 6EI

L

2 c

4EI

L

6EI

L

2 s�6EI

L

2 c

2EI

L

AE

L

c

2+

12EI

L

3 s

2 AE

L

cs� 12EI

L

3 cs

6EI

L

2 s

AE

L

s

2+

12EI

L

3 c

2 �6EI

L

2 c

Sym. 4EI

L

3

7777777777775

page 8 of 8

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

CIVL3340 L8: Direct Stiffness Method:

Truss Elements I

BY JOE GATTAS, FARIS ALBERMANI

Element Transformation

A complete structural stiffness matrix relationship can be assembled from the constituent ele-ments’ stiffness, however this requires all elements use a consistent, global coordinate system.Basic forces {S} and displacements {v} can be converted to local forces {F 0} and displace-ments {r0} with the local transformation matrix T . Local forces and displacements can then beconverted to global forces {F} and global displacements {r} with global transformation matrix[L

D

]

{v} = [T ]{r0}{F 0} = [T ]T{S}

{r0} = [LD

]{r}{F} = [L

D

]

T{F 0}

For basic-to-local, force and displacement transformations are transposition of each other. Sim-ilarly for local-to-global transformations. This is known as the contragredient law. Stiffness re-lations still exist in the local system {F 0} = [k0

]{r0} and global system {F} = [K]{r}. Stiffnesstransformations can be obtained directly from the above relations as follows.

S = kv = kTr0

T TS = T TkTr0

F 0= k0r0

k0= T TkT

F = Kr

LT

D

F 0= KLT

D

r0

LT

D

k0r0 = KLT

D

r0

K = LT

D

k0LD

The above transformations are valid for any element, but T and LD

transformation matrices varybetween truss, beam, and frame elements. Note that L

D

is an orthogonal matrix, i.e. LT

D

=

L�1D

. The following lecture first demonstrates truss element transformations and subsequently acomplete truss structure stiffness solution.

page 1 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Truss Element Force Transformations

Basic

P {S} = P

Local

Fx,j’Fy,j’

Fx,i’Fy,i’{F 0} =

8>><

>>:

F 0x,i

F 0y,i

F 0x,j

F 0y,j

9>>=

>>;=

8>><

>>:

�1

0

1

0

9>>=

>>;{S} = T TS

where T =< �1, 0, 1, 0 >

Global

Fx,jFy,j

Fx,i

Fy,i

Fx’Fy’

θ

θ

θFx

Fy

{F} =

8>><

>>:

Fx,i

Fy,i

Fx,j

Fy,j

9>>=

>>;=

2

664

c �s 0 0

s c 0 0

0 0 c �s0 0 s c

3

775 {F 0} = LT

D

F 0

where LD

=

LB

0

0 LB

�and [L

B

] =

c s�s c

�

Note: global transformation matrix LD

assembled from in-clination component matrix L

B

. Direct nodal transforma-tions are F

x

=F 0x

cos ✓ � F 0y

sin ✓ and Fy

=F 0x

sin ✓ + F 0y

cos ✓.c = cos ✓ and s = sin ✓.

Truss Element Displacement Transformations

Basic

Δ{v} = �

page 2 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Local

ui’vi’

uj’

vj’

{r0} =

8>><

>>:

u0i

v0i

u0j

v0j

9>>=

>>;, and {v} =

��1 0 1 0

{r0} = Tr0

which can be shown from � = u0j

�u0i

= �u0i

+0v0i

+u0j

+0v0j

Global

ui

vi uj

vj{r} =

8>><

>>:

ui

vi

uj

vj

9>>=

>>;=

2

664

c �s 0 0

s c 0 0

0 0 c �s0 0 s c

3

775 {r0} = LT

D

r0

Truss Element Stiffness Transformation

Basic

k[k] =

AE

L

Local

kijkji’’

kjj’

kii’[k0

] = T TkT =

AE

L

2

664

1 0 �1 0

0 0 0

1 0

Sym. 0

3

775

page 3 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Global

Kjj

Kii

KijKji

[K] = LT

D

k0LD

=

AE

L

2

664

c2 cs �c2 �css2 �cs �s2

c2 csSym. s2

3

775

Example 8.1: Truss Element Stiffness Transformations

Formulate the local and global element stiffnesses of the truss shown in Example 7.2.

5m 5m

8.66m

400kN

E=200GPaA=104 mm2

1.5A,

E

A,E

1.5A,

E

1.5A

,E

A,E

45o2

12

3

4

1

3 45

Tabulate the basic member properties. Units N-mm-MPa.

EID i j E A L ✓ AE/L c s cs

1 1 2 2⇥ 10

51⇥ 10

45⇥ 10

3 0 4⇥ 10

5 1 0 02 3 4 2⇥ 10

51⇥ 10

45⇥ 10

3 0 4⇥ 10

5 1 0 03 1 3 2⇥ 10

51.5⇥ 10

410⇥ 10

3 ⇡/3 3⇥ 10

5 0.5 0.866 0.4334 2 4 2⇥ 10

51.5⇥ 10

410⇥ 10

3 ⇡/3 3⇥ 10

5 0.5 0.866 0.4335 2 3 2⇥ 10

51.5⇥ 10

48.66⇥ 10

3 ⇡/2 3.46⇥ 10

5 0 1 0

page 4 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Truss elements have two nodes and two nodal displacements per node, so total DOF is 4 in localand global systems.

For element 1, stiffness coefficients are between nodes 1 and 2:

Local

[k0] =

AE

L

2

664

1 0 �1 0

0 0 0

1 0

Sym. 0

3

775

= 10

5

2

664

4 0 �4 0

0 0 0

4 0

Sym. 0

3

775

Global

[K] =

AE

L

2

664

c2 cs �c2 �css2 �cs �s2

c2 csSym. s2

3

775

[K1] = 10

5

2

664

4 0 �4 0

0 0 0

4 0

Sym. 0

3

775

Element 2 has the same length, orientation, and properties as element 1 so possesses the sameelement stiffness matrices except for nodes 3 and 4 instead of nodes 1 and 2.

For element 3, stiffness coefficients are between nodes 1 and 3:

Local

[k0] = 10

5

2

664

3 0 �3 0

0 0 0

3 0

Sym. 0

3

775

Global

[K3] = 10

5

2

664

0.75 1.30 �0.75 �1.302.25 �1.30 �2.25

0.75 1.30Sym. 2.25

3

775

Element 4 has the same length, orientation, and properties as element 3 so possesses the sameelement stiffness matrices except for nodes 2 and 4 instead of nodes 1 and 3.

For element 5, stiffness coefficients are between nodes 2 and 3:

Local

[k0] = 10

5

2

664

3.46 0 �3.46 0

0 0 0

3.46 0

Sym. 0

3

775

Global

[K5] = 10

5

2

664

0 0 0 0

3.46 0 �3.460 0

Sym. 3.46

3

775

page 5 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Global Stiffness Matrix

Once each element’s stiffness is expressed in the global coordinate system, a single globalstiffness matrix can be assembled. The global stiffness matrix K

S

will be a square matrix of anorder equal to the total (unrestrained plus restrained) DOF of the system, i.e. a truss structurewith 3 nodes will possess a global stiffness matrix of order 3⇥ 2 = 6.

K33

K11

2

1 2

3

1

3

3

3 K332

K222

K111 K222

K133K313 K322

K232

K121

K211

K33

K11 K22

K13K31K12K21

K32K23

Truss KS Ke

Assembly of KS

is simply done by linearly superimposing the stiffness contribution of each el-ement to each appropriate system DOF. For a truss, each element possesses a 4 ⇥ 4 globalelement stiffness matrix which is composed of four 2⇥ 2 sub-matrices K

ii

, Kjj

, Kij

= KT

ji

. Givenan element e with start node i and end node j, Ke is:

[Ke

] =

K

ii

Kij

Kji

Kjj

�

The global stiffness matrix is assembled by summing stiffness contributions from each elementat each node DOF. For the above example, a 6⇥ 6 global global element stiffness matrix can bepopulated with:

Element 1 [K1];

u1, v1 u2, v22

664

3

775

K111 K1

12 0

K121 K1

22 0

0 0 0

Plus Element 2 [K2]:

u2, v2 u3, v32

664

3

775

K111 K1

12

K121 K1

22 + K222 K2

23

0 K232 K2

33

Plus Element 3 [K3]:

u1, v1 u3, v32

664

3

775

K111 + K3

11 K112 K3

13

K121 K1

22 +K222 K2

23

K331 K2

32 K233 + K3

33

page 6 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

This gives a final global stiffness matrix of:

KS

=

u1, v1 u2, v2 u3, v32

664

3

775

K111 +K3

11 K112 K3

13

K122 +K2

22 K223

Sym. K233 +K3

33

Stiffness Equation Solution

The system applied load vector R, nodal displacement vector r, and equivalent applied fixedend action vector P can be assembled in a similar way to the global stiffness matrix, that isby summing contributions from each element to each appropriate system DOF. For example,the figure below shows the previous three element truss system with applied loads and supportreactions. There are two DOFs for each of the three nodes, so R and r are each 6⇥ 1 vectors.

2

1 2

3

1

3

100N50N

{r} =

8>>>>>><

>>>>>>:

u1

v1u2

v2u3

v3

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

0

0

u2

0

u3

v3

9>>>>>>=

>>>>>>;

{R} =

8>>>>>><

>>>>>>:

Rx,1

Ry,1

Rx,2

Ry,2

Rx,3

Ry,3

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

Rx,1

Ry,1

0

Ry,2

100

50

9>>>>>>=

>>>>>>;

The system stiffness equation is given by {R} = [KS

]{r} + {P}. From the above example, itcan be seen that for free DOFs, reaction forces are zero and displacements are unknown. Con-versely, for supported DOFs, displacements are typically zero and reaction forces are unknown.The stiffness equation can be therefore be partitioned into free and supported displacementcomponents:

⇢R

f

Rs

�=

K

ff

Kfs

Ksf

Kss

�⇢rf

rs

�+

Pf

Ps

�

If support displacements are all zero, this gives the following solution for free nodal displace-ments.

