Once the finite element formulation has been established and the global system of equations has been solved, the primary unknowns—nodal displacements—are obtained. However, in structural analysis, the quantities of engineering interest are typically strains, stresses, and internal forces. These are derived from the displacement field through kinematic relations and constitutive laws. This section summarizes the evaluation of these physical quantities for truss elements and frame elements.
In this lecture, we use these displacements to compute:
• internal forces (section forces),
• trains,
• stresses
The computation of physical quantities follows these steps:
• Extract element nodal displacements from the global displacement vector.
• Transform displacements to the local coordinate system (if necessary)
• Compute strains using displacement-strain relations.
• Compute stresses using constitutive equations.
• Evaluate internal forces using stress resultants or direct force-displacement relations.
For a truss element, only axial deformation is considered. The axial strain is:
$$\varepsilon = \frac{du}{dx}$$Using Hooke's law we can obtainted the stress:
$$\sigma = E \varepsilon$$where $E$ is Young's modulus. The axial force in the element is:
$$N = A \sigma = AE \varepsilon$$where $A$ is the cross-sectional area. For truss elements, stress is constant along the element if linear shape functions are used.
The bending moment is related to curvature by:
$$\kappa = -\frac{d^2 w}{dx^2}$$where $w(x)$ is the transverse displacement (deflection of the beam). This is the Euler-Bernoulli constitutive relation for bending.
$$M = EI \kappa$$where $I$ is the second moment of area.
For a 2D beam element, the nodal displacement vector in global coordinates is:
$$\u =\begin{bmatrix} u_1 & u_1 & u_3 & u_4 & u_5 & u_6 \end{bmatrix}^T$$Here ($u_1$,$u_2$,$u_3$) and ($u_4$,$u_5$,$u_6$) denote axial, transverse, and rotational DOFs respectively.
To perform element-level calculations, we transform it into the local coordinate system:
$$\mathbf{u}^{(loc)} = \mathbf{T} \mathbf{u}$$where $\mathbf{T}$ is the transformation matrix.The transformation ensures consistency between global geometry and local element formulation. The displacement field inside the element is approximated using shape functions.
with linear shape functions:
$$N_1^u(x) = 1 - \frac{x}{L^e}, \quad N_4^u(x) = \frac{x}{L^e}$$Transverse displacement (Euler-Bernoulli beam)
$$w(x) = N_2(x)\,u_2 + N_3(x)\, u_3 + N_5(x)\,u_5 + N_6(x)\,u_6$$Hermite interpolation is used to ensure continuity of displacement and slope.
$$N_2(x) = 1 - 3\left(\cfrac{x-x_1}{L^e}\right)^2 + 2\left(\cfrac{x-x_1}{L^e}\right)^3 \quad N_3(x) = (x-x_1)\left(1-\cfrac{x-x_1}{L^e}\right)^2$$ $$N_5(x) = 3\left(\cfrac{x-x_1}{L^e}\right)^2 -2\left(\cfrac{x-x_1}{x_2-x_1}\right)^3 \quad N_6(x) = (x-x_1)\left[\left(\cfrac{x-x_1}{L^e}\right)^2 - \cfrac{x-x_1}{L^e}\right]$$The internal forces can be written in compact form:
$$\mathbf{s}(x) = \mathbf{D} \mathbf{B}(x) \mathbf{u}^{(loc)}$$where:
• $\mathbf{B}(x)$ - strain-displacement matrix,
• $\mathbf{D}$ - material matrix:
This is the fundamental constitutive relation in finite element post-processing.
$$\mathbf{D} = \begin{bmatrix} EA & 0 \\ 0 & EI \end{bmatrix}$$At any point $x$ along the beam:
$$\mathbf{s}(x) = \begin{bmatrix} N(x) \\ M(x) \end{bmatrix}$$The strain--displacement matrix in derivative form is:
$$\mathbf{B}(x) = \begin{bmatrix} \dfrac{dN_1^u}{dx} & 0 & 0 & \dfrac{dN_4^u}{dx} & 0 & 0 \\[8pt] 0 & -\dfrac{d^2 N_2}{dx^2} & -\dfrac{d^2 N_3}{dx^2} & 0 & -\dfrac{d^2 N_5}{dx^2} & -\dfrac{d^2 N_6}{dx^2} \end{bmatrix}$$First row corresponds to axial strain, second row to curvature. Using explicit derivatives, this becomes:
$$\mathbf{B}(x) =\begin{bmatrix} -\dfrac{1}{L} & 0 & 0 & \dfrac{1}{L} & 0 & 0 \\[10pt] 0 & \dfrac{12x}{L^3} - \dfrac{6}{L^2} & \dfrac{6x}{L^2} - \dfrac{4}{L} & 0 & -\dfrac{12x}{L^3} + \dfrac{6}{L^2} & \dfrac{6x}{L^2} - \dfrac{2}{L} \end{bmatrix}$$In practice, internal forces are evaluated at several points along the element:
$$x_i \in [0, L^e]$$This allows reconstruction of force diagrams.
The normal stress due to axial force and bending:
$$\sigma(x, y) = \frac{N(x)}{A} + \frac{M(x) y}{I}$$where $y$ is the distance from the neutral axis.