Comprehensive Study Notes on Machine Elements Design

Fundamentals of Mechanical Engineering Design

• Definitions of Design Disciplines:

  • Design: The formulation of a plan for the satisfaction of a human need. A design problem is authentic, practical, and results in an actionable outcome possessing physical reality.
  • Engineering Design: The iterative decision-making process wherein scientific principles and technical tools (mathematics, computers, engineering graphics, and formal communication) are utilized to devise a component, system, or process that meets specified objectives.
  • Mechanical (Machine) Design: The design of physical objects and systems of a mechanical nature (machines, products, structures, mechanisms, and instruments) through the application of mathematics, material science, engineering mechanics, and manufacturing technologies. Machine design requires broad technical versatility rather than narrow specialization.

• Phases and Iteration in the Engineering Design Process:

  • Identification of Need: Initiates the design sequence. Recognizes a deficiency, operational defect, or emerging market demand. Needs are frequently triggered by adverse circumstances (such as excessive noise, packaging variation, or component failure) and require concise problem articulation.
  • Definition of Problem: Establishes rigorous engineering specifications, including input/output metrics, spatial and geometric constraints, mass limitations, environmental conditions, operational life, and cost boundaries.
  • Synthesis: The conceptual invention stage where alternative schemes and configurations connecting potential system elements are generated, investigated, and qualitatively compared.
  • Analysis and Optimization: Mathematical modeling, force and stress calculation, kinetic analysis, and iterative refinement. Infeasible concepts are eliminated; viable concepts are numerically optimized to maximize performance and efficiency.
  • Evaluation: Experimental validation of the finalized design, typically involving physical prototype fabrication and laboratory testing under simulated operating environments.
  • Presentation: Formal communication of the design via engineering drawings, manufacturing specifications, bills of materials, and design documentation.
  • Iteration: Feedback loops dynamically interconnect the analysis, synthesis, problem definition, and need stages, driving repeated redesign until all engineering constraints are satisfied.

• Classification of Machine Design:

  • Adaptive Design: Adapts standard, pre-existing engineering systems or products to a new operational context without altering fundamental design principles (e.g., configuring standard conveyor systems or haulage mechanisms for a new plant layout).
  • Developmental Design: Starts with an existing product or mechanism and implements modifications or improvements to enhance performance, aesthetic quality, or manufacturing economy (e.g., developing a new model of an automobile).
  • New Design: Deploys existing scientific principles to engineer an entirely novel machine or device to resolve an unaddressed challenge. Requires high creative synthesis and targeted exploratory research (e.g., specialized transport vehicles for desert or maritime shipboard environments).

• Design Methods:

  • Rational Design: Dimensions and material choices are determined mathematically from mechanics of materials, stress-strain states, and failure theories.
  • Empirical Design: Based on historical practice, experimentation, and empirical rules rather than exact theoretical analysis. For example, the tightening force of a threaded fastener is conventionally estimated as:     $P = 284\,d$     where $d$ is the bolt nominal diameter in millimeters and $P$ is the tightening load in kilograms.
  • Industrial Design: Focuses on human factors, industrial ergonomics, external aesthetics, production ease, market appeal, and integration of off-the-shelf standard components.

• Comprehensive Design Considerations:

  • Strength and Rigidity: Ensuring components sustain operational loads without plastic deformation, fracture, or excessive deflection.
  • Reliability: The statistical probability that a machine element will perform its intended function without failure over its designed service life under specified operating conditions. Compliance with regulatory standards such as OSHA (Occupational Safety and Health Administration) and EPA (Environmental Protection Agency) dictates careful evaluation of stress and strength distributions.
  • Stability and Resonance: Structural rigidity against toppling, column buckling, and torsional instability. Natural frequencies must be designed away from operating excitation frequencies to avoid resonance and uncontrolled dynamic stresses.
  • Corrosion Resistance: Electrochemical surface degradation can rarely be completely eliminated. Protective coatings, cathodic protection, and environmental isolation are selected where corrosion-resistant alloys are economically unfeasible.
  • Tribological Characteristics (Wear and Friction): Wear represents progressive mechanical or chemical removal of surface material. Regulated through contact geometry, surface hardening, and hydrodynamic/elastohydrodynamic lubrication.
  • Thermal Properties: Thermal expansion, thermal gradients, thermal fatigue, and creep resistance at elevated working temperatures.
  • Aesthetics and Ergonomics: External form, color harmony, visual balance, and control placement must comply with industrial design principles without compromising maintainability, producerability, or manufacturing cost.
  • Producerability and Production Economics: Simplicity of part geometries, minimization of high-cost materials, adherence to standard dimensions, and optimization of assembly and inspection procedures.

• Engineering Standards and Design Codes:

  • Standard: A set of technical specifications for parts, materials, dimensions, or test protocols intended to ensure interchangeability, uniformity, quality, and streamlined inventory control.
  • Code: A regulatory standard adopted by professional or governmental bodies for the analysis, design, fabrication, and erection of systems to enforce minimum levels of public safety, operational efficiency, and structural integrity.
  • Key Standards Organizations:
  • AA: Aluminum Association
  • ANSI: American National Standards Institute
  • ASM: American Society for Metals (ASM International)
  • ASME: American Society of Mechanical Engineers
  • ASTM: American Society for Testing and Materials
  • BSI: British Standards Institution
  • ISO: International Organization for Standardization
  • SAE: Society of Automotive Engineers

• Unit Systems:

  • Newton's second law dictates the physical relationship:     $F = m \times a$$[F] = [M][L][T]^{-2}$
  • Gravitational Systems: Force ($[F]$), length ($[L]$), and time ($[T]$) are selected as fundamental base units; mass ($[M]$) is a derived unit.
  • Absolute Systems (SI): Mass ($[M]$, kilogram), length ($[L]$, meter), and time ($[T]$, second) are fundamental base units. The unit of force is derived:     $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$

• Design Uncertainty, Design Factor, and Factor of Safety:

  • Sources of Uncertainty: Inhomogeneity of materials, variations in chemical composition, localized heat-treatment variations, residual stresses from welding or shrink fits, load intensity and distribution variations, simplifying assumptions in mathematical models, localized stress concentrations, environmental corrosion, and wear.
  • Deterministic Design Factor ($n_d$): Ratio of a loss-of-function parameter to the maximum allowable parameter:     $n_d = \frac{\text{loss-of-function parameter}}{\text{maximum allowable parameter}}$$n_d = \frac{\text{loss-of-function strength}}{\text{allowable stress}} = \frac{S}{\sigma_{\text{allow}}}$
  • Factor of Safety ($n$): The realized numerical ratio evaluated after standard component sizes, standard plate gauges, or stock bar diameters are assigned.

