Mechanical Properties of Solids · Lesson 4 of 7
Shear Modulus and Bulk Compression
“Compare shape change under shear with volume change under pressure, and calculate shear displacement, compression and density changes.”
• Derive G = FL/(AΔx) and choose the sheared area and face separation correctly. • Explain why an extended helical coil can be governed mainly by shear of its wire. • Define bulk modulus and compressibility with a consistent compression sign. • Apply pressure–volume relations to oceans, cubes, glass and water density, checking inconsistent data.
Shear modulus measures resistance to shape distortion
Normal stress pulls or pushes across a section. Shear stress acts tangentially and shifts one face relative to another. If a block has loaded face area A and perpendicular distance L between the fixed and displaced faces, a tangential force F gives stress F/A and shear strain Δx/L. Their ratio in the small linear elastic range defines shear modulus G, also called modulus of rigidity.
Although the algebra resembles the axial-extension equation, the physics and geometry differ. Here the loaded force is parallel to A and displacement is perpendicular to L. The relevant modulus is G, not Y. Both have units Pa, so a unit check alone cannot catch choosing the wrong one. Use a loading sketch first.
| Material | Shear modulus G / GPa |
|---|---|
| Aluminium | 25 |
| Brass | 36 |
| Copper | 42 |
| Glass | 23 |
| Iron | 70 |
| Lead | 5.6 |
| Nickel | 77 |
| Steel | 84 |
| Tungsten | 150 |
| Wood | 10 |
These representative values indicate how different materials resist shear. A high G means a small shear strain for a specified shear stress. The source offers G ≈ Y/3 as a rough comparison; it is not an exact universal relation and is not a substitute for tabulated material data. Wood and other direction-dependent materials especially require specifying how the specimen is oriented. Ordinary liquids and gases do not sustain a static shear deformation as an elastic solid does: they continue to flow under a maintained shear stress.
Problem
A square lead slab of side 0.50 m and thickness 0.10 m is fixed along its lower edge. A force of 9.0 × 10⁴ N acts tangentially on its upper narrow face. Use G = 5.6 × 10⁹ Pa.
- 1.Use loaded face area A = 0.50 × 0.10 = 0.050 m². The large square side-face area 0.25 m² is not the loaded narrow-face area.
- 2.Shear stress = 9.0 × 10⁴/0.050 = 1.80 × 10⁶ Pa.
- 3.Shear strain = τ/G = 3.214 × 10⁻⁴. The corresponding face separation is L = 0.50 m.
- 4.Δx = γL = 1.607 × 10⁻⁴ m = 0.161 mm, approximately 0.16 mm.
- 5.The support prevents gross rotation and translation while the material distorts. The small strain supports the small-angle approximation.
Problem
A cube has edge 0.10 m and one face fixed to a vertical wall. A 100 kg mass acts on its opposite face. Use G = 25 GPa to find the vertical displacement.
- 1.The load acts parallel to the fixed and loaded faces, so use shear, not axial compression.
- 2.F = mg = 980 N, A = (0.10)² = 0.010 m², and face separation L = 0.10 m.
- 3.Δx = FL/(AG) = 980(0.10)/(0.010 × 25 × 10⁹) = 3.92 × 10⁻⁷ m.
- 4.This is 0.392 μm. With g rounded to 10 m/s² it becomes 4.0 × 10⁻⁷ m.
A stretched coil can involve shear in its wire
Pulling the ends of a closely wound helical spring moves its ends apart, but the spring is not equivalent to a straight wire of that end-to-end length. The applied load mainly twists the coiled wire about its local axis. That twisting is resisted by shear in the wire, so the axial extension of the coil is governed principally by its shear modulus. The visible extension direction alone therefore does not identify Young’s modulus. The detailed torsion formula for a spring is outside this chapter; the physical distinction is sufficient here.
Bulk modulus measures all-sided compression
Uniform surrounding pressure acts normally on every part of a body’s surface. In an isotropic material the ideal response is a volume reduction without a change of geometrical shape. Bulk modulus B measures resistance to that volume change. It applies to solids, liquids and gases; it does not require a solid to preserve a fixed shear shape. Use the increase in surrounding pressure from the initial state rather than automatically inserting a final absolute pressure.
