Encyclopedia of Continuum Mechanics

Living Edition
| Editors: Holm Altenbach, Andreas Öchsner

Beams, Plates, and Shells

  • Michael KrommerEmail author
  • Yury Vetyukov
Living reference work entry
DOI: https://doi.org/10.1007/978-3-662-53605-6_1-1



In this article we shortly present the fundamental relations of the classical theories for straight and slender beams as well as for thin plates and shells. In particular, Bernoulli-Euler beams, Kirchhoff plates, and Kirchhoff-Love shells are discussed in a geometrically and physically linear regime. Further, we restrict the content to homogenous beams, plates, and shells, for which the material behavior is assumed isotropic. Our presentation rests on the use of a direct tensor notation, which avoids unnecessarily lengthy equations using specific coordinates but rather enables a compact and concise formulation.

Bernoulli-Euler Beams

We study straight and slender beams of length with a solid cross section Ω. The beam is homogenous with isotropic linear elastic material behavior. The position vector of a point P of the undeformed beam is \( \underline {R}_3 = \underline {R}(x) + \underline {z}\); x is the axial coordinate with unit base vector \( \underline K\), and \( \underline {z}\) is the position of P within the cross section. Distributed loadings per unit length are the axial force px, transverse forces \( \underline p\), and moments \( \underline m\). At the end faces x = 0, axial forces \(\bar {N}\), transverse forces \(\bar { \underline {Q}}\) and bending moments \(\bar { \underline {M}}\) are applied, see Fig. 1.
Fig. 1

Undeformed straight beam with external loadings


In the small deformation regime, we impose the Bernoulli-Euler kinematic hypothesis (see Ziegler 1995) that the cross section is rigid and it remains normal to the deformed beam axis in the course of deformation; moreover, a rotation of the cross section about the beam axis is not admitted. Hence, the displacement vector of a point P is
$$\displaystyle \begin{aligned} \underline u_3=u(x)\underline K+\underline v(x)+\underline\upvarphi(x)\times\underline z\,,\; (\%)' \equiv \partial / \partial x , \end{aligned} $$
with the axial displacement u and the transverse deflections \( \underline v\) of the beam axis and with the vector of small rotation of the cross section \( \underline \upvarphi \). According to the Bernoulli-Euler kinematic hypothesis, \( \underline \upvarphi \) accounts for the rotation due to bending, \( \underline \upvarphi = \underline K\times \underline v'\). Finally, we introduce strain measures of extension e and linearized curvatures \( \underline \kappa \) and find the strain tensor as the symmetric part of the displacement gradient tensor \((\nabla + \underline K \partial / \partial x) \underline u_3\):
$$\displaystyle \begin{aligned} \begin{array}{rcl} {} &\displaystyle &\displaystyle \underline{\underline{\varepsilon}}_3 = \varepsilon_{xx}\underline K \,\underline K,\quad \varepsilon_{xx}=e+\underline K\cdot\underline\kappa\times\underline z,\quad \notag\\ &\displaystyle &\displaystyle e=u',\underline\kappa=\underline\upvarphi'=\underline K\times\underline v''. \end{array} \end{aligned} $$

