Higher-Order Eigenvalue Problems and Beam Models, differential equations lesson

Press a ruler flat on a table with part of it sticking out over the edge, and flick the free end. Nothing like the tension of a guitar string pulls it back. It springs back because it resists bending, and that resistance leads to an equation with four derivatives in space instead of two. This lesson sets up the Euler-Bernoulli beam model, reads clamped, pinned and free ends as conditions on the shape, and finds natural frequencies and mode shapes from a fourth-order eigenvalue problem. Throughout, $x$ is position along the beam, $t$ is time, subscripts are partial derivatives, and for a shape $X(x)$ a prime means $d/dx$ .

Key terms

Euler-Bernoulli beam

Definition The Euler-Bernoulli model describes the small transverse displacement $u(x, t)$ of a slender beam on $0 \le x \le L$ . With constant flexural rigidity $EI > 0$ and constant mass per unit length $\rho A > 0$ , the unforced equation is

\rho A\,u_{tt} + EI\,u_{xxxx} = 0.

Here $E$ is Young’s modulus of the material, $I$ is the second moment of area of the cross section, $\rho$ is the density and $A$ is the cross-sectional area. The bending moment is $EI\,u_{xx}$ and the shear force is its $x$ -derivative $(EI\,u_{xx})_x$ , each up to a sign convention, and the balance of forces and moments on a short piece of the beam gives $\rho A\,u_{tt} = -(EI\,u_{xx})_{xx}$ . With constant $EI$ this is the equation above. The model assumes small deflections and a linearly elastic material, and it neglects shear deformation, rotary inertia and any axial force.

In plain words A string resists stretching, so its restoring force follows the curvature $u_{xx}$ . A beam resists bending. The curvature sets the bending moment, the change of the moment along the beam sets the shear force, and the change of the shear force is the net push on each piece, which makes four derivatives in all.

Example Measure length in units of $L$ and time in units of

T_0 = L^2\sqrt{\frac{\rho A}{EI}}.

With $EI$ in $\text{N}\,\text{m}^2$ and $\rho A$ in $\text{kg/m}$ , this is a time in seconds. In the variables $\xi = x/L$ and $\tau = t/T_0$ the equation becomes $u_{\tau\tau} + u_{\xi\xi\xi\xi} = 0$ , which is the form with $EI = \rho A = L = 1$ used below. The time scale grows like $L^2$ , so for the same material, cross section and supports, a beam twice as long has natural frequencies four times lower. A string with the same tension and density and fixed ends would have frequencies only two times lower.

Fourth-order eigenvalue problem

Definition Separating variables in the beam equation, $u = X(x)\,T(t)$ , gives $T'' + \lambda T = 0$ for the time factor and the spatial problem

\begin{gathered} EI\,X'''' = \lambda\,\rho A\,X, \\[0.4em] 0 < x < L, \end{gathered}

with two homogeneous boundary conditions at each end. This is a fourth-order eigenvalue problem, and a number $\lambda$ for which it has a nonzero solution $X$ is an eigenvalue. Multiplying by $X$ and integrating by parts twice gives

\begin{aligned} &\int_0^L X X''''\,dx \\[0.4em] &\quad = \Big[X X''' - X'X''\Big]_0^L \\[0.4em] &\qquad + \int_0^L (X'')^2\,dx. \end{aligned}

When each end is clamped, pinned or free, the bracket is zero at both ends. At a clamped end $X$ and $X'$ vanish, at a pinned end $X$ and $X''$ vanish, and at a free end $X''$ and $X'''$ vanish, so both products in the bracket are zero there. Then

\begin{aligned} &EI\int_0^L (X'')^2\,dx \\[0.4em] &\quad = \lambda\,\rho A\int_0^L X^2\,dx \end{aligned}

for a real mode shape, and the same computation with $\lvert X \rvert^2$ shows that no eigenvalue is complex. So $\lambda \ge 0$ , and $\lambda = \omega^2$ with $\omega \ge 0$ . For $\omega > 0$ the time factor is $a\cos\omega t + b\sin\omega t$ , and $\omega$ is a natural frequency. Put $\beta^4 = \rho A\,\omega^2/EI$ with $\beta > 0$ . The equation $X'''' = \beta^4X$ has characteristic roots $\pm\beta$ and $\pm i\beta$ , so

\begin{aligned} X &= c_1\cos\beta x + c_2\sin\beta x \\[0.4em] &\quad + c_3\cosh\beta x \\[0.4em] &\quad + c_4\sinh\beta x. \end{aligned}

