Analysis of Indeterminate Structure Session 05-22 Matakuliah: S0725 – Analisa Struktur Tahun: 2009

Analysis of Indeterminate Structure Session 05-22

Matakuliah : S0725 – Analisa StrukturTahun : 2009

Bina Nusantara University 3

Contents

•3 Equations Method•Flexibility Method•Slope Deflection Method•Cross Method/ Moment Distribution Method•Matrix Analysis


3 Equation Method

This method will be used for analizing the indeterminate structure ( support reaction, internal loads )


3 Equation Method

Degree of indeterminacy i = r – 3 i = 7 – 3 = 4

A

B C D

E

P1 q2P2q1

Gambar 1.1 . Struktur statis tak tentu


3 Equation Method

Pada bagian konstruksi diantara 2 perletakan yang berdekatan diberikan kebebasan untuk berputar sudut satu sama lain. Sebagai akibatnya akan timbul CELAH pada balok di tempat tumpuan ( perletakan ) sebagai akibat dari adanya beban luar. ( Gambar 1.2.a )

Principles :

llk

P1 P2q1

celahokro

kl

okr o

kl

lk

k - 1 k + 1

Gambar 1.2.a. Putran sudut akibat beban luar


3 Equation Method

Pada hakekatnya balok ini adalah menerus , utuh dan tidak boleh ada celah , maka harus ada MOMEN pada tumpuan / perletakan antara yang berfungsi mengembalikan celah tadi menjadi utuh kembali. Sebagai gaya statis kelebihan harus dipilih berupa momen-momen pada perletakan antara, umumnya momen – momen yang bekerja adalah MOMEN NEGATIF. ( Gambar 1.2.b )

Princiles :

llk

'kr'kl

Mk

lkk - 1 k + 1

Mk-1 Mk+1

Gambar 1.2.b. Momen lentur negatif


3 Equation MethodCompatibility eq :Sebagai suatu syarat kompatibilitas , joint ‘k’ merupakan rotasi yang kompatibel, sehingga persamaan kompatibilitasnya menjadi

okl + o

kr = ’kl + ’kr

Dimana : o dan ’ diambil nilai harga mutlaknya.

Sehingga harga Mk yang positif berarti bekerja sebagai momen lentur negatif , dimana ini berarti bahwa kita harus MERUBAH tanda gaya statis kelebihan yang didapatkan.


3 Equation MethodPERSYARATAN KOMPATIBILITAS :Jika tidak ingin merubah tanda tersebut maka harus dimasukkan anggapan – anggapan bahwa momen peralihan merupakan momen lentur positif, sehingga persamaan kompatibilitas menjadi :

( o

kl + o

kr ) + ( ’kl + ’kr ) = 0

Sehingga hasil-hasil momen peralihan sudah langsung berikut tandanya menyatakan MOMEN LENTUR SEBENARNYA , jika hasilnya negatif , berarti bekerja sebagai momen lentur negatif.


3 Equation MethodKarena

nilai ’kl tergantung pada Mk-1 dan Mk

nilai ’kr tergantung pada Mk dan Mk+1

m a k a ...

dari persamaan kompatibilitas di atas akan selalu didapatkan maksimum sebanyak 3 momen tumpuan antara yang terlibat dalam persamaan kompatibilitas , sehingga metoda ini disebut juga

PERSAMAAN TIGA MOMEN


3 Equation MethodDengan mennjau nilai putaran sudut akibat momen – momen peralihan persamaan kompatibilitas dapat dituliskan sebagai

’kl = Mk ( ll / 3EI ) + Mk-1 ( ll / 6EI )

’kr = Mk ( lr / 3EI ) + Mk+1 (lr / 6EI ) +

’kl + ’kr = Mk-1 ( ll / 6EI )+Mk {( ll / 3EI )+( lr / 3EI )}+

Mk+1 (lr / 6EI )


3 Equation MethodSehingga persamaan 3 momen pada tumpuan k ...

Mk-1 ( ll / 6EI )+Mk {( ll / 3EI )+( lr / 3EI )}+ Mk+1 (lr / 6EI ) +

( okl + o

kr ) = 0

Agar persamaan ini dapat digunakan secara efisien dan tepat maka diperlukan rumus – rumus dari berbagai type beban , baik dari beban luar maupun momen peralihan ( momen pada tumpuan antara ).


3 Equation Method

C a t a t a n ...

