ME-6201 AEP 3D Stress PDF
ME-6201 AEP 3D Stress PDF
ME-6201 AEP 3D Stress PDF
3-D stress-Strain
Prof. S.K.Sahoo
The stresses (normal and shear) on a inclined plane
• Let consider the general 3D state of stress
at a point specified by, x , y , z , xy , yz , zx
• Similar to 2D case, let find out the
y yy normal and shear stress on a
A arbitrarily doubly inclined plane ABC.
yx
yz zz N(l1,m1,n1)
xy
xz zx • The plane ABC is
xx specified by its normal N.
xx zy zy It is represented by its
O xz direction cosines
xy zx B x (l1,m1,n1).
yz • If N makes angles , Φ, with x,y,z axes,
zz yx then l1 = cos,m1 =cos Φ,n1 =cos .
C yy
• Values of l1,m1,n1 are not independent.
z Having relations, l12+m12+n12=1
The stresses (normal and shear) on a inclined plane
• As the elemental cube is in static equilibrium, the
tetrahedron ABCO will be in equilibrium by the
stress act in faces OAC, OAB, OBC and the
y yy resultant stress/traction on face ABC.
A
yx
yz zz N(l1,m1,n1)
xy
xz zx xx
xx zy zy
O xz
xy zx x
B • Let Area of the plane ABC is A.
yz
zz yx So, Area of OAB=An1
Area of OBC=Am1
C yy
z Area of OAC=Al1
The stresses (normal and shear) on a inclined plane
• Let the resultant stress has components Sx, Sy &
Sz in x, y, z-directions. Applying force balance
equations, we have:
y
F x 0 AS x x Al1 xy Am1 zx An1 0
A
zz
F y 0,
yx
C AS z zx Al1 yz Am1 z An1 0
yy
z
The stresses (normal and shear) on a inclined plane
S x xl1 xy m1 zx n1
• Simplifying,
S y xyl1 y m1 yz n1
y S z zxl1 yz m1 z n1
A
zz
S
zx y
n3
zy Sz
n1 • Adding normal
xz n2 components of S , S ,
Sx x y
xx Sz, we have,
O
xy yz B x
n n1 n 2 n 3
yx
C n S xl1 S y m1 S z n1
yy
z
The stresses (normal and shear) on a inclined plane
• Putting values of Sx,
Sy, Sz, we have
n xl12 y m12 z n12 2 xyl1m1 2 yz m1n1 2 zx n1l1
y
A
s
zz
zx n s2 2 n2
xz zy
S x2 S y2 S z2 n2
xx O
xy yz B x
yx
C
yy
z
Principal stresses and Principal planes in 3D stress system
• Let the plane ABC (direction cosine of l,m,n) is oriented in such a
way that the resultant stress σ has only normal component, ie, is
normal to plane ABC, ie, a principal stress.
• As σ is itself normal to
y plane ABC its components
A N(l,m,n) in x,y,z-directions Sx, Sy, Sz
can be found by
zz
zx
= n
multiplying it with its
direction cosines l, m, n.
Sy Sx l Sy m Sz n
zy
xz
Sx
xx O Sz • Also Sx, Sy, Sz can be
found from stress terms
xy yz B x S l m n
x xx xy zx
yx
S y xyl yy m yz n
C
yy S z zxl yz m zz n
z
Principal stresses and Principal planes in 3D stress system
• Subtracting first set from second ones, ( xx )l xy m zx n 0
we have,
• These are three homogeneous linear xyl ( yy )m yz n 0
equations in l,m,n. To give a non-zero
solution it is necessary that the
zxl yz m ( zz )n 0
determinant should be zero.
( xx ) xy xz l
( xx ) xy xz
( )
( ) m 0
0
xy yy yz
xy yy yz
xz xz yz ( zz ) n
yz ( zz )
• Expanding this determinant we get a 3 ( xx yy zz ) 2
cubic equation in σ as,
( xx yy yy zz zz xx xy2 yz2 zx2 )
( xx yy zz 2 xy yz zx xx yz2 yy zx2 zz xy2 ) 0
• The three roots of the cubic equation (designated as σI, σII , σIII ) are three
principal stresses and for each Principal stress associated by a principal plane.
It can be obtained by putting the value of σ (σI, σII , σIII ) in the equations and
solved separately. These planes are orthogonal to each other.
