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GHOUSIA COLLEGE OF ENGINEERING
RAMANAGARAM-562159
EXPERIMENTAL STRESS ANALYSIS
[15ME832]
Dr. MOHAMMED IMRAN
ASST PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING

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Module-2
STRAIN ANALYSIS METHOD
Introduction to rosette
A strain gage rosette is, by definition, an arrangement of two or more closely
positioned gage grids, separately oriented to measure the normal strains along
different directions in the underlying surface of the test part.
Electrical-resistance strain gages are normally employed on the free surface of a
specimen to establish the stress at a particular point on this surface. In general it is
necessary to measure three strains at a point to completely define either the stress or
the strain field, In terms of principal strains it is necessary to measure ∈ ∈ and the
direction of ∈ relative to the x-axis as given by the principal angle ∅. Conversion of
the strains in to stresses requires, in addtion, knowledge of the elastic constants E and
 of the specimen material.
Where considering only one direction of stress (x-axis) and = = 0 will
be
We know Hook’s law =
Let as assume isotropic state of stress where = = = ; = 0 then the
magnitude of stress can be established from
= = = = (1)
Rosette: It is used to determine the strain and stress at an point of interest that is
where maximum stress of seen in the critical component to be determined.
Configurations of strain rosettes
There are two common configurations of strain gauge rosettes; rectangular and
delta. A rectangular strain gauge rosette consists of more than one strain gauges; A,
B, C etc. Strain gauges A and B are placed in between strain gauges at a 45o
angle on
an XY-axis, strain gauges A and c are placed in between strain gauges at a 90o
angle
on an XY-axis such positions of strain gauge rosettes are called as rectangular
rosettes.
A delta strain gauge rosette consists of more than one strain gauges; A, B, C
etc. Strain gauges A and B are placed in between strain gauges at a 30o
angle on an
XY-axis, strain gauges A and c are placed in between strain gauges at a 60o
angle on
an XY-axis such positions of strain gauge rosettes are called as delta rosettes.

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The different types or configurations of strain rosettes are
1. Two-Element Rosette
a. Two-Element Rectangular Rosette
b. Two-Element delta Rosette
2. Three-Element Rosette
a. Three-Element Rectangular Rosette
b. Three-Element delta Rosette
3. Four-Element Rosette
Two-Element Rosette
1. Two-Element Rectangular Rosette.
A rectangular strain gauge rosette consists of more
than one strain gauges; A, B etc. Strain gauges A
and B are placed in between strain gauges at a 45o
or 90o
angle on an XY-axis such positions of strain
gauge rosettes are called as two rectangular rosettes.
2. Two-Element delta Rosette
A delta strain gauge rosette consists of more than
one strain gauges; A, B. Strain gauges A and B
are placed in between strain gauges at a 30o
or
60o
angle on an XY-axis such positions of strain
gauge rosettes are called as two delta rosettes.
Three-Element Rosette
1. Three-Element Rectangular Rosette
A rectangular strain gauge rosette consists of more
than one strain gauges; A, B and C. Strain gauges A
and B are placed in between strain gauges at a 45o
angle on an XY-axis, strain gauges A and C are placed
in between strain gauges at a 90o
angle on an XY-axis
such positions of strain gauge rosettes are called as
three rectangular rosettes.

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2. Three-Element delta Rosette
A delta strain gauge rosette consists of more
than one strain gauges; A, B and C. Strain
gauges A and B are placed in between
strain gauges at a 30o
or 1200
angle on an
XY-axis, strain gauges A and C are placed in
between strain gauges at a 60o
or 1200
angle
on an XY-axis such positions of strain gauge
rosettes are called as three delta rosettes.
Four-Element Rosette
A delta strain gauge rosette consists of more
than one strain gauges; A, B, C and D. Strain
gauges placed in between strain gauges at a
A, B, c & D angle on an XY-axis, such
positions of strain gauge rosettes are called as
four delta rosettes.
TWO-ELEMENT RECTANGULAR ROSETTE
Consider a two-element rectangular rosette similar to those illustrated in fig:1 is
mounted on the specimen with its axes coincident with the principal directions. The
angle between the two strain gauge is 450
or 900
such type of positioned strain rosettes
called rectangular rosettes. The two principal strains, ∈ and ∈ obtained from the
gages can be employed to give the principal stresses and .
Fig 1:Two-element rectangular rosette.

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=
1 −
( + ); =
1 −
( + ) (2)
These relations given the complete state of stresses at a point only.
THREE-ELEMENT ROSETTE
Fig 2:Three gage elements placed at arbitrary angles relative to the x and y axes.
In the most general case, no knowledge of the stress field or its directions is available
before the experimental analysis is conducted. Three-element rosettes are required in
these instances to completely establish the stress field. To show that three strain
measurements are sufficient, consider three strain gages aligned, along axes A, B, and
C, as shown in Fig. (2).
From Equations of stress transformation where element is rotated on z-axis
therefore stresses along z direction is zero. Using fig 2: equation is evident that
= + +
= + +
= + + (3)
The Cartesian components of strain , can be determined from a
simultaneous solution of Eq. (3). The principal strains and the principal directions can
then be established by employing Eq. Principal stresses. The results are
Using , = ± + =
= + + − +

