Lecture #3:
- Stress analysis-Triaxial in Cartesian coordinates
- Triaxial States of Stress
- General method for triaxial stress state:
References:
1- Dieter G.E., (1986), Mechanical Metallurgy, McGraw-Hill
2-Timoshenko S, Goodier J.N.,(1984), Theory of elasticity,McGraw-Hill
Company, Inc.
3- Barber J.R., (2004), Elasticity, 2nd Edition, Kluwer academic publishers.
4- Robert J. A., Vlado A. L., (2006), Mechanics of solids and materials,
Cambridge university press.
5- Hearn E.j., 1977, Mechanics of Materials, Vol.1&2., Pergamon Press,
London.
Triaxial States of Stress
Let ABC represent a principal plane (of area A) cutting througLh a unit
cube. Let represent the resultant principal stress acting on the principal
plane.
Dr.Haydar Al-Ethari
Let l, m, n, respectively, equal the direction cosines corresponding to the
angles between the stress and the x-, y-, and z-axes. As noted above, in
this problem we are assuming that the stress is perpendicular to the plane
ABC. On the inclined plane, the stress will have components that are
normal to the plane (and shear components that are parallel to the plane,
There is no shear on the plane in this example). The stress can also be
resolved into components parallel to the x-, y-, and z-axes (i.e., Sx, Sy, Sz).
– Sx = l
Area COB = Al.
– Sy = m
Area AOC = Am.
Sz = n
Area AOB = An.
2
2
2
S = σ + τ (There is no shear on the plane in this example, so S = )
S 2 = S X2 + SY2 + S Z2 (Total stress can be expressed in terms of orthogonal
components).
In all directions: ∑ F = 0 , so:
x-direction: σAl − σ xx Al − σ yx Am − σ zx An = 0
y-direction: σAm − σ yy Am − σ xy Al − σ zy An = 0
z-direction: σAn − σ zz An − σ xz Al − σ yz Am = 0
So: (σ − σ xx )l − σ yx m − σ zx n = 0
- σ xyl + (σ − σ yy )m − σ zy n = 0
- σ xz l − σ yz m + (σ − σ zz )n = 0
Or:
σ − σ xx
− σ yx
− σ zx 


 − σ xy σ − σ yy − σ zy 
 − σ xz
− σ yz σ − σ zz 

l 
m  =0
 
 n 
The solution of the determinant of the matrix on the left yields a cubic
equation in terms of σ :
Or:
The directions in which the principal stresses act are determined by
substituting 1, 2, 3 each back into:
(σ − σ xx )l − σ yx m − σ zx n = 0
- σ xyl + (σ − σ yy )m − σ zy n = 0
- σ xz l − σ yz m + (σ − σ zz )n = 0
and then solving the resulting equations simultaneously for l, m, n, (using
the relation
).
The stress invariants will be:
I1 = σ xx + σ yy + σ zz
I 2 = σ xxσ yy + σ yyσ zz + σ zz σ xx − σ xy2 − σ yz2 − σ zx2
I 3 = σ xxσ yyσ zz + 2σ xyσ yzσ zx − σ xxσ yz2 − σ yyσ zx2 − σ zz σ xy2
σ 3 − I1σ 2 + I 2σ − I 3
The quantities I1, I2, I3 are known as stress invariants because for a given
stress state they are the same in all coordinate systems. However, the
principal stresses are easily obtained in the fully three dimensional case
using the following equations:
Where:
Example 1:
Determine the principal stresses for the state of stress given below:
− 50
0 
 0
− 50 10
0  MPa

 0
− 75
0
2
2
σ 3 − (10 − 75 )σ 2 + [(10 )(− 75 ) − (− 50 ) ]σ − (− 75 )(− 50 )
=0
In this problem one there are no shear stress along the z-axes, therefore, one
of the principal stresses is σ = −75MPa since τ xz = τ zx = τ yz = τ zy = 0
Can you find the other principal stresses?
Example 2:
The state of stresses at a point is given by:
500 500 − 400 

