Download pdf - Module2 stiffness- rajesh sir

Dept. of CE, GCE Kannur Dr.RajeshKN

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Structural Analysis - III

Dr. Rajesh K. N.Asst. Professor in Civil Engineering

Govt. College of Engineering, Kannur

Stiffness Method


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• Development of stiffness matrices by physical approach –stiffness matrices for truss,beam and frame elements –displacement transformation matrix – development of total stiffness matrix - analysis of simple structures – plane truss beam and plane frame- nodal loads and element loads – lack of fit and temperature effects.

Stiffness method

Module II

Dept. of CE, GCE Kannur Dr.RajeshKN3

• Displacement components are the primary unknowns

• Number of unknowns is equal to the kinematic indeterminacy

• Redundants are the joint displacements, which are automatically specified

• Choice of redundants is unique

• Conducive to computer programming

FUNDAMENTALS OF STIFFNESS METHOD

Introduction


• Stiffness method (displacements of the joints are the primary unknowns): kinematic indeterminacy

• kinematic indeterminacy

• joints: a) where members meet, b) supports, c) free ends• joints undergo translations or rotations

• in some cases joint displacements will be known, from the restraint conditions

• the unknown joint displacements are the kinematicallyindeterminate quantities

o degree of kinematic indeterminacy: number of degrees of freedom


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• in a truss, the joint rotation is not regarded as a degree of freedom. joint rotations do not have any physical significance as they have no effects in the members of the truss

• in a frame, degrees of freedom due to axial deformations can be neglected


•Example 1: Stiffness and flexibility coefficients of a beam

Stiffness coefficients

Unit displacement

Unit action

Forces due to unit dispts– stiffness coefficients

11 21 12 22, , ,S S S S Dispts due to unit forces – flexibility coefficients

11 21 12 22, , ,F F F F


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•Example 2: Action-displacement equations for a beam subjected to several loads

1 11 12 13A A A A= + +

1 11 1 12 2 13 3A S D S D S D= + +

2 21 1 22 2 23 3A S D S D S D= + +

3 31 1 32 2 33 3A S D S D S D= + +


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1 11 1 12 2 13 3 1

2 21 1 22 2 23 3 2

1 1 2 2 3 3

......

...............................................................

n n

n n

n n n n nn n

A S D S D S D S DA S D S D S D S D

A S D S D S D S D

= + + + += + + + +

= + + + +

1 11 12 1 1

2 21 22 2 2

1 211

...

...... ... ... ... ......

...

n

n

n n nn nnn n nn

A S S S DA S S S D

S S S DA× ××

⎧ ⎫ ⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪=⎨ ⎬ ⎨ ⎬⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪⎩ ⎭⎩ ⎭⎩ ⎭

A = SD

•Action matrix, Stiffness matrix, Displacement matrix

•Stiffness coefficient ijS

{ } [ ] { } [ ] [ ]1 1A F D S F− −= ⇒ =

oIn matrix form,

{ } [ ]{ }A S D=

[ ] [ ] 1F S −=

Stiffness matrix


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Example: Propped cantilever (Kinematically indeterminate to first degree)

Stiffness method (Direct approach: Explanation using principle of superposition)

•degrees of freedom: one

• Kinematically determinate structure is obtained by restraining all displacements (all displacement components made zero -restrained structure)

• Required to get Bθ


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Restraint at B causes a reaction of MB as shown.

Hence it is required to induce a rotation of Bθ

The actual rotation at B is Bθ

2

12BwLM = −

(Note the sign convention: anticlockwise positive)

•Apply unit rotation corresponding to Bθ

Let the moment required for this unit rotation be Bm

4B

EImL

= anticlockwise


• Moment required to induce a rotation of Bθ B Bm θis

2 4 012 BwL EI

Lθ− + =

3

48B

BB

wLEI

Mm

θ∴ = − =

Bm (Moment required for unit rotation) is the stiffness coefficient here.

0B B BM m θ+ = (Joint equilibrium equation)


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Stiffnesses of prismatic members

Stiffness coefficients of a structure are calculated from the contributions of individual members

Hence it is worthwhile to construct member stiffness matrices

[ ] [ ] 1Mi MiS F −=


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Member stiffness matrix for prismatic beam member with rotations at the ends as degrees of freedom

[ ] 2 121 2Mi

EISL

⎡ ⎤= ⎢ ⎥

⎣ ⎦

[ ] [ ]

1

11 2 13 6

1 266 3

Mi Mi

L LLEI EIS F

L L EIEI EI

−

−−

−⎡ ⎤⎢ ⎥ −⎛ ⎞⎡ ⎤

= = =⎢ ⎥ ⎜ ⎟⎢ ⎥− −⎣ ⎦⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦

2 1 2 16 21 2 1 2(3)

EI EIL L

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

Verification:

