• The solution to Newton's GEXACT VALUE & FORMULA
  • The theory controlling planetary spinTHE MATHEMATICAL LAW
  • The pressure at the centre of a massEARTH'S CORE PRESSURE (calculation procedure)
  • Proof of the non-exitence of Dark MatterDOES NOT EXIST
  • The atom as Newton and Coulomb describe itNO NEED FOR A UNIFICATION THEORY
The solution to Newton's G1 The theory controlling planetary spin2 Pressure at the centre of the Earth3 Proof of the non-exitence of Dark Matter4 The atom as Newton describes it5
Useful Stuff Algebra Trig Functions Calculus Max-Min Differentiation Rules Differentiation Trig Differentiation Logs Integration Methods Standard Integrals Stiffness & Capacity Mohr's Circle Earth's Atmosphere Earth's Properties Stars & Planets Laws of Motion Solar System Orbits Planetary Spin Rydberg Atom Planck Atom Newton Atom The Atom Dark Matter? Core Pressure The Big Bang Brakes and Tyres Vehicle Impacts Speeding vs Safety Surface Finish Pressure Classes Hardness Conversion Thermodynamics Steam (properties) Heat Capacity Work Energy Power Constants

Stiffness and Capacity Formulas
(bending, torsion & axial tension)

The following table contains alternative ways to calculate the stiffness or capacity of a structural member such as a beam or bar
Refer to CalQlata's Engineering Basics technical help page on this website

Bending:

Torsion:

Axial:

E ÷ R = M ÷ I = σ ÷ y

G.θ ÷ L = T ÷ J = τ ÷ r

E = F ÷ A = σ ÷ e

The above formulas can be rearranged as follows:

E.I = M.R = σ.I.R ÷ y

G.J = T.L ÷ θ = τ.J.L ÷ θ.r

E.A = F = σ.A ÷ e

By setting the beam or bar length and/or angle of twist to '1',
the stiffness formulas become:

E.I = M.R = σ.I.R ÷ y

G.J = T = τ.J ÷ r

E.A = F = σ.A

{N.m²} or {lbf.in²}

{N.m/m/°} or {lbf.in/in/°}

{N[.m/m]} or {lbf[.in/in]}

The first term in each of the above formulas can be used to calculate the stiffness of a beam using known theoretical material and sectional properties. If the properties of a composite construction (e.g. HPHT Flexible Pipe) are calculated using this term, each material in each layer must be defined individually and summed to provide a total value.

The second term in each of the above formulas can be used to calculate the stiffness of a beam from empirical values established in physical tests per unitary dimension (e.g. Torsion: 1m of beam length twisted to 1° of angular rotation or Axial: 1m length of beam stretched by 1m)

The third term in each of the above formulas can be used to define the capacity (limiting deformation) of any beam (or bar) after either one of the first two terms has been established

The symbols used in the above formulas are described as follows:
E {Young's modulus} for material
G {shear modulus} = E ÷ 2.(1+ν) where ν = 0.3 for steel, 0.4 for plastic, 0.5 for rubber, etc.
F = axial force
I = second moment of area of beam or bar
   e.g. for a pipe or tube: I = π.(Øₒ⁴ – Øᵢ⁴) ÷ 64;  J = π.(Øₒ⁴ – Øᵢ⁴) ÷ 32;  A = π.(Øₒ² – Øᵢ²) ÷ 4
σ = SMYS for material
τ {maximum shear stress} = ≤0.5 x SMYS for material
e {strain} = δL/L (L = length of beam or bar)
R = bend radius @ neutral axis of beam
r & y = distance from neutral axis to extreme fibre
θ = twist in degrees(°) or radians (rads)

If the stiffness of an item or material is measured in N.m/m (lbf.in/in) or N (lbf), it is describing the force that would be necessary to double or halve its length.
If it is measured in N/m (lbf/in) or N (lbf), it is describing the force that would be necessary to stretch (or compress) it by one unit length.
N.m² (lbf.in²) is actually bending moment per unit deformation {N.m²/m (lbf.in²/in)}, which is also measured in N.m/m (lbf.in/in) or N (lbf).

Compound Stiffness

The total stiffness of groups of elements are added up in series or in parallel (just like resistors in an electrical circuit):

The sum of a collection of elements in series (e.g. Fig 1; k₄ & k₅) are added as follows:
k = 1 / (1/k₁ + 1/k₂ + 1/k₃)

The sum of a collection of elements in parallel (e.g. Fig 1; k₆ & k₇) are added as follows:
k = k₁ + k₂ + k₃

The addition of multiple stiffness elements

Fig 1. Stiffness of Multiple Elements

The sum of the collection of stiffness elements in Fig 1 are added as follows:
k₁₁ = k₁ + k₂ + k₃
k₁₂ = 1 / (1/k₄ + 1/k₅)
k₁₃ = k₆ + k₇
k = 1 / (1/k₁₁ + 1/k₁₂ + 1/k₁₃)

For example; the compound stiffness of a double spring assembly in the suspension of a car (e.g. one spring inside the other) must be added together in parallel (e.g. k₁₃ above).

Further Reading

You will find further reading on this subject in reference publications(1, 2, 3 & 4)

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