Equilibrium Conditions

We derive the equilibrium conditions from the principle of virtual work in the form with δA(e) being the virtual work of the external forces and Π3 the strain energy per unit volume. We adopt the common assumption of a uniaxial stress condition at bending and extension and write \(2\varPi _3=E\varepsilon _{xx}^2\) with E being Young’s modulus. Placing the axis \( \underline z= \underline 0\) in the area center, denoting by \( \underline { \underline {J}}\) the tensor of area moments and integrating over the cross section, we find the strain energy per unit length:
$$\displaystyle \begin{aligned} \varPi&=\int_\varOmega\varPi_3\mathrm{d}\varOmega=\frac{1}{2}E\varOmega e^2+\frac{1}{2}E\underline\kappa\cdot\underline{\underline{J}}\cdot\underline\kappa,\notag\\ \underline{\underline{J}}&=-\underline K\times\int_\varOmega \underline z \, \underline z\,\mathrm{d}\varOmega\times\underline K,\quad \int_\varOmega\underline z\,\mathrm{d}\varOmega=\underline 0. {} \end{aligned} $$
The work of the external forces and moments is
$$\displaystyle \begin{aligned} \delta A^{(e)} &= \int_0^\ell { \left( p_x \delta u + \underline{p} \cdot \delta \underline{v} + \underline{m} \cdot \delta \underline{\upvarphi} \right) \mathrm{d}x}\notag\\ &\quad +\left.\left( \bar{N} \delta u + \bar{\underline{Q}} \cdot \delta \underline{v} + \bar{\underline{M}} \cdot \delta \underline{\upvarphi}\right)\right|{}_0^\ell. \end{aligned} $$
Introducing the Lagrange multiplier \( \underline Q(x)\) to account for the Bernoulli-Euler kinematic constraint of vanishing transverse shear strain, \( \underline {\upvarphi } - \underline K\times \underline v' = \underline 0\), we formally consider the variations \(\delta \underline {v}\) and \(\delta \underline {\upvarphi }\) independent and demand
$$\displaystyle \begin{aligned} &\delta A^{(i)} + \delta A^{(e)} + \delta C=0,\notag\\ &\text{with:} \quad \delta C=\int_0^\ell { \underline Q \cdot \left( \delta \underline{\upvarphi} \times \underline K - \delta \underline{v}' \right) \mathrm{d}x}. \end{aligned} $$
Note that \( \underline \upvarphi - \underline K\times \underline v' = \underline 0\) is identical to \( \underline {\upvarphi } \times \underline K - \underline {v}' = \underline 0\) as \( \underline {v}' \cdot \underline K = 0\) and \( \underline {\upvarphi } \cdot \underline K = 0\) hold. The variational equation (6) along with the strain energy (4) and the virtual work of external forces (5) are equivalent to the below system of equations of equilibrium (7), constitutive relations (8), and boundary conditions (9):
$$\displaystyle \begin{aligned} \begin{array}{rcl} {} N' &\displaystyle +&\displaystyle p_x = 0 ,\quad \underline{Q}' + \underline{p} = \underline{0} ,\notag\\ \underline{M}' &\displaystyle +&\displaystyle \underline K \times \underline{Q} + \underline m = \underline{0}, \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} {} &\displaystyle &\displaystyle N = \frac{\partial\varPi}{\partial e} = E \varOmega u' ,\quad \underline{M} = \frac{\partial\varPi}{\partial\underline{\kappa}} = E \underline{\underline{J}} \cdot \underline \upvarphi', \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} {} x&\displaystyle =&\displaystyle 0,\ell:N = \bar{N} \; \; \mbox{or} \; \; u = \bar{u} ,\notag\\ \underline{Q} &\displaystyle =&\displaystyle \bar{\underline{Q}} \; \; \mbox{or} \; \; \underline{v} = \bar{\underline{v}} ,\quad \underline{M} = \bar{\underline{M}} \; \; \mbox{or} \; \; \underline{\upvarphi} = \bar{\underline{\upvarphi}}. \end{array} \end{aligned} $$
The first type of boundary conditions in (9) is denoted as dynamic boundary conditions, whereas the second type are kinematic ones. We identify N as the axial force, \( \underline {Q}\) as the transverse shear force vector, and \( \underline {M}\) as the bending moment vector. No constitutive relation exists for the shear force for a Bernoulli-Euler beam. For statically determinate problems, one finds \( \underline Q\) from the equations of equilibrium. Otherwise, with the constitutive relations and the kinematic boundary conditions, one arrives at a boundary value problem for the displacements u and \( \underline v\). Using Cartesian coordinates y and z in the directions of principal axes of \( \underline { \underline {J}}\), we obtain the differential equations
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle E \varOmega u'' + p_x = 0,\notag\\ &\displaystyle &\displaystyle E J_y w^{IV} = p_z ,\quad E J_z v^{IV} = p_y, \end{array} \end{aligned} $$
with the deflections v and w in the directions y and z; the distributed moments \( \underline m\) are omitted in these equations. The corresponding boundary conditions are directly derived from (9).

Stress Distribution

The stress tensor \( \underline { \underline {\sigma }}_3=\sigma _{xx} \underline K \, \underline K\) contains only the axial stress σxx due to extension and bending. This axial stress results from N and \( \underline M\) and follows from (2) to
$$\displaystyle \begin{aligned} \sigma_{xx} &= E \left( u' - \underline{v}'' \cdot \underline{z} \right)\notag\\ &= \frac{N}{A} - \left( \underline{\underline{J}}^{-1} \cdot \underline{M} \right) \cdot \left( \underline K \times \underline{z} \right)\notag\\ &= \frac{N}{A} + z \frac{M_y}{J_y} - y \frac{M_z}{J_z}. \end{aligned} $$
Here, we have a mean stress due to the axial force N and a distribution linear in \( \underline {z}\) due to the bending moments \( \underline {M}\). Likewise to the transverse shear force vector \( \underline Q\), the corresponding transverse shear stresses can only be computed from equilibrium conditions. The solution of this problem belongs to the field of linearized elasticity and is therefore not presented here; for a detailed treatment, we refer to Lurie (2005).

Kirchhoff Plates

We study thin plates, which are homogenous with isotropic linear elastic material behavior. The position vector \( \underline {R}_3\) of an arbitrary point P of the plate in the undeformed configuration is \( \underline {R}_3 = \underline {R} + z \underline {K}\). \( \underline {R}\) is the position of a point P0 in a plane reference surface within the plate; the mid-surface of the plate is taken as this reference surface. \( \underline {K}\) is the constant unit vector normal to the reference surface, and z is the distance of a point P from the reference surface. We apply distributed forces \( \underline p = \underline p_{\scriptscriptstyle \bot } + p_z \underline K\) (with the in-plane part \( \underline p_{\scriptscriptstyle \bot }\) and the transverse force pz) as well as distributed in-plane moments \( \underline m = \underline m_{\scriptscriptstyle \bot }\) per unit area Ω of the plate. Forces \(\bar { \underline P} = \bar { \underline P}_{\scriptscriptstyle \bot } + \bar {P}_z \underline {K}\) and in-plane moments \(\bar { \underline M} = \bar { \underline M}_{\scriptscriptstyle \bot }\) per unit length act at the boundary ∂Ω. (%) refers to the in-plane part of either a vector or a tensor.


The kinematics is determined by the Kirchhoff kinematic hypothesis; see, e.g., the textbook (Eschenauer et al., 1997). The position vector in the deformed configuration is \( \underline {r}_3 = \underline {r} + z \underline {k}\) with \( \underline r\) being the position vector of P0 in the deformed configuration and \( \underline {k}\) the unit vector normal to the deformed reference surface. Hence, any material line that was orthogonal to the reference surface in the undeformed configuration remains orthogonal to the deformed reference surface, and its length is not changed in the course of deformation. At infinitesimal deformation, the linearized increment in the unit normal vector \( \underline k- \underline K\) is an in-plane vector \( \underline \upvarphi = \underline \omega \times \underline K\) with \( \underline \omega \) being the vector of small rotation. The displacement vector can now be written as \( \underline {u}_3 = \underline {u} + z \underline \upvarphi \), in which \( \underline {u} = \underline {u}_{\scriptscriptstyle \bot }( \underline {R}) + w( \underline {R}) \underline K\) is the displacement vector of points of the reference surface, which is composed of the in-plane displacements \( \underline {u}_{\scriptscriptstyle \bot }\) and the deflection w.

We introduce the differential operator as \(\nabla _3 = \nabla + \underline K (\partial / \partial z)\) and compute the linearized strain tensor to \( \underline { \underline {\varepsilon }}_3 = \underline { \underline {e}} + z \underline { \underline {\kappa }} + ( \underline K \underline \gamma + \underline \gamma \underline K )/2\) with the reference surface strain tensor \( \underline { \underline {e}}\), the linearized curvature tensor \( \underline { \underline {\kappa }}\), and the shear strain vector \( \underline \gamma \):
$$\displaystyle \begin{aligned} \underline{\underline{e}} = \left( \nabla \underline{u}_{\scriptscriptstyle \bot} \right)^S ,\quad \underline{\underline{\kappa}} = ( \nabla \underline \upvarphi )^S, \quad \underline \gamma = \underline \upvarphi + \nabla w. \end{aligned} $$
The Kirchhoff kinematic hypothesis imposes the constraint \( \underline \gamma = \underline \upvarphi + \nabla w = \underline 0\); hence, \( \underline \upvarphi = - \nabla w\), and just the plane components remain in the strain tensor:
$$\displaystyle \begin{aligned} \underline{\underline{\varepsilon}}_3 \equiv \underline{\underline{\varepsilon}} = \underline{\underline{e}} + z \underline{\underline{\kappa}} = \left( \nabla \underline{u}_{\scriptscriptstyle \bot} \right)^S - z \nabla \nabla w. \end{aligned} $$

