Elastic Moduli and Young’s Modulus

You now know that stress and strain are proportional for small deformations, and that this proportionality is Hooke’s law. But here is the really useful part: the constant that links stress to strain is not just some abstract number. It is a property of the material itself. Measure it once for steel, and you can predict how any steel component will deform under any load, without testing that particular component. This constant is the modulus of elasticity (a number that tells you how stiff or rigid a material is). The higher the modulus, the harder it is to deform the material. The lower the modulus, the more easily it gives way.

Different types of deformation, stretching, shearing, or squeezing from all sides, each have their own modulus. In this topic, we focus on the most commonly used one: Young’s modulus, which deals with stretching and compressing along a single direction.

What the Modulus of Elasticity Tells You

Recall from the previous topic that the stress-strain curve has a straight-line region at the beginning (from O to A in Fig. 8.2), where Hooke’s law holds. Within this region, the ratio of stress to strain stays constant. That constant ratio is the modulus of elasticity.

Every material has its own characteristic value of this ratio. Steel has a very high value, meaning you need enormous stress to produce even a tiny strain. Rubber has a very low value, meaning a small stress causes a large strain. This single number lets engineers compare materials at a glance and decide which one is best suited for a given job.

Young’s Modulus: Measuring Resistance to Stretching

When a solid is pulled (tension) or pushed (compression) along its length, the relevant modulus is named after Thomas Young. It connects the tensile (or compressive) stress to the longitudinal strain.

Experiments show an important symmetry: for a given material, the strain produced has the same magnitude regardless of whether the stress is tensile or compressive. Pulling or pushing gives the same stiffness number. This means a single value of Young’s modulus covers both situations.

The definition is straightforward:

$Y = \frac{\sigma}{\varepsilon} \qquad \text{(8.7)}$

where $\sigma$ is the tensile or compressive stress and $\varepsilon$ is the longitudinal strain.

Writing stress and strain in terms of the basic quantities (force $F$, original length $L$, cross-sectional area $A$, and change in length $\Delta L$):

$\sigma = \frac{F}{A}, \qquad \varepsilon = \frac{\Delta L}{L}$

Substituting into the definition:

$Y = \frac{F / A}{\Delta L / L}$

Flipping the denominator (dividing by a fraction is the same as multiplying by its inverse):

$Y = \frac{F \times L}{A \times \Delta L} \qquad \text{(8.8)}$

This expanded form is extremely useful in numerical problems. If you know any four of the five quantities ($Y$, $F$, $L$, $A$, $\Delta L$), you can find the fifth.

Units and Dimensions

Strain ($\Delta L / L$) is a pure ratio, so it has no unit. Since Young’s modulus equals stress divided by a dimensionless number, its unit is the same as the unit of stress:

$[Y] = \text{N m}^{-2} = \text{Pa (Pascal)}$

Values for real materials are typically in the range of $10^9$ to $10^{11}$ Pa, so they are often quoted in gigapascals (GPa), where $1 \text{ GPa} = 10^9 \text{ Pa}$.

Comparing Materials: The Young’s Modulus Table

The table below lists Young’s moduli, ultimate strengths, and yield strengths for several common materials. Looking at these numbers side by side reveals clear patterns.

Table 8.1: Young’s moduli and yield strengths of some materials

Material	Density $\rho$ ($\text{kg m}^{-3}$)	Young’s modulus $Y$ ($10^9$ $\text{N m}^{-2}$)	Ultimate strength $\sigma_u$ ($10^6$ $\text{N m}^{-2}$)	Yield strength $\sigma_y$ ($10^6$ $\text{N m}^{-2}$)
Aluminium	2710	70	110	95
Copper	8890	110	400	200
Iron (wrought)	7800 - 7900	190	330	170
Steel	7860	200	400	250
Glass*	2190	65	50	—
Concrete	2320	30	40	—
Wood*	525	13	50	—
Bone*	1900	9.4	170	—
Polystyrene	1050	3	48	—

*Tested under compression

What the Numbers Reveal

A few things stand out immediately:

• Metals dominate the top of the modulus scale. Steel leads at $200 \times 10^9$ $\text{N m}^{-2}$, followed by wrought iron ($190 \times 10^9$) and copper ($110 \times 10^9$). These materials resist stretching very strongly.
• Non-metals sit much lower. Glass ($65 \times 10^9$), concrete ($30 \times 10^9$), wood ($13 \times 10^9$), bone ($9.4 \times 10^9$), and polystyrene ($3 \times 10^9$) are all far easier to deform.
• Steel is the stiffest common engineering metal. To stretch a thin steel wire of cross-sectional area $0.1 \text{ cm}^2$ by just 0.1% of its length, you need a force of about 2000 N. For the same wire dimensions and the same 0.1% stretch, aluminium needs only about 690 N, brass about 900 N, and copper about 1100 N.

This is exactly why steel is the go-to material for heavy-duty machines, bridges, railway tracks, and building frameworks. It deforms the least under a given load, keeping structures rigid and predictable.

Worked Examples

Example 8.1: Stress, Elongation, and Strain in a Steel Rod

Problem: A structural steel rod has a radius of 10 mm and a length of 1.0 m. A force of 100 kN stretches it along its length. Calculate (a) the stress, (b) the elongation, and (c) the strain on the rod. Young’s modulus of structural steel is $2.0 \times 10^{11}$ $\text{N m}^{-2}$.

Solution:

The rod is clamped at one end and the force $F$ is applied at the other end, parallel to the rod’s length.

(a) Finding the stress

Stress is force per unit cross-sectional area. The cross-section is a circle of radius $r = 10 \text{ mm} = 10^{-2} \text{ m}$:

$\text{Stress} = \frac{F}{A} = \frac{F}{\pi r^2}$

Substituting the values:

$= \frac{100 \times 10^3 \text{ N}}{3.14 \times (10^{-2})^2 \text{ m}^2}$

$= \frac{10^5}{3.14 \times 10^{-4}}$

$= 3.18 \times 10^8 \text{ N m}^{-2}$

(b) Finding the elongation

Rearranging Eq. (8.8) to isolate $\Delta L$:

$\Delta L = \frac{(F/A) \times L}{Y} = \frac{\sigma \times L}{Y}$

Substituting:

$= \frac{3.18 \times 10^8 \text{ N m}^{-2} \times 1 \text{ m}}{2.0 \times 10^{11} \text{ N m}^{-2}}$

$= 1.59 \times 10^{-3} \text{ m}$

$= 1.59 \text{ mm}$

So a 100 kN force stretches a 1 m steel rod by only about 1.6 mm, reflecting steel’s very high stiffness.

(c) Finding the strain

Strain is the fractional change in length:

$\text{Strain} = \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3} \text{ m}}{1 \text{ m}} = 1.59 \times 10^{-3} = 0.16\%$

A strain of just 0.16% confirms that the rod is well within its elastic range.

Example 8.2: Load on a Copper-Steel Composite Wire

Problem: A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the total elongation is 0.70 mm. Find the applied load.

Solution:

Since the two wires are connected in series, the same tension (equal to the applied load $W$) passes through both. They also share the same cross-sectional area $A$ because their diameters are identical.

Step 1: Relate the individual elongations

For each wire, stress equals Young’s modulus times strain:

$\frac{W}{A} = Y_c \times \frac{\Delta L_c}{L_c} = Y_s \times \frac{\Delta L_s}{L_s}$

where the subscripts $c$ and $s$ stand for copper and steel. Rearranging to find the ratio of elongations:

$\frac{\Delta L_c}{\Delta L_s} = \frac{Y_s}{Y_c} \times \frac{L_c}{L_s}$

Step 2: Substitute the known values

From Table 8.1: $Y_c = 1.1 \times 10^{11}$ $\text{N m}^{-2}$, $Y_s = 2.0 \times 10^{11}$ $\text{N m}^{-2}$, $L_c = 2.2 \text{ m}$, $L_s = 1.6 \text{ m}$.

$\frac{\Delta L_c}{\Delta L_s} = \frac{2.0 \times 10^{11}}{1.1 \times 10^{11}} \times \frac{2.2}{1.6}$

The $10^{11}$ factors cancel:

$= \frac{2.0}{1.1} \times \frac{2.2}{1.6} = 1.818 \times 1.375 = 2.5$

So the copper wire stretches 2.5 times as much as the steel wire. This makes sense: copper has a lower modulus (less stiff) and the copper wire is also longer.

Step 3: Solve for individual elongations

The total elongation is:

$\Delta L_c + \Delta L_s = 7.0 \times 10^{-4} \text{ m}$

Substituting $\Delta L_c = 2.5 \, \Delta L_s$:

$2.5 \, \Delta L_s + \Delta L_s = 7.0 \times 10^{-4}$

$3.5 \, \Delta L_s = 7.0 \times 10^{-4}$

$\Delta L_s = 2.0 \times 10^{-4} \text{ m} = 0.20 \text{ mm}$

$\Delta L_c = 5.0 \times 10^{-4} \text{ m} = 0.50 \text{ mm}$

Step 4: Calculate the load

Using the copper wire’s data (either wire gives the same answer):

$W = \frac{A \times Y_c \times \Delta L_c}{L_c}$

The cross-sectional area for a wire of radius $r = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}$:

$A = \pi r^2 = 3.14 \times (1.5 \times 10^{-3})^2 = 7.07 \times 10^{-6} \text{ m}^2$

Substituting:

$W = \frac{7.07 \times 10^{-6} \times 1.1 \times 10^{11} \times 5.0 \times 10^{-4}}{2.2}$

$= \frac{7.07 \times 10^{-6} \times 5.5 \times 10^{7}}{2.2}$

$= \frac{3.89 \times 10^2}{2.2}$

$\approx 1.8 \times 10^2 \text{ N}$

The applied load is approximately 180 N.

Example 8.3: Thighbone Compression in a Human Pyramid

Problem: In a circus act, a human pyramid is balanced entirely on the legs of a performer lying on his back (see Fig. 8.4). The total mass of all performers, tables, and plaques is 280 kg. The bottom performer’s mass is 60 kg. Each of his thighbones (femur) has a length of 50 cm and an effective radius of 2.0 cm. Find the compression in each thighbone.

Fig 8.4: Human pyramid in a circus

Solution:

Step 1: Find the mass supported by the legs

The bottom performer’s own weight is supported by the ground (his back is on the ground). His legs only support the weight of everyone above him:

$\text{Supported mass} = 280 - 60 = 220 \text{ kg}$

Step 2: Find the force on each thighbone

The weight supported by both legs together:

$W = 220 \times 9.8 = 2156 \text{ N}$

Since the load is shared equally between two thighbones:

$F = \frac{2156}{2} = 1078 \text{ N}$

Step 3: Calculate the cross-sectional area

The thighbone radius is $r = 2.0 \text{ cm} = 2.0 \times 10^{-2} \text{ m}$:

$A = \pi r^2 = 3.14 \times (2.0 \times 10^{-2})^2 = 1.26 \times 10^{-3} \text{ m}^2$

Step 4: Compute the compression

From Table 8.1, the Young’s modulus for bone is $Y = 9.4 \times 10^9$ $\text{N m}^{-2}$. The length of each thighbone is $L = 0.5 \text{ m}$.

Using $\Delta L = \frac{F \times L}{Y \times A}$:

$\Delta L = \frac{1078 \times 0.5}{9.4 \times 10^9 \times 1.26 \times 10^{-3}}$

$= \frac{539}{1.184 \times 10^7}$

$= 4.55 \times 10^{-5} \text{ m}$

That is about $4.55 \times 10^{-3}$ cm, an extremely tiny compression. The fractional change is:

$\frac{\Delta L}{L} = \frac{4.55 \times 10^{-5}}{0.5} = 9.1 \times 10^{-5} = 0.0091\%$

Even under the full weight of an entire human pyramid, each thighbone compresses by less than one-hundredth of a percent of its length. Bone may have a modest Young’s modulus compared to steel, but its generous cross-sectional area keeps the actual deformation negligible for everyday loads.