The Strain Gauge Rosette
In the practical context, the most important tool in experimental stress analysis is the
electrical resistance strain gauge. This device measures accurately the surface strain in
the direction in which it is applied.
In many cases the directions in which the strains are to be measured are known and no
difficulty exists. However in other cases there may be no clearly defined direction and the
investigator may wish to determine the greatest strain at the point.
To allow this to be accomplished strain-gauge rosettes have been developed. These multielement gauges are arranged in a given geometry as shown in Figure 4.
Figure 4.
From any three linear strain measurements it is possible to determine the complete strain
system as follows:
Consider the case shown in Figure 5 where strains are measured in directions OA, OB
and OC.
Figure 5.
The procedure is as follows:
Applying equation (1) to directions A, B and C:
εa =
1
(ε x + ε y ) + 1 (ε x − ε y )cos 2θ a + 1 γ xy sin 2θ a
2
2
2
εb =
1
(ε x + ε y ) + 1 (ε x − ε y )cos 2θ b + 1 γ xy sin 2θ b
2
2
2
εc =
1
(ε x + ε y ) + 1 (ε x − ε y )cos 2θ c + 1 γ xy sin 2θ c
2
2
2
Then, since εa, εb, εc, θa, θb and θc are known, we have three simultaneous equations in εx,
εy and γxy which we solve.
We then find θ1 and θ2 from:
γ xy
tan 2θ =
εx −εy
And substitute back into equation (1) for ε1 and ε2.
Example 1
The following results are obtained from a 600 strain gauge rosette:
Strain in direction of strain gauge A = 750 microstrain;
Strain in direction of SG B, 600 to A = 350 microstrain;
Strain in direction of SG C, 1200 to A = 100 microstrain.
Determine the principal strains and their directions.
From equation (1):
εa =
Hence:
1
(ε x + ε y ) + 1 (ε x − ε y )cos 2θ a + 1 γ xy sin 2θ a
2
2
2
ε a = 750 =
1
(ε x + ε y ) + 1 (ε x − ε y )cos 2θ a + 0
2
2
Therefore:
ε x = 750 µε
(i)
And:
ε b = 350 =
1
(ε x + ε y ) + 1 (ε x − ε y )cos120 + 1 γ xy sin 120
2
2
2
350 = 0.25ε x + 0.75ε y + 0.433γ xy
(ii)
ε c = 100 =
1
(ε x + ε y ) + 1 (ε x − ε y )cos 240 + 1 γ xy sin 240
2
2
2
100 = 0.25ε x + 0.75ε y − 0.433γ xy
(iii)
Then (ii) – (iii) gives:
250 = 0.866γ xy
Hence:
γ xy = 288.7 µε
And εx and γxy in (ii) gives:
εy =
350 − 0.25 × 750 − 125
= 50 µε
0.75
Then if εx = 750, εy = 50 and γxy = 289:
tan 2θ =
289
750 − 50
Therefore:
2θ = 22.4o and 180o + 22.4o
θ = 11.2o and 101.2o
And from equation (1):
ε 11.2 =
1
(750 + 50 ) + 1 (750 − 50 ) cos 22.4 + 1 289 sin 22.4
2
2
2
ε 11.2 = 779 µε
(ε1)
ε 101.2 =
1
(750 + 50 ) + 1 (750 − 50 ) cos 202.4 + 1 289 sin 202.4
2
2
2
ε 101.2 = 21µε
(ε2)