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
When loaded with 3000 lbs tension the three strain readings were: εA = 310 µε εB = -20 µε εC = 26 µε
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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...