Elastic modulus

The modulus of elasticity (also known as the elastic modulus, the tensile modulus, or Young's modulus) is a number that measures an object or substance's resistance to being deformed elastically (i.e., non-permanently) when a force is applied to it. The elastic modulus of an object is defined as the slope of its stress–strain curve in the elastic deformation region: A stiffer material will have a higher elastic modulus. An elastic modulus has the form.

$\lambda \ \stackrel{\text{def}}{=}\ \frac {\text{stress}} {\text{strain}}$

where stress is the force causing the deformation divided by the area to which the force is applied and strain is the ratio of the change in some length parameter caused by the deformation to the original value of the length parameter. If stress is measured in pascals, then since strain is a dimensionless quantity, the units of λ will be pascals as well. The antonym of Elasticity is "Compliance".

Since the strain equals unity for an object whose length has doubled, the elastic modulus equals the stress induced in the material by a doubling of length. While this scenario is not generally realistic because most materials will fail before reaching it, it gives heuristic guidance, because small fractions of the defining load will operate in exactly the same ratio. Thus, for steel with a Young's modulus of 30 million psi, a 30 thousand psi load will elongate a 1 inch bar by one thousandth of an inch; similarly, for metric units, a load of one-thousandth of the modulus (now measured in gigapascal) will change the length of a one-meter rod by a millimetre.

In a general description, since both stress and strain are described by second-rank tensors, including both stretch and shear components, the elasticity tensor is a fourth-rank tensor with up to 21 independent constants.

Specifying how stress and strain are to be measured, including directions, allows for many types of elastic moduli to be defined. The three primary ones are:

• Young's modulus (E) describes tensile elasticity, or the tendency of an object to deform along an axis when opposing forces are applied along that axis; it is defined as the ratio of tensile stress to tensile strain. It is often referred to simply as the elastic modulus.
• The shear modulus or modulus of rigidity (G or $\mu \,$) describes an object's tendency to shear (the deformation of shape at constant volume) when acted upon by opposing forces; it is defined as shear stress over shear strain. The shear modulus is part of the derivation of viscosity.
• The bulk modulus (K) describes volumetric elasticity, or the tendency of an object to deform in all directions when uniformly loaded in all directions; it is defined as volumetric stress over volumetric strain, and is the inverse of compressibility. The bulk modulus is an extension of Young's modulus to three dimensions.

Three other elastic moduli are Axial Modulus, Lamé's first parameter, and P-wave modulus.

Homogeneous and isotropic (similar in all directions) materials (solids) have their (linear) elastic properties fully described by two elastic moduli, and one may choose any pair. Given a pair of elastic moduli, all other elastic moduli can be calculated according to formulas in the table below at the end of page.

Inviscid fluids are special in that they cannot support shear stress, meaning that the shear modulus is always zero. This also implies that Young's modulus is always zero.

In some English texts the here described quantity is called elastic constant, while the inverse quantity is referred to as elastic modulus.

See also

• Bending stiffness
• Dynamic modulus
• Elastic limit
• Elastic wave
• Flexural modulus
• Hooke's Law
• Impulse excitation technique
• Proportional limit
• Stiffness
• Tensile strength
• Transverse isotropy

