Axial stress & strain: practice problems with step-by-step solutions

Free Axial stress & strain practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: normal stress, strain and elongation, stepped bar.

Free practice problemEasy
80 kNØ30100 GPaØ100/50800 mmσr​ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa). Both are 800 mm long and are compressed together between rigid plates by 80 kN. Find the compressive stress σr in the rod.

P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 100 GPaL = 800 mm
MPa
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What you practise

  1. Easynormal stress
  2. Mediumstrain and elongation
  3. Hardstepped bar

Key formulas

Normal stress
σ = F/A
Strain
ε = ΔL/L
Hooke’s law
σ = E·ε
Elongation
ΔL = FL/(EA)
Thermal strain
εT = α·ΔT

Learn Axial stress & strain from scratch

1What stress is

Pull on a steel rod and it resists. Whether it breaks does not depend on the force alone, but on how much material shares that force.

A thin wire and a thick cable can carry the same load. The thin wire is far more likely to fail, because each square millimetre of it carries more.

dLP

Normal stress σ (sigma) is the force per unit area of the cross-section. The force P acts along the axis of the bar, so this is called an axial load.

σ = P / A

Engineers work in newtons and millimetres. One newton per square millimetre is one megapascal:

1 N/mm² = 1 MPa

So convert kN to N first (× 1000). For a round bar of diameter d, the area is a circle:

A = π · d² / 4
Self-check A bar with a cross-section of 100 mm² carries 10 kN. What is the stress?
  1. 0.1 MPa
  2. 10 MPa
  3. 100 MPa
  4. 1000 MPa

Answer: 100 MPa. σ = 10 000 N / 100 mm² = 100 N/mm² = 100 MPa. 0.1 MPa comes from forgetting to convert kN to N.

2Strain and Hooke’s law

Under load the bar also gets a little longer. This change in length is the elongation δ (delta).

dLδP

Strain ε (epsilon) is the elongation divided by the original length. It has no unit, because it is mm per mm:

ε = δ / L

Strains in metals are tiny, so they are often given in microstrain: 1 µε = 10⁻⁶. A strain of 0.0008 is 800 µε.

For most materials, stress and strain are proportional as long as the load is not too large. This is Hooke’s law:

σ = E · ε

E is the Young’s modulus, the stiffness of the material. Steel has E ≈ 200 GPa, aluminium about 70 GPa. Since 1 GPa = 1000 MPa, steel has E = 200 000 MPa.

Combining the three formulas gives the elongation directly:

δ = P · L / (A · E)
Self-check A steel bar (E = 200 GPa) has a stress of 200 MPa. What is the strain?
  1. 0.001
  2. 0.1
  3. 1000
  4. 1

Answer: 0.001. Convert E to MPa: 200 GPa = 200 000 MPa. Then ε = 200/200 000 = 0.001 = 1000 µε. Dividing 200 by 200 without converting gives 1, which is far too large.

3The method step by step

  1. Write all values in N and mm: P in N (kN × 1000), lengths in mm (m × 1000), E in MPa (GPa × 1000).
  2. Find the area: A = π·d²/4 for a round bar.
  3. Find the stress: σ = P/A, in MPa.
  4. Find the strain: ε = σ/E. Multiply by 10⁶ if the answer is wanted in µε.
  5. Find the elongation: δ = ε·L, in mm.

Stop at the step the question asks for. Most questions use only the first two or three steps.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A steel bar (E = 200 GPa) has a diameter of 20 mm and a length of 2 m. It carries an axial tensile load of 50 kN. Find the stress, the strain and the elongation.

1

Look at the bar and convert units

Ø 20 mm2 mP = 50 kN

P = 50 kN = 50 000 N, L = 2 m = 2000 mm and E = 200 GPa = 200 000 MPa.

2

Area

A = π · 20² / 4 = 314.2 mm²
3

Stress

σ = P / A = 50 000 / 314.2 = 159.2 MPa

That is well below the strength of ordinary steel (about 250 MPa), so Hooke’s law applies.

4

Strain

ε = σ / E = 159.2 / 200 000 = 0.000796 = 796 µε
5

Elongation

δ = ε · L = 0.000796 · 2000 = 1.59 mm
6

Check the answer

Use the combined formula as a second route:

δ = P·L / (A·E) = 50 000 · 2000 / (314.2 · 200 000) = 1.59 mm

A 2 m steel bar stretching by about 1.6 mm is realistic: steel strains stay well below 0.2 %.

Answer

σ = 159.2 MPa, ε = 796 µε and δ = 1.59 mm.

5Bars with more than one segment

A stepped bar is made of segments with different diameters, and sometimes different materials.

A1​L1​A2​L2​P

Cut the bar anywhere and look at the part to the right of the cut. Only P pulls on it, so every segment carries the same force P.

The areas differ, so the stresses differ. The thinnest segment has the largest stress.

