0
$\begingroup$

I've often heard stress acts at a point and not on an entire cross-sectional area to which the load is applied normally.
Im a little confused here, Let us consider an area of cross-section A and let a load P be applied normal to A resulting in stress s.
Does this mean-

  1. s acts at a point on A and not on the entire A? If yes, why?
  2. Is the value of s the same at all points on A?
  3. Isn't A significant to s?
$\endgroup$
  • $\begingroup$ "I've often heard" -- from whom? Can you share a link, or a book reference? $\endgroup$ – TimWescott Oct 28 '19 at 17:25
  • $\begingroup$ @TimWescott I've heard and read from several sources but don't remember the names of all the sources. But I do remember reading from the book "Structures - Or why don't thing fall down" by J.E. Gordan where he stated that "stress always acts at a point in a material and is independent of the area on which the load is applied." I'll edit the question and post the extract from the book in the question after some time. $\endgroup$ – Somanna Oct 28 '19 at 18:45
0
$\begingroup$

In general, the stress can change from point to point on a surface.

If you have a flat surface with a uniform pressure load applied to it, the normal component of the stress must equal the pressure everywhere to maintain equilibrium, but there can also be shear stress components at the surface which are not constant over the whole surface.

For example, consider a cantilever beam with a uniform distributed load applied along its length. The shear stress at the surface (which is the same thing as the "axial stress" in the beam which you calculate using $My/I$) varies from zero at the free end to a maximum value at the fixed end. And when dealing with beams, you don't usually bother about the direct stress component caused by the uniform load at all, because it is small compared with the axial stress.

| improve this answer | |
$\endgroup$
0
$\begingroup$

Imagine you have a steel column as a solid square of 6-inch by 6-inch section carrying a load P sitting on a square base plate, 12-inch by 12-inch.

The normal stress on the base plate is not $ \ \sigma= \frac {P}{12^2} $

The stress due to the P on the baseplate will be a complex mixture of normal stress, shear stress and bending (moment) stress.

if we consider just the normal stress on the baseplate it follows a pattern of concentric roundish square contour lines which are degrading in intensity as they open farther. I attach a crude diagram here, approximately showing the spread of stress contours.

stress under the column

That is why we consider the stress locally just for a point. In this case, we want to make sure the maximum normal stress is below the allowed stress, not the average stress. The distribution of normal stress into the depth of the plate is a bit complicates but basically the stress opens into the depth following an exponential spiral profile.

In many of the structural members like beams, stress increases linearly vertically as the point is farther from neutral axis of the beam. There are however structural members much more sophisticated than a simple beam where the stress varies greatly on a different point of a section, eg, cranck shaft of a car engine. Some members have almost uniform stress load across their cross-section, eg, a truss tension member at a point not very near its joint.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.