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1. 1
P4 Stress and Strain Dr. A.B. Zavatsky
HT08
Lecture 8
Plane Strain and
Measurement of Strain Plane stress versus plane strain.
Transformation equations.
Principal strains and maximum shear strains.
Mohr’s circle for plane strain.
Measurement of strain and strain rosettes.

2. 2
Plane stress versus plane strain
σ
z
=0, τ
xz
=0, τ
yz
=0
σ
x
, σ
y
, τ
xy
may be
non-zero.
τ
xz
=0, τ
yz
=0
σ
x
, σ
y
, σ
z
, τ
xy
may be non-zero.
γ
xz
=0, γ
yz
=0
ε
x
, ε
y
, ε
z
, γ
xy
may be non-zero.
ε
z
=0, γ
xz
=0, γ
yz
=0
ε
x
, ε
y
, γ
xy
may be
non-zero.
Stresses
Strains
Plane StressPlane Strain
Plane stress and plane strain do not ordinarily occur simultaneously.
One exception is when σ
z
= 0 and σ
x
= -σ
y
, since Hooke’s Law gives
ε
z
= 0.

3. 3
Consider the change in length and orientation of the diagonal of a
rectangular element in the xyplane after strains ε
x
, ε
y
, and γ
xy
are applied.
x
y
y
1
x
1
Transformation Equations for Plane Strain
We want to derive equations for the normal strains ε
x1
and ε
y1
and the
shear strain γ
x1y1
associated with the x
1
y
1
axes, which are rotated counter-
clockwise through an angle θfrom the xyaxes.
ε
x
dx
ε
x
dxcosθ
θ
dx
ds dy
Diagonal increases in length in
the x
1
direction by ε
x
dx cosθ.
α
1
Diagonal rotates clockwise
by α
1
.
θ ε α
θ
ε
α
sin
sin
1
1
ds
dx
dx ds
x
x
=
=

4. 4
x
y
y
1
x
1
ε
y
dy
ε
y
dysin θ θ
dx
ds
dy
Diagonal increases in length in
the x
1
direction by ε
y
dy sinθ.
α
2
Diagonal rotates counter-
clockwise by α
2
.
x
y
y
1
x
1
γ
xy
dycosθ
θ
dx
ds dy
Diagonal increases in length in
the x
1
direction by γ
xy
dx cosθ.
α
3
Diagonal rotates clockwise
by α
3
.
θ ε α
θ
ε
α
cos
cos
2
2
ds
dy
dy ds
y
y
=
=
θ γ α
θ
γ
α
sin
sin
3
3
ds
dy
dy ds
xy
xy
=
=
γ
xy
dy
γ
xy

5. 5
The total increase in the length of the diagonal is:
θ
γ
θ
ε
θ
ε
cos sin cos )(dy dy dx ds
xy y x
+
+
=
Δ
The normal strain
ε
x1
is the change in length over the original length:
θ γ θ ε θ ε ε
cos sin cos
)(
1
ds
dy
ds
dy
ds
dx
ds
ds
xy y x x
+ + =
Δ
=
dx
dy
ds
θ θ
sin cos= =
ds
dy
ds
dx
θ
The normal strain
ε
y1
can be found by substituting
θ
+90° into the
equation for
ε
x1
.
θ θ γ θ ε θ ε ε
cos sin sin cos
2 2
1xy y x x
+ + =
So, the normal strain
ε
x1
is:

6. 6
To find the shear strain
γ
x1y1
, we must find the decrease in angle
of lines in the material that were initially along the x
1
y
1
axes.
x
y
y
1
x
1
θ
α
β
β
α
γ
+
=
11yx
θ γ θ θ ε ε α
θ γ θ θ ε θ θ ε α
θ γ θ ε θ ε α
α
α
α
α
2
2
3 2 1
sin cos sin) (
sin cos sin cos sin
sin cos sin
xy y x
xy y x
xy y x
ds
dy
ds
dy
ds
dx
− − −=
− + −=
− + −=
−
+
−
=
To find
α
, we just sum
α
1
, α
2
,
and
α
3
,
taking the direction of the
rotation into account.

