Mechanical Engineering Design · Part 1 — Basics

Deflection and Stiffness

Spring rates and the deflection of bars, shafts, and beams — by direct integration, superposition, singularity functions, and the energy method of Castigliano — extended to curved members, statically indeterminate structures, column buckling, and impact loading.

4–1Spring Rates

Stiffness is a local slope

General spring rate is \(k(y)=dF/dy\); for a linear element, \(k=F/y\). Compliance \(c=1/k\) is often easier to combine. Stiffness describes load–deformation response, whereas strength describes resistance to failure; a component may be strong enough yet too flexible for alignment or vibration requirements.

Learning objectives

Recognize axial, torsional, bending, energy, stability, and impact models.
Compute displacement, slope, spring rate, and critical load from explicit boundaries.
Combine compatible deformations and solve one redundant reaction.
Check linearity, slenderness, units, signs, and serviceability.
Iterate geometry while documenting material and standards provenance.

Physical meaning

A steep load–deflection curve is stiff; its area is stored strain energy. Tangent stiffness governs small changes about the current operating point of a nonlinear device.

4–2Tension, Compression, and Torsion

Axial and torsional compliance

A uniform bar has \(\delta=FL/(AE)\) and \(k=AE/L\). A circular shaft has \(\theta=TL/(GJ)\) and torsional rate \(GJ/L\). These relations require small elastic deformation and appropriate section properties; compression members must also be checked for buckling.

Uniform-member stiffness

\[ k_a=\frac{AE}{L},\qquad k_t=\frac{GJ}{L} \]

EXAMPLE 1Compliance of a two-material tie

A concentric tie carries a steady tensile load of 40 kN through a 400 mm steel segment with area 600 mm² and a 250 mm aluminum segment with area 300 mm². Given \(E_s=200\ \text{GPa}\) and \(E_a=70\ \text{GPa}\), find total extension and equivalent axial stiffness.

Recognition strategy

This foundation problem is governed by elastic compliance: identify the deformation mode, choose the matching stiffness relation, and preserve compatibility and boundary conditions before evaluating serviceability.

Given information

Common tensile force \(F=40\ \text{kN}\); perfect series connection with no intermediate load.
Steel: \(L_s=400\ \text{mm}\), \(A_s=600\ \text{mm}^2\), \(E_s=200000\ \text{MPa}\).
Aluminum: \(L_a=250\ \text{mm}\), \(A_a=300\ \text{mm}^2\), \(E_a=70000\ \text{MPa}\).

Model and interpretation

Each prismatic segment is an axially loaded, linearly elastic spring. Series compatibility makes total extension the sum of segment extensions. Tension and elongation are positive; joint slip, thread compliance, self-weight, temperature, and stress concentrations are neglected.

Step-by-step solution

1. Plan. Because the elements are in series, the same \(40000\ \text{N}\) passes through each segment, while their compliances \(L/(AE)\) add.

2. Formulate. Compute steel extension: \(\delta_s=40000(400)/[600(200000)]=0.1333\ \text{mm}\).

3. Calculate. Compute aluminum extension: \(\delta_a=40000(250)/[300(70000)]=0.4762\ \text{mm}\).

4. Resolve. Add compatible deformations to obtain \(\delta=0.6095\ \text{mm}\), then calculate \(k_{eq}=F/\delta=65.63\ \text{kN/mm}\).

Verification

Units: In \(FL/(AE)\), \(\text{N}\!\cdot\!\text{mm}/[(\text{mm}^2)(\text{N/mm}^2)]\) reduces to millimetres; \(F/\delta\) is \(\text{N/mm}\).

Signs and direction: Both segment extensions are positive because the applied force is tensile. A compressive load under the same convention would reverse deformation signs but retain stiffness magnitude.

Plausibility: The smaller, lower-modulus aluminum segment contributes about 78 percent of total compliance, which is consistent with its much smaller \(AE/L\) value.

Final result: \(\boxed{\delta_{total}=0.6095\ \text{mm},\quad k_{eq}=65.63\ \text{kN/mm}}\)

Engineering interpretation

Stiffness controls displacement but does not establish strength. The segment stresses are \(66.7\ \text{MPa}\) in steel and \(133.3\ \text{MPa}\) in aluminum; allowable stress, fatigue, joints, tolerance, and thermal mismatch require separate checks.