The minus sign is physical bookkeeping. Defining ΔV as final minus initial makes compression negative. If a question asks the fractional compression c, use c = −ΔV/V = Δp/B, a positive magnitude. If both versions are mixed, an otherwise correct calculation acquires an incorrect sign. For very large pressure ranges the modulus can vary and one constant is only an approximation.
| Material/state | Bulk modulus B / GPa |
|---|---|
| Aluminium (solid) | 72 |
| Brass (solid) | 61 |
| Copper (solid) | 140 |
| Glass (solid) | 37 |
| Iron (solid) | 100 |
| Nickel (solid) | 260 |
| Steel (solid) | 160 |
| Water (liquid) | 2.2 |
| Ethanol (liquid) | 0.9 |
| Carbon disulphide (liquid) | 1.56 |
| Glycerine (liquid) | 4.76 |
| Mercury (liquid) | 25 |
| Air at STP | 0.00010 |
Compare steel, water and air: their listed moduli are 1.6 × 10¹¹ Pa, 2.2 × 10⁹ Pa and about 1.0 × 10⁵ Pa. Air is roughly 1.6 million times as compressible as this steel, and about 22,000 times as compressible as water. These are representative comparisons, not an assertion that every solid is a million times stiffer than every gas. Gases have much space between molecules and compress readily; a solid’s closely coupled neighbours strongly resist a reduction in separation. Liquids also resist volume change substantially. Gas compressibility depends strongly on pressure and temperature conditions.
Ocean pressure and volume change
For a water column of height h and area A, its weight is ρAhg. Dividing by area gives the extra pressure ρgh. This simple argument assumes approximately constant density and g. It is a prerequisite for the ocean calculation, not a claim that density never changes. The elastic calculation then estimates that density/volume correction.
Problem
Estimate the fractional compression of water at 3000 m using density 1000 kg/m³, g = 10 m/s² and B = 2.2 × 10⁹ Pa.
- 1.The pressure increase relative to the surface is Δp ≈ ρgh = 1000 × 10 × 3000 = 3.0 × 10⁷ Pa.
- 2.Fractional compression is c = Δp/B = 0.0136, or 1.36%.
- 3.The signed volume strain is ΔV/V = −0.0136. The same physical change can be reported as positive compression magnitude or negative signed strain, provided the convention is stated.
- 4.The calculation uses constant density to estimate the pressure and constant B for the compression; it is a first approximation.
Pressure–volume data must be physically consistent
Problem
A problem states an initial volume of 100.0 L, a pressure increase of 100.0 atm, and a final volume of 100.5 L. What does the bulk-modulus equation reveal?
- 1.The stated ΔV is +0.5 L: the water expands even though the pressure is said to increase. With the ordinary compression model, these data are inconsistent.
- 2.Substitution gives B = −(100 × 1.013 × 10⁵)/(0.5/100) = −2.026 × 10⁹ Pa. A negative result here flags the contradictory data; it must not simply be reported as the bulk modulus of water.
- 3.If the intended final volume was 99.5 L, the contraction would instead be 0.5 L and B = +2.026 × 10⁹ Pa. This is an explicitly corrected interpretation, not a solution of the literal 100.5 L statement.
- 4.Compared with air at constant temperature near 1 atm, whose bulk modulus is about 1.013 × 10⁵ Pa, the corrected water modulus is about 20,000 times larger. Molecular spacing and coupling explain the much smaller compressibility of water.
Density and further bulk-compression applications
Problem
Surface density is 1.03 × 10³ kg/m³ and B = 2.2 × 10⁹ Pa. Estimate the density where the absolute pressure is 80 atm, taking the surface as 1 atm.
- 1.The increase from the surface is 79 atm = 8.003 × 10⁶ Pa, not 80 atm if 80 atm denotes the final absolute pressure.
- 2.c = Δp/B = 0.003638 and ρ = 1030/(1 − c) ≈ 1033.76 kg/m³.
- 3.Rounded to the precision relevant here, ρ ≈ 1.034 × 10³ kg/m³.
- 4.If 80 atm were instead intended as a pressure increase, the estimate would be 1033.81 kg/m³, giving the same rounded result. State which pressure interpretation is used.
