# Strain Gauge Angle Homework question.

1. May 29, 2012

### xXOfNiRXx

[a]1. The problem statement, all variables and given/known data[/b]

I thought it would be easier to post a picture since this way the drawing can be seen. I'm having a hard time deciding what angles to use with the relevant equations, see below.

2. Relevant equations
I know that:

Ex' = (Ex/2)*(1+cos(2*theta))+(Ey/2)*(1-cos(2*theta))+(gammaxy/2)*sin(2*theta))
Ey' = (Ex/2)*(1-cos(2*theta))+(Ey/2)*(1+cos(2*theta))-(gammaxy/2)*sin(2*theta))
Gammax'y'= -Ex*sin(2*theta)+Ey*Sin(-60)+gammaxy*cos(2*theta))

Then, back solve for Ex, Ey, and gamm. Then use: Sigmaxx = (E/(1-v2))*Exx+ (v*E*Eyy)/(1-v2).

From There, Sigmax = P/A

3. The attempt at a solution

I used -30 and 2theta = -60 for the angle in all three of the above equations. However, a friend used -30 for Ex, 60 for Ey, and 15 for gammaxy. What is the correct angle? I get 12.5KN as a final solution, while she gets 1.85KN.

I used a matrix:

Ex Ey Gammaxy = Ex' or Ey' or gammax'y' respectiveyl

.75 .25 -.433 = 270E-6
.25 .75 .433 = 10E-6
.866 -.866 .5 = 0

I find that: Ex = 250E-6 Ey=.75E-6 and gammaxy = -225E-6

I find that Sigma x = 50Mpa

and Finally P = 12.5 KN based on the above equations.

Did I make the correct assumption that I should use -30 as the angle for all three stress transformation equations? All help would be greatly appreciated. Thanks!

2. May 30, 2012

### nvn

xXOfNiRXx: You are correct; use theta = -30 deg for all of your relevant equations. Your relevant equations are correct. However, your solution and answer are currently wrong. Try again. Also, do not round your numbers so much. Hint 1: You seem to have said gammax'y' is zero, and gammaxy is nonzero; but shouldn't these two be the other way around?

(1) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 1.85 kN, not 1.85kN. See the international standard for writing units (ISO 31-0). Or see any credible text book.

(2) The unit symbol for kilonewton is kN, not KN. The unit symbol for megapascal is MPa, not Mpa. Always use correct capitalization of unit symbols.

(3) Numbers less than 1 must always have a zero before the decimal point. E.g., 0.25, not .25. See the above links.

Last edited: May 30, 2012
3. May 30, 2012

### xXOfNiRXx

Thank you for the feedback. I will correct my errors momentarily. Can you explain why gammax'y' is not zero and why gammaxy is zero?

4. May 30, 2012

### nvn

xXOfNiRXx: In your relevant equations in post 1, you currently have three equations and four unknowns. You need one more equation. Hint 2: What is gammaxy for principal axes?

Alternately, you could use hint 3, instead of hint 2. Hint 3: What is an expression for eps_y, in terms of eps_x, using Poisson's ratio, when the y faces are unconstrained and not loaded, where eps means epsilon?

Last edited: May 30, 2012
5. May 30, 2012

### xXOfNiRXx

Got it, thanks for your help. :)

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