m physical modeling / IICSM

PHYSICS / ENGINEERING / COMPUTING

Solid mechanics models

Derive stress, strain, axial deformation, torsion, beam bending, failure and stability, with twenty worked examples.

Subject library · 51 guides · derivations & worked examples

Fundamental physics

Space and flight

Materials and transport

Systems and mathematics

Marine and road vehicles

Electrical and computing

Companion site

Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

1. Stress and elasticity

Definitions & inputs. F axial force, A area, δ extension, L length, E Young modulus, ν Poisson ratio.

  1. Average force density and relative extension define stress and strain.

    σ=F/A,ϵ=δ/L\sigma=F/A,\quad\epsilon=\delta/L

  2. Apply Hooke’s law and lateral contraction.

    σ=Eϵ,ϵtrans=−νϵ\sigma=E\epsilon,\quad\epsilon_{trans}=-\nu\epsilon

  3. Substitute the definitions into the constitutive relation.

    δ=FL/(EA)\delta=FL/(EA)

Interpretation. Average stress omits local concentration near holes, contacts and load introduction.

↑ Return to definitions and contents

2. Torsion and bending

Definitions & inputs. T torque, r radius, J polar area moment, M bending moment, y distance from neutral axis, I second area moment.

  1. Linear shear variation and shear elasticity produce shaft twist.

    τ=Tr/J,θ=TL/(GJ)\tau=Tr/J,\quad\theta=TL/(GJ)

  2. Plane sections remaining plane connects curvature to bending moment.

    σx=−My/I,EIv′′=M\sigma_x=-My/I,\quad EI v''=M

  3. Integrate twice for a tip-loaded cantilever with zero base displacement and slope.

    vtip=FL3/(3EI)v_{tip}=FL^3/(3EI)

Interpretation. Shear deformation and rotary inertia matter for short or thick members and high frequencies.

↑ Return to definitions and contents

3. Energy and combined stress

Definitions & inputs. U elastic energy, σ1,σ2 principal plane stresses, σy yield stress.

  1. Integrate the linearly increasing force during quasistatic loading.

    U=12FδU=\tfrac12F\delta

  2. Form the distortional-energy stress invariant for plane stress.

    σVM=σ12−σ1σ2+σ22\sigma_{VM}=\sqrt{\sigma_1^2-\sigma_1\sigma_2+\sigma_2^2}

  3. Compare the yield criterion with the applied equivalent stress.

    ny=σy/σVMn_y=\sigma_y/\sigma_{VM}

Interpretation. A yield ratio alone does not certify fracture, fatigue or stability performance.

↑ Return to definitions and contents

4. Thermal load and stability

Definitions & inputs. α thermal expansion, ΔT temperature change, K effective-length factor, Le=KL.

  1. Free thermal expansion is an eigenstrain.

    ϵth=αΔT\epsilon_{th}=\alpha\Delta T

  2. Full axial restraint cancels total strain; heating creates compression.

    σ=−EαΔT\sigma=-E\alpha\Delta T

  3. The lowest buckling mode solves the elastic beam-column eigenproblem.

    Pcr=π2EI/(KL)2P_{cr}=\pi^2EI/(KL)^2

Interpretation. Imperfections, eccentricity, inelasticity and uncertain restraints reduce applicability of Euler’s ideal load.

↑ Return to definitions and contents

Graphical worked example

L=2 m, A=100 mm², E=200 GPa; small-strain uniform axial elastic loading. X axis: Axial force (kN). Y axis: Rod extension (mm).
L=2 m, A=100 mm², E=200 GPa; small-strain uniform axial elastic loading. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Axial stress

Definitions & inputs. F=10 kN, A=100 mm².

  1. Choose the governing model and isolate the requested quantity.

    σ=F/A\sigma=F/A

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    σ=10000/100\sigma=10000/100

  3. Evaluate the expression; the result uses the units shown.

    Result=100 MPa\mathrm{Result}=100\ {\rm MPa}

Interpretation. One N/mm² equals one MPa.

↑ Return to definitions and contents
Example 02. Axial strain

Definitions & inputs. σ=100 MPa, E=200 GPa.

  1. Choose the governing model and isolate the requested quantity.

    ϵ=σ/E\epsilon=\sigma/E

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ϵ=100/200000\epsilon=100/200000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0005 \mathrm{Result}=0.0005\ {}

Interpretation. The assumed strain is small.

↑ Return to definitions and contents
Example 03. Rod elongation

Definitions & inputs. F=10 kN,L=2 m,A=100 mm²,E=200 GPa.

  1. Choose the governing model and isolate the requested quantity.

    δ=FL/(EA)\delta=FL/(EA)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    δ=104(2)/(2×101110−4)\delta=10^4(2)/(2\times10^{11}10^{-4})

  3. Evaluate the expression; the result uses the units shown.

    Result=0.001 m\mathrm{Result}=0.001\ {\rm m}

Interpretation. The rod extends by one millimetre.