• Worked Example: Preferred Sizing of an Axial Tension Rod:

  • Problem Parameters: Tensile axial load $P = 8900\,\text{N}$, material tensile strength $S = 165.36\,\text{MPa}$, design factor $n_d = 3.0$.
  • Allowable Stress:$\sigma_{\text{allow}} = \frac{S}{n_d} = \frac{165.36 \times 10^6\,\text{Pa}}{3.0} = 55.12 \times 10^6\,\text{Pa}$
  • Cross-Sectional Area and Minimum Diameter:$A = \frac{\pi d^2}{4} = \frac{P}{\sigma_{\text{allow}}}$$d = \sqrt{\frac{4 P n_d}{\pi S}} = \sqrt{\frac{4 \times 8900 \times 3.0}{\pi \times (165.36 \times 10^6)}} = 0.0143\,\text{m} = 14.3\,\text{mm}$
  • Selection of Standard Preferred Size:
  • From standard metric stock sizes ($10, 11, 12, 14, 16, 18, 20, 22, 25\,\text{mm}$, etc.), select the next larger size: $d = 16\,\text{mm}$.
  • Realized Factor of Safety ($n$):$n = \frac{S A}{P} = \frac{(165.36 \times 10^6) \times \pi \times (16 \times 10^{-3})^2}{4 \times 8900} = 3.68$

Load and Stress Analysis Fundamentals

• Elementary Stress and Load Conventions:

  • Material Strength Symbols: $S$ = generic strength; $S_y$ = yield strength; $S_{ut}$ = ultimate tensile strength; $S_s$ or $S_{sy}$ = shear yield strength.
  • Stress Component Symbols: $\sigma$ = normal stress; $\tau$ = shear stress; $\sigma_x, \sigma_y, \sigma_z$ = Cartesian normal stresses; $\tau_{xy}, \tau_{yz}, \tau_{zx}$ = Cartesian shear stresses; $\sigma_r$ = radial stress component.
  • Beam Equilibrium and Internal Sign Conventions:
  • Positive shear force ($V$) tends to rotate the beam segment clockwise (upward on left face, downward on right face).
  • Positive bending moment ($M$) creates compression in the upper fibers and tension in the lower fibers (concave upward curvature).

• Direct and Secondary Stress Modes:

  • Axial Normal Stress (Tension and Compression):$\sigma = \pm \frac{F}{A}$
  • Bearing Stress ($\sigma_b$ or $\sigma_c$): Contact compressive stress developed between solid bodies pressed against each other:     
  • Flat column base contact: $\sigma_c = \frac{P}{A}$
  • Cylindrical pin in a member with hole (projected area): $\sigma_b = \frac{P}{t \times d}$
  • Rectangular key securing gear hub of length $L$ and width $w$ to shaft of diameter $d$: $\sigma_b = \frac{4 P}{d \times L}$
  • Direct Transverse Shear Stress:
  • Single Shear: $\tau = \frac{P}{A}$
  • Double Shear: $\tau = \frac{P}{2 A}$
  • Flexural (Bending) Stress:$\sigma = -\frac{M y}{I}$$\sigma_{\max} = \frac{M c}{I} = \frac{M}{Z}$     where $y$ is vertical distance from neutral axis, $I$ is second moment of area, $c$ is maximum fiber distance, and $Z = I/c$ is section modulus.
  • Torsional Shear Stress in Circular Members:$\frac{\tau}{r} = \frac{T}{J} = \frac{G \theta}{l}$$\tau_{\max} = \frac{T r_o}{J} = \frac{T}{Z_t}$$\theta = \frac{T l}{G J}$
  • Solid circular shaft of diameter $d$:       $J = \frac{\pi d^4}{32}$$Z_t = \frac{\pi d^3}{16}$
  • Hollow circular shaft with outer diameter $d$ and inner diameter $d_i$:       $J = \frac{\pi (d^4 - d_i^4)}{32}$$Z_t = \frac{\pi (d^4 - d_i^4)}{16 d}$

• State of Stress at a Point and Stress Transformation:

  • General 3D stress involves nine tensor components, reducible by moment equilibrium ($\tau_{xy} = \tau_{yx}$, $\tau_{yz} = \tau_{zy}$, $\tau_{zx} = \tau_{xz}$) to six independent stress components: $\sigma_x, \sigma_y, \sigma_z, \tau_{xy}, \tau_{yz}, \tau_{zx}$.
  • Plane Stress State: Occurs when stresses normal to one coordinate plane are zero ($\sigma_z = \tau_{xz} = \tau_{yz} = 0$).
  • Plane-Stress Transformation Equations: For an arbitrary cut plane inclined at angle $\phi$ counterclockwise from the $x$-axis:     $\sigma = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\phi) + \tau_{xy} \sin(2\phi)$$\tau = -\frac{\sigma_x - \sigma_y}{2} \sin(2\phi) + \tau_{xy} \cos(2\phi)$
  • Principal Stresses and Principal Planes:
  • Setting $\frac{d\sigma}{d\phi} = 0$ defines the principal planes:       $\tan(2\phi_p) = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}$
  • On principal planes, the shear stress is identically zero ($\tau = 0$).
  • Principal stresses are given by:       $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$
  • Maximum In-Plane Shear Stress:
  • Setting $\frac{d\tau}{d\phi} = 0$ defines planes of maximum shear stress:       $\tan(2\phi_s) = -\frac{\sigma_x - \sigma_y}{2 \tau_{xy}}$
  • Since $\tan(2\phi_s) = -\frac{1}{\tan(2\phi_p)}$, the planes of maximum shear stress are oriented at $\pm 45^\circ$ to the principal planes.
  • The extreme in-plane shear stress magnitude is:       $\tau_{\max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2}$
  • Normal stress acting on planes of maximum shear stress is equal and uniform:       $\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2}$