4–3Deflection Due to Bending

Curvature links load to shape

For an Euler–Bernoulli beam, \(EIy^{\prime\prime}=M(x)\) under the chapter sign convention. Differentiation links load, shear, moment, slope, and deflection; integration constants encode support conditions. Small slopes, linear elasticity, and plane sections remaining plane define the model’s domain.

Beam relation

\[ EI\frac{d^2y}{dx^2}=M(x),\qquad \theta=\frac{dy}{dx} \]

4–4Beam Deflection Methods

Select the simplest defensible method

Direct integration is transparent for few regions; singularity functions handle discontinuities; superposition uses verified linear load cases; Castigliano targets selected displacements and redundants; numerical models serve complex geometry. Method choice does not remove the need for equilibrium and boundary-condition checks.

4–5Beam Deflections by Superposition

Superposition requires linearity

Deflections from separate loads may be added only when material response is linear, geometry changes are small, and supports do not change state. Keep a common coordinate and sign convention, and confirm every tabulated case uses matching restraints, load location, and section-property definition.

EXAMPLE 2Camera-arm tip alignment

An aluminum camera arm is modeled as a 600 mm long cantilever with a 40 mm by 60 mm rectangular section, the 60 mm dimension vertical. A downward tip load of 1.50 kN acts through the section centroid. For \(E=70\ \text{GPa}\), determine tip slope and deflection and compare deflection with a 2.5 mm alignment limit.

Recognition strategy

This intermediate problem is governed by elastic compliance: identify the deformation mode, choose the matching stiffness relation, and preserve compatibility and boundary conditions before evaluating serviceability.

Given information

Length \(L=600\ \text{mm}\), load \(P=1500\ \text{N}\), modulus \(E=70000\ \text{N/mm}^2\).
Rectangle width \(b=40\ \text{mm}\), vertical depth \(h=60\ \text{mm}\).
Fixed root and free tip; downward displacement is negative.

Model and interpretation

Use Euler–Bernoulli beam theory with constant \(EI\), small slope, linear elasticity, and an ideally rigid clamp. The centroidal load causes bending about the horizontal neutral axis; shear deformation and local clamp flexibility are initially omitted.

Step-by-step solution

1. Plan. Select \(I=bh^3/12\) for the actual bending axis, giving \(I=40(60)^3/12=720000\ \text{mm}^4\).

2. Formulate. Use cantilever relations \(\theta_L=-PL^2/(2EI)\) and \(y_L=-PL^3/(3EI)\), with negative signs assigned to clockwise slope and downward motion.

3. Calculate. Substitution gives \(\theta_L=-1500(600)^2/[2(70000)(720000)]=-0.005357\ \text{rad}\).

4. Resolve. The tip deflection is \(y_L=-1500(600)^3/[3(70000)(720000)]=-2.143\ \text{mm}\), leaving \(0.357\ \text{mm}\) below the stated limit.

Verification

Units: The slope ratio is dimensionless and reported in radians; \(PL^3/(EI)\) reduces from \(\text{N}\!\cdot\!\text{mm}^3/(\text{N/mm}^2\!\cdot\!\text{mm}^4)\) to millimetres.

Signs and direction: Negative slope and deflection agree with the chosen upward-positive coordinate and a downward load. Their magnitudes, not signs, are compared with the alignment specification.

Plausibility: The small-slope ratio \(|y|/L=0.00357\) supports geometric linearization. Cubic length sensitivity explains why modest span growth would quickly consume the margin.

Final result: \(\boxed{\theta_L=-5.36\times10^{-3}\ \text{rad},\quad y_L=-2.14\ \text{mm};\ \text{limit met by }0.36\ \text{mm}}\)

Engineering interpretation

The nominal beam model passes the stated displacement limit, but a professional release should include clamp and joint compliance, load eccentricity, tolerances, vibration frequency, creep or temperature effects, and verified material modulus.

4–6Beam Deflections by Singularity Functions

One expression across beam regions

Macaulay brackets activate at load or property changes, allowing repeated integration without separate equations for every interval. Constants follow from physical boundary and continuity conditions. A stepped \(EI\) often requires piecewise curvature or compliance expressions even when moment has one compact form.

