2_kuliah stress strain heru sukanto

7/26/2019 2_kuliah Stress Strain Heru Sukanto

1/35

TEGANGAN REGANGANAKIBAT

BEBAN AKSIAL


2/35

MATERIALS UNDER AXIAL LOADING

= deformasi (pertambahanpanjang)

A = luas penampang

P = beban aksial batang

Apa yang sama ?

Resistance offered by

the material per unit

cross-sectional area


3/35

Berapa defo

?

Apanya yan

kedua spesim


4/35

STRESS STRAIN DIAGRAM

specimenextensometer

Typical tensil

specimen

ga

len


5/35

Zona Elastic dan Plast

Awal Under load

F

bostr

F

elastic + plas

Bondsstretch &

plane shea


6/35

ENGINEERING STRESS

Shearstress, t:

Area, A

Ft

Ft

Fs

F

F

Fs

t = Fs

Ao

Tensile stress, s:

original area

before loading

Area, A

Ft

Ft

s =Ft

Ao2

f

2m

Nor

in

lb=


7/35

Tensile strain: Lateral strain:

Shear strain:

ENGINEERING STRAIN

q

90

90 -qy

x qg = x/y= tan

e =

Lo

-eL= L

wo

/2

L/2

Lowo


8/35


9/35

Plastic tensile strain at failure:

DUCTILITY

Another ductility measure: 100xA

AARA%

o

fo-

=

x 100L

LLEL%

o

of-

=

Engineering tensile strain, e

Engineering

tensile

stress, s

smaller %EL

larger %EL AoAfLo


10/35

Energy to break a unit volume of material

Approximate by the area under the stress-strain

curve.

TOUGHNESS

Brittle fracture: elastic energy

Ductile fracture: elastic + plastic energy

very small toughne(unreinforced poly

Engineering tensile strain, e

Engineering

tensile

stress, s

small toughness (ceramics)

large toughness (metals)


11/35

DEFORMASI (HOOKES LAW)

Tentukan deformasi (pertambahan panjng)

seperti gambar samping akibat beban yan

ditanggungnya (E = 200 GPa) !


12/35

Batang kaku BDE ditopang oleh dua batang AB dan CD

terbuat dari aluminium (E = 70 Gpa) dg luas penampan

sedangkan batang CD terbuat dari baja (E = 200 Gpa

penampang 600 mm2. Untuk beban 30 kN seperti pada

defleksi (lendutan) di titik B, D dan E !!


13/35

POISSONS RATIO

Untuk batang tipis yg dikenai beban aksial

0=== zyx

xE

sss

e

Perpanjangan dalam arah x selalu disertai

dengan kontraksi dalam arah lainnya. Deng

asumsi bahwa material adalah isotropik

(homogen, memiliki orientasi butir yg sam

semua arah),

0= zy ee

Poissons ratio didefinisikan dg:

x

z

x

y

e

e

e

e -=-==

strainaxial

strainlateral


14/35

GENERALIZED HOOKES LAW

Elemen yg dikenai beban multi aksial, rega

normal komponen yg disebabkan kompone

tegangan bisa ditentukan dengan prinsip

superposition. Syaratnya:

1) strain is linearly related to stress

2) deformations are small

EEE

EEE

EEE

zyxz

zyxy

zyx

x

ssse

ssse

sss

e

--=

--=

--=

Didapatkan:


15/35

DILATATION: BULK MODULUS

Relative to the unstressed state, the change in vol

e)unit volumperin volume(changedilatation

21

111111

=

-

=

=

-=-=

zyx

zyx

zyxzyx

E

e

sss

eee

eeeeee

For element subjected to uniform hydrostatic pres

modulusbulk

213

213

=-

=

-=-

-=

Ek

k

p

Epe

Subjected to uniform pressure, dilatation must be

negative, therefore

210


16/35

SHEARING STRAIN

Elemen kubus yg mengalami tegangan geser a

terdeformasi menjadi genjang. Regangan gese

bersesuaian diukur dalam hal perubahan sudut

sisi,

xyxy f gt =

A plot of shear stress vs. shear strain is similar

previous plots of normal stress vs. normal stra

except that the strength values are approximate

half. For small strains,

zxzxyzyzxyxy GGG gtgtgt ===

where G is the modulus of rigidity or shear mo


17/35

EXAMPLE

Sebuah balok kotak dari bahan dg

modulus of rigidity G = 90 ksi

dilekatkan pada dua plat horisontal. Plat

bawah dipasang permanen sedangkanplat atas dikenai gaya horisontalP yg

mengakibatkan plat bergeser 0.04 in.

