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Intermediate civil engineering: 50 formulas and methods

Undergraduate civil engineering in 50 cards: beam formulas, buckling, concrete and steel design, soils, open channels, hydrology, traffic and water treatment.

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Static determinacy of a beam

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  1. 1
    Static determinacy of a beam
    Determinate if r=3+cr = 3 + c; indeterminate if r>3+cr > 3 + c
    rr is the number of reaction components and cc the number of extra condition equations, such as one per internal hinge. With r<3+cr < 3 + c the beam is unstable, and it is also unstable if all its reactions are parallel or concurrent.
  2. 2
    Static determinacy of a plane truss
    Determinate if b+r=2jb + r = 2j; indeterminate if b+r>2jb + r > 2j; unstable if b+r<2jb + r < 2j
    bb members, rr reaction components, jj joints. The degree of indeterminacy is b+r−2jb + r - 2j.
  3. 3
    Method of joints
    Isolate each joint and solve ∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0, starting at a joint with no more than two unknown member forces
  4. 4
    Method of sections
    Cut the truss through the members of interest, usually no more than three unknowns, and apply the three equilibrium equations to one side
    Taking moments about the point where two of the cut members meet gives the third force directly.
  5. 5
    Zero-force members in a truss
    At an unloaded, unsupported joint with two non-collinear members, both carry zero force; at one with three members, two of them collinear, the third carries zero force
  6. 6
    Relationship between shear force and bending moment
    dMdx=V\dfrac{dM}{dx} = V
    So the change in moment between two points equals the area under the shear diagram, and the moment peaks where the shear passes through zero.
  7. 7
    Maximum moment in a simply supported beam under a uniform load
    Mmax⁡=wL28M_{\max} = \dfrac{wL^2}{8}, at midspan
  8. 8
    Maximum moment in a simply supported beam with a central point load
    Mmax⁡=PL4M_{\max} = \dfrac{PL}{4}, at midspan
  9. 9
    Midspan deflection of a simply supported beam under a uniform load
    δ=5wL4384EI\delta = \dfrac{5wL^4}{384EI}
    EIEI is the flexural rigidity: Young’s modulus times the second moment of area.
  10. 10
    Midspan deflection of a simply supported beam with a central point load
    δ=PL348EI\delta = \dfrac{PL^3}{48EI}
  11. 11
    Elastic curve equation
    EId2vdx2=M(x)EI\dfrac{d^2v}{dx^2} = M(x)
    Integrate twice and apply the support conditions to get slopes and deflections. Valid for small deflections of linear elastic beams.
  12. 12
    Second moment of area of a rectangle about its centroid
    I=bh312I = \dfrac{bh^3}{12}
    bb is the width and hh the depth in the direction of bending, so doubling the depth makes a beam eight times stiffer.
  13. 13
    Parallel axis theorem
    I=Ic+Ad2I = I_c + Ad^2
    IcI_c is the second moment about the centroidal axis, AA the area and dd the distance between the two parallel axes.
  14. 14
    Flexure formula
    σ=MyI\sigma = \dfrac{My}{I}
    yy is the distance from the neutral axis, where bending stress is zero. In a homogeneous, linear elastic section the neutral axis passes through the centroid.
  15. 15
    Elastic section modulus
    S=IcS = \dfrac{I}{c}, so σmax⁡=MS\sigma_{\max} = \dfrac{M}{S}
    cc is the distance from the neutral axis to the extreme fiber. For a rectangle, S=bh26S = \dfrac{bh^2}{6}.
  16. 16
    Shear stress in a beam
    τ=VQIb\tau = \dfrac{VQ}{Ib}
    QQ is the first moment, about the neutral axis, of the area beyond the point, and bb the width there. In a rectangle the peak is 1.5 V/A1.5\,V/A, at the neutral axis.
  17. 17
    Euler buckling load
    Pcr=π2EI(KL)2P_{cr} = \dfrac{\pi^2 EI}{(KL)^2}
    KLKL is the effective length. With equal effective lengths, use the smaller II: the column buckles about its weaker axis.
  18. 18
    Effective length factor KK
    Pinned–pinned 1.0, fixed–free 2.0, fixed–pinned 0.7, fixed–fixed 0.5
    These are theoretical values. Design guides recommend higher ones for the fixed cases, since real connections aren’t perfectly rigid.
  19. 19
    Slenderness ratio of a column
    KLr\dfrac{KL}{r}, where r=I/Ar = \sqrt{I/A} is the radius of gyration
    Euler’s critical stress is σcr=π2E(KL/r)2\sigma_{cr} = \dfrac{\pi^2 E}{(KL/r)^2}. It applies only while that stays below the proportional limit; stocky columns yield or buckle inelastically first.
  20. 20
    Balanced section in reinforced concrete
    The tension steel starts to yield just as the extreme compression fiber of the concrete reaches its crushing strain
    The crushing strain is taken as 0.003 in ACI 318 and 0.0035 in Eurocode 2 for normal-strength concrete.
  21. 21
    Under-reinforced section
    Less steel than balanced, so the steel yields before the concrete crushes: a ductile failure, with wide cracks and large deflections as warning
    An over-reinforced section fails by sudden crushing of the concrete, so codes limit the amount of tension steel.
  22. 22
    Equivalent rectangular stress block (ACI)
    A uniform stress of 0.85fc′0.85f'_c over a depth a=β1ca = \beta_1 c from the compression face
    cc is the neutral axis depth. β1=0.85\beta_1 = 0.85 for concrete up to 28 MPa28\ \mathrm{MPa} (4000 psi4000\ \mathrm{psi}), and decreases for stronger concrete.
  23. 23
    Nominal moment strength of a singly reinforced rectangular beam
    Mn=Asfy(d−a2)M_n = A_s f_y\left(d - \dfrac{a}{2}\right), with a=Asfy0.85fc′ba = \dfrac{A_s f_y}{0.85 f'_c b}
    dd is the effective depth, from the compression face to the centroid of the tension steel. It assumes the steel has yielded.
  24. 24
    Limit states of a steel tension member
    Yielding of the gross section and fracture of the net section through the bolt holes
    Block shear rupture at the connection is checked too.
  25. 25
    Lateral-torsional buckling
    A beam bent about its strong axis deflects sideways and twists, because its compression flange isn’t braced enough
    Moment capacity falls as the unbraced length LbL_b grows; bracing the compression flange prevents it.
  26. 26
    Plastic moment of a steel section
    Mp=FyZM_p = F_y Z
    ZZ is the plastic section modulus: bh2/4bh^2/4 for a rectangle, 1.5 times its elastic SS. For wide-flange beams Z/SZ/S is typically about 1.1 to 1.2.
  27. 27
    Compact steel section
    A section whose flanges and web are stocky enough to reach the plastic moment before local buckling
    Compactness is judged from the width-to-thickness ratios of the flanges and web.
  28. 28
    Effective stress
    σ′=σ−u\sigma' = \sigma - u
    Total stress minus pore water pressure (Terzaghi); it controls soil strength and compressibility. Below the water table, u=γwzwu = \gamma_w z_w, where zwz_w is the depth below it and γw≈9.81 kN/m3\gamma_w \approx 9.81\ \mathrm{kN/m^3}.
  29. 29
    Atterberg limits
    The water contents at which a fine-grained soil changes state: shrinkage limit (solid to semisolid), plastic limit (semisolid to plastic) and liquid limit (plastic to liquid)
  30. 30
    Plasticity index
    PI=LL−PLPI = LL - PL
    The range of water content over which the soil is plastic. With LLLL, it places a fine soil on the plasticity chart.
  31. 31
    Darcy’s law
    Q=kiAQ = kiA
    kk is the hydraulic conductivity and i=Δh/Li = \Delta h / L the hydraulic gradient. Water in the pores moves faster, at the seepage velocity ki/nki/n, where nn is the porosity.
  32. 32
    Critical hydraulic gradient
    ic=γ′γw≈1i_c = \dfrac{\gamma'}{\gamma_w} \approx 1
    γ′\gamma' is the submerged unit weight. At this upward gradient the effective stress in a cohesionless soil falls to zero and it boils (the quick condition).
  33. 33
    Proctor compaction test
    Compacting soil at several water contents with a set energy to find the maximum dry density and the optimum water content
    Compaction mainly expels air. The modified test uses more energy, giving a higher maximum dry density at a lower optimum water content.
  34. 34
    Terzaghi’s bearing capacity equation for a strip footing
    qu=cNc+qNq+12γBNγq_u = cN_c + qN_q + \tfrac{1}{2}\gamma BN_\gamma
    cc is the cohesion, qq the overburden pressure at footing level and BB the width; the NN factors depend on the friction angle. The allowable pressure is quq_u divided by a factor of safety, typically 3.
  35. 35
    Manning’s equation
    V=1nR2/3S1/2V = \dfrac{1}{n} R^{2/3} S^{1/2}
    SI units; in US units the 11 becomes 1.491.49. nn is the roughness coefficient, RR the hydraulic radius and SS the energy slope, equal to the bed slope in uniform flow.
  36. 36
    Hydraulic radius
    R=APR = \dfrac{A}{P}
    Flow area over wetted perimeter. In a wide rectangular channel, R≈yR \approx y, the depth.
  37. 37
    Froude number in open-channel flow
    Fr=VgDFr = \dfrac{V}{\sqrt{gD}}
    D=A/TD = A/T is the hydraulic depth, with TT the top width (D=yD = y in a rectangular channel). Fr<1Fr < 1 is subcritical, Fr=1Fr = 1 critical and Fr>1Fr > 1 supercritical.
  38. 38
    Critical depth in a rectangular channel
    yc=(q2g)1/3y_c = \left(\dfrac{q^2}{g}\right)^{1/3}
    qq is the discharge per unit width. Specific energy E=y+V22gE = y + \dfrac{V^2}{2g} is at its minimum here: Emin⁡=1.5 ycE_{\min} = 1.5\,y_c.
  39. 39
    Hydraulic jump
    An abrupt rise in the water surface where supercritical flow changes to subcritical, dissipating energy in turbulence
    Stilling basins below spillways are designed to hold a jump.
  40. 40
    Sequent depths of a hydraulic jump in a rectangular channel
    y2y1=12(1+8Fr12−1)\dfrac{y_2}{y_1} = \dfrac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right)
    y1y_1 and Fr1Fr_1 are the depth and Froude number just upstream of the jump.
  41. 41
    Rational method for peak runoff
    Q=CiAQ = CiA
    CC is the runoff coefficient, ii the rainfall intensity for a duration equal to the time of concentration and AA the area. In SI, Q=0.278 CiAQ = 0.278\,CiA gives m3/s\mathrm{m^3/s} from mm/h\mathrm{mm/h} and km2\mathrm{km^2}. Meant for small catchments.
  42. 42
    Runoff coefficient
    The fraction of rainfall that becomes direct runoff
    Typically about 0.7 to 0.95 for pavement and roofs, and much lower for lawns and woodland.
  43. 43
    Time of concentration
    The time runoff takes to travel from the hydraulically most distant point of a catchment to the outlet
    Once rain has lasted this long, the whole catchment contributes to the flow at the outlet.
  44. 44
    Stopping sight distance
    d=vt+v22ad = vt + \dfrac{v^2}{2a}
    Reaction distance plus braking distance. AASHTO typically assumes t=2.5 st = 2.5\ \mathrm{s} and a deceleration a=3.4 m/s2a = 3.4\ \mathrm{m/s^2}.
  45. 45
    Superelevation
    Tilting a curve’s cross section toward the inside, so part of the centripetal force comes from the bank rather than tire friction
    For design, e+f=v2gRe + f = \dfrac{v^2}{gR}, where ee is the superelevation rate, ff the side friction factor and RR the curve radius.
  46. 46
    Fundamental equation of traffic flow
    q=kvq = kv
    Flow qq (vehicles per hour) equals density kk (vehicles per kilometer) times space-mean speed vv.
  47. 47
    Maximum flow in the Greenshields model
    qmax⁡=vfkj4q_{\max} = \dfrac{v_f k_j}{4}
    Speed falls linearly with density, v=vf(1−k/kj)v = v_f\left(1 - k/k_j\right), so flow peaks at half the jam density kjk_j and half the free-flow speed vfv_f.
  48. 48
    Steps of conventional drinking water treatment
    Coagulation, flocculation, sedimentation, filtration, disinfection
    Coagulants such as alum neutralize the charge on fine particles; gentle mixing during flocculation then grows them into flocs heavy enough to settle.
  49. 49
    Surface overflow rate of a settling tank
    vo=QAsv_o = \dfrac{Q}{A_s}
    AsA_s is the tank’s surface area. In an ideal tank, every particle that settles faster than vov_o is removed, whatever the depth.
  50. 50
    Activated sludge process
    Aerating wastewater with a suspended culture of microorganisms that consume organic matter, then settling the biomass in a clarifier and returning part of it to the aeration tank
    A common form of secondary treatment, after screening, grit removal and primary settling.

Frequently asked questions

Should I start with this deck?

If you know the basics of loads, supports, stress and soil types, yes. Otherwise start with Civil engineering basics.

Do the cards follow a particular design code?

They teach principles shared by most codes. Where a card quotes a code value, such as the ACI stress block, it names the code; for real design, check the code and edition that apply to you.

Which units are used?

SI units, with the US customary version noted where an equation’s constant changes, as in Manning’s equation.

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