• Worked Example: 2D Plane Stress Transformation:

  • Given Stress State: $\sigma_x = -10\,\text{MPa}$, $\sigma_y = +20\,\text{MPa}$, $\tau_{xy} = -20\,\text{MPa}$.
  • Calculation of Principal Stresses:$\sigma_{1,2} = \frac{-10 + 20}{2} \pm \sqrt{\left(\frac{-10 - 20}{2}\right)^2 + (-20)^2} = 5 \pm \sqrt{(-15)^2 + 400} = 5 \pm 25\,\text{MPa}$$\sigma_1 = +30\,\text{MPa}$$\sigma_2 = -20\,\text{MPa}$
  • Principal Directions:$\tan(2\phi_p) = \frac{2(-20)}{-10 - 20} = \frac{-40}{-30} = 1.333$$2\phi_p = 53.13^\circ, 233.13^\circ \implies \phi_p = 26.56^\circ, 116.56^\circ$

• Worked Example: Steel Bar Secured by Two Pins:

  • Configuration: Steel bar thickness $t = 5\,\text{mm}$, two pins $A$ and $B$ of diameter $d = 6\,\text{mm}$ spaced $50\,\text{mm}$ apart. Load $P = 5\,\text{kN}$ applied at free end, located $100\,\text{mm}$ from pin $B$. Rotation center $O$ is located midway between pins ($25\,\text{mm}$ from each pin; distance from $O$ to $P$ is $125\,\text{mm}$).
  • Equilibrium Analysis:
  • Moment summation about $O$:       $P \times 0.125 = (F_A + F_B) \times 0.025$$(5 \times 10^3) \times 0.125 = (F_A + F_B) \times 0.025 \implies F_A + F_B = 25\,\text{kN}$
  • Force balance in vertical direction:       $F_B - F_A = P = 5\,\text{kN}$
  • Solving simultaneously:       $F_A = 10\,\text{kN}$$F_B = 15\,\text{kN}$
  • Shear Stresses in Pins:$\tau_A = \frac{F_A}{\frac{\pi d^2}{4}} = \frac{10 \times 10^3}{\frac{\pi (0.006)^2}{4}} = 354\,\text{MPa}$$\tau_B = \frac{F_B}{\frac{\pi d^2}{4}} = \frac{15 \times 10^3}{\frac{\pi (0.006)^2}{4}} = 531\,\text{MPa}$
  • Bearing Stresses in Pins:$\sigma_{b,A} = \frac{F_A}{d \times t} = \frac{10 \times 10^3}{0.006 \times 0.005} = 333.3\,\text{MPa}$$\sigma_{b,B} = \frac{F_B}{d \times t} = \frac{15 \times 10^3}{0.006 \times 0.005} = 500.0\,\text{MPa}$

• Worked Example: Cylindrical Cantilever Bar under Combined Loading:

  • Configuration: Solid shaft diameter $d = 50\,\text{mm}$, total cantilever length $100\,\text{mm}$, rigidly fixed at wall ($x=0$). Transverse vertical force $P_y = 20\,\text{kN}$ applied at free end ($x=100\,\text{mm}$) with an offset lever arm of $50\,\text{mm}$ in $z$-direction. Cross-section investigated is plane $ABC$ located $60\,\text{mm}$ from free end ($40\,\text{mm}$ from wall).
  • Section Internal Resultants at Section ABC:
  • Torque: $T = P_y \times 50 \times 10^{-3} = (20 \times 10^3) \times 0.05 = 1.0\,\text{kNm}$
  • Bending Moment at section ($x = 40\,\text{mm}$ from wall, $60\,\text{mm}$ moment arm from load):       $M = P_y \times (60 \times 10^{-3}) = (20 \times 10^3) \times 0.06 = 1.2\,\text{kNm}$
  • Stress Calculations:
  • Section modulus in bending: $Z = \frac{\pi d^3}{32} = \frac{\pi (0.05)^3}{32} = 1.227 \times 10^{-5}\,\text{m}^3$
  • Section modulus in torsion: $Z_t = \frac{\pi d^3}{16} = \frac{\pi (0.05)^3}{16} = 2.454 \times 10^{-5}\,\text{m}^3$
  • Torsional shear stress (uniform around perimeter):       $\tau = \frac{T}{Z_t} = \frac{1.0 \times 10^3}{2.454 \times 10^{-5}} = 40.7\,\text{MPa}$
  • Bending stress at top fiber (element $B$):       $\sigma_b = \frac{M}{Z} = \frac{1.2 \times 10^3}{1.227 \times 10^{-5}} = 97.78\,\text{MPa}$
  • Elements $A$ and $C$ (on neutral axis):       $\sigma_b = 0$$\tau = 40.7\,\text{MPa}$
  • Principal Stresses at Element $B$ ($\sigma_x = 97.78\,\text{MPa}$, $\sigma_y = 0$, $\tau_{xy} = 40.7\,\text{MPa}$):       $\sigma_{1,2} = \frac{97.78}{2} \pm \sqrt{\left(\frac{97.78}{2}\right)^2 + (40.7)^2} = 48.89 \pm 63.51\,\text{MPa}$$\sigma_1 = +112.4\,\text{MPa}$$\sigma_2 = -14.62\,\text{MPa}$
  • Maximum Shear Stress at Element $B$:       $\tau_{\max} = 63.51\,\text{MPa}$

Static Failure Theories for Ductile and Brittle Materials

• Static Loading and Failure Modes:

  • Static Load: A stationary mechanical force or moment that remains constant in magnitude and direction, or increases gradually to a steady value without inducing dynamic impact or shock.
  • Failure Criteria:
  • Yielding: Excessive plastic (inelastic) strain rendering a ductile component functionally inoperable without structural severance.
  • Fracture: Complete physical separation of a component into two or more pieces, typical of brittle materials.
  • Ductility Demarcation: Materials with percent elongation at fracture $\ge 15\%$ are treated as ductile; materials with percent elongation $< 5\%$ are treated as brittle.
  • Brittle Fracture in Ductile Materials: Can occur under high-cycle fatigue, cryogenic temperatures, severe impact shock loading, triaxial tensile constraint, excessive work hardening, or severe quenching.