4–7Strain Energy

Strain energy measures elastic work

For a linear spring, \(U=F\delta/2=F^2/(2k)\). Distributed energy includes axial \(N^2/(2AE)\), torsional \(T^2/(2GJ)\), and bending \(M^2/(2EI)\) terms. Shear energy can matter in short deep beams; its correction factor must come from an identified section model.

Elastic strain energy

\[ U=\int\left(\frac{N^2}{2AE}+\frac{T^2}{2GJ}+\frac{M^2}{2EI}\right)dx \]

4–8Castigliano's Theorem

Castigliano targets a displacement

For a linearly elastic structure, displacement in a load direction is \(\partial U/\partial F_i\), and rotation in a moment direction is \(\partial U/\partial M_i\). Introduce a dummy load where no real load exists, differentiate, then set it to zero. Include every energy mode significant to the requested accuracy.

Castigliano’s theorem

\[ \delta_i=\frac{\partial U}{\partial F_i},\qquad \theta_i=\frac{\partial U}{\partial M_i} \]

4–9Deflection of Curved Members

Curved members couple geometry and compliance

Hooks, rings, and curved frames may have bending, axial, and shear energy simultaneously. Write internal resultants as functions of angle or arc length and use the correct curved geometry; a straight-member approximation is acceptable only after a sensitivity or limiting check.

4–10Statically Indeterminate Problems

Compatibility supplies the missing equation

A statically indeterminate structure has more reactions than equilibrium equations. Release one redundant, calculate displacement at the release from loads and the redundant, and enforce the actual compatibility condition. Support settlement, thermal strain, and joint compliance belong in that condition, not as afterthoughts.

Problem-solving strategy

Identify the requested motion, deformation modes, and service limit.
Draw the load model and write equilibrium before a deflection equation.
Select a method matching load continuity, geometry, and desired outputs.
Apply support, continuity, and compatibility conditions with explicit signs.
Audit dimensions, limiting cases, strain level, and neglected compliance.
Resize and repeat because \(A\), \(I\), \(J\), and buckling regime change with geometry.

4–11Compression Members — General

Compression introduces instability

A compression member can lose lateral stability at stress far below yield. Effective length represents end restraint, while radius of gyration \(k=\sqrt{I/A}\) measures section efficiency. Initial crookedness and eccentricity turn ideal bifurcation into progressive bending, so real capacity is lower than a perfect-column prediction.

4–12Long Columns with Central Loading

Euler model for slender columns

For an ideal slender column, \(P_{cr}=\pi^2EI/L_e^2\). Use the weakest bending axis and a defensible effective length. Euler predicts an elastic instability and should not be used below the material-and-restraint transition slenderness.

Euler critical load

\[ P_{cr}=\frac{\pi^2EI}{L_e^2}=\frac{\pi^2EA}{(L_e/k)^2} \]

FIG. 4.1Column design map: the Euler curve governs slender columns; the J. B. Johnson parabola (tangent at T) governs intermediate columns below (l/k)₁. (Original diagram, redrawn.)

4–13Intermediate-Length Columns with Central Loading

Intermediate columns need inelastic guidance

When slenderness falls below the Euler transition, material yielding and residual stress interact with buckling. Johnson-type parabolic relations are educational approximations with stated material and end-condition inputs; production design should use the applicable structural or machinery standard and certified properties.

4–14Columns with Eccentric Loading

Eccentricity amplifies bending

A load offset creates primary moment, and lateral deflection creates additional \(P\text{–}\Delta\) moment. Secant-column formulas assume specific restraints and elastic behavior. Small uncertainty in eccentricity can dominate peak stress near instability, so tolerances and joint alignment are design inputs.

4–15Struts or Short Compression Members

Short struts are strength-controlled

Stocky compression members usually reach crushing, yielding, or local bearing limits before Euler buckling. Check direct stress, local geometry, and material failure criteria while still screening local plate or shell buckling where thin walls exist.

4–16Elastic Stability

Stability depends on equilibrium path

Elastic stability concerns whether a small disturbance grows or decays. Eigenvalue buckling is useful for mode recognition but commonly overpredicts real capacity because it omits imperfections and nonlinearities. Nonlinear analysis and tests require calibrated imperfection amplitudes and boundary stiffness.