Tentukan a) tegangan geser bahan, b)

Gaya P yg mengenai plat atas..

SOLUTION:

Determine the average ang

deformation or shearing st

the block.

Use the definition of shear

find the forceP.

Apply Hookes law for she

and strain to find the corre

shearing stress.


18/35

Determine the average angular deform

or shearing strain of the block.

rad020.0in.2

in.04.0tan == xyxyxy ggg

Apply Hookes law for shearing stress

strain to find the corresponding shearin

stress.

p1800rad020.0psi1090 3 === xyxy Ggt

Use the definition of shearing stress to the forceP.

1036in.5.2in.8psi1800 3=== AP xyt

kips0.36=P


19/35

RELATION AMONG E, , AND G

An axially loaded slender bar will

elongate in the axial direction and

contract in the transverse directions.

= 12G

E

Components of normal and shear stra

related,

If the cubic element is oriented as in

bottom figure, it will deform into a

rhombus. Axial load also results in astrain.

An initially cubic element oriented a

top figure will deform into a rectangu

parallelepiped. The axial load produ

normal strain.


20/35

EXAMPLE

Sebuah lingkaran berdiameter d= 9 in.

digoreskan pada plat aluminium dg keteba

= 3/4 in. Gaya dikenakan pada bidang plasehingga mengakibatkan tegangan normal

= 12 ksi and sz = 20 ksi.

Bila E= 10x106psi dan = 1/3, tentukan

perubahan dari:

a) Panjang diameterAB,

b) Panjang diameter CD,

c) Tebal plat, dan

d) Volume plat.


21/35

SOLUTION:

Apply the generalized Hookes Law to

find the three components of normal

strain.

in./in.10600.1

in./in.10067.1

in./in.10533.0

ksi203

10ksi12

psi1010

1

3

3

3

6

-

-

-

=

--=

-=

--=

=

--

=

--=

EEE

EEE

EEE

zyxz

zyxy

zy

xx

ssse

ssse

ssse

Evaluate the deformation compo

9in./in.10533.0 3-== dxAB e

9in./in.10600.1 3-== dzDC e

i75.0in./in.10067.1 3--== tyt e

in108.4 3-=AB

104.14 3-=DC

in10800.0 3--=t

Find the change in volume

3

33

75.0151510067.1

/inin10067.1

==

==

-

-

eVV

e zyx eee

3in187.0=V


22/35

COMPOSITE MATERIALS

Fiber-reinforced composite materials are forme

from lamina of fibers of graphite, glass, or

polymers embedded in a resin matrix.

z

zz

y

yy

x

xx EEE

e

s

e

s

e

s===

Normal stresses and strains are related by Hook

Law but with directionally dependent moduli o

elasticity,

x

zxz

x

yxy

e

e

e

e -=-=

Transverse contractions are related by direction

dependent values of Poissons ratio, e.g.,

Materials with directionally dependent mechan

properties are anisotropic.

SAINT VENANTS PRINCIPLE


23/35

SAINT-VENANTS PRINCIPLE

Loads transmitted through rigid

plates result in uniform distribut

of stress and strain.

Saint-Venants Principle:

Stress distribution may be assu

independent of the mode of loa

application except in the imme

vicinity of load application poi

Stress and strain distributions

become uniform at a relatively l

distance from the load applicatio

points.

Concentrated loads result in larg

stresses in the vicinity of the loa

application point.


24/35

STRESS CONCENTRATION: HOLE

Discontinuities of cross section may result in high

localized or concentrated stresses. ave

max

s

s=K


25/35

STRESS CONCENTRATION: FILLET


26/35

EXAMPLE 2.12

Determine the largest axial load P

that can be safely supported by a

flat steel bar consisting of two

portions, both 10 mm thick, and

respectively 40 and 60 mm wide,

connected by fillets of radius r = 8mm. Assume an allowable normal

stress of 165 MPa.

SOLUTION:

Determine the geometric ratios a

find the stress concentration factofrom Fig. 2.64b.

Apply the definition of normal stfind the allowable load.

Find the allowable average norm

stress using the material allowab

normal stress and the stress

concentration factor.


27/35

Determine the geometric ratio

find the stress concentration f

from Fig. 2.64b.

82.1

mm40

mm850.1

mm40

mm60

=

===

K

d

r

d

D

Find the allowable average no

stress using the material allow

normal stress and the stress

concentration factor.

M7.90

82.1

MPa165maxave ===

K

ss

Apply the definition of norm

to find the allowable load.