{Rf

} = [Kff

]{rf

}+⇠⇠⇠⇠⇠[K

fs

]{rs

}+ {Pf

}{r

f

} = [Kff

]

�1({R

f

}� {Pf

})

The above examples do not show applied loads necessitating the use of equivalent applied fixedend action vector P . The next lecture will discuss truss problems in which {r

s

} 6= 0 and {P} 6= 0.

page 7 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Once free node displacements are known, support reactions are obtained with:

{Rs

} = [Ksf

]{rf

}+⇠⇠⇠⇠⇠[K

ss

]{rs

}+ {Ps

}{R

s

} = [Ksf

]{rf

}+ {Ps

}

Member design is based on basic member forces, which are obtained with transformations as:

{S} = [k]{v}= [k][T ]{r0}= [k][T ][L

D

]{r}

For a truss element, this simplifies to {Se} =

AE

L

< �c,�s, c, s > {re} where re is a 4 ⇥ 1 vectorcontaining global nodal displacements of element e.

Example 8.2: Truss Element Stiffness Solution

Solve the nodal displacements, reaction forces, and member forces of truss system shown inExample 8.1.

Element global stiffness matrices are as follows.

[K1] = [K2

] = 10

5

2

664

4 0 �4 0

0 0 0

4 0

Sym. 0

3

775

[K3] = [K4

] = 10

5

2

664

0.75 1.30 �0.75 �1.302.25 �1.30 �2.25

0.75 1.30Sym. 2.25

3

775

[K5] = 10

5

2

664

0 0 0 0

3.46 0 �3.460 0

Sym. 3.46

3

775

The global stiffness matrix has 4 ⇥ 2 = 8 DOFs so has an order of 8. [K1] gives a stiffness

contribution to nodes 1 and 2.

K1S

= 10

5

u1 v1 u2 v2 u3 v3 u4 v42

6666666664

3

7777777775

4 0 �4 0 0 0 0 0

0 0 0 0 0 0 0 0

�4 0 4 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

page 8 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

[K2] gives a stiffness contribution to nodes 3 and 4.

K1+2S

= 10

5

u1 v1 u2 v2 u3 v3 u4 v42

6666666664

3

7777777775

4 0 �4 0 0 0 0 0

0 0 0 0 0 0 0 0

�4 0 4 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 4 0 �4 0

0 0 0 0 0 0 0 0

0 0 0 0 �4 0 4 0

0 0 0 0 0 0 0 0

[K3] gives a stiffness contribution to nodes 1 and 3.

K1+2+3S

= 10

5

u1 v1 u2 v2 u3 v3 u4 v42

6666666664

3

7777777775

4.75 1.30 �4 0 �0.75 �1.30 0 0

1.30 2.25 0 0 �1.30 �2.25 0 0

�4 0 4 0 0 0 0 0

0 0 0 0 0 0 0 0

�0.75 �1.30 0 0 4.75 1.30 �4 0

�1.30 �2.25 0 0 1.30 2.25 0 0

0 0 0 0 �4 0 4 0

0 0 0 0 0 0 0 0

[K4] gives a stiffness contribution to nodes 2 and 4 and [K5

] gives a stiffness contribution tonodes 2 and 3.

K1+2+3+4+5S

= KS

= 10

5

u1 v1 u2 v2 u3 v3 u4 v42

6666666664

3

7777777775

4.75 1.30 �4 0 �0.75 �1.30 0 0

1.30 2.25 0 0 �1.30 �2.25 0 0

�4 0 4.75 1.30 0 0 �0.75 �1.300 0 1.30 5.71 0 �3.46 �1.30 �2.25

�0.75 �1.30 0 0 4.75 1.30 �4 0

�1.30 �2.25 0 �3.46 1.30 5.71 0 0

0 0 �0.75 �1.30 �4 0 4.75 1.300 0 �1.30 �2.25 0 0 1.30 2.25

Node load and displacement vectors are assembled from system information.

page 9 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

{r} =

8>>>>>>>>>><

>>>>>>>>>>:

u1

v1u2

v2u3

v3u4

v4

9>>>>>>>>>>=

>>>>>>>>>>;

=

8>>>>>>>>>><

>>>>>>>>>>:

0

0

u2

0

u3

v3u4

v4

9>>>>>>>>>>=

>>>>>>>>>>;

{R} =

8>>>>>>>>>><

>>>>>>>>>>:

Rx,1

Ry,1

Rx,2

Ry,2

Rx,3

Ry,3

Rx,4

Ry,4

9>>>>>>>>>>=

>>>>>>>>>>;

=

8>>>>>>>>>><

>>>>>>>>>>:

Rx,1

Ry,1

0

Ry,2

0

0

2.82⇥ 10

5

�2.82⇥ 10

5

9>>>>>>>>>>=

>>>>>>>>>>;

{P} =

8>>>>>>>>>><

>>>>>>>>>>:

0

0

0

0

0

0

0

0

9>>>>>>>>>>=

>>>>>>>>>>;

The stiffness equation is partitioned into free nodal displacements (u2, u3, v3, u4, v4) and fixednodal displacements (u1, v1, v2).

{Rf

} = [Kff

]{rf

}+⇠⇠⇠⇠⇠[K

fs

]{rs

}+���{Pf

}8>>>><

>>>>:

0

0

0

2.82⇥ 10

5

�2.82⇥ 10

5

9>>>>=

>>>>;

= Kff

8>>>><

>>>>:

u2

u3

v3u4

v4

9>>>>=

>>>>;

where Kff

= 10

5

u2 u3 v3 u4 v42

6664

3

7775

4.75 0 0 �0.75 �1.300 4.75 1.30 �4 0

0 1.30 5.71 0 0

�0.75 �4 0 4.75 1.30�1.30 0 0 1.30 2.25

Solve free nodal displacements.

{rf

} = [Kff

]

�1{Rf

} =

8>>>><

>>>>:

�0.419.8�2.210.9�7.8

9>>>>=

>>>>;

Solve support reactions.

{Rs

} = [Ksf

]{rf

}+⇠⇠⇠⇠⇠[K

ss

]{rs

}+���{Ps

}

8<

:

Rx,1

Ry,1

Ry,2

9=

; = 10

5

u2 u3 v3 u4 v4" #�4 �0.75 �1.30 0 0

0 �1.30 �2.25 0 0

1.30 0 �3.46 �1.30 �2.25

8>>>><

>>>>:

�0.49.8�2.210.9�7.8

9>>>>=

>>>>;

= 10

6

8<

:

�0.28�0.771.05

9=

;

page 10 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L8

Direct Stiffness Method: Truss Elements I

Solve basic member forces for each element.

Se

=

AE

L< �c,�s, c, s > {re}

S1= 4⇥ 10

5 < �1,�0, 1, 0 >

8>><

>>:

0

0

�0.40

9>>=

>>;= �1.63⇥ 10

5

S2= 4⇥ 10

5 < �1,�0, 1, 0 >

8>><

>>:

9.8�2.210.9�7.8

9>>=

>>;= 4.46⇥ 10

5

S3= 3⇥ 10

5 < �0.5,�0.866, 0.5, 0.866 >

8>><

>>:

0

0

9.8�2.2

9>>=

>>;= 8.92⇥ 10

5

S4= 3⇥ 10

5 < �0.5,�0.866, 0.5, 0.866 >

8>><

>>:

�0.40

10.9�7.8

9>>=

>>;= �3.27⇥ 10

5

S5= 3.46⇥ 10

5 < 0,�1, 0, 1 >

8>><

>>:

�0.40

9.8�2.2

9>>=

>>;= �7.73⇥ 10

5

163kN (C)

446kN (T)

892k

N (T

)

327k

N (C

)

773k

N (C

)

7.8mm10.9mm9.8mm

2.2mm

0.4mm

772kN 1056kN282kN

page 11 of 11

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

CIVL3340 L9: Direct Stiffness Method:

Truss Elements II

BY JOE GATTAS, FARIS ALBERMANI

Introduction

The first part of this lecture will examine software implementations of the direct stiffness method.An implementation of a truss solver in MATLAB is presented along with a review of commercially-available structural analysis software.

The second part of this lecture will extend the direct stiffness method for truss elements describedin the previous lecture. Three extensions are developed:

1. Use of non-zero support displacements r

s

to allow for the solution of support settlementproblems.

2. Use of equivalent applied fixed end action vector P to allow for the solution of thermalchange and fabrication error (lack-of-fit) problems.

3. Adjustment of the local-global transformation matrix to allow for the solution of truss sys-tems with inclined supports.

MATLAB Direct Stiffness Implementation

A MATLAB implementation of a direct stiffness truss solver is available for download from thecourse website. Run the script ‘A runTrussAnalysis.m’ to run the solver. The script loads an inputquestion (three examples are given) and then sequentially executes three functions to solve theinput truss system: ‘B initialTrussCalcs.m’, ‘C assembleStiffnessMatrix.m’, and ‘D solveTruss.m’.A fifth function ‘E displayTrussResults.m’ is also called to display the calculated results. The ex-ecution script, an example input, functions, and outputs are printed below.

page 1 of 14

%Written 2014-05-23 by JMG. Contact j.gattas@uq.edu.au with bugs.%Updated 2015-05-09.%Copyright 2015, The Univerersity of Queensland.