Problem
Use the listed bulk moduli to find the volume response of glass under 10 atm, a 10 cm copper cube under 7.0 × 10⁶ Pa, and 1 L of water compressed by 0.10%. Treat the stated pressures as increases.
- 1.Glass: c = (10 × 1.013 × 10⁵)/(37 × 10⁹) = 2.74 × 10⁻⁵, or 0.00274%. Its signed volume strain is the negative of that fraction.
- 2.Copper: initial volume = (10 cm)³ = 1000 cm³. c = (7.0 × 10⁶)/(140 × 10⁹) = 5.0 × 10⁻⁵, so contraction = 0.050 cm³.
- 3.Water: 0.10% means c = 0.0010. Required pressure increase = Bc = 2.2 × 10⁶ Pa.
- 4.One litre is 10⁻³ m³, so that water contraction is 10⁻⁶ m³ = 1 mL. The pressure needed for a specified fractional compression is independent of initial volume in this model.
| Loading | Response variable | Material parameter | Static applicability |
|---|---|---|---|
| Axial normal stress | ΔL/L | Young’s modulus Y | Solid with defined member geometry |
| Tangential shear stress | Δx/L | Shear modulus G | Solid under small equilibrium shear |
| Uniform pressure increase | ΔV/V | Bulk modulus B | Solid, liquid or gas under specified conditions |
This comparison separates deformation modes; it should not be read as saying that axial tension always preserves volume. A stretched solid normally also contracts sideways. Its volume response depends on that lateral strain, which is considered next. Uniform hydrostatic pressure and uniaxial loading impose different stresses even when their numerical force-per-area values match.
Quick Quiz
Quiz
In the lead-slab example, why is A = 0.050 m²?
The upper narrow face is 0.50 m wide and 0.10 m thick. Choose the area by the force-transmission surface.
What mainly governs extension of a closely wound helical spring through twisting of its wire?
A coil’s end-to-end stretching mainly twists its wire; shear rigidity resists that twisting.
An increased pressure Δp compresses a liquid by fractional amount c = 0.001. Which equation has the correct sign?
Compression makes the signed volume change negative, while B is positive.
A stable liquid is said to expand when only its surrounding pressure is increased at fixed conditions. What should you do before quoting B?
The stated data conflict with the normal positive-bulk-modulus model; report and resolve that inconsistency explicitly.
Two samples of the same liquid have different initial volumes. Under the same small pressure increase, which is equal?
ΔV/V = −Δp/B is independent of the initial sample size, while |ΔV| scales with V.
Practice Problems
- A solid block has a tangentially loaded face of area 0.020 m² and a perpendicular separation of 0.12 m between its loaded and fixed faces. Find its shear stress, shear strain and lateral displacement under an 800 N tangential force, using G = 40 GPa.
- A liquid sample initially occupies 2.0 L. Its surrounding pressure increases uniformly by 5.0 MPa, and its bulk modulus is 2.0 GPa. Find its signed volume strain, volume contraction in mL and final volume. Assume the modulus is constant over this small change.
- Equal 100 cm³ samples of two materials have bulk moduli 50 GPa and 100 GPa. Both experience a uniform pressure increase of 20 MPa. Calculate their volume contractions and compare their compressibilities. Explain the difference between absolute and fractional volume change.
- A liquid initially has density 1000 kg/m³ and bulk modulus 2.2 GPa. Estimate its density after a pressure increase of 11 MPa. Use the lesson’s constant-modulus small-compression model and conserve the sample’s mass.
- Identify the appropriate elastic parameter for (a) axial extension of a straight uniform wire, (b) extension of a closely wound coil through twisting of its wire, and (c) a small volume reduction of water under uniform pressure. Explain why knowing only a force-per-area value does not select the correct parameter.
Key Takeaways
• Choose G for tangential shape distortion and B for uniform-pressure volume change. • Choose the actual loaded face area and perpendicular face separation in a shear problem. • Bulk modulus is positive; signed compression strain is negative and compressibility is 1/B. • Check fractions versus percentages and pressure increase versus absolute pressure before calculating. • Inconsistent pressure–volume data must be identified rather than hidden by dropping a sign.