Principal-Stresses-3D-Example.pdf

Dr. Karim Heinz Muci

ME-216 Introduction to Solid Mechanics

General State of Stress at a Point: Principal Stresses and Principal Directions Example Problem

Given the following matrix of components of the stress tensor corresponding to a point in a loaded mechanical component, find the principal stresses and the principal directions.

⎡ 40 − 20 10⎤ [σ ~] = ⎢⎢− 20 − 80 5 ⎥⎥ MPa 5 60⎥⎦ ⎢⎣ 10

Solution

•

The problem statement provides the matrix of components of the stress tensor at a point in a mechanical component:

⎡ σ x [σ ~] = ⎢⎢τ yx ⎢τ zx ⎣ σ

x

σ

y

τ

xy

σ

y

τ

zy

xz ⎤

⎡ 40 − 20 10⎤ ⎥ ⎢ τ 5 ⎥ Mpa yz ⎥ = ⎢− 20 − 80 ⎥ 10 5 60 σ ⎥ ⎢ ⎥⎦ z ⎦ ⎣ τ

= 40 MPa = −80 MPa

= 60 MPa τ xy = τ yx = −20 MPa σ

z

τ

= τ zy = 5 MPa

τ

= τ xz = 10 MPa

yz

zx

Step 1: Find the Stress Invariants

• I 1 = σ x + σ y + σ z I 1 = ( 40 MPa) + ( −80 MPa) + (60 MPa) I 1 = 20 MPa

-1-



2 2 2 − τ yz − τ zx • I 2 = σ xσ y + σ yσ z + σ zσ x − τ xy

I 2 = (40 MPa)( −80 MPa) + ( −80 MPa)(60 MPa) + (60 MPa)( 40 MPa)

−( −20 MPa)2 − (5 MPa)2 − (10 MPa)2 I 2 = −6,125 MPa

2

2 2 2 − σ yτ zx − σ zτ xy • I 3 = σ xσ yσ z + 2τ xyτ yzτ zx − σ xτ yz

I 3 = ( 40 MPa)( −80 MPa)(60 MPa) + 2( −20 MPa)(5 MPa)(10 MPa)

−(40 MPa)(5 MPa) 2 − ( −80 MPa)(10 MPa) 2 − (60 MPa)( −20 MPa) 2 I 3 = −211,000 MPa

3

Another option to find the value of I 3 is: σ

x

I 3 = τ yx τ

zx

τ

xy

σ

y

τ

zy

τ

xz

τ

yz

40 − 20 10 − 20 MPa 10 MPa = − 20 MPa − 80 MPa 5 MPa = − 20 − 80 5 MPa 3 40 MPa

10 MPa

σ

z

5 MPa

60 MPa

10

5

60

Using the calculator to evaluate the determinant: I 3 = −211,000 MPa

•

3

Thus, the stress invariants are given by: I 1 = 20 MPa I 2 = −6,125 MPa

2

I 3 = −211,000 MPa

3

Step 2: Find the Principal Stresses

•

Write the cubic equation to find the principal stresses: 3

− I 1σ 2 + I 2σ − I 3 = 0

3

− (20 MPa )σ 2 + (−6,125 MPa 2 )σ − (−211,000 MPa 3 ) = 0

3

− (20 MPa )σ 2 − (6,125 MPa 2 )σ + (211,000 MPa 3 ) = 0

σ

σ

σ

•

Using the calculator, we obtain the three real roots of the cubic equation:

-2-


σ

•


= 39.331419 MPa , 64.213203 MPa , − 83.544623 MPa

Arranging the values of the principal stresses according to the convention

σ

1

≥ σ 2 ≥ σ 3 :

= 64.213203 MPa σ = 39.331419 MPa 2 σ = −83.544623 MPa 3 σ

1

•

We can do a quick check to see if the values found are correct: I 1 = σ x + σ y + σ z = σ 1 + σ 2 + σ 3 = 20 MPa 1 + σ 2

+ σ 3 = (64.213203 MPa) + (39.331419 MPa) + ( −83.544623 MPa) σ + σ + σ = 19.999999 MPa = 20 MPa  1 2 3 σ

