# Timoshenko beam theory

The Timoshenko beam theory was developed by Russian scientist and engineer of Ukrainian ethnicity Stephen Timoshenko early in the 20th century.[1][2] The model takes into account shear deformation and rotational inertia effects, making it suitable for describing the behaviour of short beams, sandwich composite beams or beams subject to high-frequency excitation when the wavelength approaches the thickness of the beam. The resulting equation is of 4th order, but unlike ordinary beam theory - i.e. Euler–Bernoulli beam theory - there is also a second order partial derivative present. Physically, taking into account the added mechanisms of deformation effectively lowers the stiffness of the beam, while the result is a larger deflection under a static load and lower predicted eigenfrequencies for a given set of boundary conditions. The latter effect is more noticeable for higher frequencies as the wavelength becomes shorter, and thus the distance between opposing shear forces decreases.

Deformation of a Timoshenko beam (blue) compared with that of an Euler-Bernoulli beam (red).

If the shear modulus of the beam material approaches infinity - and thus the beam becomes rigid in shear - and if rotational inertia effects are neglected, Timoshenko beam theory converges towards ordinary beam theory.

## Quasistatic Timoshenko beam

Deformation of a Timoshenko beam. The normal rotates by an amount $\theta_x = \varphi(x)$ which is not equal to $dw/dx$.

In static Timoshenko beam theory without axial effects, the displacements of the beam are assumed to be given by

$u_x(x,y,z) = -z~\varphi(x) ~;~~ u_y(x,y,z) = 0 ~;~~ u_z(x,y) = w(x)$

where $(x,y,z)$ are the coordinates of a point in the beam, $u_x, u_y, u_z$ are the components of the displacement vector in the three coordinate directions, $\varphi$ is the angle of rotation of the normal to the mid-surface of the beam, and $w$ is the displacement of the mid-surface in the $z$-direction.

The governing equations are the following uncoupled system of ordinary differential equations:

\begin{align} & \frac{\mathrm{d}^2}{\mathrm{d} x^2}\left(EI\frac{\mathrm{d} \varphi}{\mathrm{d} x}\right) = q(x,t) \\ & \frac{\mathrm{d} w}{\mathrm{d} x} = \varphi - \frac{1}{\kappa AG} \frac{\mathrm{d}}{\mathrm{d} x}\left(EI\frac{\mathrm{d} \varphi}{\mathrm{d} x}\right). \end{align}

The Timoshenko beam theory for the static case is equivalent to the Euler-Bernoulli theory when the last term above is neglected, an approximation that is valid when

$\frac{EI}{\kappa L^2 A G} \ll 1$

where

• $L$ is the length of the beam.
• $A$ is the cross section area.
• $E$ is the elastic modulus.
• $G$ is the shear modulus.
• $I$ is the second moment of area.
• $\kappa$, called the Timoshenko shear coefficient, depends on the geometry. Normally, $\kappa = 5/6$ for a rectangular section.

Combining the two equations gives, for a homogeneous beam of constant cross-section,

$EI~\cfrac{\mathrm{d}^4 w}{\mathrm{d} x^4} = q(x) - \cfrac{EI}{\kappa A G}~\cfrac{\mathrm{d}^2 q}{\mathrm{d} x^2}$

The bending moment $M_{xx}$ and the shear force $Q_x$ in the beam are related to the displacement $w$ and the rotation $\varphi$. These relations, for a linear elastic Timoshenko beam, are:

$M_{xx} = -EI~\frac{\partial \varphi}{\partial x} \quad \text{and} \quad Q_{x} = \kappa~AG~\left(-\varphi + \frac{\partial w}{\partial x}\right) \,.$

### Boundary conditions

The two equations that describe the deformation of a Timoshenko beam have to be augmented with boundary conditions if they are to be solved. Four boundary conditions are needed for the problem to be well-posed. Typical boundary conditions are:

• Simply supported beams: The displacement $w$ is zero at the locations of the two supports. The bending moment $M_{xx}$ applied to the beam also has to be specified. The rotation $\varphi$ and the transverse shear force $Q_x$ are not specified.
• Clamped beams: The displacement $w$ and the rotation $\varphi$ are specified to be zero at the clamped end. If one end is free, shear force $Q_x$ and bending moment $M_{xx}$ have to be specified at that end.

### Example: Cantilever beam

A cantilever Timoshenko beam under a point load at the free end.