The four end conditions form a homogeneous linear system for $c_1, \dots, c_4$ , which has a nonzero solution exactly when its determinant is zero. That determinant equation is the frequency equation. Fourth-order problems do not automatically inherit the theorems for second-order Sturm-Liouville problems. For example, an eigenvalue need not be simple, as the free end card shows.

In plain words The beam equation asks for shapes that the bending stiffness pushes back in proportion to the shape itself. Four derivatives need four conditions, two at each end, and the energy identity says that the stored bending energy can never be negative, so the squared frequency cannot be negative either.

Example With $EI = \rho A = L = 1$ , the shape $X = \sin \pi x$ has $X'''' = \pi^4 X$ , so $\lambda = \pi^4$ . The energy identity agrees. Since $(X'')^2 = \pi^4\sin^2 \pi x$ and $\int_0^1 \sin^2 \pi x\,dx = \tfrac12$ ,

\begin{gathered} \int_0^1 (X'')^2\,dx = \frac{\pi^4}{2}, \\[0.4em] \int_0^1 X^2\,dx = \frac12, \end{gathered}

and the ratio of the two integrals is $\pi^4$ , which is $\lambda$ .

The identity also says when $\lambda = 0$ can occur. Zero frequency needs $X'' = 0$ on the whole beam, so $X$ is a straight line, and that line must satisfy all four end conditions. A beam pinned at both ends allows no such line except $X = 0$ , so its lowest frequency is positive.

Clamped end

Definition At a clamped end, also called a fixed or built-in end, the support prevents both translation and rotation. The conditions at that end are

X = 0 \quad\text{and}\quad X' = 0,

and the vibration satisfies $u = 0$ and $u_x = 0$ there at all times. These are conditions on the geometry, so they keep this form when the rigidity varies.

In plain words A clamped end is held in place and kept from turning, like a beam set into a wall. Holding the end at the right height is not enough. Its slope is held too, which is a stronger restraint than holding the position alone.

Example Clamping $x = 0$ removes half of the general solution. The conditions read $c_1 + c_3 = 0$ and $\beta(c_2 + c_4) = 0$ , so every admissible shape has the form

\begin{aligned} X &= A(\cos\beta x - \cosh\beta x) \\[0.4em] &\quad + B(\sin\beta x - \sinh\beta x). \end{aligned}

The shape $\sin \pi x$ is not allowed at a clamped end at $0$ , because its slope there is $\pi$ , not $0$ . A clamped end also rules out zero frequency, since a straight line with zero height and zero slope at one point is zero everywhere.

Pinned end

Definition At a pinned end, also called a simply supported end, the support holds the displacement at zero but lets the beam rotate freely, so no bending moment acts there. The conditions at that end are

X = 0 \quad\text{and}\quad X'' = 0.

The moment condition is $EI\,X'' = 0$ , and since $EI > 0$ it is the same as $X'' = 0$ even when the rigidity varies. The slope at a pinned end is not restricted and is usually not zero.

In plain words A pinned end sits on a hinge. It cannot move up or down, but it can turn, and because the hinge exerts no bending moment on the beam, the curvature there is zero.

Example Pinning both ends of $[0, 1]$ picks out the sine modes. At $x = 0$ the conditions give $c_1 + c_3 = 0$ and $\beta^2(c_3 - c_1) = 0$ , so $c_1 = c_3 = 0$ . At $x = 1$ they give $c_2\sin\beta + c_4\sinh\beta = 0$ and $\beta^2(c_4\sinh\beta - c_2\sin\beta) = 0$ . Dividing the second by $\beta^2$ and adding the two gives $2c_4\sinh\beta = 0$ , so $c_4 = 0$ , and then $c_2\sin\beta = 0$ with $c_2 \ne 0$ forces $\beta = n\pi$ . The mode shapes are $\sin n\pi x$ for $n = 1, 2, 3, \dots$ , and their slope at $0$ is $n\pi \ne 0$ , so the beam does turn at each pin.

