Young Modulus | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Young Modulus

Stress tells us what is applied to the material. Strain tells us how much the material changes in response.

The Young modulus compares the two, so it measures the stiffness of the material, not one particular specimen.

Compare response with cause

For the same stress, a stiffer material produces less strain. Its ratio of stress to strain is therefore larger. This ratio is constant only in the linear region where stress is proportional to strain.

Definition

1 mark

What is meant by the Young modulus of a material?

Model answer: The ratio of tensile stress to tensile strain.

E=\frac{\sigma}{\varepsilon}

Symbols

$E$ = Young modulus (Pa)
$\sigma$ = tensile stress (Pa)
$\varepsilon$ = tensile strain (no unit)

A larger Young modulus means a stiffer material.
Young modulus has unit Pa because strain has no unit.
Use points from the straight proportional region only.

Swipe left or right to see the whole diagram →

In the linear region, the gradient is the Young modulus. Using a point far from the origin reduces percentage reading uncertainty.

Worked example

Read the Young modulus from a graph

A point on the straight line has stress 140 MPa and strain $2.0\times10^{-3}$ .

Convert 140 MPa to 140 × 10⁶ Pa.
$E=\sigma/\varepsilon=(140\times10^6)/(2.0\times10^{-3})$ .

Answer

E=7.0\times10^{10}\text{ Pa}=70\text{ GPa}

Material stiffness and specimen stiffness are different

The Young modulus belongs to the material. The spring constant belongs to a specimen with a particular length and area. Substituting stress and strain into the Young-modulus relation gives the link between them.

E=\frac{F/A}{x/L}=\frac{FL}{Ax}

Symbols

$F$ = tensile force (N)
$L$ = original length (m)
$A$ = cross-sectional area (m²)
$x$ = extension (m)

Within the proportional region, the specimen also obeys $k=F/x$ . Replacing $F/x$ by $k$ gives $E=kL/A$ ; rearranging gives the specimen stiffness below.

k=\frac{EA}{L}

Symbols

$k$ = stiffness of this wire (N m⁻¹)
$E$ = stiffness of its material (Pa)
$A$ = cross-sectional area (m²)
$L$ = original length (m)

For the same material and length, a thicker wire has a larger spring constant.
For the same material and area, a longer wire has a smaller spring constant.
Changing the wire dimensions changes $k$ , but not $E$ .

Your turn9702/23/M/J/24 Q3(c)

A wire has spring constant 90 N m^-1, original length 3.2 m and area $4.0\times10^{-7}\text{ m}^2$ . Determine its Young modulus. (9702/23/M/J/24 Q3(c))

Show worked answer

Use $E=kL/A$ .
$E=(90\times3.2)/(4.0\times10^{-7})=7.2\times10^8\text{ Pa}$ .

Answer

7.2\times10^8\text{ Pa}

Young Modulus

Stress tells us what is applied to the material. Strain tells us how much the material changes in response.

The Young modulus compares the two, so it measures the stiffness of the material, not one particular specimen.

Compare response with cause

Definition

1 mark

What is meant by the Young modulus of a material?

Model answer: The ratio of tensile stress to tensile strain.

E=\frac{\sigma}{\varepsilon}

Symbols

$E$ = Young modulus (Pa)
$\sigma$ = tensile stress (Pa)
$\varepsilon$ = tensile strain (no unit)

A larger Young modulus means a stiffer material.
Young modulus has unit Pa because strain has no unit.
Use points from the straight proportional region only.

Swipe left or right to see the whole diagram →

In the linear region, the gradient is the Young modulus. Using a point far from the origin reduces percentage reading uncertainty.

Worked example

Read the Young modulus from a graph

A point on the straight line has stress 140 MPa and strain $2.0\times10^{-3}$ .

Convert 140 MPa to 140 × 10⁶ Pa.
$E=\sigma/\varepsilon=(140\times10^6)/(2.0\times10^{-3})$ .

Answer

E=7.0\times10^{10}\text{ Pa}=70\text{ GPa}

Material stiffness and specimen stiffness are different

E=\frac{F/A}{x/L}=\frac{FL}{Ax}

Symbols

$F$ = tensile force (N)
$L$ = original length (m)
$A$ = cross-sectional area (m²)
$x$ = extension (m)

Within the proportional region, the specimen also obeys $k=F/x$ . Replacing $F/x$ by $k$ gives $E=kL/A$ ; rearranging gives the specimen stiffness below.

k=\frac{EA}{L}

Symbols

$k$ = stiffness of this wire (N m⁻¹)
$E$ = stiffness of its material (Pa)
$A$ = cross-sectional area (m²)
$L$ = original length (m)

For the same material and length, a thicker wire has a larger spring constant.
For the same material and area, a longer wire has a smaller spring constant.
Changing the wire dimensions changes $k$ , but not $E$ .

Your turn9702/23/M/J/24 Q3(c)

A wire has spring constant 90 N m^-1, original length 3.2 m and area $4.0\times10^{-7}\text{ m}^2$ . Determine its Young modulus. (9702/23/M/J/24 Q3(c))

Show worked answer

Use $E=kL/A$ .
$E=(90\times3.2)/(4.0\times10^{-7})=7.2\times10^8\text{ Pa}$ .

Answer

7.2\times10^8\text{ Pa}