英文版静力学课件第六章-Analysis of Structures

VECTOR

MECHANICS

FOR

ENGINEERSENGINEERING

3120STATICSCHAPTER

6ANALYSIS

OFSTRUCTURESANALYSIS

OF

STRUCTURES

-

TOPICS

IN

CHAPTER

6Introduction

Definition

of

a

Truss

Simple

TrussesAnalysis

of

Trusses

by

the

Method

of

Joints

Joints

Under

SpecialLoading

ConditionsSpace

Trusses

Sample

Problem

6.1Analysis

of

Trusses

by

the

Method

of

SectionsTrusses

Made

of

Several

Simple

Trusses

Sample

Problem

6.3Analysis

of

FramesFrames

Which

Cease

to

be

Rigid

When

Detached

From

Their

SupportsSample

Problem

6.4

MachinesANALYSIS

OF

STRUCTURES

-

INTRODUCTIONFor

the

equilibrium

of

structures

made

of

severalconnected

parts,

the

internal

forces

as

well

theexternal

forces

are

considered.In

the

interaction

between

connected

parts,

Newton’s3rd

Law

states

that

the

forces

of

action

and

reactionbetween

bodies

in

contact

have

the

same

magnitude,

same

line

of

action,

and

opposite

sense.Three

categories

of

engineering

structures

are

consideredFrames:

contain

at

least

one

multi-force

member,i.e.,memberacted

upon

by

3

or

more

forces.Trusses:

formed

from

two-force

members,

i.e.,straight

members

with

end

point

connectionsMachines:

structures

containing

moving

partsdesigned

to

transmit

and

modify

forcesDEFINITION

OF

A

TRUSSA

truss

consists

of

straight

members

connected

atjoints.

No

member

is

continuous

through

a

joint.Most

structures

are

made

of

several

trussesjoined

together

to

form

a

space

framework.

Eachtruss

carries

those

loads

which

act

in

its

plane

andmay

be

treated

as

a

two-dimensional

structure.Bolted

or

welded

connections

are

assumedto

be

pinned

together.

Forces

acting

at

thememberends

reduce

to

a

single

force

and

nocouple.

Only

two-force

members

are

considered.When

forces

tend

to

pull

the

member

apart,

itis

in

tension.

When

the

forces

tend

to

compressthe

member,

it

is

in

compression.MEMBERS

OF

A

TRUSSMembers

of

a

truss

are

slender

and

not

capable

of

supportinglarge

lateral

loads.

Loads

must

be

applied

at

the

joints.TYPICAL

TRUSS

CONFIGURATIONSSIMPLE

TRUSSES

A

rigid

truss

will

not

collapsunder

the

application

of

a

load

A

simple

truss

is

constructed

bysuccessively

adding

twomembers

and

one

connection

tothe

basic

triangular

truss.In

a

simple

truss,

m

=

2n

-

3

wherem

is

the

total

number

of

membersand

n

is

the

number

of

joints.ANALYSIS

OF

TRUSSES

BY

SIMPLE

METHOD

OF

JOINTSDismember

the

truss

and

create

afree

body

diagram

for

each

member

andpin.The

two

forces

exerted

on

each

memberare

equal,

have

the

same

line

of

action,and

opposite

sense.Forces

exerted

by

a

memberon

the

pinsor

joints

at

its

ends

are

directed

along

themember

and

equal

and

opposite.Conditions

of

equilibrium

on

the

pinsprovide

2n

equations

for

2n

unknowns.For

a

simple

truss,

2n

=

m

+

3.

May

solvefor

m

member

forces

and

3

reactionforces

at

the

supports.Conditions

for

equilibrium

for

the

entire

truss

provide

3

additional

equations

which

are

not

independent

of

the

pin

equations.JOINTS

UNDER

SPECIAL

LOADING

CONDITIONS

Forces

in

opposite

members

intersecting

intwo

straight

lines

at

a

joint

are

equal.Recognition

of

joints

under

specialloading

conditions

simplifies

a

trussanalysis.

The

forces

in

two

opposite

members

areequal

when

a

load

is

aligned

with

a

thirdmember.

The

third

member

force

is

equalto

the

load

(including

zero

load).

