14 three moment equation

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

Three Moment Equation

Theory of Structure - I


2

Lecture Outlines

Introduction

Proof of Three Moment Equation

Example


3

Introduction Developed by French Engineer Clapeyron in

1857. This equation relates the internal moments in

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

By successive application of this equation to each span of the beam, one obtains a set of equations that may be solved simultaneously for the unknown internal moments at the support.


4

Proof: Real Beam A general form of three moment equation can

be developed by considering the span of a continuous beam.

L C RML MC MC MR

P1 P2 P3 P4

WL WR

LL LR


5

Conjugate Beam (applied loads) The formulation will be based on the

conjugate-beam method. Since the “real beam” is continuous over the

supports, the conjugate-beam has hinges at L, C and R.

L’

XL XR

LL LRCL1CR1

R’

AL /EIL AR /EIR


6

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


7

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


8

L

LC

L

LL

L

LL

LLL

CLL

L

L

LL

L

L

LLL

EILM

EILM

EIxA

LLEIMLL

EIM

Lx

EIA

LCC

36

32

21

31

211)(1

21

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

RRR

CRR

R

R

RR

R

R

RRR

EILM

EILM

EIxA

LLEIMLL

EIM

Lx

EIA

LCC

36

32

21

31

211)(1

21


9

Equating

and simplifying yields

)(2121 RRLL CCCC

RR

RR

LL

LL

R

RR

R

R

L

LC

L

LL

LIxA

LIxA

ILM

IL

ILM

ILM 662

General Equation

(1)


10

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 and uniform distributed loads.


11

L C C RC R

LL

KLLLKRLR

PL PR

w

R

RR

L

LLRR

R

RRLL

L

LL

R

RR

R

R

L

LC

L

LL

ILw

ILwkk

ILPkk

ILP

ILM

IL

ILM

ILM

442

333

23

2

If the areas and centroidal distances for their M/EI diagrams are substituted in to 3-Moment equation,

(2)


12

Special Case: If the moment of inertia is constant for the

entire span, IL = IR.

44

233

3232 RRLLRRRRLLLLRRRLCLL

LwLwkkLPkkLPLMLLMLM

(3)


13

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


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


15

Substituting the values in equation 2,

ftkM

IIIIM

B

B

.5.177

05.0*425*325.025.020*150020

5.02520

33

2


16

For span AB:

kVV

F

kA

A

M

AF

BL

BL

y

y

y

B

xx

6.440754.30

;0

4.30

0)5.12(755.177)25(

;0

0;0

75 k

A B

12.5’ 12.5’

VBL

177.5k.ftAy


17

For span BC:

kVV

F

kC

C

M

BR

BR

y

y

y

B

6.1201538.2

;0

38.2

0)15(155.177)20(

;0

15 k

B C

15 ft 5 ft

VBR

177.5k.ft Cy


18

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


19

Practice Problems:

Chapter 9

Example 9-11 to 9-13 and Exercise

Structural Analysis by R C Hibbeler


20

14 three moment equation

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