7. 7
Using trigonometric identities for sin
θ
cos
θ
, sin
2
θ
, and cos
2
θ
gives
the strain transformation equations …
To find the angle
β
, we can substitute
θ
+90 into the equation for
α
,
but we must insert a negative sign, since
α
is counterclockwise and
β
is clockwise.
θ γ θ θ ε ε β
θ γ θ θ ε ε β
2
2
cos cos sin) (
)90 ( sin )90 cos( )90 sin( ) (
xy y x
xy y x
+ − −=
+ + + + − =
So, the shear strain
γ
x1y1
is:
) cos (sin
2
cos sin) (
2
) cos (sin cos sin) (2
cos cos sin) ( sin cos sin) (
2 2 11
2 2
11
2 2
11
11
θ θ
γ
θ θ ε ε
γ
θ θ γ θ θ ε ε γ
θ γ θ θ ε ε θ γ θ θ ε ε γ
β
α
γ
− − − −=
− − − −=
− − − − − −=
+
=
xy
y x
yx
xy y x yx
xy y x xy y x yx
yx

8. 8
The equations have the same form, but with different variables:
()
θ
γ
θ
ε ε γ
θ
γ
θ
ε
ε
ε
ε
ε
2cos
2
2sin
2 2
2sin
2
2cos
2 2
11
1
xy y x yx
xy y x y x
x
+
−
−=
+
−
+
+
=
Now, compare the strain transformation equations to the stress trans-
formation equations:
()
θ τ θ
σ σ
τ
θ τ θ
σ
σ
σ
σ
σ
2cos 2sin
2
2sin 2cos
2 2
11
1
xy
y x
yx
xy
y x y x
x
+
−
−=
+
−
+
+
=
y y
x x
x x
σ ε
σ ε
σ
ε
⇔
⇔
⇔
1 1
xy
xy
yx
yx
τ
γ
τ
γ
⇔
⇔
2
2
11
11

9. 9
So, the all the equations that we derived based on the stress trans-
formation equations can be converted to equations for strains ifwe
make the appropriate substitutions.
Remember that ε
z
= 0 (plane strain).
Shear strains are zero on the principal planes.
Principal stresses and principal stra ins occur in the same directions.
Maximum Shear Strains
2 2
max
2 2 2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
=
xy y x
γ ε ε γ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
−=
xy
y x
s
γ
ε ε
θ
2tan
The maximum shear strains are associated with axes at 45° to the directions
of the principal strains.
Principal Strains and Principal Angles
2 2
2,1
2 2 2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
±
+
=
xy y x y x
γ ε ε ε ε
ε
y x
xy
p
ε ε
γ
θ
−
= 2 tan

10. 10
Mohr’s Circle for Plane Strain
ε
x1
(
γ
x1y1
/2)
c
R
Plot
ε
x1
instead of
σ
x1
.
Plot (
γ
x1y1
/2) instead of
τ
x1y1
.
ε
1
ε
2
Principal strains
ε
1
,
ε
2
ε
s
γ
max
/2
Maximum shear strain
γ
max
with
associated normal strain
ε
s

11. 11
Example
An element of material in plane strain has ε
x
= 340 x 10
-6
,
ε
y
= 110 x 10
-6
,
γ
xy
= 180 x10
-6
Find the principal strains, the (in-plane) maximum shear strains,
and the strains on an element oriented at an angle
θ
=30°.
(based on Gere & Timoshenko, p 439)
Plane strain means that ε
z
= 0.
The transformation equations
with θ=30° give
ε
x1
= 360 x 10
-6
γ
x1y1
= -110 x 10
-6
Using ε
x
+ ε
y
= ε
x1
+ ε
y1
gives ε
y1
= 90 x 10
-6
x
y
ε
x
ε
y
γ
xy
Equations give
ε
1
= 371 x 10
–6
ε
2
= 79 x 10
–6
θ
p
= 19.0° and 109.0°
γ
max
= 290 x 10
–6
θ
s
= -26.0° and 64.0°

12. 12
ε
γ
/2
A (θ=0)
ε
x
= 340 x10
-6
(γ
xy
/2) = 90 x10
-6
A (θ=0)
Units on axes are strain x 10
-6
R
146 90 115
)2/ 180( )225 340(
2 2
2 2
= + =
+ − =
R
R
B (θ=90)
B (θ=90)
ε
y
= 110 x10
-6
-(γ
xy
/2) = -90 x10
-6
ε
1
ε
2
79,371
146 225
2,1
2,1
=
±
=
±
=
εε
Rc
c
225
2
110 340
2
=
+
=
+
= =
c
c
y x
avg
ε
ε
ε
2θ
p1
2θ
p2
° =
° =
−
=
0.19
05.38 2
225 340
90
2tan
1
1
1
p
p
p
θ
θ
θ
° =
° +
=
0. 109
180 2 2
2
1 2
p
p p
θθ
θ
Principal Strains
Mohr’s Circle

