Module3 direct stiffness- rajesh sir

Structural Analysis - III

Di t Stiff M th dDirect Stiffness Method

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKNDept. of CE, GCE Kannur Dr.RajeshKN

1

Module IIIModule III

Direct stiffness method

• Introduction – element stiffness matrix – rotation transformation


matrix – transformation of displacement and load vectors and stiffness matrix – equivalent nodal forces and load vectors –assembly of stiffness matrix and load vector – determination of assembly of stiffness matrix and load vector determination of nodal displacement and element forces – analysis of plane truss beam and plane frame (with numerical examples) – analysis of grid space frame (without numerical examples) grid – space frame (without numerical examples)

Dept. of CE, GCE Kannur Dr.RajeshKN

2

Introduction

• The formalised stiffness method involves evaluating the displacement transformation matrix CMJ correctlyp y

• Generation of matrix CMJ is not suitable for computer programming

H th l ti f di t tiff th d • Hence the evolution of direct stiffness method

Dept. of CE, GCE Kannur Dr.RajeshKN3


• We need to simplify the assembling process of SJ , the Jassembled structure stiffness matrix

• The key to this is to use member stiffness matrices for actions The key to this is to use member stiffness matrices for actions and displacements at BOTH ends of each member

If b di l d i h f • If member displacements are expressed with reference to global co-ordinates, the process of assembling SJ can be made simplesimple


4

Member oriented axes (local coordinates) d t t i t d ( l b l di t )and structure oriented axes (global coordinates)

Lδx

y

L

Local axes

LδL

Lδ sinLδ θLδ θ

xYYYY

cosLδ θ

θ

Global axes yGlobal axesGlobal axesθ

XXXX


Global axes

1. Plane truss member

Stiffness coefficients in local coordinates

13

2 4

y

0 0

Degrees of freedom

1

⎛ ⎞⎜ ⎟

Unit displacement

xEAL

EAL

⎜ ⎟⎝ ⎠corr. to DOF 1

0 0EA EA⎡ ⎤−⎢ ⎥

[ ]

0 0

0 0 0 0

0 0M

L L

SEA EA

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

=Member stiffness matrix in local coordinates


0 0 0 0L L⎢ ⎥

⎢⎢⎣ ⎦

⎥⎥

Transformation of displacement vector

θ

Displacements in global

22 2 cosU D θ=

p gcoordinates: U1 and U2

Displacements in local di D d D

2D

DY iU D θ

12 2 sinU D θ= − θ

1 11 12 1 2cos sinU U U D Dθ θ= + = −

coordinates: D1 and D2

1DY

11 1 cosU D θ=

21 1 sinU D θ=2 21 22 1 2sin cosU U U D Dθ θ= + = +

1 1cos sinU Dθ θ−⎧ ⎫ ⎧ ⎫⎡ ⎤

⎧ ⎫ ⎧ ⎫⎡ ⎤

1 1

2 2

cos sinsin cos

U DU D

θ θθ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ ⎩ ⎭

X 1 1

2 2

cos sinsin cos

D UD U

θ θθ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤∴ =⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦⎩ ⎭ ⎩ ⎭


{ } [ ]{ }D R U=

C id i b th d

⎧ ⎫ ⎧ ⎫

Considering both ends,

1 1

2 2

cos sin 0 0sin cos 0 00 0 i

D UD UD U

θ θθ θ

θ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥3 3

4 4

0 0 cos sin0 0 sin cos

D UD U

θ θθ θ

⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

{ } [ ]{ }LOCAL T GLOBALD R D=

[ ] [ ]R O⎡ ⎤[ ] [ ] [ ][ ] [ ]T

R OR

O R⎡ ⎤

= ⎢ ⎥⎣ ⎦

Rotation matrix


Transformation of load vector Actions in global

θcoordinates: F1 and F2

22 2 cosF A θ= 1 11 12 1 2cos sinF F F A Aθ θ= + = −22 2

12 2 sinF A θ= − θ 2A 2 21 22 1 2sin cosF F F A Aθ θ= + = +

Y

11 1 cosF A θ=

21 1 sinF A θ=1A 1 1

2 2

cos sinsin cos

F AF A

θ θθ θ

−⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ ⎩ ⎭

11 1

1 1cos sinA Fθ θ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥

X2 2sin cosA Fθ θ⎨ ⎬ ⎨ ⎬⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

{ } [ ]{ }A R F=


{ } [ ]{ }A R F=

C id i b th dConsidering both ends,

1 1

2 2

cos sin 0 0sin cos 0 0

A FA F

θ θθ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬

3 3

4 4

0 0 cos sin0 0 sin cos

A FA F

θ θθ θ

⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

{ } [ ]{ }LOCAL T GLOBALi.e., A R A=


10

Transformation of stiffness matrix

{ } [ ]{ }LOCAL M LOCALA S D=

[ ]{ } [ ][ ]{ }T GLOBAL M T GLOBALR A S R D=

{ } [ ] [ ][ ]{ }1GLOBAL T M T GLOBALA R S R D−=

[ ]{ } [ ][ ]{ }T GLOBAL M T GLOBAL

[ ] [ ]1 TR R−{ } [ ] [ ][ ]{ }GLOBAL T M T GLOBAL

{ } [ ]{ }A S D

[ ] [ ]T TR R=

{ } [ ]{ }GLOBAL MS GLOBALA S D= [ ] [ ] [ ][ ]TMS T M TS R S R=where,

Member stiffness matrix in global coordinates


[ ] [ ] [ ][ ]TS R S R=

T⎡ ⎤ ⎡ ⎤ ⎡ ⎤

[ ] [ ] [ ][ ]MS T M TS R S R=

0 0 1 0 1 0 0 00 0 0 0 0 0 0 0

Tc s c ss c s cEA

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥0 0 1 0 1 0 0 0

0 0 0 0 0 0 0 0c s c sLs c s c

−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 2c cs c cs⎡ ⎤− −⎢ ⎥ M b tiff t i 2 2