BA

1


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[ ] 11 12

21 22

3 2

2

12 6

6 4

M MMi

M M

S SS

S S

EI EIL LEI EIL L

⎡ ⎤= ⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥

= ⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

Member stiffness matrix for prismatic beam member with deflection and rotation at one end as degrees of freedom


[ ] [ ]

13 212

1

2

2 33 26 3 6

2

Mi Mi

L LL LLEI EIS F

EI LL LEI EI

−

−−

⎡ ⎤⎢ ⎥ ⎛ ⎞⎡ ⎤

= = =⎢ ⎥ ⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠⎢ ⎥⎣ ⎦

( )2

2

3

2

2

6 363

12 6

6 423

EI EIL LEI EIL

LEIL LL L

L

−⎡ ⎤= =⎢ ⎥−⎣

⎡ ⎤−⎢ ⎥⎢ ⎥⎢⎢

⎦ ⎥− ⎥⎣ ⎦

Verification:

[ ]MiEASL

=

•Truss member


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•Plane frame member

[ ]11 12 13

21 22 23

31 32 33

3 2

2

0 0

12 60

6 40

M M M

Mi M M M

M M M

S S SS S S S

S S S

EAL

EI EIL LEI EIL L

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦


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•Grid member

[ ]

3 2

2

12 60

0 0

6 40

Mi

EI EIL L

GJSL

EI EIL L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦


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•Space frame member

[ ]

3 2

3 2

2

2

0 0 0 0 0

12 60 0 0 0

12 60 0 0 0

0 0 0 0 0

6 40 0 0 0

6 40 0 0 0

Z Z

Y Y

Mi

Y Y

Z Z

EAL

EI EIL L

EI EIL LS

GJL

EI EIL L

EI EIL L

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦


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(Explanation using principle of complimentary virtual work)Formalization of the Stiffness method

{ } [ ]{ }Mi Mi MiA S D=

Here{ }MiD contains relative displacements of the k end with respect to j end of the i-th member

If there are m members in the structure,

{ }{ }{ }

{ }

{ }

[ ] [ ] [ ] [ ] [ ][ ] [ ] [ ] [ ] [ ][ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

{ }{ }{ }

{ }

{ }

11 1

22 2

33 3

0 0 0 00 0 0 00 0 0 0

0 0 0 0

0 0 0 0

MM M

MM M

MM M

MiMi Mi

MmMm Mm

SA DSA D

SA D

SA D

SA D

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

M M MA = S D


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{ } [ ]{ }M M MA S D=

[ ]MS is the unassembled stiffness matrix of the entire structure


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• Relative end-displacements in will be related to a vector of joint

displacements for the whole structure,

{ }MD

{ }JD

• If there are no support displacements specified,

{ }RD will be a null matrix

• Hence, { } [ ]{ } [ ] [ ] { }{ }

FM MJ J MF MR

R

DD C D C C

D⎧ ⎫

= = ⎡ ⎤ ⎨ ⎬⎣ ⎦⎩ ⎭

{ }JD{ }FDfree (unknown) joint displacements

{ }RDand restraint displacements consists of:

{ } [ ]{ }M MJ JD C D=

displacement transformation matrix (compatibility matrix) [ ]MJC


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• Elements in displacement transformation matrix

(compatibility matrix) [ ]MJC are found from compatibility conditions.

•Each column in the submatrix consists of member displacements caused by a unit value of a support displacementapplied to the restrained structure.

[ ]MRC

[ ]MFC• Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacementapplied to the restrained structure.

{ }MDrelate to respectively

[ ]MFC

[ ]MRC

and{ }FD

{ }RD

and


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{ }MDδ

{ } [ ]{ } [ ] [ ] { }{ }

FM MJ J MF MR

R

DD C D C C

Dδ

δ δδ

⎧ ⎫= = ⎡ ⎤ ⎨ ⎬⎣ ⎦

⎩ ⎭

• Suppose an arbitrary set of virtual displacements

is applied on the structure.

{ } { } { } { }T T T FJ J F R

R

DW A D A A

Dδ

δ δ δδ⎧ ⎫⎡ ⎤= = ⎨ ⎬⎣ ⎦ ⎩ ⎭

{ }JDδ• External virtual work produced by the virtual displacements

{ }JAand real loads is


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{ } { }TM MU A Dδ δ=

• Internal virtual work produced by the virtual (relative) end displacements { }MDδ { }MAand actual member end actions is

{ } { } { } { }T TJ J M MA D A Dδ δ=

• Equating the above two (principle of virtual work),

{ } [ ]{ }M MJ JD C D= { } [ ]{ }M M MA S D=But and

{ } [ ]{ }M MJ JD C Dδ δ=Also,

{ } { } { } [ ] [ ] [ ]{ }TT T TJ J J MJ M MJ JA D D C S C Dδ δ=Hence,

{ } [ ]{ }J J JA S D=


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[ ] [ ] [ ][ ]TJ MJ M MJS C S C=Where, , the assembled stiffness matrix for the

entire structure.