↑ Return to definitions and contents
Example 04. Lateral strain

Definitions & inputs. ν=.3, axial strain .001.

  1. Choose the governing model and isolate the requested quantity.

    ϵt=−νϵa\epsilon_t=-\nu\epsilon_a

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ϵt=−0.3(0.001)\epsilon_t=-0.3(0.001)

  3. Evaluate the expression; the result uses the units shown.

    Result=−0.0003 \mathrm{Result}=-0.0003\ {}

Interpretation. Positive axial extension produces lateral contraction.

↑ Return to definitions and contents
Example 05. Shear modulus

Definitions & inputs. E=210 GPa,ν=.3, isotropic material.

  1. Choose the governing model and isolate the requested quantity.

    G=E/[2(1+ν)]G=E/[2(1+\nu)]

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    G=210/[2(1.3)]G=210/[2(1.3)]

  3. Evaluate the expression; the result uses the units shown.

    Result=80.76923 GPa\mathrm{Result}=80.76923\ {\rm GPa}

Interpretation. E, G and ν are not independent for isotropic elasticity.

↑ Return to definitions and contents
Example 06. Bulk modulus

Definitions & inputs. E=210 GPa,ν=.3.

  1. Choose the governing model and isolate the requested quantity.

    K=E/[3(1−2ν)]K=E/[3(1-2\nu)]

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    K=210/[3(0.4)]K=210/[3(0.4)]

  3. Evaluate the expression; the result uses the units shown.

    Result=175 GPa\mathrm{Result}=175\ {\rm GPa}

Interpretation. K describes hydrostatic compression.

↑ Return to definitions and contents
Example 07. Circular shaft polar moment

Definitions & inputs. Radius r=10 mm.

  1. Choose the governing model and isolate the requested quantity.

    J=πr4/2J=\pi r^4/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    J=π(0.01)4/2J=\pi(0.01)^4/2

  3. Evaluate the expression; the result uses the units shown.

    Result=1.570796×10−8 m4\mathrm{Result}=1.570796\times10^{-8}\ {\rm m}^4

Interpretation. Hollow shafts subtract the inner radius fourth power.

↑ Return to definitions and contents
Example 08. Shaft surface shear

Definitions & inputs. T=10 N m,r=.01 m,J=πr⁴/2.

  1. Choose the governing model and isolate the requested quantity.

    τ=Tr/J\tau=Tr/J

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    τ=10(0.01)/(π10−8/2)\tau=10(0.01)/(\pi10^{-8}/2)

  3. Evaluate the expression; the result uses the units shown.

    Result=6.366198 MPa\mathrm{Result}=6.366198\ {\rm MPa}

Interpretation. Maximum shear occurs at the outer radius.

↑ Return to definitions and contents
Example 09. Shaft twist

Definitions & inputs. T=10 N m,L=1 m,G=80 GPa,J=π(0.01)⁴/2.

  1. Choose the governing model and isolate the requested quantity.

    θ=TL/(GJ)\theta=TL/(GJ)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    θ=10/[80×109(π10−8/2)]\theta=10/[80\times10^9(\pi10^{-8}/2)]

  3. Evaluate the expression; the result uses the units shown.

    Result=0.007957747 rad\mathrm{Result}=0.007957747\ {\rm rad}

Interpretation. This omits end effects near torque introduction.

↑ Return to definitions and contents
Example 10. Rectangular beam area moment

Definitions & inputs. Width b=.02 m,height h=.04 m.

  1. Choose the governing model and isolate the requested quantity.

    I=bh3/12I=bh^3/12

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    I=0.02(0.04)3/12I=0.02(0.04)^3/12

  3. Evaluate the expression; the result uses the units shown.

    Result=1.066667×10−7 m4\mathrm{Result}=1.066667\times10^{-7}\ {\rm m}^4

Interpretation. Height is measured along the bending direction.

↑ Return to definitions and contents
Example 11. Bending surface stress

Definitions & inputs. M=100 N m,y=.02 m,I=1.0666667×10⁻⁷ m⁴.

  1. Choose the governing model and isolate the requested quantity.

    ∣σ∣=My/I|\sigma|=My/I

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ∣σ∣=100(0.02)/(1.0666667×10−7)|\sigma|=100(0.02)/(1.0666667\times10^{-7})

  3. Evaluate the expression; the result uses the units shown.

    Result=18.75 MPa\mathrm{Result}=18.75\ {\rm MPa}

Interpretation. The opposite surface has the opposite stress sign.

↑ Return to definitions and contents
Example 12. Cantilever tip deflection

Definitions & inputs. F=10 N,L=.5 m,E=200 GPa,I=10⁻⁸ m⁴.

