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Advanced mechanical engineering: 50 key concepts

Upper-level mechanical engineering: fracture, fatigue, FEA, turbulence, turbomachinery, compressible flow, vibration, control, creep, composites and GD&T.

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Mode I stress intensity factor

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  1. 1
    Mode I stress intensity factor
    KI=YσπaK_I = Y\sigma\sqrt{\pi a}
    YY is a geometry factor: 1 for a central crack of length 2a2a in a wide plate. KK is in MPam\mathrm{MPa\sqrt{m}}.
  2. 2
    Plane-strain fracture toughness
    The critical value KIcK_{Ic} of the mode I stress intensity factor: a crack grows unstably once KI≥KIcK_I \ge K_{Ic}
    A valid ASTM E399 test needs a thickness and crack length of at least 2.5(KIcσy)22.5\left(\dfrac{K_{Ic}}{\sigma_y}\right)^2.
  3. 3
    The three crack loading modes
    Mode I: opening, with tension normal to the crack. Mode II: in-plane shear (sliding). Mode III: out-of-plane shear (tearing)
  4. 4
    J-integral
    A path-independent contour integral around a crack tip that gives the energy release rate, extending fracture mechanics to elastic–plastic materials
    For linear-elastic behavior J=G=K2/E′J = G = K^2/E', with E′=EE' = E in plane stress and E/(1−ν2)E/(1 - \nu^2) in plane strain.
  5. 5
    Paris law
    dadN=C (ΔK)m\dfrac{da}{dN} = C\,(\Delta K)^m
    Fatigue crack growth per cycle in the stable middle region; ΔK=Kmax⁡−Kmin⁡\Delta K = K_{\max} - K_{\min}, and mm is typically 2 to 4 for metals.
  6. 6
    S–N curve
    A plot of stress amplitude against the number of cycles to failure, usually on log axes
    Steels often level off at an endurance limit after about 10610^6 to 10710^7 cycles; aluminium alloys have no true endurance limit.
  7. 7
    Alternating and mean stress
    σa=σmax⁡−σmin⁡2\sigma_a = \dfrac{\sigma_{\max} - \sigma_{\min}}{2}, σm=σmax⁡+σmin⁡2\sigma_m = \dfrac{\sigma_{\max} + \sigma_{\min}}{2}
    The stress ratio is R=σmin⁡/σmax⁡R = \sigma_{\min}/\sigma_{\max}: −1-1 for fully reversed loading, 0 for zero-to-maximum.
  8. 8
    Goodman criterion for fatigue with a mean stress
    σaSe+σmSut=1n\dfrac{\sigma_a}{S_e} + \dfrac{\sigma_m}{S_{ut}} = \dfrac{1}{n}
    For tensile mean stress. Soderberg uses SyS_y in place of SutS_{ut} and is more conservative; the Gerber parabola is less conservative.
  9. 9
    Miner’s rule
    Failure is predicted when ∑niNi=1\displaystyle\sum \dfrac{n_i}{N_i} = 1
    nin_i is the number of cycles applied at stress level ii and NiN_i the life at that level. It ignores the order of the loads.
  10. 10
    Fatigue stress concentration factor
    Kf=1+q (Kt−1)K_f = 1 + q\,(K_t - 1)
    qq is the notch sensitivity, from 0 (the notch has no effect) to 1 (Kf=KtK_f = K_t).
  11. 11
    Stiffness matrix of a two-node bar element
    AEL[1−1−11]\dfrac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}
    It relates nodal forces to nodal displacements. The assembled global matrix is singular until supports stop rigid-body motion.
  12. 12
    h- vs p-refinement
    h-refinement uses smaller elements; p-refinement raises the polynomial order of the elements’ shape functions
    A mesh convergence study refines until the results of interest stop changing.
  13. 13
    Shear locking
    Excessive bending stiffness in fully integrated linear elements, which can’t bend without spurious shear strain
    Cures include reduced integration, incompatible-mode elements and quadratic elements.
  14. 14
