Strength of Materials - Lecture Notes / Mehmet Zor

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YIELD AND FRACTURE CRİTERIA

(tvid- 7)

7.

(for Isotropic Materials.)

Strength of Materials - Lecture Notes / Mehmet Zor

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7.1 What are Flow and Fracture Criteria? Why are they necessary?

• In the topics so far, strength calculations were made when there was only one type of stress (σ or τ) at a point.
• For example, in axial loading (tension-compression) and bending, only σ_x stress occurs and is compared with allowable normal stress (σ_all).
• In torsion and shear loading, only τ_xy occurs and is compared with allowable shear stress (τ_all).
• Now we are looking for an answer to the following question: When more than one stress type (σ_x, τ_xy, σ_y, etc.) exists at the same time at a point, which one will we compare with σ_all or τ_all? or what path will we follow?

There are multiple criteria that answer this question for isotropic materials:

• Yield criteria determine whether yielding exists for ductile materials.
• Fracture criteria allow us to check whether there is fracture in brittle materials.
• Here we will examine the 4 most important of these criteria.

7.Yield and Fracture Criteria for Isotropic Materials

?

?

σ_x , τ_{xy ,} σ_z….

Figure 7.1

Figure 7.2

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7.Yield and Fracture Criteria for Isotropic Materials

7.2.1 Tresca Criterion

( Other Name: Maximum Shear Stress Criterion)

According to this criterion, the condition for yielding at any point of a material is that the maximum shear stress at that point is equal to or greater than the shear stress at the yield limit in simple tension.

The condition for yielding is:

7.2 Yield Criteria for Ductile Materials

This criterion gives very good results in ductile materials.

We think that a 3-dimensional stress state is formed at a point like A as a result of the forces applied to the object shown on the side, which is made of the same material as the tensile sample. We pay attention to the 3D Mohr circle at this point A. Since the great circle exceeds the yield limits, yielding will occur at point A.

Mohr circle for point A

(7.1)

D₁

D₂

Figure 7.3

Figure 7.4

Strength of Materials - Lecture Notes / Mehmet Zor

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7.Yield and Fracture Criteria for Isotropic Materials

7.2.2 Von-Mises Criterion (other names: Equivalent Stress or Distortion Energy Criterion)

Yield Condition:

This criterion reduces the multiaxial stress state to the uniaxial stress state. It defines 1 equivalent stress (𝜎𝑣𝑚) instead of all stresses. This criterion says that "If this equivalent stress equals or exceeds the yield stress, yielding occurs at that point". This criterion gives excellent results in yield control of ductile materials.

Equivalent stress ((or Von-Mises Stress) calculation in terms of Principal Stresses:

Equivalent Stress (or Von-Mises Stress) calculation in terms of Stress Components:

(7.2a)

(7.2b)

* Equivalent stress formulas are derived from the concepts of strain energy and distortion energy. To understand how these are extracted, Strength of Materials books should be studied.

(7.3)

Figure 7.5

Figure 7.6

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and/or

7.3.1 Rankine Criterion (Maximum Principal Stress Criteria)

According to this criterion, in order for a material to fracture at any point, the maximum principal stress at that point must exceed the tensile strength of the material (𝜎_𝑜) or the minimum principal stress must exceed the compressive strength (𝜎_𝑜𝑐).

Fracture Condition:

In order for there to be no break at a point, the three-dimensional Mohr circle at that point must remain within the vertical red lines in the figure.

It is a criterion that gives partially good results for the breaking of Brittle Materials.

7.3. Fracture Criteria for Brittle Materials

7.Yield and Fracture Criteria for Isotropic Materials

σ_ο

σ_οc

: Tensile Strength (Tensile stress at fracture time in simple tension test )

: Compressive Strength (Compressive stress at fracture time in simple compressive test )

If at least one of the equations 7.4a and 7.4b is satisfied, a fracture occurs at that point.

If we explain this criterion from Mohr's Circle:

Simple Tensile test

(at fracture time)

Simple Compressive test (at fracture time)

3D Mohr circle of point B

(7.4a)

(7.4b)

Since the Mohr circle of point B remains within the red lines, no fracture occurs at this point.

Figure 7.7

Figure 7.8

Strength of Materials - Lecture Notes / Mehmet Zor

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7.Yield and Fracture Criteria for Isotropic Materials

7.3.2 Columb Criterion

According to this criterion, which gives very good results for brittle materials, the following inequality must be satisfied in order to avoid fracture at any point in a material.

σ₁

σ_ο

σ₃

σ_οc

: Maximum principal stress

: Minimum principal stress

: Tensile Strength

: Compressive Strength

If the 3D Mohr circle at a point remains within the tangents (red lines) of the Mohr circles at the moment of fracture of simple tension and simple compression, there will be no fracture at that point.

If we explain this criterion from Mohr's Circle:

Simple tension at fracture instant

Simple compression at fracture instant

3D Mohr circle of point D

Since the Mohr circle at point D remains between the tangents (within the red lines), no fracture occurs according to the Coulomb criterion. (However, according to the Rankine criterion, a break occurs as σ_oc is passed.)

(7.5)

Figure 7.9

Figure 7.10

Strength of Materials - Lecture Notes / Mehmet Zor

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7.Yield and Fracture Criteria for Isotropic Materials

Example 7.1:

The principal stresses at any two points a and b in a brittle material with σ₀ =100ΜPa, σ_οc=-400MPa are given in the table below. Check the breakage at these two points according to the Rankine and Columb criteria.

