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7/26/2019 2_kuliah Stress Strain Heru Sukanto
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TEGANGAN REGANGANAKIBAT
BEBAN AKSIAL
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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
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Berapa defo
?
Apanya yan
kedua spesim
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STRESS STRAIN DIAGRAM
specimenextensometer
Typical tensil
specimen
ga
len
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Zona Elastic dan Plast
Awal Under load
F
bostr
F
elastic + plas
Bondsstretch &
plane shea
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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=
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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
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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
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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)
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DEFORMASI (HOOKES LAW)
Tentukan deformasi (pertambahan panjng)
seperti gambar samping akibat beban yan
ditanggungnya (E = 200 GPa) !
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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 !!
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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
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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:
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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
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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
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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.
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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
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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.
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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.
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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
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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
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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.
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STRESS CONCENTRATION: HOLE
Discontinuities of cross section may result in high
localized or concentrated stresses. ave
max
s
s=K
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STRESS CONCENTRATION: FILLET
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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.