結(jié)構(gòu)力學(xué)-課件chapter

華南理工大學(xué)土木與交通學(xué)院STRUCTURALYSISR.

C.

HIBBELER華南理工大學(xué)土木與交通學(xué)院YSIS:CHAPTER

8:

DISPLACEMENT

METHOD

OFSLOPE-DEFLECTION

EQUATIONS2華南理工大學(xué)土木與交通學(xué)院Chapter

Outline1.2.3.4.5.ysis:

General

ProceduresDisplacement

Method

ofSlope-Deflection

Equations

ysis

of

Beamsysis

of

Frames:

No

Sidesway

ysis

of

Frames:

Sidesway3華南理工大學(xué)土木與交通學(xué)院YSIS:8.1DISPLACEMENT

METHOD

OFGENERAL

PROCEDURES4華南理工大學(xué)土木與交通學(xué)院Displacement

Method

of

ysis:General

Procedures1

f11R1

f12

R2

......

f1n

Rn

02

f21R1

f22

R2

......

f2n

Rn

0n

fn1R1

fn2

R2

......

fnn

Rn

0Force

method

and

displacement

methodIn

force

method,

much

work

is

required

to

set

up

the

compatibilityequations,

and

furthermore

each

equation

written

involves

all

theunknowns,

making

it

difficult

to

solve

the

resulting

set

of

equations

unlessa

computer

is

availableForce

method

is

therefore

limited

to

structures

which

are

not

highlyindeterminateDisplacement

method

requires

less

work

both

to

write

the

necessaryequations

for

the

solution

of

a

problem

and

to

solve

these

equations

for

theunknown

displacements

and

associated

internal

loads5華南理工大學(xué)土木與交通學(xué)院Displacement

Method

ofGeneral

Proceduresysis:General

procedures

for

displacement

methodDisplacement

method

requires

satisfying

equilibrium

equations

for

thestructuresThe

loads

are

written

in

terms

of

the

unknown

displacements

by

using

theload-displacement

relationsThe

equilibrium

equations

are

solved

for

the

displacementsOnce

the

displacements

are

obtained,

the

unknown

loads

are

determinedusing

the

load-displacement

relationsCompatibility

equations

are

satisfied

automatically6華南理工大學(xué)院Displacement

Method

ofGeneral

Proceduresysis:Degrees

of

freedomThe

displacements

ofthe

nodeson

astructureare

referred

to

as

the

degrees

of

freedoms

forthe

structureIn

three

dimensions,

each

node

on

a

frame

orbeam

can

have

at

most

three

lineardisplacements

and

three

rotationaldisplacementsIn

two

dimensions,

each

node

can

have

at

mosttwo

linear

displacements

and

one

rotationaldisplacementNodal

displacements

may

be

restricted

by

thesupports,

or

due

to

assumptions

based

on

thebehavior

of

the

structure7華南理工大學(xué)土木與交通學(xué)院8.2SLOPE

–DEFLECTION

EQUATIONS8Slope-Deflection

EquationsGeneral

case

The

slope-deflection

method

is

so

named

since

it

relatesthe

unknown

slopes

and

deflections

to

momentsof

a

memberTo

develop

t eral

form

of

theslope-deflection

equations,

we

will

consider

thetypical

span

AB

of

the

continuous

beam

when

subjected

to

arbitrary

loading

andhaving

a

constant

EIWe

wish

to

relate

the

beam’s

internal

endmomentsMAB

and

MBA

in

terms

ofits

threedegrees

of

freedom,

namely,

its

angular

displacements

A

and

B

,

andlineardisplacement

?,

and

the

applied

loadsMoments

and

angular

displacements

will

be

considered

positive

when

they

actclockwise

on

thespanThe

linear

displacement

?is

considered

positive

as

shown,

since

thisdisplacementcauses

thecord

of

the

span

to

rotate

clockwiseTheslope-deflection

equations

can

be

obtained

byusing

the

principle

ofsuperposition by

considering

separa y

themoments

developed

at

each

supportdue

to

each

of

the

displacements,

A

,

B

and

?,

and

then

the

loads9Slope-Deflection

EquationsAngular

displacement

at

A,

A-

Consider

node

A

of

the

member

to

rotate

A

while

id

node

B

is

heldfixed

and

determine

the

moment

MAB

needed

to

cause

this

displacement-

The

problem

is

equivalent

to

how

to

determine

A

due

to

the

applied

coupleMAB,

which

can

be

solved

using

the

force

method1

L2

2

L2

3

EI20

2L3M

ABVAB

0

A

VAB

f

AAABAB

VEI

1

L2MM

ABA

BMAB=primary

structure+AVAB

LVABBredundant

VABappliedABL110Slope-Deflection

EquationsAngular

displacement

at

A,

A-

Consider

node

A

of

the

member

to

rotate

A

while

id

node

B

is

heldfixed

and

determine

the

moment

MAB