• It is useful to partition into submatrices pertaining to free

(unknown) joint displacements

[ ]JS

{ }FD { }RDand restraint displacements

{ } [ ]{ } { }{ }

[ ] [ ][ ] [ ]

{ }{ }

FF FRF FJ J J

RF RRR R

S SA DA S D

S SA D⎧ ⎫ ⎧ ⎫⎡ ⎤

= ⇒ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦

[ ] [ ] [ ][ ]TFF MF M MFS C S C= [ ] [ ] [ ][ ]T

FR MF M MRS C S C=

[ ] [ ] [ ][ ]TRF MR M MFS C S C= [ ] [ ] [ ][ ]T

RR MR M MRS C S C=

Where,


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{ } [ ]{ } [ ]{ }F FF F FR RA S D S D= + { } [ ]{ } [ ]{ }R RF F RR RA S D S D= +

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−⇒ = −⎡ ⎤⎣ ⎦

{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +

• Support reactions

{ }RCArepresents combined joint loads (actual and equivalent) applied directly to the supports.

If actual or equivalent joint loads are applied directly to the supports,

Joint displacements


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{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +

• Member end actions are obtained adding member end actions calculated as above and initial fixed-end actions

{ } { } [ ][ ]{ }M ML M MJ JA A S C D= +i.e.,

{ }MLAwhere represents fixed end actions


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Important formulae:

Joint displacements:

Member end actions:

Support reactions:

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= ⎡ − ⎤⎣ ⎦




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•Problem 1

[ ]

4 2 0 02 4 0 020 0 2 10 0 1 2

MEISL

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

[ ] 2 121 2Mi

EISL

⎡ ⎤= ⎢ ⎥

⎣ ⎦Member stiffness matrix of beam member

Kinematic indeterminacy = 2


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Fixed end actions

Equivalent joint loads


Joint displacements

Free (unknown) joint displacements { }FD { }RDRestraint displacements


Joint displacements

Free (unknown) joint displacements { }FD { }RDRestraint displacements

•Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacementapplied to the restrained structure.

[ ]MFC

•Each column in the submatrix consists of member displacements caused by a unit value of a support displacementapplied to the restrained structure.

[ ]MRC


0 1 1 0

0 0 0 1

[ ]

0 01 01 00 1

M FC

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2=1 =1


1 L 1 L 0 0 1 0 0 0

1 L− 1 L− 1 L 1 L 0 0 1 L− 1 L−


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[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦

0 0 1 1 00 1 0 1 010 0 0 1 1

0 0 0 1 1

LLLL

L

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥−⎣ ⎦

[ ]

1 1 1 01 0 1 0

0 0 1 10 0 1 1

MR

L LL L

CL LL L

−⎡ ⎤⎢ ⎥−⎢ ⎥=

−⎢ ⎥⎢ ⎥−⎣ ⎦

DR1 DR2 DR3 DR4=1 =1 =1 =1

DR1 DR2 DR3 DR4=1 =1 =1 =1

DF1 DF2=1 =1


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{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ } [ ] { }1F FF FD S A−∴ =

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C=

4 2 0 0 0 00 1 1 0 2 4 0 0 1 020 0 0 1 0 0 2 1 1 0

0 0 1 2 0 1

EIL

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

6 121 2

EIL

⎡ ⎤= ⎢ ⎥

⎣ ⎦

is a null matrix, since there are no support displacements

{ }RD


{ }1

6 1 121 2 29F

EI PLDL

−⎛ ⎞⎡ ⎤ ⎧ ⎫

= ⎨ ⎬⎜ ⎟⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠

Free (unknown) joint displacements

2 201.11 818 11

011

PLE EII

PL⎧ ⎫ ⎧= =

⎫⎨⎨ ⎬

⎭ ⎩⎩⎬⎭

2

3

6 2 3 320 0 3 3L L L LEI

L L L⎡ ⎤− −

= ⎢ ⎥−⎣ ⎦

[ ] [ ] [ ][ ]TFR MF M MRS C S C=

4 2 0 0 1 1 00 1 1 0 2 4 0 0 1 0 1 02 10 0 0 1 0 0 2 1 0 0 1 1

0 0 1 2 0 0 1 1

LEIL L

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦



is a null matrix.{ }RD

Support reactions

[ ] [ ] [ ][ ]TRF MR M MFS C S C=

2

6 02 02

3 33 3

LEIL

⎡ ⎤⎢ ⎥⎢ ⎥=−⎢ ⎥⎢ ⎥− −⎣ ⎦

1 1 0 4 2 0 0 0 01 0 1 0 2 4 0 0 1 01 20 0 1 1 0 0 2 1 1 00 0 1 1 0 0 1 2 0 1

TLEI

L L

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥=

−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

{ } { } [ ]{ }R RC RF FA A S D∴ = − +


[ ] [ ] [ ][ ]TRR MR M MRS C S C=

1 1 0 4 2 0 0 1 1 01 0 1 0 2 4 0 0 1 0 1 01 2 10 0 1 1 0 0 2 1 0 0 1 10 0 1 1 0 0 1 2 0 0 1 1

