Engineering · Strength of materials

Average normal stress in tension and compression

Average normal stress is internal axial force divided by the loaded cross-sectional area; with uniform distribution it also equals local stress.

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Formula

Calculate

σ_{m} = \frac{F}{A}

σ in Pa = N/m²; F in N; A in m²

Variables & units

Variables & units
Symbol	Meaning	Unit
$σₘ$	Average normal stress in the section	Pa
$F$	Axial force normal to the section	N
$A$	Loaded cross-sectional area	m²

LaTeX

\sigma_\mathrm{m} = \frac{F}{A}

QUANTA STUDYquanta-study.de

Straight bar with equal outward tensile forces F at both ends. A cut normal to the axis marks area A. Four equal arrows schematically show a uniform tensile stress distribution. — Centred axial tension: F acts normal to cut area A. Uniform stress gives local σ = F/A; the area average also holds without uniformity.

Normal stress calculator

Find the mean normal stress, axial force or cross-sectional area of a bar, plus circular sections and change in length; tension counts as positive, compression as negative.

Calculation model

Find

Axial force in the bar N

Internal axial force in the bar, tension positive, compression negative, e.g. 18 kN tension in a flat bar.

Cross-sectional area A

Load-bearing cross-sectional area, e.g. 180 mm² for a flat bar of 12 mm × 15 mm.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for σₘ, substitute, evaluate

1. Rearrange for σ_m
$\sigma_{\mathrm m}=N/A$
2. Substitute values (in SI units)
- N = 18 kN = 18 000 N
- A = 180 mm² = 0.00018 m²
$\frac{18000}{0.00018}$
3. Result
σ_m = 100 MPa

Keep the sign and reference area. Inverse area from N=σ_m=0 is not unique.

Result: Mean axial stress, tension positive σ_m

σ_m = 100 MPa

With N = 18 kN and A = 180 mm², σ_m = 100 MPa.

Quick tips

Formula triangle: N on top, σ and A below. Cover the quantity you need and the rest is the calculation, e.g. A = N ÷ σ.
Unit shortcut: 1 MPa = 1 N/mm². Force in N divided by area in mm² gives the stress directly in MPa, e.g. 18,000 N ÷ 180 mm² = 100 MPa.

Assumptions and limits

Determine N at the chosen cut from a free-body diagram first. Tension positive, compression negative; actual positive load-bearing area. N/A is only the area average and needs no Hooke law. It also represents local stress only under uniform distribution. No local peak, allowable material stress or buckling resistance is calculated.

Derivation & background

This is the definition of average stress. The average applies to nonuniform distributions too, but only a uniform distribution makes it equal to local stress. It does not determine local peaks from notches, eccentricity or bending. In the elastic range, the material law $\sigma=E\varepsilon$ relates this stress to strain; the definition alone is not a strength check.

Exam blueprint

Validity range

F is internal axial force at the cut and A > 0 is the loaded section area. F/A gives the area average; local stress additionally requires approximately uniform distribution.

Derivation steps

The area average of local normal stress follows from the resultant internal force.

1Cut the bar normal to its axis and include all external forces on the retained portion.
2Determine internal axial force $N$ from $\sum F_x=0$ . The formula card labels this section force $F$ .
3The resultant of local normal stress is $N=\int_A\sigma\,\mathrm dA$ .
4Divide by the positive area: $\sigma_{\mathrm m}=\frac1A\int_A\sigma\,\mathrm dA=\frac NA=\frac FA$ .
5Only uniform distribution additionally gives $\sigma=\sigma_{\mathrm m}=\frac FA$ for local stress.
6Check $1\,\mathrm{N\,mm^{-2}}=1\,\mathrm{MPa}$ and the sign. Local peaks and buckling require separate models.

Rearrangements

Find mean normal stress σₘ

\sigma_\mathrm{m} = \frac{F}{A}

Preserve signs, boundary conditions and units when rearranging.

Find internal axial force

F=\sigma_{\mathrm m}A

F is internal axial force N found at the cut, not an arbitrary external load. A must be positive; tension positive and compression negative.

Find load-bearing area

A=\frac{F}{\sigma_{\mathrm m}}

Only for $\sigma_{\mathrm m}\ne0$ and $F/\sigma_{\mathrm m}>0$ . If $F=\sigma_{\mathrm m}=0$ , this balance does not determine a unique positive area.

Task variant

A centred tension bar with a 12 mm × 15 mm section carries 18 kN. Find σ_m.

Approach: first calculate $A=12\cdot15\,\mathrm{mm^2}=180\,\mathrm{mm^2}$ .
With $F=18000\,\mathrm N$ , $\sigma_{\mathrm m}=\frac{18000\,\mathrm N}{180\,\mathrm{mm^2}}=100\,\mathrm{MPa}$ .

Check: $100\,\mathrm{N\,mm^{-2}}\cdot180\,\mathrm{mm^2}=18000\,\mathrm N$ . This average does not determine local notch stress.

Common mistakes

Substituting an external force without a free-body diagram.

Determine N at the chosen cut from equilibrium; additional loads and self-weight can change it.

Equating 1 mm² with 10⁻³ m².

Square the conversion: $1\,\mathrm{mm^2}=10^{-6}\,\mathrm{m^2}$ . Directly, $1\,\mathrm{N\,mm^{-2}}=1\,\mathrm{MPa}$ .

Treating average stress as maximum local stress.

Notches, eccentric forces and bending can produce larger local stresses.

Exam context

Find internal force, section area and signs in an axial bar problem.
Distinguish average stress, elastic strain and stability assessment.