Free end

Definition At a free end with no applied load, neither a bending moment nor a shear force acts. For constant $EI$ the conditions at that end are

X'' = 0 \quad\text{and}\quad X''' = 0.

With variable rigidity the conditions are written for the moment and the shear themselves, $EI\,X'' = 0$ and $(EI\,X'')' = 0$ , and the equation becomes $(EI\,X'')'' = \lambda\,\rho A\,X$ . Since $(EI\,X'')' = (EI)'X'' + EI\,X'''$ and $EI > 0$ , an unloaded free end still gives $X'' = X''' = 0$ . With an applied end force or moment the rigidity factors stay in the conditions. Free does not mean zero displacement or zero slope, since both are left to the motion.

In plain words Nothing holds a free end. It can move and turn as it likes, and what vanishes there are the forces, the bending moment and the shear, not the position.

Example A cantilever on $[0, 1]$ with $EI = \rho A = 1$ is clamped at $0$ and free at $1$ . Start from the form that the clamped end card found and write $c = \cos\beta$ , $s = \sin\beta$ , $C = \cosh\beta$ and $S = \sinh\beta$ . The conditions $X''(1) = 0$ and $X'''(1) = 0$ become

\begin{aligned} A(c + C) + B(s + S) &= 0, \\[0.4em] A(s - S) - B(c + C) &= 0. \end{aligned}

The determinant is $-(c + C)^2 - (s + S)(s - S) = -2 - 2cC$ , using $c^2 + s^2 = 1$ and $C^2 - S^2 = 1$ . So the frequency equation is $\cos\beta\cosh\beta = -1$ , whose smallest positive root is $\beta_1 \approx 1.8751$ , and the lowest frequency is $\omega_1 = \beta_1^2 \approx 3.516$ in these units. A beam free at both ends behaves differently. The shapes $X = 1$ , a rigid translation, and $X = x$ , a small rigid tilt, both have $X'' = X''' = 0$ everywhere, so $\lambda = 0$ is an eigenvalue with two independent shapes. That cannot happen for a regular second-order Sturm-Liouville problem with separated conditions, whose eigenvalues are simple.

Beam mode shapes

Definition A beam mode shape is a nonzero solution $X_n$ of the fourth-order eigenvalue problem with its boundary conditions, and its natural frequency is $\omega_n = \sqrt{\lambda_n}$ . For $\omega_n > 0$ the vibration

\begin{gathered} u = X_n(x)\,T_n(t), \\[0.4em] T_n = a\cos\omega_n t + b\sin\omega_n t, \end{gathered}

satisfies the beam equation and all four end conditions at every time, and by linearity so does any finite sum of such vibrations. For a beam pinned at both ends of $[0, L]$ , the modes are

\begin{gathered} X_n = \sin\frac{n\pi x}{L}, \\[0.4em] \omega_n = \left(\frac{n\pi}{L}\right)^2\sqrt{\frac{EI}{\rho A}}, \end{gathered}

for $n = 1, 2, 3, \dots$ . The mode shape fixes the spatial pattern and the frequency, and the initial displacement and velocity fix the coefficients $a$ and $b$ . For any combination of clamped, pinned and free ends, two modes with different frequencies are orthogonal, $\int_0^L X_mX_n\,dx = 0$ , by the same integration by parts as in the energy identity.

In plain words A mode is a shape that vibrates in place, with every point moving in step at one frequency. For a pinned beam, mode $n$ has $n$ half-waves and a frequency proportional to $n^2$ , so doubling the mode number quadruples the frequency. For an ideal string it only doubles it.

Example With $EI = \rho A = L = 1$ and both ends pinned, take

u(x, t) = \sin(\pi x)\cos(\pi^2t).

Then $u_{tt} = -\pi^4u$ and $u_{xxxx} = \pi^4u$ , so $u_{tt} + u_{xxxx} = 0$ . At $x = 0$ and $x = 1$ the sine factor vanishes, so $u = 0$ , and $u_{xx} = -\pi^2u$ vanishes too, which is the zero-moment condition. At $t = 0$ the shape is $\sin \pi x$ , and the velocity $u_t = -\pi^2\sin(\pi x)\sin(\pi^2t)$ is zero. The next pinned modes have $\omega_2 = 4\pi^2$ and $\omega_3 = 9\pi^2$ . An ideal string fixed at both ends with the same lowest frequency $\pi^2$ would have its next frequencies at $2\pi^2$ and $3\pi^2$ .