%reset + setup clear; clc; close all;

%Question Inputs: createss trussNodes and trussMembers variables inputs_Tutorial7;

%Calculate truss lengths, member inclinations, and basic member stiffnesses [trussMembers,trussPlot] = B_initialTrussCalcs(trussNodes, trussMembers);

%Assemble global stiffness matix globalStiffness = C_assembleStiffnessMatrix(trussNodes, trussMembers);

%Solve nodal displacements + basic member forces [trussNodes, trussMembers] = D_solveTruss(trussNodes, trussMembers,globalStiffness);

Published with MATLAB® R2014b

%Assignment 4 Direct Stiffness - Question 2

P=25e3; %NE=150e03; %MPaA=4e3; %mm^2

%elementID, startNode, endNode, youngs, areatrussMembers= [1 1 2 E A 2 2 3 E A 3 4 3 E A/2 4 1 4 E A 5 1 3 E A 6 4 2 E A];

%nodeID , nodeX, nodeY, restraintX, restraintY, loadX, loadYtrussNodes=[1 0 0 1 1 0 0 2 0 4000 1 1 0 0 3 3000 4000 1 0 0 0 4 3000 0 1 0 0 -P];

Published with MATLAB® R2014b

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

page 2 of 14

function [trussMembers,trussPlot] = B_initialTrussCalcs(trussNodes, trussMembers)%[trussMembers,trussPlot] = B_initialTrussCalcs(trussNodes, trussElements) appends element lengths,%alphas, and basic stiffnesses k=EA/L to trussMembers. It also gives back%trussPlot = [xTruss,yTruss] while is the truss geometry for plotting.%%Written 2014-05-23 by Joe Gattas. Contact j.gattas@uq.edu.au with bugs.%Copyright 2014, The Univerersity of Queensland.

%set up empty data storage elementCount=size(trussMembers,1); nodeCount=size(trussNodes,1); [elementLengths,alphas, basicStiffnesses]=deal(zeros(elementCount,1)); [xTruss,yTruss] = deal(elementCount*3,1);

%calculate data for each element for counter=1:elementCount %get element start node startNode = trussMembers(counter,2); startNodeLocation = find(trussNodes(:,1)==startNode); startNodeXY=[trussNodes(startNodeLocation,2),trussNodes(startNodeLocation,3)]; %get element end node endNode = trussMembers(counter,3); endNodeLocation = find(trussNodes(:,1)==endNode); endNodeXY=[trussNodes(endNodeLocation,2),trussNodes(endNodeLocation,3)];

%add truss elements to truss plot curTrussPlotRow = (counter-1)*3+1; xTruss(curTrussPlotRow:curTrussPlotRow+2)=[startNodeXY(1);endNodeXY(1);nan]; %nan stops successive elements from being joined yTruss(curTrussPlotRow:curTrussPlotRow+2)=[startNodeXY(2);endNodeXY(2);nan];

%calculate element attributes elementLengths(counter) = sqrt((endNodeXY(2)-startNodeXY(2))^2+(endNodeXY(1)-startNodeXY(1))^2); alphas(counter) = atan2((endNodeXY(2)-startNodeXY(2)) , (endNodeXY(1)-startNodeXY(1))); basicStiffnesses(counter) = trussMembers(counter,4)*trussMembers(counter,5)/elementLengths(counter); end

%place into outputs trussMembers=[trussMembers,elementLengths,alphas, basicStiffnesses]; trussPlot=[xTruss',yTruss'];

end

function [globalStiffness] = C_assembleStiffnessMatrix(trussNodes, trussMembers)%[globalStiffness] = C_assembleStiffnessMatrix(trussNodes, trussMembers) assembles the global%stiffness matrix.%%Written 2014-05-23 by Joe Gattas. Contact j.gattas@uq.edu.au with bugs.%Copyright 2014, The Univerersity of Queensland.

%set up empty data storage elementCount=size(trussMembers,1); nodeCount=size(trussNodes,1); globalStiffness=zeros(2*nodeCount,2*nodeCount); T=[-1 0 1 0];

%calculate and insert element global stiffnesses for counter=1:elementCount %convert into local coordinates k^m = T' k T memberStiffnesLocal=T'*trussMembers(counter,8)*T; alpha = trussMembers(counter,7); %convert into global coordinates K^m = Ld' k^m Ld; L=[cos(alpha) sin(alpha); -sin(alpha) cos(alpha) ]; Ld = [L L.*0; L.*0 L ]; memberStiffnessGlobal = Ld'* memberStiffnesLocal * Ld;

%place member stiffness values in global stiffness matrix %KLoc1 = global stiffness location startNode x and y

KLoc1 = 2*(trussMembers(counter,2)-1)+[1,2]; KLoc2 = 2*(trussMembers(counter,3)-1) +[1,2]; globalStiffness(KLoc1,KLoc1)=globalStiffness(KLoc1,KLoc1)+memberStiffnessGlobal(1:2,1:2); globalStiffness(KLoc1,KLoc2)=globalStiffness(KLoc1,KLoc2)+memberStiffnessGlobal(1:2,3:4); globalStiffness(KLoc2,KLoc1)=globalStiffness(KLoc2,KLoc1)+memberStiffnessGlobal(3:4,1:2); globalStiffness(KLoc2,KLoc2)=globalStiffness(KLoc2,KLoc2)+memberStiffnessGlobal(3:4,3:4); end

end

Published with MATLAB® R2014b

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

page 3 of 14

function [trussNodes,trussMembers] = D_solveTruss(trussNodes, trussMembers,globalStiffness)%[trussNodes,trussMembers] = D_solveTrussCalcs(trussNodes, trussMembers) appends nodal displacements x and y%to trussNodes. Also replaces trussNodes restraint boolean with support reaction.%Appends basic member forces S to trussMembers.%%Written 2014-05-23 by Joe Gattas. Contact j.gattas@uq.edu.au with bugs.%Copyright 2014, The Univerersity of Queensland.

%set up empty data storage elementCount=size(trussMembers,1); nodeCount=size(trussNodes,1); nodalDisplacements = zeros(nodeCount*2,1); basicForces=zeros(elementCount,1); T=[-1 0 1 0];

%put node data in vector format restraint=reshape(trussNodes(:,4:5)',[],1); %supportReactions supportReactions = zeros(nodeCount*2,1); %appliedLoads=R appliedLoads =reshape(trussNodes(:,6:7)',[],1); %fixedEndActions=P fixedEndActions = zeros(nodeCount*2,1);

%solve free node displacements rf freeNodeLocations = find(~restraint); supportNodeLocations = find(restraint); %restrained Rf = appliedLoads(freeNodeLocations); Pf = fixedEndActions(freeNodeLocations); Kff = globalStiffness(freeNodeLocations,freeNodeLocations); %rf = Kff^-1 * (Rf-Pf); rf = Kff\(Rf-Pf); %insert free nodal displacements into all nodal displacements nodalDisplacements(freeNodeLocations) = rf;

Ksf = globalStiffness(supportNodeLocations,freeNodeLocations); Rs = Ksf * rf; supportReactions(supportNodeLocations) = Rs;

%solve basic member forces for each element for counter=1:elementCount %set up coordinate conversions matrices startNode = trussMembers(counter,2); endNode = trussMembers(counter,3);

KLoc1 = 2*(startNode-1)+[1,2]; KLoc2 = 2*(endNode-1) +[1,2]; alpha = trussMembers(counter,7); L=[ cos(alpha) sin(alpha); -sin(alpha) cos(alpha) ]; Ld = [L L.*0; L.*0 L ];

basicForces(counter) = trussMembers(counter,8)*T*Ld*nodalDisplacements([KLoc1,KLoc2]); end

%place into outputs trussMembers=[trussMembers,basicForces]; trussNodes(:,4:5) = reshape(supportReactions,2,[])'; trussNodes = [trussNodes,reshape(nodalDisplacements,2,[])'];

end

Published with MATLAB® R2014b

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

page 4 of 14

A - Define problem:

trussMembers =

1.0e+05 *

Columns 1 through 7

0.0000 0.0000 0.0000 1.5000 0.0400 0.0400 0.0000 0.0000 0.0000 0.0000 1.5000 0.0400 0.0300 0 0.0000 0.0000 0.0000 1.5000 0.0200 0.0400 0.0000 0.0000 0.0000 0.0000 1.5000 0.0400 0.0300 0 0.0001 0.0000 0.0000 1.5000 0.0400 0.0500 0.0000 0.0001 0.0000 0.0000 1.5000 0.0400 0.0500 0.0000

Columns 8 through 9

1.5000 0 2.0000 0 0.7500 0.0827 2.0000 0 1.2000 -0.1033 1.2000 0.2092

trussNodes =

1.0e+04 *

Columns 1 through 7

0.0001 0 0 0.6200 0.8267 0 0 0.0002 0 0.4000 -1.2550 1.6733 0 0 0.0003 0.3000 0.4000 -0.6200 0 0 0 0.0004 0.3000 0 1.2550 0 0 -2.5000

Columns 8 through 9

0 0 0 0 0 -0.0000 0 -0.0000

E =

150000

A =

4000

B - Calculate basic member properties:

lenghths =

4000 3000 4000 3000 5000 5000

alpha =

90.0000 0 90.0000 0 53.1301 126.8699

k =

150000 200000 75000 200000 120000 120000

C - Assemble global stiffness matrix:

globalStiffness =

1.0e+05 *

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

page 5 of 14

Columns 1 through 7

2.4320 0.5760 -0.0000 -0.0000 -0.4320 -0.5760 -2.0000 0.5760 2.2680 -0.0000 -1.5000 -0.5760 -0.7680 0 -0.0000 -0.0000 2.4320 -0.5760 -2.0000 0 -0.4320 -0.0000 -1.5000 -0.5760 2.2680 0 0 0.5760 -0.4320 -0.5760 -2.0000 0 2.4320 0.5760 -0.0000 -0.5760 -0.7680 0 0 0.5760 1.5180 -0.0000 -2.0000 0 -0.4320 0.5760 -0.0000 -0.0000 2.4320 0 0 0.5760 -0.7680 -0.0000 -0.7500 -0.5760

Column 8

0 0 0.5760 -0.7680 -0.0000 -0.7500 -0.5760 1.5180

D - Solve truss:

memberForces =

1.0e+04 *

0 0 0.8267 0 -1.0334 2.0916

supportReactions =

1.0e+04 *

0.6200 0.8267 -1.2550 1.6733 -0.6200 0 1.2550 0

displacement =

0 0 0 0 0 -0.1076 0 -0.2179

Published with MATLAB® R2014b

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

page 6 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

Commercial Structural Analysis Software

MATLAB and structural analysis software (SpaceGass and Strand7) are available in EAIT com-puter labs. Problems with any of the computer lab installations should be reported to:mailto:student-helpdesk@eait.uq.edu.au. Personal student software installations are alsoavailable but technical support is not provided by the university for these installations.

