Index / Sheet list
Engineering formulas & units

50 essential mechanical engineering formulas with units and applications

One page with the fifty formulas mechanical engineers use most, grouped by subject, with SI units, where each is used, and links to full worked examples.

On this page
  1. Strength of materials (1–10)
  2. Thermodynamics (11–20)
  3. Fluid mechanics (21–30)
  4. Manufacturing (31–40)
  5. Machine design and heat transfer (41–50)

This is the quick-reference sheet: fifty formulas that cover most hand calculations in mechanical design, from a stress check on a bracket to the power of a pump. Each table gives the formula, what it is used for and the SI units of the result. The full guides linked under each table explain every symbol, list the assumptions and work through an example.

Free PDF: 50 essential formulasSix A4 pages: every formula with a sketch, its symbols, SI units and assumptions.Download PDF
MotorBearingBearingGearTorque, TSpeed, NGear load, FP = 2πNT / 60τ = 16T / πd³
One shaft, many formulas. The motor's torque T and speed N set the power, the gear load bends the shaft between its bearings, the torque twists it, and the bearings carry the reactions.

Before you use any of these

Use consistent units: base SI (N, m, Pa, J, W, K), or N, mm and MPa for stresses. Use absolute temperature (K) and absolute pressure in gas formulas. Each formula has assumptions; the full guides list them.

Strength of materials (1–10)

#FormulaUsed forResult in
1σ=F/A\sigma = F/ATensile or compressive stress in a rod, bolt or columnPa
2σ=Eε\sigma = E\varepsilon, ε=δ/L\varepsilon = \delta/LHooke’s law: linking stress and strainPa
3δ=FL/AE\delta = FL/AEStretch or shortening of a barm
4τ=F/nA\tau = F/nAShear in pins, rivets and keysPa
5σ=EαΔT\sigma = E\alpha\Delta TStress in a part that cannot expandPa
6σb=My/I\sigma_b = My/IBending stress in beams and bracketsPa
7τ=Tr/J\tau = Tr/JShear stress in a twisted shaftPa
8θ=TL/GJ\theta = TL/GJAngle of twist of a shaftrad
9δ=FL3/48EI\delta = FL^3/48EIDeflection of a simply supported beam, central loadm
10Pcr=π2EI/(KL)2P_{cr} = \pi^2EI/(KL)^2Buckling load of a slender columnN

Full explanations and examples: strength of materials formulas.

Thermodynamics (11–20)

#FormulaUsed forResult in
11pV=mRTpV = mRTMass, pressure or temperature of a gas—
12Q−W=ΔUQ - W = \Delta UFirst law for a closed systemJ
13Q=mcΔTQ = mc\Delta THeat to warm or cool a part or a fluidJ
14W=p(V2−V1)W = p(V_2 - V_1)Work at constant pressureJ
15W=mRTln⁡(V2/V1)W = mRT\ln(V_2/V_1)Work in an isothermal processJ
16T2/T1=(p2/p1)(γ−1)/γT_2/T_1 = (p_2/p_1)^{(\gamma-1)/\gamma}Temperature after adiabatic compressionK
17Q˙−W˙=m˙(Δh+ΔV2/2+gΔz)\dot Q - \dot W = \dot m(\Delta h + \Delta V^2/2 + g\Delta z)Energy balance on turbines, compressors, nozzlesW
18η=1−TL/TH\eta = 1 - T_L/T_HCarnot (maximum) efficiency—
19COPR=QL/WCOP_R = Q_L/WPerformance of a refrigerator or heat pump—
20ΔS=mcln⁡(T2/T1)\Delta S = mc\ln(T_2/T_1)Entropy change of a solid or liquidJ/K

Full explanations and examples: thermodynamics formulas.

Fluid mechanics (21–30)

#FormulaUsed forResult in
21p=ρghp = \rho g hPressure at a depth in a liquidPa
22F2=F1A2/A1F_2 = F_1 A_2/A_1Force from a hydraulic press or cylinderN
23FB=ρfgVF_B = \rho_f g VBuoyant forceN
24Q=A1V1=A2V2Q = A_1V_1 = A_2V_2Continuity: velocity change in a ductm³/s
25p+12ρV2+ρgz=constp + \tfrac{1}{2}\rho V^2 + \rho g z = \text{const}Bernoulli: pressure and velocity along a flowPa
26Re=ρVD/μRe = \rho VD/\muLaminar or turbulent flow—
27hf=f(L/D)(V2/2g)h_f = f(L/D)(V^2/2g)Friction loss in a ductm
28f=64/Ref = 64/ReFriction factor in laminar flow—
29P=ρgQH/ηP = \rho g Q H/\etaPump powerW
30FD=12ρV2CDAF_D = \tfrac{1}{2}\rho V^2 C_D ADrag force on a bodyN

Full explanations and examples: fluid mechanics formulas.