• Maximum-Normal-Stress (MNS) Theory (Rankine's Theory):

  • States that failure occurs when the maximum principal normal stress in a part exceeds the tensile yield or ultimate strength, or the minimum principal stress falls below the compressive strength.
  • For ordered principal stresses ((\sigma_1 \ge \sigma_2 \ge \sigma_3)):     $\sigma_1 \ge S_{ut} \quad \text{or} \quad \sigma_3 \le -S_{uc}$
  • Highly inaccurate for ductile materials yielding in shear, but standard for brittle materials characterized by continuous, linear stress-strain curves terminating in abrupt tensile fracture.

• Maximum-Shear-Stress (MSS) Theory (Tresca / Guest Theory):

  • Predicts that yielding initiates when the absolute maximum shear stress in a 3D element equals or exceeds the maximum shear stress in a simple uniaxial tensile test specimen at yield:     $\tau_{\max} = \frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2} \implies \sigma_1 - \sigma_3 \ge S_y$
  • Incorporating factor of safety $n$:     $\tau_{\max} = \frac{S_y}{2 n} \implies \sigma_1 - \sigma_3 = \frac{S_y}{n}$
  • Predicts the shear yield strength to be exactly half the tensile yield strength:     $S_{sy} = 0.50\,S_y$
  • Plane Stress ($\sigma_A, \sigma_B$ with third principal stress $= 0$) Cases:
  • Case 1 ($\sigma_A \ge \sigma_B \ge 0$): $\sigma_1 = \sigma_A, \sigma_3 = 0 \implies \sigma_A \ge \frac{S_y}{n}$
  • Case 2 ($\sigma_A \ge 0 \ge \sigma_B$): $\sigma_1 = \sigma_A, \sigma_3 = \sigma_B \implies \sigma_A - \sigma_B \ge \frac{S_y}{n}$
  • Case 3 ($0 \ge \sigma_A \ge \sigma_B$): $\sigma_1 = 0, \sigma_3 = \sigma_B \implies \sigma_B \le -\frac{S_y}{n}$

• Distortion Energy (DE) Theory (von Mises–Hencky / Shear Energy Theory):

  • Asserts that plastic yielding is induced exclusively by angular distortion of the material element, and that hydrostatic stress components produce only elastic volume changes without causing plastic slip.
  • Derivation:
  • Average hydrostatic normal stress:       $\sigma_{\text{avg}} = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3}$
  • Total strain energy per unit volume for isotropic material with Young's modulus $E$ and Poisson's ratio $\nu$:       $u = \frac{1}{2E} \left[ \sigma_1^2 + \sigma_2^2 + \sigma_3^2 - 2\nu(\sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_3 \sigma_1) \right]$

• Hydrostatic volumetric strain energy:   $u_v = \frac{3 \sigma_{\text{avg}}^2}{2E}(1 - 2\nu) = \frac{1 - 2\nu}{6E}\left[ \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + 2(\sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_3 \sigma_1) \right]$

• Shear distortion energy per unit volume ($u_d = u - u_v$):   $u_d = \frac{1 + \nu}{3E} \left[ \frac{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2} \right]$

• For uniaxial tension at yield ($\sigma_1 = S_y, \sigma_2 = \sigma_3 = 0$):   $u_d = \frac{1 + \nu}{3E} S_y^2$

  • Von Mises Effective Stress ((\sigma')):$\sigma' = \sqrt{\frac{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2}}$
  • In general Cartesian coordinates:     $\sigma' = \frac{1}{\sqrt{2}} \sqrt{(\sigma_x - \sigma_y)^2 + (\sigma_y - \sigma_z)^2 + (\sigma_z - \sigma_x)^2 + 6(\tau_{xy}^2 + \tau_{yz}^2 + \tau_{zx}^2)}$
  • For 2D Plane Stress ($\sigma_z = \tau_{yz} = \tau_{zx} = 0$):     $\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2}$$\sigma' = \sqrt{\sigma_A^2 - \sigma_A \sigma_B + \sigma_B^2}$
  • Yielding occurs when:     $\sigma' \ge \frac{S_y}{n}$
  • Shear Yield Strength Prediction under Pure Shear:
  • In pure shear ($\sigma_x = \sigma_y = 0, \tau_{xy} = S_{sy}$), setting $\sigma' = S_y$ yields:       $\sqrt{3 S_{sy}^2} = S_y \implies S_{sy} = \frac{S_y}{\sqrt{3}} = 0.577\,S_y$

• Comparison of Failure Theories for Ductile Materials:

  • Extensive multi-axial testing (e.g., Ni-Cr-Mo steel, AISI 1023 steel, 2024-T4 Al, 3S-H Al) validates that the Distortion Energy ellipse accurately predicts yielding in all four quadrants.
  • The MSS hexagon lies completely enclosed within the DE ellipse, making MSS conservative by up to $15.5\%$.
  • MSS is suitable for rapid, conservative sizing, while DE is preferred when tight design margins and mass optimization are required.

• Worked Example: Stock-Size Selection of Cantilever Round Tube:

