Mechanical Engineering Design · Part 1 — Basics

Load and Stress Analysis

From free-body diagrams and beam diagrams to the full toolbox of stress analysis: plane-stress transformation and Mohr's circle, bending and transverse shear, torsion, stress concentration, pressurized and rotating cylinders, press fits, curved beams, and Hertzian contact.

3–1Equilibrium and Free-Body Diagrams

Load paths begin with isolation

A free-body diagram replaces every removed connection by its reaction and shows all applied forces, couples, dimensions, axes, and sign choices. Equilibrium, \(\sum \mathbf F=\mathbf0\) and \(\sum \mathbf M=\mathbf0\), determines external reactions before any stress formula is useful. For accelerating systems, include inertial terms rather than calling the model static.

Learning objectives

Construct auditable free-body diagrams and recover internal resultants.
Select nominal and local stress equations from loading and geometry.
Transform plane stress and verify invariants.
Check applicability, units, signs, failure modes, and data provenance.
Iterate a preliminary geometry while separating stress from strength.

Action–reaction and boundaries

Connection forces are equal and opposite on mating bodies. Pins, bearings, welds, threads, and contact patches constrain different motions; replacing them all by the same reaction model can make the equations determinate but physically wrong.

3–2Shear Force and Bending Moments in Beams

From distributed loading to critical sections

With upward distributed load \(q\) positive, \(dV/dx=q\) and \(dM/dx=V\). Point forces jump the shear diagram and applied couples jump the moment diagram. Peaks in smooth moment regions occur where \(V=0\), while discontinuities and boundaries also require inspection.

Load–shear–moment chain

\[ \frac{dV}{dx}=q,\qquad \frac{dM}{dx}=V \]

EXAMPLE 1Support reactions and peak bending demand

A simply supported packaging-machine rail spans 1.80 m and carries a downward 6.0 kN point load 0.60 m from the left pin. Determine both support reactions and the maximum internal bending moment, using upward force and sagging moment as positive.

Recognition strategy

This is a foundation load-to-stress problem: isolate the member, determine the governing internal resultant, and only then select the stress relation appropriate to the section and loading.

Given information

Span \(L=1.80\ \text{m}\), load \(P=6.0\ \text{kN}\), and load coordinate \(a=0.60\ \text{m}\).
Left support is a pin, right support is a roller; beam self-weight is neglected.
Unknowns are \(R_A\), \(R_B\), and peak \(M\).

Model and interpretation

Treat the rail as a planar, statically determinate beam. The free-body diagram has vertical reactions at both supports; no horizontal load means the horizontal pin reaction is zero. A positive internal moment causes sagging.

Step-by-step solution

1. Plan. Take moments about the left pin so its reaction is eliminated: \(R_B(1.80)-6.0(0.60)=0\), which directly exposes the right reaction.

2. Formulate. Solve \(R_B=2.00\ \text{kN}\), then apply vertical equilibrium \(R_A+R_B-6.0=0\) to obtain \(R_A=4.00\ \text{kN}\).

3. Calculate. Between the left support and load, \(V=+4.00\ \text{kN}\) and \(M(x)=4.00x\ \text{kN}\!\cdot\!\text{m}\); after the load, shear becomes \(-2.00\ \text{kN}\).

4. Resolve. Because shear changes from positive to negative at the point load, moment peaks there: \(M_{\max}=4.00(0.60)=2.40\ \text{kN}\!\cdot\!\text{m}\).

Verification

Units: A reaction has force units, while reaction times distance gives \(\text{kN}\!\cdot\!\text{m}\); each equilibrium equation is dimensionally homogeneous.

Signs and direction: Both reactions are upward and positive under the stated convention. The shear jump is \(-6.0\ \text{kN}\), exactly the applied downward point load.

Plausibility: The reactions sum to \(6.0\ \text{kN}\), and the moment returns from its positive peak to zero over the remaining span, satisfying the roller boundary.

Final result: \(\boxed{R_A=4.00\ \text{kN},\ R_B=2.00\ \text{kN},\ M_{\max}=2.40\ \text{kN}\!\cdot\!\text{m}}\)

Engineering interpretation

The calculation establishes load demand, not material strength or safety. The peak nominal bending stress still requires section geometry, local features, dynamic amplification, and a separate strength or fatigue criterion.

3–3Singularity Functions

Singularity functions as bookkeeping

Macaulay brackets encode load discontinuities in one expression: \(\langle x-a\rangle^n=0\) for \(x

Integration rule

\[ \int \langle x-a\rangle^n dx=\frac{\langle x-a\rangle^{n+1}}{n+1},\quad n\ge0 \]

3–4Stress

Stress is local demand

Stress describes internal force intensity at a material point: normal stress acts perpendicular to a cut and shear stress acts tangent to it. Stress varies with load and geometry; strength is a material or component resistance backed by test data. Never compare a force directly with a strength value.