Formula cluster

Stress, strain and beam models

Internal section force and load-bearing area determine mean stress. Strain additionally requires a material law; bending or buckling requires a separate component model.

E = σ/ε = FL₀/(A₀ΔL)

Young's modulus

F = k·x

Hooke's Law (Spring Force)

M = EIκ ≈ EI·w″

Euler–Bernoulli beam moment–curvature relation

Worked example

A bar with $A=200\,\mathrm{mm^2}$ carries internal axial force $F=20\,\mathrm{kN}$ . Approach: $\sigma_{\mathrm m}=\frac FA$ . Converting gives $A=2\cdot10^{-4}\,\mathrm{m^2}$ , hence $\sigma_{\mathrm m}=\frac{20000\,\mathrm N}{2\cdot10^{-4}\,\mathrm{m^2}}=100\,\mathrm{MPa}$ . Check: stress times area recovers 20 kN.

Applications

Estimate axial tension or compression in idealized bars.

How do you calculate with Average axial normal stress?

Here is how to work through a typical Average axial normal stress (σₘ = F/A) task step by step:

1
Task
A centred tension bar with a 12 mm × 15 mm section carries 18 kN. Find σ_m.
Solution path
1. Approach: first calculate $A=12\cdot15\,\mathrm{mm^2}=180\,\mathrm{mm^2}$ .
2. With $F=18000\,\mathrm N$ , $\sigma_{\mathrm m}=\frac{18000\,\mathrm N}{180\,\mathrm{mm^2}}=100\,\mathrm{MPa}$ .
Check: $100\,\mathrm{N\,mm^{-2}}\cdot180\,\mathrm{mm^2}=18000\,\mathrm N$ . This average does not determine local notch stress.

Understand and apply the equation

Average normal stress divides an internal axial force by the loaded cross-sectional area. The crucial questions are which force acts at the cut and whether this average also represents local stress. This guide connects free-body analysis, area calculations and original examples with the limits of the bar model.

Force and stress are different quantities

Normal force acts perpendicular to the chosen cut surface. Many mechanics texts denote the internal force by N. The Quanta equation uses F for this axial resultant at the selected cut. Determine it from equilibrium first.

\sigma_{\mathrm m}=\frac{F}{A}=\frac{N}{A}

Average normal stress σ_m is axial resultant divided by cross-sectional area. A must be positive.[1]

At fixed F, doubling area halves average stress. This describes load and geometry. Whether the material supports the stress elastically, how much the bar extends and whether other failure modes occur are separate questions.

View diagram: area and mean stress

Doubling area halves mean stress

QUANTA STUDYquanta-study.de

The curve shows only section-average stress at fixed internal force, not a strength or stability limit.

A=180\,\mathrm{mm^2}

is the first worked example.[1]

Free with a Quanta account. Your download stays ready after registration.

Derive the relation from a cut and area integral

An imaginary cut perpendicular to the bar axis exposes forces transmitted by the removed part to the retained part. The free-body diagram includes external loads, self-weight when relevant, and the internal axial force. Under static assumptions, axial forces sum to zero.[3][1]

\begin{aligned}N&=\int_A\sigma\,\mathrm dA\\\sigma_{\mathrm m}&=\frac1A\int_A\sigma\,\mathrm dA\end{aligned}

The resultant is the integral of local normal stress; dividing by A gives its area average.[1]

\sigma_{\mathrm m}=\frac NA

The average follows from the definition even for a nonuniform stress distribution.

If local stress is approximately uniform over A, it can be taken outside the integral: $N=\sigma A$ . Only with this additional assumption does $\frac NA$ also give local stress. It fits a suitable straight prismatic bar under centred axial loading, away from load introduction and disturbances.[1]

Convert units and cross-sectional areas correctly

\begin{aligned}1\,\mathrm{Pa}&=1\,\mathrm{N\,m^{-2}}\\1\,\mathrm{MPa}&=1\,\mathrm{N\,mm^{-2}}\end{aligned}

Newtons divided by square millimetres give megapascals directly.

\begin{aligned}1\,\mathrm{mm^2}&=10^{-6}\,\mathrm{m^2}\\1\,\mathrm{kN}&=1000\,\mathrm N\end{aligned}

Square the length conversion when converting area.

\begin{aligned}A_{\mathrm{Kreis}}&=\frac{\pi d^2}{4}\\A_{\mathrm{Rechteck}}&=bh\\A_{\mathrm{Ring}}&=\frac{\pi(D^2-d^2)}4\end{aligned}

Solid circle of diameter d, rectangle of width b and height h, and annulus with D > d ≥ 0.

Use the cross-sectional area that actually transmits force. A hollow section requires annular area; an opening transmits no axial force. A hole can also create local stress concentrations beyond the change in area alone.

For a known positive area, the force rearrangement is $N=\sigma_{\mathrm m}A$ ; its sign distinguishes tension from compression. The area rearrangement $A=\frac{N}{\sigma_{\mathrm m}}$ requires $\sigma_{\mathrm m}\ne0$ and a positive quotient. Nonzero axial force and its corresponding mean stress therefore have the same sign. When $N=\sigma_{\mathrm m}=0$ , any positive area remains consistent with this balance; it does not determine a unique area.[1]

Three calculations: stress, section sizing and internal force

Average stress in a tension bar

Given: An idealised bar carries centred tension $N=18\,\mathrm{kN}$ . Its rectangular area is 12 mm × 15 mm.
Find: Find σ_m in MPa and check using SI units.

