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PROPIEDADES

MECANICAS

DE LOS

MATERIALES

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ESFUERZO

Y

DEFORMACIN

EN

UN

PLANO

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124

PROPIEDADES

MECNICAS

DE

LOS

MATERIALES

*ir*

:tirli|l

ir|

Xi.:.:ili$

:ii,o.l.,ii.ii:r,,.'

iiliiit=iiu,iix*i

iiiiiiii.:rtiiri

'tt*iitiiuii

i,t:i.+;r:=

Gomentar[os

I.

t,as

ecuaciones

El

y

E2

tienen

una

estructura

muv

ca

:iiili::ri=

"f##ffi

o

a

,oo

=

Ee

=

7o

i

ro,

x

oo

x

lo_u

*

o-

voo=

E"o=

Z0

X

tO,

i

OO

i

lO-ur,

itilir+iir:ir.'irir=,,f,:.:.

,

,.

-

.,

,

-,

-,..,.,,..=-,.,.,-.,

-,ii.i.

o:

{

'..;"'.tf,tt....

:ll-r':l

:::-,,:-::r::j:rij.:rr;rjli.i+:::i:rrfii:ir'.,,=.,,-,,.,1;,'a,i..,.

==**=l-=*****I="

De

la

ecuacion

3

l2c

se

obene

e..,

a

-.:l',',*,1(,1,.i,, .1 ,).-

0

-

0'25(54'13

+

3453)

:

*i*=$$ur*=*n*i='

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always of opposite signs for uniaxial tress (i.e., if one strain is elongation, the

is contraction).

for different materials,

but

for most metals, Poisson

s

ratio has a value

l Becau e the volume of material must remain constant,

the

large

st

Poisson s ratio is 0.5. Yalues approaching this upper

limitare

found

materials such as rubber.

NSHIP BETWEEN

f G ANO

tio is related to the elaslic modulus and the shear modulus G

by

the formula

le

1

~ j j

EXAMPLE 3 1

(3.7)

test was conducted on a 1.975-in.-wide by 0.375-in.-thick specimen of a

plastic.

A

4.000-in. gage length was marked on the specimen before load applica

In

the elastic portion of the stress-strain curve al an applied load of P = 6,000 lb,

in the

gage

length was measured as 0.023 in ., and the contraction in the

55

POISSON S RATIO

The Poisson effect exhibited by

matcrials

ca

uses no additional

stresses in the lateral direction

unless the transverse deforma

tion is inhibited or prevented in

sorne manner.

measured as 0.004 in. Determine:

elastic modulus

E

so

n s

ratio

v

~

:1

L : : _ _ : : : : : : t _ _ p

modulus

G.

p - 4.000

in. ~

-

the Solution

and

the

initial measured dimensions of the bar, the normal stress

can

be

Thc

normal strain in the long itudinal (i.e., axial) direction i;

10 0

g can be com-

from the elongation in the gage length and the initial gage length. With these two

s, the elastic modulus can be calculated from Eq. (3.4). (b) From the contrac

in the width and the initial bar wid th, thc strain in the lateral (i e., transverse) direc

t:

131

can

be computed. Poisso

n

s ratio can then be computed from Eq. (3.6). (c) The

can

be calculated from Eq. (3.7).

normal stress in the plastic specimen is

J

=

longitudinal strain is

6,000 lb .

= 8,101.27 psi

( 1.975 in.) (0.375 in.)

CJong

=

(0.023 in. ) . .

(

00

. ) = 0.005750 m / m

4.

o Jl.

efo

re, the elastic modulus is

J 8,

101.27 psi) . .

E

- = (000

5

. ) = 1,408,9 16ps 1.409.000psi

f; .

750

l l . l l

Ans

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l.3m

6

(b)

The

lateral strain is

(-0.004 in.) = _ 0.002025 in. f in.

1 1at = 1.975 in.\

From Eq. (3.6), Poissons rat io can be computed as

1 1a1 =

V -

f.1ong

(-0.002025 in.f

in.)

= 0.352

(0.005750 in.f in.)

Ans.

(c) The shear modulus G is computed from Eq . (3.7) as

E 1 408,916 psi . .

e = 2( 1 +

v)

= 2( 1 +0.352) = 521,049 psi= 521,000 psi

Ans.

EXAMPLE 3 2

.