  1. Choose the governing model and isolate the requested quantity.

    δ=FL3/(3EI)\delta=FL^3/(3EI)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    δ=10(0.5)3/[3(2×1011)(10−8)]\delta=10(0.5)^3/[3(2\times10^{11})(10^{-8})]

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0002083333 m\mathrm{Result}=0.0002083333\ {\rm m}

Interpretation. Valid while deflection and rotations remain small.

↑ Return to definitions and contents
Example 13. Simply supported midpoint moment

Definitions & inputs. Centre point load F=100 N,span L=2 m.

  1. Choose the governing model and isolate the requested quantity.

    Mmax=FL/4M_{max}=FL/4

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Mmax=100(2)/4M_{max}=100(2)/4

  3. Evaluate the expression; the result uses the units shown.

    Result=50 N m\mathrm{Result}=50\ {\rm N\,m}

Interpretation. End supports transmit no bending moment in this model.

↑ Return to definitions and contents
Example 14. Elastic strain energy

Definitions & inputs. Final F=1000 N,linear extension δ=.001 m.

  1. Choose the governing model and isolate the requested quantity.

    U=Fδ/2U=F\delta/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    U=1000(0.001)/2U=1000(0.001)/2

  3. Evaluate the expression; the result uses the units shown.

    Result=0.5 J\mathrm{Result}=0.5\ {\rm J}

Interpretation. The result assumes loading from zero without dissipation.

↑ Return to definitions and contents
Example 15. Von Mises plane stress

Definitions & inputs. Principal stresses 100 and 50 MPa; third is zero.

  1. Choose the governing model and isolate the requested quantity.

    σVM=σ12−σ1σ2+σ22\sigma_{VM}=\sqrt{\sigma_1^2-\sigma_1\sigma_2+\sigma_2^2}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    σVM=10000−5000+2500\sigma_{VM}=\sqrt{10000-5000+2500}

  3. Evaluate the expression; the result uses the units shown.

    Result=86.60254 MPa\mathrm{Result}=86.60254\ {\rm MPa}

Interpretation. Principal values already account for any shear transformation.

↑ Return to definitions and contents
Example 16. Yield ratio

Definitions & inputs. Yield strength 250 MPa, equivalent applied stress 100 MPa.

  1. Choose the governing model and isolate the requested quantity.

    n=σy/σVMn=\sigma_y/\sigma_{VM}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    n=250/100n=250/100

  3. Evaluate the expression; the result uses the units shown.

    Result=2.5 \mathrm{Result}=2.5\ {}

Interpretation. Other failure modes may control.

↑ Return to definitions and contents
Example 17. Free thermal elongation

Definitions & inputs. α=12×10⁻⁶/K,L=2 m,ΔT=50 K.

  1. Choose the governing model and isolate the requested quantity.

    δ=αLΔT\delta=\alpha L\Delta T

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    δ=12×10−6(2)(50)\delta=12\times10^{-6}(2)(50)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0012 m\mathrm{Result}=0.0012\ {\rm m}

Interpretation. No mechanical load develops when expansion is unrestrained.

↑ Return to definitions and contents
Example 18. Fully restrained thermal stress

Definitions & inputs. E=200 GPa,α=12×10⁻⁶/K,ΔT=50 K.

  1. Choose the governing model and isolate the requested quantity.

    σ=−EαΔT\sigma=-E\alpha\Delta T

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    σ=−(200×109)(12×10−6)(50)\sigma=-(200\times10^9)(12\times10^{-6})(50)

  3. Evaluate the expression; the result uses the units shown.

    Result=−120 MPa\mathrm{Result}=-120\ {\rm MPa}

Interpretation. Heating produces compression; yielding or buckling may invalidate the elastic result.

↑ Return to definitions and contents
Example 19. Pinned Euler column

Definitions & inputs. E=200 GPa,I=10⁻⁸ m⁴,L=1 m,K=1.

  1. Choose the governing model and isolate the requested quantity.

    Pcr=π2EI/(KL)2P_{cr}=\pi^2EI/(KL)^2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Pcr=π2(2×1011)(10−8)P_{cr}=\pi^2(2\times10^{11})(10^{-8})

  3. Evaluate the expression; the result uses the units shown.

    Result=19739.21 N\mathrm{Result}=19739.21\ {\rm N}

Interpretation. This is an ideal bifurcation load, not an allowable load.

↑ Return to definitions and contents
Example 20. Cantilever stiffness

Definitions & inputs. E=200 GPa,I=10⁻⁸ m⁴,L=.5 m.

  1. Choose the governing model and isolate the requested quantity.

    k=3EI/L3k=3EI/L^3

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    k=3(2×1011)(10−8)/(0.5)3k=3(2\times10^{11})(10^{-8})/(0.5)^3

  3. Evaluate the expression; the result uses the units shown.

    Result=48000 N m−1\mathrm{Result}=48000\ {\rm N\,m}^{-1}

Interpretation. This maps a small transverse tip displacement to tip force.

↑ Return to definitions and contents

Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.