Introduction to Bending
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Shearing Stresses in Beams
Pure Bending of Symmetric Beams
Pure bending occurs when a beam is loaded with two equal and opposite moment couples along the same plane. Under the conditions of pure bending transverse planes stay straight and can be thought of as radial lines. Axial planes will either lengthen or contract, dependent on location, the axial plane that does not lengthen or contract is termed the neutral axis.
The neutral axis radius of curvature, \rho, and the angle of curvature, \theta, can be used to determine the strain-state of a curved beam.
\varepsilon_x=\dfrac{-y}{\rho}
Alternatively this can be expressed in terms of the maximum bending strain:
\varepsilon_x=-\dfrac{y}{y_{max}}\varepsilon_m
Where the maximum strain is:
\varepsilon_m=\left|\dfrac{y_{max}}{\rho}\right|
Using the relationship between elastic stress and strain:
\sigma_x=E\varepsilon_x
\sigma_x=-\dfrac{y}{y_{max}}\sigma_m
The neutral axis must pass through the centroid of the beam cross-section, this provides a reliable way to determine the neutral axis.
For a static beam, \displaystyle F_x=\int\sigma_x dA=0
\displaystyle\int\sigma_xdA=\int-\dfrac{y}{y_{max}}\sigma_mdA
\displaystyle -\dfrac{\sigma_m}{y_{max}}\int ydA=0
\displaystyle \int ydA=0
A reminder that y is defined by the neutral axis, therefore the first moment about the neutral axis must be 0. In other words, the neutral axis must be the centroid of the beam cross-section.
The stress can be related directly to the bending moment, M, and beam geometry by way of the cross-section moment of inertia, I. Bending moment is signed, it is positive when the positive y axis is in compression due to bending.
\sigma_x=-\dfrac{My}{I}
For a beam under bending moment, \displaystyle M=\int -y\sigma_x dA.
\displaystyle M=\int -y\left(-\dfrac{y}{y_{max}}\sigma_m\right) dA
\displaystyle M=\dfrac{\sigma_m}{y_{max}}\int y^2dA
M=\dfrac{\sigma_m}{y_{max}}I
\sigma_m=\dfrac{My_{max}}{I}
\sigma_x=-\dfrac{y}{y_{max}}\dfrac{My_{max}}{I}
\sigma_x=-\dfrac{My}{I}
Finally, the radius of curvature can be determined solely by geometry and the materials Young’s modulus.
\rho=\dfrac{EI}{M}
Transverse Cross-Sections
As with all forms of axial deformation there is a corresponding transverse plane deformation. Transverse cross-sectional planes will deform according to Poisson’s ratio:
\varepsilon_y=\varepsilon_z=-\nu\varepsilon_x
\varepsilon_y=\varepsilon_z=\nu\dfrac{y}{\rho}
The expansion and contraction of surfaces above and below the neutral surface will cause curvature in the transverse planes. The transverse curvature can be described by the anticlastic radius of curvature, \rho^\prime.
\rho^\prime=\dfrac{\rho}{\nu}
Eccentric Loading: Principle of Superposition
The principle of superposition states that:
“The net result due to multiple causes is equal to the sum of individual responses due to each individual cause.”
We can apply the principle of superposition to stress, expressed with the following equation:
\sigma_x=\sigma_{\text{axial, }x}+\sigma_{\text{bending, }x}+\sigma_{\text{torsion, }x}
Eccentric loading occurs when a tensile or compressive force is applied at a distance from the line of action that would transmit the force through the object. Under eccentric loading stresses will occur due to both tension/compression forces and bending moment. The principle of superposition can therefore be applied for a dynamically equivalent section, and therefore the stress distribution can be found.
In summary, for any general eccentric loading case we can calculate the total stress as follows:
\sigma_x=\sigma_{\text{axial, }x}+\sigma_{\text{bending, }x}
\sigma_x=\dfrac{F_x}{A}+\dfrac{M_y z}{I_y}-\dfrac{M_z y}{I_z}