Physical meaning

An internal resultant is the area integral of a stress distribution. A nominal stress intentionally smooths that distribution; a local stress resolves peaks near geometric or contact features.

3–5Cartesian Stress Components

Cartesian components and signs

At a point, three normal and three independent shear components form a symmetric stress tensor because moment equilibrium requires complementary shears to match. Plane stress applies near a free surface when out-of-plane components are negligible; it is not automatically valid in thick constrained regions.

3–6Mohr's Circle for Plane Stress

Transforming plane stress

The transformed normal and shear stresses describe the same point on a differently oriented plane. The Mohr-circle center \(C=(\sigma_x+\sigma_y)/2\) and radius \(R=\sqrt{[(\sigma_x-\sigma_y)/2]^2+\tau_{xy}^2}\) give principal stresses \(C\pm R\) and in-plane extreme shear \(R\).

Principal direction

\[ 2\phi_p=\operatorname{atan2}(2\tau_{xy},\sigma_x-\sigma_y) \]

FIG. 3.1Mohr's circle for plane stress: x-face (A) and y-face (B) define the diameter; the σ-axis intercepts are the principal stresses, and the circle's top/bottom are the extreme shears. (Original diagram, redrawn.)

EXAMPLE 2Principal stresses at a bracket surface

At a machined bracket surface, strain-derived plane stresses are \(\sigma_x=80\ \text{MPa}\), \(\sigma_y=-20\ \text{MPa}\), and \(\tau_{xy}=30\ \text{MPa}\). Find the in-plane principal stresses, maximum in-plane shear stress, and principal-plane angle measured counterclockwise from the x-face.

Recognition strategy

This is an intermediate stress-transformation problem: recognize a complete plane-stress state at one point, compute its invariants, and use the transformation angle rather than returning to external equilibrium.

Given information

Plane-stress components are \(80\), \(-20\), and \(30\ \text{MPa}\).
Tension is positive; positive \(\tau_{xy}\) follows the chapter face convention.
The requested angle identifies a plane normal, not a physical crack angle.

Model and interpretation

Use the stress element at one material point. Mohr-circle center and radius give invariant stress magnitudes; use the two-argument angle relation to retain the correct quadrant. Out-of-plane stress is zero but must be considered later for three-dimensional maximum shear.

Step-by-step solution

1. Plan. Compute the circle center \(C=(80-20)/2=30.0\ \text{MPa}\), separating the hydrostatic in-plane part from the deviatoric radius.

2. Formulate. Compute \(R=\sqrt{[(80-(-20))/2]^2+30^2}=58.31\ \text{MPa}\), which is also the maximum in-plane shear magnitude.

3. Calculate. The principal values are \(\sigma_1=C+R=88.31\ \text{MPa}\) and \(\sigma_2=C-R=-28.31\ \text{MPa}\).

4. Resolve. Evaluate \(2\phi_p=\operatorname{atan2}(60,100)=30.96^\circ\), hence \(\phi_p=15.48^\circ\); substituting this angle makes transformed shear zero.

Verification

Units: Every term under the square root is in \(\text{MPa}^2\), so the radius and all reported principal or shear stresses are in \(\text{MPa}\).

Signs and direction: The tensile principal value exceeds \(\sigma_x\), and the lower value is compressive. Reversing the shear convention would reverse the reported angle but not the principal magnitudes.

Plausibility: The invariants check: \(\sigma_1+\sigma_2=60.0\ \text{MPa}=\sigma_x+\sigma_y\), and half their difference equals \(58.31\ \text{MPa}\).

Final result: \(\boxed{\sigma_1=88.3\ \text{MPa},\ \sigma_2=-28.3\ \text{MPa},\ \tau_{\max,\mathrm{in}}=58.3\ \text{MPa},\ \phi_p=15.5^\circ}\)

Engineering interpretation

These are local elastic stresses inferred at a surface, not strengths. A design decision requires material data with provenance, uncertainty in the measured components, and an appropriate static or fatigue failure model.

3–7General Three-Dimensional Stress

Three-dimensional stress

The three principal stresses are eigenvalues of the symmetric stress tensor. Even at a plane-stress surface, the third principal stress is zero and can control the absolute maximum shear, \(\tau_{\max}=\max(|\sigma_1-\sigma_2|,|\sigma_2-\sigma_3|,|\sigma_3-\sigma_1|)/2\).

