1
1
CVEN2301 Mechanics of Solids
STRESS TRANSFORMATION
Chongmin Song
School of Civil and Environmental EngineeringThe University of New South Wales
Stress and Strain at a Point
zzyzx
yzyyx
xzxyx
στττστττσ
zzyzx
yzyyx
xzxyx
στττστττσ
σσσσ zz
σσσσ xx
σσσσ yy
ττττzy
ττττzx
ττττxz
ττττxy
ττττyxττττyz
Plane xPlane z
Plane y
y
xz
3D Stresses at a Point
Stress tensor
),,,( zyxji
jiij
=
=ττ
Complementary property of shear
2
Plane stresses 0zz zx zyσ τ τ= = =
Sign convention: Positive normal stress acts outward fromall faces and positive shear stress acts upward on the right-hand face of the element.
Plane StressesA structural member may be subjected to both normal
stress and shear stress .
Which stress do we use for design purposes?
3
Design assumption: material behaviour (elastic/plastic/in failure) is identical along the bar.
Normal stress:
Shear stress:
0=τ
Along Section Plane a-a
4
Along Section Plane b-b
N40030sin800
N8.69230cos800
=×=
=×=
�
�
V
N
: force Shear
:force Normal
Along Section Plane b-b
kPa375101.8475
8.692
kPa217101.8475
400
m101.8475)60sin/04.0(04.0
3-
3-
23-
=×
==
=×
==
×=×=
A
N
A
V
A
σ
τ
:stress Normal
:stress Shear
:area Sectional�
5
Equivalent Stress States
Stress Transformation
At a specified location under a given loading:• The material behaviour (elastic/plastic/in failure) is
uniquely determined.• The stress states of an element will change with its
orientation but are equivalent.
The equations for stress transformation provides us with unique values to be used in structural designs.
6
EXAMPLE 1
Equilibrium (of force) in x’, y’ coordinates
The answer is independent of ∆A
7
Equilibrium (of force) in x’, y’ coordinates
Stresses on plane b-b
8
Stresses on plane b-b
Equivalent stress states
9
• Stress transformation from x, y axes to x’, y’ axes• The orientation of an inclined plane will be defined using
the angle θ. • Angle θ is measured from x-axis to x’-axis in the
counterclockwise direction
Stress Transformation
σσσσ x'x'
ττττ y'x'
ττττ x'y'
σσσσ y'y'
x'
y'
System 2
θθθθ xσσσσxx
ττττyx
ττττxy
σσσσyy
x
y
System 1
θ
σσσσxx
ττττyx
ττττxy
σσσσyy
x
y
System 1
• Consider the equilibrium of a triangular segment
Stress Transformation
θ
10
Stress Transformation
θτθσσσσ
σ
θθθθ
θθθ
θθτθσθσσ
θθσθθτθθσθθτσ
2sin2cos22
2/)2cos1(cos
;2/)2cos1(sin
;cossin22sin
)cossin2(sincos
0cos)cos(sin)cos(
sin)sin(cos)sin(
'
2
2
22'
'
xyyxyx
x
xyyxx
xxy
yxyx
AA
AAA
+−
++
=
+=
−==
++=
=∆−∆−
∆−∆−∆
:stress Normal
:identitiesric trigonomet Using
ion x'-directin mEquilibriu
Stress Transformation
θτθσσ
τ
θθτθθσστ
θθσθθτθθσθθττ
2cos2sin2
)sin(coscossin)(
0sin)cos(cos)cos(
cos)sin(sin)sin(
''
22''
''
xyyx
yx
xyxyyx
xxy
yxyyx
AA
AAA
+−
−=
−+−=
=∆−∆−
∆−∆+∆
:stress Shear
ion y'-directin mEquilibriu
11
Stress Transformation
θτθσσσσ
σ
θθ
θτθσσσσ
σ
2sin2cos22
)
2sin2cos22
'
'
xyyxyx
y
xyyxyx
x
−−
−+
=
+←
+−
++
=
:90( ion y'-directin stress Normal
:ion x'-directin stress Normal
�
x
y’
x’
EXAMPLE 2
12
EXAMPLE 2
MPa15.4
)30sin()30(2cos2
5080
2
5080
2sin2cos22
MPa8.25
)30sin()30(2cos2
5080
2
5080
2sin2cos22
'
'
−=
−−−−−−+−=
−−
−+
=
−=
−+−−−++−=
+−
++
=
��
��
xy
xyyxyx
y
xy
xyyxyx
x
τ
θτθσσσσ
σ
τ
θτθσσσσ
σ
:system coordinate y', x'in Stresses
Stresses in x, y coordinate system
EXAMPLE 2
MPa8.68
)30(2cos)25(2
5080
2cos2sin2''
−=
−−+−−−=
+−
−=
�
θτθσσ
τ xyyx
yx
Note the signs and directions of stresses
13
Principal Stresses
( )
( )
121
'
90
:2180tan(2tan
2/2tan
02cos22sin22
ppp
ppp
yx
xyp
pxypyx
p
x
d
d
θθθ
θθθ
σστ
θ
θτθσσ
θθθσ
+=
+=
−=
=+−
−==
�
�
and
for exist solutions two ), As
The principal stresses represent the maximum and minimum normal stress at the point.