Jika pada perletakan ujung – ujung adalah SENDI & ROL, maka i jumlah perletakan antara. Sedangkan jika perletakan ujung adalah JEPIT, maka perletakan jepit ini diperlakukan sebagai perletakan antara , dengan MENGUBAH menjadi SENDI dan Momen Jepit sebagai gaya kelebihan . ( Gambar 1.3 )


3 Equation Method

M

Gambar 1.3. Struktur dengan perletakan Jepit yang diubah Sendi + Momen jepit


While analyzing indeterminate structures, it is necessary to satisfy (force) equilibrium, (displacement) compatibility and force-displacement relationships

Method Analysis


(a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads(b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps(c) Force-displacement requirements depend on the manner the material of the structure responds to the applied loads, which can be linear/nonlinear/viscous and elastic/inelastic; for our study the behavior is assumed to be linear and elastic

Method Analysis


(a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads

Method Analysis


(b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps

Method Analysis


(c) Force-displacement requirements depend on the manner the material of the structure responds to the applied loads, which can be linear/nonlinear/viscous and elastic/inelastic; for our study the behavior is assumed to be linear and elastic

Method Analysis


Two methods are available to analyze indeterminate structures, depending on whether we satisfy force equilibrium or displacement compatibility conditions – They are:

Force method and

Displacement Method

Method Analysis


Force Method satisfies displacement compatibility and force-displacement relationships; it treats the forces as unknowns – Two methods which we will be studying are Method of Consistent Deformation and (Iterative Method of) Moment Distribution

Method Analysis


Displacement Method satisfies force equilibrium and force-displacement relationships; it treats the displacements as unknowns – Two available methods are Slope Deflection Method and Stiffness (Matrix) method

Method Analysis


Method Analysis

Method Unknowns Equations used for solution

Coefficient of the unknowns

Forced Method

Forces Compatibility and force displacement

Flexibility Coefficients

Displacement Method

Displacements

Equilibrium and force displacement

Stiffness Coefficients


Two methods are available to analyze indeterminate structures, depending on whether we satisfy force equilibrium or displacement compatibility conditions - They are: Force method and Displacement Method

• Force Method satisfies displacement compatibility and force-displacement relationships; it treats the forces as unknowns - Two methods which we will be studying are Method of Consistent Deformation and (Iterative Method of) Moment Distribution

• Displacement Method satisfies force equilibrium and force-displacement relationships; it treats the displacements as unknowns - Two available methods are Slope Deflection Method and Stiffness (Matrix) method

Flexibility Method


FORCED METHOD

This method is also known as flexibility method or compatibility method.

In this method the degree of static

indeterminacy of the structure is

determined and the redundants are

identified.

Flexibility Method

VTU Programme


FORCED METHOD

A coordinate is assigned to each

redundant.

Thus,P1, P2 - - - - - -Pn are the redundants

at the coordinates 1,2, - - - - - n.If all the redundants are removed , the resulting structure known as released structure, is statically determinate

Flexibility Method

VTU Programme


FORCED METHODThis released structure is also known as basic determinate structure. From the principle of super position the net displacement at any point in statically indeterminate structure is some of the displacements in the basic structure

due to the applied loads and the redundants. This is known as the compatibility condition and may be expressed by the equation;

Flexibility Method

VTU Programme


FORCED METHOD

∆1 = ∆1L + ∆1R Where ∆1 - - - - ∆n = Displ. At Co-ord.at 1,2 - -n∆2 = ∆2L + ∆2R ∆1L ---- ∆nL = Displ.At Co-ord.at 1,2 - - - - -n | | | Due to aplied loads| | | ∆1R ----∆nR = Displ.At Co-ord.at 1,2 - - - - -n ∆n = ∆nL + ∆nR Due to Redudants

Flexibility Method

VTU Programme


FORCED METHOD

The above equations may be return as [∆] = [∆L] + [∆R] - - - - (1)

∆1 = ∆1L + δ11 P1 + δ12 P2 + - - - - - δ1nPn

∆2 = ∆2L + δ21 P1 + δ22 P2 + - - - - - δ2nPn

| | | | | | | | | | - - - - - (2)

∆n = ∆nL + δn1 P1 + δn2 P2 + - - - - - δnnPn

Flexibility Method

VTU Programme


FORCED METHOD∆ = [∆ L] + [δ] [P] - - - - - - (3)

[P]= [δ]-1 {[∆] – [∆ L]} - - - - - - (4)

If the net displacements at the redundants are zero then

∆1, ∆2 - - - - ∆n =0,

Then [P] = - [δ] -1 [∆ L] - - - - - -(5)

The redundants P1,P2, - - - - - Pn are Thus determined

Flexibility Method

VTU Programme


Slope Deflection• Slope deflection equations

The slope deflection equations express the member end moments in terms of rotations angles. The slope deflection equations of member ab of flexural rigidity EIab and length Lab are:



where θa, θb are the slope angles of ends a and b respectively, Δ is the relative lateral displacement of ends a and b. The absence of cross-sectional area of the member in these equations implies that the slope deflection method neglects the effect of shear and axial deformations.



