AcademyEquilibrium and Materials

Academy

Elastic Moduli

Level 1 - Physics topic page in Equilibrium and Materials.

Principle

An elastic modulus is stiffness for one deformation mode.

Notation

\(Y\)
Young modulus
\(\mathrm{Pa}\)
\(G\)
shear modulus
\(\mathrm{Pa}\)
\(B\)
bulk modulus
\(\mathrm{Pa}\)
\(\sigma,\epsilon,\tau,\gamma\)
normal and shear stress-strain variables
Pa, none
\(F,A,\Delta L,L_0\)
axial load, area, extension, and original length
\(\mathrm{N,\;m^{2},\;m,\;m}\)
\(\Delta p,\Delta V/V\)
pressure change and fractional volume change
Pa, none

Method

Derivation 1: Build Young modulus from normal stress and strain

Young modulus compares axial stress with axial strain in the linear elastic region.

000000100200300400500strainstresslarge Ysmall Y
A larger modulus gives a steeper elastic stress-strain relation.
Young modulus
\[Y=\frac{\sigma}{\epsilon}\]
Substitute axial definitions
\[Y=\frac{F/A}{\Delta L/L_0}=\frac{FL_0}{A\Delta L}\]
Rearrange for force
\[F=\frac{YA}{L_0}\Delta L\]
Axial stiffness
\[k_{\mathrm{axial}}=\frac{YA}{L_0}\]
Extension
\[\Delta L=\frac{FL_0}{YA}\]

Derivation 2: Build shear modulus for sideways deformation

Shear modulus uses tangential stress and the resulting angular distortion.

Shear modulus
\[G=\frac{\tau}{\gamma}\]

Derivation 3: Build bulk modulus for compression

Bulk modulus compares pressure change with fractional volume change. The minus sign keeps the modulus positive during compression.

Bulk modulus
\[B=-\frac{\Delta p}{\Delta V/V}\]

Rules

These are the compact results from the derivations above.

Young modulus
\[Y=\frac{\sigma}{\epsilon}=\frac{FL_0}{A\Delta L}\]
Shear modulus
\[G=\frac{\tau}{\gamma}\]
Bulk modulus
\[B=-\frac{\Delta p}{\Delta V/V}\]
Axial stiffness
\[k_{\mathrm{axial}}=\frac{YA}{L_0}\]

Examples

Question
A wire has
\[Y=2.0\times10^{11}\,\mathrm{Pa}\]
\[L=1.5\,\mathrm{m}\]
\[A=2.0\times10^{-6}\,\mathrm{m^2}\]
and load
\[400\,\mathrm{N}\]
Find extension.
Answer
\[\Delta L=\frac{FL}{YA}=\frac{400(1.5)}{(2.0\times10^{11})(2.0\times10^{-6})}=1.5\times10^{-3}\,\mathrm{m}\]

Checks

  • Choose the correct deformation mode.
  • Larger modulus means less strain.
  • Bulk modulus uses volume change.
  • Parallel members share extension.