= +
Where is the angle between the principal axis
stresses can then be computed from the principal strains by utilizing Eqs. (2).
THREE-ELEMENT RECTANGULAR ROSETTE
In actual practice, three-
fixed at specified values) are employed to provide sufficient data to completely
define the stress field. These rosettes are defined by the fixed angles as the rectangular
rosette, the delta rosette, and the tee
The three-element rectangular rosette employs gages placed at the 0
positions, as indicated in Fig. 3.
Fig 3: Gage position in a three element rectangular rosette.
For this particular rosette it is clear from Eqs. (3) that
= ; =
ℎ
Thus by measuring the strains
strain , can be quickly and simply established through the use of
equation (a).Next, by utilizing Eqs. (4), the principal strains
beestablished as
= ( + ) +
6
− − +
2 =
is the angle between the principal axis ( ) and the x axis. The principal
stresses can then be computed from the principal strains by utilizing Eqs. (2).
ELEMENT RECTANGULAR ROSETTE
-element rosettes with fixed angles (that are,
fixed at specified values) are employed to provide sufficient data to completely
define the stress field. These rosettes are defined by the fixed angles as the rectangular
rosette, the delta rosette, and the tee-delta rosette.
element rectangular rosette employs gages placed at the 00
,45
positions, as indicated in Fig. 3.
Fig 3: Gage position in a three element rectangular rosette.
For this particular rosette it is clear from Eqs. (3) that
= + + ; =
= 2 − −
Thus by measuring the strains , , the Cartesian components of
can be quickly and simply established through the use of
equation (a).Next, by utilizing Eqs. (4), the principal strains ∈
) ( − ) + (2 − − )
(4)
) and the x axis. The principal
stresses can then be computed from the principal strains by utilizing Eqs. (2).
ngles (that are, , and
fixed at specified values) are employed to provide sufficient data to completely
define the stress field. These rosettes are defined by the fixed angles as the rectangular
,450
, and 90°
Fig 3: Gage position in a three element rectangular rosette.
; (a)
components of
can be quickly and simply established through the use of
∈ and ∈ can

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= ( + ) − ( − ) + (2 − − ) (b)
And the principal angle ∅ is given by
2 =
(2 − − )
( − )
( )
The solution of equation (C) gives two values for the angle∅, namely, ∅ which refers
to the angle between thex-axis and the axis or the maximum principal strain ∈ and
∅ which is the angle between the x axis and the axis of the minimum principal strain
. It is possible to show that the principal axes can be identified by applying the
following rules
0 < ∅ < 90° ℎ > ( + )
−90° < ∅ < 0 ℎ < ( + )
∅ = 0 ℎ > =
∅ = ±90° ℎ < = (d)
Finally, the principal stresses occurring in the component canbe established
byemploying (b) (c) together with (2) to obtain
= ( )
+ ( )
( − ) + (2 − − )
= ( )
− ( )
( − ) + (2 − − ) (e)
The use of Eqs. (a) To (e) permits a determination of the Cartesian components of
strain, the principal strains and their directions, and the principal stresses by a totally
analytical approach.
THREE ELEMENT DELTA ROSETTES:
The delta rosette employs three gages placed at the 0. 120°, and 240° positions, as
indicated in Fig.4.

Figure4. Gage positions in a three
For the angular layout of the delta rosette it is clear from Eqs. (3) That
=
= +
= +
Solving Eq.(a) for ,
=
Also from the Eq. (4) the principal strains
of , gives
= ( + +
= ( + + )
(C)
The principal angle can be determined from Eq (4) as
2 =
2 √⁄
2 −
The solution of Eq. (d) gives two values for the principal angle
the rectangular rosette.
Principal angles can be identifying by applying the following rules:
0 < ∅ < 90°
−90° < ∅ < 0
∅ = 0
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Figure4. Gage positions in a three-element delta rosette.
For the angular layout of the delta rosette it is clear from Eqs. (3) That
;
− √ ;
+ √ ;
in terms of , gives
; =
1
3
[2( + ) − ] ;
=
2√3
3
( − ); ( )
Also from the Eq. (4) the principal strains can be written in terms
) +
√
( − ) + ( − ) + ( −
) −
√
( − ) + ( − ) + ( −
can be determined from Eq (4) as
√3 ( − )
(2 + + )
=
√3( − )
[ − ( + )]
The solution of Eq. (d) gives two values for the principal angle , as was the case for
Principal angles can be identifying by applying the following rules:
ℎ >
ℎ <
ℎ = > =
element delta rosette.
(a)
can be written in terms
)
)
)]
( )
, as was the case for

∅ = ±90°
Finally, the principal stresses can determine from the principal strain by employing
Eq. (2) to obtain
=
( + +
( − )
=
( )
( )
−
By employing Eq. (e) to (f), it is possible to determine the Cartesian components of
strain, the principal strain and their directions, and the principal stress from the three
observation strain made with a delta rosette.
CORRECTIONS FOR TRANSVERSE STRAIN EFFECTS
It was noted that foil-type resistance strain gages exhibit a sensitivity S, to transverse
strains. Fig: 5. shows that in certain instances this transverse sensitivity can lead to
large errors, and it is important t
different procedures for correcting data have been developed.
D
= ϵ 1 +
The first procedure requires a priori knowledge of the ratio
correction factor is evident in Eq. (b)below. Where
ϵ =
Fig5: Error as a function of transverse sensitivity factor with the biaxial strain
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ℎ = < =
Finally, the principal stresses can determine from the principal strain by employing
)
)
+
√
( + )
( − ) + ( − ) +
√
( )
( − ) + ( − ) + ( −
By employing Eq. (e) to (f), it is possible to determine the Cartesian components of
strain, the principal strain and their directions, and the principal stress from the three
observation strain made with a delta rosette.
ORRECTIONS FOR TRANSVERSE STRAIN EFFECTS
type resistance strain gages exhibit a sensitivity S, to transverse
strains. Fig: 5. shows that in certain instances this transverse sensitivity can lead to
large errors, and it is important to correct the data to eliminate this effect. Two
different procedures for correcting data have been developed.
+
ϵ
ϵ
; =
F
(1 −  )
ϵ′ =
The first procedure requires a priori knowledge of the ratio
ϵ
ϵ
of the strain
correction factor is evident in Eq. (b)below. Where
= ϵ′
(  )
ϵ
ϵ
(a)
Fig5: Error as a function of transverse sensitivity factor with the biaxial strain
ratio as a parameter.
(e)
Finally, the principal stresses can determine from the principal strain by employing
) + ( − )
) (f)
By employing Eq. (e) to (f), it is possible to determine the Cartesian components of
strain, the principal strain and their directions, and the principal stress from the three
type resistance strain gages exhibit a sensitivity S, to transverse
strains. Fig: 5. shows that in certain instances this transverse sensitivity can lead to
o correct the data to eliminate this effect. Two
D
of the strain field. The
Fig5: Error as a function of transverse sensitivity factor with the biaxial strain

