Normal stress from a centric axial force
An axial force applied centrically produces, under homogeneous loading, a normal stress that is constant over the entire cross-section:
σ = N/A
Tensile and compressive stress are not separate formulas here, just the two sign cases of this same normal stress.
Calculating tensile stress
For a centric tensile force FZ > 0:
σZ = FZ/A
Calculating compressive stress
For a centric compressive force FD, with the sign convention used here (N < 0 for compression):
σD = FD/A
Combining axial force and bending
An additional bending moment produces a linear stress contribution across the cross-section. In the linear-elastic range, the axial-force and bending contributions can be superimposed:
σ(x,y) = N/A + σB(x,y)
sigmaB is determined with the same stress solver used by the unsymmetrical-bending calculator -- no second implementation of the bending formulas.
Eccentric axial force
Instead of entering Mx and My directly, an axial force can be given with an eccentricity ex, ey. The offset acts as a lever arm, producing a statically equivalent moment:
My = N·exMx = -N·ey
Both input modes then use the same stress solver.
Example: 100 × 50 mm rectangle with N and Mx
At N = -10,000 N and Mx = 500 N·m, sigmaN = -10,000/5,000 = -2 N/mm². The bending contribution at y = 0 is -Mx*(0-25)/Ix = 500,000*25/1,041,666.667 = 12 N/mm². Total at P1: sigma = -2 + 12 = 10 N/mm² (tension), even though N is a compressive force.
Scope and limitations
The calculator supports one closed, non-self-intersecting outer contour and one polygonal inner contour in the linear-elastic range. Disconnected regions, arcs, DXF, shear stresses, buckling and plastic cross-section utilization are currently excluded.
FAQCommon questions
How do you calculate the normal stress from an axial force?
For an axial force N applied centrically, the normal stress sigma = N/A is constant over the entire cross-section, where A is the cross-sectional area.
How do you calculate tensile stress?
Tensile stress is the case sigma = FZ/A with an outward-pulling force FZ > 0. It is not a separate formula, just the normal-stress formula with a positive sign.
How do you calculate compressive stress?
Compressive stress is the case sigma = FD/A with a compressing force FD, counted here as a negative N. This is again the same normal-stress formula, just with a negative sign.
What is the difference between normal stress, tensile stress and compressive stress?
Normal stress is the umbrella term for any stress acting perpendicular to the cross-section. Tensile and compressive stress are simply the two sign cases of that same normal stress -- positive for tension, negative for compression.
When does sigma = N/A apply?
Only when the axial force acts centrically (through the area centroid) and no additional bending moment is present. As soon as a moment or eccentricity is added, sigma = N/A is only the constant part of an overall linearly varying stress distribution.
How are axial force and bending combined?
By superposition in the linear-elastic range: sigma(x,y) = N/A + sigmaB(x,y), where sigmaB is the same bending-stress contribution used by the unsymmetrical-bending calculator.
What is an eccentric axial force?
An axial force that does not act through the area centroid but at an offset point (ex, ey).
How does an eccentric force produce a bending moment?
The offset acts as a lever arm: My = N*ex and Mx = -N*ey. An eccentric force is therefore statically equivalent to the same force at the centroid plus this equivalent moment.
When does tension occur despite a compressive force?
When the bending moment (or eccentricity) is large enough that its stress contribution at one outer fibre exceeds the constant compressive part N/A. The stress at that fibre then flips sign to tension, even though the resultant force is compressive.
Where does the neutral axis lie under axial force and bending?
For pure bending it passes through the centroid. Once an axial force is added, the zero-stress line shifts parallel to that, wherever sigma = N/A + sigmaB(x,y) = 0 is satisfied -- it no longer has to pass through the centroid.
When may the superposition principle be used?
In the linear-elastic material range, as long as no stability effects (e.g. buckling) significantly affect the deformation and deflections remain small.