Timoshenko–Ehrenfest beam theory

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

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

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

${\displaystyle u_{x}(x,y,z)=-z~\varphi (x)~;~~u_{y}(x,y,z)=0~;~~u_{z}(x,y)=w(x)}$

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

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

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

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

${\displaystyle {\frac {3EI}{\kappa L^{2}AG}}\ll 1}$

where

• ${\displaystyle L}$ is the length of the beam.
• ${\displaystyle A}$ is the cross section area.
• ${\displaystyle E}$ is the elastic modulus.
• ${\displaystyle G}$ is the shear modulus.
• ${\displaystyle I}$ is the second moment of area.
• ${\displaystyle \kappa }$, called the Timoshenko shear coefficient, depends on the geometry. Normally, ${\displaystyle \kappa =5/6}$ for a rectangular section.
• ${\displaystyle q(x)}$ is a distributed load (force per length).

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

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

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

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

Derivation of quasistatic Timoshenko beam equations

From the kinematic assumptions for a Timoshenko beam, the displacements of the beam are given by ${\displaystyle u_{x}(x,y,z,t)=-z~\varphi (x,t)~;~~u_{y}(x,y,z,t)=0~;~~u_{z}(x,y,z)=w(x,t)}$ Then, from the strain-displacement relations for small strains, the non-zero strains based on the Timoshenko assumptions are ${\displaystyle \varepsilon _{xx}={\frac {\partial u_{x}}{\partial x}}=-z~{\frac {\partial \varphi }{\partial x}}~;~~\varepsilon _{xz}={\frac {1}{2}}\left({\frac {\partial u_{x}}{\partial z}}+{\frac {\partial u_{z}}{\partial x}}\right)={\frac {1}{2}}\left(-\varphi +{\frac {\partial w}{\partial x}}\right)}$ Since the actual shear strain in the beam is not constant over the cross section we introduce a correction factor ${\displaystyle \kappa }$ such that ${\displaystyle \varepsilon _{xz}={\frac {1}{2}}~\kappa ~\left(-\varphi +{\frac {\partial w}{\partial x}}\right)}$ The variation in the internal energy of the beam is ${\displaystyle \delta U=\int _{L}\int _{A}(\sigma _{xx}\delta \varepsilon _{xx}+2\sigma _{xz}\delta \varepsilon _{xz})~\mathrm {d} A~\mathrm {d} L=\int _{L}\int _{A}\left[-z~\sigma _{xx}{\frac {\partial (\delta \varphi )}{\partial x}}+\sigma _{xz}~\kappa \left(-\delta \varphi +{\frac {\partial (\delta w)}{\partial x}}\right)\right]~\mathrm {d} A~\mathrm {d} L}$ Define ${\displaystyle M_{xx}:=\int _{A}z~\sigma _{xx}~\mathrm {d} A~;~~Q_{x}:=\kappa ~\int _{A}\sigma _{xz}~\mathrm {d} A}$ Then ${\displaystyle \delta U=\int _{L}\left[-M_{xx}{\frac {\partial (\delta \varphi )}{\partial x}}+Q_{x}\left(-\delta \varphi +{\frac {\partial (\delta w)}{\partial x}}\right)\right]~\mathrm {d} L}$ Integration by parts, and noting that because of the boundary conditions the variations are zero at the ends of the beam, leads to ${\displaystyle \delta U=\int _{L}\left[\left({\frac {\partial M_{xx}}{\partial x}}-Q_{x}\right)~\delta \varphi -{\frac {\partial Q_{x}}{\partial x}}~\delta w\right]~\mathrm {d} L}$ The variation in the external work done on the beam by a transverse load ${\displaystyle q(x,t)}$ per unit length is ${\displaystyle \delta W=\int _{L}q~\delta w~\mathrm {d} L}$ Then, for a quasistatic beam, the principle of virtual work gives ${\displaystyle \delta U=\delta W\implies \int _{L}\left[\left({\frac {\partial M_{xx}}{\partial x}}-Q_{x}\right)~\delta \varphi -\left({\frac {\partial Q_{x}}{\partial x}}+q\right)~\delta w\right]~\mathrm {d} L=0}$ The governing equations for the beam are, from the fundamental theorem of variational calculus, ${\displaystyle {\frac {\partial M_{xx}}{\partial x}}-Q_{x}=0~;~~{\frac {\partial Q_{x}}{\partial x}}+q=0}$ For a linear elastic beam ${\displaystyle {\begin{aligned}M_{xx}&=\int _{A}z~\sigma _{xx}~\mathrm {d} A=\int _{A}z~E~\varepsilon _{xx}~\mathrm {d} A=-\int _{A}z^{2}~E~{\frac {\partial \varphi }{\partial x}}~\mathrm {d} A=-EI~{\frac {\partial \varphi }{\partial x}}\\Q_{x}&=\int _{A}\sigma _{xz}~\mathrm {d} A=\int _{A}2G~\varepsilon _{xz}~\mathrm {d} A=\int _{A}\kappa ~G~\left(-\varphi +{\frac {\partial w}{\partial x}}\right)~\mathrm {d} A=\kappa ~AG~\left(-\varphi +{\frac {\partial w}{\partial x}}\right)\end{aligned}}}$ Therefore the governing equations for the beam may be expressed as ${\displaystyle {\begin{aligned}{\frac {\partial }{\partial x}}\left(EI{\frac {\partial \varphi }{\partial x}}\right)+\kappa AG~\left({\frac {\partial w}{\partial x}}-\varphi \right)&=0\\{\frac {\partial }{\partial x}}\left[\kappa AG\left({\frac {\partial w}{\partial x}}-\varphi \right)\right]+q&=0\end{aligned}}}$ Combining the two equations together gives ${\displaystyle {\begin{aligned}&{\frac {\partial ^{2}}{\partial x^{2}}}\left(EI{\frac {\partial \varphi }{\partial x}}\right)=q\\&{\frac {\partial w}{\partial x}}=\varphi -{\cfrac {1}{\kappa AG}}~{\frac {\partial }{\partial x}}\left(EI{\frac {\partial \varphi }{\partial x}}\right)\end{aligned}}}$