2 2

cs s cs sEAL c cs c cs

⎢ ⎥− −⎢ ⎥=⎢ ⎥− −⎢ ⎥

Member stiffness matrix in global coordinatesfor a plane truss member

2 2cs s cs s⎢ ⎥− −⎣ ⎦

p


12

2. Plane frame memberStiffness coefficients in local coordinates

2 52

4

5

Degrees of freedom0 0 0 0EA EA⎡ ⎤

13

46

Degrees of freedom

3 2 3 2

0 0 0 0

12 6 12 60 0

L LEI EI EI EI

L L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥

[ ]2 2

6 4 6 20 0

0 0 0 0Mi

EI EI EI EIL L L LS

EA EA

⎢ ⎥⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥

Member stiffness matrix in local coordinates

3 2 3 2

0 0 0 0

12 6 12 60 0

L LEI EI EI EI

L L L L

−⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎢ ⎥


2 2

6 2 6 40 0EI EI EI EIL L L L

⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

Transformation of displacement vector

1 1

2 2

cos sin 0 0 0 0sin cos 0 0 0 0

D UD U

θ θθ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥

3 3

4 4

0 0 1 0 0 00 0 0 cos sin 0

D UD Uθ θ

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥5 5

6 6

0 0 0 sin cos 00 0 0 0 0 1

D UD U

θ θ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥

⎣ ⎦⎩ ⎭ ⎩ ⎭

{ } [ ]{ }LOCAL T GLOBALD R D=

[ ] [ ]R O⎡ ⎤

{ } [ ]{ }LOCAL T GLOBAL

[ ] [ ] [ ][ ] [ ]T

R OR

O R⎡ ⎤

= ⎢ ⎥⎣ ⎦

Rotation matrix


Transformation of load vector

{ } [ ]{ }LOCAL T GLOBALA R A={ } [ ]{ }

cos sin 0 0 0 0θ θ⎡ ⎤⎢ ⎥

[ ]

sin cos 0 0 0 00 0 1 0 0 0

R

θ θ⎢ ⎥−⎢ ⎥⎢ ⎥

= ⎢ ⎥[ ]0 0 0 cos sin 00 0 0 sin cos 0

TRθ θθ θ

= ⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥0 0 0 0 0 1⎢ ⎥⎣ ⎦


Transformation of stiffness matrix

{ } [ ]{ }GLOBAL MS GLOBALA S D=

[ ] [ ] [ ][ ]TMS T M TS R S R= Member stiffness matrix in

global coordinatesglobal coordinates

EA EA⎡ ⎤

0 0 0 00 0 0 0

c ss c

⎡ ⎤⎢ ⎥−⎢ ⎥3 2 3 2

0 0 0 0

12 6 12 60 0

EA EAL L

EI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥

[ ] 0 0 1 0 0 00 0 0 00 0 0 0

T c ss c

R

⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

[ ]2 2

6 4 6 20 0

0 0 0 0M

EI EI EI EIL L L LS

EA EAL L

⎢ ⎥⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥

Where, ,

0 0 0 0 0 1⎢ ⎥⎣ ⎦

3 2 3 2

12 6 12 60 0

6 2 6 40 0

L LEI EI EI EI

L L L LEI EI EI EI

⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎢ ⎥⎢ ⎥


2 20 0L L L L

⎢ ⎥−⎢ ⎥⎣ ⎦

Assembling global stiffness matrixg gPlane truss

2Force

21

331

3

2 4

13

Action/displacement components in local coordinates of members


4 4

3

4

3

4

2 6

1 562

514

3

Action/displacement components in 2

5

6Action/displacement components in

global coordinates of the structure


1 5

1 2 3 4Global DOF

11 12 13 14

1 1 1 1

21 22 23 24

2 2

2 2

12

M M M Ms s s ss s s s

c cs c cscs s cs sEA

− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥

1 2 3 4

[ ] 1 1 1 1

31 32 33 34

1 1 1 1

41 42 43 44

1 2 2

2 2

234

M M M M

M M M M

M

s s s ss s s s

cs s cs sEASc cs c csL

− −⎢ ⎥ ⎢ ⎥= =− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦41 42 43 44

1 1 1 1

2 2 4M M M M

s s s scs s cs s⎢ ⎥ ⎢ ⎥− − ⎣ ⎦⎣ ⎦

1 2 3 4 5 6

12

× × × ×⎡ ⎤⎢ ⎥× × × ×⎢ ⎥ C t ib ti f

[ ]

234JS

× × × ×⎢ ⎥× × × ×⎢ ⎥

= ⎢ ⎥

Contribution of Member 1 to global stiffness matrix[ ]

45

J ⎢ ⎥× × × ×⎢ ⎥⎢ ⎥⎢ ⎥


6⎢ ⎥⎣ ⎦

3 4 5 6Global DOF

11 12 13 14

2 2 2 2

21 22 23 24

34

M M M Ms s s ss s s s⎡ ⎤⎢ ⎥⎢ ⎥

3 4 5 6

[ ] 2 2

2

2 2

31 32 33 34

2 2 2 2

41 42 43 44

456

M M M M

M M M

M

M

s s s ss s s ss s s s

S = ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦2 2 2 2

6M M M M

s s s s⎣ ⎦

1 2 3 4 5 6

12

⎡ ⎤⎢ ⎥⎢ ⎥ C t ib ti f

[ ]

234JS

⎢ ⎥× × × ×⎢ ⎥

= ⎢ ⎥× × × ×⎢ ⎥

Contribution of Member 2 to global stiffness matrix

456

× × × ×⎢ ⎥⎢ ⎥× × × ×⎢ ⎥⎣ ⎦


6⎢ ⎥× × × ×⎣ ⎦

11 12 13 14 1⎡ ⎤1 2 5 6

Global DOF

[ ]

11 12 13 14

3 3 3 3

21 22 23 24

3 3 3 3

12

M M M M

M M M M

s s s ss s s s

S

⎡ ⎤⎢ ⎥⎢ ⎥[ ] 3 3

3

3 3

31 32 33 34

3 3 3 3

41 42 43 44

56

M M M M

M M M

M

Ms s s s

S = ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

41 42 43 44

3 3 3 36

M M M Ms s s s⎢ ⎥⎣ ⎦

1 2 3 4 5 6

12

× × × ×⎡ ⎤⎢ ⎥

1 2 3 4 5 6

[ ]

23

JS

⎢ ⎥× × × ×⎢ ⎥⎢ ⎥

= ⎢ ⎥

Contribution of Member 3 to global stiffness matrix[ ]