Matlab

The University of Queensland has a license for undergraduates to use MATLAB for study or re-search but not for use in commercial applications. Installation instructions are here: https://

www.its.uq.edu.au/student-matlab?pid=883. Numerous MATLAB toolboxes (including struc-tural analysis toolboxes) are available for free from File Exchange http://au.mathworks.com/

matlabcentral/fileexchange/.

Space Gass

Space Gass is a common analysis software http://www.spacegass.com/. The link for this year’sstudent version is www.spacegass.com/exes/electron/student. The student version has nosize restrictions and is therefore suitable for any sized student projects. It will expire at the endof March next year. Most first time users find that they can become productive with the programin just a few hours. It is very intuitive and comes with a comprehensive (600+ page) electronicreference manual. A suite of training videos at http://www.spacegass.com/training will getyou up to speed with SPACE GASS very quickly. If you get an ”expiry” message when you runSpacegass new version, refer to www.spacegass.com/reset.

Miscellaneous

• Instant Structural Analysis (iSA): Structural analysis software available to civil staff andresearch students.

• Microstran: An analysis software available in the EAIT computer labs but no longer com-mercially distributed.

• Oasys GSA: In-house analysis software of Arup and available commercially.

• Strand7 : Finite element analysis software available in EAIT computer labs.

• Abaqus: Finite element analysis software available to civil staff and research students.

page 7 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

Extended Truss Analysis: Support Settlement

Support settlements are non-zero boundary displacements. For a truss element, these may beintroduced directly into the nodal displacement vector r. Once partitioned, the supported nodedisplacements r

s

are therefore non-zero. The stiffness solution then becomes:

{Rf

} = [K

ff

]{rf

}+ [K

fs

]{rs

}+ {Pf

}{r

f

} = [K

ff

]

�1({R

f

}� [K

fs

]{rs

}� {Pf

})

where ({Rf

}� [K

fs

]{rs

}�{Pf

}) is the effective nodal forces. The supported node displacementsr

s

are similarly retained in the support reaction equation:

{Rs

} = [K

sf

]{rf

}+ [K

ss

]{rs

}+ {Ps

}

Extended Truss Analysis: Thermal Changes

Thermal changes in a truss element induce a change in length. For a truss element with lengthL, coefficient of thermal expansion ↵, cross section area A, and temperature change �T , thislength change is equal to:

�

T

= ↵L�T

In a statically-indeterminate truss, the thermally-induced length change is restrained so addi-tional system forces are induced. This is equal to:

P

T

= �AE

L

�

T

= �EA↵�T

Note the negative sign as this is the force required to oppose a thermal change in length. This isa force exhibited in the basic system. It can be expressed in the global system through transfor-mations described in the previous lecture. This gives:

{P} =

8>><

>>:

P

0x,i

P

0y,i

P

0x,j

P

0y,j

9>>=

>>;= L

T

D

T

T

S =

2

664

c �s 0 0

s c 0 0

0 0 c �s

0 0 s c

3

775

8>><

>>:

�1

0

1

0

9>>=

>>;P

T

= EA↵�

T

8>><

>>:

c

s

�c

�s

9>>=

>>;

Once expressed in the global system, the thermally-induced forces can be introduced into thestiffness matrix equation within the an equivalent applied fixed end action vector P . This is par-titioned into free and supported vectors to solve for r

f

and R

s

using the above equations. Basicmember force can still be calculated directly from nodal displacements but must also include thebasic thermally-induced load:

{S} = [k][T ][L

D

]{r}+ P

T

page 8 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

Extended Truss Analysis: Fabrication Errors

Fabrication errors in trusses systems induce a similar effect to themally-induced changes, i.e. achange in length is restrained so additional system forces are induced. The fabrication error �change can be incorporated directly in to the above system, with element over-size positive andunder-size negative.

{P} =

8>><

>>:

P

0x,i

P

0y,i

P

0x,j

P

0y,j

9>>=

>>;=

AE�

L

8>><

>>:

c

s

�c

�s

9>>=

>>;

Example 9.1: Truss Support Settlement and Thermal Change

Shown below is a truss structure with three elements and four nodes. It is subject to a horizontalload, a vertical load, a thermal change, and a support settlement as shown. Use the directstiffness method to calculate free node displacements, reaction forces, and member forces. Allelements have A =80mm2, E =200GPa and ↵ =1.2⇥ 10�5/�C.

3m

4m

6kNE=200GPaA=80mm2

+20 Co

5kN

2.5mm

α=1.2x10-5/ Co

All elements AE

Tabulate a definition of the system. Assign coordinate system, node numbers, element numbers,and element directions.

page 9 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

12

3 42

1

3EID i j E A L ✓ AE/L c s cs

1 1 2 2⇥ 10

5 80 3⇥ 10

3⇡ 5.33⇥ 10

3 -1 0 02 1 3 2⇥ 10

5 80 5⇥ 10

3 �2.21 3.2⇥ 10

3 -0.6 -0.8 0.483 1 4 2⇥ 10

5 80 4⇥ 10

3 �⇡/2 4⇥ 10

3 0 -1 0

Element global stiffness matrices are as follows.

[K

1] = 10

3

2

664

5.33 0 �5.33 0

0 0 0

5.33 0

Sym. 0

3

775

[K

2] = 10

3

2

664

1.15 1.54 �1.15 �1.54

2.05 �1.54 �2.05

1.15 1.54

Sym. 2.05

3

775

[K

3] = 10

3

2

664

0 0 0 0

4 0 �4

0 0

Sym. 4

3

775

Assemble global stiffness matrix.

K

S

= 10

3

u1 v1 u2 v2 u3 v3 u4 v42

6666666664

3

7777777775

6.45 1.54 �5.33 0 �1.15 �1.55 0 0 u1

6.05 0 0 �1.54 �2.05 0 �4 v1

5.33 0 0 0 0 0 u2

0 0 0 0 0 v2

1.15 1.54 0 0 u3

2.05 0 0 v3

0 0 u4

Sym. 4 v4

page 10 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

The basic thermally-induced load in element two is given by:

P

2T

= �EA↵�T = �3840

which is expressed in the global system as:

{P} = EA↵�

T

8>><

>>:

c

s

�c

�s

9>>=

>>;= 10

3

2

64

3

75

�2.30 u1

�3.07 v1

2.30 u3

3.07 v3

Node load and displacement vectors are assembled from system information. Displacementvector includes support settlement.

{r} =

8>>>>>>>>>><

>>>>>>>>>>:

u1

v1

u2

v2

u3

v3

u4

v4

9>>>>>>>>>>=

>>>>>>>>>>;

=

8>>>>>>>>>><

>>>>>>>>>>:

u1

v1

0

0

0

�2.5

0

0

9>>>>>>>>>>=

>>>>>>>>>>;

{R} =

8>>>>>>>>>><

>>>>>>>>>>:

R

x,1

R

y,1

R

x,2

R

y,2

R

x,3

R

y,3

R

x,4

R

y,4

9>>>>>>>>>>=

>>>>>>>>>>;

=

8>>>>>>>>>><

>>>>>>>>>>:

�5000

�6000

R

x,2

R

y,2

R

x,3

R

y,3

R

x,4

R

y,4

9>>>>>>>>>>=

>>>>>>>>>>;

{P} =

8>>>>>>>>>><

>>>>>>>>>>:

P

x,1

P

y,1

P

x,2

P

y,2

P

x,3

P

y,3

P

x,4

P

y,4

9>>>>>>>>>>=

>>>>>>>>>>;

= 10

3

8>>>>>>>>>><

>>>>>>>>>>:

�2.30

�3.07

0

0

2.30

3.07

0

0

9>>>>>>>>>>=

>>>>>>>>>>;

The stiffness equation is partitioned into free nodal displacements (u1, v1) and fixed nodal dis-placements (u2, v2, u3, v3, u4, v4). Solve for free nodal displacements.

{Rf

} = [K

ff

]{rf

}+ [K

fs

]{rs

}+ {Pf

}{r

f

} = [K

ff

]

�1({R

f

}� [K

fs

]{rs

}� {Pf

})

⇢u1

v1

�= 10

3

�6.45 1.54

1.54 6.05

�1

0

BBBBBB@

⇢�5000

�6000

�� 10

3

��5.33 0 �1.15 �1.54 0 0

0 0 �1.54 �2.05 0 �4

8>>>>>><

>>>>>>:

0

0

0

�2.5

0

0

9>>>>>>=

>>>>>>;

�⇢�2300

�3070

�

1

CCCCCCA

=

⇢�.74

�1.14

�

page 11 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

Solve support reactions.

{Rs

} = [K

sf

]{rf

}+ [K

ss

]{rs

}+ {Ps

}8>>>>>><

>>>>>>:

R

x,2

R

y,2

R

x,3

R

y,3

R

x,4

R

y,4

9>>>>>>=

>>>>>>;

= 10

3

2

666664

3

777775

�5.33 0

0 0

�1.15 �1.54

�1.54 �2.05

0 0

0 �4

⇢�.74

�1.14

�+ 10

3

2

666664

3

777775

5.33 0 0 0 0 0

0 0 0 0 0

1.15 1.54 0 0

2.05 0 0

0 0

Sym. 4

8>>>>>><

>>>>>>:

0

0

0

�2.5

0

0

9>>>>>>=

>>>>>>;

+

8>>>>>><

>>>>>>:

0

0

2300

3070

0

0

9>>>>>>=

>>>>>>;

= 10

3

8>>>>>><

>>>>>>:

3.93

0

1.07

1.43

0

4.57

9>>>>>>=

>>>>>>;

Solve basic member forces for each element.

