solidwork stress project

SECTION 1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION PAGE 1/5

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION (Lecture)

Def: direction cosines: The direction cosines of a vector: ˆ ˆ ˆ x y zV V V  V i j k w.r.t. the x-y-z coordinates are:

 

 

 

cos cos ,

cos cos ,

cos cos ,

x

y

z

V l x

V m y

V n z







   

  

          

V V

V V

V V

, where 2 2 2 V x y zV V V   (1.24)

FIGURE 1.19 Stress components on a tetrahedron.

If V is a unit normal vector n̂ to a plane ABC (Fig 1.19), i.e., ˆ 1n , then

ˆ ˆ ˆ ˆ ˆ ˆˆ x y zn n n l m n     n i j k i j k (1.12-A1)

and 2 2 2 1l m n   (1.25)



 

 

 

ˆ ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆ ˆˆ

QAB ABC ABC ABC

QAC ABC ABC ABC

QBC ABC ABC ABC

l m n l

l m n m

l m n n

                              

n i i j k i

n j i j k j

n k i j k k

(a)

Def: stress traction vector (or surface traction): ˆ ˆ ˆ x y zp p p  p i j k ~ unit:

2

N Pa

m

   

  (1.12-A2)

force balance on the plane ABC:

 

 

 

0 :

0 :

0 :

x x ABC x QAB xy QAC xz QBC x xy xz ABC

y y ABC xy QAB y QAC yz QBC xy y yz ABC

z z ABC xz QAB yz QAC z QBC xz yz z ABC

F p l m n

F p l m n

F p l m n

     

     

     

              

            

           







SECTION 1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION PAGE 2/5

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

thus, in scalar-component form:

x x xy xz

y xy y yz

z xz yz z

p l m n

p l m n

p l m n

  

  

  

    

      

(1.26)

in tensor-index form: i ji j ij jp n n    , 1,2,3i j  (1.12-A3)

in vector-matrix form:  ˆp n τ (1.12-A4)

Special Case: Surface Tractions in Plane Problems

In 2-D (x-y plane), 0n  , then Eq (1.26) reduces to: x x xy

y xy y

p l m

p l m

 

 

  

  (1.12-A5)

Consider now a Cartesian x′-y′-z′ coordinate system with the same origin Q as the afore-mentioned x-y-z

coordinates. If the unit normal vector n̂ is placed to coincide with the x′-axis, then a set of direction cosines:

       1 1 1, , cos , ,cos , ,cos ,l m n x x x y x z     

will be formed. By the same token, one can place n̂ to coincide with the y′- and z′-axes, respectively, to form

another two sets of direction cosines:

       2 2 2, , cos , ,cos , ,cos ,l m n y x y y y z      &        3 3 3, , cos , ,cos , ,cos ,l m n z x z y z z     

Mnemonics: The relations among the two Cartesian coordinate systems and the direction cosines are listed in

Table 1.2.

TABLE 1.2 Notation for Direction Cosines.

Note that the sets:

       

       

       

1 2 3

1 2 3

1 2 3

, , cos , ,cos , ,cos ,

, , cos , ,cos , ,cos ,

, , cos , ,cos , ,cos ,

l l l x x x y x z

m m m y x y y y z

n n n z x z y z z

       

       

      

are the respective direction cosines of

the Cartesian x-, y- and z-axes w.r.t. the Cartesian x′-y′-z′ coordinate system. By placing the unit normal vector

n̂ to coincide with the x′-, y′- and z′-axes, respectively, Eq (1.26) becomes:

1 1 1

1 1 1

1 1 1

x x xy xz

y xy y yz

z xz yz z

p l m n

p l m n

p l m n

  

  

  

    

      

2 2 2

2 2 2

2 2 2

x x xy xz

y xy y yz

z xz yz z

p l m n

p l m n

p l m n

  

  

  

    

      

3 3 3

3 3 3

3 3 3

x x xy xz

y xy y yz

z xz yz z

p l m n

p l m n

p l m n

  

  

  

    

      

(1.12-A6)

Following the same logic, we can project the components of the above-mentioned stress traction vector ˆ ˆ ˆ

x y zp p p p  i j k in the x′-, y′- and z′-axes, respectively, then take force balances along the x′-, y′- or z′-

direction to obtain:

SECTION 1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION PAGE 3/5

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

1 1 1

2 2 2

3 3 3

x x y z

x y x y z

x z x y z

p l p m p n

p l p m p n

p l p m p n









 

 

    

      

1 1 1

2 2 2

3 3 3

x y x y z

y x y z

y z x y z

p l p m p n

p l p m p n

p l p m p n







 



 

    

      

1 1 1

2 2 2

3 3 3

x z x y z

y z x y z

x x y z

p l p m p n

p l p m p n

p l p m p n







 

 



    

      

(1.27)

Substitute Eq (1.12-A6) into Eq (1.27), we get:

       

     

2 2 2

1 1 1 1 1 1 1 1 1

1 2 1 2 1 2 1 2 2 1 2 1 1 2 1 2 2 1

1 3 1 3 1 3 1 3 3 1 1 3 3 1 1 3 3 1

2 2 2

2 2 2

2

2

x x y z xy yz xz

x y x y z xy yz xz

x z x y z xy yz xz

y x y z

l m n l m m n l n

l l m m n n l m l m m n m n l n l n

l l m m n n l m l m m n m n l n l n

l m n

      

      

      

    



 

 



     

        

        

     

       

2 2 2 2 2 2

2 2 2

3 3 3 3 3 3 3 3 3

2 3 2 3 2 3 2 3 3 2 2 3 3 2 2 3 3 2

2

xy yz xz

z x y z xy yz xz

y z x y z xy yz xz

l m m n l n

l m n l m m n l n

l l m m n n l m l m m n m n l n l n

 

      

      



 

     

  

               

(1.28)

Note: Owing to the symmetry of the stress tensor: ij ji  , only 6 of the 9 stress components thus developed

are unique.

The direction cosine components in Table 1.2 can be collected to form:

direction cosine (or orthogonal rotation) matrix:   11 12 13 1 1 1

21 22 23 2 2 2

31 32 33 3 3 3

ij

l l l l m n

l l l l l m n

l l l l m n

       

             

l (1.12-A7)

Note: The direction cosine matrix is NOT symmetric: ij jil l

The scalar-component form of Eq (1.28) can then be expressed in the short-hand forms:

tensor-index notation: rs ir js ij

rs ri sj ij

l l

l l

 

 

  

 (1.29)

or vector-matrix notation:      

      

T

T

   



τ l τ l

τ l τ l (1.12-A8)

Finally, it can be proven the components of the direction cosine (or orthogonal rotation) matrix have the

following properties:

2 2 2

1 1 1

2 2 2

2 2 2

2 2 2

3 3 3

1

1

1

l m n

l m n

l m n

    

      

and

1 2 1 2 1 2

2 3 2 3 2 3

1 3 1 3 1 3

0

0

0

l l m m n n

l l m m n n

l l m m n n

   

      

     1

l l T

 (1.30)

Similarly,

2 2 2

1 2 3

2 2 2

1 2 3

2 2 2

1 2 3

1

1

1

l l l

m m m

n n n

    

      

and

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

0

0

0

l m l m l m

m n m n m n

n l n l n l

   

      

(1.12-A9)

SECTION 1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION PAGE 4/5

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

Special Case: In-plane (or 2-D) stress tensor transformation

 

cos sin 0

sin cos 0

0 0 1

ijl

 

 

   

       

l and  

cos sin 0

sin cos 0

0 0 1

T

ji l

 

 

   

      

l (1.12-A10)

Figure 1.12-A1 Two-dimensional (orthogonal) rotational transformation.

then in component notation:

 

2 2

2 2

2 2

cos sin 2 sin cos cos 2 sin 2 2 2

sin cos 2 sin cos cos 2 sin 2 2 2

sin cos sin cos cos sin sin 2 cos 2 2

c

x y x y

x x y xy xy

x y x y

y x y xy xy

y x

x y x y xy xy

z z

x z zx

              

              

              

 

 





 



 

       

       

       



 os sin

cos sin

yz

y z yz zx

  

    

            

 

(1.12-A11)

or  

2 2

2 2

2 2

cos sin 2 sin cos cos 2 sin 2 2 2

sin cos 2 sin cos cos 2 sin 2 2 2

sin cos sin cos cos sin sin 2 cos 2 2

x y x y

x x y xy xy

x y x y

y x y xy xy

x y

xy x y xy xy

              

              

             

             

             

          

cos sin

cos sin

z z

yz yz zx

zx zx yz



 

   

    

        

     

  

(1.12-A12)

Mnemonics: The equations in each pair of the relations in the 2-D Special Case are interchangeable by

switching the primed and unprimed and by replacing  with .