    Stress at a sharp re-entrant corner in a linear-elastic FE model
    A singularity: the computed peak stress keeps rising as the mesh is refined, so it has no meaning
    Model the real fillet radius, or judge the stress away from the corner.
  15. 15
    Explicit vs implicit time integration
    Explicit methods are only conditionally stable, so the time step is limited by the smallest element; implicit methods can be unconditionally stable but solve a system of equations every step
    Explicit suits short, violent events like crashes and impacts; implicit suits slower dynamics.
  16. 16
    Reynolds-averaged Navier–Stokes (RANS)
    The flow equations averaged after splitting each flow variable into a mean and a fluctuation; the leftover Reynolds stresses need a turbulence model
    The Reynolds stress tensor is −ρ ui′uj′‾-\rho\,\overline{u_i' u_j'}. Having to model it is the closure problem.
  17. 17
    Boussinesq eddy-viscosity hypothesis
    Reynolds stresses are modeled as proportional to the mean strain rate, through a turbulent (eddy) viscosity
    −ρ ui′uj′‾=μt(∂uˉi∂xj+∂uˉj∂xi)−23ρk δij-\rho\,\overline{u_i' u_j'} = \mu_t\left(\dfrac{\partial \bar{u}_i}{\partial x_j} + \dfrac{\partial \bar{u}_j}{\partial x_i}\right) - \tfrac{2}{3}\rho k\,\delta_{ij}, where kk is the turbulent kinetic energy.
  18. 18
    k–ε vs k–ω SST turbulence models
    Standard k–ε is robust in free shear flows but poor near walls and in adverse pressure gradients; SST uses k–ω near walls and blends to k–ε away from them
    SST generally predicts flow separation better.
  19. 19
    Dimensionless wall distance y+y^+
    y+=y uτνy^+ = \dfrac{y\,u_\tau}{\nu}, with friction velocity uτ=τw/ρu_\tau = \sqrt{\tau_w/\rho}
    Resolving the viscous sublayer needs the first cell at y+≈1y^+ \approx 1; wall functions expect roughly 30<y+<30030 < y^+ < 300.
  20. 20
    DNS vs LES
    DNS resolves every turbulent scale down to the Kolmogorov scale; LES resolves the large eddies and models only the small, subgrid scales
    DNS cost grows roughly as Re3\mathrm{Re}^3, which limits it to low Reynolds numbers.
  21. 21
    Euler turbomachinery equation
    w=U2Vθ2−U1Vθ1w = U_2 V_{\theta 2} - U_1 V_{\theta 1}
    Work per unit mass done on the fluid by a pump or compressor rotor. UU is the blade speed and VθV_\theta the swirl component of absolute velocity. For a turbine the sign reverses.
  22. 22
    Pump affinity laws for a fixed impeller diameter
    Q∝NQ \propto N, H∝N2H \propto N^2, P∝N3P \propto N^3
    NN is the rotational speed, so cutting speed by 20% roughly halves the power.
  23. 23
    Pump specific speed
    Ns=NQH3/4N_s = \dfrac{N\sqrt{Q}}{H^{3/4}}
    Taken at the best efficiency point. Low values suit radial (centrifugal) impellers; high values suit mixed-flow and axial ones.
  24. 24
    Condition for avoiding pump cavitation
    The net positive suction head available (NPSHa) must exceed the NPSH required by the pump
    NPSHa=pin−pvρg+V22g\mathrm{NPSH_a} = \dfrac{p_{\text{in}} - p_v}{\rho g} + \dfrac{V^2}{2g}, with the absolute inlet pressure pinp_{\text{in}} and the liquid’s vapor pressure pvp_v.
  25. 25
    Stagnation temperature of an ideal gas
    T0T=1+γ−12M2\dfrac{T_0}{T} = 1 + \dfrac{\gamma - 1}{2}M^2
    M=V/aM = V/a, with the speed of sound a=γRTa = \sqrt{\gamma R T}.
  26. 26
    Area–velocity relation for isentropic flow
    dAA=(M2−1)dVV\dfrac{dA}{A} = (M^2 - 1)\dfrac{dV}{V}
    Subsonic flow speeds up in a converging duct, supersonic flow in a diverging one, hence the converging–diverging nozzle.
  27. 27
    Choked flow
