# Drawing the shear and bending moment diagram of a 2D rigid frame [closed]

I am taking a civil engineering course that we are constantly asked to tackle questions that are not really covered in the course. I have been looking for suitable reference materials but with no success. Here is one of the questions that I am stuck at:

Find the shear and bending moment diagram of the following rigid frame:

I have tried working out the forces but they seem to be wrong after checking with a free program called "2D frame analysis."

How should I approach this problem?

• Welcome to Engineering.SE! We do not solve homework questions in the format you presented it. I assume you would like to know the solution. See this meta post to learn more. If you show a clear attempt and provide your calculations you are more likely to get an answer. Also your second question might be too broad as it is primaryly opinion based. See here. You might get some references in the comments though. Jul 30 '15 at 8:34
• I'm voting to close this question as off-topic because we have a policy of not accepting resource recommendation questions. Jul 30 '15 at 11:25
• Hi saldtch, welcome to the site. I've edited out the part of your question that, as Chris points out, is off-topic here. Please edit your question to add the details of your attempt to work out the forces, so that we can see where you went wrong (or if perhaps the free program was wrong!) - if you do this, I will open the question up for additional answers.
– Air
Jul 30 '15 at 15:43
• Thanks very much for the comment. I really should have posted my attempts. Really sorry about that! Aug 4 '15 at 9:29

This is an isostatic structure, which means it can be "trivially" solved (once one is more experienced) by hand.

The first thing to do is to calculate the reactions. In this case you can see that all forces are horizontal and only $$R_A$$ can withstand horizontal forces, so we already know that $$X_A$$ will be equal to the negative sum of all external forces (defining positive as to the right): $$X_A = -10 - 3\cdot4 = -22\text{kN}$$

To get the vertical reactions we need to calculate the moment around a given point, which must be equal to zero. I choose point $$A$$. $$M_A = -3\cdot4\cdot\frac{4}{2} - 10\cdot5 + Y_B\cdot4 = 0\\ \therefore Y_B = 18.5\text{kN}$$

Since the sum of vertical loads must be null, this means that $$Y_A = -18.5\text{kN}$$. Knowing all the reactions we can now calculate the internal forces.

Let's start at point C. Here we have a concentrated transverse force, which means we'll have a constant shear force of $$10\text{kN}$$ from C to D. At point C the bending moment will be null, while at point D it will be equal to the product of the force and its lever arm, so $$M_{D,C} = 10\cdot1 = -10\text{kNm}$$ ($$M_{D,C}$$ is the moment at D on the side of C. It is negative because I arbitrarily chose counter-clockwise rotations as positive.). Axial forces here will clearly be null.

Now let's go to point B. The reaction gives us a concentrated shear force of $$18.5\text{kN}$$ along B to D. Just as above, the bending moment at D will therefore be $$M_{D,B} = 18.5\cdot4 = +74\text{kNm}$$ (the force is trying to rotate point D counter-clockwise, therefore the moment is positive). Once at D, the shear force will become an axial force along A to D, which means this segment will experience tension of $$18.5\text{kN}$$. The same does not apply to the shear force generated at D because the support at B can't restrain that horizontal force.

Now let's go to point A. The negative vertical reaction gives us an axial tensile force which we saw before, so that tells us so far, so good. The negative horizontal force gives us a concentrated shear force which is diluted by the distributed loads along A to D. The shear force at D will be equal to the sum of the (negative) reaction with the distributed load, so $$-22 + 3\cdot4 = 10\text{kN}$$, which is equal to the concentrated force at C, which means there is no shear discontinuity at D between A and C. This is also a good sign since there is no reason for there to be a discontinuity since there are no horizontal loads or reactions coming from B. The moment at D will be equal to the partial moments due to the reaction and the distributed load, so $$M_{D,A} = -22\cdot4 + 3\cdot4\cdot\frac{4}{2} = -64\text{kNm}$$.

Note how the total moment around D is null: $$-10+74-64 = 0\text{kNm}$$. This must always be the case.

With this we have all the internal forces of the structure, as can be seen here (figures created with Ftool, a free 2D frame analysis tool. It is set to omit the bending moment sign):

• Thanks a lot for the reply! I got about 3/4 of my workings correct. That free problem seems gave me wrong answer or I just input the data with the wrong method. Since I really got not useful course material I had to do a lot of guess works. I was not very sure if the two forces would induce a moment at B. I will try to re-digest this again in a few days. Thanks very much again! Aug 4 '15 at 9:36