needed

to

cause

this

displacement-

The

problem

is

equivalent

to

how

to

determine

A

due

to

the

applied

coupleMAB,

which

can

be

solved

using

the

force

methodA

BMA

B1M

AB=ABprimary

structure+AABV

LABVBredundant

VAB

applied1ABL12ABAM

AB

M

LAB12AB4EI

2EI

V L

1EIMBAM

L

1EI4EI

LALA11Slope-Deflection

EquationsAngular

displacement

at

B,

B-

Similarly,

when

end

B

of

the

beam

rotates

to

its

final

position

while

end

Ais

held

fixed,

we

can

relate

the

applied

moment

MBA

to

the

angulardisplacement

B

and

the

reaction

moment

MAB

at

the

wallBABBBAL

2EI

ML

4EI

M12Slope-Deflection

EquationsRelative

linear

displacement,

?If

the

far

node

B

of

the

member

is

displaced

relative

to

A,

so

that

the

cordof

the

member

rotates

clockwise

and

yet

both

ends

do

not

rotate,

thenequal

but

opposite

moment

and

shear

reactions

are

developed

in

thememberMoment

MAB

and

MBA

can

be

related

to

the

displacement

?

using

the

forcemethod

M

6EI

L2ABBAM

MABABBABA13Slope-Deflection

EquationsFixed-end

moments-

The

fixed-end

moments

(FEM)AB

and

(FEM)BA

can

be

determined

using

theforce

methodMAB

(FEM

)ABMBA

(FEM

)BA14Slope-Deflection

EquationsSlope-deflection

equations-

If moments

due

to

each

displacementand

loading

are

added

together,

the

resultantmoments

at s

can

be

written

as:The

results

can

be

expressed

as

a

single

equationMN

2Ek

2N

F

3

(FEM

)NM

N

internal

moment

at

the

near-end

of

the

spanE,

k

modulus

of

elasticity

and

span

stiffness

N

,

F

near-end

and

far-end

slopes

or

angular

displacements

of

the

span

at

the

supports

span

rotation

of

its

cord

due

to

a

lineardisplacement(

FEM

)

N

fixed-end

moment

at

the

near-end

supportThe

slope-deflection

equation

is

applied

twice

for

eaember

span

(AB);that

is,

application

is

from

A

to

B

and

from

B

to

A

for

span

ABABA

BABM

2E

I

2

3

(FEM

)BAB

ABAM

2E

I

2

3

(FEM

)

L

L

L

L

15學(xué)院Pin-supported

end

spanSometimes

an

end

span

of

a

beam

or

frameis

supported

bya

pin

orroller

atits

far

endThe

moment

at

theroller

or

pin

iszeroProvided

the

angular

displacement

atthissupport

doesnothave

to

bedetermined,we

can

modify

t eral

slope-deflectionequation

so

that

it

hasto

beapplied

only

onceto

the

span

rather

than

twiceUsing

t eral

slope-deflection

equations,we

have-

From

the

second

equation,

it

can

be

seen

that

D

can

beexpressed

in

terms

ofC

andΨ-

Substitute

D

into

theequation,

we

get

the

modified

slope-deflectionequation-

This

is

onlyapplicable

for

end

span

with

far

end

pinned

or

roller

supportedSlope-Deflection

EquationsMCD

3Ek[C

]

(FEM

)CDCD

CDDC2(FEM

)

(FEM

)

1

(FEM

)11

2Ek[2C

D

3

]

(FEM

)CD

0

2Ek[2D

C

3

]

(FEM

)

DCD

=

[C

3

]

(FEM

)DC2

4EkMCD16華南理工大學(xué)土木與交通學(xué)院8.3YSIS

OF

BEAMES17華南理工大學(xué)土木與交通學(xué)院Label

all

the

supports

and

joints

(nodes)

in

order

to

identify

the

spans

ofthe

beam

or

frame

between

the

nodesCompatibility

at

the

nodes

is

maintained

provided

the

members

that

arefixed

connected

to

a

node

undergo

the

same

displacements

as

the

nodeSlope-deflection

equationsThe

slope-deflection

equations

relate

the

unknown

moments

applied

to

thenodes

to

the

displacements

of

the

nodes

for

any

span

of

the

structureApply

the

slope-deflection

equation

to

each

end

of

the

span,

therebygeneratin-

If

a

span

ato

slope-deflection

equations

for

each

spanof

a

continuous

beam

or

frame

is

pin

supported,

applythe

modified

slope-deflection

equation

only

to

the

restrained

end,

therebygenerating

one

slope-deflection

equation

for

the

spanysis

of

Beams:

General

ProceduresDegrees

of

freedom18華南理工大學(xué)土木與交通學(xué)院Equilibrium

equationsWrite

an

equilibrium

equation

for

each

unknown

degree

of

freedom

for

thestructureSubstitute

the

slope-deflection

equations

into

the

equilibrium

equationsand

solve

the

unknown

joint

displacementsysis

of

Beams:

General

Procedures19華南理工大學(xué)土木與交通學(xué)院ysis

of

BeamsExample

8.1Draw

the

shear

and

moment

diagrams

for

the

beam

where

EI

is

constant.20華南理工大學(xué)ysis

of

BeamsExample

8.1

(Solution)Degrees

of

freedomUnknown

angular

displacement,

BSlope-deflection

equations2

spans

must

be

considered

in

this

problem(FEM)AB=(FEM)BA=0

since

there

is

no

load

on

span

ABUsing

the

formulas

for

FEMs,

we

have:Since

A

and

C

are

fixed

support,

A=C=0Since

the

supports

do

not

settle

nor

are

they

displaced

up

ordown,

AB=BC=0Slope-deflection

equation-

Similarly,

we

have30

20BC30CB20(FEM

)

wL2

6(62

)

7.2

kN

m, (FEM

)

wL2

6(62

)

10.8

kN

m4NM

2E

I

2

3

(FEM

)

,

L

N

F

NABBM

2E

I

2(0)

8

B

3(0)

0

EI

10.82

33BA

BBC

BCBM

EI

,

M

2EI

-7.2

,

MB

EI

21華南理工大學(xué)ysis

of

BeamsExample

8.1

(Solution)Equilibrium

equationsConsider

the

moment

equilibrium

at

support

BHere

MBA

and

MBC

are

assumed

to

act

in

the

counterclockwise

direction

tobe

consistent

with

the

sign

convention

used

in

the

slope-deflectionequationsThe

beam

shears

contribute

negligible

moment

about

B

since

the

segmentis

of

differentiallengthThus,

we

have

7.2)

023

6.17

MBC

0M

BAEI

(

2EIBBBEI22華南理工大學(xué)土木與交通學(xué)院ysis

of

BeamsExample

8.1

(Solution)Equilibrium

equations-

Re-substituting

B

into

the

slope-deflection

equations

givesUsing

these

results,

the

shears

at of

spans

are

determined

fromequilibrium

equationsThe

free-body

diagram

of

the

entire

beam

and

the

shear

and

momentdiagrams

are

shownMCB

1.54

kN

m;

M

BA

3.09

kN

m;

3.09

kN

m;

12.86

kN

mM

ABM

BC23通華南理工大學(xué)學(xué)院24華南理工大學(xué)院252627華南理工大學(xué)土木與交通學(xué)院8.4YSIS

OF

FRAMES:

NO

SIDESWAY28華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

SideswayA

frame

will

not

sidesway

to

the

left

or

right

provided

it

is

properlyrestrainedNo

sidesway

will

occur

in

an

unrestrained

frame

provided

it

is

symmetricwith

respect

to

both

loading

and

geometryFor

both

cases

the

term

ψ

in

the

slope-deflection

equations

is

equal

to

zero,since

bending

does

not

cause

the

joints

to

have

a

linear

displacement29華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

SideswayExample

8.5Determine

the

moments

at

each

joint

of

the

frame.

EI

is

constant.30華南理工大學(xué)ysis

of

Frames:

No

SideswayExample

8.5

(Solution)Degrees

of

freedomUnknown

angular

displacements,

B

and

cSlope-deflection

equations3

spans

must

be

considered

in

this

case:

AB,

BC

and

CDFixed-end

moments96

965wL2

5wL2(FEM

)BC

80

kN

m, (FEM

)CB

80

kN

mNote

that

A

D

0

and

AB

BC

CD

0,

since

no

sidesway

will

occurSlope-deflection

equationMBA

0.333EIBMCBMDC

0.5EIC

0.25EIB

80

0.1667EIC

80,NNM

2E

I

2

3

FEM

L

N

F

M

AB

M

BC

MCD0.1667EIB

,

0.5EIB

0.25EIC0.333EIC

,31華南理工大學(xué)ysis

of

Frames:

No

SideswayExample

8.5

(Solution)Equilibrium

equations-

Considering

the

moment

equilibriumat

joints

B

and

C,

we

have:M

BA

M

BC

0-

Substituting

the

slope-deflectionequations

into

the

above

equations,we

get:0.833EIB

0.25EIC

800.833EIC

0.25EIB

80

MCD

0MCB

137.1BCEI

32華南理工大學(xué)土木與交通學(xué)院CABDExample

11.5

(Solution)Equilibrium

equations-

Re-substituting

B

and

c

into

the

slope-deflectionequations

givesysis

of

Frames:

No

SideswayM

AB

22.9

kN

m;MBA

45.7

kN

mMCB

45.7

kN

m

45.7

kN

m;

45.7

kN

m;

22.9

kN

mMBCMCDMDC33華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

Sidesway34華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

Sidesway35華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

Sidesway36華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

No

Sidesway37華南理工大學(xué)土木與交通學(xué)院8.5YSIS

OF

FRAMES:

SIDESWAY38華南理工大學(xué)土木與交通學(xué)院ysis

of

Frames:

SideswayA

frame

will

sidesway

when

it

or

the

loading

acting

on

it

is

nonsymmetricThe

loading

P

causes

an

unequal

moments

at

joint

B

and

CMBC

tends

to

displace

joint

B

to

the

rightMCB

tends

to

displace

joint

C

to

the

leftSince

MBC

>

MCB,

the

net

res