Mechanics of Materials · Chapter 14

Energy Methods

Work and strain energy, impact loads, and deflections by virtual work and Castigliano's theorem.

Chapter Objectives

Express the work of forces and couples, and the strain energy stored under axial, bending, shear, and torsional loadings.
Apply conservation of energy to deflections and to impact problems (falling and suddenly applied loads).
Find displacements and slopes with the method of virtual forces for trusses and beams.
Apply Castigliano's second theorem as an alternative energy route to the same deflections.

14.1External Work and Strain Energy

Work of a Force and of a Couple Moment (Linear Elastic)

\[ U_e = \int_0^\Delta F\,dx = \frac{1}{2}P\Delta \qquad\qquad U_e = \int_0^\theta M\,d\theta = \frac{1}{2}M\theta \]

The ½ appears because a gradually applied load grows from zero — the work equals the average force times the displacement (the triangular area under the F–x line).
If a load \(P'\) is already acting when further displacement \(\Delta'\) occurs, its additional work is the full \(P'\Delta'\) (a rectangular area) — no ½. This distinction is central to the virtual-work method.

Strain Energy

If no energy is lost to heat, the external work done by the loads converts to internal strain energy stored in the deformed body. For uniaxial normal stress the energy per unit volume is \(u = \sigma^2/2E\) (Section 3.4); for shear it is \(u = \tau^2/2G\). Integrating these densities over a member's volume for each loading type gives the working formulas of the next section.

14.2Elastic Strain Energy for Various Types of Loading

Strain Energy — Axial Load

\[ U_i = \int_0^L \frac{N^2}{2AE}\,dx \qquad\qquad U_i = \frac{N^2L}{2AE} \;\text{(constant N, A)} \]

Energy capacity rises with length and falls with stiffness: an aluminum rod stores ~3× the energy of an identical steel rod under the same load, and doubling A halves the stored energy.

Strain Energy — Bending, Transverse Shear, and Torsion

\[ U_i = \int_0^L \frac{M^2}{2EI}\,dx \qquad\quad U_i = \int_0^L \frac{f_s V^2}{2GA}\,dx \qquad\quad U_i = \int_0^L \frac{T^2}{2GJ}\,dx = \frac{T^2L}{2GJ} \]

\(f_s\) = form factor for shear, a dimensionless property of the cross-section shape (\(f_s = 6/5\) for a rectangle), arising because the shear stress VQ/It is non-uniform.
For slender beams the shear strain energy is negligible compared with bending energy — comparing the two for a rectangular cantilever shows the shear share matters only when the depth approaches the span. This is why beam deflection methods keep only the M²/2EI term.

14.3Conservation of Energy

Conservation of (Mechanical) Energy

\[ U_e = U_i \qquad\quad \text{Truss: } \frac{1}{2}P\Delta = \sum\frac{N^2L}{2AE} \qquad\quad \text{Beam: } \frac{1}{2}P\Delta = \int_0^L \frac{M^2}{2EI}\,dx \]

Valid for gradually applied loads, elastic behavior, and no heat loss.
Limitation: only one external load may act, and the method delivers only the displacement under that load, in its direction — the motivation for the more general virtual-work method.

14.4Impact Loading

Energy Analysis of Impact

Dynamic loads — one object striking another — are analyzed with conservation of energy, assuming no losses to heat, sound, or local plasticity, the struck member responds elastically, and the striking body does not rebound (maximum energy transfer). For a block of weight W falling a height h onto an elastic member (spring constant k, static deflection \(\Delta_{st} = W/k\)), equating the falling work to the stored energy yields a quadratic whose largest root is:

Maximum Impact Displacement and Impact Factor

\[ \Delta_{max} = \Delta_{st}\left[1 + \sqrt{1 + \frac{2h}{\Delta_{st}}}\right] \qquad\qquad n = \frac{\Delta_{max}}{\Delta_{st}}, \quad \sigma_{max} = n\,\sigma_{st} \]

\(n\) = the impact factor: once known, the dynamic problem becomes a static one magnified n times.
Suddenly applied load (h = 0): \(n = 2\) — a load released instantly produces twice the static displacement and stress. For a horizontally moving mass with velocity v, replace the drop energy by \(\tfrac{1}{2}(W/g)v^2\).