Manufacturing (31–40)

#FormulaUsed forResult in
31N=1000V/πDN = 1000V/\pi DSpindle speed from cutting speedrev/min
32t=L/fNt = L/fNTurning time per passmin
33MRR=1000VfdMRR = 1000VfdMaterial removal rate in turningmm³/min
34vf=fzzNv_f = f_z z NTable feed in millingmm/min
35MRR=aeapvfMRR = a_e a_p v_fMaterial removal rate in millingmm³/min
36t=(L+0.3D+c)/fNt = (L + 0.3D + c)/fNDrilling timemin
37P=FcV/60P = F_c V/60, Fc=kcfdF_c = k_c f dCutting force and powerW
38F=LtτsF = L t \tau_sBlanking and punching forceN
39BA=θ(π/180)(R+Kt)BA = \theta(\pi/180)(R + Kt)Sheet-metal bend allowancemm
40VTn=CVT^n = CTaylor tool lifemin

Full explanations and examples: manufacturing calculations.

Machine design and heat transfer (41–50)

#FormulaUsed forResult in
41P=2πNT/60P = 2\pi N T/60Power from torque and speedW
42i=N1/N2=z2/z1i = N_1/N_2 = z_2/z_1Gear ratio—
43N1D1=N2D2N_1 D_1 = N_2 D_2Speed ratio of a belt drive—
44F=kxF = kxSpring forceN
45τ=8FDKw/πd3\tau = 8FDK_w/\pi d^3Shear stress in a helical springPa
46σ=F/At\sigma = F/A_tTensile stress in a bolt (on the stress area)Pa
47L10=(C/P)3L_{10} = (C/P)^3Rating life of a ball bearing10⁶ rev
48Q˙=kAΔT/L\dot Q = kA\Delta T/LHeat conducted through a wallW
49Q˙=hAΔT\dot Q = hA\Delta THeat carried away by convectionW
50η=1−1/rγ−1\eta = 1 - 1/r^{\gamma - 1}Ideal efficiency of a petrol (Otto) engine—

These are explained with diagrams and examples in how gears work, bearings and shafts, belt and chain drives, springs and clutches, keys, couplings and fasteners, heat transfer fundamentals and Otto and Diesel cycles.

Worked example: one shaft, four formulas

A motor drives a 30 mm solid steel shaft at 1450 rev/min and delivers 5.5 kW. Find the torque, the shaft’s shear stress and the speed of the output gear if the 20-tooth pinion drives a 60-tooth gear.

T=60P2πN=60×55002π×1450=36.2 N⋅mT = \frac{60P}{2\pi N} = \frac{60 \times 5500}{2\pi \times 1450} = 36.2\ \text{N·m}τ=16Tπd3=16×36.2π×0.0303=6.8 MPa\tau = \frac{16T}{\pi d^3} = \frac{16 \times 36.2}{\pi \times 0.030^3} = 6.8\ \text{MPa}N2=N1 z1z2=1450×2060=483 rev/min,T2≈3×36.2=109 N⋅mN_2 = N_1\,\frac{z_1}{z_2} = 1450 \times \frac{20}{60} = 483\ \text{rev/min}, \qquad T_2 \approx 3 \times 36.2 = 109\ \text{N·m}
AnswerT = 36.2 N·m, τ = 6.8 MPa, output 483 rev/min at about 109 N·m (before gear losses)

The low shear stress shows that this shaft’s diameter is set by stiffness, keyways and bearing sizes, not by strength, which is common for small drive shafts.

Key points

  • Formulas 1–10 check whether a part is strong and stiff enough.
  • Formulas 11–30 cover energy, heat and flow in engines, pumps and ducts.
  • Formulas 31–40 plan machining and sheet-metal work.
  • Formulas 41–50 size drives, springs, bolts and bearings and estimate heat flow.

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