Structural Analysis and Vibrations

Natural frequency of vibrating systems

The natural frequency of a vibrating system is the frequency at which it oscillates freely after being disturbed, with no external force acting. For a single degree of freedom system with mass m and stiffness k, \omega_n = \sqrt{k/m} rad/s and f_n = \omega_n/2\pi Hz. For a continuous member the frequency depends on its geometry: the first mode of a uniform cantilever is \omega_1 = \frac{(1.875)^2}{L^2}\sqrt{\frac{EI}{\rho A}}, so \omega_1 \propto 1/L^2. Two identical beams whose lengths are in a ratio of 2 therefore have first frequencies in the inverse square ratio, i.e. 4.

For a rigid body that rotates about a pin and is restrained by a spring, the natural frequency follows from energy. Write T = \frac{1}{2}I_O\dot{\theta}^2 and V = \frac{1}{2}k_{eff}\theta^2, where k_{eff} is the spring stiffness multiplied by the square of its lever arm about the pin. Then \omega_n^2 = k_{eff}/I_O. A point mass m at distance r from the pin contributes I_O = mr^2, and a spring attached at half the rod length reduces the effective stiffness by a factor of 4.

For a uniform bar pinned at one end and held by a spring at its free end acting perpendicular to the bar, I_O = mL^2/3 and k_{eff} = kL^2, so \omega_n = \sqrt{3k/m}. Static gravity moments shift only the equilibrium position and do not change the natural frequency.

A uniform rigid bar suspended from two identical springs has two independent modes. Vertical translation gives m\ddot{x} + 2kx = 0, while rotation about the centroid gives I_G\ddot{\theta} + 2k(l/2)^2\theta = 0 with I_G = ml^2/12. The rotational frequency is \sqrt{3} times the translational frequency.

The number of degrees of freedom equals the number of independent coordinates needed to fix the configuration. A rigid wing section that can both translate vertically and rotate, supported by a linear spring and a torsional spring, is a two-degree-of-freedom plunge-pitch model.

For an undamped system M\ddot{x} + Kx = 0, the natural frequencies satisfy \det(K - \omega^2 M) = 0. With equal masses m and K = k\begin{bmatrix} 2 & -1 \\\\ -1 & 2 \end{bmatrix}, substituting \lambda = \omega^2 m/k gives (2-\lambda)^2 - 1 = 0, so \lambda = 1 or 3. The larger frequency is then \sqrt{3}\sqrt{k/m}.

Each natural frequency has a mode shape. Free motion from zero initial velocity is a combination of mode shapes whose amplitudes are the projections of the initial displacement onto them. If the initial displacement vector is proportional to a single mode shape, only that natural frequency appears; otherwise both are excited.

Beam deflection curve analysis

In small-deflection beam theory, the deflection curve satisfies EI\,y'' = M(x). Slope and deflection are obtained by integrating the moment expression and applying the boundary conditions at supports and free ends. The location of maximum deflection is found by setting dy/dx = 0 and selecting the interior root that gives the largest |y|, not a root at a free end.

When a support reaction is redundant, it is found by compatibility: the deflections of the connected members at the contact point must be equal. Using superposition, the deflection due to the applied load is added to the deflection due to the unknown reaction, and the reaction is obtained by equating the two. When both members have the same section and modulus, these properties cancel out of the reaction.

Superposition also applies to a cantilever whose tip is supported. The support force is treated as an unknown load, and its value is fixed by requiring zero deflection at the support. The bending stress is \sigma = My/I, so for two sections with identical geometry the stress ratio equals the ratio of the bending moments at the root.

The strain energy in bending is U = \int_0^L \frac{M^2}{2EI}\,dx, with transverse shear energy neglected. For a cantilever with a tip load P, M = Px, and the integration gives U = \frac{P^2L^3}{6EI}. The energy is quadratic in the load, and the numerical factor is 1/6.

For a stepped cantilever, the tip deflection under a tip load is found by integrating \int \frac{Mm}{EI}\,dx segment by segment, where m is the moment caused by a unit load at the point of interest. Each segment uses its own second moment of area I.

Thin-walled section torsion and shear center

In a single closed thin-walled cell under torque, the shear flow is uniform around the wall, q = T/(2A_m), where A_m is the area enclosed by the wall mid-line. The shear stress is \tau = q/t. In a multi-cell section, each cell carries its own shear flow, the flow in a shared wall is the difference of the flows in the two adjacent cells, and the zero-twist compatibility condition \oint \frac{q\,ds}{Gt} = 2A\theta is applied to each cell. The shear modulus G does not appear in the resulting shear stress.