Using Advanced Scientific Calculators to Find the Principal Stresses

•

Some advanced scientific calculators can find the eigenvalues of a matrix. Providing the following matrix as input to a calculator

⎡ 40 − 20 10 ⎤ [σ ~] = ⎢⎢− 20 − 80 5 ⎥⎥ MPa 5 60⎥⎦ ⎢⎣ 10

⇒

⎡ 40 − 20 10⎤ ⎢− 20 − 80 5 ⎥ ⎢ ⎥ 5 60⎥⎦ ⎢⎣ 10

and using a “built-in” function to find the eigenvalues of a matrix, the following result was obtained: Eigenvalue s

= 39.331419, 64.213203, − 83.544623

Thus, the principal stresses ( σ 1 ≥ σ 2 ≥ σ 3 ) are given by:

= 64.213203 MPa σ = 39.331419 MPa 2 σ = −83.544623 MPa 3 σ

1

Notes: − Unless stated otherwise, in a homework or exam problem you should not use “built-in” functions on your calculator to find the eigenvalues of a matrix. However, you can

-3-


− − −


always use them to check if the values that you obtained for the principal stresses are correct. In HP advanced scientific calculators (models HP48 and higher) the “built-in” function to find the eigenvalues of a matrix is “EGVL”. Check the manual of your calculator to see if it allows you to input a matrix and find its eigenvalues. When we use a calculator, we input values without units. Thus, we must keep track of the units on our own.

Step 3: Find the Principal Directions

•

Since the three principal stresses are different ( σ 1 ≠ σ 2 ≠ σ 3 ), there are only three principal directions that are mutually perpendicular to each other.

•

We proceed to find two of the principal directions and obtain the third one using one of the following relations: nˆ1 = nˆ2 × nˆ3 nˆ2 = nˆ3 × nˆ1 nˆ3 = nˆ1 × nˆ2

Remember that the unit vectors nˆ1, nˆ2 , nˆ3 corresponding to the principal directions must form a right-handed system.

•

In what follows, first we will find nˆ1 and nˆ2 . Then, we will use the relationship nˆ3 = nˆ1 × nˆ2 to find nˆ3 .

Finding the Principal Direction nˆ1

•

ˆ nˆ1 = n x1iˆ + n y1 jˆ + n z1k

•

We make use of the following equations to find nˆ1

⎡σ x − σ 1 ⎢ ⎢ τ yx ⎢ τ zx ⎣ •

τ

xy

σ

y

− σ 1

τ

zy

⎤ ⎧n x1 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ τ yz ⎥ ⎨n y1 ⎬ = ⎨0 ⎬ ⎪ ⎪ ⎪ ⎪ σ − σ ⎥ n z 1 ⎦ ⎩ z1 ⎭ ⎩0 ⎭ τ

xz

and

In expanded form, the above equations can be written as:

-4-

2

2

2

n x1 + n y1 + n z1 = 1



(σ x − σ 1 )n x1 + τ xy n y1 + τ xz n z1 = 0 yx n x1 + (σ y

τ

− σ 1 )n y1 + τ yz n z1 = 0

zx n x1 + τ zy n y1 + (σ z

τ

− σ 1 )n z1 = 0

n x21 + n y21 + n z21 = 1

Since the determinant of the matrix of coefficients is zero and the three principal stresses are different ( σ 1 ≠ σ 2 ≠ σ 3 ), only two of the first three equations are independent (i.e., one of the first three equations is redundant).

•

Substituting values: 10 − 20 ⎧ n x1 ⎫ ⎧0⎫ ⎡40 − 64.213203 ⎤ ⎢ ⎥ MPa ⎪n ⎪ = ⎪0⎪ 5 − 20 − 80 − 64.213203 ⎨ y1 ⎬ ⎨ ⎬ ⎢ ⎥ ⎪ n ⎪ ⎪0⎪ 10 5 60 − 64.213203⎥⎦ ⎢⎣ ⎩ z1 ⎭ ⎩ ⎭ 10 − 20 ⎧ n x1 ⎫ ⎧0⎫ ⎡ − 24.213203 ⎤ ⎢ ⎥ MPa ⎪n ⎪ = ⎪0⎪ 5 − 20 − 144.213203 ⎨ y1 ⎬ ⎨ ⎬ ⎢ ⎥ ⎪ n ⎪ ⎪0⎪ 10 5 ⎢⎣ − 4.213203⎥⎦ ⎩ z1 ⎭ ⎩ ⎭ 10 − 20 ⎡ − 24.213203 ⎤ ⎧ n x1 ⎫ ⎧0⎫ ⎢ ⎥ ⎪n ⎪ = ⎪0⎪ 5 − 20 − 144.213203 ⎢ ⎥ ⎨ y1 ⎬ ⎨ ⎬ 10 5 ⎢⎣ − 4.213203⎥⎦ ⎪⎩ n z1 ⎪⎭ ⎪⎩0⎪⎭

