Formula sheet

Every key formula from the finished topics, grouped by branch. Use the Print button for a handout.

Conventions used. Tension and tensile stress are positive. A positive (sagging) bending moment makes a beam curve like a smile. Unless stated otherwise, SI units are used: forces in N or kN, lengths in m or mm, stresses in Pa, MPa or GPa.

Engineering Mechanics

Resultant of a force system

Components of a force
Fx=Fcosθ,Fy=FsinθF_x = F\cos\theta, \quad F_y = F\sin\theta
Components of the resultant
Rx=∑Fx,Ry=∑FyR_x = \sum F_x, \quad R_y = \sum F_y
Magnitude of the resultant
R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}
Direction of the resultant
θ=tan−1RyRx\theta = \tan^{-1}\frac{R_y}{R_x}

Equilibrium and support reactions

Force equilibrium
∑Fx=0,∑Fy=0\sum F_x = 0, \quad \sum F_y = 0
Moment equilibrium
∑MA=0\sum M_A = 0
Equivalent load of a UDL
W=wL,acting at mid-lengthW = wL, \quad \text{acting at mid-length}
Reaction of a simply supported beam
RB=∑(load×distance from A)LR_B = \frac{\sum (\text{load} \times \text{distance from A})}{L}

Centroid and second moment of area

Centroid of a composite area
y‾=∑Aiyi∑Ai\bar{y} = \frac{\sum A_i y_i}{\sum A_i}
Second moment of a rectangle (about its own centroid)
I=bh312I = \frac{bh^3}{12}
Parallel-axis theorem
I=∑(Ii+Aidi2)I = \sum \left( I_i + A_i d_i^2 \right)

Dry friction and the friction angle

Limiting static friction
Fmax=μsNF_{\max} = \mu_s N
Kinetic friction
F=μkNF = \mu_k N
Angle of repose
tanφs=μs\tan\phi_s = \mu_s
Acceleration down a rough slope
a=g(sinθ−μkcosθ)a = g\left(\sin\theta - \mu_k\cos\theta\right)

Strength of Materials

Normal and shear stress, strain

Normal stress
σ=PA\sigma = \frac{P}{A}
Shear stress (average)
τ=VA\tau = \frac{V}{A}
Normal strain
ε=δL\varepsilon = \frac{\delta}{L}
Hooke's law
σ=Eε\sigma = E\varepsilon
Factor of safety
FoS=σyσactual\text{FoS} = \frac{\sigma_{\text{y}}}{\sigma_{\text{actual}}}

The stress-strain diagram

Elastic modulus (slope of the straight part)
E=σεE = \frac{\sigma}{\varepsilon}
Percent elongation
%elongation=Lf−L0L0×100\%\,\text{elongation} = \frac{L_f - L_0}{L_0}\times 100
Permanent strain after unloading
εp=ε−σE\varepsilon_p = \varepsilon - \frac{\sigma}{E}

Axial deformation of bars

Axial deformation
δ=PLAE\delta = \frac{PL}{AE}
Free thermal expansion
δT=αLΔT\delta_T = \alpha L \,\Delta T
Thermal stress (fully restrained bar)
σ=−EαΔT\sigma = -E\alpha\,\Delta T

Torsion of circular shafts

Polar moment, solid shaft
J=πd432J = \frac{\pi d^4}{32}
Polar moment, hollow shaft
J=π32(do4−di4)J = \frac{\pi}{32}\left(d_o^4 - d_i^4\right)
Torsion formula
τ=TρJ,τmax=TrJ\tau = \frac{T\rho}{J}, \quad \tau_{\text{max}} = \frac{Tr}{J}
Angle of twist
φ=TLGJ\phi = \frac{TL}{GJ}

Shear force and bending moment diagrams

Shear force at a section
V=∑Fup, left−∑Fdown, leftV = \sum F_{\text{up, left}} - \sum F_{\text{down, left}}
Load–shear relation
dVdx=−w\frac{\mathrm{d}V}{\mathrm{d}x} = -w
Shear–moment relation
dMdx=V\frac{\mathrm{d}M}{\mathrm{d}x} = V
Simply supported beam, central point load
Mmax=PL4M_{\text{max}} = \frac{PL}{4}
Simply supported beam, full-span UDL
Mmax=wL28M_{\text{max}} = \frac{wL^2}{8}
Cantilever, end load or full UDL
Mmax=PLorwL22M_{\text{max}} = PL \;\text{or}\; \frac{wL^2}{2}

Bending stress: the flexure formula

Flexure formula
σ=MyI\sigma = \frac{My}{I}
Maximum bending stress
σmax=McI=MZ\sigma_{\text{max}} = \frac{Mc}{I} = \frac{M}{Z}
Section modulus
Z=IcZ = \frac{I}{c}
Rectangular section
Z=bh26Z = \frac{bh^2}{6}

Beam deflection

Moment–curvature
EId2ydx2=MEI\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = M
Simply supported, central point load
δmax=PL348EI\delta_{\text{max}} = \frac{PL^3}{48EI}
Simply supported, full-span UDL
δmax=5wL4384EI\delta_{\text{max}} = \frac{5wL^4}{384EI}
Cantilever, end load
δmax=PL33EI\delta_{\text{max}} = \frac{PL^3}{3EI}
Cantilever, full-span UDL
δmax=wL48EI\delta_{\text{max}} = \frac{wL^4}{8EI}