Equilibrium Conditions

The equilibrium conditions result from the principle of virtual work analogous to Eq. (16) used for beams. Within a plane stress assumption (\( \underline { \underline {\sigma }}_3 {\equiv } \underline { \underline {\sigma }}\)) and with the plane stress Young’s modulus Y = E∕(1 − ν2), the strain energy Π3 for the isotropic plate reads
$$\displaystyle \begin{aligned} 2 \varPi_3 = Y \nu\, \mathrm{tr} \underline{\underline{\varepsilon}}\, \mathrm{tr} \underline{\underline{\varepsilon}} + Y \left( 1 - \nu \right) \underline{\underline{\varepsilon}} \cdot \cdot \, \underline{\underline{\varepsilon}}. \end{aligned} $$
Integration with respect to the thickness renders the strain energy Π per unit area as the sum of the membrane energy Πm and the bending energy Πb:
$$\displaystyle \begin{aligned} \varPi_m &= \frac{1}{2} A \left( \nu \mathrm{tr} \underline{\underline{e}}\, \mathrm{tr} \underline{\underline{e}} + \left( 1 - \nu \right) \underline{\underline{e}} \cdot \cdot \, \underline{\underline{e}} \right) \notag\\ &= {\varPi}_m (\underline{e} = \left(\nabla \underline{u}_{\scriptscriptstyle \bot} \right)^S), \notag\\ \varPi_b&= \frac{1}{2} D \left( \nu \mathrm{tr} \underline{\underline{\kappa}}\, \mathrm{tr} \underline{\underline{\kappa}} + \left( 1 - \nu \right) \underline{\underline{\kappa}} \cdot \cdot \, \underline{\underline{\kappa}} \right) \notag\\ &= {\varPi}_b ( \underline{\kappa} = ( \nabla \underline \upvarphi )^S ), {} \end{aligned} $$
with the membrane (extensional) stiffness Open image in new window and the plate (bending) stiffness D = Eh3∕12(1 − ν2). We write the virtual work of the external loadings as In the principle of virtual work, we assume the variations \(\delta \underline {u}\), δw and \(\delta \underline {\upvarphi }\) independent. Hence, similar to (6), we account for the constraint \( \underline \upvarphi + \nabla w = \underline 0\) by means of the vector-valued Lagrange multiplier \( \underline {Q}\) with the augmentation term
$$\displaystyle \begin{aligned} \delta C = - \int_{\varOmega} { \underline{Q} \cdot \left( \delta\underline \upvarphi + \nabla \delta w \right) \mathrm{d}\varOmega} \end{aligned} $$
in the principle of virtual work. The Euler equations of the augmented principle are
$$\displaystyle \begin{aligned} \nabla \cdot \underline{\underline{N}} &+ \underline{p}_{\scriptscriptstyle \bot} = \underline{0} \; , \quad \nabla \cdot \underline{Q} + p_z = 0 \; , \notag\\ \nabla \cdot \underline{\underline{M}} &- \underline{Q} + \underline{m} \times \underline K = \underline{0}, \end{aligned} $$
with constitutive relations for the in-plane force tensor \( \underline { \underline {N}}\) and the moment tensor \( \underline { \underline {M}}\)
$$\displaystyle \begin{aligned} \underline{\underline{N}} &= \frac{\partial \varPi_m}{\partial \underline{\underline{e}}} = A \nu \underline{\underline{I}}_{\scriptscriptstyle \bot} \text{tr} \underline{\underline{e}} + A \left( 1 - \nu \right) \underline{\underline{e}} \; , \notag\\ \underline{\underline{M}} &= \frac{\partial \varPi_b}{\partial \underline{\underline{\kappa}}} = D \nu \underline{\underline{I}}_{\scriptscriptstyle \bot} \text{tr} \underline{\underline{\kappa}} + D \left( 1 - \nu \right) \underline{\underline{\kappa}}, \end{aligned} $$
and the boundary conditions at ∂Ω. \( \underline {n}\) and \( \underline {s}\) are the outer unit normal vector and the tangential vector of ∂Ω, \( \underline {Q} \cdot \underline {n} + \nabla \left ( \underline { \underline {M}} \cdot \underline {n} \cdot \underline {s} \right ) \cdot \underline {s}\) is the Kirchhoff shear force, and \(\bar {M}_{n} = \left ( \bar { \underline M} \times \underline K \right ) \cdot \underline {n}\) and \(\bar {M}_{s} = \left ( \bar { \underline M} \times \underline K \right ) \cdot \underline {s}\) are the bending moment and twisting moment acting at the boundary. Moreover, at any corner point Pi of the boundary, either \([\,[ \underline { \underline {M}} \cdot \underline {n} \cdot \underline {s} - \bar {M}_s ]\,]_{P_i} = 0\) or \(w = \bar {w}\) must be satisfied. Note that our formulation does not account for concentrated transverse forces at the boundary; for a discussion of the latter, we refer the reader to, e.g., Timoshenko and Woinowsky-Krieger (1959). The transverse shear force vector \( \underline Q\) must again be computed from the equilibrium conditions as there is no constitutive relation available. The overall problem splits into two problems. The first one – the in-plane problem of a disk involving \( \underline { \underline {N}}\), \( \underline {u}_{\scriptscriptstyle \bot }\), \( \underline {p}_{\scriptscriptstyle \bot }\), and \( \underline P_{\scriptscriptstyle \bot }\) – is a problem of the two-dimensional linearized theory of elasticity, and the second one, the bending problem, results into the well-known fourth-order Kirchhoff plate equation (distributed moments \( \underline {m}\) are omitted):
$$\displaystyle \begin{aligned} D \Delta \Delta w = p_z. \end{aligned} $$
Concerning a detailed discussion of the boundary conditions and the corner forces, we must refer to the exhaustive literature on elastic plates due to space limitations, e.g., Timoshenko and Woinowsky-Krieger (1959).

