defined stress and strain defined elastic properties

Solve examples

Learn stress transformation and principal stresses

Failure theories

Last week, we:

Defined stress and strain

Defined elastic properties.

Learned stress strain relations for uniaxial case

This week, we will

Gave multiaxial stress-strain relations

GYTE öğrencileri tarafından çoğaltılmasında sakınca yoktur

Multiaxial Stress-Strain


Elastic Constants

Hydrostatic or mean stress

Bulk modulus

Volumetric strain

For an isotropic material two elastic constant are required and enough to define elastic behaviour

D =

(D=DV/V)


Copyright © 2011 Pearson Education South Asia Pte Ltd

Example:A bar made of A-36 steel (E=200 GPa, n=0.32) has the dimensions shown in Fig. 3–22. If an axial force of P = 80 kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. What is the change in volume? The material behaves elastically.



• The normal stress in the bar is

• Given that E = 200 GPa

• The axial elongation of the bar is therefore

Solution

mm/mm 108010200

100.16 6

9

6

===E

z

z

(Ans) m120101205.11080 66

z ==== mL

zz

MPa 16Pa 100.1605.01.0

1080 6

3

====A

Pz



• The contraction strains in both the x and y directions are

• The changes in the dimensions of the cross section are

Solution

(Ans) m28.105.0106.25

(Ans) m56.21.0106.25

6

6

===

===

yyy

xxx

L

L

(Ans) 106.25- 1080)32.0( - 66 ====zyx

v



Solution

To find the change in volume

V = V0 [1 + ε x + ε y + ε z]

zyx =

V

VoV

66 1028.8 10)6.256.2580( ==

V

VoV

356 1016.21028.8 .)05.0.1.0.5.1( mVoV ==


Example: Elastik modulusu E, Poisson oranı n, akma mukavemeti Sy olan aluminyumdan yapılmış L x W x t (Uzunluk X Genişlik X Kalınlık) boyutlarında bir plaka uzunluk yönünde P yükünün etkisi altındadır, genişlik yönünde ise iki ucundan ankstredir (genişlik sabit). Bu koşullar altında uzunluğundaki değişmeyi ve kalınlığındaki azalmayı hesaplayınız.

x

y

P P


A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi.

For E = 10x106 psi and n = 1/3, determine the change in:

a) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.

Example:


SOLUTION:

• Apply the generalized Hooke’s 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

zyxx

nn

nn

nn

• Find the change in volume

33

333

in75.0151510067.1

/inin10067.1

. =D=D

==D

VV

zyx

3in187.0=DV

• Evaluate the deformation components.

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

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

in.75.0in./in.10067.1 3== tyt

in.108.4 3=AB

in.104.14 3=DC

in.10800.0 3=t


AVERAGE NORMAL STRESS

A

P=


NORMAL AND SHEAR STRESS

A

Fz

Az

D

D=

D 0lim

A

F

A

F

y

Azy

x

Azx

D

D=

D

D=

D

D

0

0

lim

lim


Stress Transformation Stress depends on orientation Force P has normal and shear components on the surface mm. As the orientation of the surface mm changes, so does the normal and shear stresses.

Two arbitrary coordinate systems and an element subjected to arbitrary stresses

Stresses on the element Forces on the element

Biaxial loading (Plane stress, z=0,xz=0, yz=0)


Stress Transformation

Similarly, we can find :

Simplifying dA terms, we get


Stress Transformation

Remember

Note: σx’ + σy’ = σx + σy

Thus the sum of the normal stresses on two perpendicular plane is an invariant quantity and independent of orientation or angle θ.


Biaxial-plane stress condition • Two principal stresses, σ1 and σ2 (σ3 =0).

Principle Stresses • When there is no shear stresses acting on the planes (=0, the normal stresses acting on the planes are maximum. • These planes are called the principal planes, and stresses normal to these planes are the principal stresses σ1, σ2 and σ3 which in general do not coincide with the cartesian coordinate axes x, y, z. Directions of principal stresses are 1, 2 and 3.

Triaxial • Three principal stresses, σ1 , σ2 and σ3 , where σ1 > σ2 > σ3.