Approach: Centred tension is distributed over the specified rectangular section in the simplified bar model. Find area first, then divide axial force by that area.
$\begin{aligned}A&=bh\\{}\sigma_m&=\frac NA\end{aligned}$
Calculate area: $A=180\,\mathrm{mm^2}$ .
$A=12\cdot15\,\mathrm{mm^2}=180\,\mathrm{mm^2}$
Convert 18 kN to 18 000 N and divide by area.
$\sigma_{\mathrm m}=\frac{18000\,\mathrm N}{180\,\mathrm{mm^2}}=100\,\mathrm{MPa}$
For the SI check, use $A=1.8\cdot10^{-4}\,\mathrm{m^2}$ .
$\sigma_{\mathrm m}=\frac{18000}{1.8\times10^{-4}}\,\mathrm{Pa}=10^8\,\mathrm{Pa}$

Result: $\sigma_{\mathrm m}=+100\,\mathrm{MPa}$ with tension taken as positive.

Check: $100\,\mathrm{N\,mm^{-2}}\cdot180\,\mathrm{mm^2}=18000\,\mathrm N$ . This establishes the area average, not the absence of local peaks.

Minimum diameter for a given limit

Given: In the simplified centred tension model, $N=24\,\mathrm{kN}$ . The exercise specifies an allowable stress of 120 MPa.
Find: Find the calculated minimum diameter of a solid circular section.

Approach: The supplied allowable average tensile stress sets the required area. Solid-circle geometry converts this minimum area to a positive calculated minimum diameter.
$\begin{aligned}\sigma_m&=\frac NA\le\sigma_{\mathrm{allow}}\\{}A_{\min}&=\frac N{\sigma_{\mathrm{allow}}}\\{}d_{\min}&=\sqrt{\frac{4A_{\min}}\pi}\end{aligned}$
From $\frac NA\le\sigma_{\mathrm{allow}}$ , obtain $A\ge\frac N{\sigma_{\mathrm{allow}}}$ .
$A_{\min}=\frac{24000\,\mathrm N}{120\,\mathrm{N/mm^2}}=200\,\mathrm{mm^2}$
Solve $A=\frac{\pi d^2}{4}$ for positive diameter.
$d_{\min}=\sqrt{\frac{4A_{\min}}\pi}=\sqrt{\frac{800}{\pi}}\,\mathrm{mm}\approx15.96\,\mathrm{mm}$
Check a candidate diameter of 16 mm.
$\sigma_{\mathrm m}(16\,\mathrm{mm})=\frac{24000}{\frac{\pi\cdot16^2}{4}}\,\mathrm{MPa}\approx119.37\,\mathrm{MPa}$

Result: In the stated teaching model, 16 mm just meets the limit. Rounding down to 15 mm gives about 135.8 MPa and exceeds it.

Check: The choice assumes the specified limit and ideal geometry. Tolerances, additional loads and other checks are outside this calculation.

An additional axial load changes internal force

Given: A straight bar is supported at its left end. A +12 kN force acts rightwards at the right end and −4 kN acts leftwards at an intermediate point. Both sections have area 200 mm². Static axial load, tension positive; cuts are away from load introduction and self-weight is neglected.
Find: Find support reaction, internal force and mean stress on either side of the intermediate load.

Approach: determine the support reaction from equilibrium of the whole body. Then cut out the right-hand portion at each section; internal force is not necessarily the same as the end load everywhere.
$\begin{aligned}R_x+12\,\mathrm{kN}-4\,\mathrm{kN}&=0\\\sigma_{\mathrm m,j}&=\frac{N_j}{A}\end{aligned}$
To the right of the intermediate load, the cut-out portion contains only the 12 kN end load. To its left, the opposing 4 kN force is also part of the balance.
$\begin{aligned}R_x&=-8\,\mathrm{kN}\\N_{\mathrm{right}}&=+12\,\mathrm{kN}\\N_{\mathrm{left}}&=12\,\mathrm{kN}-4\,\mathrm{kN}=+8\,\mathrm{kN}\end{aligned}$
Same area, different internal axial force: calculate each mean stress separately.
$\sigma_{\mathrm m,left}=\frac{8000\,\mathrm N}{200\,\mathrm{mm^2}}=40\,\mathrm{MPa},\qquad\sigma_{\mathrm m,right}=60\,\mathrm{MPa}$

Result: The left support reaction is −8 kN. Both sections are in tension, but their mean stresses are 40 and 60 MPa.

Check: Internal force jumps by +4 kN at the concentrated intermediate load; equilibrium of each cut-out section confirms the results. Local stress peaks at load introduction are not determined.

Continue: strain and extension

For a homogeneous linearly elastic bar, $\sigma=E\varepsilon$ relates stress to strain. E is Young’s modulus and $\varepsilon=\frac{\Delta L}{L}$ is relative extension. Substituting uniform axial stress gives extension when N, A and E are constant along the length.[2]

\begin{aligned}\varepsilon&=\frac\sigma E\\\Delta L&=\frac{NL}{EA}\end{aligned}

This extension expression requires additional material and modelling assumptions.

In the first example, additionally suppose $L=600\,\mathrm{mm}$ and $E=200000\,\mathrm{MPa}$ . Then $\varepsilon=\frac{100\,\mathrm{MPa}}{200000\,\mathrm{MPa}}=0.0005$ and $\Delta L=\varepsilon L=0.30\,\mathrm{mm}$ . A different modulus changes strain at fixed force and area, but not the average stress $\frac NA$ .