S

e

=

AE

L

< �c,�s, c, s > {re}+ {P e

T

}

S

1= 5.33⇥ 10

3< 1, 0,�1, 0 >

8>><

>>:

�.74

�1.14

0

0

9>>=

>>;+ 0 = �3.93⇥ 10

3

S

2= 3.2⇥ 10

3< 0.6, 0.8,�0.6,�0.8 >

8>><

>>:

�.74

�1.14

0

�2.5

9>>=

>>;� 3840 = �1.78⇥ 10

3

S

3= 4⇥ 10

3< 0, 1, 0,�1 >

8>><

>>:

�.74

�1.14

0

0

9>>=

>>;+ 0 = �4.57⇥ 10

3

page 12 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

1.44mm0.74mm

1.78k

N (C)

1.43kN 4.57kN1.07kN

2.5mm

3.93kN 3.93kN (C)

4.57

kN (C

)

Extended Truss Analysis: Inclined Supports

Certain structural systems have have support restraint that is not aligned with the global co-ordinate system. So called inclined supports can be accounted for in the stiffness equation byadjusting the member stiffness of all elements connected to the inclined support to give alignedstiffness values. Stiffness values can be expressed in a nodal coordinate system x

00 � y

00 wherex

00 is aligned with the free displacement direction in the inclined roller. An inclined roller conditioncan then be the implemented in the stiffness matrix as a zero displacement in the y

00 direction.

To formulate stiffness in this coordinate system, an inclined nodal rotation matrix L

0B

is used withinclination angle ✓

0 between x

00 and the element axis. The element global transformation matrix

then becomes L

D

=

L

B

0

0 L

0B

�if the inclined support occurs at element node j, or

L

0B

0

0 L

B

�for

node i. Element global stiffness is then derived as follows.

y’’

θi

x’’yx

θ j’[K] = L

T

D

k

0L

D

[K] =

AE

L

2

664

c

2cs �cc

0 �cs

0

s

2 �c

0s �ss

0

c

02c

0s

0

Sym. s

02

3

775

where c

0= cos ✓

0 and s

0= sin ✓

0.

Example 9.2: Inclined Support Element Stiffness formulation

Shown below is a truss system with an inclined roller support at node one. Calculate the inclinednodal stiffness K11. Assume AE for all members.

page 13 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L9

Direct Stiffness Method: Truss Elements II

4m 4m

4m

130o

2 3

4

12 3

4

Element one

y’’

θ jx’’

yx

θ i’

L = 4m

✓

0i

= ⇡/6, c0 = 0.866, s0 = 0.5, c0s0 = 0.433

✓

j

= 0, c = 1, s = 0, cs = 0

[K

1] = AE

2

664

0.187 0.108 �0.217 0

0.063 �0.125 0

0.25 0

Sym. 0

3

775

Element two

y’’

θ j

x’’

yx

θ i’

L = 5.66m

✓

0i

= 5⇡/12, c0 = 0.259, s0 = 0.966, c0s0 = 0.25

✓

j

= ⇡/4, c = 0.707, s = 0.707, cs = 0.5

[K

2] = AE

2

664

0.012 0.044 �0.032 �0.032

0.165 �0.121 �0.121

0.088 0.088

Sym. 0.088

3

775

K11 = K

111 +K

211 = AE

0.199 0.152

0.152 0.228

�

page 14 of 14

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

by Joe Gattas, Faris Albermani

Introduction

The direct sti�ness method can be applied to beam elements. Beam elements are taken here to

mean elements with a rotational and vertical translational nodal degrees of freedom, an orientation

restricted to alignment with the global x axis, and no axial e�ects.

Beam Element Force Transformations

Basic

MjMi{S} =

IMi

Mj

J

Local = Global

MjMi

Fy,jFy,i{F} = {F

Õ} =

Y___]

___[

Fy,i

Mi

Fy,j

Mj

Z___̂

___\=

S

WWWU

1/L 1/L

1 0

≠1/L ≠1/L

0 1

T

XXXV {S} = T

TS

where T =

C1/L 1 ≠1/L 0

1/L 0 ≠1/L 1

D

and LD = [I]

Fy,i = Mi/L + Mj/L can be shown from

qM@j = Mi +

Mj ≠ Fy,iL and Fy,i = ≠Fy,j can be shown from

qFy = 0.

page 1 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

Beam Element Displacement Transformations

Basic

φjφi {v} =

I„i

„j

J

Local = Global

θjθivi

vj{r} = {r

Õ} =

Y___]

___[

vi

◊i

vj

◊j

Z___̂

___\

and {v} =

C1/L 1 ≠1/L 0

1/L 0 ≠1/L 1

D

{r

Õ} = Tr

Õ

Beam Element Sti�ness Transformation

Basic

kjjkiikijkjj

[k] =

2EI

L

C2 1

1 2

D

Local=Global

KjjKiiKijKji

[K] = [k

Õ] = T

TkT =

S

WWWWWWU

12EIL3

6EIL2

≠12EIL3

6EIL2

4EIL

≠6EIL2

2EIL

12EIL3

≠6EIL2

Sym. 4EIL

T

XXXXXXV

Member Loads / Fixed End Actions

Member loads, i.e. loads that don’t occur at joint (node) locations, are common in beam systems.

The sti�ness matrix is expressed in terms of nodal forces and displacements, so equivalent fixed end

action vector P must be used to account for non-nodal loads.

If a beam element is assumed restrained against all displacements, i.e. fully fixed, an applied member

load will generate a fixed end action. Fixed end moments from common member loads are listed

in handout ‘L0 - fixed end moments’. Fixed end vertical forces can be calculated from equilibrium

equations. Reversed fixed end actions can be used to replace an applied member force with nodal

forces that generate equivalent displacements. From linear superposition, multiple equivalent actions

generated at any given node may be summed to assemble the system equivalent fixed end action

page 2 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

vector.

Point load P at element mid-span.

P

P/2 P/2

PL/8PL/8

L/2 L/2

{P

P } =

Y__]

__[

Z__̂

__\

≠P/2

≠PL/8

≠P/2

PL/8

Uniformly distributed load w.

w

wL/2 wL/2

wL/12wL/12

L/2 L/2

2 2

{P

w} =

Y__]

__[

Z__̂

__\

≠wL/2

≠wL

2/12

≠wL/2

wL

2/12

Member Forces

Element member forces can be formulated in the basic or local (= global) system. Formulation in

the local system will include member bending and shear forces and is given by:

{F

Õe} =

Y___]

___[

Fy,i

Mi

Fy,j

Mj

Z___̂

___\= [K

e]{r

e} + {P

e}

The direct sti�ness method gives element forces in terms of the joint convention for positive force

direction. Beam system bending moment and shear force diagrams are typically expressed in terms

of the beam convention. A comparison between the two is shown below. It can be seen that Mi and

Fy,j should have their signs reversed when converting from a joint to a beam convention.

Joint Convention Positive Beam Convention Positive

MjMi

Fy,jFy,iMjMi

Fy,jFy,i

page 3 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

Example 10.1: Beam Element Sti�ness Method

Solve the below structure consisting of beam elements. Assume L=4m, P=3kN, E=210GPa, and

I=2.5◊10

8mm

4.

4@L/2

All EI 2P P PPL

Tabulate a definition of the system.

211 2 3

EID i j E I L

1 1 2 2.1◊10

52.5◊10

84000

2 2 3 2.1◊10

52.5◊10

84000

Element global sti�ness matrices are as follows.

[K

1] = 10

3

v1 ◊1 v2 ◊2S

WWU

T

XXV

9.8 19.7e3 ≠9.8 19.7e3

52.5e6 ≠19.7e3 26.3e6

9.8 ≠19.7e3

Sym. 52.5e6

[K

2] = 10

3

v2 ◊2 v3 ◊3S

WWU

T

XXV

9.8 19.7e3 ≠9.8 19.7e3

52.5e6 ≠19.7e3 26.3e6

9.8 ≠19.7e3

Sym. 52.5e6

Assemble the global sti�ness matrix.

[KS] = 10

3

S

U

T

VK

111 K

112 0

K

1+222 K

223

Sym. K

322

=

v1 ◊1 v2 ◊2 v3 ◊3S

WWWWWWWU

T

XXXXXXXV

9.8 19.73 ≠9.8 19.7e3 0 0 v152.5e6 ≠19.7e3 26.3e6 0 0 ◊1

19.7 0 ≠9.8 19.7e3 v2105e6 ≠19.7e3 26.3e6 ◊2

9.8 ≠19.7e3 v3Sym. 52.5e6 ◊3

page 4 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

Convert member (non-node) loads to equivalent applied node loads.

Element 1: Point load 2P at element mid-span.

[P

1] =

Y__]

__[

Z__̂

__\

P v1PL/4 ◊1

P v2≠PL/4 ◊2

= 10

3

Y___]

___[

3

3000

3

≠3000

Z___̂

___\

Element 2: Point load P at element mid-span.

[P

1] =

Y__]

__[

Z__̂

__\

P/2 v2PL/8 ◊2P/2 v3

≠PL/8 ◊3

= 10

3

Y___]

___[

1.5

1500

1.5

≠1500

Z___̂

___\

Assemble applied node load, equivalent node load, and node displacement vectors.

{r} =

Y________]

________[

v1◊1v2◊2v3◊3

Z________̂

________\

=

Y________]

________[

0

0

0

◊20

◊3

Z________̂

________\

{R} =

Y________]

________[

Fy,1M1Fy,2M2Fy,3M3

Z________̂

________\

=

Y________]

________[

Fy,1M1Fy,2

12 ◊ 10

6

Fy,3 + 3000

0

Z________̂

________\

{P} =

Y________]

________[

Py,1PM,1Py,2PM,2Py,3PM,3

Z________̂

________\

= 10

3

Y________]

________[

3

3000

4.5

≠1500

1.5

≠1500

Z________̂

________\

Solve free node displacements ◊2 and ◊3.