The

forces

in

two

members

connected

at

ajoint

are

equal

if

the

members

are

alignedand

zero

otherwise.SPACE

TRUSSES

–

3D

TRUSSESAn

elementary

space

truss

consists

of

6members

connected

at

4

joints

to

form

atetrahedron.A

simple

space

truss

is

formed

and

can

beextendedwhen3

new

members

and

1

jointare

added

at

the

same

time.In

a

simple

space

truss,

m

=

3n

-

6

wherem

is

the

number

of

members

and

n

is

thenumber

of

joints.Conditions

of

equilibrium

for

the

jointsprovide

3n

equations.

For

a

simple

truss,

3n

=m

+6

and

the

equations

can

be

solved

for

m

member

forces

and

6

support

reactions.Equilibrium

for

the

entire

truss

provides

6additional

equations

which

are

notindependent

of

the

joint

equations.EXAMPLE

PROBLEM

6.1Using

the

method

ofjoints,

determine

theforce

in

each

memberof

the

trussSTRATEGY:Based

on

a

free-body

diagram

of

theentire

truss,

solve

the

3

equilibriumequations

for

the

reactions

at

E

and

C.Joint

A

is

subjected

to

only

two

unknown

member

forces.

Determine

thesefromthe

joint

equilibrium

requirements.In

succession,

determine

unknownmemberforces

at

joints

D,

B,

and

E

fromjoint

equilibrium

requirements.All

member

forces

and

support

reactionsare

known

at

joint

C.

However,

the

joint

equilibriumrequirements

may

be

applied

to

check

theresults.EXAMPLE

PROBLEM

6.1

-

SOLUTIONMODELING

and

ANALYSISBased

on

a

free-body

diagram

of

theentire

truss,

solve

the

3

equilibriumequations

for

the

reactions

at

E

and

C.EXAMPLE

PROBLEM

6.1

–

SOLUTION

(continued)

There

are

now

only

two

unknown

memberforces

at

joint

D.

Draw

FBD

at

D

and

solve.We

now

solve

the

problem

by

movingsequentially

from

joint

to

joint

and

solvingthe

associated

FBD

for

the

unknown

forces

Joint

A

or

C

is

subjected

to

only

two

unknownmember

forces.

Draw

FBD

for

joint

A

anddetermine

the

two

unknown

forces

from

thejoint

equilibrium

requirements.EXAMPLE

PROBLEM

6.1

–

SOLUTION

(continued)There

are

now

only

two

unknown

memberforces

at

joint

B.

Assume

both

are

in

tension.

There

is

now

one

unknown

member

force

atjoint

E

(or

C).

Use

joint

E

and

assume

themember

EC

is

in

tension.EXAMPLE

PROBLEM

6.1

–

SOLUTION

(continued)REFLECT

and

THINKAll

member

forces

and

support

reactionsare

known

at

joint

C.

However,

the

jointequilibrium

requirements

may

be

appliedto

check

the

results.Using

the

computed

values

of

FCB

and

FCE,you

can

determine

the

reactions

Cx

and

Cyby

considering

the

equilibriumof

Joint

C.You

can

also

simply

use

the

computedvalues

of

all

forces

acting

on

the

joint

C(forces

in

members

and

reactions)

andcheck

that

the

joint

is

in

equilibrium.ANLYSIS

OF

TRUSSES

BY

THE

METHOD

OF

SECTIONSWhen

the

force

in

only

one

member

or

theforces

in

a

very

few

members

are

desired,the

method

of

sections

works

well.To

determine

the

force

inmember

BD,

forma

section

by

“cutting”

the

truss

at

n-n

andcreate

a

free

body

diagram

for

the

left

side.An

FBD

could

have

been

created

for

the

rightside,

but

why

is

this

a

less

desirable

choice?Think

and

discuss.Notice

that

the

exposed

internal

forces

are

all

assumed

to

be

in

tension.With

only

three

members

cut

by

thesection,

the

equations

for

staticequilibrium

may

be

applied

to

determinethe

unknownmemberforces,

including

FBD.ANLYSIS

OF

TRUSSES

BY

THE

METHOD

OF

SECTIONSppkkUsing

the

left-side

FBD,

write

one

equilibrium

equation

that

can

be

solved

tofind

FBD.

Check

your

equation

with

aneighbor;

resolve

any

differences

between

your

answers

if

you

can.Assume

that

the

initial

section

cut

wasmade

using

line

k-k.

Why

would

this

be

apoor

choice?