The slope deflection equations can also be written using the stiffness factor .

and the chord rotation


Slope Deflection



When a simple beam of length Lab and flexural rigidity EIab is loaded at each end

with clockwise moments Mab and Mba, member end rotations occur in the same direction. These rotation angles can be calculated using the unit dummy force method or the moment-area theorem


Slope Deflection



Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. be in equilibrium. Therefore,


FORCED METHOD∆ = [∆ L] + [δ] [P] - - - - - - (3)

[P]= [δ]-1 {[∆] – [∆ L]} - - - - - - (4)

If the net displacements at the redundants are zero then

∆1, ∆2 - - - - ∆n =0,

Then [P] = - [δ] -1 [∆ L] - - - - - -(5)

The redundants P1,P2, - - - - - Pn are Thus determined

Slope Deflection Method

VTU Programme


Consider portion AB of a continuous beam, shown below, subjected to a distributed load w(x) per unit length and a support settlement of at B; EI of the beam is constant.



A B

B

A

= rigid body motion = /L

=

FEMAB

FEMBA

MBA=(2EIA)/L

MAB=(4EIA)/L

A

(i) Due to externally applied loads

(ii) Due to rotation A at support A

+



MBA=(4EIB)/L

B

MAB=(2EIB)/L

+

A B(iii) Due to rotation B at support B

L

MAB=(-6EI)/L2

A B

L

MBA=(-6EI)/L2

+

(iv) Due to differential settlement of (between A and B)



BA(FEM))L

3ΔB2θA(θ

L

2EI2

L

6EIΔ

L

B4EIθ

L

A2EIθ

BA(FEM)BAMBAMBAMBA(FEM)BAM

AB(FEM))L

3ΔBθA(2θ

L

2EI2

L

6EIΔ

L

B2EIθ

L

A4EIθ

AB(FEM)ABMABMABMAB(FEM)ABM



A

B

=

(FEM)AB

(FEM)BA

(FEM)BA/2

(FEM)BA

+



+

A

MAB=(3EIA)/L

+

MAB=(3EI)/L2

MAB = [(FEM)AB - (FEM)BA/2]+(3EIA)/L -(3EI)/L2

Modified FEM at end A

= PL3/(3EI),M = PL = (3EI/L3)(L) = 3EI/L2









This method is at the core of the moment distribution method, and is also very powerful.

Consider a beam of length L, subjected to end moments (clockwise positive), and downward transverse loads either distributed or concentrated

The end slopes are θA θB.





)2...(..........M)2(L

EI2M

)1...(..........M)2(L

EI2M

FBBABA

FABAAB

FBFA M,Mare the numerical values of the fixed-end moments, e.g. wL2/8, PL/8, Pab2/L2, etc…





0MMand,0 BCBACA

kNm5.6216

PLM

andEI64

PL

8

PL

L

EI8

)2(L

EI2M

8

PL)2(

L

EI2M

BC

2

BB

BBC

BBA

Here,



(Method developed by Prof. Hardy Cross in 1932)The method solves for the joint moments in continuous beams andrigid frames by successive approximation.

Moment Distribution Method


Thus the Moment Distribution Method (also known as the Cross Method) became the preferred calculation technique for reinforced concrete structures.



The description of the moment distribution method by Hardy Cross is a little masterpiece. He wrote: "Moment Distribution. The method of moment distribution is this:• Imagine all joints in the structure held so that they cannot rotate and compute the moments at the ends of the members for this condition;•at each joint distribute the unbalanced fixed-end moment among the connecting members in proportion to the constant for each

member defined as "stiffness";



•multiply the moment distributed to each member at a joint by the carry-over factor at the end of the member and set this product at the other end of the member; •distribute these moments just

"carried over";



•repeat the process until the moments to be carried over are small enough to be neglected; and•add all moments - fixed-end moments, distributed moments, moments carried over - at each end of each member to obtain the true moment at the end." [Cross 1949:2]



1. Restrain all possible displacements.

2. Calculate Distribution Factors:

The distribution factor DFi of a member connected to any joint J is

where S is the rotational stiffness , and is given by



3. Determine carry-over factorsThe carry-over factor to a fixed end is always 0.5, otherwise it is 0.0.

4. Calculate Fixed End Moments. These could be due to in-span loads, temperature variation and/or relative displacement between the ends of a member.