  • Design Data: Material is 2014 aluminum alloy with yield strength $S_y = 276\,\text{MPa}$, design factor $n_d = 4.0$. Cantilever length $L = 120\,\text{mm}$. Loads at tip: bending force $F = 1.75\,\text{kN}$, axial tension $P = 9.0\,\text{kN}$, torque $T = 72\,\text{Nm}$.
  • Target von Mises Stress:$\sigma' \le \frac{S_y}{n_d} = \frac{276\,\text{MPa}}{4.0} = 69.0\,\text{MPa} = 0.0690\,\text{GPa}$
  • Internal Resultants at Fixed Support:
  • Axial force: $P = 9.0 \times 10^3\,\text{N}$
  • Maximum bending moment: $M = F \times L = (1.75 \times 10^3) \times 0.120 = 210\,\text{Nm}$
  • Torque: $T = 72\,\text{Nm}$
  • Cross-Sectional Formulation:$\sigma_x = \frac{P}{A} + \frac{M c}{I} = \frac{4 P}{\pi (d^2 - d_i^2)} + \frac{32 M d}{\pi (d^4 - d_i^4)}$$\tau_{xz} = \frac{T d}{2 J} = \frac{16 T d}{\pi (d^4 - d_i^4)}$
  • Trial 1: $42 \times 5\,\text{mm}$ Tube ($d = 42\,\text{mm}$, $t = 5\,\text{mm}$, $d_i = 32\,\text{mm}$):
  • $d^4 - d_i^4 = (42 \times 10^{-3})^4 - (32 \times 10^{-3})^4 = 2.063 \times 10^{-6}\,\text{m}^4$
  • $d^2 - d_i^2 = (42 \times 10^{-3})^2 - (32 \times 10^{-3})^2 = 7.40 \times 10^{-4}\,\text{m}^2$
  • Axial stress: $\frac{4 \times 9000}{\pi \times (7.40 \times 10^{-4})} = 15.486\,\text{MPa}$
  • Bending stress: $\frac{32 \times 210 \times (42 \times 10^{-3})}{\pi \times (2.063 \times 10^{-6})} = 43.547\,\text{MPa}$
  • Total normal stress: $\sigma_x = 15.486 + 43.547 = 59.033\,\text{MPa}$
  • Torsional shear stress:       $\tau_{xz} = \frac{16 \times 72 \times (42 \times 10^{-3})}{\pi \times (2.063 \times 10^{-6})} = 7.466\,\text{MPa}$
  • Von Mises equivalent stress:       $\sigma' = \sqrt{\sigma_x^2 + 3\tau_{xz}^2} = \sqrt{(59.033 \times 10^6)^2 + 3(7.466 \times 10^6)^2} = 60.43\,\text{MPa} = 0.06043\,\text{GPa}$
  • Realized Factor of Safety:       $n = \frac{S_y}{\sigma'} = \frac{276}{60.43} = 4.57 \ge 4.0 \quad (\text{Satisfactory})$
  • Trial 2: $42 \times 4\,\text{mm}$ Tube ($d = 42\,\text{mm}$, $t = 4\,\text{mm}$, $d_i = 34\,\text{mm}$):
  • $d^4 - d_i^4 = (42 \times 10^{-3})^4 - (34 \times 10^{-3})^4 = 1.775 \times 10^{-6}\,\text{m}^4$
  • $d^2 - d_i^2 = (42 \times 10^{-3})^2 - (34 \times 10^{-3})^2 = 6.08 \times 10^{-4}\,\text{m}^2$
  • Normal stress: $\sigma_x = 18.850 + 50.613 = 69.463\,\text{MPa}$
  • Torsional shear stress: $\tau_{xz} = 8.677\,\text{MPa}$
  • Von Mises equivalent stress:       $\sigma' = \sqrt{(69.463)^2 + 3(8.677)^2} = 71.07\,\text{MPa} = 0.07107\,\text{GPa}$
  • Realized Factor of Safety:       $n = \frac{276}{71.07} = 3.88 < 4.0 \quad (\text{Unsatisfactory})$

• Worked Example: Sizing a Bolt under Combined Tensile and Shear Loads:

  • Design Conditions: Axial pull $F = 10\,\text{kN}$, transverse shear load $F_s = 5\,\text{kN}$, material yield strength $S_y = 100\,\text{MPa}$.
  • Cross-Sectional Formulations:$A = \frac{\pi d^2}{4} = 0.7854\,d^2$$\sigma_x = \frac{10 \times 10^3}{0.7854\,d^2} = \frac{12.73 \times 10^3}{d^2}\,\text{N/m}^2$$\tau_{xy} = \frac{5 \times 10^3}{0.7854\,d^2} = \frac{6.366 \times 10^3}{d^2}\,\text{N/m}^2$$\sigma_y = 0$
  • 1. Maximum Principal Stress Theory:$\sigma_1 = \frac{\sigma_x}{2} + \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} = \frac{12.73 \times 10^3}{2 d^2} + \sqrt{\left(\frac{12.73 \times 10^3}{2 d^2}\right)^2 + \left(\frac{6.366 \times 10^3}{d^2}\right)^2} = \frac{15.367 \times 10^3}{d^2}$$\sigma_1 = S_y \implies \frac{15.367 \times 10^3}{d^2} = 100 \times 10^6 \implies d = 0.0124\,\text{m} = 12.4\,\text{mm}$
  • 2. Maximum Shear Stress Theory:$\sigma_2 = \frac{\sigma_x}{2} - \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} = -\frac{2.637 \times 10^3}{d^2}$$\tau_{\max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{18.004 \times 10^3}{2 d^2} = \frac{9.002 \times 10^3}{d^2}$$\tau_{\max} = \frac{S_y}{2} \implies \frac{9.002 \times 10^3}{d^2} = 50 \times 10^6 \implies d = 0.01342\,\text{m} = 13.42\,\text{mm}$
  • 3. Distortion Energy Theory:$\sigma' = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} = \frac{10^3}{d^2} \sqrt{(15.367)^2 - (15.367)(-2.637) + (-2.637)^2} = \frac{16.84 \times 10^3}{d^2}$$\sigma' = S_y \implies \frac{16.84 \times 10^3}{d^2} = 100 \times 10^6 \implies d = 0.01298\,\text{m} = 12.98\,\text{mm}$

Stress Concentrations and Geometric Discontinuities

• Mechanics of Stress Concentrations:

  • Standard elementary stress formulas ($P/A$, $M c/I$, $T r/J$) strictly assume uniform, defect-free geometries with constant cross-sections.
  • Physical components require cross-sectional discontinuities (fillets, steps, holes, keyways, grooves, threads) termed stress raisers.
  • Stress concentration describes the localized stress amplification occurring in the immediate vicinity of geometric discontinuities.

• Stress Concentration Factors:

  • Theoretical (geometric) stress concentration factor for normal stress:     $K_t = \frac{\sigma_{\max}}{\sigma_0}$
  • Theoretical stress concentration factor for shear stress:     $K_{ts} = \frac{\tau_{\max}}{\tau_0}$
  • Nominal stresses ($\sigma_0, \tau_0$) are based on net remaining cross-sectional area at the discontinuity.
  • $K_t$ and $K_{ts}$ depend exclusively on geometry and are independent of material composition.
  • Determination Methods: Experimental photoelasticity, finite element analysis (FEA), brittle lacquer coatings, electrical resistance strain-gage grids, and elasticity theory.

