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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN4

• 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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN6

•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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

• 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

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ] [ ]

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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•Grid member

[ ]

3 2

2

12 60

0 0

6 40

Mi

EI EIL L

GJSL

EI EIL L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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=

Dept. of CE, GCE Kannur Dr.RajeshKN

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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, 

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

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Important formulae:

Joint displacements:

Member end actions:

Support reactions:

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

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

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

Dept. of CE, GCE Kannur Dr.RajeshKN

Joint displacements

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

0 1 1 0

0 0 0 1

[ ]

0 01 01 00 1

M FC

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2=1 =1

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1 L 1 L 0 0 1 0 0 0

1 L− 1 L− 1 L 1 L 0 0 1 L− 1 L−

Dept. of CE, GCE Kannur Dr.RajeshKN

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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

Dept. of CE, GCE Kannur Dr.RajeshKN

{ }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

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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

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∴ = − +

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ] [ ] [ ][ ]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

−⎡ ⎤⎢ ⎥−⎢ ⎥=− − −⎢ ⎥⎢ ⎥−⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

=

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

{ }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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42

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,

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

44

[ ]

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

MEIS

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

[ ] 2 121 24Mi

EIS ⎡ ⎤= ⎢ ⎥

⎣ ⎦Member stiffness matrix of beam member

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

1 0 0 0

0 1 1 0

DF1 =1

DF2 =1 L

Dept. of CE, GCE Kannur Dr.RajeshKN

47

[ ]

1 0 00 1 00 1 00 0 1

M FC

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2 DF3=1 =1 =1

0 0 0 1

DF3 =1

Dept. of CE, GCE Kannur Dr.RajeshKN

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∴

Dept. of CE, GCE Kannur Dr.RajeshKN

{ }

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

−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−⎩ ⎭

=

Dept. of CE, GCE Kannur Dr.RajeshKN

50

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

{ }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

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪−⎩ ⎭

=

is a null matrix{ }RD

Member end actions

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

Dept. of CE, GCE Kannur Dr.RajeshKN

•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

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

[ ] 2 121 2Mi

EISL

⎡ ⎤= ⎢ ⎥

⎣ ⎦Member stiffness matrix of beam member

Dept. of CE, GCE Kannur Dr.RajeshKN

Fixed end moments

2525

Equivalent joint loads

2525

Support settlements { }RD(Restraint displacements)

A

BC D30mm 1RD

Free (unknown) joint displacements { }FD

1FD 2FD3FD

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN54

[ ] [ ] [ ]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

Dept. of CE, GCE Kannur Dr.RajeshKN

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⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ] [ ] [ ][ ]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−⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

Dept. of CE, GCE Kannur Dr.RajeshKN

{ }

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−= −⎡ ⎤⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

58

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

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= −

Dept. of CE, GCE Kannur Dr.RajeshKN

{ } { } [ ] { }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⎡ ⎤= ⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN60

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

{ } { }

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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,

Dept. of CE, GCE Kannur Dr.RajeshKN

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,

Dept. of CE, GCE Kannur Dr.RajeshKN

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

− −⎡ ⎤⎢ ⎥−⎢ ⎥

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

− − −⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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

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

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

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

{ }

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

Dept. of CE, GCE Kannur Dr.RajeshKN

66

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

{ }

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

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ]

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

Unassembled stiffness matrix

Equivalent joint loads

85.33

42.67

1

2

3

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤

= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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

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

−

⎦

⎧ ⎫⎨ ⎬⎩ ⎭

Dept. of CE, GCE Kannur Dr.RajeshKN

71

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

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

⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥

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

is a null matrix{ }RD

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

Dept. of CE, GCE Kannur Dr.RajeshKN

72

•Homework 2:

32

1

80kNm

3m3m

4m

6m

20kN mC

A

B D

Dept. of CE, GCE Kannur Dr.RajeshKN

•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

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

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

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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

Dept. of CE, GCE Kannur Dr.RajeshKN

77

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

{ }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

Dept. of CE, GCE Kannur Dr.RajeshKN

78

[ ]0.2 0.0 0.00.0 0.2 0.00.0 0.0 0.2

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

•Problem 6:

1

2

3

50 kN

80 kN

5 m

5 m5 m

4 m 4 m

3 m

3 m

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ]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

Dept. of CE, GCE Kannur Dr.RajeshKN

80

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

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

= ⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦

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

-250.651-481.771⎧

=⎫

⎨ ⎬⎩ ⎭

Joint displacements

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

-0.096 0.216⎡ ⎤

= ⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

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 ML M MF F MR RA A S C D C D= + +

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

17.70817.70897.917

⎧ ⎫⎪ ⎪⎨−⎪⎩

= ⎬− ⎪⎭

Member Forces:

Dept. of CE, GCE Kannur Dr.RajeshKN

82

•Problem 7:

[ ]0.2 0.0 0.00.0 0.5 0.00.0 0.0 0.2

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

1

2

3

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ]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

Dept. of CE, GCE Kannur Dr.RajeshKN84

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

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

= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭

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

104.167-13.228⎧

=⎫

⎨ ⎬⎩ ⎭

Joint displacements

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

0.000 0.756⎡ ⎤

=⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

{ }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 ML M MF F MR RA A S C D C D= + +

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

10.46.6-14.6

⎧ ⎫⎪ ⎪⎨ ⎬⎪

=⎪⎩ ⎭

Member Forces:

Dept. of CE, GCE Kannur Dr.RajeshKN

[ ] 0.866 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.500

MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled stiffness matrix

•Problem 8 (Homework 3):

Dept. of CE, GCE Kannur Dr.RajeshKN

1

0.8660.51

0.8660.5

[ ]MFC =

0.866 0.500 1.000 0.0000.500 -0.866

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

DF1 DF2=1 =1

Dept. of CE, GCE Kannur Dr.RajeshKN

88

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

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

= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭

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

2.887-0.773⎧ ⎫⎨⎩

= ⎬⎭

Joint displacements

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

0.158 0.591⎡ ⎤

= ⎢ ⎥⎣ ⎦

Dept. of CE, GCE Kannur Dr.RajeshKN

{ } 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 ML M MF F MR RA A S C D C D= + +

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

1.8302.8871.057

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

=

Member Forces:

Dept. of CE, GCE Kannur Dr.RajeshKN

90

•Homework 4:

AE is constant.

20 kN

1 2 3

600

450

A B C

D

1m

Dept. of CE, GCE Kannur Dr.RajeshKN

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