ENG5312 – Mechanics of Solids II

Material Property Relationships

Generalized Hooke’s Law

Assumptions:

o Isotropic, homogeneous material;

o The principal of superposition applies, i.e. linear-elastic material behaviour with small deflections.

Assume a material is exposed to three normal stresses

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σ x,

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σ y and

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σ z , which will result in three normal strains

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εx ,

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εy and

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εz .

Using Hooke’s Law, the strain in the

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x -direction due to

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σ x is:

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εx' = σ x

E

Due to Poisson’s ratio there will be normal strains in the

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x -direction due to

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σ y and

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σ z :

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εx'' =−ν

σ yE

;

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εx''' =−ν

σ zE

Using superposition, the total normal strain in the

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x -direction is the sum of the components due to stresses

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σ x ,

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σ y and

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σ z :

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εx =1Eσ x −ν σ y +σ z( )[ ] (36a)

Similarly, in the

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y and

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z-directions:

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εy =1Eσ y −ν σ x +σ z( )[ ] (36b)

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εz =1Eσ z −ν σ x +σ y( )[ ] (36c)

Note: A rectangular block will remain rectangular due to the application of

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σ x,

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σ y and

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σ z since no shear strain exists.

From experiment, it has been found that the application of a shear stress

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τ xy will cause an element to deform due to shear strain

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γxy only. Therefore, Hooke’s Law for shear stress and strain can be written as:

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ENG5312 – Mechanics of Solids II

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γxy =τ xyG

;

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γyz =τ yzG

;

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γxz =τ xzG

(37)

The generalized form of Hooke’s Law Eqs. (EQREF) can be used to determine three-dimensional deformations due to an applied stress field, or stress field due to given strains.

Relationships between E,

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ν and G

The derivation requires use of the principal stresses, generalized Hooke’s Law, and the strain transformation equations:

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G = E2 1+ν( ) (38)

Dilation

Consider an element (

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dx,dy,dz ) exposed to a stress field (

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σ x ,

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σ y and

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σ z ). The deformed dimensions of the element will be

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(1+ε x )dx ,

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(1+ε y )dy and

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(1+ε z )dz , and the change in volume (

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δV ) of the element will be:

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δV = 1+ε x( ) 1+ε y( ) 1+ε z( )dxdydz − dxdydz

Since the strains are small, the strain products can be neglected, and

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δV can be written as:

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δV = ε x +ε y +ε z( )dxdydz

Division of

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δV by the volume

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dV gives the dilation (

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e) or volumetric strain, i.e. the change in volume per unit volume.

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e = δVdV= ε x +ε y +ε z (39)

Using the generalized Hooke’s Law the dilation can be written in terms of the applied stresses:

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e =1−2νE

σ x +σ y +σ z( ) (40)

The dilation can be used to determine changes in volume due to an applied stress field.

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ENG5312 – Mechanics of Solids II

Bulk Modulus

Consider an element subjected to uniform pressure on all sides (e.g. a block submerged in water), then the normal stresses

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σ x =σ y =σ z =−p (i.e. compressive stress due to applied pressure p), and using Eq. (40):

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pe= E3(1−2ν )

= k (41)

The LHS (normal stress/dilation) is similar to

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σ /ε = E , therefore, the RHS is called a bulk modulus,

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k .

Note: For most metals,

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ν ≈1/ 3, therefore,

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k ≈ E . For a material that does not change volume,

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k→∞, therefore, the maximum theoretical value of

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ν is 0.5. During yielding a material does not change volume, therefore,

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ν =0.5 is used for plastic yielding.

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ENG5312 – Mechanics of Solids II

Theories of Failure To determine if a specific design will fail it is first necessary to determine

the maximum normal and shear stresses that occur in a member (using mechanics fundamentals and stress concentration factors). Then the principal stresses must be determined at these locations in the member. Then an appropriate failure theory for biaxial or tri-axial stress field can be applied.

Ductile Materials

For a uni-axial state of stress, the failure of a ductile material is usually specified by the initiation of yielding.

Two failure theories exist for a ductile material exposed to multi-axial stress fields: 1) maximum-shear-stress theory; and 2) maximum-distortion-energy theory.

1) Maximum-Shear Stress Theory

If a ductile material is subjected to a tension test it yields along slip planes oriented at

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≈45o to the axis of the applied load. Using Mohr’s circle for an element in the tension test sample:

At

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90o to the applied normal stress (here the principal axis) on Mohr’s circle we get the maximum shear stress. Therefore, the material is failing in shear at

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45o .

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ENG5312 – Mechanics of Solids II

Henri Tresca used this result to determine a failure theory applied to the failure of ductile materials subjected to any loading.

