Mechanical Properties of Solids · Lesson 7 of 7
Chapter Summary and Practice
“Connect the complete chapter through a formula reference, model-selection checks and mixed practice with answer guidance.”
• Select the correct stress, strain and elastic constant for a loading situation. • Recall formula conditions, dimensions and graph interpretations. • Combine force balance, compatibility and elastic deformation in source-based applications. • Detect unit, sign, material-limit and data-consistency errors in calculations.
One connected picture of a loaded solid
Begin with the forces and constraints. These determine the internal force transmitted across a chosen section. Divide by the appropriate area to identify the relevant stress, then use a material law to connect stress and strain. Finally convert fractional deformation into the requested length, lateral shift or volume change. A graph or a limiting stress tells you whether the linear elastic model is still appropriate. Geometry determines how a specimen responds even when its material constants stay unchanged.
Normal loading of a member uses Y and longitudinal strain. Tangential loading uses G and shear strain. Uniform surrounding pressure uses B and volume strain. Poisson’s ratio connects longitudinal deformation with sideways response. Gradual elastic loading stores work; the linear force–extension relation makes that work a triangular area. Engineering applications combine these response laws with strength, load paths, cross-section shape and stability.
Formula reference with conditions
| Purpose | Relation | Essential condition |
|---|---|---|
| Normal/shear stress | σ = Fnormal/A; τ = Ftangential/A | Specify the cut or loaded face; simple uniform models use average stress |
| Longitudinal strain | ε = ΔL/L | Use original length and a stated sign convention |
| Shear strain | γ = Δx/L = tan θ ≈ θ | θ in radians for the small-angle approximation |
| Axial response | Y = σ/ε; ΔL = FL/(AY) | Small linear elastic axial deformation |
| Shear response | G = τ/γ; Δx = FL/(AG) | Small elastic shear; correct face separation and area |
| Bulk response | B = −Δp/(ΔV/V) | Uniform pressure change and approximately constant B |
| Compressibility | κ = 1/B | Same material state and conditions as B |
| Poisson response | ν = −(Δd/d)/(ΔL/L) | Signed transverse and axial changes in small linear response |
| Stored energy | U = ½FΔL = AY(ΔL)²/(2L) | Gradual recoverable loading from zero force, linear law |
| Energy density | u = ½σε = ½Yε² = σ²/(2Y) | Same linear elastic model |
| Beam sag | δ = Wl³/(4bd³Y) | Specified simply supported rectangular beam, central load |
| Load capacity | Fallow = Aσallow | Allowable stress must reflect appropriate loading and design conditions |
| Quantity | SI unit | Dimensions |
|---|---|---|
| Stress, Y, G, B | Pa = N/m² | M L⁻¹ T⁻² |
| Strains and Poisson’s ratio | No unit | 1 |
| Compressibility | Pa⁻¹ | M⁻¹ L T² |
| Wire force–extension stiffness | N/m | M T⁻² |
| Elastic energy | J | M L² T⁻² |
| Elastic energy density | J/m³ | M L⁻¹ T⁻² |
Distinctions that prevent common mistakes
Elastic means recoverable; Hookean additionally means linear. Stiffness, yield strength, ultimate strength and ductility describe different properties. A high Y gives small strain at a given small stress, but it cannot prove that a high-load specimen remains elastic. A large recoverable extension of rubber does not give it a larger Y than steel. Material comparisons depend on grade and conditions; the statement that every metal is stiffer than every alloy is not a general law.
A tension F at each end gives stress F/A across the wire, not 2F/A. A junction load can make tensions in connected segments different. Identical elongations in parallel supports can arise from geometric compatibility even when their tensions are unequal. A diameter must be halved before using πr², an annular hole must be subtracted, and a centimetre-square area carries a 10⁻⁴ conversion to square metres. A strain of 0.001 is 0.1%. Use the pressure increase from the initial state; distinguish signed negative volume strain from positive compression magnitude.
Stress is not an ordinary force vector. Its description requires a specified internal plane and force component. Likewise, a bulk-modulus calculation does not model every loading with units Pa: a mountain base or a wire pulled along one axis is not under uniform pressure on every surface. Axial loading can also create lateral and volume changes; it is not generally volume preserving. Check boundary conditions and model meaning along with units.
Worked synthesis: limiting stress and energy
Problem
A steel cable has radius 1.5 cm and limiting stress 1.0 × 10⁸ Pa. Calculate the maximum force in this ideal stress-limited model.
- 1.Convert radius to 0.015 m. A = πr² = 7.069 × 10⁻⁴ m².
- 2.Fmax = Aσmax = 7.069 × 10⁴ N. This is a force, approximately 70.7 kN.
- 3.If an equivalent static hanging mass is requested, divide by g: about 7210 kg for 9.8 m/s². Do not report a mass in newtons.
- 4.This is the supplied stress-limit model. A commercial chairlift system requires additional engineering checks beyond this calculation.
Problem
A specimen remains Hookean up to a chosen working stress σw. Express its working strain and stored energy density in terms of Y and σw.
- 1.Use the material relation to get εw = σw/Y.
