Learning topic
Eccentric Compression
Calculate stress under an eccentric compressive load using axial force plus bending, locate the neutral axis, and determine the section core for no-tension loading.
Eccentric compression occurs when the line of action of a compressive force does not pass through the centroid of the cross-section. The force is statically equivalent to a centric axial force and one or two bending moments.
Eccentricity and force reduction
If the load \(N\) has eccentricities \(e_x\) and \(e_y\) relative to centroidal principal axes, then \(M_x=Ne_y\) and \(M_y=Ne_x\); the signs depend on the adopted coordinate convention. An eccentric compression problem is therefore an axial-force-plus-bending problem.
An eccentric axial force $N$, applied with eccentricities $e_x$ and $e_y$ relative to the centroid, can be replaced by a centric force and bending moments. For principal centroidal axes:
$$M_x=Ne_y,\qquad M_y=Ne_x.$$
The normal stress follows from superposition of axial loading and biaxial bending:
$$\sigma(x,y)=\frac{N}{A}-\frac{M_x}{I_x}y+\frac{M_y}{I_y}x.$$
- $N$ — axial force;
- $e_x$, $e_y$ — eccentricities relative to the centroidal axes;
- $A$ — cross-sectional area;
- $M_x$, $M_y$ — bending moments caused by eccentricity;
- $I_x$, $I_y$ — principal centroidal second moments of area.
The signs of $N$, the moments, and the coordinates must follow one consistent convention. When compression is taken as negative, the stress signs directly indicate which parts of the section are in compression or tension.
Normal-stress formula
For a section with centroidal principal axes, the normal stress at a point \((x,y)\) can be written, with a consistent sign convention, as
\[\sigma(x,y)=\frac{N}{A}+\frac{M_x y}{I_x}+\frac{M_y x}{I_y}.\]
For uniaxial eccentricity, \(M=Ne\) and the extreme stresses are
\[\sigma=\frac{N}{A}\pm\frac{M}{W}=\frac{N}{A}\left(1\pm\frac{e}{k}\right),\]
where \(A\) is area, \(W\) is the appropriate section modulus, and \(k=W/A\) is the kern distance in that direction.
Stress distribution and neutral axis
The \(N/A\) part is uniform; the bending part varies linearly. As eccentricity grows, compression on one edge rises while it falls on the other. The neutral axis is the line where \(\sigma=0\). Its position must be obtained from the complete stress expression when bending acts about two axes.
Section kern
The section kern is the region around the centroid within which a compressive force must act so that normal stresses remain compressive or zero throughout the cross-section.
For a rectangular section of width $b$ and height $h$, the kern is a rhombus with vertices at:
$$e_x=\pm\frac{b}{6},\qquad e_y=\pm\frac{h}{6}.$$
For a solid circular section of diameter $d$, the kern is a concentric circle of radius:
$$r_k=\frac{d}{8}=\frac{R}{4}.$$
- $e_x$, $e_y$ — limiting eccentricities for a rectangular section;
- $b$, $h$ — rectangle dimensions;
- $d$ — circle diameter;
- $R$ — circle radius;
- $r_k$ — kern radius.
The kern boundary is obtained from the condition that normal stress becomes zero at one extreme point. Loading outside the kern produces tensile normal stress in part of the section according to the full-section linear model.
The calculator below evaluates the kern radius or diameter for a solid circular section.
The kern is the region in which a compressive resultant must act so that no tensile stress appears anywhere in the section. It is important for masonry, soil contact and other situations where tension or separation is unacceptable.
Rectangular-section example
For a rectangular section \(b\times h\) and eccentricity along \(h\), \(A=bh\), \(W=bh^2/6\), and
\[\sigma_{max,min}=\frac{N}{bh}\left(1\pm\frac{6e}{h}\right).\]
The no-tension condition is \(|e|\le h/6\). At \(|e|=h/6\), stress at one edge is zero. For example, with \(b=200\) mm, \(h=300\) mm, \(N=120\) kN and \(e=30\) mm, the mean stress is \(2.0\) MPa, while \(\sigma_{max}=3.2\) MPa and \(\sigma_{min}=0.8\) MPa: the full section remains compressed.
When the resultant leaves the kern
If a member, foundation, or contact interface cannot transmit tension, the full-section linear distribution is no longer physically valid once tensile stress is predicted. The actual contact area becomes partial; use an appropriate no-tension contact model rather than simply accepting negative pressure.
Calculation procedure
- Locate the centroid and the principal centroidal axes.
- Determine \(N\), \(e_x\), \(e_y\), \(M_x\), and \(M_y\).
- Evaluate stress at the extreme points of the section.
- Locate the neutral axis and check the kern where tension is not allowed.
- Perform the required strength, stability, or contact-pressure check.
Common mistakes
Typical errors are using the wrong section modulus, reversing the bending-moment sign, checking only one edge under biaxial eccentricity, and applying the full-area formula after contact separation has occurred.
About this topic
Eccentric compression occurs when an axial load is applied away from the centroid of a cross-section. This section presents normal stress distribution formulas combining axial forces and bending moments. It introduces the concept of the core of a section—the central region where load application guarantees no tensile stresses develop in brittle materials.