45

JS ⎢ ⎥⎢ ⎥⎢ ⎥× × × ×⎢ ⎥

stiffness matrix


6⎢ ⎥× × × ×⎣ ⎦

Assembled global stiffness matrix

11 12 13 1411 12 13 14s s s ss s s s+ + 1⎡ ⎤

1 2 3 4 5 6

1 1 1 1

21 22 23 24

1 1 1 1

3 3 3 3

21 22 23 24

3 3 3 3

M M M M

M M M M

M M M M

M M M M

s s s s

s s s s

s s s s

s s s s

+ +

+ +

1

2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

[ ]31 32 3 11 12 13 14

2 2 2 2

21 22 23 24

3 34

1 1 1 1

41 42 43 44

M M M MM M M M

J

s s s s

s s s s

s s s s

sS

ss s

+ +=

+ +

3

4

⎢ ⎥⎢ ⎥⎢ ⎥

31 3

2 2 2 2

31 32 33

1 1

2

2

1 1

23 3 32

M M M M

M MM M

M M M M

MMs s

s s s s

s s

s s

s

s s

s

+ +

+ 34

2

33 34

3

4

5MM ss +

⎢ ⎥⎢ ⎥⎢ ⎥41 42 43 44

2

41 42 43 44

3 2 23 32 36

M M M MM M M Ms s s ss s s s+ +⎢ ⎥⎣ ⎦


Imposing boundary conditions2

p g yPlane truss example

21

1

2

0U U U U= = = =Boundary conditions are:3

311 2 5 6 0U U U UBoundary conditions are:

{ }GLOBALD

11 12 13 1411 12 13 14s s s ssF s s s+ +⎧ ⎫ U⎧ ⎫⎡ ⎤

{ }GLOBALD

3 3 3 31 1 1 1

2 21 221 22 23 24

1 1 1 1

23 24

1

2 3 3 3 3

M MM M M M

M M M M

M M

M M M M

s s s s

s s s s

sF

F

s s s

s s s s

+ +

+ +

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪

1

2

U

U

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥11 12 13 14

2 2 2 2

21 22 23 24

2

31 32 33 34

1 1 1 1

41 42 43 44

1

3

1 1 1 24 2 2

M M M M M M M

M M M

M

M M M M M

s s s s

s s s

s s s s

s s s s

F

F s

+ +=

+ +

⎪ ⎪⎨ ⎬⎪ ⎪ 4

3U

U

⎪ ⎪⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥21 1 1 1 24

5

2 2M M MM M M M M

F

F

s⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

31 32 33 34

2 2 2 2

41 42 43 44

31 32 33 34

3 3 3 3

41 42 43 44

4

5M M M MM M M Ms ss s s s s U

U

+ +⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥

⎣ ⎦⎩ ⎭Dept. of CE, GCE Kannur Dr.RajeshKN

6F⎩ ⎭41 42 43 44

2 2 2 2

41 42 43 44

3 3 3 3 6M MM M M MM Mss s s s ss s U+ +⎣ ⎦⎩ ⎭

Reduced equation system q y(after imposing boundary conditions)

33 34

1 1

43 44

11 12

2 2

21 22

3 3MM M MF U

F U

s ss s+ +⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥+ +⎩ ⎭ ⎣ ⎦⎩ ⎭1 2 1 24 4M MM M

F Us sss⎢ ⎥+ +⎩ ⎭ ⎣ ⎦⎩ ⎭

•This reduced equation system can be solved to get the unknown displacement components

3 4,U U

{ } [ ]{ }LOCAL T GLOBALD R D=•From

{ }LOCALD can be found out.

{ } { }D D=F h b


{ } { }LOCAL MiD D=•For each member,

{ } { } [ ]{ }A A S D{ } { } [ ]{ }Mi MLi Mi MiA A S D= +Member end actions

Where,

Fixed end actions on the member, { }MLiA

Member stiffness matrix, [ ]MiS

[ ]

in local coordinates

Displacement components of the member, [ ]MiD

{ } { } [ ]{ }LOCAL T GLOBALMi D R DD ==As we know,

{ } { } [ ][ ]{ }i i iM ML M i iT GLOBALA A S R D∴ = +


25

Direct Stiffness Method: Procedure

STEP 1: Get member stiffness matrices for all members [ ]MiS

STEP 2: Get rotation matrices for all members [ ]TiR

STEP 3: Transform member stiffness matrices from local coordinates into global coordinates to get [ ]MSiS[ ]MSi

STEP 4: Assemble global stiffness matrix [ ]JS

STEP 5: Impose boundary conditions to get the reduced stiffness matrix [ ]S[ ]FFS


26

STEP 6: Find equivalent joint loads from applied loads on each q j ppmember (loads other than those applied at joints directly)

STEP 7 T f b ti f l l di t i t l b l STEP 7: Transform member actions from local coordinates into global coordinates to get the transformed load vector

STEP 8: Find combined load vector by adding the above transformed load vector and the loads applied directly at joints

[ ]CA

STEP 9: Find the reduced load vector by removing members in h l d di b d di i

[ ]FCAthe load vector corresponding to boundary conditions

STEP 10: Get displacement components of the structure in global coordinates { } [ ] { }1

F FF FCD S A−=


27

{ } [ ]{ }LOCAL T GLOBALD R D=STEP 11: Get displacement components of each member in local coordinates

STEP 12: Get member end actions from

{ } { } [ ][ ]{ }Mi MLi Mi iT GLOB iALR DA A S∴ = +

{ } { } [ ][ ]R RC RF FA A S D= − +STEP 13: Get reactions from

{ }RCA represents combined joint loads (actual and equivalent) applied directly to the supports.q ) pp y pp


28

• Problem 1:

1 Member stiffness matrices in local co ordinates

1EA ⎡ ⎤⎡ ⎤

1. Member stiffness matrices in local co-ordinates (without considering restraint DOF)

[ ]1

1 001.155

0 0 0 0M

EAS L

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦0 0 0 0⎣ ⎦ ⎣ ⎦

[ ] 1 0⎡ ⎤ [ ] 1 2 0⎡ ⎤


[ ]2

1 00 0MS ⎡ ⎤

= ⎢ ⎥⎣ ⎦

[ ]3

1 2 00 0MS ⎡ ⎤

= ⎢ ⎥⎣ ⎦

2. Rotation (transformation) matrices

[ ]1

cos sin 0.5 0.866i 0 866 0 5TRθ θθ θ

−⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[ ]160 sin cos 0.866 0.5T

θ θ θ=−⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

[ ]290

0 11 0TR

θ

−⎡ ⎤= ⎢ ⎥⎣ ⎦90 1 0θ =− ⎣ ⎦

[ ]3150

0.866 0.50.5 0.866TR

θ =−

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦⎣ ⎦


30

3. Member stiffness matrices in global co-ordinates(Transformed member stiffness matrices)

[ ] [ ] [ ][ ]1 1 1 1T

MS T M TS R S R=

0.5 0.866 1 0 0.5 0.86610 866 0 5 0 0 0 866 0 51 155

T− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦0.866 0.5 0 0 0.866 0.51.155⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0 216 0 375−⎡ ⎤0.216 0.3750.375 0.649