{Rf} = [Kff ]{rf} +⇠⇠⇠⇠⇠[Kfs]{rs} + {Pf}

{rf} = [Kff ]

≠1({Rf} ≠ {Pf})

I◊2◊3

J

=

C1.05e11 0.26e11

0.26e11 0.53e11

D≠1 AI12 ◊ 10

6

0

J

≠I

≠1.5 ◊ 10

6

≠1.5 ◊ 10

6

JB

= 10

≠3I

0.139

≠0.041

J

Solve support reactions.

{Rs} = [Ksf ]{rf} +⇠⇠⇠⇠⇠[Kss]{rs} + {Ps}

Y___]

___[

Fy,1M1Fy,2

Fy,3 + 3000

Z___̂

___\= 10

6

S

WWWU

19.7 0

26.23e3 0

0 19.7

≠19.7 ≠19.7

T

XXXV 10

≠3I

0.139

≠0.041

J

+ 10

3

Y___]

___[

3

3000

4.5

1.5

Z___̂

___\

= 10

3

Y___]

___[

5.73

6640

3.70

≠0.42

Z___̂

___\, where Fy,3 + 3000 = ≠0.42e3, so Fy,3 = ≠3420

page 5 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

13.9e-5 -4.1e-5

3.70kN 3.42kN

6.64kNm

5.73kN

Solve member forces in the local system.

{F

Õe} =

Y___]

___[

Fy,i

Mi

Fy,j

Mj

Z___̂

___\= [K

e]{r

e} + {P

e}

{F

Õ1} =

Y___]

___[

Fy,1M1Fy,2M2

Z___̂

___\= 10

3

S

WWU

T

XXV

9.8 19.7e3 ≠9.8 19.7e3

52.5e6 ≠19.7e3 26.3e6

9.8 ≠19.7e3

Sym. 52.5e6

Y___]

___[

0

0

0

13.9 ◊ 10

≠5

Z___̂

___\+ 10

3

Y___]

___[

3

3000

3

≠3000

Z___̂

___\

= 10

3

Y___]

___[

5.73

6.64 ◊ 10

3

0.27

4.29 ◊ 10

3

Z___̂

___\

{F

Õ2} =

Y___]

___[

Fy,2M2Fy,3M3

Z___̂

___\= 10

3

S

WWU

T

XXV

9.8 19.7e3 ≠9.8 19.7e3

52.5e6 ≠19.7e3 26.3e6

9.8 ≠19.7e3

Sym. 52.5e6

Y___]

___[

0

13.9 ◊ 10

≠5

0

≠4.1 ◊ 10

≠5

Z___̂

___\+ 10

3

Y___]

___[

1.5

1500

1.5

≠1500

Z___̂

___\

= 10

3

Y___]

___[

3.43

7.71 ◊ 10

3

≠0.43

0

Z___̂

___\

page 6 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

-6.64kNm

4.82kNm 4.29kNm

-7.71kNm

0kNm

Shear

Bending

-0.86kNm

5.73kN3.43kN

-0.27kN0.43kN

Beam Support Settlement

Beam support settlement can be incorporated into the sti�ness solution in the same manner as

described for truss elements, i.e. with rs ”= 0. Alternatively, support settlement can be incorporated

as equivalent fixed end actions.

Vertical displacement support settlement � at element node j.

6EIΔ/L

L/2 L/2

2

Δ

12EIΔ/L3

6EIΔ/L2

12EIΔ/L3{P

�} =

Y__]

__[

Z__̂

__\

≠12EI�/L

3

≠6EI�/L

2

12EI�/L

3

≠6EI�/L

2

Example 10.2: Beam Element Support Settlement

Assemble the system equivalent fixed end action vector for the below beam structure. Incorporate

both support settlements and applied member loads into the fixed end action vector. Assume EI

for all elements.

page 7 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

L L L

w

ΔA ΔB

A B C

From left to right, nodes are numbered 1,2,3,4 and elements are numbered 1,2,3.

Element 1 has i

thnode 1 and j

thnode 2. There is a support settlement �A at the j

thnode. The

equivalent fixed end action for this is as follows.

{P

1} =

Y___]

___[

Fy,1M1Fy,2M2

Z___̂

___\= {P

�,j} = EI

Y___]

___[

≠12�A/L

3A

≠6�A/L

2A

12�A/L

3A

≠6�A/L

2A

Z___̂

___\

Element 3 has i

thnode 3 and j

thnode 4. There is a support settlement �B at the i

thnode. The

equivalent fixed end action for this is as follows. Note that the signs are reversed as in a fixed

end beam, a downwards i

thnode displacement generates the same e�ect as an upwards j

thnode

displacement.

{P

3} =

Y___]

___[

Fy,3M3Fy,4M4

Z___̂

___\= {P

�,i} = EI

Y___]

___[

12�B/L

3C

6�B/L

2C

≠12�B/L

3C

6�B/L

2C

Z___̂

___\

Element 2 has i

thnode 2 and j

thnode 3. There is a support settlement �A at the i

thnode, a

support settlement �B at the j

thnode, and an applied distributed member load w. The equivalent

fixed end action for this is as follows.

page 8 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L10

Direct Sti�ness Method: Beam Elements

{P

2} =

Y___]

___[

Fy,2M2Fy,3M3

Z___̂

___\= {P

�,i} + {P

�,j} + {P

w}

= EI

Y___]

___[

12�A/L

3B

6�A/L

2B

≠12�A/L

3B

6�A/L

2B

Z___̂

___\+ EI

Y___]

___[

≠12�B/L

3B

≠6�B/L

2B

12�B/L

3B

≠6�B/L

2B

Z___̂

___\+ EI

Y___]

___[

≠wLB/(2EI)

≠wL

2B/(12EI)

≠wLB/(2EI)

wL

2B/(12EI)

Z___̂

___\

The system equivalent fixed end action vector is the sum of member fixed end action vectors.

{P} =

Y_____________]

_____________[

Fy,1M1Fy,2M2Fy,3M3Fy,4M4

Z_____________̂

_____________\

= EI

Q

ccccccccccccca

Y_____________]

_____________[

≠12�A/L

3A

≠6�A/L

2A

12�A/L

3A

≠6�A/L

2A

0

0

0

0

Z_____________̂

_____________\

+

Y_____________]

_____________[

0

0

12�A/L

3B ≠ 12�B/L

3B ≠ wLB/(2EI)

6�A/L

2B ≠ 6�B/L

2B ≠ wL

2B/(12EI)

≠12�A/L

3B + 12�B/L

3B ≠ wLB/(2EI)

6�A/L

2B ≠ 6�B/L

2B + wL

2B/(12EI)

0

0

Z_____________̂

_____________\

+

Y_____________]

_____________[

0

0

0

0

12�B/L

3C

6�B/L

2C

≠12�B/L

3C

6�B/L

2C

Z_____________̂

_____________\

R

dddddddddddddb

page 9 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

CIVL3340 L11

Direct Stiffness Method: Frame Elements

BY JOE GATTAS, FARIS ALBERMANI

Introduction

The direct stiffness method can be extended to frame elements by combining the axial loadderivations of truss elements with the bending load derivations of beam elements. Frame ele-ments therefore have horizontal translational, vertical translational, and rotational nodal degreesof freedom.

Frame Element Force Transformations

Basic

MjMi P {S} =

8<

:

Mi

Mj

P

9=

;

Local

Mj

Mi

Fx,j’Fy,j’

Fx,i’Fy,i’

{F 0} =

8>>>>>><

>>>>>>:

F 0x,i

F 0y,i

Mi

F 0x,j

F 0y,j

Mj

9>>>>>>=

>>>>>>;

=

2

6666664

0 0 �1

1/L 1/L 0

1 0 0

0 0 1

�1/L �1/L 0

0 1 0

3

7777775{S} = T TS

where [T ] =

2

40 1/L 1 0 �1/L 0

0 1/L 0 0 �1/L 1

�1 0 0 1 0 0

3

5

Global

page 1 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

Mj

Mi

Fx,jFy,j

Fx,i

Fy,i{F} =

8>>>>>><

>>>>>>:

Fx,i

Fy,i

Mi

Fx,j

Fy,j

Mj

9>>>>>>=

>>>>>>;

=

2

6666664

c �s 0 0 0 0

s c 0 0 0 0

0 0 1 0 0 0

0 0 0 c �s 0

0 0 0 s c 0

0 0 0 0 0 1

3

7777775{F 0} = LT

D

F 0

where LD

=

LB

0

0 LB

�and [L

B

] =

2

4c s 0

�s c 0

0 0 1

3

5

Frame Element Displacement Transformations

Basic

Δ

φjφi {v} =

8<

:

�i

�j

�

9=

;

Local

θj

θi

ui’vi’

uj’

vj’{r0} =

8>>>>>><

>>>>>>:

u0i

v0i

✓i

u0j

v0j

✓j

9>>>>>>=

>>>>>>;

and {v} =

2

40 1/L 1 0 �1/L 0

0 1/L 0 0 �1/L 1

�1 0 0 1 0 0

3

5 {r0} = Tr0

Global

θj

θiuj

vj

ui

vi

{r} =

8>>>>>><

>>>>>>:

ui

vi

✓i

uj

vj

✓j

9>>>>>>=

>>>>>>;

and {r} =

2

6666664

c �s 0 0 0 0

s c 0 0 0 0

0 0 1 0 0 0

0 0 0 c �s 0

0 0 0 s c 0

0 0 0 0 0 1

3

7777775{r0} = LT

D

r0

page 2 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

Frame Element Stiffness Transformation

Basic

kjjkii kijkjj

[k] =

2

6664

4EI

L

2EI

L

0

2EI

L

4EI

L

0

0 0

AE

L

3

7775

Local

kijkji’’

kii’kjj’