σ₁ σ₃

a 50 -100

b 90 -100

According to Rankine criterion:

According to Columb criterion:

No fracture occurs at either point

No fracture occurs at point a

Fracture occurs at point b

Check for point a:

Check for point b:

Check for point a:

Check for point b:

Strength of Materials - Lecture Notes / Mehmet Zor

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7.Yield and Fracture Criteria for Isotropic Materials

According to this criterion, fracture occurs at loads that are outside the envelope curves of the Mohr circles at the instant of fracture in simple tension, simple compression and complete shear. It is a criterion that gives very good results for brittle materials.

* As can be seen, in brittle materials, the necessity of examining the principal stresses in all three criteria emerges.

* Principal stresses can be calculated, but an additional separate evaluation is required for the Columb and Mohr criteria.

7.3.3 Mohr Criterion (For Informational Purposes)

Figure 7.11

Strength of Materials - Lecture Notes / Mehmet Zor

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Example 7.2

At the point where the plane stress state is given in the figure, Check for damage (yield or fracture)

a-) For aluminum material, (σ_yield= 200MPa, τ_yield= 100MPa)

b-) For wooden material, (σ_{tension-fracture}= σ_o=160MPa, σ_{compressive-fracture}= σ_oc=-400MPa )

.

Principal Stresses:

7.Yield and Fracture Criteria for Isotropic Materials

For the criteria, principal stresses and maximum shear stress must be calculated. We will use the Mohr circle for this.

Radius of Mohr's Circle :

Mohr circle in the x-y plane

Solution:

Figure 7.12

Figure 7.13

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We see that the Mohr circle we drew on the x-y plane is the largest circle. Therefore :

We drew the 3D Mohr circle on the same figure. (we added 2 more circles with dashed lines)

Or we could make the following comment without drawing the 3D Mohr circle:

Now we can move on to the solution of the options:

a-)

Since aluminum is a ductile material, it is controlled according to yield criteria.

1-) According to Tresca criterion

We find the equivalent stress from equation 7.2a:

2-) Yield control of aluminum according to von mises criterion:

..>>

7.Yield and Fracture Criteria for Isotropic Materials

Figure 7.14

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b-) Since wood is a brittle material, it is checked according to fracture criteria.

1-) According to Rankine criterion:

(no fracture occurs.)

Attention: One of the biggest mistakes we make as engineers is to use Tresca or Von-Mises yield criteria in brittle materials and Rankine or Columb fracture criteria in ductile materials. For this reason, it is necessary to know very well from the beginning whether the material is ductile or brittle.

(fracture occurs.)

7.Yield and Fracture Criteria for Isotropic Materials

Or the same result can be found by using principal stresses (from equation 7.2b):

…. >1

2-) According to Columb criterion:

(Note: In the 8th topic, there will be examples where yield and fracture criteria are used. )

If we substitute the values:

(no yield ocurs)

Strength of Materials - Lecture Notes / Mehmet Zor

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7.4 Why didn't we use these criteria for single type loadings?

7.Yield and Fracture Criteria for Isotropic Materials

For example: Let's consider a rod subjected to tensile loading only in the x direction.

F

F

x

All other stress components are zero.

Now let's prove that condition (I) complies with the yield criteria:

According to Von mises criterion:

Yield condition:

As can be seen, the yield condition (I) that we used before for single type loading is compatible with the Von-mises criterion.

(I)

We obtained the same condition as (I).

Figure 7.15

Strength of Materials - Lecture Notes / Mehmet Zor

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7.Yield and Fracture Criteria for Isotropic Materials

F

F

x

Acording to Tresca criterion:

At yielding instant:

If we draw the Mohr circle when there is only 𝜎_𝑥 (simple tension):

At any instant :

Yield condition according to Tresca

Therefore, the yield condition (I) on the previous page is also compatible with the Tresca criterion.

Notice that in the 3D Mohr circle, there is only this circle, and the other two circles are points.

Homework: Try to prove in a similar way that single type loading complies with all yield and fracture criteria.

We achieved the same condition.

We have proven this equation, which we have used before in ductile materials, here as well.

Figure 7.15

Figure 7.16

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A plane stress state occurs at the Q point of an object that is exposed to the influence of external loads and is in static equilibrium, and the stress values ​​are shown in the figure.

Accordingly, taking the safety coefficient n = 2,

a-) Yielding and safety control if the material is steel,

b-) If the material is Alumina, check for fracture and safety.

Steel (ductile) : σ_-yield = 400MPa, τ_-yield =200MPa

Alümina AL₂O₃ (brittle) : σ_{-fracture-tension} = 262MPa, σ_{-fracture-compression} = -2600MPa

Question7.3

7.Yield and Fracture Criteria for Isotropic Materials

Answers:

a-) no yielding occurs and safe

b-) no fracture occurs but unsafe.

Figure 7.17

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Answers:b-) σ_max =-51.46MPa, σ_min =-118.54MPa, θ_p=31.71^o , c-) σ_x’ =-111.93MPa, τ_x’y’ =19.98MPa, d-) τ_max =59.27MPa, e-) Yield occurs, f-) no fracture occurs.

7.Yield and Fracture Criteria for Isotropic Materials

Question 7.4* (2017-Final) For the point where the plane stress state is shown, a-10) Draw the Mohr Circle. b-5) Determine the principal stresses and their directions. c-10) Calculate the stress components in the shaded m-m plane. d-) Determine the maximum shear stress. According to the correct one of the Columb or Tresca criteria: e-10) Perform the yielding control for the ductile material with a yield stress of 100MPa. f-10) Perform the fracture control for the brittle material with tensile strength of 200MPa and compression strength of 400MPa.

Figure 7.18