SECTION 1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION PAGE 5/5

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

Example: 2-D Cartesian vs. polar coordinates

If the 2-D primed coordinate system  ,x y  is the polar coordinate system  ,r  , then

 

2 2

2 2

2 2

cos sin 2 sin cos cos 2 sin 2 2 2

sin cos 2 sin cos cos 2 sin 2 2 2

sin cos sin cos cos sin sin 2 cos 2 2

x y x y

r x y xy xy

x y x y

x y xy xy

y x

r x y xy xy





              

              

              

       

  

      

       



(1.12-A13)

or

 

2 2

2 2

2 2

cos sin 2 sin cos cos 2 sin 2 2 2

sin cos 2 sin cos cos 2 sin 2 2 2

sin cos sin cos cos sin sin 2 cos 2 2

r r x r r r

r r y r r r

r xy r r r

    

    

   

              

              

              

       

  

      

      



(1.12-A14)

which are the same as Eqs (1.18) and (1.9-A1) in Sec 1.9.

Large Deformation/Finite Elasticity Theory (Option)

Following the principles of Small Deformation/Infinitesimal Elasticity Theory, the previous definitions for the

stress tensor and traction vector do not make a distinction between the deformed and undeformed (or

reference) configurations of the body since such a distinction only leads to small modifications that are

considered higher-order effects and are normally neglected. However, for Large Deformation/Finite Elasticity

Theory, sizeable differences exist between these configurations and several stress tensors have been defined.

For instance, both directions of the force and surface normal in the true (Cauchy) stress tensor is defined

based on the deformed configuration. Other examples of stress tensors defined in Finite Elasticity are:

Kirchhoff stress tensor, the 1 st & 2

nd Piola-Kirchhoff stress tensors and Biot stress tensor. In addition,

there are also several incremental stress updates, also called stress rates: Jaumann, Green-Naghdi, Oldroyd,

Trusdell, convective, etc. It should be noted that since the physical meaning of the integral of stress with

respect to strain is the strain energy, the selection of proper pair of stress and strain tensors should obey the

rule of stress objectivity. That is, the superimposition of rigid-body motion on the deformed configuration

should not alter the stress state and should produce no extra strain energy.

SECTION 1.13 THREE-DIMENSIONAL PRINCIPAL & MAXIMUM SHEARING STRESSES PAGE 1/3

ME 54100: ADVANCED STRESS ANALYSIS ME I4200: APPLIED STRESS ANALYSIS

CHAPTER 1 – ANALYSIS OF STRESS

1.13 THREE-DIMENSIONAL PRINCIPAL & MAXIMUM SHEARING STRESSES (Lecture)

Let ij j p jn n  or   ˆ ˆτ n np or

x xy xz

pyx y yz

zx zy z

l l

m m

n n

  

   

  

           

              

(1.13-A1)

 stress tensor eigenvalue problem:

0

0

0

px xy xz

pyx y yz

pzx zy z

l

m

n

   

   

   

           

               

(1.31)

non-trivial solution, i.e.,

0

0

0

l

m

n

       

           

, we must have:

     3 2

1 2 3det det det 0

px xy xz

ij p ij p p p p pyx y yz

pzx zy z

I I I

   

          

   

   

                

Iτ (1.32 & 33)

here  

         1 2 3

1 2 3

eigenvalues: , , the extreme stationary values of stresses

ˆ ˆ ˆ ˆeigenvector: , ,

p

p

or      



th

th

principal stresses

principal stress directions

n n

p

p n n

In the cubic equation, Eq (1.33),  1 2 3, ,I I I are the three fundamental stress invariants:

Def : 1 st stress invariant: 1 1 2 3x y zI            (1.34a)

Def: 2 nd

stress invariant:

 

 

 