Explore further: engineering and true stress

Tensile tests often use the original area A₀. Engineering stress is then $\frac{N}{A_0}$ . True stress uses the current area A. At small deformation the difference is small; with substantial necking, the definition must be explicit.[2]

\begin{aligned}\sigma_{\mathrm{tech}}&=\frac N{A_0}\\\sigma_{\mathrm{wahr}}&=\frac NA\end{aligned}

The same force is referred to different reference areas.

For an original numerical example, let $N=10000\,\mathrm N$ , $A_0=100\,\mathrm{mm^2}$ and $A=80\,\mathrm{mm^2}$ . Engineering stress is 100 MPa and the current-area stress is 125 MPa. A nonuniform neck can additionally have a multiaxial local stress state.

When $\frac NA$ alone is insufficient

Eccentric loading can add bending, so normal stress is not constant over the section.
Notches, holes and section changes can create local peaks. The average is not an upper bound.
Average compressive stress remains calculable, but slender bars can buckle. Low compressive stress does not rule out instability.
The definition $\frac NA$ does not require Hooke’s law. The elastic extension formula does require a suitable material model.
An allowable value is an input. It is not automatically yield stress, tensile strength or a code-derived design value.

Five exercises with solutions

Calculate · Foundation

A bar carries 7.5 kN with $A=250\,\mathrm{mm^2}$ . Find average tensile stress.

Show hint

Use newtons for force.

Learning goal: Calculate mean tensile stress using kilonewtons and square millimetres.

Show solution and calculation

Approach: divide internal axial force by load-bearing area.
$\begin{aligned}\sigma_m&=\frac{7500}{250}\,\frac{\mathrm N}{\mathrm{mm^2}}\\{}&=30\,\frac{\mathrm N}{\mathrm{mm^2}}\\{}&=30\,\mathrm{MPa}\text{.}\end{aligned}$ Check: 30 N/mm² times 250 mm² recovers 7500 N.

\sigma_{\mathrm m}=\frac{7500}{250}\,\mathrm{MPa}=30\,\mathrm{MPa}

Calculate · Application

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σ_m.

Show hint

Subtract the inner circle area from the outer circle area.

Learning goal: Use the actual load-bearing annular area for a hollow section.

Show solution and calculation

Approach: $A=\frac\pi4(D^2-d^2)$ , then $\sigma_{\mathrm m}=\frac NA$ .
$A=\frac{\pi(400-256)}4\,\mathrm{mm^2}=36\pi\,\mathrm{mm^2}\approx113.10\,\mathrm{mm^2}$ .
$\sigma_m\approx70.74\,\mathrm{MPa}$ .

Check: the annulus is smaller than the outer circle, so stress exceeds that in a solid bar with the same outside diameter.

\sigma_{\mathrm m}=\frac{8000}{36\pi}\,\mathrm{MPa}\approx70.74\,\mathrm{MPa}

Interpret · Application

Compressive force is $N=-12\,\mathrm{kN}$ and area is 300 mm². Does σ_m alone establish buckling resistance?

Show hint

Keep the sign and distinguish stress from stability.

Learning goal: Distinguish negative mean compressive stress from a stability conclusion.

Show solution and calculation

Approach: $\sigma_{\mathrm m}=\frac{-12000\,\mathrm N}{300\,\mathrm{mm^2}}$ .
$\sigma_{\mathrm m}=-40\,\mathrm{MPa}$ .
No: buckling assessment also needs length, end constraints, stiffness and section inertia.

Check: the negative sign agrees with the prescribed compressive force.

\sigma_{\mathrm m}=\frac{-12000}{300}\,\mathrm{MPa}=-40\,\mathrm{MPa}

Interpret · Application

A straight stepped bar is loaded statically in centred tension by 6.0 kN. No additional axial load acts between its two sections; areas are 100 mm² and 300 mm². Find mean normal stress in each section away from the transition. Does the thicker section also carry three times the internal axial force?

Show hint

Find internal axial force with a section cut. With no extra load between cuts it is the same in both sections.

Learning goal: Distinguish equal axial force from different mean stresses in a stepped tensile bar.

Show solution and calculation

Approach: divide the same internal axial force of +6000 N by each load-bearing area.
Mean tensile stresses are +60 MPa and +20 MPa.
Axial force is not three times greater; only area changes.

Check: stress times area yields 6000 N in both sections. Near the transition, local stress distribution can differ from these section averages.

\begin{aligned}N_1=N_2&=+6000\,\mathrm N\\\sigma_{\mathrm m,1}&=\frac{6000\,\mathrm N}{100\,\mathrm{mm^2}}=+60\,\mathrm{MPa}\\\sigma_{\mathrm m,2}&=\frac{6000\,\mathrm N}{300\,\mathrm{mm^2}}=+20\,\mathrm{MPa}\\\sigma_{\mathrm m,1}A_1&=\sigma_{\mathrm m,2}A_2\end{aligned}

[1]

Error analysis · Application

Two homogeneous prismatic bars each have area 200 mm² and carry 10 kN of static centred tension. Lengths agree but Young’s moduli are 200 GPa and 100 GPa. Both remain linearly elastic. A solution claims equal normal stress also means equal strain. Check both stresses and strains; load-introduction regions are excluded.

Show hint

Stress follows from force and area. Strain additionally requires the material law with E.

Learning goal: Avoid interpreting mean normal stress as strain without a material law.

Show solution and calculation

Approach: first find the equal stress of 50 MPa, then divide by each bar’s own Young’s modulus.
Strains are 0.00025 and 0.00050.
The second bar therefore strains twice as much.