TL LEI

L L L

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

− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3

12 6 12 06 4 6 0212 6 18 60 0 6 6

LL L LEI

LL

−⎡ ⎤⎢ ⎥−⎢ ⎥=− − −⎢ ⎥⎢ ⎥−⎣ ⎦


2

2

2 6 02 0 02

33 3 11833 3

PPL LEI PL

L EIPP

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

2

2

02 20 002

3 3318 33 33

3

P PPL PL

PL EI PEI LP P

PP P

⎧ ⎫−⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= − + = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭− −⎩ ⎭ ⎩ ⎭ ⎪ ⎪⎩ ⎭

{ } { } [ ]{ }R RC RF FA A S D∴ = − +

2

310

323

PPL

P

P

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

=


41


{ }2

3 4 2 0 0 0 03 2 4 0 0 1 0 02

2 0 0 2 1 1 0 19 182 0 0 1 2 0 1

MPL EI PLA

L EI

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

2

3 4 2 0 0 03 2 4 0 0 02

2 0 0 2 1 09 182 0 0 1 2 1

PL EI PLL EI

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

3 03 0

2 19 92 2

PL PL⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

11

130

PL⎧ ⎫⎪ ⎪−⎪ ⎪⎨ ⎬⎪⎪

=⎪⎪⎩ ⎭

is a null matrix{ }RD

Member end actions

{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +


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3

0 00 0 0 2 0 6 4 6 01 1 0 0 4 0 6 2 6 02

0 0 0 2 0 0 3 31 1 1 1 2 0 0 3 3

0 0 1 1

L LL L L

L LEIL L LL

L L

⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦⎢ ⎥− −⎣ ⎦

[ ] [ ][ ] [ ]

2 2 2

2 2

3 2

6 6 2 3 32 0 0 3 3

6 0 12 6 12 022 0 6 4 6 03 3 12 6 18 63 3 0 0 6 6

FF FR

RF RR

L L L L L LL L L L

S SL LEIS SL L L L L

L L LL L

⎡ ⎤− −⎢ ⎥−⎢ ⎥⎢ ⎥ ⎡ ⎤−

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

[ ] [ ] [ ][ ]TJ MJ M MJS C S C=

0 00 0 0 4 2 0 0 0 0 1 1 01 1 0 0 2 4 0 0 0 1 0 1 01 2 1

0 0 0 0 0 2 1 0 0 0 1 11 1 1 1 0 0 1 2 0 0 0 1 1

0 0 1 1

L LL L

LEIL LL L L

L

⎡ ⎤⎢ ⎥ −⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎢ ⎥− −⎣ ⎦

Alternatively, if the entire [SJ] matrix is assembled at a time,


43

•Problem 2:

AB C D

40 kN10 kN/m

2 m4 m 4 m 2 m

100 kN

Kinematic indeterminacy = 3 (Not considering joint D in the overhanging portion)

Degrees of freedom

AB C D

DF2DF1 DF3


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[ ]

2 1 0 01 2 0 020 0 2 140 0 1 2

MEIS

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦


[ ] 2 121 24Mi

EIS ⎡ ⎤= ⎢ ⎥



45

Fixed end actions

Equivalent joint loads + actual joint loads

AB

13.33 kNm 13.33

20 kN 20 kN

BC

20 20

20 kN 20 kN

AB

13.33 kNm 6.67

20 kN 20 +20 = 40 kN 20 +100 = 120 kN

2x100 – 20 = 180 kNm


1 0 0 0

0 1 1 0

DF1 =1

DF2 =1 L


47

[ ]

1 0 00 1 00 1 00 0 1

M FC

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2 DF3=1 =1 =1

0 0 0 1

DF3 =1


48

{ } [ ] { }1F FF FD S A−∴ =

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C=

1 0 0 2 1 0 0 1 0 00 1 0 1 2 0 0 0 1 020 1 0 0 0 2 1 0 1 040 0 1 0 0 1 2 0 0 1

T

EI⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 0.5 00.5 2 0.50 0.5 1

EI⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

is a null matrix.{ }RD∴


{ }

11 0.5 0 13.33

1 0.5 2 0.5 6.670 0.5 1 180

FDEI

−−⎛ ⎞⎡ ⎤ ⎧ ⎫⎪ ⎪⎜ ⎟⎢ ⎥= −⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟ −⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠

Joint displacements

43.43560.33210.6

1EI

−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−⎩ ⎭

=


50


{ }43.43560.33210.6

13.33 2 1 0 0 1 0 013.33 1 2 0 0 0 1 02 120 0 0 2 1 0 1 0420 0 0 1 2 0 0 1

MEIA

EI

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

−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−

⎣⎩

⎭ ⎦ ⎣ ⎦⎭

⎩

02525200

kNm

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪−⎩ ⎭

=


Member end actions

{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +


•Problem 3

Analyse the beam. Support B has a downward settlement of 30mm. EI=5.6×103 kNm2

[ ]

4 2 0 0 0 02 4 0 0 0 00 0 2 1 0 020 0 1 2 0 060 0 0 0 4 20 0 0 0 2 4

MEIS

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


[ ] 2 121 2Mi

EISL

⎡ ⎤= ⎢ ⎥



Fixed end moments

2525


2525

Support settlements { }RD(Restraint displacements)

A

BC D30mm 1RD

Free (unknown) joint displacements { }FD

1FD 2FD3FD


53

BA C D

1 1FD =

0 1 1 0 0 0

and consist of member displacements due to unit displacements on the restrained structure.[ ]MFC [ ]MRC


[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦

0 0 0 1 31 0 0 1 31 0 0 1 60 1 0 1 60 1 0 00 0 1 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

31 2 1

1 1 11FF F RDD D D

= = ==

A

BC D1 1RD =

1 3− 1 3− 1 6 1 6 0 0


55

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C=

0 0 0 4 2 0 0 0 0 0 0 01 0 0 2 4 0 0 0 0 1 0 01 0 0 0 0 2 1 0 0 1 0 00 1 0 0 0 1 2 0 0 0 1 030 1 0 0 0 0 0 4 2 0 1 00 0 1 0 0 0 0 2 4 0 0 1

T

EI

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

6 1 01 6 2

30 2 4

EI⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


[ ] [ ] [ ][ ]TFR MF M MRS C S C=

0 0 0 4 2 0 0 0 0 1 31 0 0 2 4 0 0 0 0 1 31 0 0 0 0 2 1 0 0 1 60 1 0 0 0 1 2 0 0 1 630 1 0 0 0 0 0 4 2 00 0 1 0 0 0 0 2 4 0

T

EI

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

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3 21 2

30

EI−⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭


{ }

16 1 0 0 3 21 6 2 25 1 2 0.03

3 30 2 4 25 0

EI EI−

−⎛ ⎞ ⎡ ⎤⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎣ ⎦

20 4 2 0 0.0453 56004 24 12 25 0.015

116 32 12 35 25 0

EI

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

3

16423 108

116 5.6 106

0.0075830.00050.007 311

−⎧ ⎫⎪ ⎪= =⎨ ⎬× × ⎪ ⎪

−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩⎩ ⎭⎭

20 4 2 0 843 4 24 12 25 28

1162 12 35 25 0

EI

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

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦


58


Support reactions

[ ]

4 2 0 0 0 0 0 0 02 4 0 0 0 0 1 0 00 0 2 1 0 0 1 0 0

1 3 1 3 1 6 1 6 0 00 0 1 2 0 0 0 1 030 0 0 0 4 2 0 1 00 0 0 0 2 4 0 0 1

EI

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

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

[ ] [ ] [ ][ ]TRF MR M MFS C S C=

[ ]3 2 1 2 03

EI= −


{ } { } [ ] { }0.007583

0 3 2 1 2 0 0.0005 0.033 2

0.0031R

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

(Support reaction corresponding to DR . i.e., reaction at B)

( )0.0116 0.045 0.0334 62 53 3

.3EI EI= − = − −=

[ ] [ ] [ ][ ]TRR MR M MRS C S C=

[ ]

4 2 0 0 0 0 1 32 4 0 0 0 0 1 30 0 2 1 0 0 1 6

1 3 1 3 1 6 1 6 0 00 0 1 2 0 0 1 630 0 0 0 4 2 00 0 0 0 2 4 0

EI

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

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

2EI⎡ ⎤= ⎢ ⎥⎣ ⎦



{ } { }

0 4 2 0 0 0 0 0 0 0 1 30 2 4 0 0 0 0 1 0 0 1 3

0.0075830 0 0 2 1 0 0 1 0 0 1 62 0.0005 0.030 0 0 1 2 0 0 0 1 0 1 66

0.003125 0 0 0 0 4 2 0 1 0 025 0 0 0 0 2 4 0 0 1 0

MEIA

−⎛⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎧ ⎫⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪= + + −⎜⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎩ ⎭⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝

⎞⎟⎟⎟⎟⎟⎟

⎜ ⎟⎜ ⎟⎠

83.755.455.440.3

40.30

⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪

⎨ ⎬−⎪⎪⎪⎩

=⎪⎪⎪⎭

Member end actions


61

[ ] [ ][ ] [ ]