•


− 24.213203n x1 − 20 n y1 + 10 n z1 = 0 − 20 n x1 − 144.213203n y1 + 5 n z1 = 0 10 n x1 + 5 n y1 − 4.213203n z1 = 0 Remember that only two of these three equations are independent. The third equation required to find the values of n x1 , n y1 , n z1 is: 2

2

2

n x1 + n y1 + n z1 = 1

which is a non-linear equation.

-5-


•


To avoid having to solve a system of equations involving a non-linear equation, we follow the procedure describe below. However, if you want, you can solve that system of equations to find n x1 , n y1 , n z1 :

− 24.213203 n x1 − 20 n y1 + 10 n z1 = 0⎫ ⎪ − 20 n x1 − 144.213203 n y1 + 5 n z1 = 0 ⎬ → Choose two of these three equations 10 n x1 + 5 n y1 − 4.213203 n z1 = 0 ⎪ ⎭ 2

2

2

n x1 + n y1 + n z1 = 1

•

Let: r

)

(

)

ˆ = ( N n ) iˆ + N n jˆ + ( N n ) k ˆ N 1 = N 1nˆ1 = N 1 n x1iˆ + n y1 jˆ + n z1k 1 x1 1 y1 1 z1 r

ˆ N 1 = N x1iˆ + N y1 jˆ + N z1k

Therefore: r

2

2

2

N 1 = N 1 = N x1 + N y1 + N z1

r

→ Magnitude of N 1

N x1

N x1 = N 1n x1

→ n x1 =

N y1 = N 1n y1

→ n y1 =

N z1 = N 1n z1

→ n z1 =

N 1

N y1 N 1

N z1 N 1 r

Note that, based on the above definitions, N 1 is a vector parallel to the unit vector nˆ1.

•

Multiply the first equations of the system of equations by N 1 : N 1( −24.213203n x1 − 20 n y1 + 10 n z1 ) = N 1(0) = 0 N 1( −20 n x1 − 144.213203n y1 + 5 n z1 ) = N 1(0) = 0 N 1(10 n x1 + 5 n y1 − 4.213203n z1) = N 1(0) = 0

− 24.213203 N x1 − 20 N y1 + 10 N z1 = 0 − 20 N x1 − 144.213203 N y1 + 5 N z1 = 0 10 N x1 + 5 N y1 − 4.213203 N z1 = 0

-6-



We can write the above equations in matrix form as follows: 10 − 20 ⎡ − 24.213203 ⎤ ⎧ N x1 ⎫ ⎧0⎫ ⎢ ⎥ ⎪ N ⎪ = ⎪0⎪ 5 − 20 − 144.213203 ⎢ ⎥ ⎨ y1 ⎬ ⎨ ⎬ 10 5 ⎢⎣ − 4.213203⎥⎦ ⎪⎩ N z1 ⎪⎭ ⎪⎩0⎪⎭ Notice that the above expression corresponds to

⎡σ x − σ 1 ⎢ ⎢ τ yx ⎢ τ zx ⎣ •

τ

xy

σ

y

− σ 1

τ

zy

⎤ ⎧ N x1 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ τ yz ⎥ ⎨ N y1 ⎬ = ⎨0 ⎬ ⎪ ⎪ ⎪ ⎪ σ − σ ⎥ N z 1 ⎦ ⎩ z1 ⎭ ⎩0 ⎭ τ

xz

r

Select an arbitrary value different than zero for one of the components of N 1 and obtain the other two using two of the three equations. In this step you must be careful. Inspect the r system of equations to make sure that you are not assigning a value to a component of N 1 that can be obtained directly from one of the three equations. Take N x1 = 1 and use two of three equations to find N y1 and N z1 :