Check: equal original lengths also make its extension twice as large. Stress has units while strain is dimensionless; these quantities must not be equated.

\begin{aligned}\sigma_1=\sigma_2&=\frac{10\,000\,\mathrm N}{200\,\mathrm{mm^2}}=50\,\mathrm{MPa}\\\varepsilon_1&=\frac{50\,\mathrm{MPa}}{200\,000\,\mathrm{MPa}}=0.00025\\\varepsilon_2&=\frac{50\,\mathrm{MPa}}{100\,000\,\mathrm{MPa}}=0.00050\\\frac{\varepsilon_2}{\varepsilon_1}&=2\end{aligned}

[2][1]

A dependable calculation sequence

Choose the cut and obtain internal axial force from a complete free-body diagram.
Calculate force-transmitting area with correct dimensions and units.
Calculate $\sigma_{\mathrm m}=\frac NA$ , state the sign convention and check $\sigma_{\mathrm m}A=N$ .
Then assess whether uniform stress, the material model and additional checks fit the case.

Begin with section equilibrium rather than an arbitrary external load. Divide the resulting axial force by the actual load-bearing area and state the sign. A material law is additionally required for strain; local stress peaks and buckling require their own models.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added force and area rearrangements with sign and zero cases; linked related formulas and retained existing content. The engineering-stress formula is now entirely typeset in mathematics without changing its meaning.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

Cite this reference page

Quanta. Average axial normal stress. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/normalspannung-zug

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Frequently asked questions about Average axial normal stress

Which force belongs in the stress formula?+

Use the internal axial force at the section being considered. Obtain it from equilibrium of a cut-out portion of the bar. The formula card labels it F; section-force diagrams often use N. It remains constant between cuts if no additional axial load acts there. An intermediate load can change it even when area is unchanged. In the third example, a 12 kN end load and an opposing 4 kN intermediate load produce 8 and 12 kN in different sections. Then divide by each load-bearing area. The sign distinguishes tension from compression; an arbitrary external load cannot replace section equilibrium.

How do I calculate with square millimetres?+

You may calculate entirely in newtons and square millimetres because $1\,\mathrm{N\,mm^{-2}}=1\,\mathrm{MPa}$ . For example, 20 kN over 200 mm² gives 100 MPa after converting force. In SI, the same force is 20000 N and area is $2\cdot10^{-4}\,\mathrm{m^2}$ , giving $10^8\,\mathrm{Pa}$ . Both routes must agree. Area conversion is squared: one square millimetre equals one millionth of a square metre. For a hollow bar, only the annulus carries load. Subtract the inner circular area before dividing force by area. Finally check that stress times area recovers the original internal force.

Is mean stress also the largest local stress?+

No. The formula gives an area average because internal force is the integral of local stress over the section. In approximately uniform axial loading of a prismatic bar away from load introduction, that average can describe local stress. A notch, section transition, support zone or eccentric load can produce a very different distribution. A larger local value is then possible even though internal force is unchanged. In the stepped-bar exercise, the calculated section averages therefore apply away from the transition. The transition region requires an appropriate local model, such as experimentally or numerically determined stress concentrations.

Does equal stress imply equal strain?+

Only if the relevant material response also agrees. Mean stress first follows from force and area. For a homogeneous bar under uniform linear elastic loading, additionally use $\varepsilon=\frac\sigma E$ . Two bars at the same 50 MPa stress but with Young’s moduli of 200 and 100 GPa therefore have strains of 0.00025 and 0.00050. With equal original lengths, the second extends twice as much. Strain is dimensionless; stress is not. Changing modulus changes strain under prescribed force and area. Under prescribed displacement, however, the internal axial force can also change.

Does the stress calculation establish a safe design?+

A stress calculation alone does not establish a safe structure. You need relevant load cases, allowable material stress and a justified safety basis. Local stress peaks, fatigue and additional stress components may require further assessments. A slender compression member can also buckle even when its mean stress is below a material limit. Length, support constraints, Young’s modulus and second moment of area then belong to the stability model. For a simple tensile sizing exercise with a specified allowable stress, use $A_{\min}=\frac F{\sigma_{\mathrm{allow}}}$ . The actual selected section must provide at least this area; the calculation does not replace other assessments required by the problem.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]MIT OpenCourseWare, Unified Engineering: Materials and Structures, axial stress and structural members.

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Average normal stress in tension and compression

Average normal stress is internal axial force divided by the loaded cross-sectional area; with uniform distribution it also equals local stress.

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Symbol

Meaning

Unit

σₘ

Average normal stress in the section

F

Axial force normal to the section

A

Loaded cross-sectional area

m²

Derivation & background

Exam blueprint

Validity range

F is internal axial force at the cut and A > 0 is the loaded section area. F/A gives the area average; local stress additionally requires approximately uniform distribution.

Derivation steps

The area average of local normal stress follows from the resultant internal force.

1Cut the bar normal to its axis and include all external forces on the retained portion.
2Determine internal axial force $N$ from $\sum F_x=0$ . The formula card labels this section force $F$ .
3The resultant of local normal stress is $N=\int_A\sigma\,\mathrm dA$ .
4Divide by the positive area: $\sigma_{\mathrm m}=\frac1A\int_A\sigma\,\mathrm dA=\frac NA=\frac FA$ .
5Only uniform distribution additionally gives $\sigma=\sigma_{\mathrm m}=\frac FA$ for local stress.
6Check $1\,\mathrm{N\,mm^{-2}}=1\,\mathrm{MPa}$ and the sign. Local peaks and buckling require separate models.

How do you calculate with Average axial normal stress?

Here is how to work through a typical Average axial normal stress (σₘ = F/A) task step by step:

Task

A centred tension bar with a 12 mm × 15 mm section carries 18 kN. Find σ_m.