[k0] = T TkT =

2

6666666666664

AE

L

0 0

�AE

L

0 0

12EI

L

36EI

L

2 0

�12EI

L

36EI

L

2

4EI

L

0

�6EI

L

22EI

L

AE

L

0 0

12EI

L

3�6EI

L

2

Sym. 4EI

L

3

7777777777775

Global

KijjiKKii

Kjj

[K] = LT

D

k0LD

=

2

6666666666664

AE

L

c2 + 12EI

L

3 s2 AE

L

cs� 12EI

L

3 cs �6EI

L

2 s �AE

L

c2 � 12EI

L

3 s2 �AE

L

cs+ 12EI

L

3 cs �6EI

L

2 s

AE

L

s2 + 12EI

L

3 c2 6EI

L

2 c�AE

L

cs+ 12EI

L

3 cs �AE

L

s2 � 12EI

L

3 c2 6EI

L

2 c

4EI

L

6EI

L

2 s�6EI

L

2 c 2EI

L

AE

L

c2 + 12EI

L

3 s2 AE

L

cs� 12EI

L

3 cs 6EI

L

2 s

AE

L

s2 + 12EI

L

3 c2 �6EI

L

2 c

Sym. 4EI

L

3

7777777777775

page 3 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

Member Loads and Fixed End Actions

Member loads in the local system can be formulated as described previously for beam elements.They must then be transformed into the global system prior to summation into the global fixedend action vector P .

{P e} = LT

D

{P 0}

Member Forces

Element member forces can be formulated in the local system to include bending, shear, andaxial forces. Caution should be taken for sign conventions with shear and bending loads, asdescribed previously for beam member forces.

Example 11.1: Frame Element Direct Stiffness Example

A frame structure is shown below. For all members E=104 MPa, I=103 mm4, and A=10mm2.Assume L=100mm, P=10N, and w=0.24N/mm. Using the DIRECT STIFFNESS METHOD, cal-culate member forces, support reactions, and draw the bending moment diagram showing allimportant values.

L

All A,E,I

P 2P

PL

0.75L

L/2 L/2

w

Define element and node numbers. Tabulate the basic member properties. Units N-mm-MPa.

page 4 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

12 1

3

2

EID i j E I A L ✓ c s cs c2 s2

1 2 1 104 103 102 100 0 1 0 0 1 02 1 3 104 103 102 125 -0.644 0.8 -0.6 -0.48 0.64 0.36

Tabulate element stiffness components.

EID EA/L 12EI/L36EI/L2

4EI/L 2EI/L

1 1000 120 6x103 4x105 2x105

2 800 61.44 3.84x103 3.2x105 0.6x105

Assemble element global stiffness matrices. Frame elements have two nodes and three nodaldisplacements per node, so total DOF is 6.

For element 1, stiffness components are between nodes 2 and 1:

[K1] =

u2 v2 ✓2 u1 v1 ✓12

666664

3

777775

1000 0 0 �1000 0 0 u2

120 6000 0 �120 6000 v24⇥ 10

50 �6000 2⇥ 10

5 ✓21000 0 0 u1

120 �6000 v1Sym. 4⇥ 10

5 ✓1

page 5 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

For element 2, stiffness components are between nodes 1 and 3:

[K2] =

u1 v1 ✓1 u3 v3 ✓32

666664

3

777775

534 �354.5 2.3⇥ 10

3 �534.1 354.5 2.3⇥ 10

3 u1

327.3 3.1⇥ 10

3354.5 �327.3 3.1⇥ 10

3 v1320⇥ 10

3 �2.3⇥ 10

3 �3.1⇥ 10

3160⇥ 10

3 ✓1534.1 �354.5 �2.3⇥ 10

3 u3

327.3 �3.1⇥ 10

3 v3Sym. �3.1⇥ 10

3320⇥ 10

3 ✓3

Assemble the global stiffness matrix. The global stiffness matrix has 3 ⇥ 3 = 9 DOF so has anorder of 9.

[KS

] = 10

3

" #K1+211 K1

12 K213

K122 0

Sym. K333

=

u1 v1 ✓1 u2 v2 ✓2 u3 v3 ✓32

666666666664

3

777777777775

1534.1 �354.5 2304 �1000 0 0 �534.1 354.5 2304 u1

447.3 �2928 0 �120 �6000 354.5 �327.3 3072 v17.2⇥ 10

50 6000 2⇥ 10

5 �2304 �3072 1.6⇥ 10

5 ✓11000 0 0 0 0 0 u2

120 6000 0 0 0 v24⇥ 10

50 0 0 ✓2

534.1 �354.5 �2304 u3

327.3 �3072 v3Sym. 3.2⇥ 10

5 ✓3

Convert member (non-node) loads to equivalent applied node loads.

Element 1: Uniformly distributed load w downwards (negative).

[P 1] = [P 01

] =

8>>>>><

>>>>>:

9>>>>>=

>>>>>;

0 u2

wL/2 v2wL2/12 ✓2

0 u1

wL/2 v1�wL2/12 ✓1

=

8>>>>>><

>>>>>>:

0

12

200

0

12

�200

9>>>>>>=

>>>>>>;

page 6 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

Element 2: Downwards vertical (global system) point load P at element mid-span. Can bereformulated into local system as P 0

y

= P cos ✓ = �16 and P 0x

= P sin ✓ = 12.

{P 02} =

8>>>>><

>>>>>:

9>>>>>=

>>>>>;

�P 0x

/2 u1

�P 0y

/2 v1�P 0

y

L/8 ✓1�P 0

x

/2 u3

�P 0y

/2 v3P 0y

L/8 ✓3

=

8>>>>>><

>>>>>>:

�6

8

250

�6

8

�250

9>>>>>>=

>>>>>>;

Global system equivalent applied loads calculated from {P e} = LT

D

{P 0}.

[L2D

] =

2

6666664

0.8 �0.6 0 0 0 0

0.6 0.8 0 0 0 0

0 0 1 0 0 0

0 0 0 0.8 �0.6 0

0 0 0 0.6 0.8 0

0 0 0 0 0 1

3

7777775

{P 2} = LT

D

{P 02} =

8>>>>>><

>>>>>>:

0

10

250

0

10

�250

9>>>>>>=

>>>>>>;

Assemble applied node load, equivalent node load, and node displacement vectors.

{r} =

8>>>>>>>>>>>><

>>>>>>>>>>>>:

u1

v1✓1u2

v2✓2u3

v3✓3

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

=

8>>>>>>>>>>>><

>>>>>>>>>>>>:

u1

v1✓10

0

0

0

0

0

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

{R} =

8>>>>>>>>>>>><

>>>>>>>>>>>>:

Fx,1

Fy,1

M1

Fx,2

Fy,2

M2

Fx,3

Fy,3

M3

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

=

8>>>>>>>>>>>><

>>>>>>>>>>>>:

0

�10

�1000

Fx,2

Fy,2

M2

Fx,3

Fy,3

M3

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

{P} =

8>>>>>>>>>>>><

>>>>>>>>>>>>:

Px1

Py,1

PM,1

Px,2

Py,2

PM,2

Px,3

Py,3

PM,3

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

=

8>>>>>>>>>>>><

>>>>>>>>>>>>:

0

22

50

0

12

200

0

10

�250

9>>>>>>>>>>>>=

>>>>>>>>>>>>;

Solve free node displacements u1, v1, and ✓1.

page 7 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

{Rf

} = [Kff

]{rf

}+⇠⇠⇠⇠⇠[K

fs

]{rs

}+ {Pf

}{r

f

} = [Kff

]

�1({R

f

}� {Pf

})8<

:

u1

v1✓1

9=

; =

2

41534.1 �354.5 2304

447.3 �2928

7.2⇥ 10

5

3

5�1 0

@

8<

:

0

�10

�1000

9=

;�

8<

:

0

22

50

9=

;

1

A=

8<

:

�0.0203�0.0994�0.0018

9=

;

Solve support reactions.

{Rs

} = [Ksf

]{rf

}+⇠⇠⇠⇠⇠[K

ss

]{rs

}+ {Ps

}8>>>>>><

>>>>>>:

Fx,2

Fy,2

M2

Fx,3

Fy,3

M3

9>>>>>>=

>>>>>>;

=

2

6666664

�1000 0 0

0 �120 6000

0 �6000 2⇥ 10

5

�534.1 354.5 �2304

354.5 �327.3 �3072

2304 3072 1.6⇥ 10

5

3

7777775

8<

:

�0.0203�0.0994�0.0018

9=

;+

8>>>>>><

>>>>>>:

0

12

200

0

10

�250

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

20.2613.14436.5�20.2640.86�889.5

9>>>>>>=

>>>>>>;

0.099mm0.020mm

0.0018

436Nmm

13.1N20.3N

890Nmm

40.9N20.3N

page 8 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

Solve member forces in the local system.

{F 0e} =

8>>>>>><

>>>>>>:

F 0x,i

F 0y,i

M 0i

F 0x,j

F 0y,j

M 0j

9>>>>>>=

>>>>>>;

= Le

D

F e

= Le

D

([Ke

]{re}+ {P e})

for Element 1, LD

= [I] so {F 01} = {F 1}.

{F 01} =

8>>>>>><

>>>>>>:

F 0x,2

F 0y,2

M 02

F 0x,1

F 0y,1

M 01

9>>>>>>=

>>>>>>;

=

2

6666664

1000 0 0 �1000 0 0

120 6000 0 �120 6000

4⇥ 10

50 �6000 2⇥ 10

5

1000 0 0

120 �6000

Sym. 4⇥ 10

5

3

7777775

8>>>>>><

>>>>>>:

0

0

0

�0.0203�0.0994�0.0018

9>>>>>>=

>>>>>>;

+

8>>>>>><

>>>>>>:

0

12

200

0

12

�200

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

20.313.1436.4�20.310.9

�323.6

9>>>>>>=

>>>>>>;

For Element 2, LD

6= [I] so {F 02} = LD

{F 2}.