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The termϵ is the apparent strain, and the correction factor CF is given by
CF = ϵ′
(  )
ϵ
ϵ
(b)
It is possible to correct the strain gage for this transverse sensitivity by adjusting its
gage factor. The corrected gage factor F∗
which should be dialed into the measuring
instrument is
F∗
= F
ϵ
ϵ
(  )
(c)
Correction for the cross-sensitivity effect when the strain field is unknown is more
involved and requires the experimental determination of strain in both the x and y
directions. If ϵ′ and ϵ′ are the apparent strains recorded in the x and y directions,
respectively. Then from Eq. (a) it is evident that
ϵ′ = (  )
ϵ + ϵ ; ϵ′ = (  )
ϵ + ϵ (d)
Where the unprimed quantities ϵ andϵ are the true strains, Solving Eqs. (d) Forϵ
andϵ gives
ϵ =
(  )
ϵ′ − ϵ′ ; ϵ =
(  )
ϵ′ − ϵ′ (e)
Equation (e) gives the true strainsϵ andϵ in terms of the apparent strains ϵ′ and
ϵ′ Correction equations for transverse strains in two- and three-element rosette.
Correction equations for transverse sensitivity effect in three-element rosette.
Transverse sensitivity effect for three element of rectangular strain gage as follows.
=
(1 −  )
(1 − K )
( − )
=
(1 −  )
(1 − K )
( − )
=
(1 −  )
(1 − K )
− ( + − )
Where , and are indicated strains and ЄA ,ЄB , ЄC are corrected strains
Transverse sensitivity effect for three element of delta strain gage as follows.
=
(1 −  )
(1 − K )
1 + − ( + )

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=
(1 −  )
(1 − K )
1 + − ( + )
=
(1 −  )
(1 − K )
1 + − ( + )
Where , and are indicated strains and ЄA ,ЄB , ЄC are corrected strains
THE STRESS GAGE:
The transverse sensitivity which was shown in the previous section to result in errors
in strain measurements can be employed to produce a special-purpose transducer
known as a stress gage.
The stress gage looks very much like a strain gages except that its grid is designed to
give a select value of Kt, so it the output ∆R/R is proportional to the stress along the
axis of the gage. The stress gage serves a very useful purpose when a stress
determination in a particular direction is the ultimate objective of the analysis, for it
can be obtained with a single gage rather than a three-element rosette.
The principle upon which a stress gage is based is exhibited in the following
derivation. The output of a gage ∆R/R as expressed by Equation given below is
D
= (ϵ + ϵ ) (a)
The relationship between stress and strain for a plane state of stress is given by
Equations
=
1
( − ) =
1
( − ) ( )
Substituting Equation (a) and(b) yields
D
= ( − ) + ( − )
D
= (1 − ) + ( − ) (c)
Examination of Eq. (c) indicates that the output of the gage DR/R will be independent
of , if = . It can also be shown that the axial sensitivity Fa of a gage is related to
the alloy sensitivity Fa, by the expression
= ( )
(d)

Substituting Eq. (d) into Eq.
Since the factor ( n)
is a constant for a given gage alloy and specimen material, the
gage: output in terms of DR/R
In practice the stress gage is made with a V
in Fig. (6a) below Further analysis of the stress gage is necessary to understand its
operation in a strain field which is unknown and in which the strain gage is placed in
an arbitrary direction. Consider the placement of the gage, as shown in Fig. (6b) along
an arbitrary .x axis which is at some unknown angle
corresponding to .The grid elements an: at a known angle
Fig:(6a)
The strain along the top grid clement is given by a modified form of Eq. (3) as
ϵ = ( +
The strain along the lower
ϵ = ( +
SummingEqs. (f) and(g) and expanding the cosine terms yield
ϵ + ϵ =
Note from the Mohr’s strain circles given by
+
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into Eq. (c) and letting Kt=  leads to
= ( n)
D
(e)
is a constant for a given gage alloy and specimen material, the
R/R is linearly proportional to stress.
In practice the stress gage is made with a V-type grid configuration. As shown
in Fig. (6a) below Further analysis of the stress gage is necessary to understand its
operation in a strain field which is unknown and in which the strain gage is placed in
rbitrary direction. Consider the placement of the gage, as shown in Fig. (6b) along
an arbitrary .x axis which is at some unknown angle with the principal axis
The grid elements an: at a known angle relative to the x axis.
Fig(6b) the stress gage relative to the x
axis and the principal axis corresponding
to
The strain along the top grid clement is given by a modified form of Eq. (3) as
( + ) + ( − ) 2( − )
The strain along the lower grid element is
( + ) + ( − ) 2( + )
and expanding the cosine terms yield
= ( + ) + ( − ) 2 2
Note from the Mohr’s strain circles given by
= + (i)
is a constant for a given gage alloy and specimen material, the
type grid configuration. As shown
in Fig. (6a) below Further analysis of the stress gage is necessary to understand its
operation in a strain field which is unknown and in which the strain gage is placed in
rbitrary direction. Consider the placement of the gage, as shown in Fig. (6b) along
with the principal axis
relative to the x axis.
the stress gage relative to the x
axis and the principal axis corresponding
The strain along the top grid clement is given by a modified form of Eq. (3) as
(f)
(g)
(h)

