Module 11: Strain Energy

Energy stored in a material under deformation.
When a structural member is subjected to a load, it deforms. This deformation requires work to be done on the member. As long as the material remains within its elastic limit, this work is stored internally as elastic strain energy, much like a compressed spring storing energy. Understanding strain energy is vital for evaluating a material's capacity to absorb impact loads without failing.

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Strain Energy

When a gradually applied external force PP causes an elastic deformation δ\delta, the work done by the force is converted entirely into internal strain energy (UU). If the load is removed, the material releases this energy as it returns to its original shape.
For a linear elastic material, the load-deformation graph is a straight line, and the stored strain energy is equal to the area under this triangle: U=12PδU = \frac{1}{2}P\delta.

Strain Energy Formula

Quantifying stored energy
For a member subjected to an axial load PP, the total stored strain energy (UU) can be expressed using Hooke's Law for deformation (δ=PLAE\delta = \frac{PL}{AE}).
U=12Pδ=12P(PLAE)=P2L2AE U = \frac{1}{2} P \delta = \frac{1}{2} P \left( \frac{PL}{AE} \right) = \frac{P^2 L}{2AE}
Where:
  • UU = Strain Energy (Joules, N-m, or lb-in)
  • PP = Axial Load
  • LL = Length of the member
  • AA = Cross-sectional area
  • EE = Modulus of Elasticity

Strain Energy

The internal energy stored in a structural material as a result of deformation caused by external loads.
Key Takeaways
  • Strain energy (UU) is the internal work stored due to deformation.
  • For an axially loaded linear elastic member, U=P2L/2AEU = P^2L / 2AE.

Strain Energy Density

Energy per unit volume
Instead of total energy, engineers often evaluate materials based on their strain energy density (uu), which is the strain energy stored per unit volume of the material. This allows for direct comparison between different materials, independent of the member's physical size.
The strain energy density can be found by dividing total energy UU by the volume V=A⋅LV = A \cdot L:
u=UV=P2L2AEAL=P22A2E u = \frac{U}{V} = \frac{\frac{P^2 L}{2AE}}{AL} = \frac{P^2}{2A^2E}
Since normal stress is σ=PA\sigma = \frac{P}{A}, we can express strain energy density in terms of stress:
u=σ22E u = \frac{\sigma^2}{2E}
Key Takeaways
  • Strain energy density (u=σ2/2Eu = \sigma^2 / 2E) normalizes stored energy by volume, allowing direct comparison of different materials.

Modulus of Resilience and Toughness

Material properties for energy absorption
Two critical material properties derived from the stress-strain diagram define a material's capacity to absorb energy.

Modulus of Resilience (uru_r)

The modulus of resilience is the maximum amount of strain energy density a material can absorb without suffering any permanent, plastic deformation. It is exactly equal to the area under the elastic (linear) portion of the stress-strain curve up to the yield point (σy\sigma_y).
ur=σy22E u_r = \frac{\sigma_y^2}{2E}
Materials with a high modulus of resilience (like spring steel) can absorb a lot of energy and perfectly snap back to their original shape.

Modulus of Toughness (utu_t)

The modulus of toughness is the total amount of strain energy density a material can absorb before it fractures or ruptures completely. It is equal to the entire area under the stress-strain curve, from origin to the rupture point.
Tough materials (like structural steel) are usually ductile, meaning they undergo massive plastic deformation before breaking, absorbing huge amounts of energy in the process. Brittle materials (like glass or unreinforced concrete) have a very low modulus of toughness because they cannot deform plastically.

Modulus of Toughness

A measure of a material's ability to absorb total energy and deform plastically before fracturing. It is represented by the entire area under the stress-strain diagram.

Elastic Strain Energy Simulation

Calculations

Total Deformation ($\delta$): 2.000 mm
Total Strain Energy ($U$): 100.00 Joules
Strain Energy Density ($u$): 100000.00 J/m³

Load-Deformation Curve (Area = Strain Energy)

Loading chart...

The total Strain Energy ($U$) is the physical work done by the axial load as it deforms the member. In the linear elastic region, this is represented by the triangular area under the load-deformation curve:

U=12PδU = \frac{1}{2} P \delta
Key Takeaways
  • Modulus of Resilience is the max energy absorbed without plastic deformation (area under elastic region).
  • Modulus of Toughness is the total energy absorbed before rupture (total area under curve).

Impact Loading

When loads are applied dynamically
Until now, we have assumed loads are applied slowly and gradually (static loading). When a load is applied suddenly—such as a weight dropped onto a beam or a vehicle striking a barrier—the resulting dynamic stress and deformation are significantly higher than if the exact same load were placed gently.
This is solved using the principle of conservation of energy: the kinetic energy or potential energy of the falling object is completely converted into internal strain energy within the structural member at the point of maximum deflection.
For a mass WW dropped from a height hh onto a member, causing a maximum dynamic deflection δmax\delta_{max}:
Work Done by Weight = Internal Strain Energy Stored
W(h+δmax)=12Pmaxδmax W(h + \delta_{max}) = \frac{1}{2} P_{max} \delta_{max}
By solving this relationship, engineers calculate an impact factor (nn), which is the ratio of dynamic stress to static stress.

