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Physics / Chapter 12: Equilibrium and Elasticity / Section 12-6

Elasticity — Question 7

Section 12-6 · Physics pages 1884–1885

Question 7 Bulk compression changes volume three times as fast as each length An isotropic solid cube initially has side 0.10 m and bulk modulus B = 50 GPa. Its surrounding pressure increases uniformly by 30 MPa. Use the small-strain law ∆V/V = −∆p/B and equal strain along all three axes. ∆p Section through a cube under pressure on all six faces, including front and back. (a) Find the volume change and approximate change of each side. (b) Explain why the axial stress is not three times the applied pressure.
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Question 7 - Solution Bulk compression changes volume three times as fast as each length ∆p (a) Derivation and calculation The initial volume is V = (.10)3 = .00100 m3. Thus ∆V V = −30 × 106 50 × 109 = −6.00 × 10−4, ∆V = −6.00 × 10−7 m3 . Equal small axial strains give ϵℓ= 1 3 ∆V V = −2.00 × 10−4, ∆ℓ= −2.00 × 10−5 m = −20.0 µm . (b) Result and interpretation Each face pair carries its own normal compressive stress: ∆σxx = ∆σyy = ∆σzz = −∆p = −30 MPa . The factor of three comes from volume geometry, ∆V/V ≃ϵx + ϵy + ϵz = 3ϵℓ, not from adding perpendicular stresses into one axial component. Physical check Increasing pressure decreases every side length and the volume, consistent with the negative strain signs.

Question 7

Bulk compression changes volume three times as fast as each length

An isotropic solid cube initially has side 0.10m0.10\,\mathrm m and bulk modulus B=50GPaB=50\,\mathrm{GPa}. Its surrounding pressure increases uniformly by 30MPa30\,\mathrm{MPa}. Use the small-strain law ΔV/V=−Δp/B\Delta V/V=-\Delta p/B and equal strain along all three axes.

Diagram shown in the original SVG worksheet.

  1. Find the volume change and approximate change of each side.

  2. Explain why the axial stress is not three times the applied pressure.

Question 7 - Solution

Bulk compression changes volume three times as fast as each length

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The initial volume is V=(.10)3=.00100V=(.10)^3=.00100 m3^3. Thus ΔVV=−30×10650×109=−6.00×10−4,\frac{\Delta V}{V}=-\frac{30\times10^6}{50\times10^9}=-6.00\times10^{-4}, ΔV=−6.00×10−7m3.\boxed{\Delta V=-6.00\times10^{-7}\ \mathrm{m^3}}. Equal small axial strains give ϵℓ=13ΔVV=−2.00×10−4,Δℓ=−2.00×10−5m=−20.0μm.\epsilon_\ell=\frac13\frac{\Delta V}{V}=-2.00\times10^{-4},\quad \boxed{\Delta\ell=-2.00\times10^{-5}\ \mathrm m=-20.0\ \mu\mathrm m}.

(b) Result and interpretation

Each face pair carries its own normal compressive stress: Δσxx=Δσyy=Δσzz=−Δp=−30MPa.\boxed{\Delta\sigma_{xx}=\Delta\sigma_{yy}=\Delta\sigma_{zz}=-\Delta p=-30\ \mathrm{MPa}}. The factor of three comes from volume geometry, ΔV/V≃ϵx+ϵy+ϵz=3ϵℓ,\Delta V/V\simeq\epsilon_x+\epsilon_y+\epsilon_z=3\epsilon_\ell, not from adding perpendicular stresses into one axial component.

Physical check

Increasing pressure decreases every side length and the volume, consistent with the negative strain signs.

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