shear and moment diagrams with applied moments

#Shear and moment diagrams with applied moments free

(Truss members only have axial forces: compression or tension.) Trusses do NOT have shear force and bending moment diagrams.Pinned/roller ends do NOT have moment reactions, but they can have externally applied moment. BMD is continuous. AFDs and SFDs may not be continuous.If shear is zero, bending moment is constant (can also be zero).Shear = rate of change of moment (a.k.a derivative or slope of moment).But before we do that, we should understand the following basics: The best way to learn AFDs, SFDs, and BMDs is to practice drawing them with a few examples. As shown in the the figure below, this means that the compression side of the member represents positive bending moment. Keeping a consistent sign convention is extremely important! We are going to define the positive sign convention as: To be consistent with our convention, BMDs are drawn on the compression side of the member. They help determine the material, size, and type of a member given a set of loads it can support without structural failure.

#Shear and moment diagrams with applied moments free

Results obtained with Ftool, a free 2D frame analysis tool.Axial, shear, and bending moment diagrams (AFD, SFD, and BMD) show the internal forces and moments along a structural member. That's also standard convention: always draw the bending moment on the side which is under tension.

Note that though the bending moment plot is above the beam, it has a negative value. Both of these are negative in our convention, so we get negative bending moment throughout the structure as well. However, the force creates a clockwise rotation to the right of the structure, which is resisted by the counter-clockwise moment at the support to the left. The downward force will obviously put BC in tension, which is positive in our convention. Since that's the case throughout the structure (being pushed down by a force to the right and pushed up by the support to the left), our shear will be entirely positive. We can see that a downward force to the right or an upward force to the left are both considered positive in shear. This shows the direction of positive internal forces depending on whether you're looking at the left or right side of the section. Below we see the standard sign convention: Now, lastly, we need to talk about signs. Basically, as far as the bending moment is concerned, that vertical step at BC is entirely irrelevant: the bending moment is basically identical to what it'd've been without BC (i.e. So, the bending moment at D is zero, at C, 30 kNm, at B, also 30 kNm, and at A it's the same as the reaction, 70 kNm. And that also applied for the beam itself! The bending moment is always 10 times the horizontal distance from the desired point to D.

The bending moment is always 10 times the horizontal lever arm.

In that case, the horizontal lever arm from A to D would be six meters, so the support's bending reaction would be 60 kNm. Indeed, think about what the reaction moment would be if AB were only 3 m long. That's not a coincidence, it always holds true. Notice this is also what you get from simply calculating the bending moment caused by D on C (just as you did to calculate the moment reaction at A). Now, something to keep in mind when trying to figure this sort of thing out is the fundamental equation of Euler-Bernouilli Beam Theory: Thankfully you've already done that correctly, so we can move on from there. So, in such problems, the first thing to do is calculate the support reactions.