L7 Structural efficiency

Ajouté : 2026-05-31

[00:00:00]last week we talked a lot about perpendicular forces forces that are applied along the length of a member a distance away from a support and we talked about the most basic types of structural elements used for resisting these perpendicular forces beams and girders we saw how applying a load a distance away from the support of a beam or girder created bending and Shear forces in the member we also saw how the bending forces translated into compression and tension stresses in the member we spent some time analyzing these stresses by defining this new geometric property called the section modulus and we have both an elastic section modulus and a plastic section modulus and we talked about the differences between those last week this section modulus allows us to convert a bending moment Force into maximum tension and compression stresses similar to how we take an applied axial load and divided by the cross-sectional area of the member to get us attention stress with all of that in mind we asked ourselves briefly how we could improve the efficiency of our beams and girders to resist these perpendicular forces and we found that by concentrating more of the area at the top and bottom of the member we could increase the efficiency that is we could achieve a greater capacity with less area than a conventional solid rectangular shape so here we achieved a slightly higher moment of inertia which is related to section modulus but we used less area now let's take this conversation around structural efficiency further let's remind ourselves that structural efficiency has a number of benefits to the client as well as to the environment all structural elements require raw material and energy to produce and so selection of a structural system plays a key role in the overall sustainability of the project so let's talk more about this idea of structural efficiency ultimately structural efficiency starts with simplifying the load path if we have a load somewhere up above ground level we want to give that load a short easy direct path through our structural elements down to the ground of course you can imagine how this goal very quickly comes into conflict with the architect's goals as the architect is trying to create rooms and Open Spaces the larger these spaces the more load we have to redirect around them and the less efficient the structural system becomes however this is not necessarily A Bad Thing a building cannot be a forest of columns after all as much as any structural engineer would love if that were the case so let's focus on these elements that span above the ground level as we've mentioned already with bending elements the more area we can concentrate at the top and bottom of the member the better so last week we saw how a wide flange shape is more efficient than a solid rectangle now you can see I have a comparison between a w18 by 35 and a W12 by 35.

[00:03:45]both have the same area and therefore they also have the same weight in pounds per linear foot so they have the same amount of material but the w18 which is just six inches deeper than the W12 has 30 percent more capacity than the W12 so when it comes to structural efficiency for elements resisting perpendicular forces deeper is better a lot better especially when you start to consider deflection now let's take this conversation further what if we have even longer spans say the architect wants to create a lecture hall or an Atrium space how can we improve the efficiency of these long-spanning members let's start by looking deeper into bending stresses in comparison to axial stresses so here we have a beam cut somewhere along its length with an internal moment and you can see I've drawn both the elastic and plastic stress profiles inside the member you can see from these profiles that there is quite a bit of area in that member that is not being fully utilized and we talked about that last week as well that was why we decided to shape our Beam with wide flanges at top and bottom so we could move more area into the regions with higher stress but let's look at the axial stress profile notice how in this case the entire cross-sectional area is utilized equally there is no region of zero stress so members that carry axial loading are inherently going to be more efficient than those carrying bending stresses so let's quickly compare a 20-foot w16 by 26 Beam with a perpendicular load applied at mid-span to a 20-foot w16 by 26 column with a tension load applied at the top of the member and let's try to calculate the maximum Force we would need to apply to make the member yield in each case so this is the same member the same length but with different loads applied and let's start with bending we learned last week that the maximum internal moment for a simply supported Beam with a point load at Mid span will occur at mid-span and it will be equal to P times L over four so that's our internal moment demand so let's start there p is what we are looking for L is 12 feet divided by 4. so this we can simplify to three feet times p we also know that stress is equal to the internal moment divided by the plastic section modulus and we know that yield stress for a typical wide flange section is always 50 Kips per square inch so now let's rearrange this formula to give us m is equal to f y times Z and now let's plug in our values three feet times p is equal to 50 Kips per square inch times 44.2 inches cubed and just to show you where I got that plastic section modulus it came from the aisc shapes database and so for a w16 by 26 you can see it has an area of 7.68 inches squared and if I continue scrolling to the right that has a plastic section modulus of 44.2 inches cubed so now we have everything defined except for the one variable that we are looking for let me rearrange this one more time we make sure our units are consistent and here we get 61.4 tips so if we apply a force greater than 61.4 Kips at Mid span of this beam we will create yielding in the member and that member would fail now let's compare that to this scenario in week two we learned that our stress is equal to our applied force over the cross-sectional area and so this calculation is much simpler here we rearrange and we get that our applied force equal to f y times the area we plug in our values and we get that the point load that we would need to apply in this case to get the member to yield is 384 kips so that's clearly a huge difference in this case we only need to apply a force greater than 61.4 to create yielding in the member in this case we need 384 Kips to create yielding so axial loaded members are going to be much more efficient than members designed for bending forces so with that in mind is there a way to configure a beam in such a way that when we apply a perpendicular load somewhere along its length this scenario here we actually generate axial stresses inside the member rather than bending stresses it may sound strange at first but that is exactly what trusses are designed to do trusses and joists are horizontal members that resist perpendicular loads like beams and girders but unlike beams and girders they do not resist these forces through bending rather they optimize the layout and orientation of the materials used to create Pathways for tension and compression forces to flow through the member so we're creating a more direct load path through the element and we are utilizing more of the material to resist the forces so trusses and joists are more efficient than beams and girders but can we be even more efficient than that well there is still one small problem with trusses and that is all the connections we need to design and fabricate between all these different elements what if we didn't need those connections between the different axial components of the Truss what if we designed a system that could transfer a force across a perpendicular distance without creating bending or tension stresses what if the system could carry forces across a perpendicular distance purely through compression well that system would be called an arch arches take perpendicularly applied loads and span them across a distance to a support using only compression so at any point along this Arch if it were a true Arch you could actually slice through the element and it would not change the behavior of the arch the axial force would still flow down to the support and we'll dig a little deeper into arches in a separate video later but can we be even more efficient than an arch can we create larger spans with even less material what if we designed a system that could transfer a force across a perpendicular distance without creating either bending or compression forces after all its compression forces that tend to be trickier than tension forces when we start to consider buckling so what if this system could carry forces across a perpendicular distance purely through tension and that would be called a catenary cable system cables if they are true catenary Cable Systems will take loads applied across a span and transfer them to a support purely through tension so let's review in terms of efficiency when we talk about long spans the order goes like this first we have themes and girders next we have trusses next we have arches and finally cables and this is an order of increasing efficiency in the next few videos we will be concentrating more on these three structural elements their behavior and the procedures for analyzing them