θτθσσσσ
σ 2sin2cos22' xy
yxyxx +
−+
+=
• Orientation of the planes of maximum and minimum normal stress (principal planes)
• Normal stress
Principal Stresses
( )
2
2
1
2
2
1
2/
22cos
2/2sin
0 0
xyyxyx
p
xyyx
xyp
xyyx
τσσσσ
θ
τσσ
τθ
τσσ
+
−
−=
+
−=
>>− and case the gConsiderin
pxyyxyx
x θθθτθσσσσ
σ =+−
++
= with ;2sin2cos22'
Normal stress acting on principal planes
2
2
2
2
2
2
2/
22cos
;2
/2sin
xyyxyx
p
xyyx
xyp
τσσσσ
θ
τσσ
τθ
+
−
−−=
+
−−=
( ) 2/2tan
yx
xyp σσ
τθ
−=
14
Principal Stresses
2
2
2
2
2
2
111
22
2/
22
2sin2cos22
xyyxyx
xyyx
xyyxyx
pxypyxyx
τσσσσ
τσσ
τσσσσ
θτθσσσσ
σ
+
−+
+=
+
−
+
−+
+=
+−
++
=
Maximum in-plane normal stress
Minimum in-plane normal stress
2
2
222
22
2sin2cos22
xyyxyx
pxypyxyx
τσσσσ
θτθσσσσ
σ
+
−−
+=
+−
++
=
Principal Stresses
2
2
2,1 22 xyyxyx τ
σσσσσ +
−±
+=
Maximum/ Minimum in-plane normal stress (σ1 ≥ σ2)
Shear stress on the principal planes
02cos2sin2
2cos2sin2
;2
2cos
2sin2tan
'' =+−
−=
=−
⇒−
==
pxypyx
yx
pxypyx
yx
xy
p
pp
θτθσσ
τ
θτθσσ
σστ
θθ
θ As
No shear stress acts on the principal planes!
15
σσσσxx = 3
τ = 2τ = 2τ = 2τ = 2
τ = 2τ = 2τ = 2τ = 2
σσσσyy =1
x
y
Find the principal stresses and show their sense on a properly oriented element.
Unit: MPa
EXAMPLE 3
MPa
MPa
MPa
xy
y
x
2
1
3
=
==
τσσ
EXAMPLE 3Principal Direction:
x'
y'
θ = 31.72 θ = 31.72 θ = 31.72 θ = 31.72 deg x
4.24 MPa
-0.24 MPa
( ) 22
422tan ==
−=
yx
xyp σσ
τθ
MPa24.4
)72.312sin(2)72.312cos(2
13
2
13
2sin2cos22'
=
×+×−++=
+−
++
=
��
θτθσσσσ
σ xyyxyx
x
MPa24.0
)72.312sin(2)72.312cos(2
13
2
13
2sin2cos22'
−=
×−×−−+=
−−
−+
=
��
θτθσσσσ
σ xyyxyx
y
MPa
MPa
MPa
xy
y
x
2
1
3
=
==
τσσ
MPa24.0
MPa24.4
2
1
−==
σσ
�72.31=pθ
16
Maximum Shear Stress
( )( )
psps
yx
xyp
pxy
yxs
sxysyx
s
yx
xyyx
yx
d
d
θθθθ
σστ
θθτ
σσθ
θτθσσ
θθθτ
θτθσσ
τ
+=⇒+=
−=−=
−−=
=−−
−==
+−
−=
�� 45;2902
2/2tan ;
2tan
12/2tan
02sin22cos22
2cos2sin2
''
''
stress shear maximum of plane the of nOrientatio
:stress Shear
The plane for maximum shear stress is orientated 45°from principal planes.
2
2
2cos2sin2
2cos2sin
avg''
22
''
yxyx
xyyx
xyyx
yx
ss
σσσσσ
τσσ
τ
θτθσσ
τ
θθ
+===
+
−=
+−
−=
stress shear maximum of
planes the on stress normal the Similarly,
stress shear Maximum
:into and for values the ngSubstituti
max
Maximum Shear Stresses
( )xy
yxs τ
σσθ
2/2tan
−−=
17
EXAMPLE 4
EXAMPLE 4
02
02
00
avg
222
=+
=
±=+=+
−=
−===
yx
xyyx
xyyx
σσσ
τττσσ
τ
ττσσ
stress normal Average
stress shear Maximum
Stresses
max
Ductile materials (e.g. mild steel) often fail due to shear stress
18
EXAMPLE 4
θ θ
EXAMPLE 4
Brittle materials (e.g. cast iron) often fail due to normal stress
Stress element
19
20
Principal stresses
Which of the following statement is incorrect?