Each segment stretches on its own, with its own A, L and E. The total elongation is the sum:

δ = P·L1 / (A1·E1) + P·L2 / (A2·E2)
Self-check A stepped bar carries 20 kN. Segment 1 has A = 400 mm², segment 2 has A = 100 mm². What is the largest stress in the bar?
  1. 200 MPa
  2. 250 MPa
  3. 40 MPa
  4. 50 MPa

Answer: 200 MPa. Both segments carry 20 kN. The thin one has 20 000/100 = 200 MPa. 50 MPa is the thick segment, 40 MPa wrongly adds the areas, and 250 MPa wrongly adds the two stresses.

6Common mistakes

  • Forgetting to convert kN to N, which makes the stress 1000 times too small.
  • Using E in GPa next to a stress in MPa. Write E in MPa (GPa × 1000).
  • Using π·d² or π·d instead of π·d²/4 for the area of a circle.
  • Leaving L in metres while δ should come out in millimetres.
  • Splitting P between the segments of a stepped bar. Each segment carries the full P.
Self-check A bar with A = 30 mm² carries P = 6 kN. What is the stress?
  1. 0.2 MPa
  2. 180 MPa
  3. 200 MPa
  4. 5 MPa

Answer: 200 MPa. σ = 6000 N / 30 mm² = 200 MPa. 0.2 MPa uses 6 instead of 6000 N; 5 MPa divides the area by the force; 180 multiplies them.

7How exam questions look

  • Easy: normal stress. A round bar of diameter d carries P; find σ = P/A.
  • Medium: strain and elongation. You also get L and the material’s E; find ε (in µε) or δ (in mm).
  • Hard: a stepped bar with two segments, possibly of different materials. Find the total elongation or the largest stress.

Write the units next to every number. Almost all wrong answers in this topic are off by a factor of 1000.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
300 kNØ30120 GPaØ100/50500 mmσr​ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa). Both are 500 mm long and are compressed together between rigid plates by 300 kN. Find the compressive stress σr in the rod.

P = 300 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 120 GPaL = 500 mm

Find σr (MPa)

Show hints
  1. Rigid plates: rod and tube shorten by the same amount, so ε is the same in both.
  2. ε = P/ΣEA.
  3. σ = E·ε.
Show the answer and the step-by-step solution

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

Ar = π·30²/4 = 706.9 mm²
At = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·5890 = 8.482 × 10⁸ N

Common strain

ε = 300000/8.482 × 10⁸ = 0.0003537

Stress in the rod

σr = 200·10³·0.0003537 = 70.74 MPa

Answer

σr = 70.74 MPa
Example 2Medium
Ø 20 mmBrass2.5 mP = 10 kN

Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).

P = 10 kNd = 20 mmL = 2.5 mE = 100 GPa

Find ε (µε)

Show hints
  1. Chain three relations: σ = P/A, ε = σ/E, δ = ε·L.
  2. Convert E to MPa (GPa × 1000) so it matches stress in MPa, and L to mm.
  3. σ = 31.83 MPa.
Show the answer and the step-by-step solution

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·20²/4 = 314.2 mm²
σ = P/A = 10·1000/314.2 = 31.83 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 31.83/100000 = 0.0003183 = 318.3 µε

Answer

ε = 318.3 µε
Example 3Hard
40 kN12300 mm500 mmØ30Ø30σ2​ = ?

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaP = 40 kN

Find σ2 (MPa)

Show hints
  1. The joint moves by the same amount for both segments: one stretches, the other shortens.
  2. k = E·A/L; P = k₁δ + k₂δ.
  3. k₁ = 471239 N/mm, k₂ = 98960 N/mm.
Show the answer and the step-by-step solution

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Stiffnesses

k₁ = 200000·706.9/300 = 471239 N/mm
k₂ = 70000·706.9/500 = 98960 N/mm

Share of segment 2

F2 = 40000·98960/570199 = 6942 N

Stress

σ2 = 6942/706.9 = 9.821 MPa (compression)

Answer

σ2 = 9.821 MPa

Questions and answers

What is Axial stress & strain?

Pull on a steel rod and it resists. Whether it breaks does not depend on the force alone, but on how much material shares that force.

A thin wire and a thick cable can carry the same load. The thin wire is far more likely to fail, because each square millimetre of it carries more.

How do I solve Axial stress & strain problems step by step?

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

What are common mistakes in Axial stress & strain?
  • The two materials do not share the stress equally: they share the strain. The stiffer one is more highly stressed.
  • The tube carries part of the load.
  • That is the stress in the tube.
What do the Axial stress & strain exercises cover?
  • Easy: normal stress
  • Medium: strain and elongation
  • Hard: stepped bar

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Continue with

Mohr’s circle & stress transformationTorsion of shafts

Calculators

Axial stress and strain calculatorThermal expansion calculator

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Mechanics of MaterialsMachine DesignFE Mechanical exam practice

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Printable worksheet

Axial stress & strain worksheet with answers
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