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 ${\displaystyle x}$ direction is positive towards right and the ${\displaystyle z}$ direction is positive upward. Following normal convention, we assume that positive forces act in the positive directions of the ${\displaystyle x}$ and ${\displaystyle z}$ axes and positive moments act in the clockwise direction. We also assume that the sign convention of the stress resultants (${\displaystyle M_{xx}}$ and ${\displaystyle Q_{x}}$) is such that positive bending moments compress the material at the bottom of the beam (lower ${\displaystyle z}$ coordinates) and positive shear forces rotate the beam in a counterclockwise direction.

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

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

and

${\displaystyle P+Q_{x}=0\implies Q_{x}=-P\,.}$

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

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

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

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

The second equation can then be written as

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

Integration and application of the boundary condition ${\displaystyle w=0}$ at ${\displaystyle x=L}$ gives

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

The axial stress is given by

${\displaystyle \sigma _{xx}(x,z)=E\,\varepsilon _{xx}=-E\,z\,{\frac {d\varphi }{dx}}=-{\frac {Pxz}{I}}={\frac {M_{xx}z}{I}}\,.}$

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

${\displaystyle u_{x}(x,y,z,t)=-z~\varphi (x,t)~;~~u_{y}(x,y,z,t)=0~;~~u_{z}(x,y,z,t)=w(x,t)}$

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

Starting from the above assumption, the Timoshenko beam theory, allowing for vibrations, may be described with the coupled linear partial differential equations:[8]

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

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

where the dependent variables are ${\displaystyle w(x,t)}$, the translational displacement of the beam, and ${\displaystyle \varphi (x,t)}$, 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,

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

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[9][10]