{F 2} =

8>>>>>><

>>>>>>:

Fx,1

Fy,1

M1

Fx,3

Fy,3

M3

9>>>>>>=

>>>>>>;

=

2

6666664

534 �354.5 2.3e3 �534.1 354.5 2.3e3327.3 3.1e3 354.5 �327.3 3.1e3

320e3 �2.3e3 �3.1e3 160e3534.1 �354.5 �2.3e3

327.3 �3.1e3Sym. �3.1e3 320e3

3

7777775

8>>>>>><

>>>>>>:

�0.0203�0.0994�0.0018

0

0

0

9>>>>>>=

>>>>>>;

+

8>>>>>><

>>>>>>:

0

10

250

0

10

�250

9>>>>>>=

>>>>>>;

{F 02} = LD

{F 2} =

8>>>>>><

>>>>>>:

28.7�4.5�678.1�40.720.5

�890.1

9>>>>>>=

>>>>>>;

page 9 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L11

Direct Stiffness Method: Frame Elements

-436Nmm

Shear

Bending -323Nmm

79.5Nmm

-890Nmm

396Nmm

678Nmm

Axial

13.1N

-10.9N-4.5N

-20.5N

20.3N (C) 28.7N (C)40.7N (C)

page 10 of 10

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

CIVL3340 L12 Structural Symmetry

BY JOE GATTAS, FARIS ALBERMANI

A 2D structure is symmetric with respect to an axis if the reflection of the structure about theaxis is identical in terms of geometry, section properties, material properties, supports, and thestructure itself.

Symmetry and Anti-Symmetry

Symmetric structures can be subjected to symmetric and anti-symmetric loading. Symmetricloads are equal when reflected about the symmetric structure centreline. Anti-symmetric loadsare the opposite when reflected about the symmetric structure centreline, i.e. an anti-symmetricload is a symmetric load times negative one.

Symmetric Loading

P P

LC

M

LC

LC

M

w w

Anti-Symmetric Loading

P

PLC

M

LC

LC

Mw

w

page 1 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

Any general load can be decomposed into symmetric and anti-symmetric components.

P

PLC

P

LC

P2P

LC

+==

Symmetric Structures Under Symmetric Loading

A symmetric structure under symmetric loading has a response that is symmetric. This meansthe structure will neither rotate nor deflect perpendicular to the axis of symmetry. Analysis cantherefore be conducted on one half of the structure, with a fixed roller boundary condition im-posed at the axis of symmetry.

LC ΔmΔm Δm==

This system reduction is of significant practical benefit for large structures, as computation timecan be significantly reduced. It also allows boundary conditions to be introduced into systems,which reduced the kinematic redundancy and thus solver time.

==w ==LCLC

Elements or applied loads that are coincident with axis of symmetry must be halved in the equiv-alent half-system.

page 2 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

==

LC

L L

E,A,I

E,A,I

E,A,I

E,A,I E,A,I

L

E,A,I

E,A/2,I/2E,A,I

PPP

w w w

Symmetric Structures Under Anti-Symmetric Loading

A symmetric structure under anti-symmetric loading has a response that is anti-symmetric. Thismeans the structure will not deflect along the axis of symmetry. Analysis can therefore be con-ducted on one half of the structure, with a pinned roller boundary condition imposed at the axisof symmetry.

LC −ΔmΔm == Δm

==

LC

L L

E,A,I

E,A,I

E,A,I

E,A,IE,A,I

L

E,A,I

E,A/2,I/2E,A,I

PPP

P

P

P

page 3 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

Example 12.1: Symmetric Truss Structure

A truss structure composed of six nodes and eleven elements is shown below. Assume P=50kN,E=150 GPa, and A=0.004m2. Using the DIRECT STIFFNESS METHOD, calculate memberforces, support reactions, and nodal displacements.

3m

4m

P

A,E

A,EA,E

A,E

A,E

A,E

A,E

A,E

A,E

A,E

A,E

3m

The structure is symmetric and so the following half-system will possess an equivalent structuralresponse. Note that pinned rollers and not fixed rollers are used as the symmetric boundaryconditions, as truss elements transmit zero bending moments.

3m

4m

P/2

A,E

0.5A,E

A,E

A,E

A,E

A,E

page 4 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

This is the same structure that was solved in Tutorial 7. The half-system solution is shown belowon the left. The full structure has a symmetric response, shown on the right.

Deflections

0.22mm

0.11mm

==

Reaction and member forces

10.3k

N (C)

20.9kN (T)

8.3k

N (T

)

8.3kN

6.2kN

16.7kN12.5kN

6.2kN

12.5kN 8.3kN

6.2kN

8.3kN

6.2kN

12.5 kN 16.7kN 16.7kN

10.3k

N (C)

20.9kN (T)

16.6

kN (T

)

20.9k

N (T)

10.3kN (C)

12.5 kN

==

page 5 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

Example 12.2: Double-Symmetric Frame Structure

Shown below is a four-element frame structure. Use structural symmetry and the direct stiffnessmethod to simplify and solve the system. Assume w=5kN/m, all the members have E=210GPa,I=4000cm4, and A=0.5cm2.

4m 4m

3m

3m

w

ww

w

Through double-symmetry, the problem is equivalent to a quarter-system with one element andtwo nodes.

4m

3mw w

1

2

1

Element one has ith node 1 and jth node 2. Element information is as follows.

L ✓ c s cs c2 s2 EA/L 12EI/L3 6EI/L2 4EI/L 2EI/L

5000 0.644 0.8 0.6 0.48 0.64 0.36 2100 806.4 2.02e6 6.72e9 3.36e9

page 6 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

For a single element system, the global stiffness matrix is equal to the element global stiffnessmatrix.

[KS

] = [K1] =

u1 v1 ✓1 u2 v2 ✓22

666664

3

777775

1.63e3 6210 �1.21e6 �1.63e3 �621 �1.21e6 u1

1.27e3 1.61e6 �621 �1.27e3 1.61e6 v16.72e9 1.21e6 �1.61e6 3.36e9 ✓1

1.63e3 621 1.21e6 u2

1.27e3 �1.61e6 v2Sym. 6.72e9 ✓2

The uniformly distributed load w downwards (negative) can be converted to an equivalent appliednodal load in the local system.

[P 01] =

8>>>>><

>>>>>:

9>>>>>=

>>>>>;

0 u01

wL/2 v01wL2/12 ✓1

0 u02

wL/2 v02�wL2/12 ✓2

=

8>>>>>><

>>>>>>:

012.5e310.4e6

012.5e3�10.4e6

9>>>>>>=

>>>>>>;

Global system equivalent applied loads calculated from {P e} = LT

D

{P 0}.

[LD

] =

2

6666664

0.8 0.6 0 0 0 0�0.6 0.8 0 0 0 00 0 1 0 0 00 0 0 0.8 0.6 00 0 0 �0.6 0.8 00 0 0 0 0 1

3

7777775

{P 1} = LT

D

{P 01} =

8>>>>>><

>>>>>>:

�7.5e310e310.4e6�7.5e310e3

�10.4e6

9>>>>>>=

>>>>>>;

Assemble applied node load, equivalent node load, and node displacement vectors.

page 7 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

{r} =

8>>>>>><

>>>>>>:

u1

v1✓1u2

v2✓2

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

u1

000v20

9>>>>>>=

>>>>>>;

{R} =

8>>>>>><

>>>>>>:

Fx,1

Fy,1

M1

Fx,2

Fy,2

M2

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

0Fy,1

M1

Fx,2

0M2

9>>>>>>=

>>>>>>;

{P} =

8>>>>>><

>>>>>>:

Px1

Py,1

PM,1

Px,2

Py,2

PM,2

9>>>>>>=

>>>>>>;

=

8>>>>>><

>>>>>>:

�7.5e310e310.4e6�7.5e310e3

�10.4e6

9>>>>>>=

>>>>>>;

Solve free node displacements u1 and v2.

{Rf

} = [Kff

]{rf

}+⇠⇠⇠⇠⇠[K

fs

]{rs

}+ {Pf

}{r

f

} = [Kff

]�1({Rf

}� {Pf

})⇢u1

v2

�=

1.63e3 �621�621 1.27e3

��1 ✓⇢00

��

⇢�7.5e310e3

�◆=

⇢1.97�6.90

�

Solve support reactions.

{Rs

} = [Ksf

]{rf

}+⇠⇠⇠⇠⇠[K

ss

]{rs

}+ {Ps

}8>><

>>:

Fy,1

M1

Fx,2

M2

9>>=

>>;=

2

664

621 �1.27e3�1.21e6 �1.61e6�1.63e3 621�1.21e6 �1.61e6

3

775

⇢1.97�6.90

�+

8>><

>>:

10.0e310.4e6�7.5e3�10.4e6

9>>=

>>;=

8>><

>>:

20.0e319.2e6�15.0e3�1.67e6

9>>=

>>;

Solve member forces in the local system.

{F 0e} =

8>>>>>><

>>>>>>:

F 0x,i

F 0y,i

M 0i

F 0x,j

F 0y,j

M 0j

9>>>>>>=

>>>>>>;

= LD

{F 1} = LD

([K1]{r1}+ {P 1}) =

8>>>>>><

>>>>>>:

12e316e319.2e6�12e39e3

�1.67e6

9>>>>>>=

>>>>>>;

page 8 of 9

SCHOOL OF CIVIL ENGINEERING

CIVL3340 L12

Structural Symmetry

6.90mm

1.97mm

1.67kNm15kN

20kN19.2kNm

Displacements Reactions & Axial

Bending Shear

12kN (C)

-1.67kNm

-19.2kNm

-9kN

16kNFull System ResponseDisplacements Bending

page 9 of 9