−
Substituting Eq. (i) and (j) into (h) gives
ϵ + ϵ
ϵ
ϵ +
If the gage manufactured so that
And Eq. (k) becomes
ϵ + ϵ
Substituting (l) into Eq. (e) gives
Where ϵ + ϵ is the average strain indicated by the two elements of the gage
is equal to (DR /R)/ .
The gage reading will give
by ( n)
obtain . The stress gage will thus give
However, it does not give any data regarding
Moreover, may not be the most important stress since itmay differ
appreciablyfrom . If the directions of the principal stresses areknownthe
may be used more effectively by choosing the x axis tocoin side with the principal
axis, corresponding to so that
conventional single-element strain gage canbe employed as a stress g
Fig: 7 A single element strain gage employed as a stress gage when the principal
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= ( − ) 2 (j)
Substituting Eq. (i) and (j) into (h) gives
ϵ = + + − 2
+ ϵ = 2 −
+ ϵ = 2 − (k)
If the gage manufactured so that is equal to arctan √ , ℎ
= =
1
1 +
= 2 − (l)
Substituting (l) into Eq. (e) gives
= ( n)
ϵ + ϵ (m)
is the average strain indicated by the two elements of the gage
The gage reading will give ϵ + ϵ , and it is only necessary to multiply.this
The stress gage will thus give directly; with single gage.
However, it does not give any data regarding or theprincipal angle
may not be the most important stress since itmay differ
If the directions of the principal stresses areknownthe
may be used more effectively by choosing the x axis tocoin side with the principal
so that = .In fact, whenprincipal directions are known, a
element strain gage canbe employed as a stress gage.
Fig: 7 A single element strain gage employed as a stress gage when the principal
directions are known.
is the average strain indicated by the two elements of the gage
and it is only necessary to multiply.this
directly; with single gage.
or theprincipal angle
may not be the most important stress since itmay differ
If the directions of the principal stresses areknownthe stress gage
may be used more effectively by choosing the x axis tocoin side with the principal
.In fact, whenprincipal directions are known, a
age.
Fig: 7 A single element strain gage employed as a stress gage when the principal

This adaption is possible if the gage is located along a line which makes an angle
with respect to the principal axis, as shown fig 7:
symmetrical about the principal axis; hence it is clear that
and Eq. (m) reduces to
The value is recorded on the strain gage and converted to
multiplying by E / (1–v), this procedure reduces the number of gages necessary if
only the value of is to be determined. The saving of a gage is of particular
importance in dynamic work when the instrumentation required becomes Complex
and the number of available channels of recording equipment is limited.
PLANE-SHEAR GAGES OR TORQUE GAGE
Consider two strain gages
axis, as shown in Fig. 8. The strains along the gage axes are given by a modified form
of Eqs. (a), below as
Figure 8 Positions of gagesA and
Using equations
′ ′ = +
′ ′ = +
′ ′ = +
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This adaption is possible if the gage is located along a line which makes an angle
with respect to the principal axis, as shown fig 7:In this case thestrains will be
symmetrical about the principal axis; hence it is clear that
ϵ = ϵ = ϵ
= = ( n)
ϵ (n)
is recorded on the strain gage and converted to or
v), this procedure reduces the number of gages necessary if
is to be determined. The saving of a gage is of particular
importance in dynamic work when the instrumentation required becomes Complex
the number of available channels of recording equipment is limited.
SHEAR GAGES OR TORQUE GAGE
Consider two strain gages A and B oriented at angles with respect to the x
axis, as shown in Fig. 8. The strains along the gage axes are given by a modified form
Figure 8 Positions of gagesA and B for measuring
+
+
+ ( + )
This adaption is possible if the gage is located along a line which makes an angle
ase thestrains will be
or directly by
v), this procedure reduces the number of gages necessary if
is to be determined. The saving of a gage is of particular
importance in dynamic work when the instrumentation required becomes Complex
the number of available channels of recording equipment is limited.
with respect to the x
axis, as shown in Fig. 8. The strains along the gage axes are given by a modified form

15
′ ′ = ′ ′ = ′ ′ = 0 (a)
=
+
+
=
+
+ (b)
From Eq.(b) the shear strain is
=
( ) ( )
(c)
If gages A and B are oriented such that
= (d)
Then EQ. (c) reduce to
=
( )
(e)
Since the cosine is an even function, = − satisfies Eq. (d).Thus, the shearing
strain is proportional to the difference between normal strains experienced by
gage A and B when they are oriented with respect to the x axis as shown in fig. 10
The angle = − can he arbitrary: however, for the angle = 4 eq. (e) reduces
simply to
=
(f)
Equation (f) indicates that the shearing strain , can be measured with a two element
rectangular rosette by orienting the gages at 45° and - 45° with respect to the x axis is
and connecting one gage in arm R1and the other in to arm R4Wheatstone bridge. The
subtraction − will be performed automatically the bridge, and the output will
give , directly.
The stress intensity factor gage (KI):
Consider a two-dimensional body with a single-ended through crack as
shown in the fig: 9 below. The stability of this crack is determined by the opening-
mode stress intensity factor KI. If the specimen is fabricated from a brittle material,
the crack will be initiated when
KI > KIC (1)

It is possible to determine K
more strain gages near the crack tip. To show an effective approach to this
measurement, consider a series of representation, using three terms, the three
representation of the strain field is
∈ =
∈ =
=
Where A0, B0, andA1 are known coefficients which depend on the geometry of the
specimen and loading. We known that A
Eq. (2) could be determine the known coefficient
gages were placed at approximate positions in the near field region.
However, the numbers of gages are required for the determination of A
can be reduced to one considering a gage oriented at an angle α and positioned along
16
Fig: 9 single edge crack
It is possible to determine KI as a function of loading on a structure by placing one or
rain gages near the crack tip. To show an effective approach to this
measurement, consider a series of representation, using three terms, the three
representation of the strain field is
( − ) − ( + ) +
+ ( − ) + ( + )
( − ) − ( + ) −
+ ( − ) − ( + )
+
are known coefficients which depend on the geometry of the
specimen and loading. We known that A0 and KI are related by
=
√
(3)
Eq. (2) could be determine the known coefficient A0, B0, and A1 if three or more strain
gages were placed at approximate positions in the near field region.
However, the numbers of gages are required for the determination of A
can be reduced to one considering a gage oriented at an angle α and positioned along
as a function of loading on a structure by placing one or
rain gages near the crack tip. To show an effective approach to this
measurement, consider a series of representation, using three terms, the three-term
(2)
are known coefficients which depend on the geometry of the
if three or more strain
However, the numbers of gages are required for the determination of A0 or KI
can be reduced to one considering a gage oriented at an angle α and positioned along