- 2.Insert this into u = ½σε: uw = σw²/(2Y). Multiply by the specimen volume for total stored energy.
- 3.At fixed working strain, higher Y gives larger energy density. At fixed working stress, higher Y gives smaller strain and smaller energy density. Always specify what is held fixed.
- 4.The relation estimates recoverable elastic energy only while the material actually remains in its linear range.
Selected answer guidance and checking strategy
| Exercise idea | Guidance/result using stated chapter data |
|---|---|
| Equal-extension wire comparison | Ys/Yc = 1.79, about 1.8 |
| Graph reading | Y = 75 GPa; yield stress about 300 MPa |
| A/B comparison | A has both larger initial slope and larger fracture stress |
| Hanging segments | Steel: 1.50 × 10⁻⁴ m; brass: 1.33 × 10⁻⁴ m |
| Aluminium cube | 3.92 × 10⁻⁷ m using g = 9.8 m/s² |
| Hollow columns | Compression strain magnitude 7.22 × 10⁻⁷ |
| Copper strip | Strain 0.001393 = 0.1393% with Y = 110 GPa |
| Equal-tension copper/iron supports | dc/di = √(190/110) ≈ 1.31 |
| Rotating wire | About 1.87 mm, conditional on linear elasticity at the required stress |
| Water pressure–volume data | Literal final 100.5 L is inconsistent; if corrected to 99.5 L, B = 2.026 GPa |
| Water density | About 1.034 × 10³ kg/m³ |
| Glass compression | 2.74 × 10⁻⁵ = 0.00274% |
| Copper cube contraction | 0.050 cm³ with B = 140 GPa |
| Water compression 0.10% | Pressure increase 2.2 × 10⁶ Pa |
Use the earlier worked calculations to check the method, not just the answer. Recompute a suspicious printed result from the stated data and preserve units throughout. Numerical differences caused by g = 9.8 versus 10 m/s² or rounded tabular constants are small and should be identified. Factor-of-ten errors, lost percentages, a revolutions-versus-radians mistake, or an expansion under an asserted pressure increase require a substantive correction, not a rounding explanation.
Check your understanding
Quiz
A specimen recovers after unloading but follows a nonlinear stress–strain curve. Which description is best?
Recovery and linearity are distinct; elastomers illustrate that distinction.
Two identical-area wire specimens carry the same small stress. Which specimen has the smaller longitudinal strain?
Within the linear range ε = σ/Y. Stress, rather than force alone, is fixed in this comparison.
The pressure on water is increased and its volume falls by 0.10%. What is its signed volume strain?
0.10% = 0.0010, and a volume decrease has a negative sign.
For a linearly loaded wire starting from zero force, doubling final force multiplies stored energy by
Extension doubles as well, so U = ½FΔL increases by four.
Which check must precede using a circular-motion load in ΔL = FL/(AY)?
The load is determined by dynamics, and the deformation law additionally requires its material assumptions to hold.
Mixed source-based practice
Practice Problems
- 8.1: Two different wires have equal extension under the same load. Derive the ratio of their Young’s moduli before substituting the stated lengths and areas.
- 8.2(a,b): Read the initial slope and the approximate yield stress from the numerical graph. Explain why these are different quantities.
- 8.3(a,b): Compare initial stiffness and fracture stress for A and B on identical graph scales.
- 8.4(a,b): Explain why steel has a much larger Young’s modulus than rubber, and why stretching a closely wound coil can involve its wire’s shear modulus.
- 8.5: For the steel–brass hanging assembly, draw separate force balances for the lower mass and the junction before finding each elongation.
- 8.6: For the wall-fixed aluminium cube, identify loaded area, face separation and load direction, then calculate the shear displacement.
- 8.7: Find the strain of each equally loaded hollow steel column using annular material area and the given radii.
- 8.8: Convert the rectangular copper cross-section to m² and calculate strain as a fraction and a percentage using the stated Y.
- 8.9: Find the force supported by a 1.5 cm radius steel cable when stress is limited to 10⁸ Pa.
- 8.10: For a symmetric rigid bar on copper and iron wires, combine equal extension with the requested equal tension and derive the diameter ratio.
- 8.11: At the bottom of the vertical circle, calculate tension using angular speed converted from revolutions per second. Then estimate extension and check the elastic assumption.
- 8.12: Check the sign of the stated water volume change under pressure. Explain why the literal data are inconsistent and state any corrected interpretation explicitly.
- 8.13: Derive the density change from mass conservation and the estimated volume contraction. Distinguish final absolute pressure from pressure increase.
- 8.14: Give the glass volume contraction as both a fraction and a percentage.
- 8.15: Find the contraction of the copper cube using the bulk modulus in the chapter table.
- 8.16: Convert 0.10% to fractional compression and calculate the required pressure increase in water.
Key Takeaways
• Choose the loading model, then solve forces and compatibility, then calculate stress and strain. • Attach assumptions and valid material ranges to formulas rather than using them as universal identities. • Treat percentages, areas, angular-speed conversion and compression signs carefully. • Use independent calculation and physical checks to resolve inconsistent numerical data.
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Elasticity in Engineering and Nature
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