Categories: Physics.

Euler-Bernoulli law

Euler-Bernoulli beam theory concerns itself with the bending of beams (e.g. the metal beams used in large buildings), subject to certain simplifying assumptions, which are generally valid for beams that are narrow, i.e. longitudinally much larger than transversely.

Consider a beam of length LL, placed upright on the z=0z = 0 plane, above the origin. If we pull the top of this beam in the postive yy-direction, we assume that it bends uniformly, i.e. with constant radius of curvature RR. We also assume that the bending is shear-free: if we treat the beam as a bundle of elastic strings, then there is no friction between them.

The central string has its length unchanged (i.e. still LL), while an arbitrary non-central string is extended or compressed to LL'. The Cauchy strain tensor element uzzu_{zz} is then:

uzz=LLL\begin{aligned} u_{zz} = \frac{L' - L}{L} \end{aligned}

Because the bending is uniform, the central string is an arc with radius RR and central angle θ\theta, where L=θRL = \theta R. The non-central string has L=θRL' = \theta R', where RR' is geometrically shown to be R=RyR' = R - y, with yy being the yy-coordinate of that string at the beam’s base. So:

uzz=RRR=yR\begin{aligned} u_{zz} = \frac{R' - R}{R} = - \frac{y}{R} \end{aligned}

By assumption, there are no shear stresses and no forces acting on the beam’s sides, so the only nonzero component of the Cauchy stress tensor σ^\hat{\sigma} is σzz\sigma_{zz}, given by Hooke’s law:

σzz=Euzz\begin{aligned} \sigma_{zz} = E u_{zz} \end{aligned}

Where EE is the elastic modulus of the material. By Hooke’s inverse law, the other nonzero strain components are as follows, where ν\nu is Poisson’s ratio:

uxx=uyy=νEσzz=νuzz=νyR\begin{aligned} u_{xx} = u_{yy} = - \frac{\nu}{E} \sigma_{zz} = - \nu u_{zz} = \nu \frac{y}{R} \end{aligned}

For completeness, we turn the strain tensor u^\hat{u} into a full displacement field u\va{u}:

ux=νxyRuy=z22R+νy2x22Ruz=yzR\begin{aligned} \boxed{ u_x = \nu \frac{x y}{R} \qquad u_y = \frac{z^2}{2 R} + \nu \frac{y^2 - x^2}{2 R} \qquad u_z = - \frac{y z}{R} } \end{aligned}

By integrating the above strains uii=ui/iu_{ii} = \ipdv{u_i}{i}, we get the components of u\va{u}:

ux=νxyR+fx(y,z)uy=νy22R+fy(x,z)uz=yzR+fz(x,y)\begin{aligned} u_x = \nu \frac{x y}{R} + f_x(y, z) \qquad u_y = \nu \frac{y^2}{2 R} + f_y(x, z) \qquad u_z = - \frac{y z}{R} + f_z(x, y) \end{aligned}

Where fxf_x, fyf_y and fzf_z are integration constants, which we find by demanding that the off-diagonal strains uiju_{ij} are zero. Starting with uxz=0u_{xz} = 0:

0=uxz=12(uxz+uzx)=12(fxz+fzx)\begin{aligned} 0 = u_{xz} = \frac{1}{2} \Big( \pdv{u_x}{z} + \pdv{u_z}{x} \Big) = \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big) \end{aligned}

Here, only fxf_x may depend on zz, and only fzf_z may depend on xx. This equation thus tell us:

fx(y,z)=zg(y)fz(x,y)=xg(y)\begin{aligned} f_x(y, z) = z \: g(y) \qquad \quad f_z(x, y) = - x \: g(y) \end{aligned}

Where g(y)g(y) is an unknown integration constant. Moving on to uxy=0u_{xy} = 0:

0=12(uxy+uyx)=12(νxR+fxy+fyx)\begin{aligned} 0 = \frac{1}{2} \Big( \pdv{u_x}{y} + \pdv{u_y}{x} \Big) = \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big) \end{aligned}

Only fxf_x may contain yy, so its yy-derivative must be a constant, so g(y)=Cyg(y) = C y. Therefore:

fx(y,z)=Cyzfy(x,z)=νx22RCxz+h(z)fz(x,y)=Cxy\begin{aligned} f_x(y, z) = C y z \qquad f_y(x, z) = - \nu \frac{x^2}{2 R} - C x z + h(z) \qquad f_z(x, y) = - C x y \end{aligned}

Where h(z)h(z) is an unknown integration constant. Finally, we put everything in uyz=0u_{yz} = 0:

