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In a tensile test the peak load was 69kN, but the point of failure load was 60kN. Why is it lower?

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closed as unclear what you're asking by grfrazee, hazzey, wwarriner, Mahendra Gunawardena, Algo Apr 20 '16 at 6:47

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That is due to necking.

When a bar is under tension, two opposing mechanisms take place:

  • The bar loses cross-sectional area since it attempts to retain its volume while being stretched (see Poisson's ratio).
  • Once the bar begins to suffer plastic deformations, it undergoes stress hardening and it's elastic modulus rises.

So long as the increase in elastic modulus is greater than the loss of area, the stress (or force) measured will increase. At a certain point, however, the material starts to lose area faster than it hardens, which means that the force required to continue deforming the element starts to diminish, until the element reaches its ultimate deformation capacity and snaps.

Now, if you'd been measuring the force and dividing by the element's original cross-sectional area to get stress results, then these would have the same profile as the force results: increasing up to the peak load and then dropping to point of failure. This is known as engineering stress.

If, however, you use special equipment to measure the actual cross-sectional area as it changes during the test, then you'll see that the stress actually increased during the entire test due to stress hardening. This is known as true stress.

Here's a graph showing the difference between engineering and true stress. The red line represents engineering stress and the blue, true stress.

enter image description here
Source: Wikipedia

As the name implies, engineers tend to use engineering stress for the simple reason that it is infinitely easier to measure and design with. And as the diagram above demonstrates, the difference between engineering and true stress in the elastic range where we spend 95% of our time is very small.

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