• Elasticity Analysis of Openings in Tension Plates:

  • Elliptical Discontinuity in Infinite Plate: For an elliptical hole with semi-major axis $b$ perpendicular to tensile stress $\sigma_1$ and semi-minor axis $a$ parallel to loading:     $\sigma_{\max} = \sigma_1 \left(1 + 2\,\frac{b}{a}\right)$
  • Circular hole ($a = b$):       $\sigma_{\max} = 3\,\sigma_1 \implies K_t = 3.0$
  • Transverse crack ($b \gg a$): $K_t \to \infty$, producing extreme localized stress.
  • Longitudinal crack ($b \ll a$): $\sigma_{\max} \to \sigma_1 \implies K_t \to 1.0$.

• Mitigation of Stress Concentrations:

  • Providing large transition fillet radii or compound elliptical fillets rather than sharp re-entrant corners.
  • Introducing relief notches or multiple smaller adjacent notches to smooth force flow paths.
  • Adding peripheral unloading grooves or counterbores around shoulder transitions.

• Worked Example: Stepped Shaft under Torsion and Bending:

  • Shaft Dimensions: Shoulder diameter $D = 40\,\text{mm}$, neck diameter $d = 30\,\text{mm}$, fillet radius $r = 6\,\text{mm}$.
  • Geometric Ratios:$\frac{r}{d} = \frac{6}{30} = 0.20$$\frac{D}{d} = \frac{40}{30} = 1.33$
  • (a) Pure Torsion of $T = 100\,\text{Nm}$:
  • From theoretical stress concentration chart for stepped shaft in torsion: $K_{ts} \approx 1.17$.
  • Nominal shear stress:       $\tau_0 = \frac{16 T}{\pi d^3} = \frac{16 \times 100}{\pi \times (0.030)^3} = 18.86\,\text{MPa}$
  • Peak shear stress:       $\tau_{\max} = K_{ts} \times \tau_0 = 1.17 \times 18.86\,\text{MPa} = 22.07\,\text{MPa}$
  • (b) Pure Bending of $M = 150\,\text{Nm}$:
  • From theoretical stress concentration chart for stepped shaft in bending: $K_t \approx 1.35$.
  • Nominal bending stress:       $\sigma_0 = \frac{32 M}{\pi d^3} = \frac{32 \times 150}{\pi \times (0.030)^3} = 56.59\,\text{MPa}$
  • Peak bending stress:       $\sigma_{\max} = K_t \times \sigma_0 = 1.35 \times 56.59\,\text{MPa} = 76.4\,\text{MPa}$

Design for Fatigue Strength and Fluctuating Loads

• Fatigue Failure Mechanism:

  • Dynamic or rotating loading produces alternating internal stress cycles. Progressive micro-crack initiation occurs at surface defects, inclusions, or geometric discontinuities.
  • Micro-cracks propagate under cyclic plastic deformation, leaving beach marks while reducing the load-carrying cross section.
  • Final failure occurs abruptly when the residual cross section can no longer sustain the peak cycle load, producing brittle-appearing catastrophic fracture with zero macroscopic plastic warning.

• Standard Moore Rotating-Beam Test and S-N Curve:

  • Standard specimen length $2l = 85\,\text{mm}$, neck diameter $d = 7.5\,\text{mm}$, contoured transition radius $R = 250\,\text{mm}$, mirror polished to remove circumferential scratches.
  • Subjected to pure bending ($M = w l$) with zero transverse shear; specimen rotation induces completely reversed normal stress cycles:     $\sigma = \frac{32 M}{\pi d^3}$
  • S-N Diagram Characteristics:
  • Low-Cycle Fatigue: $1 \le N \le 10^3$ cycles.
  • High-Cycle Fatigue: $10^3 \le N \le 10^6$ cycles.
  • Endurance Limit ($S_e'$): Stress level below which steels sustain infinite cycles ($N > 10^6$) without failure.
  • Non-ferrous alloys (aluminum, copper, magnesium) do not exhibit a distinct horizontal knee; fatigue strength continues declining with cycle count.

• Endurance Limit Estimation for Steels:

  • Unadjusted laboratory endurance limit $S_e'$ based on minimum tensile strength $S_{ut}$:     $S_e' = \begin{cases} 0.50\,S_{ut} & S_{ut} \le 1400\,\text{MPa} \\ 700\,\text{MPa} & S_{ut} > 1400\,\text{MPa} \end{cases}$

• Finite-Life S-N Power-Law Formulation ($10^3 \le N \le 10^6$):

  • The high-cycle segment follows a power relationship:     $S_f = a N^b$$\log(S_f) = \log(a) + b \log(N)$
  • Boundary conditions: at $N = 10^3$, $S_f = f S_{ut}$; at $N = 10^6$, $S_f = S_e$.
  • Mathematical coefficients:     $b = -\frac{1}{3} \log\left(\frac{f S_{ut}}{S_e}\right)$$a = \frac{(f S_{ut})^2}{S_e}$
  • Permissible cycles to failure for completely reversed applied stress $\sigma_a$:     $N = \left(\frac{\sigma_a}{a}\right)^{\frac{1}{b}}$
  • Fatigue strength fraction $f$ is obtained as a function of tensile strength $S_{ut}$. For $S_{ut} \le 70\,\text{kpsi}$ ($\approx 490\,\text{MPa}$), use conservative value $f = 0.90$.