Ri gid bar BC is supported by a pin at A and a 100-mm

wide X 6-mm-th i

ck

aluminum

[E

= 70 GPa; Y. = 22.5 X

10-

6

/C; v = 0.33] a

ll

oy bar at B A strain gagc affixed to

the surface of the aluminum bar is used to measure its

longitud inal strain. Before load

P

is applied to the rig

id

bar at C, the strain gage measures

ze

ro longitudinal strain

at an ambient temperature of20 C. After load

Pis

applied

to the rigid bar at C and the temperature drops to 10 C,

a l

ong

itudinal strain

of

+ 2,400

i:

is measured in the

aluminum bar. Determine:

( 1)

---

IOOmm

Strain

gag e

1 0m

p

(a) the str

ess

in member

(l

).

(b) the magnitude

of

l

oad

P

(c) the change in the width

of

the aluminum bar (i.e., the

100-mm dimension).

Plan the

Solution

T hi s problem illustrates sorne misconceptions common in applying Hooke s Law and

Poisson s ratio, particu larly when temperature change is a factor in the ana lysis .

SOLUTION

(a) Since the e lastic modulus E and the longitudinal strain

i

are given in the problem, one

might be tempted to compute the normal stress in aluminum bar (l) from Hooke s

Law

[Eq.

(3.4)]:

[

l ,000 MPal [ 1 mm/ mm ] = 168 MPa

a

1

=

E

1

e

1

= (

70

GPa)(2.400

i:

t

GPa 1,000,000

T

hi

s calculation is not corr

ec

t fo r the normal str

ess

in mem

be

r

1).

Why is it

incorrect?

From Eq. (2.7), the total strain

1 t

o wi in an object includes a portion due to stress

f. ,, and a portion due to temperature change

r

The strain gage affixed to member

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Compression

B

a)

FIGUR 15 12 Mohr s failure criterion (planc stress

.

I

I

I

I

I

I

I

uuc

Experimental data from

1cnsion and

ompression

te

st

s

b)

Mohr

s

failure criterion for a plane stress state rnay be represented on a graph

of

principal stresses

in

the

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598

Plan the Solution

The principal stresses will

be

determined

fo

r the given state

of

stress. Using th

ese

stres

ses, the maximum-shear-stress theory and the max imum-distortion-energy theory wi

ll

be used to inves

ti

gate the potential

fo

r failure.

SOLUTION

The

principal s

tr

esses can

be calc

ulated from the stress transformati

on

e

qua

tions [Eq

(12.12)] or from Mohrs ci rcle, as discussed in Section 12.9. Equation 12.12 will be

used here. From the stress el

emen

t, the values to

be

used in the stress transformation

eq

uations are

ax=

+ 75 MPa,

ay=

O MPa, and r,y =

+90 MP

a.

The

in-plane principal

stresses are calculated as

api p2

ax a

y

= 2

2

x ; Y r Y

75MPa +O MP

a /

(75MPa-OMPa)

2

(

9

)

2

1

2

+ OMPa

=

135.0MP

a,

60

.0MPa

.. - 1mum-Shectr-Stn '-'

1ht.1Q

Since

aP

is positi

ve

and

ap

2

is negative, failure will occur

if

lap

-

a

p2I

2 .

ay

For

the

principal stresses existing in the component:

lap1 - ap2I = j135.0 MPa - (

-60.0

MPa)I = 195.0 MPa < 270 MPa

Therefore, the

co

mponent does n

ot

fai l according to the maximum-sh

ea

r-stress theory.

The

factor

of

safety associated with this state

of

str

ess

can

be

ca

lc

ul

ated as

l qllli

dcnt Stn

270

MPa

=

.385

FS = l95.0MPa

Ans.

The

Mises equi valent stress aM associated with the maximum-distortion-energy the

ory

can

be ca

lcu lated from Eq. ( 15.8) for the plane stress state consi

der

ed here .

- [ 2 2

l

/2

aM - apl - a,,1

ap2

ap2

=

[

135.0MPa)

2

- ( 135.0MPa)(-60 .0MPa) +

60.0MPa

)

2

]

1

/

2

= 1

73.0MPa

e

1 .: in

.um-llasto

t n l

n r

(., l'hc:

.

r

3 I

.; tor

cif

S1rt.\

Ans.

The

factor of safety for the maximum-distortion-energy theory can be calcul ated from

the Mises equi valent stress:

270MPa

= 1.561

FS = l73.0 MPa

According to the maximum-shear-stress theory, the component does not fai

l.

Ans

.