−⎡ ⎤= ⎢ ⎥−⎣ ⎦


[ ]2

0 1 1 0 0 11 0 0 0 1 0

T

MSS− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0 0⎡ ⎤= ⎢ ⎥0 1= ⎢ ⎥⎣ ⎦

[ ]3

0.866 0.5 1 0 0.866 0.510 5 0 866 0 0 0 5 0 8662

T

MSS− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

[ ]3 0.5 0.866 0 0 0.5 0.8662MS ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0.375 0.217⎡ ⎤⎢ ⎥0.217 0.125

= ⎢ ⎥⎣ ⎦


32

4. Global stiffness matrix

0 216 0 0 375 0 375 0 0 217⎡ ⎤

(by assembling transformed member stiffness matrices)

[ ] 0.216 0 0.375 0.375 0 0.2170.375 0 0.217 0.649 1 0.125FFS

+ + − + +⎡ ⎤= ⎢ ⎥− + + + +⎣ ⎦

0.591 0.1580 158 1 774

−⎡ ⎤= ⎢ ⎥⎣ ⎦0.158 1.774 ⎢ ⎥−⎣ ⎦

This is the reduced global stiffness matrix, since restraint DOF were not considered. Hence, boundary conditions are automatically incorporated automatically incorporated.


5. Loads

{ }0

5FCA ⎧ ⎫= ⎨ ⎬−⎩ ⎭

6. Joint displacements6. Joint displacements

0 772⎧ ⎫{ } [ ] { }1

F FF FCD S A−=0.7722.89

−⎧ ⎫= ⎨ ⎬−⎩ ⎭


34

7. Member forces


Member 1

{ } { } [ ][ ]{ }

{ } { } [ ][ ]{ }1 1 1 1 1TM M G LM L BAL OR DA A S= + [ ][ ]{ }1 1 1T GLOBALM R DS=

1 0.5 0.86601 155

0.772⎡ ⎤ −⎡ ⎤⎢ ⎥= ⎢ ⎥⎢−⎧

⎥⎫

⎨ ⎬1.833

=⎧ ⎫⎨ ⎬1.155

0.866 0.50 0 2.89⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦⎨ ⎬−⎩ ⎭ 0⎨ ⎬

⎩ ⎭


35

{ } [ ][ ]{ }R DA SM b 2 { } [ ][ ]{ }2 2 2 2T GLOBALM M R DA S=Member 2

0.7722

1 0 0 10 0 1 0 .89

−⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦

−⎧ ⎫⎨

⎦ −⎩⎣⎬⎭

2.890⎧

=⎫

⎨ ⎬⎩ ⎭

Member 3

{ } [ ][ ]{ }3 3 3 3T GLOBALM M R DA S=

1 2 0 0.866 0.50 0 0.5 0.866

0.7722.89

− −⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥−⎣ ⎦

−⎧ ⎫⎨

⎣ ⎦⎬−⎩ ⎭

1.0570

=⎧ ⎫⎨ ⎬⎩ ⎭⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭


36

• Problem 2 :

10kN m5

48

4m

0kN mCB

20kN 24

6 794m

1

9

4m 1

2A EI is constant.

1

2

3Displacements in global co-ordinates


37

global co ordinates

2 56

9 21 4

3 64

626 9 2

5

1 1

Ki ti 1Kinematic indeterminacy

DOF in local co-ordinates

12

33


38

0 0 0 0

12 6 12 6

EA EAL L

EI EI EI EI

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥

3 2 3 2

2 2

12 6 12 60 0

6 4 6 20 0

EI EI EI EIL L L LEI EI EI EIL L L L

⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

Member stiffness matrix [ ]

0 0 0 0

12 6 12 6

MiL L L LS

EA EAL L

EI EI EI EI

⎢ ⎥=⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥

of a 2D frame member in local coordinates

3 2 3 2

2 2

12 6 12 60 0

6 2 6 40 0

EI EI EI EIL L L LEI EI EI EIL L L L

⎢ ⎥− − −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦L L L L⎢ ⎥⎣ ⎦


39

1. Member stiffness matrices in local co-ordinates ( ith t id i t i t DOF)

6Local DOFMember 1

(without considering restraint DOF)

[ ] 4EIS ⎡ ⎤= ⎢ ⎥ [ ]1EI EI= =

6Global DOF

[ ]1MSL

= ⎢ ⎥⎣ ⎦[ ]1EI EI

Member 2

6 9Global DOF3 6Local DOF

Member 2

[ ]4 2EI EI

L LS

⎡ ⎤⎢ ⎥⎢ ⎥

6 9Global DOF

0.5EI EI⎡ ⎤[ ]2 2 4ML LSEI EIL L

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

0.5EI EI⎡ ⎤

= ⎢ ⎥⎣ ⎦


40


In this case, transformation matrices are:

[ ] [ ]1 1TR = corresponding to local DOF 6

[ ]2

1 00 1TR ⎡ ⎤

= ⎢ ⎥⎣ ⎦

corresponding to local DOFs 3 & 6⎣ ⎦


[ ]1MSS EI= [ ]2

0.50.5MS

EI EIS

EI EI⎡ ⎤

= ⎢ ⎥⎣ ⎦


⎣ ⎦

4 A bl d ( d d d) l b l tiff t i

6 9Gl b l DOF

4. Assembled (and reduced) global stiffness matrix

[ ] 0.5EI EIS

IE⎡ + ⎤= ⎢ ⎥

6 9Global DOF

2 0.5EI ⎡ ⎤

= ⎢ ⎥[ ]0.5FFS

EI EI= ⎢ ⎥⎣ ⎦ 0.5 1

EI= ⎢ ⎥⎣ ⎦


5. Loads20

13.33 13.33B

C

2013.33 20

13.33BC20

B 20 20C C20

Fi d d ti A

Combined (Eqlt.+ actual) joint loads

A Fixed end actions A (Loads in global co-ordinates)

{ }13.33

13 33FCA−⎧ ⎫

= ⎨ ⎬⎩ ⎭

Loads corresponding to global DOF 6, 9:


43

{ }13.33FC ⎨ ⎬

⎩ ⎭Loads corresponding to global DOF 6, 9:

6 Joint displacements

11 431 ⎧ ⎫

6. Joint displacements

{ } [ ] { }1F FF FCD S A−=

11.4319.04

1EI

−⎧=

⎫⎨ ⎬⎩ ⎭


44

7. Member end actions


: fixed end actions for member i{ }MLiA

Member 1

{ } { } [ ][ ]{ }1 1 1 1 1TM M G LM L BAL OR DA A S= +

{ } [ ]{ }DA S { } { }4 10 11 43EI⎡ ⎤{ } [ ]{ }1 1 1ML M GLOBALDA S= + { } { }0 11.43

11.43L EI

⎡ ⎤= + −⎢ ⎥⎣ ⎦= −

This is the member end action corresponding to local DOF 6 f M b 1 i b d h d f


45

of Member 1. i.e., member end moment at the top edge of Member 1.

Member 2{ } { } [ ]{ }2 2 2 2M ML M GLOBALA A DS= +

Member 2

13.33 11.4313.33 1

0.59.04

10.5

EI EIEI EI EI

−⎧ ⎫ ⎧ ⎫+⎡ ⎤

= ⎢ ⎥⎣

⎨ ⎬ ⎨−⎩ ⎭ ⎩⎦⎬⎭

13.33 1.91−⎧ ⎫ ⎧ ⎫+⎨ ⎬ ⎨= ⎬

11.42=⎧ ⎫⎨ ⎬

Th th b ti di t l l DOF 3

13.33 13.325+⎨ ⎬ ⎨

⎩ ⎭⎬− ⎩ ⎭ 0⎨ ⎬

⎩ ⎭

These are the member actions corresponding to local DOFs 3 and 6 of Member 2. i.e., member end moments of Member 2.


46

11.43

0B 0BC11.43

Member end moments

A


47

• Problem 3 :

10 kips0.24 kips/in.

20 kips1000 kips-in

75 in

100 in 50 in 50 in100 in 50 in 50 in


48

2 5

13

46

8

79

8

Global DOF7


49

21

3

56Local DOF

4

6Local DOF


50

4

Free DOF


51

1. Member stiffness matrices in local co-ordinates ( ith t id i t i t DOF)(without considering restraint DOF)

0 0

12 6

XEAL

EI EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

1000 0 0⎡ ⎤⎢ ⎥[ ]1 3 2

12 60

6 40

Z ZM

Z Z

EI EISL LEI EI

⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−

5

0 120 60000 6000 4 10

⎢ ⎥= −⎢ ⎥− ×⎢ ⎥⎣ ⎦

20L L

⎢ ⎥−⎣ ⎦

0 0

12 6

XEAL

EI EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

800 0 0⎡ ⎤⎢ ⎥[ ]2 3 2

12 60

6 40

Z ZM

Z Z

EI EISL LEI EI

⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥

5

0 61.44 38400 3840 3.2 10

⎢ ⎥= ⎢ ⎥×⎢ ⎥⎣ ⎦


52

20L L

⎢ ⎥⎣ ⎦


[ ]cos sin 0

i 0R⎡ ⎤⎢ ⎥

θ θθ θ[ ] sin cos 0

0 0 1TR ⎢ ⎥= −⎢ ⎥

⎢ ⎥⎣ ⎦

θ θ

Member 1 Member 2

[ ]2

0.8 0.6 00.6 0.8 0R

−⎡ ⎤⎢ ⎥= ⎢ ⎥[ ]1

1 0 00 1 0R⎡ ⎤⎢ ⎥= ⎢ ⎥ [ ]

0 0 1⎢ ⎥⎢ ⎥⎣ ⎦

[ ]1

0 0 1⎢ ⎥⎢ ⎥⎣ ⎦


53


[ ] [ ] [ ][ ]1 1 1 1T

MS T M TS R S R=

1 0 0 1000 0 0 1 0 00 1 0 0 120 6000 0 1 0

T⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

5

0 1 0 0 120 6000 0 1 00 0 1 0 6000 4 10 0 0 1

= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ×⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

5

1000 0 00 120 6000

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥50 6000 4 10− ×⎢ ⎥⎣ ⎦


54

[ ]0.8 0.6 0 800 0 0 0.8 0.6 00 6 0 8 0 0 61 44 3840 0 6 0 8 0

T

S− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥[ ]25

0.6 0.8 0 0 61.44 3840 0.6 0.8 00 0 1 0 3840 3.2 10 0 0 1

MSS ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥×⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0.8 0.6 0 640 480 00 6 0 8 0 36 86 49 15 3840

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥

5

0.6 0.8 0 36.86 49.15 38400 0 1 2304 3072 3.2 10

= −⎢ ⎥ ⎢ ⎥×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

534.12 354.51 2304354 51 327 32 3072

−⎡ ⎤⎢ ⎥= −⎢ ⎥

5

354.51 327.32 30722304 3072 3.2 10

⎢ ⎥×⎢ ⎥⎣ ⎦


4. Assembled (reduced) global stiffness matrix

1000 534.12 0 354.51 0 2304+ − +⎡ ⎤

. sse b ed ( educed) g oba st ess at x

[ ]5 5

0 354.51 120 327.32 6000 30720 2304 6000 3072 4 10 3.2 10

FFS⎡ ⎤⎢ ⎥= − + − +⎢ ⎥

+ − + × + ×⎢ ⎥⎣ ⎦0 2304 6000 3072 4 10 3.2 10+ + × + ×⎢ ⎥⎣ ⎦

1534 12 354 51 2304⎡ ⎤

5

1534.12 354.51 2304354.51 447.32 2928

−⎡ ⎤⎢ ⎥− −⎢⎢

= ⎥⎥52304 2928 7.2 10− ×⎢⎣ ⎦⎥


56

5. Loads⎧ ⎫

{ }010JA

⎧ ⎫⎪ ⎪= −⎨ ⎬⎪ ⎪

Actual joint loads1000⎪ ⎪−⎩ ⎭

{ }022A

⎧ ⎫⎪ ⎪⎨ ⎬Equivalent joint loads { } 22

50EA = −⎨ ⎬

⎪ ⎪−⎩ ⎭

Equivalent joint loads

0⎧ ⎫⎪ ⎪{ } { } { } 32

1050FC J EA A A ⎪ ⎪= + = −⎨ ⎬

⎪ ⎪−⎩ ⎭

Hence, combined joint loads


⎩ ⎭

6. Joint displacements

{ } [ ] { }1F FF FCD S A−={ } [ ] { }F FF FCD S A

1

{ }

11534.12 354.51 2304 0354.51 447.32 2928 32FD

−− ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥∴ = − − −⎨ ⎬⎢ ⎥ ⎪ ⎪52304 2928 7.2 10 1050