Think,

then

discuss

with

aneighborNotice

that

any

cut

may

be

chosen,

so

longas

the

cut

creates

a

separated

sectionSo,

for

example,

this

cut

with

line

p-p

is

acceptable.TRUSSES

MADE

OF

SEVERAL

SIMPLE

TRUSSESnon-rigidrigid

Compound

trusses

are

staticallydeterminant,

rigid,

and

completelconstrained.Truss

contains

a

redundant

memberand

is

statically

indeterminate.

Additional

reaction

forces

may

benecessary

for

a

rigid

truss.

Necessary

but

insufficient

conditfor

a

compound

truss

to

be

staticaldeterminant,

rigid,

and

completelconstrained,EXAMPLE

PROBLEM

6.3Determine

the

force

in

membersFH,

GH,

and

GI.STRATEGY:Draw

the

FBD

for

the

entiretruss.

Apply

the

equilibriumconditions

and

solve

for

thereactions

at

A

and

L.Make

a

cut

throughmembers

FH,

GH,

and

GI

andtake

the

right-hand

section

aa

free

body

(the

left

sidewould

also

be

good).3.

Apply

the

conditions

forstatic

equilibrium

todetermine

the

desiredmember

forces.EXAMPLE

PROBLEM

6.3

-

SOLUTIONTake

the

entiretruss

as

a

freebody.

Apply

theconditions

forstatic

equilibriumto

solve

for

thereactions

at

Aand

L.EXAMPLE

PROBLEM

6.3

–

SOLUTION

(continued)Make

a

cut

through

members

FH,Apply

the

conditions

for

statequilibrium

to

determine

theGH,

and

GI

and

take

the

right-hand

desired

member

forces.section

as

a

free

body.

Draw

thisFBD..EXAMPLE

PROBLEM

6.3

-

SOLUTIONREFLECT

and

THINK:Resolve

a

force

into

components

to

include

it

in

theequilibrium

equations.

By

first

sliding

this

force

aloline

of

action

to

a

more

strategic

point,

toeliminate

oof

its

components

from

a

moment

equilibrium

equation.ANALYSIS

OF

FRAMES

Frames

and

machines

are

structures

with

at

least

onemultiforce

member.

Frames

are

designed

to

supportloads

and

are

usually

stationary.

Machines

containmoving

parts

and

are

designed

to

transmit

and

modifyforces.

A

freebodydiagram

of

the

complete

frame

is

used

todetermine

the

external

forces

acting

on

the

frame.Internal

forces

are

determined

by

dismembering

the

framand

creating

free-body

diagrams

for

each

component.Forces

on

two

force

members

have

known

lines

ofaction

but

unknown

magnitude

and

sense.Forces

on

multi-force

members

have

unknownmagnitude

and

line

of

action.

They

must

berepresented

with

two

unknown

components.Forces

between

connected

components

are

equal,have

the

same

line

of

action,

and

opposite

sense.FRAMES

WHICH

CEASE

TO

BE

RIGIDWHEN

DETACHED

FROM

THEIR

SUPPORT

Some

frames

may

collapse

if

removed

from

theirsupports.

Such

frames

can

not

be

treated

as

rigidbodies.

A

free-body

diagram

of

the

complete

frame

indicatesfour

unknown

force

components

which

can

not

bedetermined

from

the

three

equilibrium

conditions.

The

frame

must

be

considered

as

two

distinct,

butrelated,

rigid

bodies.With

equal

and

opposite

reactions

at

the

contactpoint

between

members,

the

two

free-bodydiagrams

indicate

6

unknown

force

components.Equilibrium

requirements

for

the

two

rigidbodies

yield

6

independent

equations.EXAMPLE

PROBLEM

6.4Members

ACE

and

BCD

areconnected

by

a

pin

at

C

and

by

thelink

DE.

For

the

loading

shown,determine

the

force

in

link

DE

andthe

components

of

the

forceexerted

at

C

onmember

BCD.STRATEGY:Create

a

free-body

diagram

for

thecomplete

frame

and

solve

for

the

supportreactions.Define

a

free-body

diagram

for

memberBCD.

The

force

exerted

by

the

link

DE

has

aknown

line

of

action

but

unknownmagnitude.

It

is

determined

by

summingmoments

about

C.With

the

force

on

the

link

DE

known,

thesum

of

forces

in

the

x

and

y

directions

maybe

used

to

find

the

force

components

at