The shear centre is the point through which a transverse load must act to produce bending without twist. For a section symmetric about an axis, it lies on that axis. Cutting a closed section open changes the shear-flow path, so the shear centre of the open section moves, while the centroid remains unchanged because a cut of zero thickness removes no area.

The position of the shear centre is found by taking moments of the shear-flow resultant about a convenient point and equating them to the moment of the applied shear load. The moment of a constant wall flow q about a point is q times the integral of x\,dy - y\,dx along that wall, and the sum over the closed contour equals -2A for a closed cell.

For a thin-walled closed section limited by an allowable shear stress \tau_{allow}, the maximum torque is T_{max} = 2A_m t\,\tau_{allow}. Sections with the same wall thickness and material therefore carry torques in proportion to their enclosed mid-line areas. A semicircular nose that adds enclosed area therefore increases torque capacity.

Method of joints for truss forces

The method of joints isolates each pin as a free body and applies \sum F_x = 0 and \sum F_y = 0 at it. A pin can carry only two unknown member forces, so the solution starts at a joint with at most two unknowns and proceeds through the truss. Member forces are assumed to be tensile, and a negative result indicates compression. A member that pushes on its joint is in compression.

A roller reaction acts perpendicular to the support surface. At a roller joint, the horizontal equilibrium therefore involves only the member forces and applied loads, which provides a quick way to find a member force that is connected to that joint.

The support reactions are often found first from global equilibrium, and the internal force of a single member can then be obtained by a section cut through that member, with the forces resolved along the member direction.

Taking moments about one support eliminates the other reaction and gives a support force directly. Vertical equilibrium at a joint with a vertical member then fixes that member force, and horizontal equilibrium gives the horizontal member force. For a joint with a horizontal load F and a member inclined at 45^\circ whose horizontal component must balance F, the member force has magnitude \sqrt{2}F.

Euler buckling of columns

The elastic Euler load of a slender column is P_{cr} = \frac{\pi^2 EI_{min}}{(KL)^2}, where K is the effective length factor set by the end conditions. Standard values are K = 1 for pinned-pinned, K = 0.7 for fixed-pinned, K = 0.5 for fixed-fixed and K = 2 for fixed-free. Buckling occurs about the axis of least second moment of area. For a rectangular section w \times t with t \le w, I_{min} = wt^3/12.

To compare buckling with yielding, equate the Euler load to the yield load P_y = \sigma_y A. For a simply supported column of rectangular section, equating these loads gives \frac{\pi^2 E t^2}{12L^2} = \sigma_y, so the thickness at which the two failure loads coincide is found directly.

The elastic flexural buckling load increases with the flexural rigidity EI and decreases with the square of the effective length. It depends on the end conditions through K. It does not depend on the material density, because the elastic formula contains only E, I, L and K.

Logarithmic decrement of underdamped vibration

For an underdamped system, the logarithmic decrement is the natural logarithm of the ratio of two successive peak amplitudes, \delta = \ln\frac{x_n}{x_{n+1}} = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}}. Inverting gives \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}}. Peaks separated by m cycles differ by the factor e^{m\delta}, so if the ratio of two peaks one cycle apart is known, \delta = \ln of that ratio. The damping coefficient is then c = \zeta c_c, with c_c = 2m\omega_n and \omega_n = \sqrt{k/m}.

For a free response written as x(t) = Ae^{-\sigma t}\sin(\omega_d t + \phi), the decay rate is \sigma = \zeta\omega_n and the damped frequency is \omega_d = \omega_n\sqrt{1-\zeta^2}. Hence \zeta = \sigma/\sqrt{\sigma^2 + \omega_d^2} can be computed directly from the two coefficients of the response.

For a system released from rest at x_0 with no initial velocity, the displacement after n complete damped periods is governed by the exponential factor e^{-2\pi n\zeta/\sqrt{1-\zeta^2}}. At t = 2T_d the sine term vanishes and the cosine term equals 1, so the amplitude is x_0 e^{-2\zeta\omega_n T_d}.

Second moment of area of thin-walled section

For a thin-walled section, the second moment of area about a centroidal axis is found by summing each wall's own second moment and its area times the square of its distance from the axis, I = \sum (I_{own} + Ad^2). The own-inertia terms are very small for thin walls and are often neglected. Each flange then contributes Ad^2, and a web centred on the axis contributes th^3/12 for web height h.

In a thin-walled I or box-type section, the flanges carry most of the bending stiffness, so I \approx \sum A_f y_f^2 is a good approximation. The bending stress is \sigma = My/I, and the maximum normal stress occurs at the outermost fibres, where y is largest.