Hello! You already know that the stiffness may result from the tension, as in the case of cables, or from the elasticity of the structure itself, as in the case of rails. In this video, we will start from the bending equation for beams to determine the eigenfrequencies and eigenmodes for such a structure. Let us write again the bending equation for beams. In a beam, inertia will be balanced with the variation of the shear force N. This shear force is related to the variation for the moment M. And we know that the moment M is related to the curvature d^2 y / d x^2. Combining these equations we get the bending equation for beams: m d^2 y/dt^2 + EI d^4 y/dx^4 = 0. Let us now calculate the bending modes of a straight beam. We first rewrite the bending equation for an infinite beam. The derivations w.r.t time and space lead to the characteristic equation -m omega^2+EI k^4=0. The solution wavenumbers k_N may be either purely real or purely imaginary. They depend on the square root of the eigenfrequencies omega_N and the ratio between the mass m and the product between the Young's modulus E and the geometrical moment of inertia I. Since the k_N are real or imaginary, the general form of the eigenmodes combines sine and cosine functions as well as hyperbolic sines and cosines. The mode shapes are thus very different from that of other types of structures! Let us now determine the four unknown constants A,B, C and D! We first rewrite the bending equation for an infinite beam and the characteristic equation. We then need to consider a finite beam to include boundary conditions! If we consider a clamped-free beam, the boundary conditions are: At x=0, the motion y and the rotation dy/dx are zero. At x=L, the shear force N and the moment M are zero. All the spatial derivatives of y may then be identified! We then get a very simple equation cos kL cosh kL + 1 = 0. The eigenmodes of a clamped-free beam may be easily determined from this equation. This plot shows variation of two functions: -cos(x) and the inverse of cosh(x). We can notice that there is infinite of solutions, that k_1 equals 1.9 and k_N is very close to (2N-1)pi/2 for higher modes. The eigenfrequencies omega_N can be derived from the k_N values. Let us now consider the eigenmodes: The first bending mode gives a global bending in the same direction for the whole beam The second bending mode leads to one node and one antinode. An additional vibration node is located at the left clamped end of the beam. As you can see, these functions are not sinusoidal at all. The third mode gives two nodes, plus one at the left end, and two antinodes in bending You have just considered a first set of boundary conditions for the clamped-free case. What happens for different boundary conditions? Let us consider a free-free condition for a straight beam. We will now have zero shear force and bending moment at both ends of the beam! The equation to solve is a bit different: cos kL cosh kL - 1 = 0. As shown by this graphical solution , it changes the whole process since k_1=0, it is a bit strange since it corresponds to omega_1 equals zero Hertz. Its means an infinitely slow motion. The explanation for this is that we get a rigid body motion due to the free-free boundary conditions! For higher modes, the k_N values are very close to (2N-1)pi/2 but the mode shapes are very different than in the clamped-free case! Let us now consider the free-free eigenmodes. The first eigenmode gives a global translation of the beam since it is a rigid body mode related to a zero eigenfrequency. The second free-free bending mode leads to two nodes and one antinode. We can now imagine the free-free flying spaghetti! The third free-free mode gives three nodes and two antinodes in bending and so on, for the fourth mode and the higher modes. From both configurations, clamped-free and free-free, we evidenced the strong influence of the boundary conditions. Starting from a single beam, we may consider several beams assembled in a so-called frame structure! This radar tower located in the Ecole Polytechnique campus is a very nice example of frame structures. The general shape is cylindrical but it is an assemblage of many prismatic reinforced concrete beams! The mass, stiffness and inertias of the beams are gathered in mass and stiffness matrices to solve a large eigenvalue problem! Here is the fifth bending mode of the tower. As you can see, the entire tower is bending as a whole! The local bending of each beam may be small but it depends on the stiffness contrast between the posts and the circular floors! Here is now the tenth eigenmode corresponding to a torsional motion and showing several nodes and antinodes at the scale of the entire tower. Finally the 14th mode corresponds to more complex vibration kinematics leading to strong changes in the tower cross-section! To summarize this video on beam modes - we determined the bending modes of a beam, examined the influence of the boundary conditions on the modes, and obtained the modes shapes which are much different in bending than in tension! More complex structures can be easily modeled by the Finite Element Method, but it is very important to know the beam dynamics behind!