14.5Principle of Virtual Work

Principle · Virtual Work (Bernoulli, 1717) — Method of Virtual Forces

To find the displacement \(\Delta\) at a point where no load acts: apply an imaginary virtual unit force P′ = 1 there (in the direction of Δ) before the real loads; it creates internal virtual loads u in the body's elements. Applying the real loads then displaces the point by Δ and deforms each element by dL. Conserving the virtual energy:

\[ 1\cdot\Delta = \sum u\cdot dL \qquad\qquad 1\cdot\theta = \sum u_\theta\cdot dL \]

For a slope θ, the virtual load is a unit couple M′ = 1. The full P′Δ (no ½) appears because the virtual load is already acting when the real displacement occurs. The equation is a statement of compatibility; displacement in any direction at any point is obtainable — overcoming the one-load limitation of conservation of energy.

14.6Method of Virtual Forces Applied to Trusses

Virtual Work — Truss Joint Displacement

\[ 1\cdot\Delta = \sum \frac{n\,N\,L}{A\,E} \]

\(n\) = internal force in each member from the virtual unit load alone; \(N\) = member force from the real loads; \(L, A, E\) per member
Temperature change contributes \(\Sigma\,n\,\alpha\,\Delta T\,L\); fabrication errors contribute \(\Sigma\,n\,\Delta L\) — virtual work handles them all in one sum.
Sign convention: tension positive in both n and N; a positive result means Δ acts in the direction of the unit load.

14.7Method of Virtual Forces Applied to Beams

Virtual Work — Beam Displacement and Slope

\[ 1\cdot\Delta = \int_0^L \frac{m\,M}{EI}\,dx \qquad\qquad 1\cdot\theta = \int_0^L \frac{m_\theta\,M}{EI}\,dx \]

\(m\) (or \(m_\theta\)) = internal moment from the virtual unit force (or unit couple) alone; \(M\) = internal moment from the real loads
Use the same x-coordinates and regions for m and M; the moments multiply, so consistent sign conventions matter. Shear deformation is neglected (slender beams).

14.8Castigliano's Theorem

Theorem · Castigliano's Second Theorem (1879)

For a body of linear elastic material under constant temperature, the displacement at a point in the direction of a force P acting there equals the partial derivative of the total strain energy with respect to that force:

\[ \Delta = \frac{\partial U_i}{\partial P} \qquad\qquad \theta = \frac{\partial U_i}{\partial M'} \]

If no force acts at the point of interest, apply a fictitious force there, differentiate, then set it to zero. Differentiating under the integral sign gives the working forms:

\[ \text{Truss: } \Delta = \sum N\left(\frac{\partial N}{\partial P}\right)\frac{L}{AE} \qquad\qquad \text{Beam: } \Delta = \int_0^L M\left(\frac{\partial M}{\partial P}\right)\frac{dx}{EI} \]

Comparing with Sections 14.6–14.7: \(\partial N/\partial P\) plays exactly the role of n, and \(\partial M/\partial P\) the role of m — Castigliano's theorem and the method of virtual forces are two routes to the same equations.

Physical meaning

Strain energy is recoverable deformation work: the area beneath a force-displacement or moment-rotation curve. It is a scalar measure accumulated through the member, while displacement is directional and must be tied to a point and coordinate direction.

Problem-solving strategy

Choose energy only after identifying the desired displacement, loading history, and whether linear elasticity is credible.
Draw real-load free bodies and write internal \(N\), \(M\), \(T\), or \(V\) functions over each region.
Select conservation of energy for a compatible single load, virtual work for a targeted displacement, or Castigliano for derivatives of total energy.
Use a unit virtual force or couple in the requested positive direction and retain sign consistently in the products.
Check energy units, compatibility, impact assumptions, stress limits, and whether neglected shear energy is small.