2

1 2 2 3 3

2

1

2 2

1

2

x xy y yz z zx

ii jj ij ji

yx y zy z xz x

x y y z z x xy yx yz zy zx

xy yz x

xz

z

I      

         

           

    

 





    

  

  

 

  



 (1.34b)

Def: 3 rd

stress invariant:

 

   

 

3

1

2 2 2

2 3

1 1 1 det det

3 6

2

2

x xy xz

ij yx y yz ij jk ki ij ji kk ii jj kk

zx zy z

x y z xy yz zx yx zy xz x yz zy y zx xz z xy

x yz y xz z xyxy y

x

z

y

zx

I

  

            

  

      

        

          

  

       

     

 



 

σ

(1.34c)

Def: (3-D) maximum shearing stress: 1 2 2 3 3 1

max max 0 2 2 2

      

       

  (1.13-A2)

SECTION 1.13 THREE-DIMENSIONAL PRINCIPAL & MAXIMUM SHEARING STRESSES PAGE 2/3

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Special Case: Plane-Stress Problems

plane-stress condition: 0z xz zx yz zy           3

3 3 0n   &      3 3 3

1 2 ˆ ˆˆ n n n i j (1.13-A3)

Eq (1.31) becomes

 

 

1

2

0

0

p

px xy

p pyx y

n

n

  

  

          

      

    2 2det 0 px xy

p p xyx y x y pxy y

         

  

        

  (1.13-A4)



1 1

2

3

2

2 1 2

0

x y

x xy

x y xy

yx y

I

I

I

   

      

 

     

   



  

Note: Typo in textbook (1.35)

2 2

2 11 1,2 2

2 2

21 2 1

max 2

: 2 2 2 4

: 2 2 4

x y x y

xy

x y

xy

I I I

I I

     

    

                   

        

principal stresses

maximum shearing stress

(1.20 & 22)

   

   

1

1

21 : tan two values

2

1 : tan two values

2 2

xy

p

x y

y x

s

xy

 

 

  







 

 

   

principal stress directions

maximum shearing stress directions

(1.19 & 21)

FIGURE 1.13-A1 The principal directions p and the maximum shear directions s .

SECTION 1.13 THREE-DIMENSIONAL PRINCIPAL & MAXIMUM SHEARING STRESSES PAGE 3/3

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CHAPTER 1 – ANALYSIS OF STRESS

Eigendecomposition & Diagonalization (Self-Study)

Def : eigen matrix:        1 2 3ˆ ˆ ˆn n n n   

(1.13-A5)

then       1 n n n

T or

  (1.13-A6)

&      1

2

3

0 0

0 0

0 0

n τ n







         

(1.13-A7)

Example 1.7 Three-Dimensional Stress in a Machine Component (Self-Study)

The stress tensor at a point in a machine element with respect to a Cartesian coordinate system is given by the

following array:

 

50 10 0

10 20 40 MPa

0 40 30

x xy xz

yx y yz

zx zy z

  

  

  

       

          

τ (f)

Determine the state of stress and  1 2 3, ,I I I for an x′, y′, z′ coordinate system defined by rotating x, y through an

angle of θ  45° counterclockwise about the z-axis (Fig 1.21a).

FIGURE 1.21 Direction cosines for 45   .

SECTION 1.14 NORMAL AND SHEAR STRESSES ON AN OBLIQUE PLANE PAGE 1/2

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CHAPTER 1 – ANALYSIS OF STRESS

1.14 NORMAL AND SHEAR STRESSES ON AN OBLIQUE PLANE (Self-Study)

Normal & Shear Tractions

Figure 1.14-A1 Comparison of general and principal stress states

p

(a) (b)

FIGURE 1.22 (a) Element in triaxial stress state; (b) traction vector decomposition.

in a general coordinate system (x-y-z or ˆ ˆ ˆ- -i j k ):  τ x xy xz

ij xy y yz

zx zy z

  

   

  

   

       

~ 6 stress components

in principal coordinate system (      1 2 3ˆ ˆ ˆ- -n n n ):  

1

2

3

0 0

0 0

0 0

ij



 



   

      

 ~ 3 stress components

Note: For the principal coordinate system, all shearing stresses vanish and thus the state includes only normal

stresses, which are the principal stresses themselves. That is to say, under transformation to principal

axes, the matrix form of the stress tensor will reduce to a diagonal form with the principal stresses as its

diagonal components.