− 24.213203(1) − 20 N y1 + 10 N z1 = 0 − 20 (1) − 144.213203 N y1 + 5 N z1 = 0 − 20 N y1 + 10 N z1 = 24.213203 − 144.213203 N y1 + 5 N z1 = 20 10⎤ ⎧ N y1 ⎫ ⎧24.213203⎫ − 20 ⎡ ⎬ ⎢ − 144.213203 5 ⎥ ⎨ N ⎬ = ⎨ 20 ⎭ ⎣ ⎦ ⎩ z1 ⎭ ⎩ Solving for N y1 and N z1 (using the calculator) we obtain: N y1 = −5.881238 × 10

−2

N z1 = 2.303696

• N x1 = 1 , N y1 = −5.881238 ×10−2 , N z1 = 2.303696 − N 1 = N x1 + N y1 + N z1 = (1) + ( −5.881238 × 10 ) + ( 2.303696) = 2.512066 2

2

2

2

2 2

-7-

2


n x1 =

n y1 = n z1 =

•

N x1 N 1

N y1 N 1 N z1 N 1

=

1 = 0.398079 2.512066

− 5.881238 × 10−2

= =


2.512066 2.303696 2.512066

n x 1 = 0 . 398079

= −0.023412

= 0.917052 n y1 = −0.023412 ,

,

n z1 = 0.917052

ˆ nˆ1 = n x1iˆ + n y1 jˆ + n z1k

(

)

ˆ nˆ1 = ± 0.398079 iˆ − 0.023412 jˆ + 0.917052 k

In the above result, we added the “ ± ” to the solution that we found for nˆ1 because the nonlinear equation (which is quadratic) admits two solutions (one positive and one negative). Notice that the two possible solutions for nˆ1 are 180 o apart, that is, they correspond to the same direction but opposite sense.


•

ˆ nˆ 2 = n x 2iˆ + n y 2 jˆ + n z 2 k

•

We make use of the following equations to find nˆ2

⎡σ x − σ 2 ⎢ ⎢ τ yx ⎢ τ zx ⎣ •

τ

xy

σ

y

− σ 2

τ

zy

⎤ ⎧ n x 2 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ τ yz ⎥ ⎨n y 2 ⎬ = ⎨0⎬ ⎪ ⎪ ⎪ ⎪ σ − σ ⎥ n z 2 ⎦ ⎩ z 2 ⎭ ⎩0⎭ τ

xz

and


(σ x − σ 2 )n x 2 + τ xy n y 2 + τ xz n z 2 = 0 yx n x 2

+ (σ y − σ 2 )n y 2 + τ yz n z 2 = 0

zx n x 2

+ τ zy n y 2 + (σ z − σ 2 )n z 2 = 0

τ

τ

n x22 + n y2 2 + n z22 = 1

-8-

2 2 2 n x 2 + n y 2 + n z 2 = 1



Since the determinant of the matrix of coefficients is zero and the three principal stresses are different ( σ 1 ≠ σ 2 ≠ σ 3 ), only two of the first three equations are independent (i.e., one of the first three equations is redundant).

•

Substituting values: 10 − 20 ⎧ n x 2 ⎫ ⎧0⎫ ⎡40 − 39.331419 ⎤ ⎢ ⎥ MPa ⎪n ⎪ = ⎪0⎪ 5 − 20 − 80 − 39.331419 ⎨ y 2 ⎬ ⎨ ⎬ ⎢ ⎥ ⎪ n ⎪ ⎪0⎪ 10 5 60 − 39.331419 ⎥⎦ ⎢⎣ ⎩ z 2 ⎭ ⎩ ⎭ 10 − 20 ⎧n x 2 ⎫ ⎧0⎫ ⎡0.668581 ⎤ ⎢ − 20 ⎥ MPa ⎪n ⎪ = ⎪0⎪ 5 − 119.331419 ⎨ y 2 ⎬ ⎨ ⎬ ⎢ ⎥ ⎪ n ⎪ ⎪0⎪ 5 20.668581⎥⎦ ⎢⎣ 10 ⎩ z 2 ⎭ ⎩ ⎭