Solution path

Approach: first calculate $A=12\cdot15\,\mathrm{mm^2}=180\,\mathrm{mm^2}$ .
With $F=18000\,\mathrm N$ , $\sigma_{\mathrm m}=\frac{18000\,\mathrm N}{180\,\mathrm{mm^2}}=100\,\mathrm{MPa}$ .

Check: $100\,\mathrm{N\,mm^{-2}}\cdot180\,\mathrm{mm^2}=18000\,\mathrm N$ . This average does not determine local notch stress.

Understand and apply the equation

Force and stress are different quantities

\sigma_{\mathrm m}=\frac{F}{A}=\frac{N}{A}

Average normal stress σ_m is axial resultant divided by cross-sectional area. A must be positive.[1]

View diagram: area and mean stress

Doubling area halves mean stress

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The curve shows only section-average stress at fixed internal force, not a strength or stability limit.

A=180\,\mathrm{mm^2}

is the first worked example.[1]

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Derive the relation from a cut and area integral

\begin{aligned}N&=\int_A\sigma\,\mathrm dA\\\sigma_{\mathrm m}&=\frac1A\int_A\sigma\,\mathrm dA\end{aligned}

The resultant is the integral of local normal stress; dividing by A gives its area average.[1]

\sigma_{\mathrm m}=\frac NA

The average follows from the definition even for a nonuniform stress distribution.

Convert units and cross-sectional areas correctly

\begin{aligned}1\,\mathrm{Pa}&=1\,\mathrm{N\,m^{-2}}\\1\,\mathrm{MPa}&=1\,\mathrm{N\,mm^{-2}}\end{aligned}

Newtons divided by square millimetres give megapascals directly.

\begin{aligned}1\,\mathrm{mm^2}&=10^{-6}\,\mathrm{m^2}\\1\,\mathrm{kN}&=1000\,\mathrm N\end{aligned}

Square the length conversion when converting area.

\begin{aligned}A_{\mathrm{Kreis}}&=\frac{\pi d^2}{4}\\A_{\mathrm{Rechteck}}&=bh\\A_{\mathrm{Ring}}&=\frac{\pi(D^2-d^2)}4\end{aligned}

Solid circle of diameter d, rectangle of width b and height h, and annulus with D > d ≥ 0.

Three calculations: stress, section sizing and internal force

Average stress in a tension bar

Given: An idealised bar carries centred tension $N=18\,\mathrm{kN}$ . Its rectangular area is 12 mm × 15 mm.
Find: Find σ_m in MPa and check using SI units.

Approach: Centred tension is distributed over the specified rectangular section in the simplified bar model. Find area first, then divide axial force by that area.
$\begin{aligned}A&=bh\\{}\sigma_m&=\frac NA\end{aligned}$
Calculate area: $A=180\,\mathrm{mm^2}$ .
$A=12\cdot15\,\mathrm{mm^2}=180\,\mathrm{mm^2}$
Convert 18 kN to 18 000 N and divide by area.
$\sigma_{\mathrm m}=\frac{18000\,\mathrm N}{180\,\mathrm{mm^2}}=100\,\mathrm{MPa}$
For the SI check, use $A=1.8\cdot10^{-4}\,\mathrm{m^2}$ .
$\sigma_{\mathrm m}=\frac{18000}{1.8\times10^{-4}}\,\mathrm{Pa}=10^8\,\mathrm{Pa}$

Result: $\sigma_{\mathrm m}=+100\,\mathrm{MPa}$ with tension taken as positive.

Check: $100\,\mathrm{N\,mm^{-2}}\cdot180\,\mathrm{mm^2}=18000\,\mathrm N$ . This establishes the area average, not the absence of local peaks.

Minimum diameter for a given limit

Given: In the simplified centred tension model, $N=24\,\mathrm{kN}$ . The exercise specifies an allowable stress of 120 MPa.
Find: Find the calculated minimum diameter of a solid circular section.

Approach: The supplied allowable average tensile stress sets the required area. Solid-circle geometry converts this minimum area to a positive calculated minimum diameter.
$\begin{aligned}\sigma_m&=\frac NA\le\sigma_{\mathrm{allow}}\\{}A_{\min}&=\frac N{\sigma_{\mathrm{allow}}}\\{}d_{\min}&=\sqrt{\frac{4A_{\min}}\pi}\end{aligned}$
From $\frac NA\le\sigma_{\mathrm{allow}}$ , obtain $A\ge\frac N{\sigma_{\mathrm{allow}}}$ .
$A_{\min}=\frac{24000\,\mathrm N}{120\,\mathrm{N/mm^2}}=200\,\mathrm{mm^2}$
Solve $A=\frac{\pi d^2}{4}$ for positive diameter.
$d_{\min}=\sqrt{\frac{4A_{\min}}\pi}=\sqrt{\frac{800}{\pi}}\,\mathrm{mm}\approx15.96\,\mathrm{mm}$
Check a candidate diameter of 16 mm.
$\sigma_{\mathrm m}(16\,\mathrm{mm})=\frac{24000}{\frac{\pi\cdot16^2}{4}}\,\mathrm{MPa}\approx119.37\,\mathrm{MPa}$

Result: In the stated teaching model, 16 mm just meets the limit. Rounding down to 15 mm gives about 135.8 MPa and exceeds it.

Check: The choice assumes the specified limit and ideal geometry. Tolerances, additional loads and other checks are outside this calculation.