⎢ ⎥ ⎪ ⎪− × −⎢ ⎥ ⎩ ⎭⎣ ⎦

0.02060.09936−⎧ ⎫

⎪ ⎪= −⎨ ⎬0.099360.001797

⎨ ⎬⎪ ⎪−⎩ ⎭




1000 0 0 1 0 0 0.0206⎡ ⎤ ⎡ ⎤ −⎧ ⎫⎪ ⎪{ }

5

0 120 6000 0 1 0 0.09930 6000 4 10 0

60 1 0.001797

MLiA⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + −⎢ ⎥ ⎢ ⎥

− ×⎢ ⎥ ⎢ ⎥

⎧ ⎫⎪ ⎪−⎨

⎣ ⎦ ⎣⎪⎩⎦

⎬⎪− ⎭0 6000 4 10 0 0 1 0.001797⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎩⎦ ⎭

8. Reactions

{ } { } [ ][ ]R RC RF FA A S D= − +


• Problem 4:

20kN/

6m2m2m

40kN 20kN/m

16kNm

2m

80kN2m

I1=I3=I2=I2

A1=A3=


60

A2=

1 2 31 2 3

4

C ti itConnectivity:

Member Node 1 Node 21 1 22 2 33 2 4


61

3 2 4

46

5

46

Free DOF


62

1. Member stiffness matrices in local co-ordinates ( ith t id i t i t DOF)(without considering restraint DOF)

50 0

3.5 10 0 012 6

XEAL

EI EI

⎡ ⎤⎢ ⎥

⎡ ⎤×⎢ ⎥⎢ ⎥⎢ ⎥[ ]1 3 2

12 60 0 6562.5 131250 13125 350006 40

Z ZM

Z Z

EI EISL LEI EI

⎢ ⎥⎢ ⎥= − = −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥ ⎣ ⎦

⎢ ⎥− 20L L

⎢ ⎥⎣ ⎦

⎡ ⎤

[ ]5

0 03.5 10 0 0

12 6

X

Z Z

EAL

EI EI

⎡ ⎤⎢ ⎥

⎡ ⎤×⎢ ⎥⎢ ⎥⎢ ⎥[ ]2 3 2

2

12 60 0 3888.89 11666.670 11666.67 46666.676 40

Z ZM

Z Z

EI EISL LEI EI

⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦


63

2L L⎢ ⎥⎣ ⎦

⎡ ⎤

[ ]5

0 03.5 10 0 0

12 60 0 6 62 1312

X

Z Z

EAL

EI EIS

⎡ ⎤⎢ ⎥

⎡ ⎤×⎢ ⎥⎢ ⎥⎢ ⎥[ ]3 3 2

2

12 60 0 6562.5 131250 13125 350006 40

Z ZM

Z Z

EI EISL LEI EI

⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦2L L⎢ ⎥⎣ ⎦


64


0 1 0−⎡ ⎤1 0 0⎡ ⎤ 1 0 0⎡ ⎤

[ ]390

0 1 01 0 00 0 1

Rθ =−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[ ]10

1 0 00 1 00 0 1

Rθ =

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[ ]20

0 1 00 0 1

Rθ =

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Member 1

0 0 1⎢ ⎥⎣ ⎦Member 2

0 0 1⎢ ⎥⎣ ⎦Member 3

⎣ ⎦


65


[ ] [ ] [ ][ ]1 1 1 1T

MS T M TS R S R=

53.5 10 0 00 6562.5 13125

⎡ ⎤×⎢ ⎥= −⎢ ⎥

0 13125 35000⎢ ⎥⎢ ⎥−⎣ ⎦


66

5T ⎡ ⎤⎡ ⎤ ⎡ ⎤[ ]

5

2

1 0 0 3.5 10 0 0 1 0 00 1 0 0 3888.89 11666.67 0 1 0

T

MSS⎡ ⎤×⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 1 0 11666.67 46666.67 0 0 1⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

53.5 10 0 00 3888.89 11666.67

⎡ ⎤×⎢ ⎥= ⎢ ⎥

0 11666.67 46666.67⎢ ⎥⎢ ⎥⎣ ⎦


67

[ ]5

3

0 1 0 3.5 10 0 0 0 1 01 0 0 0 6562.5 13125 1 0 0

T

MSS⎡ ⎤− × −⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥

0 0 1 0 13125 35000 0 0 1⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

50 1 0 0 3.5 10 01 0 0 6562 5 0 13125

⎡ ⎤− ×⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥1 0 0 6562.5 0 13125

0 0 1 13125 0 35000

⎢ ⎥⎢ ⎥= − ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

6562.5 0 13125⎡ ⎤⎢ ⎥50 3.5 10 0

13125 0 35000

⎢ ⎥= ×⎢ ⎥⎢ ⎥⎣ ⎦


68

⎣ ⎦

4. Assembled (and reduced) global stiffness matrix. sse b ed (a d educed) g oba st ess at x

5 53.5 10 3.5 10 6562.5 0 0 0 0 0 13125⎡ ⎤× + × + + + + +⎢ ⎥[ ] 50 0 0 6562.5 3888.89 3.5 10 13125 11666.67 0

