Moment Distribution

47
Moment Distribution ERT 348 Controlled Environment Design 1 Ms Siti Kamariah Binti Md Sa’at School of Bioprocess Engineering, UniMAP [email protected]

description

Equivalent Frame Design

Transcript of Moment Distribution

Page 1: Moment Distribution

Moment Distribution

ERT 348 Controlled Environment Design 1

Ms Siti Kamariah Binti Md Sa’atSchool of Bioprocess Engineering,

[email protected]

Page 2: Moment Distribution

Introduction

Such structures are indeterminate, i.e. there are more unknown variables than can be solved using only the three equations of equilibrium.

This section deals with continuous beams and propped cantilevers.

An American engineer, Professor Hardy Cross, developed a very simple, elegant and practical method of analysis for such structures called Moment Distribution.

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Indeterminate structures

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Bending (Rotational) Stiffness

A fundamental relationship which exists in the elastic behaviour of structures and structural elements is that between an applied force system and the displacements which are induced by that system,

Force = Stiffness x DisplacementP = k x δ

Stifness, k = P/δwhen δ = 1.0 (i.e. unit displacement) the

stiffness is: ‘the force necessary to maintain a UNIT displacement, all other displacements being equal to zero.’

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Bending (Rotational) Stiffness

The displacement can be a shear displacement, an axial displacement, a bending (rotational) displacement or a torsional displacement, each in turn producing the shear, axial, bending or torsional stiffness.

When considering beam elements in continuous structures using the moment distribution method of analysis, the bending stiffness is the principal characteristic which influences behaviour.

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Example:

The force (MA) necessary to maintain this displacement can be shown (e.g. Using McCaulay’s Method) to be equal to (4EI)/L.

If the bending stiffness of the beam is equal to (Force/1.0), therefore k = (4EI)/L.

This is known as the absolute bending stiffness of the element.

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General Principles & Definition

Member stiffness factor

Joint stiffness factor The total stiffness factor of joint A is

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

L

EIK

4

10000100050004000 KKT

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General Principles & Definition

Member relative stiffness factor Quite often a continuous beam or a frame will

be made from the same material E will therefore be constant

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

L

IKR

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Carry-Over Moment

When the beam element deforms due to the applied rotation at end A, an additional moment (MB) is also transferred by the element to the remote end if it has zero slope (i.e. is fixed).

The moment MB is known as the carry-over moment.

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Carry-Over Moment

Carry-over (CO) factor

Solving for and equating these eqn,

The moment M at the pin induces a moment of M’ = 0.5M at the wall

In the case of a beam with the far end fixed, the CO factor is +0.5

AA 2

; 4

L

EIM

L

EIM BA

AB MM 5.0

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Carry-Over Moment

Therefore be stated that ‘if a moment is applied to one end of a beam then a moment of the same sense and equal to half of its value will be transferred to the remote end provided that it is fixed.’

If the remote end is ‘pinned’, then the beam is less stiff and there is no carry-over moment.

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Pinned End

‘the stiffness of a pin-endedbeam is equal to ¾ × the stiffness of a

fixed-end beam.’

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Free Bending Moment

When a beam is free to rotate at both ends, no bending moment can develop at the supports, then the bending moment diagram resulting from the applied loads on the beam is known as the Free Bending Moment Diagram.

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Fixed Bending Moment

When a beam is fixed at the ends (encastre) such that it cannot rotate, i.e. zero slope at the supports, then bending moments are induced at the supports and are called Fixed-End Moments.

The bending moment diagram associated only with the fixed-end moments is called the Fixed Bending Moment Diagram.

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Fixed Bending Moment

+

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Example calculation 1:

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1

2

1 2

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Propped Cantilever

The fixed-end moment for propped cantilevers (i.e. one end fixed and the other end simply supported) can be derived from the standard values given for encastre beams as follows.

Consider the propped cantilever, which supports a uniformly distributed load as indicated.

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Propped Cantilever

The structure can be considered to be the superposition of an encastre beam with the addition of an equal and opposite moment to MB

applied at B to ensure that the final moment at this support is equal to zero.

