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Which is more accurate-true strain or engineering strain and why?
Wiki User
∙ 13y ago
Short answer:
strain engineering = change in L / original L
true strain = ln(1+strain engineering)
Engineering strain is the change in length divided by the original length, so that a 1 inch part strained 50% or .5 in/in would become 1.5 in or if strained -50% or -.5 in/in would become .5 inches. But these two strains are not the same amount of deformation since as a material is stretched further the change in length is distributed over a longer length for positive values and over a smaller length for larger values. Consider progressing from the now 1.5 in. (50%) strained part and continuing to 100% and the .5 in. (-50%) strained part and continuing to -100%. The next change in length is distributed over 1.5 in. and .5 inches respectively despite this the equation considers this change relative to the same original length of 1 inch. True strain is the change in length divided by the instantaneous length integrated from the original length to the instantaneous length. This resolves to the equation above.
Engineering Stress is more of an approximation. As stress levels increase, the actual cross sectional area of the object will change due to the force (think of a rubber band getting thinner as it gets stretched out).
Since stress is force divided by area the stress changes as a product of two variables. If you think of it that way, you are thinking of true stress.
Engineering stress holds the cross sectional area constant at its original value.
So if you look at a stress strain diagram, engineering stress levels off at the ultimate strength but true stress continues to climb because it is being divided by a smaller and smaller number as the object is stretched to the point of failure.
Wiki User
∙ 14y ago
Wiki User
∙ 15y agoStrain is the deflection divided by the original length. It is unitless, and usually reported as a percentage. As an example, lets say you have a 1.000 long specimen. Under tensile load, it measures 1.020 in the same units (inches, centimeters, etc...). The deflection is .020 in those units. The strain is .020/1.020 = .0196, which is unitless. This would normally be reported as a %, so the strain = 1.96%.
Wiki User
∙ 13y agotrue strain
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Without getting into all the math, the engineering strain utilizes the initial length of the specimen in the calculation, the true strain utilizes the instantaneous length of the specimen.Getting into the math:strain engineering = change in L / original Ltrue strain = ln(1+strain engineering)Engineering strain is the change in length divided by the original length, so that a 1 inch part strained 50% or .5 in/in would become 1.5 in or if strained -50% or -.5 in/in would become .5 inches. But these two strains are not the same amount of deformation since as a material is stretched further the change in length is distributed over a longer length for positive values and over a smaller length for larger values. Consider progressing from the now 1.5 in. (50%) strained part and continuing to 100% and the .5 in. (-50%) strained part and continuing to -100%. The next change in length is distributed over 1.5 in. and .5 inches respectively despite this the equation considers this change relative to the same original length of 1 inch. True strain is the change in length divided by the instantaneous length integrated from the original length to the instantaneous length. This resolves to the equation above.
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Without getting into all the math, the engineering strain utilizes the initial length of the specimen in the calculation, the true strain utilizes the instantaneous length of the specimen.Getting into the math:strain engineering = change in L / original Ltrue strain = ln(1+strain engineering)Engineering strain is the change in length divided by the original length, so that a 1 inch part strained 50% or .5 in/in would become 1.5 in or if strained -50% or -.5 in/in would become .5 inches. But these two strains are not the same amount of deformation since as a material is stretched further the change in length is distributed over a longer length for positive values and over a smaller length for larger values. Consider progressing from the now 1.5 in. (50%) strained part and continuing to 100% and the .5 in. (-50%) strained part and continuing to -100%. The next change in length is distributed over 1.5 in. and .5 inches respectively despite this the equation considers this change relative to the same original length of 1 inch. True strain is the change in length divided by the instantaneous length integrated from the original length to the instantaneous length. This resolves to the equation above.
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