3–8Elastic Strain

Elastic strain compatibility

For isotropic linear elasticity, \(\epsilon_x=[\sigma_x-\nu(\sigma_y+\sigma_z)]/E\) and \(\gamma_{xy}=\tau_{xy}/G\). These relations require small, recoverable strain and known properties at service temperature; plasticity, anisotropy, creep, or damaged material needs another model.

3–9Uniformly Distributed Stresses

Uniform stress models

Axial stress \(\sigma=F/A\) and direct shear \(\tau=V/A\) are average values. They are useful away from load introduction and abrupt geometry changes. Bearing stress is a projected-area nominal measure, not the detailed contact pressure distribution.

Axial deformation companion

\[ \sigma=\frac{F}{A},\qquad \delta=\frac{FL}{AE} \]

3–10Normal Stresses for Beams in Bending

Normal stress in straight-beam bending

Under small elastic curvature with plane sections remaining plane, \(\sigma_x=-My/I\). The neutral axis passes through the centroid for homogeneous sections, and the extreme magnitude is \(Mc/I=M/Z\). This nominal formula does not include notches, residual stress, or nonlinear yielding.

Elastic bending stress

\[ \sigma_x=-\frac{My}{I},\qquad |\sigma|_{\max}=\frac{|M|}{Z} \]

3–11Shear Stresses for Beams in Bending

Transverse shear flow

Beam shear follows \(\tau=VQ/(Ib)\), where \(Q\) is the first moment of the area outside the evaluation line and \(b\) is local width. For a rectangle, shear is parabolic and peaks at the neutral axis; thin webs therefore carry much of an I-section’s shear.

Beam shear

\[ \tau=\frac{VQ}{Ib},\qquad \tau_{\max,\mathrm{rect}}=\frac{3V}{2A} \]

FIG. 3.2Transverse shear in a rectangular beam: zero at the outer fibers, parabolic, maximum 3V/2A at the neutral axis. (Original diagram, redrawn.)

3–12Torsion

Torsion and section choice

For circular shafts in Saint-Venant torsion, \(\tau=T\rho/J\) and \(\theta=TL/(GJ)\). Closed thin-wall sections carry shear flow efficiently; open and rectangular sections require their own torsion constants. Do not substitute polar moment \(J\) for a noncircular torsion constant without justification.

Circular-shaft torsion

\[ \tau_{\max}=\frac{Tr}{J},\qquad \theta=\frac{TL}{GJ} \]

3–13Stress Concentration

Stress concentration and iteration

A theoretical factor \(K_t=\sigma_{\max}/\sigma_{\mathrm{nom}}\) comes from specified geometry and elastic boundary conditions. It is not a universal material constant. Static ductile yielding, brittle fracture, and fatigue use concentration information differently; later failure chapters supply those decisions.

Problem-solving strategy

Trace loads through supports and draw free bodies.
Find reactions and internal \(N,V,M,T\) at candidate sections.
Select a stress distribution consistent with geometry and assumptions.
Apply sourced concentration or contact data only within its stated range.
Audit equilibrium, dimensions, signs, limiting behavior, and governing location.
Resize, recompute geometry ratios, and repeat until requirements converge.

EXAMPLE 3Iterating a shoulder diameter for local stress

A nonrotating stepped steel shaft carries a static bending moment of \(1.20\ \text{kN}\!\cdot\!\text{m}\) at a shoulder. A chart value \(K_t=1.60\) is supplied for the proposed diameter ratio and fillet ratio. With allowable local bending stress \(120\ \text{MPa}\), determine the minimum small diameter and select a whole-millimetre trial size. State what remains outside this calculation.

Recognition strategy

This is a professional load-to-stress problem: isolate the member, determine the governing internal resultant, and only then select the stress relation appropriate to the section and loading.

Given information

Moment \(M=1.20\times10^6\ \text{N}\!\cdot\!\text{mm}\), supplied theoretical factor \(K_t=1.60\).
Allowable local static stress is \(120\ \text{N/mm}^2\).
Solid circular section; trial diameter must be a whole millimetre.

Model and interpretation

The nominal surface bending stress is \(32M/(\pi d^3)\), and the elastic local peak is \(K_t\) times nominal. The supplied chart factor is conditional on geometry; after choosing a diameter, its ratios must be recomputed and the chart revisited.

Step-by-step solution

1. Plan. Set the local-stress requirement \(K_t(32M)/(\pi d^3)\le120\ \text{MPa}\), keeping demand distinct from the allowable value.