Consider Fig 1.22, an element in a triaxial stress state:    

   

1 2 3

0 0 0

x y z

xy xz yz

     

  

  



, the traction vector p

on a surface with a unit normal vector ˆ ˆ ˆˆ l m n  n i j k , from Eq (1.26), becomes:

   1 2 3x y z p p p l m n   (a)

SECTION 1.14 NORMAL AND SHEAR STRESSES ON AN OBLIQUE PLANE PAGE 2/2

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With the aid of Fig 1.22b, p can also be decomposed into a normal traction vector σ and a shear traction

vector τ as:

 p σ τ with Eq (1.12-A4)  

2 2

ˆ

Pythagorean theorem

    

 

normal traction :

shear traction :

σ p n

τ p σ (1.14-A1)

Eq (a) into Eq (1.14-A1a)  2 2 2 2 2 2 2 2 2

1 2 3p l m n         (1.36)

Eq (1.28a)  2 2 2 2x y z xyl m n      = yzlm  xzmn  ln

 2 2 2

1 2 3l m n      (1.37)

Alternatively, Eqs (1.14-A1a) & (1.12-A3)  2 2 2

1 2 3i i ij jp n n l m n       =

Substitute Eq (1.37) into Eq (1.36), we get:

    2

2 2 2 2 2 2 2 2 2 2 2 2

1 2 3 1 2 3p l m n l m n               (1.38)

Since Eq (1.25): 2 2 2 1l m n          1 2

2 2 22 2 2 2 2 2

1 2 2 3 3 1l m m n n l              

(1.39)

Note: Eq (1.39) indicates that if the principal stresses are all equal: 1 2 3    , the shear stress  vanishes,

regardless of the choices of the direction cosines:  , ,l m n .

Finally, for a general stress state (Fig 1.14-A1a), we have:

 

     

2 2 2

1 2 2 2 2

2

2x y z xy yz xz

x xy xz xy y yz xy yz z

l m n lm mn ln

l m n l m n l m n

      

          

       

             

=

(1.40 & 41)

Lamé’s Stress Ellipsoid (Option)

FIGURE 1.23 Stress ellipsoid.

22 2

1 2 3

1yx z pp p

  

           

      (1.39)

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 1/7

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1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES (Lecture)

Spherical Stress Tensor

Def: mean (or average or hydrostatic) stress:

1 2 3 1

3 3 3

x y z

m

I     

      (1.14-S1)

Def: spherical stress tensor:

0 0

0 0

0 0

m

ij m ij m

m

p



  



   

      

(1.14-S2)

where Kronecker delta ij is defined as:

Def:  

   

1 0 0 1 if

1 00 0 if

0 0 1

unit identity matrixIij

i j no sum or

i j 

     

       

(1.14-S3)

fundamental spherical stress invariants

Def: 1 st spherical stress invariant:

1 1 2 3 13x y z mI p p p p p p I        (1.14-S4a)

Def: 2 nd

spherical stress invariant:

2

2 2 1

1 2 2 3 3 1

0 00

0 00

3 3

x zy

y xz

x y y z z x m

p pp I

p pp

I p p p p p p p p p p p p 

  

       

(1.14-S4b)

Def: 3 rd

spherical stress invariant:

3 3 1

3 1 2 3

0 0

det 0 0 27

0 0

x

ij y x y z m

z

p I

I p p p p p p p p

p

        (1.14-S4c)

Note: As presented in next topic: Chapter 2: Strain & Material Properties, the 2 nd

spherical stress

invariant 2

21 2 3

3 m

I I   is directly proportional to the dilatational component of strain energy.

