Continuing on my discussion of steels from
this post, I first want to introduce a few basic physics/engineering terms.
Force and Acceleration:
The concept of a force is something that I think that everyone is familiar with, but not everyone is familiar with the formal definition. A force (F) (Note 1), from a physics perspective, is defined as a mass (m) times an acceleration (A). That is, F = m * A. Acceleration is the change in velocity (V) over time (t). This is written as A = dV/dt, which can be read as acceleration equals the time derivative of velocity. If acceleration is constant, we can simplify this to:
A = (V1 - V0) / (T1 - T0) where 1 is the final velocity/time, and 0 is the initial velocity/time. For example, if we have a car that is merging onto a highway and takes 10 seconds to go from the 70 kph (~43.5 mph) to 100 kph (~62 mph), then we can determine that the (constant or average) acceleration was A = (70 - 45) kph / 10s = 25 kph / 10 s = 2.5 kph/s.
Now, let's convert the units to get something more sensible:
A = 2.5 * 1000 m / s * 3600 s = ~0.694 m / s^2 = ~0.7 meters per second squared. This is roughly 0.07 G's of acceleration, where G = 9.81 m/s^2.
Note 1 - The System International (SI) units for force is the Newton, defined as 1 N = 1 kg * m / s^2, stated as one newton equals one kilogram meter per second squared.
A 2023 Honda Civic has a curb weight of approximately 1875 kg (~3000 lbs). If we want to look at the maximum force of this car hitting a concrete barrier, we can use the above equations to come up with a value. Let's look at two cases, in the first the car crumples and comes to a standstill over 1 second. In the second case, the car is more rigid and comes to a stop over 1/10th of a second. The civic is traveling at 40 kph (~25 mph) in each case.
Fcrumple = m * A = m * (V1 - V0) / (T1 - T0) = 1875 kg * (0 - 40) kph / (1 - 0) s, where 40 kph = 11.1 m/s, so
Fcrumple = 1875 kg * (0 - 11.1 m/s) / (1 - 0) s = 1875 kg * (-11.1 m/s) / (1s) = -11.1 * 1875 kg / s^2 = -20812.5 kg / s^2 = 20812.5 N
The negative sign indicates that the force is against the direction of travel, which makes sense. For this case, we can easily find the average or constant acceleration by re-writing F = m * A as A = m / F, so A = (-20812.5 kg * m / s^2) / 1875 kg = -11.1 m / s^2 = -1.13 G's
Again, the negative sign makes sense as this is a deceleration.
Now, for the rigid car case, in which the car comes to a complete stop in 0.1 seconds, we use the above equations substituting t1 = 0.1 seconds, and we get:
F = -208125 N
A = -111.1 m / s^2 = -11.3 G's
This is why crumple zones are so important for low and moderate speed collisions. The crumpling action increases the total time that the car spends decelerating (slowing down), which greatly reduces the forces and deceleration. 11G's is sufficient to cause trained jet fighter pilots to blackout.
Stresses:
In continuum mechanics, a sub-branch of engineering and physics, stresses in solids are broken into two related kinds, Axial Stresses and Shear Stresses. Axial stresses are those that happen in the direction of concern, and shear stresses occur perpendicular to the direction of concern, as shown below:
It is possible to have pure axial stresses, pure shear stresses, or a combination of the two. In fact, the stresses are related, and you can do some matrix math to find both the line (direction) of maximum force in which shear stresses will be zero, or the perpendicular in which shear stresses are maximized. A visual technique for doing so is called the Mohr's Circle, shown below:
Sigma (σ, alt-229) is the symbol used for axial stresses, while Tau (τ, alt-231) is used for shear stresses. For this discussion, we will looking at axial stresses only (Note 2). In pure axial stress the equation is:
σ = F / A where F is the force and A is the cross sectional area perpendicular to the direction of the force. The SI units for stress is the Pascal (Pa), where 1 Pa = 1 N / m^2. Typically in metals, we are looking at stresses in the megapascal range, or 10^6 Pa, abbreviated as MPa.