2. Formulate. Rearrange before substitution: \(d_{\min}=[32K_tM/(\pi\sigma_{\mathrm{allow}})]^{1/3}\). This exposes the cube-root sensitivity to moment and concentration factor.

3. Calculate. Substitute to obtain \(d_{\min}=[32(1.60)(1.20\times10^6)/(\pi(120))]^{1/3}=54.62\ \text{mm}\).

4. Resolve. Select \(d=55\ \text{mm}\); the resulting nominal stress is \(73.46\ \text{MPa}\), and the predicted local peak is \(117.5\ \text{MPa}\).

5. Review. Recompute the actual shoulder and fillet ratios for the 55 mm choice, re-read \(K_t\) from an identified source, and iterate if the factor changes materially.

Verification

Units: Using \(M\) in \(\text{N}\!\cdot\!\text{mm}\) and \(d^3\) in \(\text{mm}^3\) produces \(\text{N/mm}^2=\text{MPa}\), as required.

Signs and direction: The bending surface has opposite tensile and compressive sides; the reported \(117.5\ \text{MPa}\) is a positive magnitude for the more critical side under the chosen moment sense.

Plausibility: Rounding diameter upward lowers stress, and the recalculated \(117.5\ \text{MPa}\) is below \(120\ \text{MPa}\). The cube law explains the modest margin.

Final result: \(\boxed{d_{\min}=54.6\ \text{mm};\ \text{select }d=55\ \text{mm},\ \sigma_{\mathrm{local}}\approx117.5\ \text{MPa}}\)

Engineering interpretation

This is an elastic static screening result, not a released shaft design. Fatigue notch sensitivity, residual stress, surface finish, tolerances, yielding, deflection, manufacturing fillet capability, inspection, and applicable company or industry rules remain to be verified.

3–14Stresses in Pressurized Cylinders

Pressure-vessel membrane and thick-wall models

Thin-wall hoop and longitudinal stresses follow force equilibrium when \(t/r\) is small; thick cylinders require Lamé radial and tangential distributions. Pressure must be consistently gauge or absolute according to the boundary loading, and closed-end axial stress must not be omitted.

Thin closed cylinder

\[ \sigma_h=\frac{pd}{2t},\qquad \sigma_l=\frac{pd}{4t} \]

FIG. 3.3Internal pressure only: tangential stress is tensile and greatest at the bore; radial stress equals −p_i at the bore and vanishes at the outer surface. (Original diagram, redrawn.)

3–15Stresses in Rotating Rings

Rotating rings

Centrifugal body force produces tensile hoop stress in a rotating ring. The model depends on density, angular speed, radius, and whether the ring is thin or radially thick. Balance quality and overspeed transients create loads beyond the ideal steady, axisymmetric solution.

3–16Press and Shrink Fits

Press and shrink fits

Interference produces contact pressure through elastic radial compliance of both members. Geometry, moduli, Poisson ratios, surface finish, temperature, and tolerance extremes govern the fit. Verify assembly force, yielding at the bore, fretting, and torque capacity using identified fit and surface data.

3–17Temperature Effects

Temperature effects

Free thermal strain is \(\epsilon_T=\alpha\Delta T\). Stress appears only when expansion is restrained or temperature varies through the body. A fully restrained uniform bar has \(\sigma=-E\alpha\Delta T\) under compression-positive restraint reaction assumptions; real joint compliance reduces this ideal value.

Thermal strain and full restraint

\[ \epsilon_T=\alpha\Delta T,\qquad \sigma_{\mathrm{restrained}}=-E\alpha\Delta T \]

3–18Curved Beams in Bending

Curved-beam bending

In strongly curved members, the neutral axis does not coincide with the centroid and normal stress is hyperbolic rather than linear. Curved-beam formulas require the correct section integral and radius definitions; using straight-beam \(Mc/I\) at a hook throat can underpredict local demand.

3–19Contact Stresses

Elastic contact stress

Hertz contact models smooth, frictionless, elastic bodies with small contact patches. Effective curvature and modulus determine patch size and peak pressure; subsurface shear can govern pitting. Edge loading, roughness, lubrication, coatings, plasticity, and misalignment require vendor data, standards, simulation, or test evidence.

3–20Summary

Chapter recap

Equilibrium converts service loads into section resultants; constitutive and geometry relations convert resultants into stresses.
Nominal demand, local demand, and material strength are distinct quantities.
Transformation, concentration, pressure, torsion, curved-beam, and contact models each have explicit limits.
Design is iterative because changing geometry changes reactions, section properties, and chart parameters.

Before moving on

You should be able to draw a complete free body, locate critical sections, select and apply a stress model, transform plane stress, and defend units, signs, assumptions, source data, and verification needs.