The maximum-shear-stress theory (or Tresca yield criterion) states: yielding of a ductile material begins when the maximum shear stress in the material reaches the shear stress in the material that causes yielding in a simple axial tension test, or:

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τ absmax = σ Y

2

Where

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σY is determined from a tension test.

The maximum shear-stress-theory requires

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τ absmax for the applied

state of stress, therefore, the principal stresses are required. Remember, if the principal stresses have the same sign then

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τ absmax

occurs out of plane:

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τ absmax = σ max

2

And if the principal stresses have opposite signs

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τ absmax occurs in

plane:

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τ absmax = σ max −σ min

2

The maximum-shear-stress theory can be stated as follows when

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σ1 and

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σ 2 have the same sign:

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σ1 =σ Y or

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σ1 =σ Y (42a)

or as follows when

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σ1 and

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σ 2 have opposite signs:

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σ1 −σ 2 =σ Y (42b)

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ENG5312 – Mechanics of Solids II

Plot the point (

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σ1,σ 2). If it falls within the hexagon the material will not fail. If the point falls on the boundary or outside the hexagon, the material will fail.

2) Maximum-distortion-energy theory

A material stores energy internally as it is deformed by an external loading. The strain-energy density (

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u) is the energy stored pre unit volume, and when a material is subjected to uni-axial stress:

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u = 12σε

Consider an element subjected to the three principal stresses

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σ1,

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σ 2 and

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σ 3, then using superposition the total strain-energy density is:

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u = 12σ 1ε1 +

12σ 2ε 2 +

12σ 3ε 3

And using Hooke’s Law, Eq. (36):

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u = 12E

σ 12 +σ 1

2 +σ 12 −2ν σ 1σ 2 +σ 1σ 3 +σ 2σ 3( )[ ] (43)

The strain-energy density can be perceived as being composed of a part due to volume change, and another part due to change of shape. Defining

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σ avg = (σ 1 +σ 2 +σ 3) / 3, then

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σ avg which acts on all faces of an element will cause the same principal strains in all directions (i.e.

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σ avg causes a change in volume but no change in shape). The remaining portion of the stress will cause the change in shape, i.e.

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(σ 1 −σ avg ),

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(σ 2 −σ avg ) and

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(σ 3 −σ avg ).

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ENG5312 – Mechanics of Solids II

Experiments have shown that materials do not yield when subjected to uniform stress (e.g.

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σ avg above). Therefore, it was proposed that a ductile material will yield when the distortion energy per unit volume of a material subjected to simple tension, i.e. maximum-distortion-energy theory.

To determine the distortion energy, substitute

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(σ 1 −σ avg ),

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(σ 2 −σ avg ) and

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(σ 3 −σ avg ) into Eq. (43):

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ud =1+ν6E

σ 1 −σ 2( )2 + σ 2 −σ 3( )

2 + σ 3 −σ 1( )2

[ ] (44)

For plane stress

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σ 3 = 0 and:

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ud =1+ν3E

σ 12 −σ 1σ 2 +σ 2

2( )

For a uni-axial tension test

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σ1 =σ Y ,

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σ 2 =σ 3 = 0:

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ud( )Y =1+ν3E

σ Y2

For the maximum-distortion-energy theory,

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ud = ud( )Y , therefore:

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σ12 −σ 1σ 2 +σ 2

2 =σ Y2 (45)

Which is the equation of an ellipse on the

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σ1,σ 2 axes:

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ENG5312 – Mechanics of Solids II

Brittle Materials

In a uni-axial stress field the failure of brittle materials is specified by fracture.

In tension tests, brittle materials fail when the normal stress reaches the ultimate stress (

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σ ult).

In torsion tests, brittle materials fail due to a maximum tensile stress (at

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45o) to the applied shear, and the maximum tensile stress is approximately the same as the normal stress required for failure in a tension test.

1) Maximum-normal stress theory

This theory states that a brittle material will fail when the maximum principal stress (sigma1) reaches the ultimate normal stress the material can endure in a pure tension test.

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σ1 =σ ult (46)

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σ 2 =σ ult

Graphically

2) Mohr’s failure criterion

Applies when the tension and compression properties of a brittle material differ.

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ENG5312 – Mechanics of Solids II

Tests performed to determine: 1)

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σ ult( ) t in a tension test; 2)

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σ ult( )c in a compression test; and 3)

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τ ult in a torsion test.

Three Mohr’s circles are drawn: 1)

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σ1 =σ 2 = 0,

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σ 3 = σ ult( )c ; 2)

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σ1 = σ ult( ) t ,

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σ 2 =σ 3 = 0; and 3)

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τ ult .

Construct a failure envelope, i.e. a curve tangent to the three circles.

To check if a material will fail: draw the Mohr’s circle for the state of stress. If the circle is contained within the failure envelop there will be no failure. If the circle is tangent to, or crosses the failure envelope then failure will occur.

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ENG5312 – Mechanics of Solids II 34