    Once a nozzle’s throat reaches Mach 1, lowering the back pressure no longer increases the mass flow rate
    The critical pressure ratio is p∗p0=(2γ+1)γ/(γ−1)≈0.528\dfrac{p^*}{p_0} = \left(\dfrac{2}{\gamma + 1}\right)^{\gamma/(\gamma - 1)} \approx 0.528 for air.
  28. 28
    What changes across a normal shock
    The flow goes from supersonic to subsonic; static pressure, temperature and density rise, stagnation pressure falls, and stagnation temperature stays the same
    The process is adiabatic but irreversible, so entropy increases.
  29. 29
    Natural frequencies of a multi-degree-of-freedom system
    The roots of det⁡([K]−ω2[M])=0\det\left([K] - \omega^2[M]\right) = 0
    Each ωi\omega_i has a mode shape {ϕi}\{\phi_i\} with ([K]−ωi2[M]){ϕi}={0}\left([K] - \omega_i^2[M]\right)\{\phi_i\} = \{0\}.
  30. 30
    Orthogonality of mode shapes
    {ϕi}T[M]{ϕj}=0\{\phi_i\}^T[M]\{\phi_j\} = 0 and {ϕi}T[K]{ϕj}=0\{\phi_i\}^T[K]\{\phi_j\} = 0 for i≠ji \ne j
    It lets modal analysis turn the coupled equations of motion into independent single-degree-of-freedom ones.
  31. 31
    Rayleigh (proportional) damping
    [C]=α[M]+β[K][C] = \alpha[M] + \beta[K]
    It keeps the modal equations uncoupled; the damping ratio of mode ii is ζi=α2ωi+βωi2\zeta_i = \dfrac{\alpha}{2\omega_i} + \dfrac{\beta\omega_i}{2}.
  32. 32
    Undamped dynamic vibration absorber
    A small spring–mass attached to a machine and tuned so ka/ma\sqrt{k_a/m_a} equals the forcing frequency, which stops the machine vibrating at that frequency
    It splits the original resonance into two new ones on either side, so it only works near the tuned frequency.
  33. 33
    PID control law
    u(t)=Kpe(t)+Ki∫0te(τ) dτ+Kddedtu(t) = K_p e(t) + K_i\displaystyle\int_0^t e(\tau)\,d\tau + K_d\dfrac{de}{dt}
    The integral term removes steady-state error; the derivative term adds damping by reacting to how fast the error changes.
  34. 34
    Transfer function
    The Laplace transform of a linear time-invariant system’s output divided by that of its input, with zero initial conditions
    For mx¨+cx˙+kx=f(t)m\ddot{x} + c\dot{x} + kx = f(t): X(s)F(s)=1ms2+cs+k\dfrac{X(s)}{F(s)} = \dfrac{1}{ms^2 + cs + k}.
  35. 35
    Stability condition for a continuous-time LTI system
    Every pole of the closed-loop transfer function has a negative real part, in the left half of the s-plane
    The Routh–Hurwitz criterion checks this from the characteristic polynomial without finding the roots.
  36. 36
    Gain margin and phase margin
    Gain margin: how much the loop gain can rise before instability, read where the phase is −180°. Phase margin: how much extra phase lag is tolerable, read where the gain is 1 (0 dB)
    Rules of thumb ask for at least 6 dB of gain margin and 30° to 60° of phase margin.
  37. 37
    Percent overshoot of an underdamped second-order system
    %OS=100 e−ζπ/1−ζ2\%\mathrm{OS} = 100\,e^{-\zeta\pi/\sqrt{1 - \zeta^2}}
    For a step input to ωn2s2+2ζωns+ωn2\dfrac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}. A damping ratio of 0.7 gives about 5%.
  38. 38
    Archard wear equation
    V=K FN sHV = K\,\dfrac{F_N\,s}{H}
    Worn volume VV grows with the normal load FNF_N and sliding distance ss and falls with the hardness HH; KK is a dimensionless wear coefficient.
  39. 39
    Stribeck curve
    The friction coefficient plotted against the Hersey number ηN/P\eta N/P, showing boundary, mixed and hydrodynamic lubrication
    η\eta is the viscosity, NN the speed and PP the load per unit area. Friction is lowest around the start of full-film lubrication.