− 20 10 ⎡0.668581 ⎤ ⎧ n x 2 ⎫ ⎧0⎫ ⎢ − 20 ⎥ ⎪n ⎪ = ⎪0⎪ 5 − 119.331419 ⎢ ⎥ ⎨ y 2 ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ 5 20.668581⎥⎦ ⎪ ⎢⎣ 10 ⎩ n z 2 ⎭ ⎩0⎭ •


0.668581n x 2 − 20 n y 2 + 10 n z 2 = 0

− 20 n x 2 − 119.331419 n y 2 + 5 n z 2 = 0 10 n x 2 + 5 n y 2 + 20.668581n z 2 = 0 Remember that only two of these three equations are independent. The third equation required to find the values of n x 2 , n y 2 , n z 2 is 2 2 2 n x 2 + n y 2 + n z 2 = 1

which is a non-linear equation.

•

To avoid having to solve a system of equations involving a non-linear equation, we follow the procedure describe below. However, if you want, you can solve that system of equations to find n x 2 , n y 2 , n z 2 : 0.668581 n x 2 − 20 n y 2 + 10 n z 2 = 0 ⎫

⎪ − 20 n x 2 − 119.331419 n y 2 + 5 n z 2 = 0⎬ → Choose two of these three equations 10 n x 2 + 5 n y 2 + 20.668581 n z 2 = 0 ⎪ ⎭

-9-



2 2 2 n x 2 + n y 2 + n z 2 = 1

•

Let: r

)

(

)

ˆ = ( N n )iˆ + N n jˆ + ( N n ) k ˆ N 2 = N 2 nˆ 2 = N 2 n x 2iˆ + n y 2 jˆ + n z 2 k 2 x 2 2 y 2 2 z 2 r ˆ N = N iˆ + N jˆ + N k 2

x 2

y 2

z 2

Therefore: r

2

2

2

N 2 = N 2 = N x 2 + N y 2 + N z 2 N x 2 = N 2 n x 2

→ n x 2 =

N y 2 = N 2n y 2

→ n y 2 =

N z 2 = N 2 n z 2

→ n z 2 =

r

→ Magnitude of N 2

N x 2 N 2

N y 2 N 2

N z 2 N 2 r

Note that, based on the above definitions, N 2 is a vector parallel to the unit vector nˆ2 .

•

Multiply the first equations of the system of equations by N 2 : N 2 (0.668581n x 2 − 20 n y 2 + 10 n z 2 ) = N 2 (0) = 0 N 2 ( −20 n x 2 − 119.331419 n y 2 + 5 n z 2 ) = N 2 (0) = 0 N 2 (10 n x 2 + 5 n y 2 + 20.668581n z 2 ) = N 2 (0)

0.668581 N x 2 − 20 N y 2 + 10 N z 2 = 0

− 20 N x 2 − 119.331419 N y 2 + 5 N z 2 = 0 10 N x 2 + 5 N y 2 + 20.668581 N z 2 = 0 We can write the above equations in matrix form as follows: 10 − 20 ⎡0.668581 ⎤ ⎧ N x 2 ⎫ ⎧0⎫ ⎢ − 20 ⎥ ⎪ N ⎪ = ⎪0⎪ 5 − 119.331419 ⎢ ⎥ ⎨ y 2 ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ 5 20.668581⎥⎦ ⎪ ⎢⎣ 10 ⎩ N z 2 ⎭ ⎩0⎭ Notice that the above expression corresponds to

- 10 -


⎡σ x − σ 2 ⎢ ⎢ τ yx ⎢ τ zx ⎣ •


τ σ

y

⎤ ⎧ N x 2 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ τ yz ⎥ ⎨ N y 2 ⎬ = ⎨0⎬ ⎪ ⎪ ⎪ ⎪ σ − σ ⎥ N z 2 ⎦ ⎩ z 2 ⎭ ⎩0⎭ τ

xy

xz

− σ 2

τ

zy

r

Select an arbitrary value different than zero for one of the components of N 2 and obtain the other two using two of the three equations. In this step you must be careful. Inspect the r system of equations to make sure that you are not assigning a value to a component of N 2 that can be obtained directly from one of the three equations. Take N x 2 = 1 and use two of three equations to find N y 2 and N z 2 :

