Ex19Principal Stress, Stress Intensity, AndVonMisesStress

7/28/2019 Ex19Principal Stress, Stress Intensity, AndVonMisesStress

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Exercise 19 Principal Stresses, Stress Intensity, and von Mises Stress 1

Exercise 19

Principal Stress, Stress Intensity, andvon Mises Stress

19-1 Introduction

We have introduced the definitions of stresses, one of the most important concepts in studying Mechanics ofMaterials. Why are stresses so important? An obvious reason is that a material may fail at a location when the stress

at that location reaches a certain critical stress value.

Ductile versus Brittle MaterialsWe usually conduct a uniaxial tensile test and plot a stress-strain relation such as 16-3[7] to determine the critical

stress value. According to the type of stress-strain relation, we may classify materials into two categories: ductile

materials and brittle materials. For a ductile material [1], the material exhibits a large amount of strain before it

fractures [2] while, for a brittle material [3], the material's fracture strain is relatively small [4]. Fracture strain is a

measure of ductility. There are essential differences between these two types of materials, namely failure points and

failure types.

Stress

Strain

y

Stress

Strain

f

[1] Ductilematerial. [3] Brittle

material.

[2] Fracturepoint.

[4] Fracturepoint.

[5] Yieldpoint.

Failure Point: Yield Point or Fracture Point?Mild steel is a typical ductile material. For ductile materials, there often exists an obvious yield point [5], beyond which

the deformation would be too large so that the material is no longer reliable or functional; the failure is accompanied

by excess deformation. Therefore, for ductile material, we are most concerned about whether the material reaches

the yield point. In a uniaxial tensile test, the yield point is characterized by a yield stress y

[5]. This is the critical

stress we want to compare with. But, with which stress we want to compare? X

? Y?

Z?

XY?

YZ?

ZX?

Cast iron and ceramics are two examples of brittle materials. For brittle materials, there usually exist no obviousyield point, and we are concerned about the fracture point. In a uniaxial tensile test, the fracture point is characterized

by a fracture stress f

[4]. This is the critical stress we want to compare with. But, again, with which stress we want

to compare? X

? Y

? Z

? XY

? YZ

? ZX

?


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2 Copyright by Huei-Huang Lee

Principal StressesIn Exercise 18, we've constructed a Mohr's circle for a special case, namely plane stress state (18-3). The concepts can

be extended to a general case. In general, it is possible to set up a local coordinate system (X-Y-Z) such that all shear

stress components vanish (XY

= YZ=

ZX= 0 ), and, among three of the normal stresses (

X

,Y,

Z), one of them

reaches a maximum and another of them reaches a minimum. These three normal stresses are called the principal

stresses of the stress state, and the corresponding three axes are called principal axes. The maximum principal stress isdenoted by 1, the middle principal stresses by

2, and the minimum principal stresses by

3. These quantities can

be reported from simulation results.

Failure Criterion for Brittle MaterialsAs mentioned, the failure of brittle materials is due to fracture (rather than yielding) and the fracture is in turn due to

an excessive tensile stress. Therefore, we may state a failure criterion for brittle materials as follows. At a certain

point of a body, if the maximum principal stress reaches the fracture tensile strength f

(which may be obtained from

a uniaxial tensile test) of the material, it will fail. In short, a point of material fails if

1 f (1)

Maximum Shear Criterion for Ductile MaterialsAlso mentioned, the failure of ductile materials is initiated by a yielding and the yielding is in turn due to an excessive

shear stress. Therefore, we may state a failure criterion for ductile materials as follows. At a certain point of a body, if

the maximum shear stress reaches the yielding shear stress y

of the material, it will fail. In short, a point of material

fails if

max

y

(2)The left-hand-side max can be reported from simulation results. The right-hand-side y can be obtained from auniaxial tensile test. In a uniaxial tensile test, when the yielding occurs, the stress state is as shown [6-8], therefore

y=

y

2 (3)

and Eq. (2) can be written as

max

y

2 (4)

Failure Type: Tensile Failure or Shear Failure?The fracture of brittle materials is mostly due to tensile failure; the yielding of ductile materials is mostly due to shear

failure. The tensile failure of brittle material is easier to understand, because the failure always occurs after cracking,

which is due to tensile stresses. Note that in a compressive test of concretes, which is a brittle material, the cracking

of concrete is due to tensile stresses, not compressive stresses; compressive stresses seldom cause a material failure.

The shear failure of ductile materials can be justified in a standard uniaxial tensile test, in which the failure isaccompanied by a necking process and a cone-shape breaking surface. It is important to note that a material may fail

due to a mix-up of both mechanisms.