A)The principal stresses represent the maximum and minimum normal stress at the point
B)When the state of stress is represented by the principal stresses, no shear stress will act on the element
C)When the state of stress is represented in terms of the maximum in-plane shear stress, no normal stress will act on the element
D)For the state of stress at a point, the maximum in-plane shear stress usually associated with the average normal stress.
Quiz
21
Mohr’s Circle of StressA graphical solution for plane stress transformation Often convenient and easy to use. There are several ways of drawing a Mohr’s circle
)3( 2cos2sin2
)2(2sin2cos22
)1(2sin2cos22
''
'
'
θτθσσ
τ
θτθσσσσ
σ
θτθσσσσ
σ
xyyx
yx
xyyxyx
y
xyyxyx
x
+−
−=
−−
−+
=
+−
++
=
Equation for stress transformation:
Mohr’s Circle of Stress
( )
( ) ( )
2
2
avg
22''
2avg'
2
22
''
2
'
2222
''
'
2;
2
22
12sin2cos)()(
)3( 2cos2sin2
)4(2sin2cos22
xyyxyx
xyyx
yxx
xyyx
yxyx
x
xyyx
yx
xyyxyx
x
R
R
τσσσσ
σ
τσστσσ
τσσ
τσσ
σ
θθ
θτθσσ
τ
θτθσσσσ
σ
+
−=
+=
=+−
+
−=+
+−
=++
+−
−=
+−
=+
−
constants) known are , ,( with
as Rewritten
using and 3 Eq.4 Eq.
Rewrite Eq. 1
22
Mohr’s Circle of Stress
( ) ( ) 22''
2avg'
avg
'''
)0,
R
R
yxx
yxx
=+− τσσ
στσ
( at centre and
radius having circle Mohr' a on is ) ,(
Sign Convention:• σ is positive to
the right, and • τ is positive
downward.
Mohr’s Circle of Stress
xyyxyx
xyyxxx
ττσσθττσσθ
−======
'''
'''
90
0
and , :G
and , : Apoints Reference
�
�
A rotation of θ of the x’ axis correspondsto a rotation 2θ on the circle in the same direction
23
Mohr’s Circle of StressDraw a Mohr’s circle• Coordinates σ, τ (positive downward)• Centre of circle C (σavg, 0)• Reference point A (σx, τxy)• Connect CA (θ=0o) and determine R• Sketch the circle
σ
τ
C
Aθ=0
o
Mohr’s Circle of StressPrincipal stress• σ1 at point B on circle• Angle 2θp1 measured from reference line CA to CB• σ2 at point D on circle; θp1 and θp2 are 90o apart• Sketch a stress element with the principal axis (+x’) rotate
by θp1 from the reference axis (+x)
σ
τ
C
A
2θθθθp1
B
D
24
Mohr’s Circle of StressMaximum shear• τmax at point E or F• Angle 2θs1 measured from reference line CA to CE or CF• Sketch a stress element
σ
τ
C
A2θθθθs1
B
D
E
F
Mohr’s Circle of StressStresses on Arbitrary Plane• Measure an angle 2θ on the circle from the reference line
CA in the same direction as angle θ• Determine the coordinates of P
σ
τ
C
A
B
DP
25
MPa60
MPa12
−==
−=
xy
y
x
τσσ
( )
�
�
5.22
45612
6tan2
plane principal ofn Orientatio
MPa49.1449.86
MPa49.2649.8
stresses Principle
MPa49.86)612(
circle theof Radius
(-12,-6)A point initial The
0) (-6, Ccenter The
MPa6
2/)012(2/
2
12
2
1
22
avg
=
=−
=
−=−−==−=
=+−=
−=
+−=+=
−
p
p
yx
R
θ
θ
σσ
σσσ
(MPa)σ
26
MPa6
MPa12
MPa8
−=
=−=
xy
y
x
τσσ
( )
)0(CA from measured
clockwisecounter 60)2(30at
CP line radial a Draw
MPa66.116)10(
circle theof Radius
(-8,-6)A point initial The
0) (2, Ccenter The
MPa22/)128(2/
22
avg
�
��
=
=
=+=
=+−=+=
θ
σσσ
R
yx
27
MPa66.504.29sin66.11
MPa20.804.29cos66.112
04.2996.3060
96.3010
6tan
circle theofgeometry theFrom
''
'
1-
==
−=−=
=−=
==
�
�
���
�
yx
x
τ
σ
ψ
φ
Determine the coordinate of point P(σx’, τx’y’)
• Face DE is 60o clockwise from the x-axis.
• Stresses represented by point Q
MPa66.504.29sin66.11
MPa2.1204.29cos66.112
circle theofgeometry theFrom
''
'
−=−=
=+=�
�
yx
x
τ
σ