0.668581(1) − 20 N y 2 + 10 N z 2 = 0

− 20 (1) − 119.331419 N y 2 + 5 N z 2 = 0 − 20 N y 2 + 10 N z 2 = −0.668581 − 119.331419 N y 2 + 5 N z 2 = 20 10⎤ ⎧ N y 2 ⎫ ⎧− 0.668581⎫ − 20 ⎡ ⎬ ⎢ − 119.331419 5 ⎥ ⎨ N ⎬ = ⎨ 20 ⎣ ⎦ ⎩ z 2 ⎭ ⎩ ⎭ Solving for N y 2 and N z 2 (using the calculator) we obtain: N y 2 = −0.185988

N z 2 = −0.438833

• N x 2 = 1 , N y 2 = −0.185988 , N z 2 = −0.438833 2

2

2

2

2

2

N 2 = N x 2 + N y 2 + N z 2 = (1) + ( −0.185988) + ( −0.438833) = 1.107775 n x 2 =

n y 2 = n z 2 =

•

N x 2 N 2

N y 2 N 2 N z 2 N 2

=

1 1.107775

= 0.902710

=

− 0.185988

=

− 0.438833

1.107775 1.107775

n x 2 = 0.902710 ,

= −0.167893 = −0.396139

n y 2 = −0.167893 ,

n z 2 = −0.396139

- 11 -



ˆ nˆ 2 = n x 2iˆ + n y 2 jˆ + n z 2 k

(

)

ˆ nˆ2 = ± 0.902710 iˆ − 0.167893 jˆ − 0.396139 k

In the above result, we added the “ ± ” to the solution that we found for nˆ2 because the nonlinear equation (which is quadratic) admits two solutions (one positive and one negative). Notice that the two possible solutions for nˆ2 are 180 o apart, that is, they correspond to the same direction but opposite sense.


•

ˆ nˆ3 = n x 3iˆ + n y 3 jˆ + n z 3k

•

Although we can use the following equations to find nˆ3

⎡σ x − σ 3 ⎢ ⎢ τ yx ⎢ τ zx ⎣

τ

xy

σ

y

− σ 3

τ

zy

⎤ ⎧n x 3 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ τ yz ⎥ ⎨n y 3 ⎬ = ⎨0⎬ ⎪ ⎪ ⎪ ⎪ σ − σ ⎥ n z 3 ⎦ ⎩ z 3 ⎭ ⎩0⎭ τ

xz

and

2

2

2

n x 3 + n y 3 + n z 3 = 1

we will determine the unit vector nˆ3 using the following expression: nˆ3 = nˆ1 × nˆ2

•

It is important to note that if we decide to find nˆ3 following the same procedure that was employed to determine nˆ1 and nˆ2 , if necessary we need to adjust the sense of nˆ3 to make sure that nˆ3 = nˆ1 × nˆ2 .

•

nˆ3 = nˆ1 × nˆ2

)

)

ˆ × n iˆ + n jˆ + n k ˆ nˆ3 = n x1iˆ + n y1 jˆ + n z1k x 2 y 2 z 2

ˆ+ nˆ3 = n x1n x 2iˆ × iˆ + n x1n y 2iˆ × jˆ + n x1n z 2iˆ × k ˆ+ n y1n x 2 jˆ × iˆ + n y1n y 2 jˆ × jˆ + n y1n z 2 jˆ × k ˆ × iˆ + n n k ˆ ˆ ˆ ˆ n z1n x 2k z1 y 2 × j + n z1n z 2k × k

- 12 -



ˆ − n n jˆ − n n k ˆ ˆ ˆ ˆ nˆ3 = n x1n y 2 k x1 z 2 y1 x 2 + n y1n z 2i + n z1n x 2 j − n z1n y 2i ˆ nˆ3 = ( n y1n z 2 − n z1n y 2 ) iˆ + ( n z1n x 2 − n x1n z 2 ) jˆ + ( n x1n y 2 − n y1n x 2 ) k

