A sheet for example has 3 degrees of freedom. These are Y, X and the pairs or acting moments. An isostatic beam (iso = equal) has 3 degrees of freedom and in turn, the links remove 3 degrees of freedom. An example of a simply supported beam, that is isostatic, is with a double bond (restricts 2 degrees of freedom, X, Y) and a simple link (restricts 1 degree of freedom, Y) Another case of an isostatic beam would be a cantilever recessed beam. Embedding removes three degrees of freedom, X, Y and par. Therefore it is isostatic. Now, a beam with an articulated support and at the same time, an embedment exceeds the 3 degrees of freedom of that beam. For example, on one end a recess and on the other, a simple link. The embedment removes 3 GL and the single bond removes 1 GL. Therefore, the beam has 3 GL but has restricted 4. That is why, this beam is HYPERstatic (hyper = large, much) Hyperstatic beams are a bit more complicated to calculate, you need to use the 3-moment theorem plus the 3 general equations of statics. I give you as an example some beams that I bothered to draw, so that you realize The first beam is simply supported with a left overhang. The second is simply supported the third one is embedded in point A the fourth is supported by B and embedded in A, is HYPERTATIC and must be calculated with the fourth equation of the statics, of the three moments, since when taking the moments at any point you will be left with two unknowns. The fifth beam is supported by A, B and C, is HYPERTATIC and must be calculated with the three moment theorem. equation of the three moments: M1 .L1 + 2. M2 . (L1 + L2) + M3 .L3 = -6 {(Area1.a1) / l1} - -6 {(area2.a2) / L2} M1 = moment at point 1 L1 = distance between point 1 and link 2 Area 1 = area of the bending moment diagram between 1 and 2 a1 = distance between the center of gravity of area 1 and point 1. It is drawn by integration or by Varignon's theorem If you have any questions, ask me if you want to upload photos of any exercise