Stress Invariants
• The cubic equations of σ can be written as, 3 J1 2 J 2 J 3 0
• Where, J1 xx yy zz J 2 xx yy yy zz zz xx xy2 yz2 zx2
xx xy yy yz xx xz
xy yy yz zz xz zz
J 3 ( xx yy zz 2 xy yz zx xx yz2 yy zx2 zz xy2 )
xx xy zx
xy yy yz • J1, J2, J3 are known as the first,
zx yz zz second, and third invariants of stress
respectively.
s2 2 n2
• Shear stress is: S x2 S y2 S z2 n2
I2l22 II2 m22 III
2 2
n2 ( I l22 II m22 III n22 ) 2
• Assuming: I II III
• The solutions are:
1 1 II III
l3 0, m3 , n3 Maximum shear stress
2 2 2
1 1 I III
l3 , m3 0, n3 Maximum shear stress
2 2 2
1 1 I II
l3 , m3 , n3 0 Maximum shear stress
2 2 2
• Each one of the three planes of maximum shear is inclined at 450
to two of the principal directions and parallel to the third one.
• When, I II III
• No shear stress exists on any plane and
state of stress is hydrostatic, where S I II III
Mean and Deviatoric Stresses
• The mean stress, also called hydrostatic stress is,
1 1 1 1
S m J1 ii ( xx yy zz ) ( I II III )
3 3 3 3
• In tensor notation, S0 0 • This tensor indicate, on any arbitrary
0 S 0
plane the stress resultant will be
0 0 S normal and equal to S
• When the mean/hydrostatic stress components are subtracted from the
stress tensor components, deviatoric stress are obtained
xx xy zx S 0 0 xx S xy zx S xx S xy S zx
xy yy yz 0 S 0 xy yy S yz S xy S yy S yz Sij
zx yz zz 0 0 S zx yz zz S S zx S yz S zz
• It can be expressed as, Where,
3l 0m 0n 0 3l 0 l 0
2
0l 2m 4n 0 m
2m 4n 0 5
0l 4m 8n 0 1
4m 8n 0 n
5
Similarly, putting the values for σII , σIII , we can
find the other two associated principal planes l 2 m2 n2 1
Example: xx 80 Mpa; yy 60 Mpa; zz 20 Mpa
Given: Stress components xy 20 Mpa; xz 40 Mpa; yz 10 Mpa
Find: Invariants, Principal Values, max. shear and oct. shear stress.
Find normal stress on a plane normal to iˆ 2 ˆj kˆ ,ie, direction cosines are:
1 2 1
, ,
Answer: First Invariant 6 6 6
Characteristic Equation:
3 160 2 5500 0 0
max 12 1 3 55 Mpa
( 2 160 5500) 0
2 oct 19 1 2 2 3 3 1
2 2 2
1 110 Mpa
2 50 Mpa 44.97Mpa
σ1 =
00
3
S x2 S y2 S z2 n2
l 2 m2 n2 1 I2l 2 II2 m 2 III
2 2
n ( I l 2 II m 2 III n 2 ) 2
Substituting, m2 1 l 2 n2
So, n l I II 1 l n III n
2 2 2 2
2
n
n l 2 II I II
III II
l
2
s
2 2
I
2
II 2
II n 2
2
III 2
II l
2
I II II n III II
2
2
s2 II n III n
Substituting value of n2 in
2 2
we can get, l
s
II I III I
2 s2 III n I n 2 s2 I n II n
Similarly, we can get, m , n
III II I II I III II III
From first equation of l2
n II n III s2 l 2 I II I III
2 2
III III
or n II s l I II I III II
2 2
2 2
It is a equation of a circle
( x a ) 2 y 2 r 2 centre at (a,0) of radius r
Thus, if l,m,n are given for a particular plane, n & s lie on the circle
and a radius
l 2 I II I III II III
2
Instruction to draw Mohr’s Circle
1. Mark the points P1, P2, P3 on the n axis where OP1 I , OP2 II , OP3 III .
2. Draw circles on diameter P1P2, P2P3 & P1P3 having centres at C1, C2, C3 at
s
T2
T3 Q2 S2 T1
P
R3
R2 S3 Q3
O
P3 P1 n
C2 P2 N C3 C1
III
n
II
I