Thus: n x 3 = n y1n z 2 − n z1n y 2 n y 3 = n z1n x 2 − n x1n z 2

Formulas to find the components of nˆ3

n z 3 = n x1n y 2 − n y1n x 2

Another approach to perform the cross product is using determinants: iˆ

ˆ k

jˆ

nˆ3 = nˆ1 × nˆ2 = n x1

n y1

n z1 = iˆ

n x 2

n y 2

n z 2

n y1

n z1

n y 2

n z 2

− jˆ

n x1

n z1

n x 2

n z 2

ˆ + k

n x1

n y1

n x 2

n y 2

ˆ nˆ3 = ( n y1n z 2 − n z1n y 2 ) iˆ + ( n z1n x 2 − n x1n z 2 ) jˆ + ( n x1n y 2 − n y1n x 2 ) k

•

(

)

ˆ = ± 0.398079 iˆ − 0.023412 jˆ + 0.917052 k ˆ nˆ1 = n x1iˆ + n y1 jˆ + n z1k

(

)

ˆ = ± 0.902710 iˆ − 0.167893 jˆ − 0.396139 k ˆ nˆ 2 = n x 2iˆ + n y 2 jˆ + n z 2 k

We will use the values corresponding to the “+” solution for nˆ1 and nˆ2 in order to find the components of nˆ3 . Then we will apply “ ± ” to the solution for obtained for nˆ3 . n x 3 = n y1n z 2 − n z1n y 2 = ( −0.023412)( −0.396139) − (0.917052)( −0.167893) = 0.163241 n y 3 = n z1n x 2 − n x1n z 2 = (0.917052)(0.902710) − (0.398079)( −0.396139) = 0.985527 n z 3 = n x1n y 2 − n y1n x 2 = (0.398079)( −0.167893) − ( −0.023412)(0.902710) = −0.045700

ˆ nˆ3 = n x 3iˆ + n y 3 jˆ + n z 3k

)

ˆ nˆ3 = ± 0.163241iˆ + 0.985527 jˆ − 0.045700k

Answer to the Problem

- 13 -



The principal stresses and the principal directions corresponding to:

⎡ 40 − 20 10⎤ [σ ~] = ⎢⎢− 20 − 80 5 ⎥⎥ MPa 5 60⎥⎦ ⎢⎣ 10 are:

)

64.213203 MPa

ˆ nˆ1 = ± 0.398079 iˆ − 0.023412 jˆ + 0.917052 k

σ

= 39.331419 MPa

ˆ nˆ 2 = ± 0.902710 iˆ − 0.167893 jˆ − 0.396139 k

σ

= −83.544623 MPa

ˆ nˆ3 = ± 0.163241iˆ + 0.985527 jˆ − 0.045700 k

1 =

σ

2

3

)

)

Additional Information Useful for Solving This Type of Problems

Using Advanced Scientific Calculators to Find the Principal Directions

•



Some advanced scientific calculators can find the eigenvectors of a matrix. In this regard, the following points must be kept in mind. Usually the eigenvectors provided by the calculator are not unit vectors. Thus, the calculator r r r provides N 1 , N 2 , N 3 instead of nˆ1 , nˆ2 , nˆ3 . However, we can easily find nˆ1 , nˆ2 , nˆ3 as follows: r

nˆ1 =

N 1 N 1

N x1 ˆ N y1 ˆ N z1 ˆ i+ j + k N 1 N 1 N 1

where

N 1 = N x1 + N y1 + N z1

=

N x 2 ˆ N y 2 ˆ N z 2 ˆ i+ j + k N 2 N 2 N 2

where

N 2 = N x 2 + N y 2 + N z 2

=

N x 3 ˆ N y 3 ˆ N z 3 ˆ i+ j + k N 3 N 3 N 3

where

N 3 = N x 3 + N y 3 + N z 3

=

r

nˆ2 =

N 2

nˆ3 =

N 3

N 2 r



N 3

2

2

2

2

2

2

2

2

2

In general, the calculator will not provide the eigenvectors is the correct order. Thus, you r need to make sure that you select for N 1 the eigenvector corresponding to the eigenvalue

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ME-216 Introduction to Solid Mechanics r

r

1 , for N 2 the eigenvector corresponding to the eigenvalue

σ

corresponding to the eigenvalue 

2 , and for N 3 the eigenvector

σ

3.

