L4 Axial Forces

Added: 2026-05-31

[00:00:00]now let's talk about axial forces axial forces as you can assume from the name are forces acting along the axis of a member when we pull along the axis of a member this is called tension when we push along the axis of a member this is called compression basic review from structures 1.

[00:00:25]now let's add to that discussion by talking about how members fail when subjected to tension and compression forces a tension member will experience yielding or rupture depending on the material properties and any holes or deficiencies in the member a compression member on the other hand fails by crushing if it is a stout short member or buckling if it is a long slender member let's focus specifically on tension failures first when we pull on a member that member experiences both stress and strain stress is the amount of force that is applied over the cross-sectional area of the member strain is the amount that the member will elongate strain is often expressed as a percentage or in inches per inch we are taking a measure of the change in length divided by the original length now we can take these two variables and measure them as we pull on a member then we can plot the values on a graph and obtain a stress strain curve the stress strain curve helps us understand the material's behavior under loading and helps us to predict how certain structural members made from that material will perform so when we pull on a member and plot its stress versus strain we end up with a graph that looks like this of course as we apply more load stress increases as does strain initially this increase happens linearly both stress and strain increase at a constant rate this portion of the curve is called the elastic region and if we apply a load that generates a stress and strain that remains in this portion of the curve when we release the member the values will return to their original location on the graph in other words we apply the load stress increases and the member starts to elongate then if we release the member before it reaches this point here then the member will return to its original length however if we continue to add load to the member Beyond a certain point then the member will enter the plastic portion of the Curve and we call that point of no return the yield stress after passing the yield stress the member will start to experience permanent or plastic deformation and if we release the member it will no longer return to its original length so that would look like this we add load which increases stress and strain we push it beyond the yield stress and then if we were to release the force from the member it would return to this part of the graph here and now we have some permanent strain in the member some permanent amount of elongation you can observe this yourself if you take a paper clip and bend it with a small amount of force the paperclip returns to its original shape with a larger Force however it will deform permanently and it will no longer return to its original shape for the most part we will want to make sure that our structural members remain in this elastic portion of the Curve however we can and do design for some plastic deformation with certain materials like Steel as you can see with this curve there is still a fair amount of strength left in the material beyond the yield point before it reaches rupture there is also more strain that the material can undergo before rupture the more a material can stretch the more ductile it will be which as we mentioned previously is very useful for seismic design under seismic loading energy will be able to dissipate as parts of the building deform plastically we may also have materials that have very high strength but low ductility as seen here low ductility is also referred to as a brittle material and we've used that term a few times already high strength brittle materials can be dangerous because they give us very little warning before they rupture a ductal material will stretch before it breaks or ruptures and we will be able to observe that stretching a brittle material on the other hand hardly stretches at all before suddenly snapping so as I mentioned earlier for the most part in this course we will be looking at the elastic portion of this curve and one of the Key Properties we will need to know is the modulus of elasticity which is equal to the slope of the linear portion of our stress strain curve modulus of elasticity in other words is equal to stress divided by strain in the elastic region of the Curve and the way that we check that a member will not yield given a load as well as that material's yield stress is quite simple we take the yield stress multiply by the cross-sectional area and check if that force is greater than our applied load now moving on to compression failures most columns will fail and buckling before they fail due to crushing and so we will only focus on buckling in this course and here we have a lovely example of a column buckling taken from my own apartment being a structural engineer can make life interesting sometimes because you notice a lot more things anyway let's look at our formula for buckling also known as Euler's buckling formula the modulus of elasticity which we've just defined the geometry of the column specifically the moment of inertia or in simpler terms the resistance to bending the length of the member and its end conditions you can imagine with this column here if we had embedded this column into the foundation rather than resting it on top of the concrete then that column would be restrained against rotation at the base and therefore the column would not be able to buckle as easily in other words rather than having this condition here we might have this condition instead in this condition the base of the column is allowed to rotate and you can see that here that at the base there is some curvature with a fixed condition at the base this column would have to be perpendicular to the ground where it meets the foundation and so there would be less curvature or less of a chance for it to buckle now let's look at a full example using this formula let's say we have a wood column with dimensions of 12 inches and three inches when we get into talking about wood we won't be able to use whole Dimensions like this as dimensional Lumber does not actually come and hold dimensions but more on that later so you can see here that we have our modulus of elasticity e already defined and let's copy over our buckling formula so we have e and we have l which is the height here call that l and say we wanted to find the critical buckling load in both the X and the Y directions it's important to note here that when we're talking about buckling we're talking about the twisting that happens around these axes so buckling for the X direction will be buckling out of plane this way and buckling in the y direction will be buckling out of plane this way so with that in mind let's now calculate our I values or our moment of inertia so for buckling about the y-axis we will have our width b equal to 12 inches and our height H equal to three inches and so that will equal 27 inches to the fourth power next with our X Direction we will have three inches for our width and 12 inches for our height cubed and then divided by 12.