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(

y, 0)

(0,0)

(

y, 0)

(

y, 0)

[6] Mohr's circle (stressstate) in a uniaxial

tensile test when theyielding occurs.

Tresca Criterion for Ductile MaterialsIt is easy to show that, for any stress state,

max

=

1

3

2 (5)

Substitution of (5) into (4) yields

1

3

2

y

2

or

1

3

y (6)

The quantity on the left-hand-side (1

3) is called the stress intensity, which the Workbench can report for you

on your request. The criterion (6) is called the Tresca Criterion, first proposed by Henri Tresca (1814-1885), a French

mechanical engineer, in 1864.

Von Mises Criterion for Ductile MaterialsCriteria (4) or (6) are not accurate enough for predicting the yielding of many ductile materials, particularly, metals. A

more sophisticated theory, called von Mises criterion, often predicts yielding of metals more accurately than criteria (4)

or (6). The von Mises theory is also based on shear failure, but derived from an energy consideration. The von Mises

criterion states that a material will begin to yield if its deviatoricstrain energyexcesses a certain criterion. A detailed

account of the theory will be presented after we introduce the notion ofstrain energy. Here, we simply state the

criterion as follows: a point of material yields if

1

2

1

2( )

2

+ 2

3( )

2

+ 3

1( )

2

y (7)

The quantity on the left-hand-side is called von Mises stress and is denoted by e; In ANSYS it is also referred to

as equivalent stress,

e=

1

2

1

2( )

2

+ 2

3( )

2

+ 3

1( )

2

(8)

[7] The maximumnormal stress is called

the yield strength.

[8] The maximum shearstress is called the

yielding shear strength.


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Which Criterion to Use?For a brittle material, we usually use the maximum principal stress criterion (1). For a ductile material, the maximum

shear stress criterion (4), or equivalently Tresca criterion (6), or von Mises criterion (7) can be used. As a general rule,

you should use the one that most accurately predicts the yielding of the material; that usually depends on the material

type. Many experiments shows that von Mises criterion (7) is the best among these criteria in predicting the yielding

of ductile metals.

Purpose of this ExerciseIn this exercise, we'll continue Exercise 18. We'll insert result objects and evaluate the principal stresses, the maximum

shear stress, the stress intensity, and the von Mises stress at the origin of . Note that these

stresses are independent of rotation of the coordinate system.

It should be noted that the material used in the C-bar is structural steel; it has a yield strength of 250 MPa and afracture strength (also called an ultimate tensile strength) of 460 MPa. A structural steel is classified as a ductile

material.

The principal stresses, from 18-10[3], are

1= 37.326 MPa (9)

2= 0.29805 MPa (10)

3= 5.1372 MPa (11)

The maximum shear stress, from 18-10[6], is

max= 21.232 MPa (12)

The stress intensity is

1

3= 37.326 (5.1372) = 42.463 MPa (13)

The von Mises stress is

e=

1

2

1

2( )

2

+ 2

3( )

2

+ 3

1( )

2

=1

237.326 + 0.29805( )

2

+ 0.29805+ 5.1372( )2

+ 5.1372 37.326( )2

= 40.262 MPa

(14)


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[1] Launch

[2] Open theproject "CBar,"

saved in Exercise 18.

[3] Double-click to start up

.

19-2 Start Up

19-3 Insert Result Objects

[1] With highlighted, insertthese six result

objects respectively.

[2] The insertedresult objects.


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19-4 View the Results

[1] Click .

[2] Highlight to view the

maximum principalstresses.

[3] Highlight to view the

maximum shearstresses.

[4] Highlight

toview the stress

intensities.

[5] Highlight

to view the von Misesstresses.


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[1] Select all sixobjects and right-click-select .This cleans up the

results.

[5] Click .

19-6 View the Results

[1] Highlight. Thenumerical value isconsistent with that

in Eq. 19-1(9).

[2] Highlight .

The numerical valueis consistent with

that in Eq. 19-1(10).

19-5 Re-scope the Result Objects [2] After clearing up,the objects turn toyellow thunders.

[3] Select for.

[4] Select .Note that, without

clearing data,Workbench prohibitsyou to redefine the

scope.


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[3] Highlight. Thenumerical value is

consistent with that

in Eq. 19-1(11).

[5] Highlight . The

numerical value isconsistent with that

in Eq. 19-1(13).

[4] Highlight. Thenumerical value is

consistent with thatin Eq. 19-1(12).

[6] Highlight.

The numerical valueis consistent withthat in Eq. 19-1(14).

Wrap UpClose , save the project, and exit Workbench.

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Ex19Principal Stress, Stress Intensity, AndVonMisesStress