Advanced Civil Engineering Handbook › Chapter 2

Chapter 2 of 14 · 41 formulas · 76 one-liners

Strength of Materials — formulas & one-liners

Stress–strain, elastic constants, thermal stress, SFD/BMD, bending and shear stress, torsion, springs, columns, cylinders, principal stresses, theories of failure, strain energy and beam deflections — the standard mechanics of solids formulas (code-independent).

2.1 Simple stress, strain and elastic constants

Elongation of a prismatic bar

δ = \frac{P L}{A E}

P = axial load; L = length; A = area; E = Young's modulus

Elongation of a tapered circular bar

δ = \frac{4 P L}{π E d _{1} d _{2}}

d₁, d₂ = end diameters; L = length

Elongation of a hanging bar under self-weight

δ = \frac{γ L ^{2}}{2 E} = \frac{W L}{2 A E}

γ = unit weight of material; W = total self-weight

Young's modulus and modulus of rigidity

E = 2 G (1 + μ)

G = modulus of rigidity; μ = Poisson's ratio

Young's modulus and bulk modulus

E = 3 K (1 - 2 μ)

K = bulk modulus

Combining both: E = 9KG/(3K + G)

Volumetric strain under triaxial stress

ε_{v} = \frac{( σ _{x} + σ _{y} + σ _{z} ) ( 1 - 2 μ )}{E}

σ_x, σ_y, σ_z = normal stresses (tension +)

Uniaxial bar: ε_v = ε(1 − 2μ); hydrostatic p: ε_v = 3p(1 − 2μ)/E

Hooke's law (stress proportional to strain) holds only up to the limit of proportionality.
A mild steel stress–strain curve shows proportional limit, elastic limit, upper and lower yield points, ultimate stress and breaking point.
Necking begins at the ultimate stress, after which engineering stress falls although true stress keeps rising.
A homogeneous isotropic material has 4 elastic constants (E, G, K, μ) of which only 2 are independent.
Poisson's ratio of an isotropic material can theoretically be at most 0.5; it is about 0.25–0.33 for metals, about 0.1–0.2 for concrete and nearly 0.5 for rubber.
When μ = 0.5 the material is incompressible and its bulk modulus becomes infinite.
E is about 200 GPa for steel, about 70 GPa for aluminium, and G for steel is about 80 GPa.
Elongation of a bar due to its own weight is half the elongation caused by an equal load applied at its end.
Factor of safety is the ratio of ultimate (or yield) stress to working stress.
Ductile materials fail in tension with a cup-and-cone fracture, while brittle materials such as cast iron break on a flat plane with little elongation.
Fatigue is failure under repeated stress below the static strength; the endurance limit is the stress below which failure does not occur for infinite cycles.
Creep is the slow increase in strain with time under constant stress.

2.2 Thermal stress and composite bars

Thermal stress in a fully restrained bar

σ = E α Δ T

α = coefficient of thermal expansion; ΔT = temperature change

Thermal stress when the support yields by δ

σ = \frac{E ( α Δ T L - δ )}{L}

δ = gap or yield allowed; L = length

Load sharing in a composite bar (equal strain)

P = σ_{1} A_{1} + σ_{2} A_{2}, \frac{σ _{1}}{E _{1}} = \frac{σ _{2}}{E _{2}}

Suffixes 1, 2 = the two materials

Compatibility for a heated composite bar

\frac{σ _{1}}{E _{1}} + \frac{σ _{2}}{E _{2}} = (α_{1} - α_{2}) Δ T

α₁ > α₂; equilibrium σ₁A₁ = σ₂A₂ (1 in compression, 2 in tension)

Thermal stress in a fully restrained bar is independent of its length and cross-sectional area.
A restrained bar develops compressive stress when heated and tensile stress when cooled.
In a heated composite bar, the material with the higher coefficient of expansion goes into compression and the other into tension.
Coefficient of thermal expansion is about 12 × 10⁻⁶ /°C for steel, 17 × 10⁻⁶ /°C for copper and 23 × 10⁻⁶ /°C for aluminium.
In a composite bar under axial load, stress in each material is proportional to its modulus of elasticity.

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2.3 Shear force, bending moment and bending stress

Load–shear–moment relations

\frac{d V}{d x} = - w, \frac{d M}{d x} = V

w = load intensity (downward +); V = shear force; M = bending moment

Flexure (bending) equation

\frac{M}{I} = \frac{σ}{y} = \frac{E}{R}

I = moment of inertia about NA; y = distance from NA; R = radius of curvature

Section modulus of a rectangle

Z = \frac{I}{y _{ma x}} = \frac{b d ^{2}}{6}

b = width; d = depth

Solid circle πd³/32; hollow circle π(D⁴ − d⁴)/(32D)

Parallel axis theorem

I = I_{G} + A h^{2}

I_G = MI about centroidal axis; h = distance between axes

Rectangle I_G = bd³/12; triangle I_G = bh³/36; circle I_G = πd⁴/64

Bending moment is maximum where shear force is zero or changes sign.
The change in bending moment between two sections equals the area of the SFD between them.
Under point loads the SFD is stepped and the BMD linear; under UDL the SFD is linear and the BMD parabolic; under uniformly varying load the SFD is parabolic and the BMD cubic.
A concentrated couple causes a sudden jump in the BMD but no change in the SFD.
A point of contraflexure is where the bending moment changes sign.
Simply supported beam: maximum BM is WL/4 for a central point load W and wL²/8 for a UDL w.
Cantilever: maximum BM is WL for an end load and wL²/2 for a UDL, both at the fixed end.
A simply supported beam carrying load varying from zero to w has reactions wL/6 and wL/3, and maximum BM wL²/(9√3) at L/√3 from the zero end.
A propped cantilever with UDL has prop reaction 3wL/8 and fixing moment wL²/8.
A fixed beam with UDL has end moments wL²/12 and mid-span moment wL²/24; with a central point load both are WL/8.
In simple bending, the neutral axis passes through the centroid of the section and bending stress varies linearly from zero there to maximum at the extreme fibre.
The strongest rectangular beam that can be cut from a circular log has depth = √2 × width; the stiffest has depth = √3 × width.
A beam of uniform strength has the same maximum bending stress at every section, obtained by varying its width or depth.

2.4 Shear stress in beams

Shear stress at a level in a beam section

τ = \frac{V A y ˉ}{I b}

V = shear force; A = area above the level; ȳ = its centroid distance from NA; b = width at the level

Maximum shear stress in a rectangular section

τ_{ma x} = \frac{3 V}{2 b d} = 1.5 τ_{a v g}

τ_avg = V/(bd)

Shear stress distribution across a rectangular section is parabolic, with maximum at the neutral axis and zero at the top and bottom fibres.
For a solid circular section the maximum shear stress is 4/3 times the average shear stress.
For a triangular section the maximum shear stress is 1.5 times the average and occurs at mid-height, while at the neutral axis it is 4/3 times the average.
For a square section with one diagonal vertical, the maximum shear stress is 9/8 times the average and occurs at d/8 from the neutral axis.
In an I-section the web carries most of the shear force, and shear stress jumps abruptly at the flange–web junction.

2.5 Torsion and springs

Torsion equation

\frac{T}{J} = \frac{τ}{r} = \frac{G θ}{L}

T = torque; J = polar MI; τ = shear stress at radius r; θ = angle of twist (rad)

Solid shaft J = πd⁴/32; hollow J = π(D⁴ − d⁴)/32

Maximum shear stress in a solid shaft

τ_{ma x} = \frac{16 T}{π d ^{3}}

d = shaft diameter

Polar modulus Z_p = πd³/16

Power transmitted by a shaft

P = \frac{2 π N T}{60}

P in W; N = speed (rpm); T = torque (N·m)

Equivalent bending moment and torque (combined M and T)

M_{e} = \frac{1}{2} (M + M^{2} + T^{2}), T_{e} = M^{2} + T^{2}

M = bending moment; T = torque

Deflection of a closely coiled helical spring

δ = \frac{64 W R ^{3} n}{G d ^{4}}

W = axial load; R = mean coil radius; n = number of turns; d = wire diameter

Stiffness k = Gd⁴/(64R³n); wire shear stress τ = 16WR/(πd³)

Wahl's stress correction factor

K_{w} = \frac{4 C - 1}{4 C - 4} + \frac{0.615}{C}

C = D/d = spring index

Bending stress in a semi-elliptic (carriage) leaf spring

σ = \frac{3 W L}{2 n b t ^{2}}

W = central load; L = span; n = number of leaves; b, t = leaf width and thickness

Central deflection δ = 3WL³/(8Enbt³)

In a circular shaft under torsion, shear stress is zero at the centre and maximum at the outer surface.
GJ is the torsional rigidity, and GJ/L is the torsional stiffness of a shaft.
A hollow shaft is stronger and stiffer than a solid shaft of the same material, length and weight.
Shafts in series carry the same torque and their angles of twist add; shafts in parallel have the same twist and their torques add.
Under pure torsion a ductile shaft fails on a plane perpendicular to its axis, while a brittle shaft fails along a 45° helix.
The wire of a closely coiled helical spring is mainly under torsion; in an open coiled spring it is under both bending and torsion.
For springs in series 1/k = 1/k₁ + 1/k₂, and for springs in parallel k = k₁ + k₂.

2.6 Columns and struts

Euler's crippling load

P_{e} = \frac{π ^{2} E I}{L _{e}^{2}}

I = least moment of inertia; L_e = effective length

L_e: both ends hinged L; both fixed L/2; one fixed one free 2L; one fixed one hinged L/√2

Rankine's formula in working form

P_{R} = \frac{σ _{c} A}{1 + a ( \frac{L _{e}}{k} ) ^{2}}, a = \frac{σ _{c}}{π ^{2} E}

k = least radius of gyration = √(I/A); a = Rankine's constant

From 1/P_R = 1/P_c + 1/P_e. Typical a: mild steel 1/7500, wrought iron 1/9000, cast iron 1/1600, timber 1/750

Limiting slenderness ratio for Euler's formula

(\frac{L _{e}}{k})_{l im} = π \frac{E}{σ _{c}}

σ_c = crushing (or yield) stress

≈ 79 for mild steel with E = 200 GPa and σ_c = 320 MPa

Stress under eccentric load on a rectangular section

σ = \frac{P}{A} (1 \pm \frac{6 e}{b})

e = eccentricity along dimension b

No tension anywhere if e ≤ b/6 (middle third rule)

A column buckles about the axis having the least moment of inertia, i.e., the least radius of gyration.
Slenderness ratio is the effective length divided by the least radius of gyration.
Euler's theory assumes an initially straight, homogeneous, linearly elastic column under truly axial load that fails by buckling alone.
Euler's formula is valid only for long columns; for short columns it gives unrealistically high loads, since P_e tends to infinity as length tends to zero.
Rankine's formula is empirical and covers short and long columns because it combines crushing and buckling.
The core (kern) of a rectangular section is a rhombus with diagonals b/3 and d/3.
The core of a solid circular section is a concentric circle of diameter d/4.

2.7 Thin and thick cylinders

Hoop and longitudinal stress in a thin cylinder

σ_{h} = \frac{p d}{2 t}, σ_{l} = \frac{p d}{4 t}

p = internal pressure; d = internal diameter; t = wall thickness

With joint efficiency: σ_h = pd/(2tη_long), σ_l = pd/(4tη_circ)

Volumetric strain of a thin cylinder

\frac{δ V}{V} = \frac{p d}{4 tE} (5 - 4 μ)

μ = Poisson's ratio

Thin sphere: δV/V = (3pd/4tE)(1 − μ)

Lamé's equations for a thick cylinder

σ_{r} = A - \frac{B}{r ^{2}}, σ_{θ} = A + \frac{B}{r ^{2}}

σ_r = radial, σ_θ = hoop stress at radius r (tension +); A, B from boundary conditions

Internal pressure p: σ_r = −p at inner radius, 0 at outer radius

Maximum hoop stress in a thick cylinder (internal pressure only)

σ_{θ, ma x} = p \frac{r _{o}^{2} + r _{i}^{2}}{r _{o}^{2} - r _{i}^{2}}

r_i, r_o = inner and outer radii

Occurs at the inner surface

A cylinder is treated as thin when its wall thickness is less than about one-twentieth of its diameter.
In a thin cylinder the hoop stress is twice the longitudinal stress, so a longitudinal joint is more critical than a circumferential one.
A thin sphere under internal pressure has stress pd/(4t) in every direction, half the hoop stress of a cylinder of the same diameter and thickness.
In a thick cylinder the hoop and radial stresses vary across the wall, and both are numerically greatest at the inner surface.
Compound cylinders made by shrink-fitting reduce the peak hoop stress by putting the inner tube in initial compression.

2.8 Principal stresses, Mohr's circle and theories of failure

Principal stresses in plane stress

σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

σ_x, σ_y = normal stresses; τ_xy = shear stress

Mohr's circle: centre ((σ_x + σ_y)/2, 0); radius = the square-root term = τ_max in plane

Orientation of principal planes

tan 2 θ_{p} = \frac{2 τ _{x y}}{σ _{x} - σ _{y}}

θ_p = angle of principal plane from the x-plane

Von Mises (distortion energy) criterion, plane stress

σ_{1}^{2} + σ_{2}^{2} - σ_{1} σ_{2} = σ_{y}^{2}

σ₁, σ₂ = principal stresses; σ_y = yield stress in simple tension

Principal planes carry no shear stress, and they are mutually perpendicular.
Planes of maximum shear stress lie at 45° to the principal planes and carry normal stress equal to the average (σ₁ + σ₂)/2.
In pure shear τ, the principal stresses are +τ and −τ acting on planes at 45°.
Mohr's circle for uniaxial tension passes through the origin, and for equal biaxial (hydrostatic) stress it shrinks to a point.
The maximum principal stress theory (Rankine) suits brittle materials.
The maximum shear stress theory (Tresca/Guest) and the distortion energy theory (von Mises/Hencky) suit ductile materials.
Under Tresca theory the material yields when the maximum shear stress reaches σ_y/2, i.e., σ₁ − σ₃ = σ_y.
Yield stress in pure shear is 0.5σ_y by Tresca theory and 0.577σ_y (σ_y/√3) by von Mises theory.
In the σ₁–σ₂ plane, the Tresca yield locus is a hexagon inscribed in the von Mises ellipse.
The maximum principal strain theory is credited to St. Venant.

2.9 Strain energy and deflection of beams

Strain energy in axial loading

U = \frac{σ ^{2}}{2 E} V = \frac{P ^{2} L}{2 A E}

V = volume of bar

σ²/(2E) is the strain energy per unit volume (modulus of resilience at elastic limit)

Stress due to a load falling through height h

σ = \frac{P}{A} [1 + 1 + \frac{2 A E h}{P L}]

P = falling load; h = height of fall; A, L = bar area and length

For h much larger than the extension: σ ≈ √(2EPh/(AL))

Castigliano's theorem (deflection)

δ_{i} = \frac{\partial U}{\partial P _{i}}

U = total strain energy; δ_i = deflection at and along load P_i

Bending strain energy U = ∫M² dx/(2EI)

Simply supported beam, central point load: max deflection

δ_{ma x} = \frac{W L ^{3}}{48 E I}

W = load; L = span; EI = flexural rigidity

End slope = WL²/(16EI)

Simply supported beam, UDL: max deflection

δ_{ma x} = \frac{5 w L ^{4}}{384 E I}

w = load per unit length

End slope = wL³/(24EI)

Cantilever, point load at free end: tip deflection

δ = \frac{W L ^{3}}{3 E I}

W = end load

Tip slope = WL²/(2EI)

Cantilever, UDL over full length: tip deflection

δ = \frac{w L ^{4}}{8 E I}

w = load per unit length

Tip slope = wL³/(6EI)

A suddenly applied load produces twice the stress and twice the extension of the same load applied gradually.
Proof resilience is the maximum strain energy stored up to the elastic limit; toughness is the energy absorbed up to fracture (area under the whole stress–strain curve).
Strain energy stored in shear is τ²/(2G) per unit volume.
Strain energy stored in a solid circular shaft under torsion is τ_max²/(4G) per unit volume of the shaft.
Maxwell's reciprocal theorem: deflection at A due to a unit load at B equals deflection at B due to a unit load at A.
The differential equation of the elastic curve is EI d²y/dx² = M.
Mohr's first moment-area theorem: the change in slope between two points equals the area of the M/EI diagram between them.
Mohr's second moment-area theorem: the tangential deviation equals the first moment of the M/EI diagram area about the point considered.
In the conjugate beam method a fixed end becomes free, a free end becomes fixed, and simple end supports remain unchanged.
Maximum deflection of a fixed beam is WL³/(192EI) under a central point load and wL⁴/(384EI) under UDL.
A fixed beam deflects one-fourth as much as a simply supported beam under a central point load and one-fifth as much under a UDL.
For the same total load, a simply supported beam deflects 5/8 as much under a UDL as under a central point load.

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