EXAMPLE 1Energy stored in a tension tie

A uniform steel tension tie carries a gradually applied \(60 \text{kN}\) load. Its length is \(1.80 \text{m}\), area is \(600 \text{mm}^2\), and \(E=200 \text{GPa}\). Determine its axial extension and stored elastic strain energy.

Recognition strategy

The load is axial, gradual, and within a stated linear-elastic material model, so either \(U=P^2L/(2AE)\) or the force-displacement work area applies.

Given information

Use \(P=60000 \text{N}\), \(L=1.80 \text{m}\), \(A=600\times10^{-6} \text{m}^2\), and \(E=200\times10^9 \text{Pa}\).

Model and interpretation

The axial free-body has a constant tensile internal resultant \(N=P\). Extension is positive; the tie is prismatic, centrally loaded, and assumed far enough from grips that average axial deformation is representative.

Step-by-step solution

1. Plan. First obtain \(AE=(600\times10^{-6})(200\times10^9)=1.20\times10^8 \text{N}\), the axial stiffness numerator used in the deformation relation.

2. Formulate. Use compatibility and Hooke's law through \(\delta=PL/(AE)\), which relates the constant internal force to the uniform axial strain.

3. Calculate. Substitute to get \(\delta=60000(1.80)/(1.20\times10^8)=9.00\times10^{-4} \text{m}=0.900 \text{mm}\).

4. Resolve. Compute recoverable work as \(U=\tfrac12P\delta=\tfrac12(60000)(9.00\times10^{-4})=27.0 \text{J}\), equivalently \(P^2L/(2AE)\).

Verification

Units: Multiplying newtons by metres gives joules, so \(27.0 \text{N}\!\cdot\!\text{m}\) is a valid strain-energy result.

Signs and direction: Tension and positive extension were selected together; energy is positive regardless of whether the axial force convention labels the member force positive or negative.

Plausibility: The force-displacement graph is triangular for gradual loading, so average force \(30 \text{kN}\) times \(0.900 \text{mm}\) independently gives \(27.0 \text{J}\).

Final result: \(\boxed{\delta=0.900 \text{mm},\quad U=27.0 \text{J}}\).

Engineering interpretation

The energy is modest because the tie is stiff, but it can be released if a connection fails. Check actual stress, net section, grip effects, and yielding separately; stored energy does not itself prove strength adequacy.

EXAMPLE 2Cantilever tip displacement by virtual work

A prismatic cantilever is \(2.40 \text{m}\) long with \(E=200 \text{GPa}\) and \(I=18.0\times10^{-6} \text{m}^4\). A downward \(8.00 \text{kN}\) force acts at its free end. Use a unit virtual force to find the vertical tip displacement, neglecting shear deformation.

Recognition strategy

The desired displacement lies at a loaded point, but the virtual-force integral directly targets its direction and exposes the moment distribution. Slender-beam bending energy is the selected model.

Given information

Take \(x\) from the free end toward the fixed support, \(L=2.40 \text{m}\), \(P=8000 \text{N}\), and a downward unit virtual force.

Model and interpretation

The real and virtual free bodies give \(M=-Px\) and \(m=-x\) for the same sign convention. Euler-Bernoulli kinematics assume small rotations, constant \(EI\), and negligible shear strain energy.

Step-by-step solution

1. Plan. Place the unit virtual load downward at the tip, so its internal moment is \(m=-x\); this establishes that a positive virtual-work result corresponds to downward displacement.

2. Formulate. From the real-load free body write \(M=-8000x\), valid over the entire span because no intermediate loads or changes in stiffness occur.