${\displaystyle EI~{\cfrac {\partial ^{4}w}{\partial x^{4}}}+m~{\cfrac {\partial ^{2}w}{\partial t^{2}}}-\left(J+{\cfrac {EIm}{\kappa AG}}\right){\cfrac {\partial ^{4}w}{\partial x^{2}~\partial t^{2}}}+{\cfrac {mJ}{\kappa AG}}~{\cfrac {\partial ^{4}w}{\partial t^{4}}}=q(x,t)+{\cfrac {J}{\kappa AG}}~{\cfrac {\partial ^{2}q}{\partial t^{2}}}-{\cfrac {EI}{\kappa AG}}~{\cfrac {\partial ^{2}q}{\partial x^{2}}}}$

Derivation of combined Timoshenko beam equation

The equations governing the bending of a homogeneous Timoshenko beam of constant cross-section are ${\displaystyle {\begin{aligned}(1)&&\quad m~{\frac {\partial ^{2}w}{\partial t^{2}}}&=\kappa AG~\left({\frac {\partial ^{2}w}{\partial x^{2}}}-{\frac {\partial \varphi }{\partial x}}\right)+q(x,t)~;~~m:=\rho A\\(2)&&\quad J~{\frac {\partial ^{2}\varphi }{\partial t^{2}}}&=EI~{\frac {\partial ^{2}\varphi }{\partial x^{2}}}+\kappa AG~\left({\frac {\partial w}{\partial x}}-\varphi \right)~;~~J:=\rho I\end{aligned}}}$ From equation (1), assuming appropriate smoothness, we have ${\displaystyle {\begin{aligned}(3)&&\quad {\frac {\partial \varphi }{\partial x}}&={\cfrac {q}{\kappa AG}}-{\cfrac {m}{\kappa AG}}~{\frac {\partial ^{2}w}{\partial t^{2}}}+{\frac {\partial ^{2}w}{\partial x^{2}}}\\(4)&&\quad {\cfrac {\partial ^{3}\varphi }{\partial x^{3}}}&={\frac {1}{\kappa AG}}{\frac {\partial ^{2}q}{\partial x^{2}}}-{\frac {m}{\kappa AG}}~{\cfrac {\partial ^{4}w}{\partial x^{2}\partial t^{2}}}+{\cfrac {\partial ^{4}w}{\partial x^{4}}}\\(5)&&\quad {\cfrac {\partial ^{3}\varphi }{\partial x\partial t^{2}}}&={\frac {1}{\kappa AG}}{\frac {\partial ^{2}q}{\partial t^{2}}}-{\frac {m}{\kappa AG}}~{\cfrac {\partial ^{4}w}{\partial t^{4}}}+{\cfrac {\partial ^{4}w}{\partial x^{2}\partial t^{2}}}\end{aligned}}}$ Differentiating equation (2) gives ${\displaystyle {\begin{aligned}(6)&&\quad J~{\frac {\partial ^{3}\varphi }{\partial x\partial t^{2}}}&=EI~{\frac {\partial ^{3}\varphi }{\partial x^{3}}}+\kappa AG~\left({\frac {\partial w^{2}}{\partial x^{2}}}-{\frac {\partial \varphi }{\partial x}}\right)\end{aligned}}}$ Substituting equation (3), (4), (5) into equation (6) and rearrange, we get ${\displaystyle {\begin{aligned}EI~{\cfrac {\partial ^{4}w}{\partial x^{4}}}+m~{\frac {\partial ^{2}w}{\partial t^{2}}}-\left(J+{\cfrac {mEI}{\kappa AG}}\right)~{\cfrac {\partial ^{4}w}{\partial x^{2}\partial t^{2}}}+{\cfrac {mJ}{\kappa AG}}~{\cfrac {\partial ^{4}w}{\partial t^{4}}}=q+{\cfrac {J}{\kappa AG}}~{\frac {\partial ^{2}q}{\partial t^{2}}}-{\cfrac {EI}{\kappa AG}}~{\frac {\partial ^{2}q}{\partial x^{2}}}\end{aligned}}}$

However, it can easily be shown that this equation is incorrect. Consider the case where q is constant and does not depend on x or t, combined with the presence of a small damping all time derivatives will go to zero when t goes to infinity. The shear terms are not present in this situation, resulting in the Euler-Bernoulli beam theory, where shear deformation is neglected.