FF FR

RF RR

S SS S

⎡ ⎤= ⎢ ⎥⎣ ⎦

6 1 0 3 21 6 2 1 20 2 4 033 2 1 2 0 3 2

E I−⎡ ⎤

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

2 0 0 20 1 1 0 0 0 4 0 0 20 0 0 1 1 0 2 1 0 1 20 0 0 0 0 1 1 2 0 1 231 3 1 3 1 6 1 6 0 0 0 4 2 0

0 2 4 0

EI

−⎡ ⎤⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎢ ⎥⎣ ⎦

[ ] [ ] [ ][ ]TJ MJ M MJS C S C=

4 2 0 0 0 0 0 0 0 1 30 1 1 0 0 0 2 4 0 0 0 0 1 0 0 1 30 0 0 1 1 0 0 0 2 1 0 0 1 0 0 1 60 0 0 0 0 1 0 0 1 2 0 0 0 1 0 1 631 3 1 3 1 6 1 6 0 0 0 0 0 0 4 2 0 1 0 0

0 0 0 0 2 4 0 0 1 0

EI

−⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ −⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

Alternatively, if the entire [SJ] matrix is assembled at a time,


62

[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦

0 0 0 1 3 1 1 3 0 01 0 0 1 3 0 1 3 0 01 0 0 0 0 1 6 1 6 00 1 0 0 0 1 6 1 6 00 1 0 0 0 0 1 3 1 30 0 1 0 0 0 1 3 1 3

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥

−⎢ ⎥⎣ ⎦[ ] [ ] [ ][ ]TJ MJ M MJS C S C=

0 0 01 3 1 1 3 0 0 4 2 0 0 0 0 0 0 01 3 1 1 3 0 01 0 01 3 0 1 3 0 0 2 4 0 0 0 0 1 0 01 3 0 1 3 0 01 0 0 0 0 1 6 1 6 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 00 1 0 0 0 1 6 1 6 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 030 1 0 0 0 0 1 3 1 3 0 0 0 0 4 2 0 1 0 0 0 0 1 30 0 1 0 0 0 1 3 1 3 0 0 0 0 2 4 0 0 1

T

EI

− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎢ ⎥

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

1 30 0 0 1 3 1 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

−⎢ ⎥⎣ ⎦

Alternatively, if ALL possible support settlements are accounted for,


63

[ ] [ ][ ] [ ]

FF FR

RF RR

S SS S

⎡ ⎤= ⎢ ⎥⎣ ⎦

0 1 1 0 0 00 0 0 1 1 0 2 0 0 2 4 2 0 00 0 0 0 0 1 4 0 0 2 2 2 0 0

1 3 1 3 0 0 0 0 2 1 0 0 0 1 2 1 2 01 0 0 0 0 0 1 2 0 0 0 1 2 1 2 031 3 1 3 1 6 1 6 0 0 0 4 2 0 0 0 2 20 0 1 6 1 6 1 3 1 3 0 2 4 0 0 0 2 20 0 0 0 1 3 1 3

EI

⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦⎢ ⎥

− −⎣ ⎦

6 1 0 2 2 3 2 1 2 01 6 2 0 0 1 2 3 2 20 2 4 0 0 0 2 22 0 0 4 3 2 4 3 0 02 0 0 2 4 2 0 033 2 1 2 0 4 3 2 3 2 1 6 01 2 3 2 2 0 0 1 6 3 2 4 30 2 2 0 0 0 4 3 4 3

EI

− −⎡ ⎤⎢ ⎥−⎢ ⎥

−⎢ ⎥⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− − − −⎢ ⎥⎢ ⎥− − −⎢ ⎥

− − −⎣ ⎦


64

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

10

6 1 0 0 2 2 3 2 1 2 0 01 6 2 25 0 0 1 2 3 2 2 0.03

3 30 2 4 25 0 0 0 2 2 0

0

EI EI−⎡ ⎤⎧ ⎫⎢ ⎥⎪ ⎪− −⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥= − − − −⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪⎜ ⎟ −⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦

20 4 2 0 0.0453 56004 24 12 25 0.015

116 32 12 35 25 0

20 4 2 843 4 24 12 3

1162 12 35 25

EI

EI

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

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

16423 108

116671

EI

−⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

0.0075830.00050.0031

−⎧ ⎫⎪ ⎪⎨ ⎬⎪⎩

=⎪⎭

Joint displacements


65

{ }

0 2 0 0 4 3 2 4 3 0 0 00.0075830 2 0 0 2 4 2 0 0 00.00050 3 2 1 2 0 4 3 2 3 2 1 6 0 0.03

3 30.003150 1 2 3 2 2 0 0 1 6 3 2 4 3 0

50 0 2 2 0 0 0 4 3 4 3 0

REI EIA

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

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


{ }

0 28.373 74.6670 28.373 1120 21.653 8450 19.973 9.333

46.29483.62762.34779.3136.5650 13.44 0

RA

−⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪= − + + =−⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪−⎨ ⎬⎪ ⎪⎪ ⎪⎪⎩ ⎭⎭ ⎪