4–17Shock and Impact

Impact converts energy into peak deformation

For a weight \(W\) dropping height \(h\) onto a linear spring, energy gives \(\tfrac12k\delta^2=W(h+\delta)\). The positive physical root determines maximum compression and \(F_{max}=k\delta\). This idealization neglects mass distribution, damping, contact duration, plasticity, and wave propagation.

Ideal drop onto a linear spring

\[ \delta=\frac{W}{k}\left(1+\sqrt{1+\frac{2hk}{W}}\right) \]

EXAMPLE 3Preliminary diameter for a slender actuator rod

A pin-ended solid steel actuator rod has unsupported length 2.00 m and must carry a concentric compressive service load of 25.0 kN with a buckling design factor of 2.5. Given \(E=200\ \text{GPa}\) and \(S_y=250\ \text{MPa}\), find the Euler minimum diameter, select the next practical whole-even-millimetre diameter, and verify the regime.

Recognition strategy

This is a professional stability-sizing problem: the long, pin-ended compression member cues an Euler trial, followed by a slenderness-regime check and iteration after selecting a manufacturable diameter.

Given information

Required critical load is \(n_dP=2.5(25.0)=62.5\ \text{kN}\).
Pin-ended effective length \(L_e=2000\ \text{mm}\), supplied \(E=200000\ \text{MPa}\), \(S_y=250\ \text{MPa}\).
Solid round trial section with \(I=\pi d^4/64\) and \(k=d/4\).

Model and interpretation

Treat the rod as initially straight, homogeneous, elastic, concentrically loaded, and ideally pin-ended. Euler buckling is a stability limit, not material crushing strength. The transition slenderness is checked after choosing geometry because the governing model depends on that choice.

Step-by-step solution

1. Plan. Impose \(P_{cr}=\pi^2EI/L_e^2\ge62500\ \text{N}\) and substitute \(I=\pi d^4/64\).

2. Formulate. Solve symbolically: \(d_{min}=[64(62500)(2000)^2/(\pi^3(200000))]^{1/4}=40.08\ \text{mm}\).

3. Calculate. Select the next specified whole-even size, \(d=42\ \text{mm}\), for which \(I=152745\ \text{mm}^4\).

4. Resolve. Recalculate \(P_{cr}=\pi^2(200000)(152745)/(2000)^2=75.38\ \text{kN}\), corresponding to \(P_{cr}/P=3.02\).

5. Review. Verify \(L_e/k=2000/(42/4)=190.5\), above \(\sqrt{2\pi^2E/S_y}=125.7\); Euler is therefore the selected idealized regime.

Verification

Units: With \(E I/L^2\), units are \((\text{N/mm}^2)(\text{mm}^4)/\text{mm}^2=\text{N}\); radius of gyration and slenderness use consistent millimetres.

Signs and direction: Compression is taken as a positive load magnitude for buckling. The lateral buckling direction is indeterminate for a perfect round rod; imperfections choose the actual direction.

Plausibility: Increasing diameter from \(40.08\) to \(42\ \text{mm}\) raises critical load through the fourth power. Direct stress is only about \(18.0\ \text{MPa}\), consistent with stability governing before yield.

Final result: \(\boxed{d_{Euler}=40.08\ \text{mm};\ \text{select }42\ \text{mm},\ P_{cr}=75.38\ \text{kN},\ P_{cr}/P=3.02}\)

Engineering interpretation

This is preliminary sizing. Actual end restraint, side load, straightness tolerance, residual stress, corrosion allowance, dynamic actuation, connection eccentricity, applicable machinery rules, supplier stock, and a validated imperfection model must be reviewed before release.

Chapter recap

Stiffness is load divided by displacement only in a linear regime; compliance is often additive.
Boundary and compatibility conditions are as important as the differential or energy equation.
Beam, Castigliano, column, and impact models answer different physical questions and have different limits.
Strength, stiffness, stability, fatigue, and reliability remain separate design checks.

Before moving on

You should be able to choose a deformation method, calculate and sign displacement or slope, combine compliance, solve a redundant reaction, classify a column, and explain which material, standard, vendor, tolerance, or test evidence remains necessary.