Stress Distribution

The distribution of the stress through the thickness is computed from the plane stress assumption \( \underline { \underline {\sigma }} = Y \nu \, \underline { \underline {I}}_{\scriptscriptstyle \bot } \mathrm {tr} \underline { \underline {\varepsilon }} + Y(1 - \nu ) \underline { \underline {\varepsilon }}\). Keeping in mind the constitutive relation for the in-plane force tensor \( \underline { \underline {N}}\) and for the moment tensor \( \underline { \underline {M}}\)
$$\displaystyle \begin{aligned}\begin{array}{l}\displaystyle \underline{\underline{N}} = A \nu \underline{\underline{I}}_{\scriptscriptstyle \bot} \text{tr} \underline{\underline{e}} + A \left( 1 - \nu \right) \underline{\underline{e}} , \\ {} \displaystyle \underline{\underline{M}} = D \nu \underline{\underline{I}}_{\scriptscriptstyle \bot} \text{tr} \underline{\underline{\kappa}} + D \left( 1 - \nu \right) \underline{\underline{\kappa}} , \end{array}\end{aligned} $$
the distribution of the plane stress tensor follows to
$$\displaystyle \begin{aligned} \underline{\underline{\sigma}} = \frac{1}{h} \underline{\underline{N}} + z\frac{12}{h^3} \underline{\underline{M}}. {} \end{aligned} $$
The shear stress vector \( \underline {\tau }\) is post-computed from the equilibrium conditions without imposing the plane stress assumption. With \( \underline { \underline {\sigma }}_3 = \underline { \underline {\sigma }} + \underline {\tau } \, \underline K + \underline K \, \underline {\tau } + \sigma _{33} \underline K \, \underline K\), we have the equilibrium conditions
$$\displaystyle \begin{aligned} \nabla \cdot \underline{\underline{\sigma}} + \partial \underline{\tau} / \partial z + \underline b_{\scriptscriptstyle \bot} = \underline 0, \end{aligned} $$
from which we find the shear stress vector \( \underline {\tau }\) to
$$\displaystyle \begin{aligned} \begin{array}{rcl} \underline{\tau} &\displaystyle =&\displaystyle \frac{3}{2 h} \underline Q \left(1 - \left( \frac{2 z}{h}\right)^2 \right) \notag\\ &\displaystyle +&\displaystyle \int_{-h/2}^{z}{ \left( \frac{\underline{p}_{\scriptscriptstyle \bot}}{h} + \zeta \frac{12}{h^3} \underline{m} - \underline b_{\scriptscriptstyle \bot} \right) \mathrm{d}\zeta}, {} \end{array} \end{aligned} $$
because the top and bottom surfaces of the plate at z = ±h∕2 are shear stress free. The first term accounts for the transverse shear force and the integral vanishes, if the in-plane part of the body force vector \( \underline b_{\scriptscriptstyle \bot }\) is linearly distributed through the thickness. In the latter case, the shear stress has a maximum at the thickness location of the reference surface (the plate mid-surface) with z = 0, which is 1.5 times the mean transverse shear force. For an asymptotic validation of the theory of thin plates, we refer to Eliseev (2006) and Vetyukov (2014).

