Strain Energy – Tension, Torsion, Bending & Shear

Strain energy is the potential energy stored in a material as a result of elastic deformation. The external work done in deforming an elastic member from its unstressed condition is stored as strain energy and is recoverable when the load is removed, provided the material remains within its elastic range.

Need help choosing a beam calculation? See the Beam Design and Analysis Guide for guidance on beam types, loading, bending, shear, deflection and selecting the appropriate RoyMech calculation method.

Strain energy methods are useful in structural and mechanical engineering for calculating deformation and deflection and for analysing members subjected to axial load, torsion, bending and shear.

Nomenclature
Axial Tension and Compression
Pure Torsion
Direct Shear
Beam Bending
Transverse Shear in Beams
Bending and Shear Energy Comparison
Summary of Strain Energy Equations

Nomenclature

Symbol	Description	SI Unit
A	Cross-sectional area	m²
b	Section width	m
c	Distance from neutral axis to outer fibre	m
E	Young's modulus	Pa (N/m²)
F	Axial or applied force	N
G	Shear modulus (modulus of rigidity)	Pa (N/m²)
h	Section height	m
I	Second moment of area	m⁴
J	Torsion constant; polar second moment of area for circular shafts	m⁴
K	Transverse-shear strain-energy factor used on this page	Dimensionless
k	Linear or rotational stiffness, as applicable	N/m or N m/rad
l	Member length	m
M	Bending moment	N m
R	Radius of curvature of neutral surface	m
T	Torque	N m
U	Strain energy	J (N m)
V	Transverse shear force	N
x	Distance along member or beam	m
z	Distance from neutral axis	m
ε	Normal strain	Dimensionless
γ	Shear strain	Dimensionless
δ	Linear displacement	m
θ	Angular displacement or angle of twist	rad
σ	Normal stress	Pa (N/m²)
τ	Shear stress	Pa (N/m²)

The equations on this page assume linear-elastic material behaviour and small deformation unless otherwise stated.

Strain Energy in Axial Tension and Compression

For a uniform prismatic member subjected to an axial tensile or compressive force, the member behaves elastically as an axial spring.

Prismatic member subjected to axial tension or compression

Derivation of strain energy for axial tension or compression

k = \frac{F}{δ} and E = \frac{σ}{ε} = \frac{F l}{A δ} therefore k = \frac{A E}{l}

U = \frac{F δ}{2} = \frac{F^{2}}{2 k} = \frac{F^{2} l}{2 A E}

For a constant axial force, cross-sectional area and Young's modulus, the strain energy is therefore:

U = F²l / (2AE)

Strain Energy in Pure Torsion

A shaft subjected to torque stores elastic strain energy as it twists. For a uniform shaft the torsional stiffness depends upon the shear modulus, torsion constant and shaft length.

Shaft subjected to pure torsion showing torque and angle of twist

Derivation of strain energy for a shaft subjected to pure torsion

k = \frac{T}{θ} and G = \frac{T l}{J θ} therefore k = \frac{G J}{l}

U = \frac{T θ}{2} = \frac{T^{2}}{2 k} = \frac{T^{2} l}{2 G J}

For circular shafts, J is the polar second moment of area. For non-circular sections the appropriate torsion constant must be used.

Strain Energy in Direct Shear

A member subjected to direct shear stores strain energy through elastic shear deformation.

Element subjected to direct shear force and shear deformation

k = \frac{F}{δ} and G = \frac{τ}{γ} = \frac{F l}{A δ} therefore k = \frac{A G}{l}

U = \frac{F δ}{2} = \frac{F^{2}}{2 k} = \frac{F^{2} l}{2 A G}

For a uniform state of direct shear, the shear stress and strain are constant over the area. Where the stress varies across the section, the strain energy is obtained by integrating the local energy density.

U = \frac{F δ}{2} = \frac{τ A δ}{2} = \frac{A}{2} \int γ τ d z = \frac{b x}{2 G} \int τ^{2} d z

Strain Energy in Beam Bending

When a beam bends elastically, strain energy is stored as a result of the normal stresses produced by the bending moment.

Differential element of a beam subjected to pure bending

Derivation of strain energy for an elastic beam subjected to bending

d U = \frac{M d θ}{2} and d s = R d θ

d U = \frac{M d s}{2 R}

\frac{M}{I} = \frac{E}{R} therefore d U = \frac{M^{2} d s}{2 E I}

For small beam deflections, ds ≈ dx, giving:

U = \int \frac{M^{2} d x}{2 E I}

Constant Bending Moment

If the bending moment is constant along a member of length l, the integration reduces to:

Strain energy of a beam subjected to a constant bending moment

U = \frac{M^{2} l}{2 E I}

Simply Supported Beam with Central Load

The following example illustrates a bending moment that varies with position. The simply supported beam carries a central load of 2V, so each support reaction is V.

Strain energy calculation for a simply supported beam carrying a central load

U = \int_{0}^{l} \frac{M^{2} d x}{2 E I} = 2 \int_{0}^{\frac{l}{2}} \frac{M^{2} d x}{2 E I}

U = \frac{V^{2}}{E I} \int_{0}^{\frac{l}{2}} x^{2} d x = \frac{V^{2} l^{3}}{24 E I}

Strain Energy Due to Transverse Shear

A beam subjected to transverse loading generally develops both bending moment and transverse shear force. The bending moment produces longitudinal normal stress, while the transverse shear force produces a shear-stress distribution through the section.

