Why we needed to relate internal resistive forces with area (stress)?

Question

When we load a body, say a prismatic bar, axially, internal resistive forces are developed within it to oppose the elongation or compression. What was the need of relating these internal resistive forces with area to define a quantity like stress. Wasn't these internal resistive forces enough?

Answer 1

A good reading about that is the: History of strength of materials by Timoshenko.

It describes the history behind Galileo Galilei and Hooke until the era of Cauchy, and the intermediate steps

Galileo (sometime between 1592 and 1610) figured out that for tension, when you double the cross-section the load to break also doubles. But run into problem with Bending (second moment of area wasn't "invented" by then.

The story fast forwards to Hook (1678) (there are a few mention for Euler and others). Hooke realised that every material essentailly behaves as a spring. Therefore the greater the displacement the greater the force in a component.

$$ F= kX$$

However, Hooke never really tried to decouple the Geometry from the material. And nowadays we know that the spring constant k can be derived (at least in an bar of isotropic material ) from:

$$K = \frac{E\cdot A}{L}$$

• K is the spring constant
• E is the Young's modulus (material)
• A is the cross-section (geometry)
• L is the Length. (geometry)

This discovery is really important because by decoupling the effects its possible to predict the behaviour more easily. If a different material is selected then you only need to change the modulus E. If you double the cross-section then only A needs to change.

Young "patented" the Youngs modulus (in 1807) (although Leonard Euler had conceived it already in 1727).

Finally, 200 years after Hooke, Cauchy in 1820s proposed the idea that you can divide the force by the cross-section to get a quantity representative of the material.

(I'll refresh my notes and come back to this when I get my hands on the book).

Answer 2

First courses on engineering statics tend to start from the forces where the structure is loaded and constrained, and at the interface between components, because in simple situations (for example statically determinate structures) the forces can be calculated independently of the internal stresses in the structure. The stresses can then by found (approximately) from the forces.

However in general the stress and strain distribution is are more important. As a simple example, consider a rectangular plate of material with tensile forces applied to each end. You can easily calculate the average tensile stress as "force/area".

However, consider what happens if there is a hole cut in the bar for some reason. Assume there are no forces applied around the circumference of the hole.

It should be obvious that the stress in the material near the hole will change in some way, and the change might be important. In fact, for a single small hole, the maximum stress is increased to 3 times the average stress. For a larger hole comparable with the width of the bar, or for several holes close together, the stress pattern is even more complicated.

Another example is a rectangular section cantilever beam. You can calculate the approximate stress distribution in the structure by finding the applied loads and moments and then, making some assumptions about the deformed shape of the beam.

For many purposes those assumptions are good enough, but in fact they are not even self-consistent. For example, suppose you know the material has non-zero Poisson's ratio (because you have measured it, not from some theoretical argument!). When the beam bends, you assume there is an axial tensile stress on one surface and a compressive stress on the opposite surface.

Therefore, because of Poisson's ratio, the width of the beam on the surface with tensile stress will decrease and the width on the opposite surface will increase. That means the beam section is no longer a rectangle, and in fact it is curved from one side to the other.

Google for "anticlastic curvature* for pictures - for example https://www.researchgate.net/figure/The-phenomenon-of-anticlastic-curvature-arises-from-Poisson-effects-When-a-beam-is-bent_fig4_11057987

Again because of Poisson's ratio, the volume of the material in tension is different from the volume in compression. Therefore the neutral axis of the beam is no longer at its geometrical center when the beam is bent.

Also, because of the shear stress through the depth of the beam, the assumption that a plane section remains plane is also false, and in fact the original plane section deforms into an S-shape viewed from the side.

None of this could be calculated just by considering the forces applied to the beam and then working out the "average" stress in the material.

Answer 3

In the mechanics of material, the most important strength parameters are measured and expressed in terms of "stress" rather than force.

For instance, the "yield" strength of a Grade 60 steel is 60 ksi, which is a well-known standard in the engineering community. However, how much force a piece of grade 60 steel can sustain at yield is unknown until the area is made known, that is $F_y = f_y*A$.

For your case, if the steel bar has a cross-sectional area of 1 in^2, and it is stretched by a tension force of 50 kips, will it yield if the bar is made of 1) grade 60 steel and 2) grade 40 steel? In addition, if the bar is 3' long, what is the elongation of each case?

The general equations for stress and deflection are $\sigma_a = F/A$, and $\delta = \dfrac{PL}{EA}$, respectively.

$1)$ For grade 60 bar, $f_y = 60 ksi > \sigma_a = 50 kips/1 in^2 = 50 ksi$, thus the stress is below the yield point, more load can be applied before the bar begins to yield.

For steel, the "Young's Modulus (E)" is 29,000 ksi, so the elongation of the bar:

$\delta = \dfrac{50 ksi*3 ft*12 in/ft}{29000 ksi*1 in^2} = 0.062"$

$2)$ For grade 40 bar, $f_y = 40 ksi < \sigma_a = 50 kips/1 in^2 = 50 ksi$, thus the bar has stretched beyond yield. For such condition, the general equation can't be used to calculate the elongation, since the Young's Modulus ($E = \dfrac{\sigma}{\epsilon}$) is valid in the elastic range only

Hope this explains why the area of an element is required in defining the strength of the element.