5. Do distribution cycles for all joints simultaneously Each cycle consists of two steps:1. Distribution of out of balance moments Mo,2.Calculation of the carry over moment at the far end of each member.

The procedure is stopped when, at all joints, the out of balance moment is a negligible value. In this case, the joints should be balanced and no carry-over moments are calculated.

6. Calculate the final moment at either end of each member.

This is the sum of all moments (including FEM) computed during the distribution cycles.



Consider the continuous beam ABCD, subjected to the given loads,as shown in Figure below. Assume that only rotation of joints occurat B, C and D, and that no support displacements occur at B, C andD. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D.




15 kN/m 10 kN/m

8 m 6 m 8 m

A B C DI I I

3 m

PROBLEMS


Step I

The joints B, C and D are locked in position before any load is applied on the beam ABCD; then given loads are applied on the beam. Since the joints of beam ABCD are locked in position, beams AB, BC and CD acts as individual and separate fixed beams, subjected to the applied loads; these loads develop fixed end moments.

8 m

-80 kN.m -80 kN.m15 kN/m

A B

6 m

-112.5kN.m 112.5 kN.m

B C 8 m

-53.33 kN.m10 kN/m

C D

150 kN53.33 kN.m

3 m



Step I

The joints B, C and D are locked in position before any load is applied on the beam ABCD; then given loads are applied on the beam. Since the joints of beam ABCD are locked in position, beams AB, BC and CD acts as individual and separate fixed beams, subjected to the applied loads; these loads develop fixed end moments.



8 m

-80 kN.m -80 kN.m15 kN/m

A B

6 m

-112.5kN.m 112.5 kN.m

B C 8 m

-53.33 kN.m10 kN/m

C D

150 kN53.33 kN.m

3 m



In beam ABFixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.mFixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m

In beam BCFixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62 = -112.5 kN.mFixed end moment at C = + (Pab2)/l2 = + (150)(3)(3)2/62 = + 112.5 kN.mIn beam AB

Fixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.mFixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m



Step II

Since the joints B, C and D were fixed artificially (to compute the the fixed-end moments), now the joints B, C and D are released and allowed to rotate. Due to the joint release, the joints rotate maintaining the continuous nature of the beam. Due to the joint release, the fixed end moments on either side of joints B, C and D act in the opposite direction now, and cause a net unbalanced moment to occur at the joint



15 kN/m 10 kN/m

8 m 6 m 8 m

A B C DI I I

3 m

150 kN

Released moments -80.0 -112.5 +53.33 -53.33+112.5

Net unbalanced moment+32.5 -59.17 -53.33



Step III

These unbalanced moments act at the joints and modify the joint moments at B, C and D, according to their relative stiffnesses at the respective joints. The joint moments are distributed to either side of the joint B, C or D, according to their relative stiffnesses. These distributed moments also modify the moments at the opposite side of the beam span, viz., at joint A in span AB, at joints B and C in span BC and at joints C and D in span CD. This modification is dependent on the carry-over factor (which is equal to 0.5 in this case); when this carry over is made, the joints on opposite side are assumed to be fixed.



Step IV

The carry-over moment becomes the unbalanced moment at the joints to which they are carried over. Steps 3 and 4 are repeated till the carry-over or distributed moment becomes small.



Step V

Sum up all the moments at each of the joint to obtain the joint moments.



Stiffness and Carry-over Factors

Stiffness = Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions

A

MAMB

A BA

RA RB

L

E, I – Member properties

A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Find A and MB.