The following derivation considers the strain energy associated with the transverse shear force.

Shear Stress in the Beam

Consider a differential slice of beam of length dx. The bending moment changes by V dx across the slice.

Differential beam element used to derive transverse shear stress from the change in bending stress

The longitudinal bending stress at distance z from the neutral axis is:

σ_{x} = \frac{M z}{I}

Across the differential length dx, the bending moment increases from M to M + V dx. The corresponding increase in longitudinal stress is therefore:

d σ_{x} = \frac{V d x z}{I}

The increase in longitudinal force over the part of the section from z = z₁ to the outer fibre z = c is balanced by the longitudinal shear stress.

F_{a} = \frac{V d x}{I} \int_{z_{1}}^{c} b z d z = τ_{x z} b d x

Here b is the local section width. For a rectangular section it is constant; for other sections it may vary through the section.

Solving for the shear stress:

τ_{x z} = \frac{V}{I} \int_{z_{1}}^{c} z d z

Rectangular Section

For a rectangular section of constant width b, the transverse shear-stress distribution is parabolic. The maximum shear stress occurs at the neutral axis and the shear stress reduces to zero at the outer fibres.

τ_{x z} = \frac{3 V}{2 A} [1 - {(\frac{z}{c})}^{2}]

Substituting this shear-stress distribution into the strain-energy expression for direct shear and integrating across the section gives:

U = \frac{b x}{2 G} \int_{- c}^{c} {[\frac{3 V}{2 b h} (1 - {(\frac{z}{c})}^{2})]}^{2} d z

For a solid rectangular section, c = h/2 and A = bh. The resulting transverse-shear strain energy for a length x subject to a constant shear force V is:

U = \frac{3 V^{2} x}{5 G b h}

Shear Energy Factors for Different Sections

The effect of a non-uniform shear-stress distribution can be represented using a section-dependent factor K.

U = \frac{K V^{2} x}{2 G A}

Section	K	Area used
Solid rectangle	1.20	Total section area
Solid circular section	1.11	Total section area
Thin-wall round tube	2.00	Total section area
Box section	1.00	Web area only
Structural section	1.00	Web area only

Where the shear force varies along the beam, the corresponding general expression is:

U = \int \frac{K V^{2}}{2 G A} d x

The factor K used here is a strain-energy multiplier. Some references instead define a shear correction factor using the reciprocal convention. The definition should therefore be checked before transferring values between formulae.

Comparison of Bending and Transverse Shear Strain Energy

For the simply supported rectangular beam considered above, of length l and carrying a central transverse load of 2V, the strain energy due to bending can be compared with that due to transverse shear.

Bending

U_{b} = \frac{V^{2} l^{3}}{24 E I} = \frac{V^{2} l^{3}}{2 E b h^{3}}

Transverse shear

U_{s} = 2 [\frac{3 V^{2} l}{10 G b h}] = \frac{3 V^{2} l}{5 G b h}

For many engineering materials, G is approximately 0.4E. Using this approximation:

U_{s} \approx \frac{3 V^{2} l}{2 E b h}

\frac{U_{b}}{U_{s}} \approx \frac{{(l / h)}^{2}}{3}

For example, if l/h = 10:

\frac{U_{b}}{U_{s}} \approx \frac{10^{2}}{3} \approx 33

The bending strain energy is therefore approximately 33 times the transverse-shear strain energy for this particular beam and loading case. This illustrates why shear deformation is often relatively small for slender beams.

Transverse-shear deformation should not automatically be neglected for deep beams, short members, sandwich structures or other arrangements where shear deformation is significant.

Summary of Strain Energy Equations

The principal strain-energy expressions for the loading cases considered on this page are summarised below.

Loading	Constant Variables	General Case
Axial	$U = \frac{F^{2} l}{2 E A}$	$U = \int \frac{F^{2}}{2 E A} d x$
Bending	$U = \frac{M^{2} l}{2 E I}$	$U = \int \frac{M^{2}}{2 E I} d x$
Torsion	$U = \frac{T^{2} l}{2 G J}$	$U = \int \frac{T^{2}}{2 G J} d x$
Direct Shear	$U = \frac{F^{2} l}{2 A G}$	$U = \int \frac{F^{2}}{2 A G} d x$
Transverse Shear	$U = \frac{K V^{2} l}{2 G A}$	$U = \int \frac{K V^{2}}{2 G A} d x$

Use of Strain Energy Methods

Strain energy provides an alternative route to displacement and deflection calculations. Instead of obtaining displacement directly from the governing differential equation, an energy expression can be formed from the internal forces and moments acting throughout the member.

This approach is particularly useful in energy methods such as Castigliano's theorem and in structures or machine components where several forms of elastic deformation contribute to the overall displacement.

For further methods of obtaining beam displacement from strain energy, see Beam Energy Methods.

Related Beam References

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Strain Energy in Beams and Engineering Components

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