For a cantilever beam, one boundary is clamped while the other is free. Let us use a right handed coordinate system where the $x$ direction is positive towards right and the $z$ direction is positive upward. Following normal convention, we assume that positive forces act in the positive directions of the $x$ and $z$ axes and positive moments act in the clockwise direction. We also assume that the sign convention of the stress resultants ($M_{xx}$ and $Q_x$) is such that positive bending moments compress the material at the bottom of the beam (lower $z$ coordinates) and positive shear forces rotate the beam in a counterclockwise direction.

Let us assume that the clamped end is at $x=L$ and the free end is at $x=0$. If a point load $P$ is applied to the free end in the positive $z$ direction, a free body diagram of the beam gives us

$-Px - M_{xx} = 0 \implies M_{xx} = -Px$

and

$P + Q_x = 0 \implies Q_x = -P\,.$

Therefore, from the expressions for the bending moment and shear force, we have

$Px = EI\,\frac{d\varphi}{dx} \qquad \text{and} \qquad -P = \kappa AG\left(-\varphi + \frac{dw}{dx}\right) \,.$

Integration of the first equation, and application of the boundary condition $\varphi = 0$ at $x = L$, leads to

$\varphi(x) = -\frac{P}{2EI}\,(L^2-x^2) \,.$

The second equation can then be written as

$\frac{dw}{dx} = -\frac{P}{\kappa AG} - \frac{P}{2EI}\,(L^2-x^2)\,.$

Integration and application of the boundary condition $w = 0$ at $x = L$ gives

$w(x) = \frac{P(L-x)}{\kappa AG} - \frac{Px}{2EI}\,\left(L^2-\frac{x^2}{3}\right) + \frac{PL^3}{3EI} \,.$

The axial stress is given by

$\sigma_{xx}(x,z) = E\,\varepsilon_{xx} = -E\,z\,\frac{d\varphi}{dx} = -\frac{Pxz}{I} = \frac{M_{xx}z}{I} \,.$

## Dynamic Timoshenko beam

In Timoshenko beam theory without axial effects, the displacements of the beam are assumed to be given by

$u_x(x,y,z,t) = -z~\varphi(x,t) ~;~~ u_y(x,y,z,t) = 0 ~;~~ u_z(x,y,z,t) = w(x,t)$

where $(x,y,z)$ are the coordinates of a point in the beam, $u_x, u_y, u_z$ are the components of the displacement vector in the three coordinate directions, $\varphi$ is the angle of rotation of the normal to the mid-surface of the beam, and $w$ is the displacement of the mid-surface in the $z$-direction.

Starting from the above assumption, the Timoshenko beam theory, allowing for vibrations, may be described with the coupled linear partial differential equations:[3]

$\rho A\frac{\partial^{2}w}{\partial t^{2}} - q(x,t) = \frac{\partial}{\partial x}\left[ \kappa AG \left(\frac{\partial w}{\partial x}-\varphi\right)\right]$
$\rho I\frac{\partial^{2}\varphi}{\partial t^{2}} = \frac{\partial}{\partial x}\left(EI\frac{\partial \varphi}{\partial x}\right)+\kappa AG\left(\frac{\partial w}{\partial x}-\varphi\right)$

where the dependent variables are $w(x,t)$, the translational displacement of the beam, and $\varphi(x,t)$, the angular displacement. Note that unlike the Euler-Bernoulli theory, the angular deflection is another variable and not approximated by the slope of the deflection. Also,

• $\rho$ is the density of the beam material (but not the linear density).
• $A$ is the cross section area.
• $E$ is the elastic modulus.
• $G$ is the shear modulus.
• $I$ is the second moment of area.
• $\kappa$, called the Timoshenko shear coefficient, depends on the geometry. Normally, $\kappa = 5/6$ for a rectangular section.
• $q(x,t)$ is a distributed load (force per length).
• $m := \rho A$
• $J := \rho I$

These parameters are not necessarily constants.

For a linear elastic, isotropic, homogeneous beam of constant cross-section these two equations can be combined to give[4][5]

$EI~\cfrac{\partial^4 w}{\partial x^4} + m~\cfrac{\partial^2 w}{\partial t^2} - \left(J + \cfrac{E I m}{k A G}\right)\cfrac{\partial^4 w}{\partial x^2~\partial t^2} + \cfrac{m J}{k A G}~\cfrac{\partial^4 w}{\partial t^4} = q(x,t) + \cfrac{J}{k A G}~\cfrac{\partial^2 q}{\partial t^2} - \cfrac{EI}{k A G}~\cfrac{\partial^2 q}{\partial x^2}$