Kirchhoff-Love Shells

Below we briefly present the basics of a linear theory of thin shells. In contrast to a thin plate in Fig. 2, the reference surface of a shell is a curved surface. We assume the same types of external loading as in the plate theory; yet, we do not split the distributed force \( \underline p\) and the force \(\bar { \underline P}\) applied at the boundary into an in-plane part and an out-of-plane part. Likewise, this applies to the displacement vector \( \underline u\) of points P0 of the curved reference surface. Also, the Kirchhoff kinematic hypothesis and the plane stress assumption are assumed to hold for a thin shell as well; then, the corresponding shell theory is denoted as Kirchhoff-Love shell theory.
Fig. 2

Plate kinematics


The kinematics of a thin shell is based on the standpoint of Koiter (1970), which was further developed by Eliseev (2010) and his co-workers. For a detailed discussion, see the textbooks Eliseev (2006) and Vetyukov (2014). Two strain tensors are introduced for a thin shell – the reference surface strain tensor \( \underline { \underline {e}}\) and the curvature tensor \( \underline { \underline {\kappa }}\); both must be invariant for a rigid body motion. This condition is fulfilled, if the metric coefficients of the reference surface remain constant. In a geometrically linear setting, we satisfy this requirement by the strain tensors
$$\displaystyle \begin{aligned} &\underline{\underline{e}} = \left( \nabla \underline{u} \right)_{\scriptscriptstyle \bot}^S , \notag\\ &\underline{\underline{\kappa}} = ( \nabla \underline \upvarphi )_{\scriptscriptstyle \bot} - \underline{\underline{B}} \cdot (\nabla \underline u)^T = - (\nabla \nabla \underline u \cdot \underline K)_{\scriptscriptstyle \bot}. \end{aligned} $$
where ∇ is the differential operator of the undeformed reference surface (for its definition in the general case of differential geometry of a parameterized surface, see Vetyukov 2014), \( \underline {u}\) is the displacement vector of points of this reference surface, \( \underline \upvarphi \) is the in-plane increment of the unit normal vector, \( \underline { \underline {B}} = - \nabla \underline K\) the second metric tensor of the undeformed reference surface, and (%) refers to that part of a tensor, which lies in the tangential plane of the reference surface in the undeformed configuration. The two relations for \( \underline { \underline {\kappa }}\) follow from the Kirchhoff kinematic constraint \( \underline \upvarphi + \nabla \underline u \cdot \underline K = \underline 0\). The above relations reduce to the ones for Kirchhoff plates for \( \underline { \underline {B}} = \underline { \underline {0}}\).

Constitutive Relations

The constitutive relations are assumed to be the same as in plate theory. The membrane and bending energies are stated in (15), and the constitutive relations are
$$\displaystyle \begin{aligned} \underline{\underline{N}} &= \frac{\partial \varPi_m}{\partial \underline{\underline{e}}} = A \nu \underline{\underline{A}} \text{tr} \underline{\underline{e}} + A \left( 1 - \nu \right) \underline{\underline{e}} ,\notag\\ \underline{\underline{M}} &= \frac{\partial \varPi_b}{\partial \underline{\underline{\kappa}}} = D \nu \underline{\underline{A}} \text{tr} \underline{\underline{\kappa}} + D \left( 1 - \nu \right) \underline{\underline{\kappa}}, \end{aligned} $$
if the strain tensors of the shell theory are inserted. \( \underline { \underline {A}} = \underline { \underline {I}} - \underline K \, \underline K= \underline { \underline {I}}_{\bot }\) is the first metric tensor of the undeformed reference surface. Hence, \( \underline { \underline {N}}\) and \( \underline { \underline {M}}\) are symmetric tensors in the tangential plane of the undeformed reference surface.