An additional axial load changes internal force

Given: A straight bar is supported at its left end. A +12 kN force acts rightwards at the right end and −4 kN acts leftwards at an intermediate point. Both sections have area 200 mm². Static axial load, tension positive; cuts are away from load introduction and self-weight is neglected.
Find: Find support reaction, internal force and mean stress on either side of the intermediate load.

Approach: determine the support reaction from equilibrium of the whole body. Then cut out the right-hand portion at each section; internal force is not necessarily the same as the end load everywhere.
$\begin{aligned}R_x+12\,\mathrm{kN}-4\,\mathrm{kN}&=0\\\sigma_{\mathrm m,j}&=\frac{N_j}{A}\end{aligned}$
To the right of the intermediate load, the cut-out portion contains only the 12 kN end load. To its left, the opposing 4 kN force is also part of the balance.
$\begin{aligned}R_x&=-8\,\mathrm{kN}\\N_{\mathrm{right}}&=+12\,\mathrm{kN}\\N_{\mathrm{left}}&=12\,\mathrm{kN}-4\,\mathrm{kN}=+8\,\mathrm{kN}\end{aligned}$
Same area, different internal axial force: calculate each mean stress separately.
$\sigma_{\mathrm m,left}=\frac{8000\,\mathrm N}{200\,\mathrm{mm^2}}=40\,\mathrm{MPa},\qquad\sigma_{\mathrm m,right}=60\,\mathrm{MPa}$

Result: The left support reaction is −8 kN. Both sections are in tension, but their mean stresses are 40 and 60 MPa.

Check: Internal force jumps by +4 kN at the concentrated intermediate load; equilibrium of each cut-out section confirms the results. Local stress peaks at load introduction are not determined.

Continue: strain and extension

\begin{aligned}\varepsilon&=\frac\sigma E\\\Delta L&=\frac{NL}{EA}\end{aligned}

This extension expression requires additional material and modelling assumptions.

Explore further: engineering and true stress

\begin{aligned}\sigma_{\mathrm{tech}}&=\frac N{A_0}\\\sigma_{\mathrm{wahr}}&=\frac NA\end{aligned}

The same force is referred to different reference areas.

When $\frac NA$ alone is insufficient

Eccentric loading can add bending, so normal stress is not constant over the section.
Notches, holes and section changes can create local peaks. The average is not an upper bound.
Average compressive stress remains calculable, but slender bars can buckle. Low compressive stress does not rule out instability.
The definition $\frac NA$ does not require Hooke’s law. The elastic extension formula does require a suitable material model.
An allowable value is an input. It is not automatically yield stress, tensile strength or a code-derived design value.

Five exercises with solutions

Calculate · Foundation

A bar carries 7.5 kN with $A=250\,\mathrm{mm^2}$ . Find average tensile stress.

Show hint

Use newtons for force.

Learning goal: Calculate mean tensile stress using kilonewtons and square millimetres.

Show solution and calculation

Approach: divide internal axial force by load-bearing area.
$\begin{aligned}\sigma_m&=\frac{7500}{250}\,\frac{\mathrm N}{\mathrm{mm^2}}\\{}&=30\,\frac{\mathrm N}{\mathrm{mm^2}}\\{}&=30\,\mathrm{MPa}\text{.}\end{aligned}$ Check: 30 N/mm² times 250 mm² recovers 7500 N.

\sigma_{\mathrm m}=\frac{7500}{250}\,\mathrm{MPa}=30\,\mathrm{MPa}

Calculate · Application

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σ_m.

Show hint

Subtract the inner circle area from the outer circle area.

Learning goal: Use the actual load-bearing annular area for a hollow section.

Show solution and calculation

Approach: $A=\frac\pi4(D^2-d^2)$ , then $\sigma_{\mathrm m}=\frac NA$ .
$A=\frac{\pi(400-256)}4\,\mathrm{mm^2}=36\pi\,\mathrm{mm^2}\approx113.10\,\mathrm{mm^2}$ .
$\sigma_m\approx70.74\,\mathrm{MPa}$ .

Check: the annulus is smaller than the outer circle, so stress exceeds that in a solid bar with the same outside diameter.

\sigma_{\mathrm m}=\frac{8000}{36\pi}\,\mathrm{MPa}\approx70.74\,\mathrm{MPa}

Interpret · Application

Compressive force is $N=-12\,\mathrm{kN}$ and area is 300 mm². Does σ_m alone establish buckling resistance?

Show hint

Keep the sign and distinguish stress from stability.

Learning goal: Distinguish negative mean compressive stress from a stability conclusion.

Show solution and calculation

Approach: $\sigma_{\mathrm m}=\frac{-12000\,\mathrm N}{300\,\mathrm{mm^2}}$ .
$\sigma_{\mathrm m}=-40\,\mathrm{MPa}$ .
No: buckling assessment also needs length, end constraints, stiffness and section inertia.

Check: the negative sign agrees with the prescribed compressive force.

\sigma_{\mathrm m}=\frac{-12000}{300}\,\mathrm{MPa}=-40\,\mathrm{MPa}

Interpret · Application

A straight stepped bar is loaded statically in centred tension by 6.0 kN. No additional axial load acts between its two sections; areas are 100 mm² and 300 mm². Find mean normal stress in each section away from the transition. Does the thicker section also carry three times the internal axial force?

Show hint

Find internal axial force with a section cut. With no extra load between cuts it is the same in both sections.

Learning goal: Distinguish equal axial force from different mean stresses in a stepped tensile bar.

Show solution and calculation

Approach: divide the same internal axial force of +6000 N by each load-bearing area.
Mean tensile stresses are +60 MPa and +20 MPa.
Axial force is not three times greater; only area changes.

