Reference Pages
Mechanical Engineering

    Scroll back to top

    Bending

    Transverse loads applied to a horizontal straight beam will cause it to deflect primarily up or down. This type of deformation is referred to as bending. A beam in bending will develop normal (tensile and compressive) bending stresses throughout the beam. The magnitude of these stresses depends on the internal bending moment in the beam as well as the beam's cross-section geometry.

    Stress Tensor from Pure Bending

    For pure bending loading conditions, the stress tensor can be simplified as shown below, depending on whether bending occurs around the y or z axis.

    Bending stress tensor.
    $$ \begin{align} T= \begin{bmatrix} \frac{M_zy}{I_z} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ T = \begin{bmatrix} \frac{M_yz}{I_y} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \ \end{align} $$

    Boundary Conditions

    Below are common support conditions and loading conditions that are used to model beams.

    Pin supports allow rotation but not translation; fixed supports restrain both translation and rotation of the beam at that location.

    Statically Determinate Beams
    Statically Indeterminate Beams
    Loadings

    Pure Bending

    Take a flexible strip, such as a thin ruler, and apply equal forces with your fingers as shown. Each hand applies a couple or moment (equal and opposite forces a distance apart). The couples of the two hands must be equal and opposite. Between the thumbs, the strip has deformed into a circular arc. For the loading shown here, just as the deformation is uniform, so the internal bending moment is uniform,

    Interactive Visualization

    The following interactive visualization allows you to explore how changing the applied bending moment affects the axial strain and bending deformation of a beam. Left: bending strain. Right: deflection magnitude. Move the slider to change the load in both views.

    Material: \( E = 2 \text{ GPa}, \ \nu = 0.25 \). Geometry: \( L = 2 \text{ m} \), width = 0.24 m, height = 0.32 m, \( I = \frac{bh^3}{12} = 6.55 \times 10^{-4} \text{ m}^4 \). Tip deflection: \( |w(L)| = \frac{FL^3}{3EI} \).

    Drag to rotate · scroll to zoom · move the slider to change the load. Orange arrow = applied load direction. Dashed gray = original position. Color and reported values are physical; the displayed motion is magnified.

    Geometry of Deformation

    Bending diagram
    Assumptions:
    • Plane sections remain plane \( \rightarrow \)no shear stress/strains.
      $$ \gamma_{xy} = \gamma_{xz} = 0\ $$
      Therefore:
      $$ \tau_{xy} = \tau_{xz} = 0\ $$
      Also, traction free boundary conditions yields...
      $$ \sigma_y = \sigma_z = \tau_{yz} = 0\ $$
    • There is a Neutral axis between the top and the bottom where the length does not change.
      $$ \sigma_x = 0 \rm\ and \rm\ \epsilon_x = 0\ $$
    • The beam deforms into a circular arc where the top surface (\( AB \)) is in compression
      $$ \sigma_x, \varepsilon_x <0 $$
      , and the bottom surface (\( AB ^\prime \)) is in tension
      $$ \sigma_x, \varepsilon_x >0 $$
      .
    • Any point in the beam is in a state of uniaxial normal stress.
    • Finding stresses is a statically indeterminate problem.

    Stress-Strain Variations

    The magnitudes of stress and strain vary along the cross section. The magnitudes increase as the point of intrest moves away from the neutral axis. The maximum is at the point farthest away from the neutral axis.
    This variation also changes as more moads are added.

    Material Behavior: linear elastic beams

    Elastic range: bending moment is such that the normal stresses remain below the yield strength.
    Hooke’s law combined with equilibrium.
    $$ \int_A y dA =0\ $$
    Moment-curvature equation.
    $$ M(x) = \frac{E(x)I_z(x)}{\rho(x)}\ $$
    Constitutive Relationship:
    $$ L_{y_i} = L_{NAi} = L_{NAf} = \rho\theta\ $$
    $$ L_{y_f} = (\rho - y)\theta\ $$
    $$ \varepsilon_x = \frac{(\rho-y)\theta - \rho\theta}{\rho\theta}\ $$
    $$ \varepsilon_x = \frac{-y\theta}{\rho\theta} = \frac{-y}{\rho}\ $$
    $$ \sigma_x = E\varepsilon_x = -\frac{Ey}{\rho}\ $$
    Force Equilibrium:
    $$ \Sigma F_x = 0\ $$
    $$ \Sigma F_x = \int dF\ $$
    $$ \Sigma F_x = \int \sigma_x dA\ $$
    $$ \Sigma F_x = \int\frac{-Ey}{\rho}dA\ $$
    $$ \Sigma F_x = \frac{-E}{\rho}\int y dA = 0\ $$
    Moment Equilibrium:
    $$ \Sigma M_z = -M - \int y dF = 0\ $$
    $$ \Sigma M_z = -M - \int y\sigma_x dA\ $$
    $$ \Sigma M_z = -M - \int y (\frac{-Ey}{\rho}) dA\ $$
    $$ \Sigma M_z = -M - + \frac{E}{\rho} \int y^2 dA = 0\ $$

    Maximum Normal Stress

    From equilibrium: the centroid is located at \( \bar{y}=0 \), i.e., the neutral axis passes through the centroid of the section.
    Elastic flextural formula.
    $$ \sigma_x (x,y) = - \frac{M(x)y}{I_z(x)}\ $$
    $$ M_z = \frac{EI_z}{\rho}\ $$
    $$ M_z = \frac{-\sigma_x}{y}I_z\ $$
    To evaluate the maximum absolute normal stress, denoting “c” the largest distance from the neutral surface, we use:
    Absolute maximum stress.
    $$ \sigma_{max} = \frac{|M|c}{I_z}\ $$
    Note: the ratio I/c depends only upon the geometry of the cross section. This ratio is called the elastic section modulus and is denoted by S.
    Absolute maximum stress.
    $$ \sigma_{max} = \frac{|M|}{S}\ $$
    Elastic section modulus.
    $$ S = \frac{I}{c}\ $$

    Heads up! - Extra

    Composite beams builds on this content.

    Recall that \( \epsilon_x = -\frac{y}{\rho} \) does not depend on the material properties of the beam, and is based only on the assumptions of geometry done so far.

    In non-homogeneous beams, we can no longer assume that the neutral axis passes through the centroid of the composite section. We should now determine that location…
    After obtaining the TRANSFORMED CROSS SECTION, we get
    $$ \int_{A_t}y d A_t = 0\ $$
    Therefore, the neutral axis passes through the centroid of the transformed cross section.
    Note: the widening (\( n > 1 \)) or narrowing (\( n < 1 \)) must be done in a direction parallel to the neutral axis of the section, since we want y-distances to be the same in the original and transformed section, so that the distance y in the flexural formula is unaltered.
    $$ \sigma_1 = -\frac{My}{I_t}\ $$
    $$ \sigma_2 = -\frac{nMy}{I_t}\ $$

    Eccentric Axial Loading in a Plane of Symmetry

    Equilibrium.
    $$ F=P \ \ \text{and} \ \ M=Pd\ $$
    Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due to a pure bending moment. Validity requires stresses below proportional limit (elastic region), deformations have negligible effect on geometry, and stresses not evaluated near points of load application.