To test materials for stress limits, we use Tensile Test Coupons, shown below:
Jaws are clamped or fixtures threaded onto the thicker sections, and the specimens are pulled. As axial stress is the force over the cross sectional area, the stress in the thicker section will be lower than in the thinner section because the force is constant through the piece. No shear stresses occur in the thinner section, and these are sized to ensure that this is section is where any elongation will occur. More on this testing later.
Strain:
Strain is the deformation of a material that occurs due to a stress. This is shown as Epsilon (ε, alt-238) and is defined as the final length at a given stress divided by the initial length. That is:
ε = (L1 - L0) / L1 where L0 is the initial length and L1 is the final length. Note that this is a unitless value, as if we measure a part before and after, our units would be cm / cm (in / in) and they cancel out. Strain is often listed as a percentage (%).
Stress-Strain Relationship
A plot of stress and strain for a carbon or low alloy steel will look something like this:
Points "B" and "C" in the above diagram represent two of the critical values we test for in metals and use in engineering design. Metals, when a force is applied, will undergo a fully recoverable elongation at stresses below the yield strength of the material. This is point "B". If we were to pull a tensile test specimen to the strain at point "B", then removed any applied stress, the metal would shrink back to its original size. This is called elastic deformation, as like elastic bands in your under clothes, it returns to the original shape and size.
If we continue to pull the piece piece, we will enter the plastic region, called such because a portion of the deflection is no longer recoverable. A metal stretched above the yield will have a permanent change in the dimensions. As we continue to pull on the metal, eventually we will hit a peak stress value we called the Ultimate Tensile Stress, because if the stress were to remain at this level, the part would continue to stretch to failure without requiring any additional load be applied (Note 3)
Now, if we consider the physics of this testing, you may see a flaw in how the data is being recorded above. When we pull on a part and stretch it, the volume of the part must remain the same. This means that the cross sectional area will decrease as we pull on the part. Poisson's Ratio is is the ratio of transverse strain to axial strain that occurs during deformation, and is around 0.28 for most steels. We can use Poisson's Ratio for a material and correct for the change in area that will occur, which leaves us a True Stress-Strain Curve, that looks like this:
Basically, on the true stress-strain curve, we have one straight line in the elastic, and a 2nd straight line in the plastic zone. There will always be a bit of a curvature at the transition due to the non-uniformity of the chemistry in the metal I discussed in my previous post.
In the elastic region, you will note that the image defines E = ▲σ/▲ε, where delta (▲, alt-30) means a differential. E is Young's Modulus, which is relatively invariant in steels, ranging from 190 to 215 GPa (GPa = 10^9 Pa). This means that to strain any steel to 1% will take approximately the same amount of stress, as long as we are below yield.
In my next post, I will show some graphs of carbon steel yield strength, ultimately tensile strength, and elongation to failure plotted against carbon content, and talk a bit about how the steels behave differently based on their carbon content.
Note 2 - Pure Shear Stress
An example of an application with pure shear stress is a bolt or shaft with two u-bolts, one mounted to a bracket and on side and one to a bracket on the opposite side. If you pull on the brackets, you will create a shear stress in the shaft.
In fact, this exact thing occurred at the
Hyatt Regency Hotel in 1981. The hotel had 3 suspended causeways, and the original drawing plans called for through-length threaded rods to go from the roof straight down to the lowest causeway. The construction company revised this to use separate rods with a 4" spacing between them. This caused sufficient additional shear stresses in the box beams that they failed during a very busy conference. 114 people were killed.
Note 3 - Strain Hardening
If we stress a material above the yield strength, we enter the plastic region. However, Young's modulus still applies, and the plastic region is actually made up of both the same elastic modulus we measured earlier, as well has the non-recoverable plastic deformation. One way to increase the strength of a steel is to deform it in the plastic region, then remove all of the applied stresses. An example of this is cold working plate to reduce the thickness. What you do in essence is increase the yield strength of the material, but at a tradeoff in ductility. Excessive cold working can cause steels to become very brittle.