Shell elements are widely used in finite element analysis for modelling thin and moderately thick structures such as plates, slabs, walls, tanks, and curved surfaces.The key idea of a shell formulation is the reduction of a 3D continuum into a 2D mid-surface representation, while still capturing:
• in-plane (membrane) behavior,
• out-of-plane bending behavior,
• transverse shear deformation.
This makes shell elements computationally efficient compared to full 3D solid elements.
A standard Reissner--Mindlin shell element has five degrees of freedom per node:
$$\{u, v, w, \theta_x, \theta_y\}$$where:
• $u, v$ are in-plane displacements,
• $w$ is the transverse displacement,
• $\theta_x, \theta_y$ are rotations of the normal vector.
The Reissner--Mindlin hypothesis assumes that normals to the mid-surface remain straight but not necessarily perpendicular after deformation.The displacement field is expressed as:
$$u(x,y,z) = u_0(x,y) + z \, \theta_y(x,y)$$ $$v(x,y,z) = v_0(x,y) - z \, \theta_x(x,y)$$ $$w(x,y,z) = w_0(x,y)$$This formulation allows transverse shear deformation to be included.The total strain field is decomposed into three parts. Membrane strains describe in-plane deformation:
$$\varepsilon_x = \frac{\partial u}{\partial x}, \quad \varepsilon_y = \frac{\partial v}{\partial y}, \quad \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}$$Bending strains are associated with rotations of the normal:
$$\kappa_x = -\frac{\partial \theta_x}{\partial x}, \quad \kappa_y = -\frac{\partial \theta_y}{\partial y}$$ $$\kappa_{xy} = -\left( \frac{\partial \theta_x}{\partial y} + \frac{\partial \theta_y}{\partial x} \right)$$Transverse shear strains
$$\gamma_{xz} = \theta_y + \frac{\partial w}{\partial x}, \quad \gamma_{yz} = -\theta_x + \frac{\partial w}{\partial y}$$For an isotropic linear elastic material, the shell response is split into three independent parts.
Membrane behavior (plane stress)
$$\boldsymbol{\sigma}_m = \mathbf{D}_m \boldsymbol{\varepsilon}_m$$where:
$$\mathbf{D}_m = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix}$$with:
$$\mathbf{D}_b = \frac{E t^2}{12(1-\nu^2)} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix}$$where:
$$\mathbf{D}_s = \kappa_s G \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad G = \frac{E}{2(1+\nu)}$$and $\kappa_s \approx \frac{5}{6}$ is the shear correction factor.
The finite element stiffness matrix of a shell element is obtained by integration over the mid-surface:
$$\mathbf{K}_e = \int_A \left( \mathbf{B}_m^T \mathbf{D}_m \mathbf{B}_m + \mathbf{B}_b^T \mathbf{D}_b \mathbf{B}_b + \mathbf{B}_s^T \mathbf{D}_s \mathbf{B}_s \right) t \, dA$$where:
• $A$ is the mid-surface area,
• $t$ is the shell thickness,
• $\mathbf{B}_m$ relates displacements to membrane strains,
• $\mathbf{B}_b$ relates rotations to bending curvatures,
• $\mathbf{B}_s$ relates displacements/rotations to shear strains.
The shell stiffness can be interpreted as a combination of three mechanisms:
• membrane stiffness (in-plane stretching),
• bending stiffness (out-of-plane curvature),
• shear stiffness (transverse deformation).
Scaling of stiffness contributions:
$$K_{\text{membrane}} \sim \frac{E t}{L}, \quad K_{\text{shear}} \sim \frac{G t}{L}, \quad K_{\text{bending}} \sim \frac{E t^3}{L^3}$$This shows that bending stiffness is highly sensitive to thickness.
An alternative to the Reissner--Mindlin formulation is the classical Kirchhoff--Love shell theory, which is valid for thin shells.
In Kirchhoff--Love theory, normals to the mid-surface:
• remain straight,
• remain perpendicular to the deformed mid-surface,
• do not experience transverse shear deformation.
This implies:
$$\gamma_{xz} = \gamma_{yz} = 0$$Because shear deformation is neglected:
• rotations are fully defined by displacement gradients,
• fewer degrees of freedom are required,
• formulation is accurate only for thin shells
Rotations are not independent variables:
$$\theta_x = -\frac{\partial w}{\partial y}, \quad \theta_y = \frac{\partial w}{\partial x}$$Thus the displacement field reduces to:
$$u = u_0(x,y) - z \frac{\partial w_0}{\partial x}, \quad v = v_0(x,y) - z \frac{\partial w_0}{\partial y}$$