0=12(uyz+uzy)=12(fyzzR+fzy)=12( ⁣ ⁣2Cx+dhdzzR)\begin{aligned} 0 = \frac{1}{2} \Big( \pdv{u_y}{z} + \pdv{u_z}{y} \Big) = \frac{1}{2} \Big( \pdv{f_y}{z} - \frac{z}{R} + \pdv{f_z}{y} \Big) = \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big) \end{aligned}

Only the first term contains xx, so to satisfy this equation, we must set C=0C = 0. The remaining terms then tell us that h(z)=z2/(2R)h(z) = z^2 / (2 R). Therefore:

fx=0fy=νx22R+z22Rfz=0\begin{aligned} f_x = 0 \qquad f_y = - \nu \frac{x^2}{2 R} + \frac{z^2}{2 R} \qquad f_z = 0 \end{aligned}

Inserting this into the components uxu_x, uyu_y and uzu_z then yields the full displacement field.

In any case, the beam experiences a bending torque with an xx-component TxT_x given by:

Tx=AyσzzdA=ERAy2dA\begin{aligned} T_x = - \int_A y \sigma_{zz} \dd{A} = - \frac{E}{R} \int_A y^2 \dd{A} \end{aligned}

Where AA is the cross-section. Th above integral is known as the area moment, and is typically abbreviated by II. This brings us to the Euler-Bernoulli law:

Tx=EIRIAy2dA\begin{aligned} \boxed{ T_x = - \frac{E I}{R} } \qquad \quad I \equiv \int_A y^2 \dd{A} \end{aligned}

The product EIE I is called the flexural rigidity, i.e. the beam’s “stiffness”. For a small deformation, i.e. a large radius of curvature RR, the law can be approximated by:

TxEId2ydz2\begin{aligned} T_x \approx - E I \dvn{2}{y}{z} \end{aligned}

Slender rods

A beam that is very thin in the transverse directions (xx and yy in this case), can be approximated as a single string or rod y(z)y(z). Each infinitesimal piece (dy,dz)(\dd{y}, \dd{z}) of the rod exerts forces FyF_y and FzF_z on the next piece, and is feels external forces-per-length KyK_y and KzK_z, e.g. gravity. In order to have equilibrium, the total force must be zero:

0=Fy(z+dz)Fy(z)+Ky(z)dz0=Fz(z+dz)Fz(z)+Kz(z)dz\begin{aligned} 0 &= F_y(z + \dd{z}) - F_y(z) + K_y(z) \dd{z} \\ 0 &= F_z(z + \dd{z}) - F_z(z) + K_z(z) \dd{z} \end{aligned}

Rearranging these relations yields these equations for the internal forces FyF_y and FzF_z:

dFydz=KydFzdz=Kz\begin{aligned} \boxed{ \dv{F_y}{z} = - K_y } \qquad \quad \boxed{ \dv{F_z}{z} = - K_z } \end{aligned}

Meanwhile, the rod also feels a torque with xx-component TxT_x, where equilibrium entails:

0=Tx(z+dz)Tx(z)+Fz(z)dyFy(z)dz\begin{aligned} 0 = T_x(z + \dd{z}) - T_x(z) + F_z(z) \dd{y} - F_y(z) \dd{z} \end{aligned}

This can be rearranged to get a differential equation for TxT_x, namely:

dTxdz=FyFzdydz\begin{aligned} \boxed{ \dv{T_x}{z} = F_y - F_z \dv{y}{z} } \end{aligned}

If FzF_z and dy/dz\idv{y}{z} are small, the last term can be dropped. These equations are widely applicable, but there is one especially important application, so much so that it is usually what is meant by “Euler-Bernoulli law”: the shape of a laterally loaded rod.

Consider a beam along the zz-axis, carrying a lateral load KyK_y, e.g. its own weight AgρA g \rho or more. Assuming there is no other load Kz=0K_z = 0 and FyFzF_y \ll F_z, the above equations become:

Tx=EId2ydz2dTxdz=FydFydz=Ky\begin{aligned} T_x = - E I \dvn{2}{y}{z} \qquad \dv{T_x}{z} = F_y \qquad \dv{F_y}{z} = - K_y \end{aligned}

Which we can simply substitute into each other, eventually leading to:

Ky=d2dz2(EId2ydz2)\begin{aligned} \boxed{ K_y = \dvn{2}{}{z}\Big( E I \dvn{2}{y}{z} \Big) } \end{aligned}

This is often referred to as the Euler-Bernoulli law as well. Typically the flexural rigidity EIEI is a constant in zz, in which case we can reduce the equation to:

Ky=EId4ydz4\begin{aligned} K_y = E I \dvn{4}{y}{z} \end{aligned}

Which is clearly solved by a fourth-order polynomial, given some boundary conditions.


  1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.