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I have an exercise where I should calculate and draw a bending moment diagram for a frame structure. The frame structure is similar to the one below, image taken from http://structx.com:

enter image description here

I would like to know, why the bending moment diagrams of the vertical beams are formulated as they are? Is there any analogy I could use from the basic simply or fixed supported beam diagrams?

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The graph below shows a fixed-end frame subjected to a concentrated load in the mid-span of the horizontal member and the resulting deflected shape.

enter image description here

Let's draw tangent lines on the deflected shapes at joints B, we note that both the curved segments of BF & BE are deflected away from the respective tangent lines with the angle of rotation $\theta = 0$, which is the same as a cantilever beam, so we identify these surfaces are subjected to tension. The same phenomenon is noted at the support joint A (with respect to the vertical column), and at the load point.

Now we have identified all surfaces subjected to tension with the realization that the tension is caused/countered by a moment, therefore we can place the moment diagram accordingly - draw the moment diagram on the tension side of a member, which is the typical convention. (See graph below)

On the graph above, the points E, F, G, and H are inflection points, at which, M = 0, and the directional sign changes on both sides of the inflection point. (At the support points of the columns, there is a pair of inward lateral forces, as shown on the first picture of your graph, that restricts the outward tendency of the legs. The lateral force produces a moment in the direction counter the moment at the support, and the internal moments of the column eventually cancel out at the inflection joint.)

enter image description here

Note, both the frame height to width ratio (h/L) and the ratio of the moment of inertia of the column vs the beam affects the moment intensities at the joints, nevertheless, for a framing column with consistent geometries throughout the height, the lower-end (support) moment of the column is always one-half of the moment at the upper-end beam-column joint.

I guess you are taking the class "Structural Analysis I", and prematurely wondering into the topics of the class of the next level. As it is essential to learn the classic methods before attempting to predict/estimate the results using simplification, I won't go beyond what I've offered thus far, so as not to confuse you or conflict with your future education.

Information (for this case only):

enter image description here

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  • $\begingroup$ Many thanks for this explanation. I would like to know, how can I calculate the bending moments in this case to the bending moment diagram? I'd like to learn to do it by hand. $\endgroup$
    – Kolumbo
    Sep 20 at 16:19
  • $\begingroup$ It is the job of your school, to which I am not qualified for a position in it :) The best you can do now is having an understanding of the diagram, and get your hands on an engineering table, such as the one added to the end of my answer. $\endgroup$
    – r13
    Sep 21 at 0:08
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Let's start from the horizontal beam on top. if the joint at the ends of this beam where a pin joint you would get a maximum moment at the center $$M_{c}= \frac{PL}{4} $$

But because they are fixed after the beam deflects under the load the corner joint rotates a little and eventually stops at equilibrium. Both beam and the column bend a bit with the joint rotation. At this position, the moments on the beam end and column top are equal and the moment at the center of the beam is reduced.

Now Half of this moment will be transferred to the base plate (carryover factor), due to the elastic properties of the column.

The moment diagram you get is the summation of all these moment transfers and deformations.

There are many ways to calculate how the moment gets redistributed, but the Hardy Cross method is the easiest for hand calculation of the small frames. [Wiki sourceframe internal forces

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