Deviatoric Stress Tensor

Def: deviatoric stress tensor:

x m xy xz

ij ij ij ij m ij yx y m yz

zx zy z m

s p

   

       

   

   

         

(1.14-S5)

Notes: a. total stress = spherical stress + deviatoric stress: ij ij ijp s   (1.14-S6)

b. The spherical stress ij m ijp   is an isotropic stress tensor. That is, its components are the same

and equal to the mean stress m in all coordinate systems and the principal spherical stress

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 2/7

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CHAPTER 1 – ANALYSIS OF STRESS

directions are arbitrary.

c. Since the spherical stress ij m ijp   is an isotropic stress tensor. That is, its components are the

same in all coordinate systems and principal directions are arbitrary. Thus, the principal directions

of the deviatoric stress tensor ij ij ij ij m ijs p       are the same as those of the stress tensor ij

itself.

fundamental deviatoric stress invariants

Def: 1 st deviatoric stress invariant:

     1 3x y z x m y m z m x y z mJ s s s                      

From Eq (1.14-S1)  1 0J  (1.14-S7a)

Def: 2 nd

deviatoric stress invariant:

       

     

2 2 2 2 2 2

2

2 2 2 2

1 2 2 3 3 1 1 2

1 6

6

1 1

6 3

x y y z z x xy yz zxJ

I I

        

     

            

          

(1.14-S7b)

Def: 3 rd

deviatoric stress invariant:     3

3 1 2 3 1 1 2 3

2 1

27 3 m m mJ I I I I            (1.14-S7c)

(a) (b)

FIGURE 1.24 (a) An octahedron containing 8 octahedral stress planes;

(b) On an octahedral plane: l m n  i.e., 1cos 54.74       .

Octahedral Stresses

Def: principal stress space: A 3-D stress space (Note: not a physical space) with the three principal stresses

 1 2 3 , ,   (that is, their principal directions:

 1 n̂ ,

 2 n̂ ,

 3 n̂ ) as the coordinate axes.

Def: In the principal stress space, a plane whose normal vector makes equal angles with each of the principal

axes (i.e. having direction cosines equal to 1

cos54.74 3

  ) is called an octahedral stress plane. There

are a total of eight octahedral stress planes, as shown in Fig 1.24. The shear and normal components of

the stress tensor on these planes are called octahedral shear stress oct and octahedral normal stress

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oct , respectively:

      2 2 2

oct 1 2 2 3 3 1 2

1 2 3 1 oct

1 2

3 3

3 3 m

J

I

      

    

       

 

     

(1.43 & 44)

Von Mises & Tresca Stresses

Def: von Mises (or effective or equivalent) stress:

       

     

2 2 2

vM

2 2 2

1 2 2 3 3 1 oct 2

1 6

2

1 3 3

2 2

x y y z z x xy yx yz zy zx xz

J

            

      

        

       

(1.14-S8)

Def: Tresca stress:  Tr max 1 2 2 3 3 12 max , ,             (1.14-S9)

Notes: a. It can be proven that: Tr vM

max oct

 

 

 

 (1.14-S10)

b. vM (or equivalently, oct and 2J ) and Tr (or equivalently, max ) play significant roles in Failure

by Yielding (Ch 4).

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 4/7

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Example 1.6 Three-Dimensional Stress in a Hub (Lecture)

A steel shaft is to be force fitted into a fixed-ended cast-iron hub. The shaft is subjected to a bending moment

M, a torque T, and a vertical force P. Suppose that at a point Q in the hub, the stress field is, as shown below,

represented by the matrix:

19 4.7 6.45

4.7 4.6 11.8 MPa

6.45 11.8 8.3

x xy xz

ij yx y yz

zx zy z

  

   

  

        

           

a. Determine the principal stresses  1 2 3, ,   and their corresponding principal directions       1 2 3ˆ ˆ ˆ, ,n n n .

b. Prove eigendecomposition and diagonalization numerically by performing the matrix multiplication:

             1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ ˆn n n n n n T

       

.

c. Obtain the associated spherical and deviatoric stress tensors.

d. Find the invariants of the original, spherical and deviatoric stress tensors, respectively.

e. Find the maximum shear, von Mises, Tresca and octahedral normal and shear stresses.

f. Plot the 3-D Mohr’s circle and indicate the locations of the principal stresses  1 2 3, ,   and the maximum

shear max .

FIGURE 1.20 (a) Hub-shaft assembly. (b) Element in three-dimensional stress.

Solution

a. Principal stresses and directions.