the ′
axis as shown in the
∈ ′ ′is obtained from equation (1)and equation (n)
Fig: 10 Rotated co
∈ =
+ ( +
+
=
−
+
The coefficient of the B0 term is eliminated by selecting the angle α as
= −
Next, the coefficient of A1
tan
2
= −
By the proper placement of a single
satisfy Eq. (6) and (7), the strain
KI by
17
axis as shown in the below figure 10, for the rotated co-ordinates, the strain
is obtained from equation (1)and equation (n)
Fig: 10 Rotated co-ordinate system positioned at point P.
− +
( )
+ −
−
+
( )
term is eliminated by selecting the angle α as
=
−
+
(
1 vanishes if the angle is selected as
− cot 2 (7)
By the proper placement of a single strain gage with angles α and
satisfy Eq. (6) and (7), the strain∈ ′ ′, is related directly to the stress intensity factor
ordinates, the strain
ordinate system positioned at point P.
(4)
( )
determined to
, is related directly to the stress intensity factor

18
∈ =
√
− + ( )
The choice the angle α and depending only on Poisson’s ratio, as indicated in the
table below.
Table: Angle α and as a function of Poisson’s ratio
, ,
0.250 73.74 63.43
0.300 65.16 61.29
0.333 60.00 60.00
0.400 50.76 57.69
0.500 38.97 54.74
SOLVED PROBLEMS ON MODULE-2
PROBLEMS ON THREE ELEMENT STRAIN ROSETTE
1. A rectangular strain rosette is bonded at a critical point onto the surface of a
structural member. When the structural member is loaded, the strain gauge shows
the following readings.
Є0 = +850μm/m, Є45 = -50μm/m, Є90 = -850μm/m
The gauge factor and the cross sensitivity of the gauges are 2.80 and 0.06
respectively.
(i) Find the actual strains
(ii) Find the magnitudes and directions of corrected principal strains.
Poisson’s ratio of the material of the strain gauge is 0.285.
Given:∈ =∈ = 850 μm/m
∈ =∈ = −50 μm/m
∈ =∈ = −850 μm/m
F = 2.80
K = 0.06
ϑ = 0.285
Rectangular rosette
To Find:
1) Corrected strain:
∈ =? ; ∈ =? ; ∈ =?

19
2) Principal strain & direction:
∈ , =? ; 2∅ =?
Solution:
1) Corrected strain:
Formula:
ϵ =
(1 −  k )
(1 − K )
(ϵ̂ − k ϵ̂ )
ϵ =
(1 −  k )
(1 − K )
ϵ̂ − k (ϵ̂ + ϵ̂ − ϵ̂ )
ϵ =
(1 −  k )
(1 − K )
(ϵ̂ − k ϵ̂ )
Substituting given data in these equations we get:
=
(1 − (0.285)(0.06))
(1 − (0.06) )
(850 − (0.06 × −850)) = . /
=
(1 − (0.285)(0.06))
(1 − (0.06) )
(−50 − (0.06)(850 + (−850) − (−50)))
= − . /
=
(1 − (0.285)(0.06))
(1 − (0.06) )
(−850 − (0.06 × 850)) = − . /
: ϵ , ϵ &ϵ are the corrected strain
2) Principal strain & their direction:
Formula:
ϵ = (ϵ + ϵ ) + (ϵ − ϵ ) + (2ϵ − ϵ − ϵ )
ϵ = (ϵ + ϵ ) − (ϵ − ϵ ) + (2ϵ − ϵ − ϵ )
And the principal angle ∅ is given by
tan2ϕ =
(2ϵ − ϵ − ϵ )
(ϵ − ϵ )
Substituting corrected values of strain in these equations we get:
=
(888.79 + (−888.79) ) + (888.79 − (−888.79)) + (2(−52.28) − 888.79 − (−888.79))
= . /

20
=
(888.79 + (−888.79) ) − (888.79 − (−888.79)) + (2(−52.28) − 888.79 − (−888.79))
= − . /
Direction:
tan2ϕ =
2(−52.28) − 888.79 − (−888.79)
888.79 − (−888.79)
2ϕ = tan (−0.05882) = −3.348
= − .
2. A three element rectangular rosette is mounted on a steel component with E =
204 GPa and υ = 0.3. The manufacturers gauge factor F and the cross-sensitivity
Kt of this type of gauge is known to be 2.8 and 0.06 respectively. The readings
corresponding to the three gauges as indicated on a strain meter with the gauge
factor control set at 2.8 are
A = +700μ strain, b = -50μ strain, C = -700μ strain
(a) Find the actual strains ЄA,ЄB , ЄC.
(b) Find the magnitude of corrected principal stresses and their direction.
(c) What is the error if indicated strains AB, Care used to calculate the
principal stresses
Rectangular rosette
Given:∈ = 700 μm/m
∈ = −50 μm/m
∈ = −700 μm/m
E = 204 GPa = 2.04 × 10 N/mm
F = 2.80
K = 0.06
ϑ = 0.3
ϑ = 0.26
To Find:
1) Corrected strain:

21
∈ =? ; ∈ =? ; ∈ =?
2) Principal stress & direction by using corrected strain:
σ , =? ; 2∅ =?
3) Principal stress by using given strain & also % error between given and actual
stresses:
σ , =?
Solution:
1) Corrected strain:
Formula:
ϵ =
(1 −  k )
(1 − K )
(ϵ̂ − k ϵ̂ )
ϵ =
(1 −  k )
(1 − K )
ϵ̂ − k (ϵ̂ + ϵ̂ − ϵ̂ )
ϵ =
(1 −  k )
(1 − K )
(ϵ̂ − k ϵ̂ )
Substituting given data in these equations we get:
=
(1 − (0.26)(0.06))
(1 − (0.06) )
(700 − (0.06 × −700)) = . /
=
(1 − (0.26)(0.06))
(1 − (0.06) )
(−50 − (0.06)(700 + (−700) − (−50)))
= − . /
=
(1 − (0.26)(0.06))
(1 − (0.06) )
(−700 − (0.06 × 700)) = − . /
: ϵ , ϵ &ϵ are the corrected strain
2) Principal stress & their direction by using corrected strain:
Formula:
= ( )
+ ( )
( − ) + (2 − − )
=
+
2(1 − )
−
1
2(1 + )
( − ) + (2 − − )
And the principal angle ∅ is given by
tan2ϕ =
(2ϵ − ϵ − ϵ )
(ϵ − ϵ )
Substituting corrected values of strain in these equations we get:

22
= (2.04 × 10 )
( . ( . ))
( . )
+ ( . )
733.063 − (−733.063) + 2(−52.316) − 733.063 − (−733.063)
= . × /
= (2.04 × 10 )
(733. ( . ))
( . )
− ( . )
733.063 − (−733.063) + 2(−52.316) − 733.063 − (−733.063)
= − . × /
Direction:
tan2ϕ =
2(−52.316) − 733.063 − (−733.063)
733.063 − (−733.063)
2ϕ = −4.08
= − .
3) Principal stress & their direction by using given strain:
= (2.04 × 10 )
( ( ))
( . )
+ ( . )
700 − (−700) + 2(−50) − 700 − (−700)
= . × /
= (2.04 × 10 )
( ( ))
( . )
− ( . )
700 − (−700) + 2(−50) − 700 − (−700)
= − . × /
To find % error between the corrected and given stress
% error =
Corrected stress − Given stress value
Corrected stress
× 100
% error =
( . × ) − ( . × )
( . × )
× 100
% error = 0.0453 × 100
% error = 4.53 %
Note: 4.53 % error between the corrected and given stress

23
3. A delta strain rosette bonded onto the surface of a structural member, made of
aluminum, yields the following strain when the structure is loaded,
Є0 = +500μm/m, Є120 = -250μm/m and Є240 = 250μm/m.
Given that Kt = - 0.07 & υ0 = 0.285. Determine the magnitudes and directions of
principal strain at the point where the strain rosette is bonded, also determine the
principal stresses if young’s modulus for aluminum is 80 GPa and Poisson’s ratio
is 0.3.
Given: ∈ =∈ = 500 μm/m
∈ =∈ = −250 μm/m
∈ =∈ = 250 μm/m
F = 2.80
K = −0.07
ϑ = 0.285
E = 80GPa = 0.8 × 10 N/mm
ϑ = 0.3
Delta rosette
To Find:
1) Corrected strain:
∈ =? ; ∈ =? ; ∈ =?
2) Principal strain & direction:
∈ , =? ; 2∅ =?
3) Principal stress:
σ , =?
Solution:
1) Corrected strain:
Formula:
=
(1 −  )
(1 − K )
1 + − ( + )

24
=
(1 −  )
(1 − K )
1 + − ( + )
=
(1 −  )
(1 − K )
1 + − ( + )
Substituting given data in these equations we get:
=
(1 − (0.285)(−0.07))
(1 − (−0.07) )
1 +
(−0.07)
3
(500) −
2
3
(−0.07)(−250 + 250)
= . /
=
(1 − (0.285)(−0.07))
(1 − (−0.07) )
1 +
(−0.07)
3
(−250) −
2
3
(−0.07)(500 + 250)
= − . /
=
(1 − (0.285)(−0.07))
(1 − (−0.07) )
1 +
(−0.07)
3
(250) −
2
3
(−0.07)(500 + (−250)
= . /
2) Principal strain & direction:
Formula:
= ( + + ) +
√
( − ) + ( − ) + ( − )
= ( + + ) −
√
( − ) + ( − ) + ( − )
The principal angle can be given by
2 =
√3( − )
[ − ( + )]
Substituting corrected values of strain in these equations we get
=
1
3
(500.52 + (−214.372) + 262.187) +
√2
3
(500.52 − (−214.372)) + ((−214.372) − 262.187) + (262.187 − 500.52)
= . /

25
= (500.52 + (−214.372) + 262.187) −
√
(500.52 − (−214.372)) + ((−214.372) − 262.187) + (262.187 − 500.52)
= − . /
Direction:
2 =
√3(500.52 − (−214.372))
[ (500.52) − ((−214.372) + 262.187)]
= 40.89
= .
3) Principal stress:
Formula:
=
( )
( )
+
√
( )
( − ) + ( − ) + ( − )
=
( + + )
( − )
−
√
( + )
( − ) + ( − ) + ( − )
Substituting corrected values of strain in these equations we get:
= (0.8 × 10 )
(500.52 + (−214.372) + 262.187)
3(1 − 0.3)
+
√2
3(1 + 0.3)
(500.52 − (−214.372)) + ((−214.372) − 262.187) + (262.187 − 500.52)
= . × /
= (0.8 × 10 )
(500.52 + (−214.372) + 262.187)
3(1 − 0.3)
−
√2
3(1 + 0.3)
(500.52 − (−214.372)) + ((−214.372) − 262.187) + (262.187 − 500.52)
= − . × /
PROBLEMS ON TWO ELEMENT STRAIN ROSETTE
4. A two element rectangular rosette was used to determine the two principal stresses at a
point . if = 860 / = −390 / , Find and and take young’s
modulus E = 207GPa and Poisson’s ratio = 0.3
Given: = 860 /
= −390 /
E = 207GPa =2.07 × 10
= 0.3
To find: Principal stresses
=?and =?
Formula:

26
=
1 −
( + ) =
1 −
( + )
Substituting values of strain in these equations we get
=
2.07 × 10
1 − (0.3)
[860 + 0.3(−390)] = . ×
=
2.07 × 10
1 − (0.3)
[−390 + 0.3(860)] = . ×
5. The following apparent strain data were obtained with two element rectangular
rosettes.
Rosettes number ′
( ) ′ ( )
1.
2.
600
-200
300
700
Determine the true strain and if = 0.01. In each case, determine the
error which would have occurred if the cross sensitivity of the gage had been
neglected.
1. Rosettes number one
Given data: ′
= 600
′ = 300
= 0.01
Assume = 0.285
Formula:
ϵ =
(1 −  )
(1 − K )
ϵ′ − ϵ′ ; ϵ =
(1 −  )
(1 − K )
ϵ′ − ϵ′
Substituting given values of strain in these equations, we gettrue strain
ϵ =
1 − (0.285 × 0.01)
(1 − (0.01) )
[600 − (0.01 × 300)] = 595.35
ϵ =
1 − (0.285 × 0.01)
(1 − (0.01) )
[300 − (0.01 × 600)] = 293.17
To find % error between the corrected or true and given strain

27
% error =
Corrected strain value − Given strain value
Corrected strain value
× 100
% error =
(595.35) − ( )
(595.35)
× 100
% error = 0.00781 × 100
% error = 0.781 %
Note: 0.781% error between the corrected and given stress
2. Rosettes number two
Given data: ′
= −200
′ = 700
= 0.01
Assume = 0.285
Formula:
ϵ =
(1 −  )
(1 − K )
ϵ′ − ϵ′ ; ϵ =
(1 −  )
(1 − K )
ϵ′ − ϵ′
Substituting given values of strain in these equations, we gettrue strain
ϵ =
1 − (0.285 × 0.01)
(1 − (0.01) )
[−200 − (0.01 × 700)] = −206.42
ϵ =
1 − (0.285 × 0.01)
(1 − (0.01) )
[700 − (0.01 × −200)] = 696.04
To find % error between the corrected or true and given strain
% error =
Corrected strain value − Given strain value
Corrected strain value
× 100
% error =
(696.04 ) − ( )
(696.04 )
× 100
% error = 0.00563 × 100
% error = 0.563 %
Note: 0.563% error between the corrected and given stress

28
ASSIGNMENT PROBLEMS On Module-1-Part-A
1. Define a Strain rosette and mention the different types of strain rosette
configurations
2. Explain the construction of the three elements Delta rosette and derive the
expressions for the principal stresses and their orientations in terms of strain
measurement readings.
3. Explain the construction of the three elements rectangular rosette and derive
the expressions for the principal stresses and their orientations in terms of
strain measurement readings.
4. A rectangular strain gauge rosette is bonded at a critical point onto the surface of a
structural member. When the structural member is loaded, the strain gauges show
the following reading:
ε0 = 850 µm/m, ε45 = -50 µm/m, ε90 = -850 µm/m
The gauge factor and cross sensitivity of the gauges are 2.80 and 0.06
respectively. Find:
 Actual strains
 Magnitude and directions of principal strains.
 The error if indicated strains ε0,ε45, ε90 are used to calculate the principal
stresses.
Given E = 200GPa and Poisson’s ratio of the material of the strain gauge is 0.285.
5. The observations made with a delta rosette mounted on a steel specimen are
єA=400µm/m; єB=-200µm/m; єC=200µm/m. Determine the principal strains &
principal stresses & the principal angles ф1& ф2
6. The following observations were made with a delta rosette mounted on a steel

29
specimen εA = 460 µm/m ; εB = -200 µm/m ; εC = 200 µm/m Determine the
principal strain, the principal stresses and their orientations. Take µ = 0.3, E =
200×10
3
N/m
2
7. The following readings of strain were obtained on a three-element rectangular
strain rosette mounted on a Aluminum for which E=70GPa , ν =0.3 , εa= +285 µ
strains εB= +65 µ strains εC= 102 µ strains
Determine:
 The Principal stresses and its direction
 The Principal strains and its direction
 The maximum shear stress
8. Three strain gauges are applied to an area; at a point in such a manner that gauge
“B” makes a +ve 30
0
with the gauge “A” and gauge “C” makes an angle of 45
0
with gauge “B”. The strains obtained are as follows.
εA= -600 µm/m, εB= -400 µm/m, εC= 400 µm/m
Take E= 2X 10
5
N/mm
2
& Poisson’s ratio µ= 0.3. Calculate principal stresses,
strains and their directions.
9. A rectangular rosette mounted on the surface of a structural member indicates the
following reading, when the member is stressed ε0 = +500 strains ε45 =
+50 strains ε90 = -500 strains. Modulus of elasticity (E)=200×10
9
N/m
2
,
Poisson’s ratio (µ)= 0.30. Gauge factor and cross sensitivity of the strain gauge
are 2.80 and 0.06 respectively.
Determine:
 Actual strains along 0°, 45°, 90° directions.
 Principal strains and maximum shear strain.
 Principal stresses and maximum shear stress.
 Directions of principal stress.
10.A rectangular strain rosette is bonded at a critical point onto the surface of a
structural member. When the structural member is loaded, the strain gauge shows
the following readings.
Є0 = +850μm/m, Є45 = -50μm/m, Є90 = -850μm/m
The gauge factor and the cross sensitivity of the gauges are 2.80 and 0.06
respectively.