• Marin Modifying Factors for Real Components:$S_e = k_a k_b k_c k_d k_e k_f S_e'$

  • Surface Condition Factor ($k_a$):$k_a = a S_{ut}^b$
  • Ground finish: $a = 1.58$, $b = -0.085$
  • Machined or cold-drawn: $a = 4.51$, $b = -0.265$
  • Hot-rolled: $a = 57.7$, $b = -0.718$
  • As-forged: $a = 272$, $b = -0.995$
  • Size Modification Factor ($k_b$):
  • For bending and torsion of round members:       $k_b = \begin{cases} 1.24\,d^{-0.107} = \left(\frac{d}{7.62}\right)^{-0.107} & 2.79\,\text{mm} \le d \le 51\,\text{mm} \\ 1.51\,d^{-0.157} & 51\,\text{mm} < d \le 254\,\text{mm} \end{cases}$
  • For non-rotating rectangular profiles of depth $h$ and width $b$: use effective equivalent diameter $d_e = 0.808\sqrt{h b}$.
  • For axial loading: no size gradient exists; $k_b = 1.0$.
  • Load Modification Factor ($k_c$):$k_c = \begin{cases} 1.0 & \text{Bending} \\ 0.85 & \text{Axial} \\ 0.59 & \text{Torsion} \end{cases}$
  • Temperature Modification Factor ($k_d$):$k_d = \begin{cases} 1.0 & T \le 350^\circ\text{C} \\ 0.53 & 500^\circ\text{C} < T \le 5000^\circ\text{C} \end{cases}$
  • Reliability Factor ($k_e$):
  • 50% ($z_a = 0$): $k_e = 1.000$
  • 90% ($z_a = 1.288$): $k_e = 0.897$
  • 95% ($z_a = 1.645$): $k_e = 0.868$
  • 99% ($z_a = 2.326$): $k_e = 0.814$
  • 99.9% ($z_a = 3.091$): $k_e = 0.753$
  • 99.99% ($z_a = 3.719$): $k_e = 0.702$
  • 99.999% ($z_a = 4.265$): $k_e = 0.659$
  • 99.9999% ($z_a = 4.753$): $k_e = 0.620$
  • Miscellaneous-Effects Factor ($k_f$): Aggregate factor accounting for residual stresses, corrosion pits, metal spraying, plating, and fretting.

• Fatigue Stress Concentration and Notch Sensitivity:

  • Fatigue notch sensitivity $q$ quantifies notch effects on fatigue strength:     $K_f = 1 + q (K_t - 1)$$K_{fs} = 1 + q_{\text{shear}} (K_{ts} - 1)$
  • When $q = 0$, material is completely insensitive ($K_f = 1.0$); when $q = 1.0$, material is fully sensitive ($K_f = K_t$).
  • Fatigue stress concentration is applied to both alternating and nominal midrange stresses in elastic conditions:     $\sigma_a = K_f \sigma_{a,0}$$\sigma_m = K_f \sigma_{m,0}$

• Fluctuating Stress Terminology and Waveforms:

  • Minimum stress: $\sigma_{\min}$
  • Maximum stress: $\sigma_{\max}$
  • Stress range: $\sigma_r = \sigma_{\max} - \sigma_{\min}$
  • Midrange (mean) stress component: $\sigma_m = \frac{\sigma_{\max} + \sigma_{\min}}{2}$
  • Alternating (amplitude) stress component: $\sigma_a = \left|\frac{\sigma_{\max} - \sigma_{\min}}{2}\right|$
  • Stress ratio: $R = \frac{\sigma_{\min}}{\sigma_{\max}}$
  • Amplitude ratio: $A = \frac{\sigma_a}{\sigma_m}$
  • Fully reversed stress cycle: $R = -1.0, \sigma_m = 0$
  • Repeated stress cycle: $R = 0, \sigma_{\min} = 0$

• Fatigue Failure Criteria for Fluctuating Stress:

  • Soderberg Line: Uses yield strength to prevent all gross plastic yielding:     $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{n}$
  • Modified Goodman Line: Linear relation based on ultimate tensile strength:     $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n}$
  • Gerber Parabolic Relation: Parabolic curve matching average experimental test points:     $\frac{n \sigma_a}{S_e} + \left(\frac{n \sigma_m}{S_{ut}}\right)^2 = 1$
  • ASME-Elliptic Line: Elliptic formulation incorporating material yield strength:     $\left(\frac{n \sigma_a}{S_e}\right)^2 + \left(\frac{n \sigma_m}{S_y}\right)^2 = 1$
  • Langer First-Cycle Yield Line: Guards against first-cycle yielding on the forward stroke:     $\sigma_a + \sigma_m = \frac{S_y}{n_y}$
  • Radial Load Line: Intersects the failure envelope with a constant operational slope:     $r = \frac{\sigma_a}{\sigma_m}$

• Worked Example: Fatigue Life of a Stepped Transmission Shaft:

  • Shaft Specifications: Simply supported in bearings at $A$ and $D$ (span $550\,\text{mm}$). Stationary transverse point load $F = 6.8\,\text{kN}$ at $x = 325\,\text{mm}$ from $A$. Diameters: $32\,\text{mm}$ from $A$ to $B$ ($250\,\text{mm}$), $38\,\text{mm}$ from $B$ to $C$ ($175\,\text{mm}$), $35\,\text{mm}$ from $C$ to $D$ ($125\,\text{mm}$). Bearing necks are $30\,\text{mm}$. All shoulder fillet radii $r = 3\,\text{mm}$. Material: AISI 1050 Cold-Drawn Steel ($S_{ut} = 690\,\text{MPa}$, $S_y = 580\,\text{MPa}$, $H_B = 197$).
  • Reactions and Bending Moment at Critical Shoulder B:
  • Equilibrium of external forces:       $R_1 \times 0.550 = 6.8 \times 10^3 \times (0.550 - 0.325) = 6.8 \times 10^3 \times 0.225$$R_1 = \frac{6.8 \times 10^3 \times 0.225}{0.550} = 2.782\,\text{kN}$
  • Moment at shoulder $B$ ($x = 0.250\,\text{m}$):       $M_B = R_1 \times 0.250 = (2.782 \times 10^3) \times 0.250 = 695.5\,\text{Nm}$
  • Endurance Limit Calculations:
  • Unadjusted rotary-beam endurance limit:       $S_e' = 0.50 \times 690 = 345\,\text{MPa}$
  • Surface factor ($k_a$) for machined/cold-drawn surface:       $k_a = 4.51 \times (690)^{-0.265} = 0.798$
  • Size factor ($k_b$) for $d = 32\,\text{mm}$ in bending:       $k_b = \left(\frac{32}{7.62}\right)^{-0.107} = 0.858$
  • Other factors: $k_c = k_d = k_e = k_f = 1.0$
  • Corrected endurance limit:       $S_e = 0.798 \times 0.858 \times 1.0 \times 1.0 \times 1.0 \times 1.0 \times 345 = 236.2\,\text{MPa}$
  • Stress Concentration and Dynamic Reversing Stress:
  • Geometric ratios at $B$: $D/d = 38/32 = 1.1875$, $r/d = 3/32 = 0.09375$.
  • From stepped shaft bending chart: $K_t \approx 1.65$.
  • Notch sensitivity for $r = 3\,\text{mm}$ and $S_{ut} = 690\,\text{MPa}$: $q = 0.84$.
  • Fatigue stress concentration factor:       $K_f = 1 + q (K_t - 1) = 1 + 0.84(1.65 - 1) = 1.546 \approx 1.55$
  • Section modulus of $32\,\text{mm}$ section:       $Z = \frac{\pi d^3}{32} = \frac{\pi (0.032)^3}{32} = 3.217 \times 10^{-6}\,\text{m}^3$
  • Reversing bending stress:       $\sigma_{\text{rev}} = K_f \frac{M_B}{Z} = 1.55 \times \frac{695.5}{3.217 \times 10^{-6}} = 335.1\,\text{MPa}$
  • Because $\sigma_{\text{rev}} = 335.1\,\text{MPa} > S_e = 236.2\,\text{MPa}$, the part has a finite fatigue life.
  • Finite Fatigue Life Estimation:
  • For $S_{ut} = 690\,\text{MPa} = 100\,\text{kpsi}$, fatigue strength fraction $f = 0.844$.
  • S-N parameters:       $b = -\frac{1}{3} \log\left(\frac{0.844 \times 690}{236.2}\right) = -\frac{1}{3} \log(2.465) = -0.1306$$a = \frac{(0.844 \times 690)^2}{236.2} = \frac{(582.36)^2}{236.2} = 1435.8\,\text{MPa}$
  • Predicted cycles to failure:       $N = \left(\frac{\sigma_{\text{rev}}}{a}\right)^{\frac{1}{b}} = \left(\frac{335.1}{1435.8}\right)^{-\frac{1}{0.1306}} = (0.2334)^{-7.657} = 6.8 \times 10^4\,\text{cycles} = 68 \times 10^3\,\text{cycles}$