Important

Even if an object is dropped from zero height (h=0h=0) but released suddenly, the impact factor is exactly 2. This means a suddenly applied load causes twice the stress and twice the deformation of a gradually applied static load of the same weight.
Key Takeaways
  • Sudden or dynamic loads cause significantly higher stresses and deflections than gradual static loads.
  • A load applied suddenly (even dropped from zero height) produces exactly twice the stress of a static load.

Specific Formulas for Strain Energy

Calculating energy stored under different types of loading
The general concept of strain energy (UU) applies to all types of elastic deformation. By substituting the specific relationships between load and deformation for different loading conditions, we can derive explicit formulas for the strain energy stored in a member.
1. Strain Energy for Axial Loading
For a uniform bar of length LL, cross-sectional area AA, and Modulus of Elasticity EE, subjected to an axial load PP, the deformation is δ=PLAE\delta = \frac{PL}{AE}. Substituting this into the work equation (U=12PδU = \frac{1}{2}P\delta) gives:
U=P2L2AE U = \frac{P^2 L}{2 A E}
2. Strain Energy for Bending (Flexure)
When a beam bends under a moment MM, the strain energy is stored as internal work done by the bending moments across the length of the beam. The total strain energy is found by integrating over the beam's length:
U=∫0LM22EIdx U = \int_{0}^{L} \frac{M^2}{2 EI} dx
Where:
  • MM = Bending moment as a function of position xx
  • EE = Modulus of Elasticity
  • II = Moment of Inertia of the cross-section
3. Strain Energy for Torsion
For a circular shaft of length LL, polar moment of inertia JJ, and Modulus of Rigidity GG, subjected to a twisting moment (torque) TT, the angle of twist is θ=TLJG\theta = \frac{TL}{JG}. The strain energy is the work done by the torque:
U=T2L2JG U = \frac{T^2 L}{2 J G}

Impact Loading

Strain energy is critical when analyzing impact loading (e.g., a weight dropped onto a beam or a collar on a rod). Instead of a slowly applied static load, an impact load applies kinetic energy suddenly. By equating the external kinetic energy of the falling object to the internal strain energy capacity of the member, engineers can calculate the much higher "dynamic" stresses (σdynamicσ_{dynamic}) and deformations (δdynamic\delta_{dynamic}) caused by impact. A suddenly applied load (dropped from zero height) produces exactly twice the stress and deflection of the same load applied statically.
Key Takeaways
  • Axial Strain Energy: U=P2L/(2AE)U = P^2 L / (2AE).
  • Bending Strain Energy: U=∫M2/(2EI)dxU = \int M^2 / (2EI) dx.
  • Torsional Strain Energy: U=T2L/(2JG)U = T^2 L / (2JG).
  • Strain energy concepts are essential for solving impact loading problems, where dynamic stresses significantly exceed static stresses.

Castigliano's Theorem

Energy methods for deflection
The Italian engineer Carlo Alberto Castigliano developed powerful energy methods for structural analysis. His second theorem states that the partial derivative of the total strain energy (UU) of any structure with respect to an applied load (PP) is equal to the displacement (δ\delta) at the point of application and in the direction of that load.
δi=∂U∂Pi \delta_i = \frac{\partial U}{\partial P_i}
This theorem provides an incredibly elegant way to find deflections in complex structures (like curved beams or frames) where geometric methods like Double Integration become mathematically unwieldy.
Key Takeaways
  • Castigliano's theorem (δi=∂U/∂Pi\delta_i = \partial U / \partial P_i) uses strain energy derivatives to find structural deflections.

Maxwell-Betti Reciprocal Theorem

Symmetry of influence coefficients
The reciprocal theorem states that for any linear elastic structure, the deflection at point AA caused by a unit load at point BB is exactly equal to the deflection at point BB caused by a unit load at point AA (δAB=δBA\delta_{AB} = \delta_{BA}).
This profound property of linear elastic systems is heavily utilized in advanced structural analysis matrix methods.
Key Takeaways
  • The Maxwell-Betti reciprocal theorem states that δAB=δBA\delta_{AB} = \delta_{BA} for linear elastic structures.

Castigliano's Second Theorem Applications

Deflection of structures using energy

Applying Castigliano's Theorem

Castigliano's Second Theorem (Δ=∂U/∂P\Delta = \partial U / \partial P) is a highly versatile tool for finding deflections in complex structures like trusses, beams, and frames.
For Trusses (Axial Loads): The internal strain energy is U=∑(N2L/2AE)U = \sum (N^2 L / 2AE). The theorem becomes Δ=∑[N(∂N/∂P)L/AE]\Delta = \sum [N (\partial N / \partial P) L / AE], where NN is the internal axial force in each member. For Beams (Bending Moments): The internal strain energy is U=∫(M2/2EI)dxU = \int (M^2 / 2EI) dx. The theorem becomes Δ=∫[M(∂M/∂P)/EI]dx\Delta = \int [M (\partial M / \partial P) / EI] dx, where MM is the internal bending moment equation.
If no actual load exists at the point where deflection is desired, a "fictitious" dummy load QQ is applied, the partial derivative ∂U/∂Q\partial U / \partial Q is taken, and then QQ is set to zero.
Key Takeaways
  • Castigliano's theorem can find deflections in trusses using internal axial forces (NN) and in beams using internal bending moments (MM).
  • A "dummy load" method is used when no actual physical load is applied at the point of interest.
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