For the stated stress tensor, the stress tensor eigenvalue problem, Eq (1.13-A1), becomes:

19 4.7 6.45

4.7 4.6 11.8

6.45 11.8 8.3

x xy xz

pyx y yz

zx zy z

l l l

m m m

n n n

  

   

  

                   

                          

The eigenvalues (i.e., principal stresses) and the corresponding eigenvectors (i.e., principal directions) can be

obtained using the Matlab command: eig. The results are:

1

2

3

11.6178

9.0015 MPa

25.3163







       

           

&      

1 2 3

1 2 3

1 2 3

1 2 3

0.0266 0.6209 0.7834

ˆ ˆ ˆ 0.8638 0.3802 0.3306

0.5031 0.6855 0.5262

n n n

l l l

m m m

n n n

             

           

(a1)

Verification. Substituting the above direction-cosine matrix into the stress tensor eigenvalue problem,

Eq (1.31), the following relations should be satisfied.

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 5/7

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CHAPTER 1 – ANALYSIS OF STRESS

For 1 11.6178 MPa  :

 

 

 

1 1 1

1 1 1

1 1 1

19 11.6178 4.7 6.45 0

4.7 4.6 11.6178 11.8 0

6.45 11.8 8.3 11.6178 0

l m n

l m n

l m n

                

  1

1 1 1 ˆ ˆ ˆ ˆ ˆ ˆˆ 0.0266 0.8638 0.5031l m n      n i j k i j k Note: 2 2 2

1 1 1 0l m n   (a2)

For 2 9.0015 MPa   :

 

 

 

2 2 2

2 2 2

2 2 2

19 9.0015 4.7 6.45 0

4.7 4.6 9.0015 11.8 0

6.45 11.8 8.3 9.0015 0

l m n

l m n

l m n

                

  2

2 2 2 ˆ ˆ ˆ ˆ ˆ ˆˆ 0.6209 0.3802 0.6855l m n     n i j k i j k Note: 2 2 2

2 2 2 0l m n   (a3)

For 3 25.3163 MPa   :

 

 

 

3 3 3

3 3 3

3 3 3

19 25.3163 4.7 6.45 0

4.7 4.6 25.3163 11.8 0

6.45 11.8 8.3 25.3163 0

l m n

l m n

l m n

                

  3

3 3 3 ˆ ˆ ˆ ˆ ˆ ˆˆ 0.7834 0.3306 0.5262l m n      n i j k i j k Note: 2 2 2

3 3 3 0l m n   (a4)

b. Eigendecomposition and diagonalization.

             1 2 3 1 2 3

0.0266 0.8638 0.5031 19 4.7 6.45 0.0266 0.6209 0.7834

ˆ ˆ ˆ ˆ ˆ ˆ 0.6209 0.3802 0.6855 4.7 4.6 11.8 0.8638 0.3802 0.3306

0.7834 0.3306 0.5262 6.45 11.8 8.3 0.5031 0.6855 0.5262

n n n n n n T

                    

                

       

               1

1 2 3 1 2 3

2

3

11.6178 0 0 0 0

ˆ ˆ ˆ ˆ ˆ ˆ 0 9.0015 0 MPa 0 0

0 0 25.3163 0 0

n n n n n n T







             

              

(b1)

c. Spherical and deviatoric stress tensors.

From Eq (1.14-S2), we have:

mean (or average or hydrostatic) stress: 19 4.6 8.3

7.5667 MPa 3 3

x y z

m

   

         (c1)

The spherical stress tensor can be obtained using Eq (1.14-S2):

0 0 7.5667 0 0

0 0 0 7.5667 0 MPa

0 0 0 0 7.5667

m

ij m

m

p







       

             

(c2)

The deviatoric stress tensor can be obtained using Eq (1.14-S5):

11.4333 4.7 6.45

4.7 12.1667 11.8 MPa

6.45 11.8 0.7333

x m xy xz

ij yx y m yz

zx zy z m

s

   

   

   

        

             

(c3)

d. Invariants.