  40. 40
    Specific film thickness (lambda ratio)
    λ=hmin⁡Rq12+Rq22\lambda = \dfrac{h_{\min}}{\sqrt{R_{q1}^2 + R_{q2}^2}}
    Roughly: λ<1\lambda < 1 means boundary lubrication, 1 to 3 mixed, above 3 a full film. RqR_q is each surface’s RMS roughness.
  41. 41
    Three stages of creep
    Primary: the strain rate falls. Secondary: a steady, minimum strain rate. Tertiary: the rate accelerates until rupture
    Creep matters in metals above about 0.4 Tm0.4\,T_m, where TmT_m is the absolute melting temperature.
  42. 42
    Norton power law for steady-state creep
    ε˙=Aσne−Q/RT\dot{\varepsilon} = A\sigma^n e^{-Q/RT}
    QQ is the activation energy for creep and nn the stress exponent, typically 3 to 8 for dislocation creep in metals.
  43. 43
    Larson–Miller parameter
    LMP=T (C+log⁡10tr)\mathrm{LMP} = T\,(C + \log_{10} t_r)
    TT is the absolute temperature, trt_r the rupture time in hours and C≈20C \approx 20. It trades time against temperature to extrapolate creep-rupture data.
  44. 44
    Longitudinal modulus of a unidirectional composite
    E1=VfEf+VmEmE_1 = V_f E_f + V_m E_m
    The rule of mixtures, assuming fiber and matrix strain equally (iso-strain).
  45. 45
    Transverse modulus of a unidirectional composite
    1E2=VfEf+VmEm\dfrac{1}{E_2} = \dfrac{V_f}{E_f} + \dfrac{V_m}{E_m}
    The inverse rule of mixtures, assuming equal stress (iso-stress). It is only a rough estimate; Halpin–Tsai is more accurate.
  46. 46
    Why laminates are usually made symmetric
    The coupling matrix [B][B] is zero, so in-plane loads don’t cause bending or twisting
    A symmetric laminate also doesn’t warp as it cools after curing.
  47. 47
    Datum reference frame (3-2-1 principle)
    Three mutually perpendicular planes set up from datum features: the primary datum touches at least three points, the secondary two and the tertiary one
    Together they constrain all six degrees of freedom of the part.
  48. 48
    Maximum material condition (MMC)
    The size at which a feature contains the most material: the smallest hole or the largest shaft
    A geometric tolerance applied at MMC gains a bonus equal to the feature’s departure from MMC.
  49. 49
    Position tolerance zone for a hole
    A cylinder of the stated diameter, centered on the true position set by basic dimensions from the datums, that must contain the hole’s axis
    A round zone gives about 57% more area than the square zone of an equivalent ± tolerance.
  50. 50
    Rule #1 (envelope principle) in ASME Y14.5
    When only a size tolerance is given, a feature’s surface may not go beyond a boundary of perfect form at MMC
    So the size tolerance also limits form errors such as bow and out-of-roundness. ISO standards default to the independency principle instead.

Frequently asked questions

What background does this deck assume?

The core undergraduate courses, as in the intermediate deck: strength of materials, dynamics and vibrations, thermodynamics, fluid mechanics and heat transfer, plus linear algebra and Laplace transforms.

When is linear-elastic fracture mechanics valid?

When the plastic zone at the crack tip is small compared with the crack and the part. For tough, ductile materials, elastic–plastic methods such as the J-integral or crack tip opening displacement are used instead.

Which GD&T standard do the cards follow?

ASME Y14.5, common in North America. ISO standards share most of the symbols but differ in some defaults, such as the independency principle in place of Rule #1.

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