30
(iii) Find the actual strains
(iv) Find the magnitudes and directions of corrected principal strains.
Poisson’s ratio of the material of the strain gauge is 0.285.
11.A three element rectangular strain rosette bonded onto a machine component as
shown in figure. Yield strains indicated as shown below. Determine the magnitude
and direction of principal strains.
ЄA = +800μm/m, ЄB = -80μm/m, ЄC = -1000μm/m.
K = 0.06 and υ0 = 0.28.
12.A delta strain rosette bonded onto the surface of a structural member, made of
aluminum, yields the following strain when the structure is loaded,
Є0 = +500μm/m, Є120 = -250μm/m and Є240 = 250μm/m.
Given that Kt = - 0.07 & υ0 = 0.285. Determine the magnitudes and directions
of principal strain at the point where the strain rosette is bonded, also
determine the principal stresses if young’s modulus for aluminum is 80 GPa
and Poisson’s ratio is 0.3.
13.A three element delta rosette is bonded at a point onto the surface of a machine
element to determine the magnitudes and directions of strains at the point. The
strains indicated by a strain indicator are as follows.
Є0 = +600μm/m, Є120 = +300μm/m and Є240 = -600μm/m
Given that the gauge factor of the strain gauges is 2, and the Poisson’s ratio of
the material of the strain gauge is 0.28 and the cross sensitivity of the strain
gauge is 0.05. Determine the magnitudes and directions of the principal strains
at the point on the surface of the machine element.

31
14.A three element rectangular strain gauge rosette is bonded on the surface of
machine component as shown in figure. Yield strain as indicated below when the
machine component is under load
ЄA = 500μm/m, ЄB = -250μm/m
ЄC = 250μm/m E = 2.1 x 105
N/mm2
υ = 0.28. Determine the magnitude and directions of the principal stresses at
the point ‘O’ on the machine component. Assume the manufacturers gauge
factor and cross sensitivity as 2.8 and 0.06 respectively.
15.A three element delta rosette is bonded onto the surface of a machine element
made of aluminum for strain measurement. Strain gauge A is along X-axis and
strain gauge B and C are oriented along directions at angles of 1200
and 2400
from
X axis measured in anticlockwise direction. Strains measured are as follows
ЄA = 750μm/m, ЄB = -250μm/m, ЄC = +300μm/m
Other data supplied is Kt = - 0.07
υ0 = 0.30
υal = 0.33
Eal = 72 GPa
Find the magnitudes of the principal strains and the principal stresses and
orientations of the principal planes.
16.A three element rectangular rosette is mounted on a steel component such that the
gauges are separated by 450
with E = 200 GPa and υ = 0.3. The manufacturers
gauge factor F of this type of gauge is known to be 2.8. The reading
corresponding to the three gauges as indicated on a strain meter with gauge factor
control set at 2.8 are
ЄA = 1000μ strain, ЄB = -100μ strain, ЄC = -1000μ strain

32
Find the magnitude of principal stresses and their directions.
17. A three element delta rosette bonded onto a machine element yields strains as
shown below.
ЄA = 600μm/m, Єb = -300μm/m, ЄC = +300μm/m
Kt = -0.07 , υ 0 = 0.30, υ al = 0.33, Eal = 71.3 GPa
Find magnitudes and direction of principal strains and stresses.
18. A three element rectangular rosette is mounted on a steel component with E =
204 GPa and υ = 0.3. The manufacturers gauge factor F and the cross-sensitivity
Kt of this type of gauge is known to be 2.8 and 0.06 respectively. The readings
corresponding to the three gauges as indicated on a strain meter with the gauge
factor control set at 2.8 are
A = +700μ strain, b = -50μ strain, C = -700μ strain
(a) Find the actual strains ЄA ,ЄB , ЄC.
(b) Find the magnitude of corrected principal stresses and their direction.
(c) What is the error if indicated strains AB, Care used to calculate the
principal stresses
19. A three element rectangular rosette is fixed to the fuselage of a jet airliner near a
window-opening. At altitudes above 3000m the airliner is pressurized to simulate

33
atmospheric conditions prevailing at 3000m altitude. The altitude at which the
airliner cruises is 12,000m. The strain indicator readings with the airliner on the
ground and cruising at 12,000m altitude are as follows
Gauge 00
Gauge 450
Gauge 900
Airliner on the ground 13 μm/m 60 μm/m 35
μm/m
At 12,000m altitude; 500 μm/m 360 μm/m -120μm/m
Determine the magnitude and direction of the principal stresses if the modulus of the
elasticity of the material, E = 70 GPa and the Poisson’s ratio υ = 0.33.
20. A three element delta rosette serves to find the state of stress at a point O on the
surface of a stressed – aluminum component. The measured strains referred to a
given direction Ox are:
Direction Strain
00
500 μm/m
1200
-250μm/m
2400
+250μm/m
kt= - 0.07, υ0 = 0.285, υal= 0.33, Eal = 70 GPa
Find the actual principal strains and stresses and their directions with respect to Ox.
21. A three element rectangular strain rosette is fixed on the inside surface of a steel
pressure vessel near an inspection hole. The strain indicator readings before and
after pressurizing the vessel to a pressure of 6 MPa were as follows:
Gauge 00
Gauge 450
Gauge 900
Before pressurizing 15 μm/m 50 μm/m 35
μm/m
After pressurizing -285 μm/m +200 μm/m +430
μm/m

34
Gauge factor F = 2, kt= - 0.03, υ0 = 0.205, υsteel= 0.3; Esteel= 200 Gpa
Determine the magnitude and directions of the principal stresses.
22.A three-element rectangular rosette was used to measure the strains at a point in a steel
component. The observed strains are
ЄA = 850μm/m, Єb = -1200μm/m, ЄC = 1000μm/m
Neglecting the transverse sensitivity effects, determine the principal stresses and their
directions.
23. The following observations are made with a three-element rectangular rosette mounted
on an aluminum component.
A = 900μm/m, B = 310μm/m, C = -200μm/m
Determine the principal strains, principal stresses and principal stress directions.
K= - 0.04, υ0 = 0.285, υal= 0.33
24.A two element rectangular rosette was used to determine the two principal stresses at a
point shown in fig. below. if = 860 / = −390 / , Find and
and take young’s modulus E = 207GPa and Poisson’s ratio = 0.3
25.The following apparent strain data were obtained with two element rectangular
rosettes.
Rosettes number ( ) ′ ( )
3.
4.
5.
6.
600
-200
1,200
600
300
700
400
-300
Determine the true strain and if = 0.01. In each case, determine the
error which would have occurred if the cross sensitivity of the gage had been
neglected.