13. 13
B (θ=90)
A (θ=0)
ε
x
= 340 x10
-6
(γ
xy
/2) = 90 x10
-6
A (θ=0)
B (θ=90)
ε
y
= 110 x10
-6
-(γ
xy
/2) = -90 x10
-6
2θ
s1
(
γ
max
/2)
292
146 )2/ (
max
max
=
=
=
γ
γ
R
° −=
−=
−
−
= −
−
=
0.26
95.51 2
)05.38 90( ) 2 90( 2
1
1
1 1
s
s
p s
θ
θ
θ
θ
2θ
s2
2θ
p1
c
R
ε
γ
/2Units on axes are strain x 10
-6
Maximum Shear
225
=
=c
s
ε
° =
° +
=
0.64
90 2 2
2
1 2
s
p s
θθ
θ

14. 14
ε
γ
/2Units on axes are strain x 10
-6
Strains when θ= 30°
B (θ=90)
A (θ=0)
ε
x
= 340 x10
-6
(γ
xy
/2) = 90 x10
-6
A (θ=0)
B (θ=90)
ε
y
= 110 x10
-6
-(γ
xy
/2) = -90 x10
-6
c
R
C (θ=30)
D (θ=30+90)
2θ
p1
2θ
° = ° =60 2 30
θ
θ
360
)05.38 60 cos( 146 225
) 2 2 cos(
1
1
1 1
=
− + =
−
+
=
x
x
p x
Rc
ε
ε
θ
θ
ε
110
55 )2/ (
)05.38 60 sin( 146 )2/ (
) 2 2 sin( )2/ (
11
11
11
1 11
−=
−=
− −=
−
−
=
yx
yx
yx
p yx
R
γ
γ
γ
θ
θ
γ
90
)05.38 60 cos( 146 225
) 2 2 cos(
1
1
1 1
=
− − =
−
−
=
y
y
p y
Rc
ε
ε
θ
θ
ε

15. 15
x
y
x
1
y
1
Principal Strains
ε
1
= 371 x 10
–6
ε
2
= 79 x 10
–6
θ
p1
= 19.0°
θ
p2
= 109.0°
x
y
x
1
y
1
Maximum Shear Strain
γ
max
= 290 x 10
–6
θ
smax
= -26.0°
ε
s
= 225 x 10
–6
ε
1
ε
2
θ
p2
θ
p1
θ
smax
γ
max
ε
s
ε
s
No shear strains

16. 16
x
y
x
1
y
1
Strains when θ= 30°
ε
x1
= 360 x 10
-6
ε
y1
= 90 x 10
-6
γ
x1y1
= -110 x 10
-6
θ
ε
x1
ε
y1
γ
x1y1

17. 17
Measurement of Strain
• It is very difficult to measure normal and shear stresses in a body,
particularly stresses at a point.
• It is relatively easy to measure th e strains on the surface of a body
(normal strains, that is, not shear strains).
• From three independent measurements of normal strain at a point,
it is possible to find principa l strains and their directions.
• If the material obeys Hooke’s L aw, the principal strains can be
used to find the principal stresses.
• Strain measurement can be direct(using electrical-type gauges
based on resistive, capacitive, inductive, or photoelectric
principles) or indirect (using optical methods, such as photo-
elasticity, the Moiré technique, or holographic interferometry).

18. 18
Resistance Strain Gauges
• Based on the idea that the resist ance of a metal wire changes
when the wire is subjected to mechanical strain (Lord Kelvin,
1856). When a wire is stretched, a longerlength of smaller
sectioned conductor results.
• The earliest strain gauges were of the “unbonded” type and
used pillars, separated by the gauge length, with wires stretched
between them.
L
o
• Later gauges were “bonded”, with the resistance element applied
directly to the surface of the strained member.
wire grid
backing
backing
expanded
view
bonded to surface

19. 19
During the 1950s, foil-type gauges began to replace the wire-type.
The foil-type gauges typically consis t of a metal film element on a
thin epoxy support and are made using printed-circuit techniques.
Foil-type gauges can be made in a number of configurations
(examples from www.vishay.com):
single element
planar
three-element
rosette
(0°-45°- 90°)
solder tabs
for wires
alignment
marks
Performance of bonded metallic strain gauges depends on: grid material and configuration, backing material, bonding material and method, gauge protection, and associated electrical circuitry.
Gauge length is typically around 1 mm.