Equilibrium Conditions

The equilibrium conditions follow again from the principle of virtual work (6) with
$$\displaystyle \begin{aligned} \delta A^{(e)} &= \int_{\varOmega} { \left(\underline{p} \cdot \delta \underline{u} + \underline{m} \times \underline K \cdot \delta \underline{\upvarphi} \right) \mathrm{d}\varOmega} \notag\\ &\quad + \int_{\partial \varOmega} { \left( \bar{\underline P} \cdot \delta \underline{u} + \bar{\underline M} \times \underline K \cdot \delta \underline{\upvarphi} \right) \mathrm{d}s}, \notag\\ \delta A^{(i)} &= - \int_{\varOmega} { \left( \underline{\underline{N}} \cdot \cdot \, \delta \underline{\underline{e}} + \underline{\underline{M}}\cdot \cdot \, \delta \underline{\underline{\kappa}} \right) \mathrm{d}\varOmega}, \notag\\ \delta C &= - \int_{\varOmega} { \underline Q \cdot \left( \delta \underline \upvarphi + \nabla \delta \underline u \cdot \underline K \right) \mathrm{d}\varOmega}, \end{aligned} $$
with the shear force vector \( \underline Q\) as a Lagrange multiplier. We derive the Euler equations
$$\displaystyle \begin{aligned} &\nabla \cdot \underline{\underline{T}} + \underline p = \underline 0 \quad \text{and} \notag\\ &(\nabla \cdot \underline{\underline{M}}) \cdot \underline{\underline{A}}- \underline Q + \underline{m} \times \underline K = \underline 0, \end{aligned} $$
with the abbreviation \( \underline { \underline {T}} = \underline { \underline {N}} - \underline { \underline {M}} \cdot \underline { \underline {B}} \ + \underline Q \underline K\) and the boundary conditions We omit the discussion of corner forces for thin shells, but rather refer to Vetyukov (2014) for details. For a plane reference surface, \( \underline { \underline {A}} = \underline { \underline {I}}_{\scriptscriptstyle \bot }\) and \( \underline { \underline {B}} = \underline { \underline {0}}\) hold, and we recover the plate equations. Likewise to plates, the transverse shear force vector \( \underline Q\) must be computed from the equilibrium conditions.

Stress Distribution

The through-the-thickness distribution of the stresses is computed from the stress recovery relations (22) and (24), which were derived for plates but are also commonly accepted for thin isotropic shells.