− 300 300 


− 1000
MPa
Calculate the stress invariants, magnitude and direction of the principal
stresses.
Solution:
I 1 = σ xx + σ yy + σ zz = 500 − 300 − 1000 = −800 MPa
I 2 = σ xxσ yy + σ yy σ zz + σ zz σ xx − σ xy2 − σ yz2 − σ zx2
=500(-300)+(-300)(-1000)+(-1000)(500)-(500)2-(300) 2-(-400)2 = -850GPa
I 3 = σ xxσ yy σ zz + 2σ xyσ yz σ zx − σ xxσ yz2 − σ yyσ zx2 − σ zz σ xy2
=500(-300)(-1000)+2(500)(300)(-400)-(500)(300)2-(-300)(-400)2
-(-1000)(500)2 = 283 106 MPa
σ 3 − I 1σ 2 + I 2σ − I 3 = 0
So
σ 3 + 800σ 2 − 85 ∗ 10 4 σ − 283 ∗ 10 6 = 0
Using Newton-Raphson method: σ i +1 = σ i −
f ′(σ ) = 3σ 2 + 1600σ − 85 ∗ 10 4
f (σ i )
f ′(σ i )
where:
OR sub. in the following equations:
Where:
get that:
= 29.16 0
1
=773MPa
σ 2 = −283.9 MPa
Determination of the directions:
For 1 = 773MPa, sub. in the following equations:
(σ − σ xx )l − σ xy m − σ xz n = 0
- σ xyl + (σ − σ yy )m − σ yz n = 0
- σ xz l − σ yz m + (σ − σ zz )n = 0
(773 − 500)l − 500m + 400n = 0
− 500l + (773 − (− 300))m − 300n = 0
− (− 400)l − 300m + (773 − (− 1000))n = 0
σ 3 = −1289.14MPa
 273 − 500 400 
− 500 1073 − 300


 400 − 300 1773 
Or
Remember that (cofactor)ij = (− 1)i + j M ij
Determining the cofactor of the 1st row elements:
A1 = 1073(1773)-(-300)(-300) = 1812429
B1 = (− 500 ∗1773 − 400 ∗ −300)(− 1)1+2 = 766500
C1 = −500 ∗ −300 − 1073 ∗ 400 = −279200
K=
A12 + B12 + C12
So l =
A1
= 0.9119
K
; m = 0.3856 and n = -0.1405
Using the same method, we can find that:
and n2 = 0.497
l 2 = −0.2717 , m2 = 0.824
l 3 = 0.3074 , m3 = -0.4151 and n3 = 0.8562
General method for triaxial stress state:
In deriving the equations used in the previous example/method, we assumed
that the stress on the inclined plane was a principal stress (i.e., it was
perpendicular to the plane and there was no shear). That was easy.
Now we will consider the case where the total stress is not principal.
Let l, m, n, respectively, equal the direction cosines corresponding to the
angles between the normal to the plane and the x-, y-, and z-axes. In this
problem the total (resultant) stress on the plane S is not necessarily coaxial
with the normal stress. Therefore:
S2 = 2 + 2
The total (resultant) stress can be resolved into:
S x = σ xx l + σ xy m + σ xz n
S y = σ xyl + σ yy m + σ yz n
S z = σ zxl + σ zy m + σ zz n
Where:
S 2 = S X2 + SY2 + S Z2
We can now write an expression for the normal stress on the plane as:
σ normal = S X l + SY m + S Z n
Or
σ = σ normal = σ xx l 2 + σ yy m 2 + σ zz n 2 + 2lmσ xy + 2mnσ xy + 2nlσ xz
We can now evaluate the magnitude of the shear stress on this plane using
the relationship:
2
τ n = S 2 − σ normal
The magnitude and directions of the shear stress components lying on the
plane are determined by resolving the stress components onto a set of
principal axes.
The maximum shear stress occurs when: τ max =
σ1 − σ 3
, where σ1 > σ 2 > σ 3
2
Example:
An oblique plane its normal makes angles of 300, 71.50 with y, z axis. Its
resultant stress acts in a direction makes angles of 430 , 750 with x, y axis
and equals 216MPa. If xy= 23.2MPa, xz=57MPa, yz= -3MPa, find the
stress tensor, the normal and the shear components of the resultant stress and
their directions.
Solution: y
For the resultant stress, S :
l = cos 43
m = cos 75
For the plane:
m = cos 30
n = cos 71.5
But, in general:
l 2 + m 2 + n 2 = 1 ; so:
for the resultant stress: n=0.631
for the plane:
l = 0.387
S X = Sl = 216 ∗ cos 43 = 158MPa
S Z = Sn = 216 ∗ 0,631 = 136.3MPa
SY = Sm = 216 ∗ cos 75 =55.9MPa
But:
S x = σ xxl + σ xy m + σ xz n
S y = σ xy l + σ yy m + σ yz n
S z = σ zxl + σ zy m + σ zz n
Sub. get that: σ xx = 309.6MPa
σ yy = 55.4 MPa
σ zz = 370MPa
σ normal = S X l + SY m + S Z n = 158 ∗ 0.387 + 55.9 ∗ cos 30 + 136.3 cos 71.5 = 152.6 MPa
So:
2
τ n = S 2 − σ normal
= 152.9MPa
Assume the direction cosines of the shear stress are ls , ms ns , then
S x = σ n l + τ n ls
ms =
so
ls =
1
(S x − σ nl ) = 158 − 152.9 ∗ 0.387 = 0.647 or
τn
152.9
1
(S y − mσ n ) = - 0.5
τn
ns =
1
(S z − nσ n ) = 0.576
τn