• Worked Example: Material Selection under Fluctuating Flexure:

  • Parameters: Flexural stress fluctuates between $\sigma_{\max} = +300\,\text{MN/m}^2$ and $\sigma_{\min} = -150\,\text{MN/m}^2$. Design factor $n = 2.0$. Material properties: $S_y = 0.55\,S_{ut}$, $S_e = 0.50\,S_{ut}$.
  • Mean and Alternating Components:$\sigma_m = \frac{300 + (-150)}{2} = 75\,\text{MN/m}^2$$\sigma_a = \frac{300 - (-150)}{2} = 225\,\text{MN/m}^2$
  • 1. Modified Goodman Criterion:$\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n}$$\frac{225}{0.50\,S_{ut}} + \frac{75}{S_{ut}} = \frac{1}{2.0}$$\frac{450 + 75}{S_{ut}} = 0.50 \implies S_{ut} = \frac{525}{0.50} = 1050\,\text{MN/m}^2$
  • 2. Soderberg Criterion:$\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{n}$$\frac{225}{0.50\,S_{ut}} + \frac{75}{0.55\,S_{ut}} = \frac{1}{2.0}$$\frac{450 + 136.36}{S_{ut}} = 0.50 \implies S_{ut} = \frac{586.36}{0.50} = 1172.72\,\text{MN/m}^2$
  • 3. Gerber Criterion:$\frac{n \sigma_a}{S_e} + \left(\frac{n \sigma_m}{S_{ut}}\right)^2 = 1$$\frac{2(225)}{0.50\,S_{ut}} + \left(\frac{2(75)}{S_{ut}}\right)^2 = 1$$\frac{900}{S_{ut}} + \frac{22500}{S_{ut}^2} = 1 \implies S_{ut}^2 - 900\,S_{ut} - 22500 = 0$$S_{ut} = \frac{900 + \sqrt{(-900)^2 - 4(1)(-22500)}}{2} = \frac{900 + \sqrt{810000 + 90000}}{2} = \frac{900 + 948.68}{2} = 924.34\,\text{MN/m}^2$

• Worked Example: Sizing Circular Bar under Fluctuating Tensile Load:

  • Loading: Alternating axial force varying from $F_{\min} = 200\,\text{kN}$ to $F_{\max} = 500\,\text{kN}$. Material has $S_{ut} = 900\,\text{MPa}$, $S_e = 700\,\text{MPa}$. Factors of safety applied: $n_{ut} = 3.5$ on ultimate strength, $n_e = 4.0$ on endurance limit. Fatigue stress concentration factor $K_f = 1.65$.
  • Force and Nominal Stress Quantities:$F_m = \frac{500 + 200}{2} = 350\,\text{kN}$$F_a = \frac{500 - 200}{2} = 150\,\text{kN}$$A = \frac{\pi d^2}{4} = 0.7854\,d^2$$\sigma_m = \frac{F_m}{A} = \frac{350 \times 10^3}{0.7854\,d^2} = \frac{445.63 \times 10^3}{d^2}\,\text{N/m}^2$$\sigma_a = K_f \frac{F_a}{A} = 1.65 \times \frac{150 \times 10^3}{0.7854\,d^2} = \frac{315.13 \times 10^3}{d^2}\,\text{N/m}^2$
  • Design via Modified Goodman Equation:$\frac{\sigma_a}{\frac{S_e}{n_e}} + \frac{\sigma_m}{\frac{S_{ut}}{n_{ut}}} = 1$$\frac{315.13 \times 10^3}{d^2 \times \left(\frac{700 \times 10^6}{4.0}\right)} + \frac{445.63 \times 10^3}{d^2 \times \left(\frac{900 \times 10^6}{3.5}\right)} = 1$$\frac{1}{d^2} \left[ \frac{315.13 \times 10^3}{175 \times 10^6} + \frac{445.63 \times 10^3}{257.14 \times 10^6} \right] = 1$$\frac{1}{d^2} \left[ 1.801 \times 10^{-3} + 1.733 \times 10^{-3} \right] = 1$$\frac{3.534 \times 10^{-3}}{d^2} = 1 \implies d = \sqrt{3.534 \times 10^{-3}} = 0.05945\,\text{m} = 59.45\,\text{mm}$
  • Select standard nominal diameter: $d = 60\,\text{mm}$.