Mohr's Circle Solution for the Strain Gauge Rosette
Quick and Dirty Mohr’s Circle Solution for the Strain Gauge Rosette
A 3 gauge rosette is attached to a simple tension bar. The three gauges of the rosette are at 45 degrees in relation to each other but the rosette is not aligned with the strap. The strap is 1.00 inches wide and 0.25 inches thick and is loaded with 3000 lbs tension with the force aligned with the long axis of the bar. The material is steel with: E = 29E6 psi and ν = 0.3
Theory:
The theory is that the stress in the bar is uniaxial with the principal stresses being equal to P/A and zero. The strains are biaxial with the maximum being P/AE and the minimum being – νP/AE. The first principal stress and strain will be aligned with the force and the long axis of the bar.
σ1 = P/A = 3000/(1.00x0.25) = 12,000 psi σ2 = 0 psi ε1 = P/AE = σ1/E = 12,000/29E6 = 0.000414 in/in = 414 µε ε2 = – ν P/AE = – ν σ1/E = – ν ε1 = -0.3x414 = 124 µε
Test Data:
When loaded with 3000 lbs tension the three strain readings were: εA = 310 µε εB = -20 µε εC = 26 µε
10/27/2008
Page 1 of 4
Solution:
Gauges A and C are perpendicular and since angles are doubled on Mohr’s circle, they will plot 180 degrees apart on the circle and define a diameter. Since A and C form a diameter of the circle, the center of the circle will be half way between them. Center at (310 + 26)/2 = 168 µε Find the offset from the center to gauges A and B 310 - 168 = 142 µε -20 - 168 = -188 µε These can be used to find the radius Radius = [ 1422 + (-188)2 ]1/2 = 235 µε Now we can find the principal strains ε1 = center + radius = 168 + 235 = 404 µε ε2 = center – radius = 168 – 235 = -67 µε Using the center and the radius, draw Mohr’s circle and draw three lines to represent the gauge readings. Consider a diameter of the circle connecting A and C. This diameter will have a perpendicular extending to B. There is only one way that this can be arraigned such that the counter clockwise sense, ABC, of the rosette is preserved. In our case the
A 3 gauge rosette is attached to a simple tension bar. The three gauges of the rosette are at 45 degrees in relation to each other but the rosette is not aligned with the strap. The strap is 1.00 inches wide and 0.25 inches thick and is loaded with 3000 lbs tension with the force aligned with the long axis of the bar. The material is steel with: E = 29E6 psi and ν = 0.3
Theory:
The theory is that the stress in the bar is uniaxial with the principal stresses being equal to P/A and zero. The strains are biaxial with the maximum being P/AE and the minimum being – νP/AE. The first principal stress and strain will be aligned with the force and the long axis of the bar.
σ1 = P/A = 3000/(1.00x0.25) = 12,000 psi σ2 = 0 psi ε1 = P/AE = σ1/E = 12,000/29E6 = 0.000414 in/in = 414 µε ε2 = – ν P/AE = – ν σ1/E = – ν ε1 = -0.3x414 = 124 µε
Test Data:
When loaded with 3000 lbs tension the three strain readings were: εA = 310 µε εB = -20 µε εC = 26 µε
10/27/2008
Page 1 of 4
Solution:
Gauges A and C are perpendicular and since angles are doubled on Mohr’s circle, they will plot 180 degrees apart on the circle and define a diameter. Since A and C form a diameter of the circle, the center of the circle will be half way between them. Center at (310 + 26)/2 = 168 µε Find the offset from the center to gauges A and B 310 - 168 = 142 µε -20 - 168 = -188 µε These can be used to find the radius Radius = [ 1422 + (-188)2 ]1/2 = 235 µε Now we can find the principal strains ε1 = center + radius = 168 + 235 = 404 µε ε2 = center – radius = 168 – 235 = -67 µε Using the center and the radius, draw Mohr’s circle and draw three lines to represent the gauge readings. Consider a diameter of the circle connecting A and C. This diameter will have a perpendicular extending to B. There is only one way that this can be arraigned such that the counter clockwise sense, ABC, of the rosette is preserved. In our case the