3 moments equation

7/28/2019 3 moments equation

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Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan

Three MomentEquation

Theory of Structure - I


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Department of Civil EngineeringUniversity of Engineering and

Technology, Taxila, Pakistan

2

Lecture Outlines

Introduction

Proof of Three Moment Equation

Example


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3

Introduction

Developed by French Engineer Clapeyron in1857.

This equation relates the internal moments in

a continuous beam at three points of supportto the loads acting between the supports.

By successive application of this equation

to each span of the beam, one obtains a setof equations that may be solvedsimultaneously for the unknown internalmoments at the support.


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4

Proof: Real Beam

A general form of three moment equation can

be developed by considering the span of a

continuous beam.

L C R

ML MC MC MR

P1 P2 P3 P4

WL WR

LL LR


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Conjugate Beam (applied

loads)

The formulation will be based on the

conjugate-beam method.

Since the real beam is continuous over thesupports, the conjugate-beam has hinges at

L, C and R.

L

XL XR

LL LRCL1CR1

R

AL /EIL AR /EIR


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Conjugate Beam (internal

moments)

Using the principle of superposition, the M /

EI diagram for the internal moments is

shown.

L LL LRCL2CR2

R

ML /EIL

MC /EIL

MC /EIR

MR /EIR


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In particular AL/EIL and AR/EIR represent the

total area under their representative M / EI

diagrams; and xL and xR locate their

centroids.

Since the slope of real beam is continuous

over the center support, we require the shear

forces for the conjugate beam.

)(2121 RRLL

CCCC


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L

LC

L

LL

L

LL

LL

L

C

LL

L

L

L

L

L

L

L

LL

EI

LM

EI

LM

EI

xA

LLEI

MLL

EI

M

Lx

EI

A

LCC

36

3

2

2

1

3

1

2

11)(

121

Summing moments about point L for left

span, we have

Summing moments about point R for the

right span yields

R

RC

R

RR

R

RR

RR

R

CRR

R

R

R

R

R

R

R

RR

EI

LM

EI

LM

EI

xA

LLEIMLL

EIM

Lx

EIA

LCC

36

32

21

31

211)(1

21


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Equating

and simplifying yields

)(2121 RRLL

CCCC

RR

RR

LL

LL

R

RR

R

R

L

L

CL

LL

LI

xA

LI

xA

I

LM

I

L

I

LM

I

LM 662

General Equation

(1)


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Eq. Modification for point load

and uniformly distributed load

Summation signs have been added to the

terms on the right so that M/EI diagrams for

each type of applied load can be treated

separately.

In practice the most common types of

loadings encountered are concentrated anduniform distributed loads.


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L C C RC R

LL

KLLLKRLR

PL PRw

R

RR

L

LL

RR

R

RR

LL

L

LL

R

RR

R

R

L

L

C

L

LL

I

Lw

I

Lwkk

I

LPkk

I

LP

I

LM

I

L

I

LM

I

LM

442

33

3

2

3

2

If the areas and centroidal distances for their

M/EI diagrams are substituted in to 3-Moment

equation,

(2)


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Special Case:

If the moment of inertia is constant for the

entire span, IL = IR.

44

2

33

3232 RRLL

RRRRLLLLRRRLCLL

LwLwkkLPkkLPLMLLMLM

(3)


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Example:

Determine the reactions at the supports for

the beam shown. The moment of inertia of

span AB is one half that of span BC.

15k3 k/ft

I0.5 I

25 ft 15 ft 5 ft

A CB


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Department of Civil Engineering

University of Engineering and


14

ML = 0

LL = 25ft

IL = 0.5I PL = 0

wL = 3k/ft

kL = 0

MC = MB

LR = 20ft

IR = I PR = 15k

wR = 0

kR = 0.25

MR = 0

Data


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Substituting the values in equation 2,

f tkM

IIIIM

B

B

.5.177

05.0*4

25*3

25.025.0

20*15

00

20

5.0

25

20

33

2


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For span AB:

kV

VF

kA

AM

AF

BL

BL

y

y

y

B

xx

6.44

0754.30;0

4.30

0)5.12(755.177)25(;0

0;0

75 k

A B

12.5 12.5

VBL

177.5k.ftAy


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For span BC:

kV

V

F

kC

C

M

BR

BR

y

y

y

B

6.12

01538.2

;0

38.2

0)15(155.177)20(

;0

15 k

B C

15 ft 5 ft

VBR

177.5k.ft Cy


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A free body diagram of the differential

segment of the beam that passes over roller

at B is shown in figure.

kB

B

F

y

y

y

2.57

06.126.44

0

By

44.6 k 12.6 k

177.5k.ft 177.5k.ft


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Practice Problems:

Chapter 9

Example 9-11 to 9-13 and Exercise

Structural Analysis by R C Hibbeler


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3 moments equation