Stress transformation and Mohr's circle

Stress on a plane at angle θ
σθ=σx+σy2+σx−σy2cos2θ+τxysin2θ\sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta
Shear stress on that plane
τθ=−σx−σy2sin2θ+τxycos2θ\tau_\theta = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta
Principal stresses
σ1,2=σx+σy2±(σx−σy2)2+τxy2\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}
Principal direction
tan2θp=2τxyσx−σy\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}
Maximum shear stress
τmax=(σx−σy2)2+τxy2\tau_{\text{max}} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}

Thin-walled pressure vessels

Cylinder, hoop stress
σh=pd2t\sigma_h = \frac{pd}{2t}
Cylinder, longitudinal stress
σL=pd4t\sigma_L = \frac{pd}{4t}
Sphere (any direction)
σ=pd4t\sigma = \frac{pd}{4t}
Thin-wall condition
dt≥20\frac{d}{t} \ge 20

Buckling of columns (Euler)

Euler critical load
Pcr=π2EI(KL)2P_{\text{cr}} = \frac{\pi^2 EI}{(KL)^2}
Radius of gyration
r=IAr = \sqrt{\frac{I}{A}}
Slenderness ratio
λ=KLr\lambda = \frac{KL}{r}
Critical stress
σcr=π2Eλ2\sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2}

Structural Analysis

Trusses: method of joints

Joint equilibrium
∑Fx=0,∑Fy=0\sum F_x = 0, \quad \sum F_y = 0
Determinacy check (m members, r reactions, j joints)
m+r=2jm + r = 2j
Force components of a member
Fx=Fcosθ,Fy=FsinθF_x = F\cos\theta, \quad F_y = F\sin\theta

Trusses: method of sections

Moment equilibrium about a point on the section
∑Mpoint=0\sum M_{\text{point}} = 0
Vertical force equilibrium of the cut part
∑Fy=0\sum F_y = 0
Number of unknowns on the cut
at most 3\text{at most } 3

Fluid Mechanics

Hydrostatic pressure and force on surfaces

Pressure at depth h
p=ρghp = \rho g h
Resultant force on a vertical rectangle
F=12ρgh2bF = \frac{1}{2}\rho g h^2 b
Centre of pressure (above the base)
yb=h3y_b = \frac{h}{3}
Overturning moment about the base
M=F·h3M = F\cdot\frac{h}{3}

Continuity and Bernoulli's equation

Continuity (incompressible)
A1v1=A2v2=QA_1 v_1 = A_2 v_2 = Q
Bernoulli's equation
p1ρg+v122g+z1=p2ρg+v222g+z2\frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2
Venturi pressure drop
p1−p2=12ρ(v22−v12)p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right)

Thermodynamics

The first law and ideal gas processes

First law (closed system)
Q=ΔU+WQ = \Delta U + W
Ideal gas law
PV=nRTPV = nRT
Isothermal work
W=P1V1lnV2V1W = P_1 V_1 \ln\frac{V_2}{V_1}
Adiabatic relation
PVγ=constant,γ=cpcvP V^{\gamma} = \text{constant}, \quad \gamma = \frac{c_p}{c_v}
Adiabatic work
W=P1V1−P2V2γ−1W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}

The Otto cycle

Compression ratio
r=v1v2r = \frac{v_1}{v_2}
Temperatures after adiabatic processes
T2=T1rγ−1,T4=T3r1−γT_2 = T_1 r^{\gamma - 1}, \quad T_4 = T_3 r^{1 - \gamma}
Heat added at constant volume
qin=cv(T3−T2)q_{\text{in}} = c_v (T_3 - T_2)
Thermal efficiency
η=1−1rγ−1\eta = 1 - \frac{1}{r^{\gamma - 1}}
Net work
wnet=ηqinw_{\text{net}} = \eta\, q_{\text{in}}

Electrical Circuits

Ohm's law and resistor networks

Ohm's law
V=IRV = IR
Power
P=VI=I2R=V2RP = VI = I^2 R = \frac{V^2}{R}
Series resistance
Req=R1+R2+R3R_{\text{eq}} = R_1 + R_2 + R_3
Parallel resistance
1Req=1R1+1R2+1R3\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

RC circuit transient response

Time constant
τ=RC\tau = RC
Charging voltage
vC(t)=V(1−e−t/τ)v_C(t) = V\left(1 - e^{-t/\tau}\right)
Discharging voltage
vC(t)=Ve−t/τv_C(t) = V\,e^{-t/\tau}
Current while charging
i(t)=VRe−t/τi(t) = \frac{V}{R}\,e^{-t/\tau}
Energy stored
E=12CvC2E = \tfrac{1}{2} C v_C^2

Series RLC resonance

Inductive reactance
XL=2πfLX_L = 2\pi f L
Capacitive reactance
XC=12πfCX_C = \frac{1}{2\pi f C}
Impedance
|Z|=R2+(XL−XC)2|Z| = \sqrt{R^2 + (X_L - X_C)^2}
Resonant frequency
f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}
Quality factor and bandwidth
Q=1RLC,BW=f0QQ = \frac{1}{R}\sqrt{\frac{L}{C}}, \quad \text{BW} = \frac{f_0}{Q}
Phase angle
φ=tan−1XL−XCR\varphi = \tan^{-1}\frac{X_L - X_C}{R}