0 0 13125 13125 11666.67 0 35000 46666.67 35000FFS ⎢ ⎥= + + + + × − + +⎢ ⎥

⎢ ⎥+ + − + + + +⎣ ⎦

706562 5 0 13125⎡ ⎤706562.5 0 131250 360451.39 1458.33

⎡ ⎤⎢ ⎥= −⎢ ⎥

13125 1458.33 116666.67⎢ ⎥

−⎢ ⎥⎣ ⎦


605. Loads

2020 kNm 60 60

202060 60

40

4040

40

Fixed end actions


7040

Fixed end actions

80kN

0+16=16 kNm0 16 16 kNm

40kN

Combined (equivalent + actual) joint loads


71

( q ) j

0⎧ ⎫ 0⎧ ⎫ 0⎧ ⎫0⎧ ⎫

⎪ ⎪000

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪

02020

⎧ ⎫⎪ ⎪−⎪ ⎪−⎪ ⎪

02020

⎧ ⎫⎪ ⎪−⎪ ⎪−⎪ ⎪

20200

⎪ ⎪−⎪ ⎪−⎪ ⎪

⎪ ⎪00

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪

4080

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪−⎪ ⎪

4080

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪−⎪ ⎪

{ }060

60RCA

⎪ ⎪⎪ ⎪⎪ ⎪= −⎨ ⎬⎪ ⎪⎪ ⎪

{ }16

;00

JA⎪ ⎪−⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪

{ }0

;060

EA⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪−

{ } { } { }16

;060

C J EA A A⎪ ⎪−⎪ ⎪= + = ⎨ ⎬⎪ ⎪⎪ ⎪−

4080

⎪ ⎪⎪ ⎪⎪ ⎪−⎪ ⎪⎪ ⎪

000

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪

606040

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪

40⎧ ⎫⎪ ⎪

606040

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪

40⎪ ⎪−⎩ ⎭

800

⎪ ⎪−⎪ ⎪

⎪ ⎪⎩ ⎭

040

⎪ ⎪⎪ ⎪⎪ ⎪−⎩ ⎭

{ } 8016

FCA ⎪ ⎪= −⎨ ⎬⎪ ⎪−⎩ ⎭

8040

⎪ ⎪−⎪ ⎪

⎪ ⎪−⎩ ⎭


72

[ ] [ ] { }1D S A−=6. Joint displacements

1706562.5 0 13125 40−⎡ ⎤ ⎧ ⎫

⎪ ⎪

[ ] [ ] { }F FF FCD S A=6. Jo t d sp ace e ts

0 360451.39 1458.33 8013125 1458.33 116666.67 16

⎡ ⎤ ⎧ ⎫⎪ ⎪⎢ ⎥= − −⎨ ⎬⎢ ⎥ ⎪ ⎪− −⎢ ⎥⎣ ⎦ ⎩ ⎭⎢ ⎥⎣ ⎦ ⎩ ⎭

5 9 60.142 10 0.646 10 0.160 10 40 − − −⎡ ⎤× − × − × ⎧ ⎫⎢ ⎥9 5 7

6 7 5

400.646 10 0.277 10 0.348 10 80

160 160 10 0 348 10 0 859 10 − − −

− − −

⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪= − × × × −⎨ ⎬⎢ ⎥

⎪ ⎪⎢ ⎥ −× × × ⎩ ⎭⎣ ⎦ 160.160 10 0.348 10 0.859 10 ⎢ ⎥− × × × ⎩ ⎭⎣ ⎦

40.594 10 −⎧ ⎫×⎪ ⎪3

3

0.222 100.147 10

−

−

⎪ ⎪= − ×⎨ ⎬⎪ ⎪− ×⎩ ⎭


73

⎩ ⎭

Summary


Summary

• Introduction – element stiffness matrix – rotation transformation matrix – transformation of displacement and load vectors and stiffness matrix – equivalent nodal forces and load vectors –assembly of stiffness matrix and load vector – determination of assembly of stiffness matrix and load vector determination of nodal displacement and element forces – analysis of plane truss beam and plane frame (with numerical examples) – analysis of grid space frame (without numerical examples) grid – space frame (without numerical examples)


75

• Problem X:

2

6

1 35

6

4

5


76

0 0EA EA⎡ ⎤

[ ]1

0 0 1 0 1 00 0 0 0 0 0 0 01

M

L L

S

⎡ ⎤−⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥

[ ]1 1 0 1 020 00 0 0 0

0 0 0 0

M EA EAL L

⎢ ⎥ −⎢ ⎥−⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦0 0 0 0⎢ ⎥⎣ ⎦

[ ] [ ] [ ]2 3 4S S S= = =[ ] [ ] [ ]2 3 4M M MS S S

1 0 1 00 0 0 01

−⎡ ⎤⎢ ⎥⎢ ⎥

1 0 1 00 0 0 01

−⎡ ⎤⎢ ⎥⎢ ⎥[ ]5

0 0 0 011 0 1 02.83

0 0 0 0

MS ⎢ ⎥=−⎢ ⎥⎢ ⎥⎣ ⎦

[ ]6

0 0 0 011 0 1 02.83

0 0 0 0

MS ⎢ ⎥=−⎢ ⎥⎢ ⎥⎣ ⎦


77

⎣ ⎦ ⎣ ⎦

Rotation matrices

cos sin 0 0 0 1 0 0sin cos 0 0 1 0 0 0θ θθ θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

[ ]190

sin cos 0 0 1 0 0 00 0 cos sin 0 0 0 1TR

θ

θ θθ θ=

⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 sin cos 0 0 1 0θ θ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

0 1 0 0−⎡ ⎤⎢ ⎥

1 0 0 00 1 0 0⎡ ⎤⎢ ⎥

[ ]390

1 0 0 00 0 0 1TR

θ =−

⎢ ⎥⎢ ⎥=

−⎢ ⎥⎢ ⎥

[ ]20

0 1 0 00 0 1 00 0 0 1

TRθ =

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦ 0 0 1 0⎢ ⎥

⎣ ⎦0 0 0 1⎢ ⎥⎣ ⎦


[ ]

1 1 0 01 1 0 0

0 707R

⎡ ⎤⎢ ⎥−⎢ ⎥=[ ]

1 0 0 00 1 0 0

R

−⎡ ⎤⎢ ⎥−⎢ ⎥ [ ]5

450.707

0 0 1 10 0 1 1

TRθ =

⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

[ ]4180 0 0 1 0

0 0 0 1

TRθ =

⎢ ⎥=−⎢ ⎥

⎢ ⎥−⎣ ⎦⎣ ⎦

[ ]

1 1 0 01 1 0 0

−⎡ ⎤⎢ ⎥⎢ ⎥[ ]6

45

1 1 0 00.707

0 0 1 10 0 1 1

TRθ =−

⎢ ⎥=−⎢ ⎥

⎢ ⎥⎣ ⎦0 0 1 1⎣ ⎦


79

[ ] [ ] [ ][ ]TS R S R[ ] [ ] [ ][ ]1 1 1 1MS T M TS R S R=

0 1 0 0 1 0 1 0 0 1 0 01 0 0 0 0 0 0 0 1 0 0 01

T −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥=

0 0 0 1 1 0 1 0 0 0 0 120 0 1 0 0 0 0 0 0 0 1 0

⎢ ⎥ ⎢ ⎥ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0 0 0 0⎡ ⎤0 1 0 0 0 1 0 1− −⎡ ⎤ ⎡ ⎤0 1 0 110 0 0 02

⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥

1 0 0 0 0 0 0 010 0 0 1 0 1 0 12

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=

− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 1 0 1⎢ ⎥−⎣ ⎦0 0 1 0 0 0 0 0⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


1 0 1 0−⎡ ⎤

[ ]2

1 0 1 00 0 0 011 0 1 02MSS

−⎡ ⎤⎢ ⎥⎢ ⎥=−⎢ ⎥1 0 1 020 0 0 0

⎢ ⎥⎢ ⎥⎣ ⎦

[ ]

0 1 0 0 1 0 1 0 0 1 0 01 0 0 0 0 0 0 0 1 0 0 01

T

S

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥[ ]3 0 0 0 1 1 0 1 0 0 0 0 120 0 1 0 0 0 0 0 0 0 1 0

MSS ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0 1 0 0 0 1 0 11 0 0 0 0 0 0 01

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

0 0 0 00 1 0 11⎡ ⎤⎢ ⎥1 0 0 0 0 0 0 01

0 0 0 1 0 1 0 120 0 1 0 0 0 0 0

⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 1 0 110 0 0 020 1 0 1

⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦


81

0 0 1 0 0 0 0 0−⎣ ⎦ ⎣ ⎦ 0 1 0 1−⎣ ⎦

1 0 1 0−⎡ ⎤⎢ ⎥

[ ]4

0 0 0 011 0 1 02

0 0 0 0

MSS⎢ ⎥⎢ ⎥=−⎢ ⎥⎢ ⎥⎣ ⎦0 0 0 0⎣ ⎦

1 1 0 0 1 0 1 0 1 1 0 0T −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

[ ]5

1 1 0 0 0 0 0 0 1 1 0 010.707 0.7070 0 1 1 1 0 1 0 0 0 1 12.83MSS

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥=

−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥0 0 1 1 0 0 0 0 0 0 1 1⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 1 0 0 1 1 1 11 1 0 0 0 0 0 0

0.17660 0 1 1 1 1 1 1

− − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥

1 1 1 11 1 1 1

0.17661 1 1 1

− −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥0 0 1 1 1 1 1 1

0 0 1 1 0 0 0 0− − −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 1 1 11 1 1 1

− −⎢ ⎥⎢ ⎥− −⎣ ⎦


1 1 0 0 1 0 1 0 1 1 0 0T⎡ ⎤ ⎡ ⎤ ⎡ ⎤

[ ]6

1 1 0 0 1 0 1 0 1 1 0 01 1 0 0 0 0 0 0 1 1 0 010.707 0.7070 0 1 1 1 0 1 0 0 0 1 12.83MSS

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=

− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥0 0 1 1 1 0 1 0 0 0 1 12.830 0 1 1 0 0 0 0 0 0 1 1⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 1 1 11 1 1 1

− −⎡ ⎤⎢ ⎥

1 1 0 0 1 1 1 11 1 0 0 0 0 0 0

− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ 1 1 1 1

0.1771 1 1 1

1 1 1 1

⎢ ⎥− −⎢ ⎥=− −⎢ ⎥⎢ ⎥− −⎣ ⎦

1 1 0 0 0 0 0 00.177

0 0 1 1 1 1 1 10 0 1 1 0 0 0 0

⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥=− −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ 1 1 1 1⎣ ⎦0 0 1 1 0 0 0 0⎣ ⎦ ⎣ ⎦


Assembled global stiffness matrix

1 10 0.1766 0 0 0.1766 0 0 0.1766 0.1766 02 2

1 1

+ + + + − − −⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

g

1 10 0 0.1766 0 0.1766 0 0.1766 0.1766 0 02 2

1 10 0 0 0.1766 0 0 0.1766 0 0.1766 0.17662 2

1 1

+ + + + − − −

+ + + − − −

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

[ ]1 10 0 0 0.1766 0 0.1766 0 0 0.1766 0.17662 2

1 10.1766 0.1766 0 0 0.1766 0 0 0.2 2

JS− + − + + −

=− − − + + + + 1766 0 0

1 10 1766 0 1766 0 0 0 0 0 1766 0 0 1766 0

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥1 10.1766 0.1766 0 0 0 0 0.1766 0 0.1766 0

2 21 10 0.1766 0.1766 0 0 0 0.1766 0 0 0.17662 2

1 10 0 0 1766 0 1766 0 0 0 0 1766 0 0 1766

⎢ ⎥− − + + + + −⎢ ⎥⎢ ⎥⎢ − − + + + − ⎥⎢ ⎥⎢ ⎥

+ + +⎢ ⎥0 0 0.1766 0.1766 0 0 0 0.1766 0 0.17662 2

− − + − + +⎢ ⎥⎣ ⎦

1 2 8 0D D D= = =Boundary conditions are:


1 2 8 0D D DBoundary conditions are:

0 .677 -0 .177 -0 .500 0 .000 -0 .177 ⎡ ⎤⎢ ⎥

Reduced stiffness matrix

[ ] -0.177 0 .677 0 .000 0 .000 0 .177 -0.500 0 .000 0 .677 0 .177 0 .000FFS =

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

[ ] 0 .000 0 .000 0 .177 0 .677 0 .000 -0.177 0 .177 0 .000 0 .000 0 .677

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦


85

5⎧ ⎫⎪ ⎪

{ }00FCA

⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪

In this problem, the reduced load vector (in global coords) can be directly written as

00

⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

{ } [ ] { }1F FF FCD S A−=

24.144⎧ ⎫⎪ ⎪

4.829 1.000 3.829 -1.000 1.000 5⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪ 5.000

19.144

⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪

=⎪

1.000 1.793 0.793 -0.207 -0.207 3.829 0.793 4.622 -1.207 0.793=

00

⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎨ ⎬⎢ ⎥ ⎪ ⎪ -5.000

5.000

⎪⎪⎪

⎪⎪⎪⎩ ⎭

-1.000 -0.207 -1.207 1.793 -0.207 0 1.000 -0.207 0.793 -0.207 1.793 0

⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎣ ⎦ ⎩ ⎭


⎣ ⎦ ⎩ ⎭

Assignment


Module3 direct stiffness- rajesh sir

Engineering

Transcript of Module3 direct stiffness- rajesh sir