Check yourself

With $EI = \rho A = L = 1$ and both ends pinned, a beam starts at rest with displacement $\sin \pi x + \tfrac12\sin 3\pi x$ . Write $u(x, t)$ and check the boundary conditions.

Show the answer

Both terms are pinned mode shapes, with $\omega_1 = \pi^2$ and $\omega_3 = 9\pi^2$ . Starting at rest leaves only the cosine factors, so

\begin{aligned} u &= \sin(\pi x)\cos(\pi^2t) \\[0.4em] &\quad + \tfrac12\sin(3\pi x)\cos(9\pi^2t). \end{aligned}

For the second term $u_{tt}$ and $u_{xxxx}$ are $-81\pi^4$ and $81\pi^4$ times the term, so each term solves $u_{tt} + u_{xxxx} = 0$ and so does the sum. At $x = 0$ and $x = 1$ both sines vanish, so $u = 0$ , and $u_{xx}$ is a combination of the same two sines, so it vanishes as well. At $t = 0$ the displacement is $\sin \pi x + \tfrac12\sin 3\pi x$ and $u_t = 0$ .

The shape $X(x) = x$ has $X'' = X''' = 0$ everywhere. For which of these supports on $[0, 1]$ is it a mode with $\omega = 0$ ? Pinned at $0$ and free at $1$ , clamped at $0$ and free at $1$ , free at both ends, pinned at both ends.

Show the answer

The shape must meet both conditions at each end. Pinned at $0$ needs $X(0) = 0$ and $X''(0) = 0$ , which hold, and free at $1$ needs $X''(1) = X'''(1) = 0$ , which hold. So the beam pinned at $0$ and free at $1$ has this zero-frequency mode, a rigid rotation about the pin. Clamped at $0$ needs $X'(0) = 0$ , but $X'(0) = 1$ , so the cantilever does not. Free at both ends works, since all four conditions involve only $X''$ and $X'''$ . Pinned at both ends fails, since $X(1) = 1 \ne 0$ . The mistake to avoid is reading a free end as $X = 0$ . A free end restricts only $X''$ and $X'''$ .

A beam with $EI = \rho A = 1$ on $[0, 1]$ is clamped at $x = 0$ and pinned at $x = 1$ . Derive the frequency equation and compare the lowest frequency with the value $\pi^2$ for a beam pinned at both ends.

Show the answer

The clamped end leaves $X = A(\cos\beta x - \cosh\beta x) + B(\sin\beta x - \sinh\beta x)$ . With $c = \cos\beta$ , $s = \sin\beta$ , $C = \cosh\beta$ and $S = \sinh\beta$ , the pinned conditions $X(1) = 0$ and $X''(1) = 0$ read

\begin{aligned} A(c - C) + B(s - S) &= 0, \\[0.4em] A(c + C) + B(s + S) &= 0. \end{aligned}

The determinant is $(c - C)(s + S) - (s - S)(c + C) = 2(cS - sC)$ , so the frequency equation is $\sin\beta\cosh\beta = \cos\beta\sinh\beta$ . A root cannot have $\cos\beta = 0$ , because then the left side would be $\pm\cosh\beta \ne 0$ , so the equation is the same as $\tan\beta = \tanh\beta$ . Its smallest positive root is $\beta_1 \approx 3.9266$ , so

\omega_1 = \beta_1^2 \approx 15.42,

compared with $\pi^2 \approx 9.87$ for the beam pinned at both ends. Clamping one end makes the support stiffer and raises the lowest frequency. Zero frequency is impossible here because a clamped end allows no straight-line mode.

Higher-Order Eigenvalue Problems and Beam Models

Key terms

Euler-Bernoulli beam

Fourth-order eigenvalue problem

Clamped end

Pinned end

Free end

Beam mode shapes

Check yourself

How this lesson connects

Builds on