The Timoshenko equation predicts a critical frequency ${\displaystyle \omega _{C}=2\pi f_{c}={\sqrt {\frac {\kappa GA}{\rho I}}}.}$ For normal modes the Timoshenko equation can be solved. Being a fourth order equation, there are four independent solutions, two oscillatory and two evanescent for frequencies below ${\displaystyle f_{c}}$. For frequencies larger than ${\displaystyle f_{c}}$ all solutions are oscillatory and, as consequence, a second spectrum appears.[11]

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:

${\displaystyle \int _{A}\tau dA=\kappa AG(\varphi -{\frac {\partial w}{\partial x}})}$ .

The shear coefficient depends on Poisson's ratio. The attempts to provide precise expressions were made by many scientists, including Stephen Timoshenko,[12] Raymond D. Mindlin,[13] G. R. Cowper,[14] N. G. Stephen,[15] J. R. Hutchinson[16] etc. (see also the derivation of the Timoshenko beam theory as a refined beam theory based on the variational-asymptotic method in the book by Khanh C. Le[17] leading to different shear coefficients in the static and dynamic cases). In engineering practice, the expressions by Stephen Timoshenko[18] are sufficient in most cases. In 1975 Kaneko[19] published an excellent review of studies of the shear coefficient. More recently new experimental data show that the shear coefficient is underestimated.[20][21]

Corrective shear coefficients for homogeneous isotropic beam according to Cowper - selection.

Cross section	Coefficient
	${\displaystyle {\frac {10(1+\nu )}{12+11\nu }}}$
	${\displaystyle {\frac {6(1+\nu )}{7+6\nu }}}$
	${\displaystyle {\frac {12(1+\nu )a^{2}(3a^{2}+b^{2})}{(40+37\nu )a^{4}+(16+10\nu )a^{2}b^{2}+\nu b^{4}}}}$
	${\displaystyle {\frac {1+\nu }{1.305+1.273\nu }}}$
	${\displaystyle {\frac {6(1+\nu )(1+m^{2})^{2}}{(7+6\nu )(1+m^{2})^{2}+(20+12\nu )m^{2}}}}$, where ${\displaystyle m={\frac {b}{a}}}$
	${\displaystyle {\frac {2(1+\nu )}{4+3\nu }}}$
	${\displaystyle {\frac {20(1+\nu )}{48+39\nu }}}$
	${\displaystyle {\frac {10(1+\nu )(1+3m)^{2}}{(12+72m+150m^{2}+90m^{3})+\nu (11+66m+135m^{2}+90m^{3})+10n^{2}((3+\nu )m+3m^{2})}}}$, where ${\displaystyle m={\frac {bt_{1}}{ht_{2}}}}$ and ${\displaystyle n={\frac {b}{h}}}$
	${\displaystyle {\frac {10(1+\nu )(1+3m)^{2}}{(12+72m+150m^{2}+90m^{3})+\nu (11+66m+135m^{2}+90m^{3})+30n^{2}(m+m^{2})+5\nu n^{2}(8m+9m^{2})}}}$, where ${\displaystyle m={\frac {2bt_{1}}{ht_{2}}}}$ and ${\displaystyle n={\frac {b}{h}}}$
	${\displaystyle {\frac {10(1+\nu )(1+4m)^{2}}{(12+96m+276m^{2}+192m^{3})+\nu (11+88m+248m^{2}+216m^{3})+30n^{2}(m+m^{2})+10\nu n^{2}(4m+5m^{2}+m^{3})}}}$, where ${\displaystyle m={\frac {bt_{1}}{ht_{2}}}}$ and ${\displaystyle n={\frac {b}{h}}}$

where ${\displaystyle \nu }$ is Poisson's ratio.

• Plate theory
• Sandwich theory