Principle Stresses

Plane stress:

To determine yield we will need maximum shear stress, max

at

Example: , then

and

Be aware: 1> 2 > 3 , therefore 400 >0 > -100

max=250 MPa


Mohr Circle

A: x, -xy) B: y, xy)

• Normal stresses are plotted along the x axis, shear stresses along the y axis. • A point on Mohr’s circle gives the magnitude and direction of the normal and shear stresses on any plane in the physical element. • A shear stress causing a clockwise rotation about any point in the physical element is plotted above the horizontal axis of the Mohr’s circle. • The stresses on the planes normal to the x and y axes are plotted as points A and B. • The shear stress is zero at points D and E, representing the values of the principal stresses σ1 and σ2 respectively. • The Angle between σx and σ1 on Mohr’s circle is 2θ.


Mohr Circle Example:


Triaxial principle stresses

where I1, I2, and I3

are called stress invariants.

The maximum principal shear stress, τmax is given by

τmax =


Common Triaxial Cases

(d ) Torsion (e) Hydrostatic compresion GYTE öğrencileri tarafından çoğaltılmasında

sakınca yoktur

Failure Criteria

1 Sut or 3 -Suc => Failure

The Maximum Normal Stress Theory (Rankine)

• States that failure occurs whenever the largest principal stress at any point

reaches a value equal to the strength of the specimen.

1 Syt or 3 -Syc => Failure

where Syt and -Syc are the tensile and compressive yield strengths, respectively.

• This theory does not agree with experiments and fails to predict yield in metals.

For example, metals do not yield under hydrostatic tension or compression.

•This theory can be used to predict failure of brittle materials.


Failure Criteria The Maximum Shear Stress Theory (Treasca)

States that yielding begins whenever the maximum shear stress becomes equal to the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield.

=> Failure


Failure Criteria The Distortion Energy Theory (von Mises)

Observation: Hydrostatic pressure does not cause yilding.

Equate the energies that produces angular distortation in the simple tension test and the given stress state.

=> Failure


Comparison of Failure Criteria

Failure occurs when stress combinations fall outside the envolope for applicable theory.

Maximum shear stress theory gives more conservative results. For pure shear loading DE predicts y=0.577 Sy MSS predicts y=0.5Sy

MNS

Failure envolope for plane stress case.


Comparison of Failure Criteria

Failure occurs when stress combinations fall outside the envolope for applicable theory.

Maximum shear stress theory gives more conservative results. For pure shear loading DE predicts y=0.577 Sy MSS predicts y=0.5Sy

Failure envolope for plane stress case.

(a) Comparison of the

Rankine, von Mises,

and Tresca criteria.

(b) Comparison of

failure criteria with test

+ cast iron

Mechanical Behavior of Materials, Hosford


Example: A region on the surface of a 6061-T4 aluminum alloy

component has strain gages attached, which indicate the following

stresses:

σ11 = 70MPa, σ22 = 120MPa, σ12 = 60MPa.

Determine the yielding for both the Tresca and von Mises

criteria, given that σ0 =150 MPa (the yield stress).

Solution: We first have to establish the principal stresses. This is easily

accomplished by a Mohr circle construction or by its analytical expression

(the equation of a circle):


Solution (continued)

According to Tresca, max =(160 − 0)/2=80 MPa.

The value max =80 MPa exceeds the Tresca criterion (σ0/2=75 MPa)

and the alloy would be unsafe.

The von Mises criterion gives

σe = 147 MPa

σe < σ0= 150 MPa therefore, the material does not yield.

Plainly, the Tresca criterion is more conservative than von Mises.


Bir uçak seyahatiniz sırasında şans! eseri uçağı tasarlayan

mühendislerden birinin yanına oturuyorsunuz. Mühendis size tasarım

sırasında von Mises kriterini kullandıklarını söylüyor. Tresca kriterini

kullanmış olmalarını tercih eder miydiniz? Açıklayın.

Example:

Answer:

We prefer Tresca criterian. Tresca allows lower stresses in design

(means thicker parts for the same load) therefore reducing the risc of

exceeding the strength of the material.


Example:


Answer:

Lower surface Upper surface

max =/2 =y

=x=500

=Sy/2 =500/2

Both surfaces yield at the same time.


• Design uncertainties mean we do not push the limit. • Factor of safety, N

N

y

working

=

Often N is between 1.2 and 4

• Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5.

Design or Safety Factors

4

0002202 /d

N,

5

N

y

working

= 1045 plain

carbon steel: y = 310 MPa

TS = 565 MPa

F = 220,000N

d

L o

d = 0.067 m = 6.7 cm


Plastic Deformation


defined stress and strain defined elastic properties

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