The invariants of the stress tensor can be obtained from Eqs (1.34a,b,c):

1 1 2 3 22.7 MPaI        (d1)

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 6/7

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CHAPTER 1 – ANALYSIS OF STRESS

  2

2 1 2 2 3 3 1 170.8 MPaI           (d2)

  3

3 1 2 3 2,647.5 MPaI     (d3)

The invariants of the spherical stress tensor can be obtained from Eqs (1.14-S4a,b,c):

1 1 22.7 MPaI I   (d4)

  2

21 2 171.7633 MPa

3

I I    (d5)

  3

31 3 433.2253 MPa

27

I I    (d6)

The invariants of the deviatoric stress tensor can be obtained from Eqs (1.14-S7a,b,c):

1 0J  (d7)

        2 2 2 2

2 1 2 2 3 3 1

1 342.5758 MPa

6 J             

  (d8)

      3

3 1 2 3 488.5897 MPam m mJ           (d9)

e. Maximum shear, von Mises, Tresca & octahedral stresses.

The maximum shear stress can be obtained from Eq (1.13-A2):

1 2 2 3 3 1

max max , , 18.467 MPa 2 2 2

      

       

  (e1)

The octahedral stresses can be obtained from Eqs (1.43 & 44):

oct 2

oct

2 octahedral shear stress: 15.1124 MPa

3

octahedral normal stress: 7.5667 MPam

J

 

  

    

(e2)

The von Mises (or effective or equivalent) stress can be obtained from Eq (1.14-S8):

vM oct

3 32.0582 MPa

2    (e3)

The Tresca stress can be obtained from Eq (1.14-S8):

Tr max2 36.9341MPa    (e4)

SECTION 1.14a SPHERICAL, DEVIATORIC, OCTAHEDRAL, VON MISES & TRESCA STRESSES PAGE 7/7

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CHAPTER 1 – ANALYSIS OF STRESS

f. 3-D Mohr’s circles.

 max

=1.8.467 MPa

 1 =11.6178 MPa

2 =9.0015 MPa

3 =23.3163 MPa





SECTION 1.15 MOHR’S CIRCLES IN THREE DIMENSIONS PAGE 1/3

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CHAPTER 1 – ANALYSIS OF STRESS

1.15 MOHR’S CIRCLES IN THREE DIMENSIONS (Option)

FIGURE 1.25 Triaxial state of stress: (a) wedge; (b) planes of maximum shear stress.

FIGURE 1.26 (a–c) Views of elements in triaxial stresses on different principal axes;

(d) Mohr’s circles for three-dimensional stress.

SECTION 1.15 MOHR’S CIRCLES IN THREE DIMENSIONS PAGE 2/3

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CHAPTER 1 – ANALYSIS OF STRESS

Equations of Three Mohr’s Circles for Stress:

direction cosines: 2 2 2 1l m n  

normal traction: 2 2 2

1 2 3l m n     

shear traction:       2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 3 1 2 2 3 3 1l m n l m m n l n                   

(a)

Solve Eq (a) 

  

  

  

  

  

  

2

2 32

1 2 1 3

2

3 12

2 3 2 1

2

1 22

3 1 3 2

0

0

0

l

m

n





  

   

   

   

  

   

 

     

     

   

         

(Note: typos in textbook) (1.46)

Without loss in generality, assume 1 2 3    , we have:

Mohr’s circles of stresses:

  

  

  

2

2 3

2

3 1

2

1 2

0

0

0

   

   





   

     

        

(Note: typos in textbook) (b)



     

     

     

2 2 22 1 1 2 3 23 2 32 4max

2 2 22 1 1 1 3 13 1 32 4max

2 2 22 1 1 1 2 12 1 22 4max







     

     

     

         

         

        

(Note: typos in textbook) (1.47)

absolute maximum shearing stress:     1 3 max 13 max 2a

   

   (1.45)

SECTION 1.15 MOHR’S CIRCLES IN THREE DIMENSIONS PAGE 3/3

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CHAPTER 1 – ANALYSIS OF STRESS

EXAMPLE 1.8 Analysis of Three-Dimensional Stresses in a Member

The state of stress on an element of a structure is illustrated in Fig. 1.27a. Using Mohr’s circle, determine

a. the principal stresses,

b. the maximum shearing stresses; Show results on a properly oriented element; Also,

c. apply the equations developed in Sec 1.14 to calculate the octahedral stresses.

FIGURE 1.27 Example 1.8. (a) Element in three-dimensional stress;

(b) Mohr’s circles of stress; (c) stress element for 26.56p  .

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