Check: stress times area yields 6000 N in both sections. Near the transition, local stress distribution can differ from these section averages.

\begin{aligned}N_1=N_2&=+6000\,\mathrm N\\\sigma_{\mathrm m,1}&=\frac{6000\,\mathrm N}{100\,\mathrm{mm^2}}=+60\,\mathrm{MPa}\\\sigma_{\mathrm m,2}&=\frac{6000\,\mathrm N}{300\,\mathrm{mm^2}}=+20\,\mathrm{MPa}\\\sigma_{\mathrm m,1}A_1&=\sigma_{\mathrm m,2}A_2\end{aligned}

[1]

Error analysis · Application

Two homogeneous prismatic bars each have area 200 mm² and carry 10 kN of static centred tension. Lengths agree but Young’s moduli are 200 GPa and 100 GPa. Both remain linearly elastic. A solution claims equal normal stress also means equal strain. Check both stresses and strains; load-introduction regions are excluded.

Show hint

Stress follows from force and area. Strain additionally requires the material law with E.

Learning goal: Avoid interpreting mean normal stress as strain without a material law.

Show solution and calculation

Approach: first find the equal stress of 50 MPa, then divide by each bar’s own Young’s modulus.
Strains are 0.00025 and 0.00050.
The second bar therefore strains twice as much.

Check: equal original lengths also make its extension twice as large. Stress has units while strain is dimensionless; these quantities must not be equated.

\begin{aligned}\sigma_1=\sigma_2&=\frac{10\,000\,\mathrm N}{200\,\mathrm{mm^2}}=50\,\mathrm{MPa}\\\varepsilon_1&=\frac{50\,\mathrm{MPa}}{200\,000\,\mathrm{MPa}}=0.00025\\\varepsilon_2&=\frac{50\,\mathrm{MPa}}{100\,000\,\mathrm{MPa}}=0.00050\\\frac{\varepsilon_2}{\varepsilon_1}&=2\end{aligned}

[2][1]

A dependable calculation sequence

Choose the cut and obtain internal axial force from a complete free-body diagram.
Calculate force-transmitting area with correct dimensions and units.
Calculate $\sigma_{\mathrm m}=\frac NA$ , state the sign convention and check $\sigma_{\mathrm m}A=N$ .
Then assess whether uniform stress, the material model and additional checks fit the case.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Added force and area rearrangements with sign and zero cases; linked related formulas and retained existing content. The engineering-stress formula is now entirely typeset in mathematics without changing its meaning.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

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Quanta. Average axial normal stress. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/normalspannung-zug

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Frequently asked questions about Average axial normal stress

Which force belongs in the stress formula?+

How do I calculate with square millimetres?+

Is mean stress also the largest local stress?+

Does equal stress imply equal strain?+

Does the stress calculation establish a safe design?+

Average normal stress in tension and compression

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Stress, strain and beam models

Worked example

Applications

How do you calculate with Average axial normal stress?

Force and stress are different quantities

Doubling area halves mean stress

Derive the relation from a cut and area integral

Convert units and cross-sectional areas correctly

Three calculations: stress, section sizing and internal force

Average stress in a tension bar

Minimum diameter for a given limit

An additional axial load changes internal force

Continue: strain and extension

Explore further: engineering and true stress

When NA\frac NAAN​ alone is insufficient

Five exercises with solutions

A bar carries 7.5 kN with A=250 mm2A=250\,\mathrm{mm^2}A=250mm2. Find average tensile stress.

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σm.

Compressive force is N=−12 kNN=-12\,\mathrm{kN}N=−12kN and area is 300 mm2. Does σm alone establish buckling resistance?

A dependable calculation sequence

Sources for this explanation

Frequently asked questions about Average axial normal stress

Subject references and further reading

Related formulas

More formulas: Mechanics, strength and materials

Retain Average axial normal stress for exams

Average normal stress in tension and compression

Variables & units

Normal stress calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Stress, strain and beam models

Worked example

Applications

How do you calculate with Average axial normal stress?

Force and stress are different quantities

Doubling area halves mean stress

Derive the relation from a cut and area integral

Convert units and cross-sectional areas correctly

Three calculations: stress, section sizing and internal force

Average stress in a tension bar

Minimum diameter for a given limit

An additional axial load changes internal force

Continue: strain and extension

Explore further: engineering and true stress

When NA\frac NAAN​ alone is insufficient

Five exercises with solutions

A bar carries 7.5 kN with A=250 mm2A=250\,\mathrm{mm^2}A=250mm2. Find average tensile stress.

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σm.

Compressive force is N=−12 kNN=-12\,\mathrm{kN}N=−12kN and area is 300 mm2. Does σm alone establish buckling resistance?

A dependable calculation sequence

Sources for this explanation

Frequently asked questions about Average axial normal stress

Subject references and further reading

Related formulas

More formulas: Mechanics, strength and materials

Retain Average axial normal stress for exams

When $\frac NA$ alone is insufficient

A bar carries 7.5 kN with $A=250\,\mathrm{mm^2}$ . Find average tensile stress.

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σ_m.

Compressive force is $N=-12\,\mathrm{kN}$ and area is 300 mm². Does σ_m alone establish buckling resistance?

When $\frac NA$ alone is insufficient

A bar carries 7.5 kN with $A=250\,\mathrm{mm^2}$ . Find average tensile stress.

A hollow bar of outside diameter 20 mm and inside diameter 16 mm carries 8 kN. Find σ_m.

Compressive force is $N=-12\,\mathrm{kN}$ and area is 300 mm². Does σ_m alone establish buckling resistance?