σ

The calculators typically provide only one of the two possible solutions for nˆ1 , nˆ2 , nˆ3 . If that is the case, we write: N y1 ⎛ N N ˆ ⎞ ⎟⎟ nˆ1 = ± ⎜⎜ x1 iˆ + jˆ + z1 k N N N 1 1 ⎠ ⎝ 1 N y 2 ⎛ N N nˆ2 = ± ⎜⎜ x 2 iˆ + jˆ + z 2 N 2 N 2 ⎝ N 2

ˆ ⎞⎟ k ⎟

⎠

N y 3 ⎛ N N ˆ ⎞ ⎟⎟ nˆ3 = ± ⎜⎜ x 3 iˆ + jˆ + z 3 k N N N 3 3 ⎠ ⎝ 3 

If necessary, you must adjust the sense of nˆ3 so that nˆ3 = nˆ1 × nˆ2 (i.e., such that nˆ1 , nˆ2 , nˆ3 form a right-handed system of unit vectors).



It must be pointed out that the vectors N 1 , N 2 , N 3 provided by the calculator may be

r

r

r

different than the ones that we find doing the calculations by hand: Their direction is the same but their magnitude and/or sense may be different.

•

Providing the following matrix as input to a calculator

⎡ 40 − 20 10 ⎤ [σ ~] = ⎢⎢− 20 − 80 5 ⎥⎥ MPa 5 60⎥⎦ ⎢⎣ 10

⎡ 40 − 20 10⎤ ⎢− 20 − 80 5 ⎥ ⎢ ⎥ 5 60⎥⎦ ⎢⎣ 10

⇒

and using a “built-in” function to find the eigenvectors of a matrix, the results presented in the following table were obtained:

1 64.213203

Eigenvectors 2 39.331419

3 − 83.544623

0.434085

1

0.165638

N y

− 2.552958 × 10−2

− 0.185988

1

N z

1

− 0.438833

− 4.637152 × 10−2

Eigenvalue N x

- 15 -


•

Based on those results and the formulas presented at the beginning of this section, the values shown in the following table were obtained:

1 1.090450

Principal Directions 2 1.107775

3 1.014685

0.398079

0.902710

0.163241

n y

− 0.023412

− 0.167893

0.985528

n z

0.917053

− 0.396139

− 0.045700

N n x

•


Therefore:

)

ˆ nˆ1 = ± 0.398079 iˆ − 0.023412 jˆ + 0.917053 k

)

ˆ nˆ 2 = ± 0.902710 iˆ − 0.167893 jˆ − 0.396139 k

)

ˆ nˆ3 = ± 0.163241iˆ + 0.985528 jˆ − 0.045700 k

•

Checking to see if the sense of nˆ3 is correct:

)

)

ˆ × 0.902710 iˆ − 0.167893 jˆ − 0.396139 k ˆ nˆ1 × nˆ 2 = 0.398079 iˆ − 0.023412 jˆ + 0.917053 k ˆ nˆ1 × nˆ2 = 0.163241 iˆ + 0.985528 jˆ − 0.045700 k nˆ3 = nˆ1 × nˆ2

Note:



If nˆ1 × nˆ2 is equal to − nˆ3 instead of nˆ3 , we simply multiply the result that was initially obtained for nˆ3 by − 1 (i.e., we change the “ ± ” by “ m ”).

Notes: − Unless stated otherwise, in a homework or exam problem you should not use “built-in” functions on your calculator to find the eigenvectors of a matrix. However, you can always use them to check if the values that you obtained for the principal directions are correct. − In HP advanced scientific calculators (models HP48 and higher) the “built-in” function to find the eigenvalues and eigenvectors of a matrix is “EGV”. − Check the manual of your calculator to see if it allows you to input a matrix and find its eigenvalues and eigenvectors.

- 16 -



− When we use a calculator, we input values without units. Thus, we must keep track of the units on our own. For the case of the eigenvectors we don’t need to worry about the units but for the case of the eigenvalues we do.

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Principal-Stresses-3D-Example.pdf

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