Question

For a 60 degree strain rosette shown in the figure, prove that the sum of the three measure- ments is independent of the orientation of the rosette, i.e., Ei + €2 + €3 = 3€ avg (1) where €1, €2, and Ez are the normal strains measured by the gages 1, 2, and 3, respectively. In addition, Eavg is the abcissa (x-coordinate) of the center of the corresponding Mohr's circle. 600 Figure 3: Problem 4

Answer 1

Evaluating $\displaystyle \epsilon$ along all three rosettes individually starting from first rosette using standard equation of Mohr's Circle

$\displaystyle \epsilon$1=$\frac{\varepsilon x+\varepsilon y}{2}$+$\frac{\varepsilon x-\varepsilon y}{2}\cos 2\Theta$+$\frac{\gamma xy}{2}\sin 2\Theta$ -Eq (1)

$\displaystyle \epsilon$2=$\frac{\varepsilon x+\varepsilon y}{2}$+
EL - Ey cos(20 + 120)
+$\frac{\gamma xy}{2}\sin (2\Theta+120)$

Now expanding $cos (2\Theta+120)$ and using cos and sin identities

sin (20 + 120)

We get;

$\displaystyle \epsilon$2=$\frac{\varepsilon x+\varepsilon y}{2}$+
EL - ey (cos120cos20 - sin 120 sin2e)
+$\frac{\gamma xy}{2}(sin2\Theta cos120+cos2\Theta sin120)$

Putting values of sin and cosine we get;

$\displaystyle \epsilon$2=$\frac{\varepsilon x+\varepsilon y}{2}$+$\frac{\varepsilon x-\varepsilon y}{2}(-\frac{1}{2}cos 2\Theta-\frac{\sqrt{3}}{2}sin2\Theta )$+$\frac{\gamma xy}{2}(sin2\Theta \frac{-1}{2}+cos20 \frac{\sqrt{3}}{2})$ - Eq (2)

And Now for $\displaystyle \epsilon$ 3 we get;

$\displaystyle \epsilon$3=$\frac{\varepsilon x+\varepsilon y}{2}$+
EL - Ey cos(20 +240)
+$\frac{\gamma xy}{2}\sin (2\Theta+240)$

Now expanding $cos (2\Theta+240)$ and $sin (2\Theta+240)$ using cos and sin identities

We get;

$\displaystyle \epsilon$3=$\frac{\varepsilon x+\varepsilon y}{2}$+$\frac{\varepsilon x-\varepsilon y}{2}(cos240 cos 2\Theta-sin240sin2\Theta )$+$\frac{\gamma xy}{2}(sin2\Theta cos240+cos2\Theta sin240)$

Putting values of sin and cosine we get;

$\displaystyle \epsilon$3=$\frac{\varepsilon x+\varepsilon y}{2}$+$\frac{\varepsilon x-\varepsilon y}{2}(-\frac{1}{2}cos 2\Theta+\frac{\sqrt{3}}{2}sin2\Theta )$+$\frac{\gamma xy}{2}(sin2\Theta \frac{-1}{2}-cos20 \frac{\sqrt{3}}{2})$ - Eq(3)

Now adding Eq(1), Eq(2), Eq (3)

We get:

$\displaystyle \epsilon$1+$\displaystyle \epsilon$2+$\displaystyle \epsilon$3=$3(\frac{\varepsilon x+\varepsilon y}{2})$+0+0

and as $\frac{\varepsilon x+\varepsilon y}{2}$ =$\displaystyle \epsilon$avg

We get;

$\displaystyle \epsilon$1+$\displaystyle \epsilon$2+$\displaystyle \epsilon$3=$\displaystyle \epsilon$avg