In: Mechanical Engineering
Tensile testing
can someone tell me how to calcualte the true strain, engineering strain, true stress and engineering stress of the following data o got from tensile testing machine. I assumed area to be constant during the test
i need to draw the curves of the true and engineeing in excel please
TESTNUM | POINTNUM | TIME | POSIT | FORCE | EXT | Area |
5368 | 1 | 0.156 | -5E-05 | 347.6111 | -0.00138 | 0.0628 |
5368 | 2 | 0.58 | -5E-05 | 348.3641 | 0 | 0.0628 |
5368 | 3 | 0.777 | 0.0007 | 363.0201 | 0.021787 | 0.0628 |
5368 | 4 | 0.875 | 0.001375 | 398.4846 | 0.041394 | 0.0628 |
5368 | 5 | 0.975 | 0.002025 | 438.6508 | 0.060276 | 0.0628 |
5368 | 6 | 1.086 | 0.002675 | 477.715 | 0.079158 | 0.0628 |
5368 | 7 | 1.175 | 0.00335 | 515.8242 | 0.098766 | 0.0628 |
5368 | 8 | 1.276 | 0.004 | 552.9417 | 0.117647 | 0.0628 |
5368 | 9 | 1.374 | 0.00465 | 590.1509 | 0.136529 | 0.0628 |
5368 | 10 | 1.483 | 0.0053 | 625.4869 | 0.155411 | 0.0628 |
5368 | 11 | 1.579 | 0.00595 | 659.7577 | 0.174292 | 0.0628 |
5368 | 12 | 1.678 | 0.006575 | 693.257 | 0.192448 | 0.0628 |
Engineering stress: The formula for finding engineering stress = P/A; whrere P is the reading from load dial and A is the initial area. For eg, at time 0.777, the engineering stress = 363.0201/0,0628 = 5780.57
True stress: The formula for finding engineering stress = P/A; whrere P is the reading from load dial and A is the area.at that particular time instant. Since the area is same at all times, Engineering stress = True stress
True strain: The formula for true stain = ln(L2/L1); where L2 is the length at a particular instant of time and L1 length is at the previous instant of time
Engineerig Strain: Engineering Strain is calculated as dL/L1 where L1: initial length of specimen. And final length=L1+dL