[00:10:45]and so our moment of inertia in the X Direction will be 432 inches to the fourth and you can see that the i in the X direction is much larger than the i in the y direction and so we would refer to that bending about the x-axis as the strong axis of the member next let's define our end conditions and let's say for Simplicity that both of these ends are pinned and so looking at our chart pinning both ends of the column will give us a k value of one and so we have k equals one now lastly you'll notice here that we have two beams framing into this column and so as this member tries to buckle in the y direction or the weak axis Direction it will be braced by these two beams and so rather than buckling like this as I drew earlier this column will not be able to buckle along its full height in the weak axis Direction rather if it were to buckle it would buckle only from here to the beam and again for Simplicity let's assume that this connection between the beam and the column is a pinned connection and so our length for the y direction buckling would be six feet assuming that these beams are at Mid height of the column and so our critical buckling in the y direction will be equal to pi squared times 1900 KSI times 27 inches to the fourth over six feet times 12 inches over feet squared and it's important that we convert to inches here as all of our other units are in inches and so we get a value of 97.7 kips now let's calculate the critical buckling in the X Direction so our critical buckling load in the X direction will be equal to pi squared times 1900 KSI times 432 inches to the fourth divided by the full height because we do not have bracing about the x-axis times 12 inches per foot squared and I've left out the K value from both of these formulas but that's because we all know that 1 times L is L so no need to write it out but if this K value were not equal to one don't forget to include it in the formula and so our critical buckling in the X direction is equal to about 390 kips and so our critical buckling load for this column will be the lesser of these two values which is our critical buckling in the y direction so our critical buckling is 97.7 kips now let's wrap up this discussion with one final are example this question States a rod of a i s i 1020 structural steel has a diameter of 3 inches and is subjected to a tension force of 75 kips and it has a diameter of three inches the modulus of elasticity for aisi 1020 Steel is 30 times 10 to the sixth PSI what is the strain well we know that this value here this modulus of elasticity is equal to stress over strain and so we can solve for strain pretty easily and get that strain is equal to stress over the modulus of elasticity we also know how to solve for stress given a force we simply take the force divided by the cross-sectional area and lastly that area is defined by this diameter value that we were given and so our area is equal to pi r squared note that we were given the diameter and so we will need to take pi times three inches divided by 2 squared and so we get 7.07 inches squared now we plug that into our formula for stress we have 75 000 Pounds divided by 7.07 inches squared and from that we get 10 600 pounds per square inch and now we plug this value into our strain equation working our way back up and we get 10 600 pounds per square inch divided by 30 times 10 to the sixth pounds per square inch and recall that strain is a dimensionless value and so we should have the same dimensions on top and bottom of this equation and indeed we do so they will cancel out and our strain is 0.000 I've run out of room here three five and so this value is equal to the amount that this member elongated the change in its length divided by its original length so strain is equal to change in length over the original length and since we've been working in inches here with these equations we can also Express this value in inches per inch and so our answer is a okay so that does it for our discussion of static equilibrium and axial forces next week we will continue our discussion of structural analysis with perpendicular forces namely bending and Shear forces so that's it I hope you have a great week good morning good afternoon and good night and I'll see you next time