Further reading

Conversion formulas
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas.
	$K=\,$	$E=\,$	$\lambda=\,$	$G=\,$	$\nu=\,$	$M=\,$	Notes
$(K,\,E)$	$K$	$E$	$\tfrac{3K(3K-E)}{9K-E}$	$\tfrac{3KE}{9K-E}$	$\tfrac{3K-E}{6K}$	$\tfrac{3K(3K+E)}{9K-E}$
$(K,\,\lambda)$	$K$	$\tfrac{9K(K-\lambda)}{3K-\lambda}$	$\lambda$	$\tfrac{3(K-\lambda)}{2}$	$\tfrac{\lambda}{3K-\lambda}$	$3K-2\lambda\,$
$(K,\,G)$	$K$	$\tfrac{9KG}{3K+G}$	$K-\tfrac{2G}{3}$	$G$	$\tfrac{3K-2G}{2(3K+G)}$	$K+\tfrac{4G}{3}$
$(K,\,\nu)$	$K$	$3K(1-2\nu)\,$	$\tfrac{3K\nu}{1+\nu}$	$\tfrac{3K(1-2\nu)}{2(1+\nu)}$	$\nu$	$\tfrac{3K(1-\nu)}{1+\nu}$
$(K,\,M)$	$K$	$\tfrac{9K(M-K)}{3K+M}$	$\tfrac{3K-M}{2}$	$\tfrac{3(M-K)}{4}$	$\tfrac{3K-M}{3K+M}$	$M$
$(E,\,\lambda)$	$\tfrac{E + 3\lambda + R}{6}$	$E$	$\lambda$	$\tfrac{E-3\lambda+R}{4}$	$\tfrac{2\lambda}{E+\lambda+R}$	$\tfrac{E-\lambda+R}{2}$	$R=\sqrt{E^2+9\lambda^2 + 2E\lambda}$
$(E,\,G)$	$\tfrac{EG}{3(3G-E)}$	$E$	$\tfrac{G(E-2G)}{3G-E}$	$G$	$\tfrac{E}{2G}-1$	$\tfrac{G(4G-E)}{3G-E}$
$(E,\,\nu)$	$\tfrac{E}{3(1-2\nu)}$	$E$	$\tfrac{E\nu}{(1+\nu)(1-2\nu)}$	$\tfrac{E}{2(1+\nu)}$	$\nu$	$\tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)}$
$(E,\,M)$	$\tfrac{3M-E+S}{6}$	$E$	$\tfrac{M-E+S}{4}$	$\tfrac{3M+E-S}{8}$	$\tfrac{E-M+S}{4M}$	$M$	$S=\pm\sqrt{E^2+9M^2-10EM}$ There are two valid solutions. The plus sign leads to $\nu\geq 0$. The minus sign leads to $\nu\leq 0$.
$(\lambda,\,G)$	$\lambda+ \tfrac{2G}{3}$	$\tfrac{G(3\lambda + 2G)}{\lambda + G}$	$\lambda$	$G$	$\tfrac{\lambda}{2(\lambda + G)}$	$\lambda+2G\,$
$(\lambda,\,\nu)$	$\tfrac{\lambda(1+\nu)}{3\nu}$	$\tfrac{\lambda(1+\nu)(1-2\nu)}{\nu}$	$\lambda$	$\tfrac{\lambda(1-2\nu)}{2\nu}$	$\nu$	$\tfrac{\lambda(1-\nu)}{\nu}$	Cannot be used when $\nu=0 \Leftrightarrow \lambda=0$
$(\lambda,\,M)$	$\tfrac{M + 2\lambda}{3}$	$\tfrac{(M-\lambda)(M+2\lambda)}{M+\lambda}$	$\lambda$	$\tfrac{M-\lambda}{2}$	$\tfrac{\lambda}{M+\lambda}$	$M$
$(G,\,\nu)$	$\tfrac{2G(1+\nu)}{3(1-2\nu)}$	$2G(1+\nu)\,$	$\tfrac{2 G \nu}{1-2\nu}$	$G$	$\nu$	$\tfrac{2G(1-\nu)}{1-2\nu}$
$(G,\,M)$	$M - \tfrac{4G}{3}$	$\tfrac{G(3M-4G)}{M-G}$	$M - 2G\,$	$G$	$\tfrac{M - 2G}{2M - 2G}$	$M$
$(\nu,\,M)$	$\tfrac{M(1+\nu)}{3(1-\nu)}$	$\tfrac{M(1+\nu)(1-2\nu)}{1-\nu}$	$\tfrac{M \nu}{1-\nu}$	$\tfrac{M(1-2\nu)}{2(1-\nu)}$	$\nu$	$M$