Support reactions


66


{ }

0 4 2 0 0 0 0 0 0 0 1 3 1 1 3 0 00 2 4 0 0 0 0 1 0 0 1 3 0 1 3 0 0

0.0075830 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 02 0.00050 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 06

0.003125 0 0 0 0 4 2 0 1 0 0 0 0 1 3 1 325 0 0 0 0 2 4 0 0 1 0 0 0

MEIA

−⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫

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

00

0.0300

1 3 1 3

⎛ ⎞⎡ ⎤⎧ ⎫⎜ ⎟⎢ ⎥⎪ ⎪⎜ ⎟⎢ ⎥⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪−⎜ ⎟⎨ ⎬⎢ ⎥

⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥ ⎩ ⎭⎜ ⎟−⎣ ⎦⎝ ⎠

{ }

83.755.455.440.3

40.30

MA

⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪

⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

=

Member end actions


[ ]

2 4 1 4 0 0 0 01 4 2 4 0 0 0 00 0 6 4.8 3 4.8 0 0

20 0 3 4.8 6 4.8 0 00 0 0 0 4 8 2 80 0 0 0 2 8 4 8

MS EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

•Problem 4



85.33

42.67

1

2

3


68

• consists of member displacements due to unit displacements on the restrained structure.

[ ]MFC

1 1FD =2 1FD =

[ ]

1 00 10 00 10 10 0

MFC

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2=1 =1


69

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ } [ ] { }1F FF FD S A−= since there are no support displacements.

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C=

2 4 1 4 0 0 0 0 1 01 4 2 4 0 0 0 0 0 1

1 0 0 0 0 0 0 0 6 4.8 3 4.8 0 0 0 02

0 1 0 1 1 0 0 0 3 4.8 6 4.8 0 0 0 10 0 0 0 4 8 2 8 0 10 0 0 0 2 8 4 8 0 0

EI

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤

= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


70

8 44 8

1 0 0 0 0 0 0 200 1 0 1 1 0 0 108

0 80 4

EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤

= ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

2 11 92

EI ⎡ ⎤= ⎢ ⎥

⎣ ⎦

{ } [ ] { }1

1 2 1 01 9 85.3332F FF F

EID S A−

− ⎛ ⎞⎡ ⎤ ⎧ ⎫∴ = = = ⎨ ⎬⎜ ⎟⎢ ⎥

⎣ ⎦ ⎩ ⎭⎝ ⎠

36 4 0 341.3328 14 8 85.333 682.66

10.034272

91203 . 784 0EI EIEI

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

−

⎦

⎧ ⎫⎨ ⎬⎩ ⎭


71


0 2 4 1 4 0 0 0 0 1 00 1 4 2 4 0 0 0 0 0 1

42.667 0 0 6 4.8 3 4.8 0 0 0 1 10.0391285.333 0 0 3 4.8 6 4.8 0 0 0 0 20.078

0 0 0 0 0 4 8 2 8 0 10 0 0 0 0 2 8 4 8 0 0

EIEI

⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥

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


0 00 120.47

42.667 200.78185.333 401.568

0 160.6240 80.312

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

= +⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

015.05967.76535.138

20.07810.039

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎨= ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

Member end actions


72

•Homework 2:

32

1

80kNm

3m3m

4m

6m

20kN mC

A

B D


•Problem 5

[ ]

2 1 0 0 0 01 2 0 0 0 00 0 2 1 0 020 0 1 2 0 00 0 0 0 2 10 0 0 0 1 2

MEISL

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦



74

[ ]MFC =

0 0 11 0 11 0 00 1 00 1 10 0 1

LL

LL

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

12

3

DF1 DF2 DF3=1 =1 =1


75

2 1 0 0 0 0 0 0 11 2 0 0 0 0 1 0 1

0 1 1 0 0 00 0 2 1 0 0 1 0 020 0 0 1 1 00 0 1 2 0 0 0 1 0

1 1 0 0 1 10 0 0 0 2 1 0 1 10 0 0 0 1 2 0 0 1

LL

EIL

L L L LLL

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[ ] [ ] [ ][ ]TFF MF M MFS C S C=

1 0 32 0 3

0 1 1 0 0 02 1 020 0 0 1 1 01 2 0

1 1 0 0 1 10 2 30 1 3

LL

EIL

L L L LLL

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

2

4 1 32 1 4 3

3 3 12

LEI LL

L L L

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


76

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ }

1

2

4 1 3 02 1 4 3 0

3 3 12F

LEID LL

L L L P

−⎛ ⎞ ⎧ ⎫⎡ ⎤

⎪ ⎪⎜ ⎟⎢ ⎥= ⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎩ ⎭⎣ ⎦⎝ ⎠

{ } [ ] { }1F FF FD S A−= , since there are no support displacements.