In contrast to beams and plates, no simple equations can be derived from the set of governing equations, unless one makes assumptions concerning the geometry and deformation of the specific shell. The simplest case is the axisymmetric deformation of a circular cylindrical shell, which we shortly discuss here. We set Open image in new window from the very beginning. The undeformed thin shell is a cylinder with radius a of the reference surface, length , and constant thickness h. The differential operator of the reference surface is \(\nabla = a^{-1} \underline e_{\phi } (\partial / \partial \phi ) + \underline e_{x} (\partial / \partial x)\); x and ϕ are the coordinates on the surface. \( \underline e_x\) is the unit vector in axial direction, \( \underline e_{\phi }\) the unit vector in circumferential direction, and \( \underline e_r\) the one in radial direction; the formulas
$$\displaystyle \begin{aligned} \partial \underline e_{x} / \partial \phi = \underline 0,\ \ \partial \underline e_{\phi}/\partial \phi = - \underline e_r ,\ \ \partial \underline e_r/\partial \phi = \underline e_{\phi}, \end{aligned} $$
apply. Due to the rotational symmetry of the problem, we have
$$\displaystyle \begin{aligned} \underline K &= \underline e_r \; , \quad \underline{\underline{A}} = \underline e_x \underline e_x + \underline e_{\phi} \underline e_{\phi} \; , \notag\\ \underline{\underline{B}} &= - a^{-1} \underline e_{\phi} \underline e_{\phi} ,\quad \underline u = u \underline e_x + w \underline e_r ,\notag\\ \underline \upvarphi &= \upvarphi \underline e_x \; , \quad \underline p = p_x \underline e_x + p_r \underline e_r , \\ \underline Q &= q \underline e_x \; , \quad \underline{\underline{N}} = n_{\phi \phi} \underline e_{\phi} \underline e_{\phi} + n_{xx} \underline e_x \underline e_x \; , \\ \underline{\underline{M}} &= m_{\phi \phi} \underline e_{\phi} \underline e_{\phi} + m_{xx} \underline e_x \underline e_x. \end{aligned} $$
We obtain the equilibrium conditions
$$\displaystyle \begin{aligned} &\nabla \cdot \underline{\underline{T}} + \underline p = (n_{xx}^{\prime} + p_x ) \underline e_x \notag\\ &\quad + ( q' - a^{-1} ( n_{\phi \phi} + a^{-1} m_{\phi \phi} ) + p_r) \underline e_r = \underline 0 , \\ &\quad (\nabla \cdot \underline{\underline{M}}) \cdot \underline{\underline{A}} - \underline Q = ( m_{xx}^{\prime} - q) \underline e_x = \underline 0, \end{aligned} $$
with (%) = (∂x), the kinematic constraint as well as the strain tensors
$$\displaystyle \begin{aligned} \underline 0 &= \underline \upvarphi + \nabla \underline u \cdot \underline K = (\upvarphi + w' ) \underline e_x \ \rightarrow \ \upvarphi = - w' , \\ \underline{\underline{e}} &= \left( \nabla \underline{u} \right)_{\bot}^S = u' \underline e_{x} \underline e_x + a^{-1} w \underline e_{\phi} \underline e_{\phi} , \\ \underline{\underline{\kappa}} &= ( \nabla \underline \upvarphi )_{\bot} - \underline{\underline{B}} \cdot (\nabla \underline u)^T = \upvarphi' \underline e_{x} \underline e_x + a^{-2} w \underline e_{\phi} \underline e_{\phi} \notag\\ &= - w'' \underline e_{x} \underline e_x + a^{-2} w \underline e_{\phi} \underline e_{\phi} , \end{aligned} $$
and the constitutive relations
$$\displaystyle \begin{aligned} \underline{\underline{N}} &= A (u' + \nu a^{-1} w)\underline e_{x} \underline e_x \notag\\ &\quad + A (a^{-1} w + \nu u' ) \underline e_{\phi} \underline e_{\phi} , \\ \underline{\underline{M}} &= D (- w'' + \nu a^{-2} w) \underline e_x \underline e_x \notag\\ &\quad + D (a^{-2} w - \nu w'') \underline e_{\phi} \underline e_{\phi} . \end{aligned} $$


From the equilibrium of moments, we have \(m_{xx}^{\prime } - q = 0\); differentiating with respect to x and inserting the constitutive relation, we obtain
$$\displaystyle \begin{aligned} D w^{{IV}} - D \nu a^{-2} w'' + q' = 0. \end{aligned} $$
With q′ = a−1(nϕϕ + a−1mϕϕ) − pr, the constitutive relations for nϕϕ and mϕϕ and with u′ = A−1nxx − νa−1w
$$\displaystyle \begin{aligned} &a^2 h^2 \left( w'' - 2 \nu \frac{w}{a^2} \right)'' + \left(12 (1 - \nu^2 ) + \frac{h^2}{a^2} \right) w \notag\\ &\quad = a^2 h^2 \left( \frac{p_r}{D} - \nu \frac{n_{xx}}{a D} \right) \end{aligned} $$
holds. Typically, the second term in each of the two brackets on the left-hand side is assumed negligible, such that we eventually end up with For nxx = 0 this equation is denoted as boiler formula in the literature. The two equations and the three boundary conditions at the each of the two ends at x = 0, complete the 6th-order linear boundary value problem.Open image in new window



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© Springer-Verlag GmbH Germany, part of Springer Nature 2019

Authors and Affiliations

  1. 1.Institute of Mechanics and MechatronicsWienAustria

Section editors and affiliations

  • Franz G. Rammerstorfer
    • 1
  • Melanie Todt
    • 2
  • Isabella C. Skrna-Jakl
    • 3
  1. 1.Institut für Leichtbau und Struktur-BiomechanikTU WienWienAustria
  2. 2.Institut für Leichtbau und Struktur-BiomechanikTU WienWienAustria
  3. 3.Institut für Leichtbau und Struktur-BiomechanikTU WienWienAustria