Your Equation

Library / Mechanical / MEC-85

Mechanical

Strain energy stored in an axially loaded member

Energy stored in a uniform bar carrying a constant axial force, within the elastic range.

Ref
MEC-85
Rev
v1
Field
Mechanical
Result
U [J]

A bar stretched by an axial force F stores elastic energy equal to the area under the force–extension line: U = ½Fδ, with δ = FL/(EA). Substituting gives U = F²L/(2EA), so the energy rises with the square of the force. It is the energy returned if the load is removed, and the basis of energy methods for deflection. The inputs are F in kN, L in m, E in GPa and A in mm², with the result in joules.

Energy stored in a uniform bar carrying a constant axial force, within the elastic range. It equals half the force times the extension.

Calculator

Solve for
J
kN
0.11,000
m
0.120
GPa
1500
mm²
1010,000
  • Axial stress, F/A100 MPa
  • Extension, FL/(EA)1 mm
  • Strain energy per unit volume25 kJ/m³

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Strain energy (J) current value 25 J
FAxial forceKey driver
1e-410,000
EYoung’s modulus
105,000
ACross-sectional area
1.251,250
LLength
1.25250
  • A 2 m bar of 500 mm² loaded with 50 kN stretches 1 mm and stores 25 J, which is the same as ½ F δ.
  • Doubling the load quadruples the stored energy (to 100 J) while doubling the area halves it: a thicker member stores less for the same load.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (J)Local stepChange in result
F Axial force0.1 to 1,000 kN1e-410,00010,000+10%+21 %
E Young’s modulus1 to 500 GPa105,0004,990+10%−9.09 %
A Cross-sectional area10 to 10,000 mm²1.251,2501,249+10%−9.09 %
L Length0.1 to 20 m1.25250248.8+10%+10 %

Sweep of Axial force

Axial force (kN)Strain energy (J)
0.11e-4
0.31620.001
10.01
3.1620.1
101
31.6210
100100
316.21,000
1,00010,000

Sweep one variable

Strain energy (J)Hover, touch, or use the arrow keys on the chart
0200040006000800010k 0.11101001000 Axial force (kN), log scale now 25

Variables and how each one contributes

U

Strain energy

Result · J

What the equation gives you. Everything else feeds it.

F

Axial force

kN · typical 0.1 to 1,000

The axial force, in kN. Raises the strain energy with its square: doubling the force quadruples the energy. Raising it by 5 kN (to 55 kN) moves the result by +21 %, from 25 to 30.25 J.

L

Length

m · typical 0.1 to 20

The length of the member, in m. Raises the strain energy in direct proportion: a longer bar stores more energy at the same force. Raising it by 0.25 m (to 2.25 m) moves the result by +12.5 %, from 25 to 28.13 J.

E

Young’s modulus

GPa · typical 1 to 500

Young’s modulus of the material, in GPa. Lowers the strain energy in inverse proportion: a stiffer bar stretches less and stores less energy. Raising it by 10 GPa (to 210 GPa) moves the result by −4.76 %, from 25 to 23.81 J.

A

Cross-sectional area

mm² · typical 10 to 10,000

The cross-section area, in mm². Lowers the strain energy in inverse proportion: a thicker bar stretches less and stores less energy. Raising it by 50 mm² (to 550 mm²) moves the result by −9.09 %, from 25 to 22.73 J.

Worked example with the values above

  1. Force in newtons: F = 50 kN = 50,000 N; area 500 mm² = 5×10−4 m²; E = 2×1011 Pa.
  2. Numerator F²L = 2.5×109 × 2 = 5×109; denominator 2EA = 2×108.
  3. U = 25 J.

Assumptions and limits

  • A linear elastic, homogeneous and isotropic material, with strains small.
  • Stresses in MPa with moduli in GPa, giving strain in microstrain.
  • Uniform axial stress over the length and a prismatic member, with no buckling.

Common mistakes

  • Mixing units: kN with mm² and GPa without conversion.
  • Using it above the elastic limit.
  • Adding energies of different load cases directly: energies from separate loads do not add, but forces do.

Statistics and uncertainty

Because energy rises with F², a 2 % error in the force gives a 4 % error in the energy, so the load is the most important input to know accurately.

Where it is used

  • Energy stored in a bolt, tie or spring-like bar, and impact loading.
  • Energy methods for deflection and for sizing a member to absorb energy.

Maths you need

  • Stress and strain: force per area and relative change in length
  • Elastic modulus: stress divided by strain in the elastic range

Related formulas

Practise this formula