2

13 3 3 03 13 3 0

843 3 5

LL LEI

L L L P

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

233

845

PLEI

L

−⎧ ⎫⎪ ⎪−⎨ ⎬⎪⎩

=⎪⎭

Joint displacements


77


{ }2

2 1 0 0 0 0 0 0 11 2 0 0 0 0 1 0 1

30 0 2 1 0 0 1 0 02 30 0 1 2 0 0 0 1 0 84

50 0 0 0 2 1 0 1 10 0 0 0 1 2 0 0 1

M

LL

EI PLAL EI

LLL

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

⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2

2 1 0 0 0 0 51 2 0 0 0 0 20 0 2 1 0 0 320 0 1 2 0 0 3840 0 0 0 2 1 20 0 0 0 1 2 5

EI PLL EI

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥−⎪ ⎪⎢ ⎥

= ⎨ ⎬⎢ ⎥ −⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

{ } [ ][ ]{ }M M MF FA S C D=

2

12992984

912

EI PLL EI

⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪

= ⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

433314

34

PL

⎧ ⎫⎪ ⎪⎪ ⎪−

=⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

Member end actions


78

[ ]0.2 0.0 0.00.0 0.2 0.00.0 0.0 0.2

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


•Problem 6:

1

2

3

50 kN

80 kN

5 m

5 m5 m

4 m 4 m

3 m

3 m


[ ]MFC =

0.8 -0.6-0.8 0.60.8 0.6

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1cos 3

6.87=0.8

36.87

1

23

0.8

10.6

53.13

cos 53.13=0.6

DF1 DF2=1 =1


80

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ }2.930 1.302 -501.302 5.208 -80FD ⎧ ⎫⎡ ⎤

= ⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦


-250.651-481.771⎧

=⎫

⎨ ⎬⎩ ⎭

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C= 0.384 -0.096

-0.096 0.216⎡ ⎤

= ⎢ ⎥⎣ ⎦


0.2 0.0 0.0 0.8 -0.6250.651

0.0 0.2 0.0 -0.8 0.6481.771

0.0 0.0 0.2 0.8 0.6

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


[ ][ ]{ }M MF FS C D=

17.70817.70897.917

⎧ ⎫⎪ ⎪⎨−⎪⎩

= ⎬− ⎪⎭

Member Forces:


82

•Problem 7:

[ ]0.2 0.0 0.00.0 0.5 0.00.0 0.0 0.2

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


1

2

3


[ ]MFC =

0.600 0.8000.000 -1.000-0.600 0.800

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1

53.13

5 m

cos 5

3.13=

0.6

0.6

1

053.13

5 m

036.87

cos 36.87=0.8

cos 36.87=0.8 0.8

DF1 DF2=1 =1


{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ }6.944 0.000 15.0000.000 1.323 -10.000FD ⎡ ⎤ ⎧ ⎫

= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭


104.167-13.228⎧

=⎫

⎨ ⎬⎩ ⎭

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C= 0.144 0.000

0.000 0.756⎡ ⎤

=⎢ ⎥⎣ ⎦


{ }0.2 0.0 0.0 0.600 0.800

104.1670.0 0.5 0.0 0.000 -1.000

-13.2280.0 0.0 0.2 -0.600 0.800

MA⎡ ⎤⎡ ⎤

⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦


{ } [ ][ ]{ }M M MF FA S C D=

10.46.6-14.6

⎧ ⎫⎪ ⎪⎨ ⎬⎪

=⎪⎩ ⎭

Member Forces:


[ ] 0.866 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.500

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


•Problem 8 (Homework 3):


1

0.8660.51

0.8660.5

[ ]MFC =

0.866 0.500 1.000 0.0000.500 -0.866

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2=1 =1


88

{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦

{ }0.577 -0.155 5-0.155 1.732 0FD⎡ ⎤ ⎧ ⎫

= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭


2.887-0.773⎧ ⎫⎨⎩

= ⎬⎭

Joint displacements

[ ] [ ] [ ][ ]TFF MF M MFS C S C= 1.774 0.158

0.158 0.591⎡ ⎤

= ⎢ ⎥⎣ ⎦


{ } 0.866 0.000 0.000 0.866 0.500

2.887 0.000 1.000 0.000 1.000 0.000

-0.773 0.000 0.000 0.500 0.500 -0.866

MA⎡ ⎤⎡ ⎤

⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦


{ } [ ][ ]{ }M M MF FA S C D=

1.8302.8871.057

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

=

Member Forces:


90

•Homework 4:

AE is constant.

20 kN

1 2 3

600

450

A B C

D

1m


91

• Development of stiffness matrices by physical approach –stiffness matrices for truss, beam and frame elements –displacement transformation matrix – development of total stiffness matrix - analysis of simple structures – plane truss beam and plane frame- nodal loads and element loads – lack of fit and temperature effects.

Stiffness method

Summary