### Axial effects

If the displacements of the beam are given by

$u_x(x,y,z,t) = u_0(x,t)-z~\varphi(x,t) ~;~~ u_y(x,y,z,t) = 0 ~;~~ u_z(x,y,z) = w(x,t)$

where $u_0$ is an additional displacement in the $x$-direction, then the governing equations of a Timoshenko beam take the form

\begin{align} m \frac{\partial^{2}w}{\partial t^{2}} & = \frac{\partial}{\partial x}\left[ \kappa AG \left(\frac{\partial w}{\partial x}-\varphi\right)\right] + q(x,t) \\ J \frac{\partial^{2}\varphi}{\partial t^{2}} & = N(x,t)~\frac{\partial w}{\partial x} + \frac{\partial}{\partial x}\left(EI\frac{\partial \varphi}{\partial x}\right)+\kappa AG\left(\frac{\partial w}{\partial x}-\varphi\right) \end{align}

where $J = \rho I$ and $N(x,t)$ is an externally applied axial force. Any external axial force is balanced by the stress resultant

$N_{xx}(x,t) = \int_{-h}^{h} \sigma_{xx}~dz$

where $\sigma_{xx}$ is the axial stress and the thickness of the beam has been assumed to be $2h$.

The combined beam equation with axial force effects included is

$EI~\cfrac{\partial^4 w}{\partial x^4} + N~\cfrac{\partial^2 w}{\partial x^2} + m~\frac{\partial^2 w}{\partial t^2} - \left(J+\cfrac{mEI}{\kappa AG}\right)~\cfrac{\partial^4 w}{\partial x^2 \partial t^2} + \cfrac{mJ}{\kappa AG}~\cfrac{\partial^4 w}{\partial t^4} = q + \cfrac{J}{\kappa AG}~\frac{\partial^2 q}{\partial t^2} - \cfrac{EI}{\kappa A G}~\frac{\partial^2 q}{\partial x^2}$

### Damping

If, in addition to axial forces, we assume a damping force that is proportional to the velocity with the form

$\eta(x)~\cfrac{\partial w}{\partial t}$

the coupled governing equations for a Timoshenko beam take the form

$m \frac{\partial^{2}w}{\partial t^{2}} + \eta(x)~\cfrac{\partial w}{\partial t} = \frac{\partial}{\partial x}\left[ \kappa AG \left(\frac{\partial w}{\partial x}-\varphi\right)\right] + q(x,t)$
$J \frac{\partial^{2}\varphi}{\partial t^{2}} = N\frac{\partial w}{\partial x} + \frac{\partial}{\partial x}\left(EI\frac{\partial \varphi}{\partial x}\right)+\kappa AG\left(\frac{\partial w}{\partial x}-\varphi\right)$

and the combined equation becomes

\begin{align} EI~\cfrac{\partial^4 w}{\partial x^4} & + N~\cfrac{\partial^2 w}{\partial x^2} + m~\frac{\partial^2 w}{\partial t^2} - \left(J+\cfrac{mEI}{\kappa AG}\right)~\cfrac{\partial^4 w}{\partial x^2 \partial t^2} + \cfrac{mJ}{\kappa AG}~\cfrac{\partial^4 w}{\partial t^4} + \cfrac{J \eta(x)}{\kappa AG}~\cfrac{\partial^3 w}{\partial t^3} \\ & -\cfrac{EI}{\kappa AG}~\cfrac{\partial^2}{\partial x^2}\left(\eta(x)\cfrac{\partial w}{\partial t}\right) + \eta(x)\cfrac{\partial w}{\partial t} = q + \cfrac{J}{\kappa AG}~\frac{\partial^2 q}{\partial t^2} - \cfrac{EI}{\kappa A G}~\frac{\partial^2 q}{\partial x^2} \end{align}

A caveat to this Ansatz damping force (resembling viscosity) is that, whereas viscosity leads to a frequency-dependent and amplitude-independent damping rate of beam oscillations, the empirically measured damping rates are frequency-insensitive, but depend on the amplitude of beam deflection.

## Shear coefficient

Determining the shear coefficient is not straightforward (nor are the determined values widely accepted, i.e. there's more than one answer); generally it must satisfy:

$\int_A \tau dA = \kappa A G \varphi\,$

The shear coefficient depends on the Poisson's ratio. The attempts to provide precise expressions were made by many scientists, including Stephen Timoshenko, Raymond D. Mindlin, G. R. Cowper, John W. Hutchinson, etc. In engineering practice, the expressions by Stephen Timoshenko[6] are sufficient in most cases.

For solid rectangular cross-section,

$\kappa = \cfrac{10(1+\nu)}{12+11\nu}$

For solid circular cross-section,

$\kappa = \cfrac{6(1+\nu)}{7+6\nu}$