Using method of consistent deformations

L

A

A

MA

B

L

fAA

A

B

1

EI

LM AA 2

2

EI

Lf AA 3

3



Applying the principle of consistent deformation,

L

MRfR A

AAAAA 2

30

EI

LM

EI

LR

EI

LM AAAA 42

2

L

EIMkhence

L

EIM

A

AAA

4;

4

Stiffness factor = k = 4EI/L




Considering moment MB,

MB + MA + RAL = 0MB = MA/2= (1/2)MA

Carry - over Factor = 1/2


Distribution Factor

Distribution factor is the

ratio according to which an externally applied unbalanced moment M at a joint is apportioned to the various members mating at the joint



Distribution Factor moment M

A CB

D

I1

L1I3

L3

I2

L2

A

D

B CMBA

MBC

MBD

At joint BM - MBA-MBC-MBD = 0

M



Distribution Factor


i.e., M = MBA + MBC + MBD

MFDMK

KM

MFDMK

KM

Similarly

MFDMK

KKM

K

M

KKK

M

KKK

L

IE

L

IE

L

IE

BDBD

BD

BCBC

BC

BABA

BBABA

BDBCBAB

BBDBCBA

B

).(

).(

).(

444

3

33

2

22

1

11



Modified Stiffness Factor

The stiffness factor changes when the far end of the beam is simply-supported.

AMA

A B

RA RB

L



Modified Stiffness Factor

As per earlier equations for deformation, given in Mechanics of Solids text-books.

fixedAB

A

AAB

AA

K

L

EI

L

EIMK

EI

LM

)(4

3

4

4

333



Solution of that problems above

Fixed end moments

mkNwl

MM

mkNwl

MM

mkNwl

MM

DCCD

CBBC

BAAB

.333.5312

)8)(10(

12

.5.1128

)6)(150(

8

.8012

)8)(15(

12

22

22




Stiffness Factors (Unmodified Stiffness)

EIEI

K

EIEIEI

K

EIEI

L

EIKK

EIEI

L

EIKK

DC

CD

CBBC

BAAB

5.08

4

5.08

4

8

4

667.06

))(4(4

5.08

))(4(4




Distribution Factors

00.1

4284.0500.0667.0

500.0

5716.0500.0667.0

667.0

5716.0667.05.0

667.0

4284.0667.05.0

5.0

0.0)(5.0

5.0

DC

DC

DC

CDCB

CD

CD

CDCB

CB

CB

BCBA

BC

BC

BCBA

BA

BA

wallBA

BA

AB

K

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

stiffnesswall

EI

KK

KDF



Solution of that problems aboveMoment Distribution Table

Joint A B C D

Member AB BA BC CB CD DC

Distribution Factors 0 0.4284 0.5716 0.5716 0.4284 1

Computed end moments -80 80 -112.5 112.5 -53.33 53.33Cycle 1

Distribution 13.923 18.577 -33.82 -25.35 -53.33

Carry-over moments 6.962 -16.91 9.289 -26.67 -12.35Cycle 2

Distribution 7.244 9.662 9.935 7.446 12.35

Carry-over moments 3.622 4.968 4.831 6.175 3.723Cycle 3

Distribution -2.128 -2.84 -6.129 -4.715 -3.723

Carry-over moments -1.064 -3.146 -1.42 -1.862 -2.358Cycle 4

Distribution 1.348 1.798 1.876 1.406 2.358

Carry-over moments 0.674 0.938 0.9 1.179 0.703Cycle 5

Distribution -0.402 -0.536 -1.187 -0.891 -0.703

Summed up -69.81 99.985 -99.99 96.613 -96.61 0moments



Solution of that problems aboveComputation of Shear Forces

8 m 3 m 3 m 8 mI I I

15 kN/m 10 kN/m150 kN

AB C

D


Simply-supported 60 60 75 75 40 40

reaction

End reaction

due to left hand FEM 8.726 -8.726 16.665 -16.67 12.079 -12.08

End reaction

due to right hand FEM -12.5 12.498 -16.1 16.102 0 0

Summed-up 56.228 63.772 75.563 74.437 53.077 27.923 moments


Solution of that problems aboveComputation of Shear Forces



Solution of that problems aboveShear Force and Bending Moment Diagrams

56.23

3.74 m

75.563

63.77

52.077

74.437

27.923

2.792 m

S. F. D.



Solution of that problems aboveShear Force and Bending Moment Diagrams

-69.80698.297

35.08

126.704

-96.613

31.693

Mmax=+38.985 kN.mMax=+ 35.59 kN.m

3.74 m84.92

-99.985

48.307

2.792 m

B. M. D




Simply-supported bending moments at center of span

Mcenter in AB = (15)(8)2/8 = +120 kN.m

Mcenter in BC = (150)(6)/4 = +225 kN.m

Mcenter in AB = (10)(8)2/8 = +80 kN.m


The section will discuss moment distribution method to analyze beams and frames composed of non prismatic members.