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Example calculation 2:

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Solutions:

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Example: Solution

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Example: Solution

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Distribution Factor

Distribution Factor (DF) That fraction of the total resisting moment

supplied by the member is called the distribution factor (DF)

K

KDF

K

K

M

MDF

i

iii

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Consider 1 case

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Total stiffness at the support = ∑K = KBA + KBC

The moment absorbed by beam BA

The moment absorbed by beam AB

Stiffness of span BA = KBA = (I1/L1)Stiffness of span BC = KBC = (I2/L2)

K

KMM

K

KMM

AB

BA

x

x

applied2

applied1

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Example Calculation 1:

6.0)60(4)40(4

)60(4

4.0)60(4)40(4

)40(4

/)10)(60(44

)10)(240(4

/)10)(40(43

)10)(120(4

466

466

EE

EDF

EE

EDF

mmmEE

K

mmmEE

K

BC

BA

BC

BA

L

EIK

4

K

KDF

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Solution:

kNmwL

FEM

kNmwL

FEM

CB

BC

800012

)(

800012

)(

2

2

Note that the above results could also have been obtained if the relative stiffness factor is used

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Solution:

As a result, portions of this moment are distributed in spans BC and BA in accordance with the DFs of these spans at the joint

Moment in BA is 0.4(8000) = 3200NmMoment in BC is 0.6(8000) = 4800NmThese moment must be carried over since

moments are developed at the far ends of the span

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Solution

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Determine the internal moment at each support of the beam. The moment of inertia of each span is indicated.

Example Calculation 2 (Example 12.2)

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

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A moment does not get distributed in the overhanging span ABSo the distribution factor (DF)BA =0

Span BC is based on 4EI/L since the pin rocker is not at the far end of the beam

Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

EE

K

EE

K

CD

BC

)10(3203

)10)(240(4

)10(3004

)10)(300(4

66

66

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Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

NmwL

FEM

NmwL

FEM

NmmNFEM

DFDF

EE

EDF

DFDF

CB

BC

BA

DCCD

CB

BABC

200012

)(

200012

)(

4000)2(2000)( overhang, toDue

0 ;516.0

484.0320300

300

101)(1

2

2

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The overhanging span requires the internal moment to the left of B to be +4000Nm.

Balancing at joint B requires an internal moment of –4000Nm to the right of B.

-2000Nm is added to BC in order to satisfy this condition.

The distribution & CO operations proceed in the usual manner.

Since the internal moments are known, the moment diagram for the beam can be constructed.

Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

Page 37: Moment Distribution

Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

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Stiffness-Factor Modifications

Member pin supported at far end As shown the applied moment M rotates end

A by an amt To determine , the shear in the conjugate

beam at A’ must be determined

L

EIM

EI

LV

LLEI

MLVM

A

AB

33'

03

2

2

1)(' 0'

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Stiffness-Factor Modifications

Member pin supported at far end (cont’d) The stiffness factor in the beam is

The CO factor is zero, since the pin at B does not support a moment

By comparison, if the far end was fixed supported, the stiffness factor would have to be modified by ¾ to model the case of having the far end pin supported

L

EIK

3

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Stiffness-Factor Modifications

Symmetric beam & loading The bending-moment diagram for the beam

will also be symmetric To develop the appropriate stiffness-factor

modification consider the beam Due to symmetry, the internal

moment at B & C are equal Assuming this value to

be M, the conjugate beam for span BC is shown

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Stiffness-Factor Modifications

Symmetric beam & loading (cont’d)

Moments for only half the beam can be distributed provided the stiffness factor for the center span is computed

L

EIK

L

EIM

EI

MLV

LL

EI

MLVM

B

BC

2

2

2'

02

)(' - 0'

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Stiffness-Factor Modifications

Symmetric beam with asymmetric loading Consider the beam as shown The conjugate beam for its center span BC is

shown Due to its asymmetric loading, the internal

moment at B is equal but opposite to that at C

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Stiffness-Factor Modifications

Symmetric beam with asymmetric loading Assuming this value to be M, the slope at

each end is determined as follows:

L

EIK

L

EIM

EI

MLV

LL

EI

MLL

EI

MLV

M

B

B

C

6

6

6'

0622

1

6

5

22

1)(' -

0'

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Determine the internal moments at the supports of the beam shown below. The moment of inertia of the two spans is shown in the figure.

Example Calculation 3 (Example 12.4)

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

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The beam is roller supported at its far end C. The stiffness of span BC will be computed on

the basis of K = 3EI/L We have:

Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

EE

L

EIK

EE

L

EIK

BC

AB

)10(1804

)10)(240(33

)10(1603

)10)(120(44

66

66

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Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution

1180

180

5294.0180160

180

4706.0180160

160

0160

160

E

EDF

EE

EDF

EE

EDF

E

EDF

CB

BC

BA

AB

NmwL

FEM BC 120008

)4(6000

8)(

22

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The forgoing data are entered into table as shown.The moment distribution is carried out.By comparison, the method considerably simplifies the distribution.The beam’s end shears & moment diagrams are shown.

Solution

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution