Shear Modulus and Bulk Modulus

Young’s modulus tells you how a material resists being stretched or compressed along a single direction. But solids do not always get pulled lengthwise. Sometimes a force pushes sideways across the surface, twisting the shape without changing the volume. Other times, pressure squeezes a body uniformly from every direction, shrinking its volume without distorting its shape. Each of these situations needs its own elastic modulus. That is where the shear modulus and the bulk modulus come in.

Shear Modulus: Resistance to Shape Distortion

Picture a thick book lying on a table. If you press your palm against the top cover and push it sideways while the bottom cover stays put, the pages fan out and the book’s rectangular cross-section turns into a parallelogram. The book has been sheared: its shape changed, but its volume stayed roughly the same. Solids behave the same way under tangential forces.

The shear modulus (also called the modulus of rigidity) measures how strongly a material fights this kind of sideways distortion. It is represented by $G$ and defined as the ratio of shearing stress to the corresponding shearing strain.

The Formula Step by Step

Start from the definitions you already know. Shearing stress is the tangential force $F$ divided by the area $A$ of the face on which it acts. Shearing strain is the angle $\theta$ through which the body deforms (or equivalently, the lateral displacement $\Delta x$ divided by the height $L$ of the face perpendicular to the displacement).

Putting these together:

$G = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{\sigma_s}{\theta}$

Writing stress and strain in terms of force, area, displacement, and height:

$G = \frac{F/A}{\Delta x / L}$

Dividing by a fraction flips it, giving the expanded working form:

$G = \frac{F \times L}{A \times \Delta x} \qquad \text{(8.10)}$

Since for small deformations the ratio $\Delta x / L$ equals the angle $\theta$ (in radians), you can also write:

$G = \frac{F}{A \times \theta} \qquad \text{(8.11)}$

Rearranging to express shearing stress directly in terms of $G$ and $\theta$:

$\sigma_s = G \times \theta \qquad \text{(8.12)}$

This last form is handy when you know the modulus and want to find how much stress a given angular deformation produces.

Units

Strain ($\theta$) is dimensionless (it is an angle in radians), so the unit of $G$ is the same as the unit of stress:

$[G] = \text{N m}^{-2} = \text{Pa}$

Values are typically in the gigapascal range for metals and engineering solids.

How Shear Moduli Compare Across Materials

The table below lists shear moduli for several common materials. Notice the enormous range, from soft lead all the way up to tungsten.

Table 8.2: Shear moduli ($G$) of some common materials

Material	$G$ ($10^9$ N $\text{m}^{-2}$ or GPa)
Aluminium	25
Brass	36
Copper	42
Glass	23
Iron	70
Lead	5.6
Nickel	77
Steel	84
Tungsten	150
Wood	10

A useful rule of thumb: for most common materials, the shear modulus is roughly one-third of Young’s modulus, i.e., $G \approx Y/3$. You can verify this approximately using the Young’s modulus values from Table 8.1 in the previous topic. The pattern makes sense physically: shearing involves sliding atomic planes past each other, which is generally easier than pulling atoms apart along a bond direction, so $G$ comes out smaller than $Y$.

Worked Example 8.4: Shearing a Lead Slab

Problem: A square lead slab has a side of 50 cm and a thickness of 10 cm. Its lower edge is riveted to the floor. A shearing force of $9.0 \times 10^4$ N is applied on its narrow face (the 50 cm $\times$ 10 cm face), parallel to the long side. By how much does the upper edge get displaced?

Fig 8.5: Lead slab fixed at the bottom and sheared by a horizontal force on its narrow face

Solution:

The force acts parallel to the narrow face. The area of that face is:

$A = 50 \text{ cm} \times 10 \text{ cm} = 0.5 \text{ m} \times 0.1 \text{ m} = 0.05 \text{ m}^2$

Calculate the shearing stress:

$\sigma_s = \frac{F}{A} = \frac{9.0 \times 10^4 \text{ N}}{0.05 \text{ m}^2} = 1.80 \times 10^6 \text{ N m}^{-2}$

The shearing strain equals $\Delta x / L$. Rearranging the strain formula in terms of stress and modulus:

$\frac{\Delta x}{L} = \frac{\text{stress}}{G}$

Now solve for the displacement $\Delta x$. The height $L$ over which the deformation occurs is the side of the slab (50 cm = 0.5 m), and the shear modulus of lead from Table 8.2 is $5.6 \times 10^9$ N $\text{m}^{-2}$:

$\Delta x = \frac{\text{stress} \times L}{G} = \frac{1.80 \times 10^6 \times 0.5}{5.6 \times 10^9}$

$= \frac{0.90 \times 10^6}{5.6 \times 10^9}$

$= 1.6 \times 10^{-4} \text{ m} = 0.16 \text{ mm}$

The upper edge shifts by just 0.16 mm. Even lead, which is among the softest common metals, deforms only a fraction of a millimetre under this substantial force. Stiffer materials like steel or tungsten would show even smaller displacements for the same load.

Bulk Modulus: Resistance to Volume Change

Now consider a completely different scenario. Instead of a force pushing sideways on one face, imagine pressure acting uniformly from every direction on the entire surface of the body. This is exactly what happens to an object submerged in a fluid: the fluid pushes inward on every square centimetre of the object’s surface with equal pressure. This uniform inward pressure is hydraulic stress (equal in magnitude to the fluid pressure), and it compresses the body, reducing its volume without changing its shape.

The bulk modulus quantifies how well a material resists this kind of uniform compression. It is represented by $B$ and defined as the ratio of hydraulic stress (pressure $p$) to the resulting volume strain ($\Delta V / V$):

$B = \frac{-p}{\Delta V / V} \qquad \text{(8.13)}$

Why the Negative Sign?

When you increase the external pressure ($p$ is positive), the body shrinks ($\Delta V$ is negative). The ratio $p / (\Delta V / V)$ would come out negative. But a modulus should be a positive number telling you “how stiff” the material is. The negative sign in front cancels the negative $\Delta V$, keeping $B$ positive for every real material in equilibrium.

Units

Since volume strain ($\Delta V / V$) is a dimensionless ratio, the bulk modulus carries the same unit as pressure:

$[B] = \text{N m}^{-2} = \text{Pa}$

Compressibility: The Flip Side of Bulk Modulus

Sometimes it is more natural to talk about how easy it is to compress something rather than how hard. The quantity that captures this is compressibility, denoted $k$. It is simply the reciprocal of the bulk modulus:

$k = \frac{1}{B} = -\frac{1}{\Delta p} \times \frac{\Delta V}{V} \qquad \text{(8.14)}$

Compressibility tells you the fractional change in volume you get for each unit increase in pressure. A large $k$ means the material gives way easily; a small $k$ means it holds its ground.

Bulk Moduli Across Solids, Liquids, and Gases

Unlike Young’s modulus and shear modulus (which apply only to solids), the bulk modulus is defined for all three states of matter, because anything with a volume, solid, liquid, or gas, can be squeezed.

Table 8.3: Bulk moduli ($B$) of some common materials

Material	$B$ ($10^9$ N $\text{m}^{-2}$ or GPa)
Solids
Aluminium	72
Brass	61
Copper	140
Glass	37
Iron	100
Nickel	260
Steel	160
Liquids
Water	2.2
Ethanol	0.9
Carbon disulphide	1.56
Glycerine	4.76
Mercury	25
Gases
Air (at STP)	$1.0 \times 10^{-4}$

The pattern across the three columns is dramatic:

• Solids have bulk moduli in the range of tens to hundreds of GPa. Nickel leads at 260 GPa, followed by steel (160 GPa) and copper (140 GPa). These materials barely budge in volume even under enormous pressure. The atoms in a solid are locked tightly together by strong interatomic bonds, leaving almost no room for further compression.
• Liquids drop by a factor of roughly 10 to 100 compared to solids. Water sits at 2.2 GPa and mercury at 25 GPa. Liquid molecules are close together but not rigidly bonded, so they can be squeezed a bit more than solids.
• Gases plummet to an entirely different scale. Air at standard conditions has a bulk modulus of only $1.0 \times 10^{-4}$ GPa, which is about a million times smaller than that of a typical solid. Gas molecules are far apart with weak interactions, leaving plenty of empty space to compress into.

The molecular picture makes this gradient intuitive. Tight atomic coupling in solids resists volume change most; moderate coupling in liquids resists it less; and the near-absence of coupling in gases means they compress readily. This is why solids are called essentially incompressible in everyday engineering, while gases are highly compressible and their compressibility varies with both pressure and temperature.

Worked Example 8.5: Compression at the Ocean Floor

Problem: The average depth of the Indian Ocean is about 3000 m. Calculate the fractional compression $\Delta V / V$ of water at the bottom, given that the bulk modulus of water is $2.2 \times 10^9$ N $\text{m}^{-2}$. (Take $g = 10$ m $\text{s}^{-2}$.)

Solution:

First, find the pressure exerted by the 3000 m water column above the bottom layer. Use the hydrostatic pressure formula $p = h \rho g$, where $h$ is the depth, $\rho$ is the density of water ($1000 \text{ kg m}^{-3}$), and $g$ is the acceleration due to gravity:

$p = 3000 \text{ m} \times 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2}$

$= 3 \times 10^7 \text{ N m}^{-2}$

That is 30 million pascals, roughly 300 times atmospheric pressure, a colossal squeeze.

Now apply the bulk modulus definition. Rearranging $B = -p / (\Delta V / V)$ to solve for the fractional compression:

$\frac{\Delta V}{V} = \frac{p}{B}$

(The magnitude is taken; the negative sign just tells us the volume decreases.)

$\frac{\Delta V}{V} = \frac{3 \times 10^7}{2.2 \times 10^9}$

Divide numerator and denominator:

$= 1.36 \times 10^{-2} = 1.36\%$

Even under the weight of a 3 km column of water, the fractional volume reduction is only about 1.4%. Water is remarkably incompressible. This is one reason hydraulic systems (brakes, presses, jacks) use liquids as the working fluid: the liquid transmits pressure almost without losing volume.

Putting It All Together: Three Moduli at a Glance

You have now met all three elastic moduli. Each one handles a different kind of deformation. The summary table below lines them up side by side so you can see the pattern clearly.

Table 8.4: Stress, strain, and the three elastic moduli compared

Property	Young’s Modulus ($Y$)	Shear Modulus ($G$)	Bulk Modulus ($B$)
Type of stress	Tensile or compressive (two equal and opposite forces perpendicular to opposite faces)	Shearing (two equal and opposite forces parallel to opposite faces, with zero net force and zero net torque)	Hydraulic (uniform pressure perpendicular to every part of the surface)
Type of strain	Longitudinal strain $\Delta L / L$ (change in length along the force direction)	Shear angle $\theta$ (pure shape distortion)	Volume strain $\Delta V / V$ (change in volume)
Changes shape?	Yes	Yes	No
Changes volume?	No	No	Yes
Formula	$Y = \frac{F \times L}{A \times \Delta L}$	$G = \frac{F}{A \times \theta}$	$B = \frac{-p}{\Delta V / V}$
Applies to	Solids only	Solids only	Solids, liquids, and gases

The key insight from this table is that shape change and volume change are independent. Tensile/compressive stress and shearing stress alter the shape but not the volume. Hydraulic stress alters the volume but not the shape. Real-world loading can combine all three, which is why engineers need all three moduli to fully describe a material’s elastic behaviour.

Notice also the last row. Young’s modulus and shear modulus are meaningful only for solids, because fluids cannot maintain a definite shape and therefore cannot sustain tensile or shearing stress. Bulk modulus, however, applies to everything, any substance that occupies a volume responds to uniform pressure.