MOMENT DISTRIBUTION METHOD FOR NONPRISMATIC MEMBER


First the procedure to obtain the necessary carry-over factors, stiffness factors and fixed-end moments will be outlined. Then the

use of values given in design tables will be illustrated. Finally the analysis of statically indeterminate structures using the moment distribution method will be outlined


MOMENT DISTRIBUTION METHOD FOR NONPRISMATIC MEMBER




Use moment-area method to find the stiffness and carry-over factors of the non-prismatic beam.

A

B

PA MB

MA

A

AABAA KP )( AABB

AABA

MCM

KM

CAB= Carry-over factor of moment MA from A to B




A

B

MA=CBAMB

=CBAKBMB(KB)MB=CABMA

=CABKA

MA(KA)

A (= 1.0)MA B (= 1.0)

AB

(a) (b)

MB


Use of Betti-Maxwell’s reciprocal theorem requires that the work done by loads in case (a) acting through displacements in case (b) is equal to work done by loads in case (b) acting through displacements in case (a)

BA

BABA

KK

MMMM

BAAB CC

)0.0()0.1()1()0(





















Stiffness-Factor Modification











































Symmetric Beam and Loading



Symmetric Beam with Antisymmetric Loading









Moment Distribution for frames: No sidesway













Moment Distribution for frames: sidesway























Matrix Analysis

D= δ x P

D=PL3/3EI = δ x P

δ=L3 /3EI

δ=Flexibility Coeff.

P=K x D

P=K x PL3 / 3EI

K=3EI / L3

K=Stiffness Coeff.

FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD.

VTU Programme


Matrix AnalysisFLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD.

D= ML / EI

D= δ x M=ML/EI

δ=L / EI

δ=Flexibility Coeff.

M=K x D =K x ML/EI

K=EI / L

K=Stiffness Coeff.

δ X K= 1

VTU Programme


Matrix AnalysisFLEXIBILITY AND STIFFNESS MATRICES : TWO CO-ORDINATE SYSTEM

VTU Programme


Matrix AnalysisFLEXIBILITY AND STIFFNESS MATRICES : TWO CO-ORDINATE SYSTEM

δ11=L3 / 3EI δ21=L2 / 2EI

δ12=δ21 =L2 / 2EI δ22=L / EI

Unit Force At Co-ord.(1)

Unit Force At Co-ord.(2)

VTU Programme


STIFFNESS MATRIX

Unit Displacement at (1)

K11=12EI / L3

K21= – 6EI / L2Forces at Co-ord.(1) & (2)

Unit Displacement at (2) Forces at Co-ord.(1) & (2)

K12= – 6EI / L2

K22=4EI / L

=P1=K11D1+K12D2

P2=K21D1+K22D2

P1

P2

K11 K12

K21 K22

D1

D2

K = 12EI / L3– 6EI / L2

– 6EI / L24EI / L

Matrix Analysis

VTU Programme


Develop the Flexibility and stiffness matrices for frame ABCD with reference to Coordinates shown

The Flexibility matrix can be developed by applying unit force successively at coordinates (1),(2) &(3) and evaluating the displacements at all the coordinates

δij = ∫ mi mj / EI x ds

δij =displacement at I due to unit load at j

Matrix Analysis

VTU Programme


Unit Load at (1) Unit Load at (2) Unit Load at (3)

Portion DC CB BA

I I 4I 4I

Origin D C B

Limits 0 - 5 0 - 10 0 - 10

m1 x 5 5 - x

m2 0 x 10

m3 - 1 -1 -1

Matrix Analysis

VTU Programme


δ11 = ∫m1.m1dx / EI = 125 / EI

δ21 = δ12 = ∫m1.m2dx / EI = 125 / 2EI

δ31 = δ13 = ∫m1.m3dx / EI = -25 / EI

δ22 = ∫m2.m2dx / EI = 1000 / 3EI

δ23 = δ32 = ∫m2.m3dx / EI = -75 / 2EI

δ33 = ∫m3.m3dx / EI = 10 / EI

δ = 1 / 6EI

750 375 -150375 2000 -225

-150 -225 60

Matrix Analysis

VTU Programme


INVERSING THE FLEXIBILITY MATRIX [ δ ]

THE STIFENESS MATRIX [ K ] CAN BE OBTAINED

K = EI

0.0174 0.0028 0.0541

0.0028 0.0056 0.0282

0.0541 0.0282 0.3412

Matrix Analysis

VTU Programme

Documents

Analysis of Indeterminate Structure Session 05-22 Matakuliah: S0725 – Analisa Struktur Tahun: 2009