20. 20
It is possible to derive an equation relating strain
ε
and the change
in resistance of the gauge
Δ
R:
R
R
F
Δ
=
1
ε
F= gauge factor (related to Poisson’s ratio and resistivity)
R = resistance of the gauge
This is a resistance change of 0.0002%, meaning that something more
sensitive than an ohmmeter is required to measure the resistance change.
Some form of bridge arrangement (such as a Wheatstone bridge) is most
widely used.
R
1
tension
R
2
compression
(Perry & Lissner)
cantilever
A typical strain gauge might have F = 2.0 and R = 120
Ω
and be
used to measure microstrain(10
-6
).
Ω = = =Δ
−
00024 .0) 120 )(0.2)( 10(
6
RF R
ε

21. 21
Strain Rosettes and
Principal Strains and Stresses
A “0°-60°-120°” strain gauge rosette is bonded to the surface of a
thin steel plate. Under one loading condition, the strain measure-
ments are
ε
A
= 60
με
,
ε
B
= 135
με
,
ε
C
= 264
με
.Find the principal
strains, their orientations, and the principal stresses.
A
B
C
120
o
60
o
x
We can use more than one approach
to find the principal stresses:
transformation equations alone,
Mohr’s circle alone, or a combination.
(Based on Hibbeler, ex. 15.20 & 15.21)

22. 22
Transformation equations
ε
A
= 60
με
,
θ
A
= 0°
ε
B
= 135
με
,
θ
B
= 60°
ε
C
= 264
με
,
θ
C
= 120°
From the measured strains, find
ε
x
,
ε
y
, and
γ
xy
.
xy y x C
xy y x C
xy y x B
xy y x B
x A
xy y x A
γ ε ε ε
γ ε ε ε
γ ε ε ε
γ ε ε ε
ε ε
γ ε ε ε
433.0 75.0 25.0 264
120 cos 120 sin 120 sin 120 cos 264
433.0 75.0 25.0 135
60 cos 60 sin 60 sin 60 cos 135
60
0cos 0sin 0 sin 0 cos 602 2
2 2
2 2
− + = =
° ° +° +° = =
+ + = =
° ° +° +° = =
= =
° ° +° +° = =
3 equations, 3 unknowns
Solve to find
ε
x
= 60
με
,
ε
y
= 246
με
,
γ
xy
= -149
με

23. 23
Use
ε
x
,
ε
y
, and
γ
xy
in the equations for principal strains to
find
ε
1
= 272
με, θ
p1
= -70.6°,
ε
2
= 34
με, θ
p2
= 19.4°.
Alternatively, use
ε
x
,
ε
y
, and
γ
xy
to construct the Mohr’s circle
for (in-plane) strains and fi nd principal strains and angles.
ε
γ
/2
2
θ
p1
2
θ
p2
ε
2
ε
1
A
A:
ε
x
= 60
με
(
γ
xy
/2)= -74.5
με
D
D:
ε
y
= 246
με
(
γ
xy
/2)= +74.5
με
c
c = (60+246)/2 = 153
με
R
R= 119
με

24. 24
A
B
C
-70.6
o
19.4
o
x
ε
2
ε
1
) (
1
) (
1
2
2
x y y
y x x
E
E
νε ε
ν
σ
νε ε
ν
σ
+
−
=
+
−
=
To find the principal stresses, us e Hooke’s Law for plane stress
(
σ
z
= 0)
ε
x
=
ε
1
= 272 x 10
-6
ε
y
=
ε
2
= 34 x 10
-6
E= 210 GPa ν
= 0.3
So, the principal stresses are:
σ
x
=
σ
1
= 65 MPa
σ
y
=
σ
2
= 26 MPa

25. 25
Mohr’s Circle
ε
A
= 60
με
,
ε
B
= 135
με
,
ε
C
= 264
με
.
A
B
C
120
o
60
o
x
ε
γ
/2
ε
A
ε
B
ε
C
ε
A
= 60 = c + R cos 2
θ
c?
R?2
θ
3 equations, 3 unknowns
240
o
ε
C
= 264 = c + R cos 2(
θ
+120)
120
o
ε
B
= 135 = c + R cos 2(
θ
+60)
Solve the equations to get c= 153, R= 119, and 2θ= 141.3°
When you solve for 2θ, you may get –38.7°. But we have drawn the diagram
above such that 2θis positive, so you should take 2 θ= -38.7°+ 180°= 141.3°.
Next, draw the Mohr’s circle and find principal strains as before.
Finally, find principal stresses using Hooke’s Law.