N103.36

.90mm10mm40

3=

== aveAP s

36=P

ELASTOPLASTIC MATERIALS


28/35

ELASTOPLASTIC MATERIALS

Previous analyses based on assumpt

linear stress-strain relationship, i.e.,

stresses below the yield stress

Assumption is good for brittle materwhich rupture without yielding

If the yield stress of ductile materials

exceeded, then plastic deformations

Analysis of plastic deformations is

simplified by assuming an idealized

elastoplastic material

Deformations of an elastoplastic mat

are divided into elastic and plastic ra

Permanent deformations result from

loading beyond the yield stress

PLASTIC DEFORMATIONS


29/35

PLASTIC DEFORMATIONS

Elastic deformation while max

stress is less than yield stressK

AAP ave

maxss ==

Maximum stress is equal to the

stress at the maximum elastic

loadingK

AP YY

s=

At loadings above the maximu

elastic load, a region of plastic

deformations develop near the

As the loading increases, the p

region expands until the sectio

a uniform stress equal to the yi

stressY

YU

PK

AP

=

=s


30/35

RESIDUAL STRESSES

When a single structural element is loaded uniformly

beyond its yield stress and then unloaded, it is permanently

deformed but all stresses disappear. This is not the general

result.

Residual stresses also result from the uneven heating or

cooling of structures or structural elements

Residual stresses will remain in a structure after

loading and unloading if

- only part of the structure undergoes plastic

deformation

- different parts of the structure undergo different

plastic deformations

EXAMPLE


31/35

EXAMPLE

Sebuah batang silinder ditempatkan

didalam tabung dengan panjang yang

sama. Pada ujung batang dan tabung

dipasang plat penopang yang kokoh

dan ujung satunya dijepit. Beban tarikdikenakan pada plat penopang secara

perlahan hingga mencapai 5,7 kips dan

kemudian dihilangkan.

a) Gambarkan diagram beban-

defleksi dari gabungan batang-

tabungb) Tentukan elongasi maksimum

c) Tentukan bagian yg mengalami

deformasi permanen

d) Hitung tegangan sisa pada batang

dan tabung

ksi36

psi1030

in.075.0

,

6

2

=

=

=

rY

r

r

E

A

ksi45

1015

in100.0

,

6

=

=

=

tY

t

t

E

A

E ample


32/35

a) draw a load-deflection diagram for the rod

tube assembly

136in.30psi1030

psi1036

kips7.2in075.0ksi36

6

3

,

,,

2,,

=

===

===

LE

L

AP

rY

rYrYY,r

rrYrY

se

s

1009in.30

psi1015

psi1045

kips5.4in100.0ksi45

6

3

,

,,

2,,

=

===

===

LE

L

AP

tY

tYtYY,t

ttYtY

se

s

tr

tr PPP

==

=

Example

b c) determine the maximum elongation and permane

Example


33/35

b,c) determine the maximum elongation and permane

at a load ofP= 5.7 kips, the rod has reached the

plastic range while the tube is still in the elastic r

in.30psi1015

psi1030

ksi30in0.1

kips0.3kips0.3kips7.27.5

kips7.2

6

3

t

2t

,

===

===

=-=-=

==

LE

L

A

P

PPP

PP

t

tt

t

t

rt

rYr

se

s

160max == t

the rod-tube assembly unloads along a line parall

to 0Yr

in.106.4560

in.106.45in.kips125

kips7.5

slopein.kips125in.1036

kips5.4

3maxp

3max

3-

-

-

-==

-=-=-=

==

=

m

P

m

4.14 =p

Example

Example


34/35

calculate the residual stresses in the rod and t

calculate the reverse stresses in the rod and tub

caused by unloading and add them to the max

stresses.

ksi2.7ksi8.2230

ksi69ksi6.4536

ksi8.22psi10151052.1

ksi6.45psi10301052.1

in.in.1052.1in.30

in.106.45

,

,

63

63

33

=-==

-=-==

-=-==

-=-==

-=-

=

=

-

-

--

tttresidual

rrrresidual

tt

rr

.

E

E

L

sss

sss

es

es

e

Example

Batang kaku BDE ditopang oleh dua batang AB dan CD. Ba


35/35

g p g g

terbuat dari aluminium (E = 70 Gpa) dg luas penampang 5

sedangkan batang CD terbuat dari baja (E = 200 Gpa) de

penampang 600 mm2. Untuk beban 30 kN seperti pada ga

defleksi (lendutan) di titik B, D dan E !!

A circle of diameter d= 9 in. is scribed on an unstressed aluminum plate of th

Forces acting in the plane of the plate later cause normal stresses sx = 12 ksi

ForE= 10x106psi and = 1/3, determine the change in:

a) the length of diameterAB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.

2_kuliah stress strain heru sukanto

Documents