3. Calculate. Integrate \(\Delta=\int_0^L mM/(EI),dx=\int_0^{2.40}8000x^2/[(200\times10^9)(18.0\times10^{-6})]dx\).

4. Resolve. Evaluate the integral as \(PL^3/(3EI)=8000(2.40)^3/[3(200\times10^9)(18.0\times10^{-6})]=0.0102 \text{m}\).

Verification

Units: The integrand \(mM/(EI)\) has units of inverse length, so integration gives length; \(0.0102 \text{m}\) is correctly a displacement.

Signs and direction: Both \(m\) and \(M\) are negative under the selected moment convention, making their product positive and confirming displacement in the downward unit-load direction.

Plausibility: The result is \(10.2 \text{mm}\); it matches the known cantilever limit \(PL^3/(3EI)\), and doubling span would increase it eightfold.

Final result: \(\boxed{\Delta_{tip}=10.2 \text{mm downward}}\).

Engineering interpretation

Virtual work is especially efficient when the requested displacement is not aligned with a real load. The result should still be checked against deflection limits, maximum bending stress, vibration requirements, and the validity of neglecting shear deformation.

EXAMPLE 3Drop load on an elastic isolation mount

A \(500 \text{N}\) instrument package drops \(50.0 \text{mm}\) onto a linear elastic isolation mount with \(k=50.0 \text{kN/m}\). Neglect damping, rebound, and local contact deformation. Determine maximum compression and maximum mount force using energy conservation.

Recognition strategy

A finite drop height means static formulas alone are insufficient. The supplied linear stiffness permits an impact-energy balance from release through maximum compression, where velocity is momentarily zero.

Given information

Use \(W=500 \text{N}\), \(h=0.0500 \text{m}\), \(k=50000 \text{N/m}\), and define downward compression as positive.

Model and interpretation

The package free body does positive gravitational work through \(h+\delta\); the mount stores \(\tfrac12k\delta^2\). This ideal model assumes a single linear spring and does not represent rebound, damping, plasticity, or force-wave effects.

Step-by-step solution

1. Plan. Find static compression \(\delta_{st}=W/k=500/50000=0.0100 \text{m}\), which supplies a useful scale for the dynamic response.

2. Formulate. Use \(\delta_{max}=\delta_{st}[1+\sqrt{1+2h/\delta_{st}}]\), derived by equating \(W(h+\delta)\) to \(\tfrac12k\delta^2\).

3. Calculate. Substitute the drop ratio: \(\delta_{max}=0.0100[1+\sqrt{1+2(0.0500)/0.0100}]=0.0432 \text{m}\).

4. Resolve. Calculate the peak spring force at maximum compression: \(F_{max}=k\delta_{max}=50000(0.0432)=2.16 \text{kN}\), or \(4.32W\).

Verification

Units: The energy equation contains \(\text{N}\!\cdot\!\text{m}\) on both sides, while \(k\delta\) gives newtons; the displacement and force units are therefore consistent.

Signs and direction: Downward displacement, gravitational work, and spring compression are all positive by definition. The mount force on the package is upward at peak compression, opposing the selected positive direction.

Plausibility: The impact factor \(4.32\) exceeds the sudden-load limit of \(2\) because the package falls before contact; setting \(h=0\) in the formula would recover that limiting case.

Final result: \(\boxed{\delta_{max}=43.2 \text{mm}}\) and \(\boxed{F_{max}=2.16 \text{kN}}\).

Engineering interpretation

The drop converts a small static load into more than four times the static mount force. A real isolation system needs stroke clearance, rated travel, damping and rebound characterization, package fragility limits, and contact-force validation.

Chapter recap

Relate gradual external work to recoverable strain energy only within an elastic, compatible model.
Build real and virtual internal-resultant functions on matching coordinates before integrating \(mM/(EI)\) or \(nNL/(AE)\).
Use Castigliano or virtual work to target a displacement, and distinguish directional displacement from scalar energy.
Before moving on: check units, load history, signs, material limits, neglected deformation modes, and dynamic-model assumptions.