I know what you are thinking: the Integral – I hate calculus, and rightfully so. The shear force for calculations of the applied 6 kip load would be 4 kips that the left most part of the beam must support. We use a point load to simplify the mathematics. When a load is applied, it is usually over a small span of the beam as a distributed load or 2 point loads and not a single point load. In either case, neither situation gives us the 6 kip load. If we zoom into the load enough (figure below), we will find out that the load is either distributed over a short length or applied at 2 (or more) smaller point loads. Think about it, when the load is applied it can’t actually be applied at a single point. In the example above, what shear load do I use in my calculation, is it the 6 kip load or the 4 kip? Decisions, decisions. What is the Shear Force?Įarly in my career, I was confused about what my shear load was from a point load. Here we end up with an rectangle with area of 20 kip-in (4 kip x 5 in) and a rectangle of -20 kip-in (-2 kip x 10 in). Right now, we could do a quick check to see if our moments will equal 0 at each end by making sure that the area above and below the curve is equal. (Note: this check does not exist for all beam types.) The reaction at each end of a simply supported beam must equal 0. At reaction R2, the shear force is increased to 0 kip. This line will remain at -2 kip for the rest of the beam length. This is shown as a vertical line with length of 6 kip. This will lower the 4 kip value of R 1 by 6 kip resulting in -2 kip. Since there are no other forces acting in this span, it is a horizontal line.Ī 5 inches, the 6 kip point is applied in the opposite direction of R1. Working from left to right, I have placed the vertical shear load R 1 from 0 in to 5 in. I have dropped a vertical line at the point load so they line up. Next we will begin to draw the shear diagram directly below the problem. I then summed the vertical forces to find R 2. For a case as simple as this I did a sum of the moments at R 2 to find R 1. Simply supported beam with point load of 6 kip (6,000 lb) It is perfectly permissible to work from right to left as well. Common practice is to work from left to right and I have shown that in my diagram with the x dimension pointing right. With a simply supported beam, it really doesn’t matter which way we go. We will then need to decide which way we will run our coordinate system for calculating the shear and moment. This will provide a built in check more on that later. This means that our moment diagram must be equal to 0 at each end. ![]() In this example, we have a simply supported beam where vertical forces are supported at the ends, but moment cannot be supported. The first step in creating a shear diagram is assessing the problem. Let’s get started on creating the shear diagram. This additional step could result in adding moment when you should have subtracted and thus getting the wrong answer. The major downside is that when integrating, you will need to invert when integrating the moment. On the other hand, the alternative convention will show the moments below the line, as it would naturally deflect. This is great, but it leaves you with thought that the deflection of your beam is upward when it is not. On the one hand, the standard convention allows you to integrate the shear diagram easily to determine the moment. I have used the standard convention my entire career, but I see why the alternative has its benefits as well. I’ll be honest, I’m torn on which convention to use.
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