Dimensional Formula of Stress: [M L⁻¹ T⁻²] with Derivation

The dimensional formula of stress is [M¹ L⁻¹ T⁻²]; in compact form, [ML⁻¹T⁻²]. Stress measures the restoring force that builds up inside a stretched, squeezed or sheared body, shared over each unit of area, so it carries the dimensions of force divided by area.

Force supplies [MLT⁻²] and area supplies [L²]; dividing turns the single power of length into L⁻¹. Its SI unit is the pascal, one newton per square metre, which is why stress, pressure and every elastic modulus look identical when written as brackets. The stress formula guide covers tensile, compressive and shear cases with numbers.

Quick facts

Dimensional formula
[M¹ L⁻¹ T⁻²]
Defining formula
$\sigma = \frac{F}{A}$
SI unit
pascal (Pa) = N m⁻² = kg m⁻¹ s⁻²
CGS unit
dyne/cm², with 1 Pa = 10 dyne/cm²
Kinds
Tensile, compressive, shearing, hydraulic
Same bracket
Pressure, Young's modulus, bulk modulus, energy density

Finding the dimensional formula of stress from σ = F/A

Stress is restoring force over cross sectional area: $\sigma = \frac{F}{A}$ .
Force, from Newton's second law: $[F] = [MLT^{-2}]$ .
Area is length squared: $[A] = [L^{2}]$ .
Divide, subtracting the length powers 1 − 2: $[\sigma] = [ML^{1-2}T^{-2}] = [ML^{-1}T^{-2}]$ .

Notice the length power. Force has L to the plus one, but dividing by an area pulls it down to minus one. Students who stop at [MLT⁻²] have written the bracket of force, not stress, and lose the mark straight away.

Pascal, newton per square metre and the CGS unit

Write the bracket in SI base units: kg m⁻¹ s⁻². Group kg m s⁻² as a newton and the leftover m⁻² stays, giving N/m², named the pascal. Engineers quote stress in MPa (10⁶ Pa) or GPa (10⁹ Pa), since everyday materials take huge stresses.

For CGS, reuse the exponents from the bracket. Mass appears once on top, so bring in 1000 g; length appears once below, so divide by 100 cm; time stays as seconds. One pascal becomes 1000 ÷ 100 = 10 g cm⁻¹ s⁻², and that equals 10 dyne/cm².

Example: a steel wire of diameter 3.0 mm carries a hanging load of 500 N. Find the tensile stress.

Radius is half the diameter, 1.5 mm, i.e. 1.5 × 10⁻³ m. Area: $A = \pi r^{2} = \pi (2.25 \times 10^{-6}) = 7.07 \times 10^{-6}\ \text{m}^{2}$ .

$\sigma = \frac{500}{7.07 \times 10^{-6}} = 7.07 \times 10^{7}\ \text{Pa}$ , roughly 71 MPa.

Halve the diameter before squaring. Using 3.0 mm as the radius makes the answer four times too small.

Quantities that carry the bracket of stress

Pressure, moduli and energy density share this bracket
Quantity	Defining relation	Reason the bracket agrees
Pressure	$P = F/A$	Normal force over area
Young's modulus	$Y = \sigma / \varepsilon$	Strain has no dimension
Bulk modulus	$B = -\Delta P / (\Delta V/V)$	Volume ratio has no dimension
Modulus of rigidity	Shear stress over shear strain	Angle has no dimension
Energy density	Energy per volume	[ML²T⁻²] over [L³]

Every elastic modulus is stress divided by a strain, and every strain is a ratio of two like quantities. That is why the moduli share the bracket of stress. Strain itself is covered under the strain formula.

A quick consistency test

Example: is $Y = \frac{FL}{A\,\Delta L}$ dimensionally sound?

$\frac{[MLT^{-2}][L]}{[L^{2}][L]} = [ML^{2-3}T^{-2}] = [ML^{-1}T^{-2}]$ . The two lengths, original and extension, cancel each other, leaving force over area. The formula passes.

Example: elastic energy stored per unit volume is claimed to be $\frac{1}{2}\,\text{stress} \times \text{strain}$ . Check it.

Stress is [ML⁻¹T⁻²] and strain has no dimension, so the product is [ML⁻¹T⁻²]. Energy per volume is $\frac{[ML^{2}T^{-2}]}{[L^{3}]}$ , the same thing. The claim is consistent.

Stress bracket mistakes to watch for

Giving the bracket of force. Stress divides by area, so L drops to minus one.
Giving stress a dimension for strain. Strain is change in length over length and has none, so stress and modulus share a bracket.
Calling stress and pressure identical ideas. They share dimensions, but stress is an internal restoring response while pressure is applied by a fluid or a surface.
Forgetting the minus on time. The power of T stays at minus two, inherited from force.

To compare stress with tension, torque and the other mechanics brackets, look through the dimensional formulas collection grouped by chapter.

Frequently asked questions

What is the dimensional formula of stress?

Stress is force over area, so its bracket is [M¹ L⁻¹ T⁻²]. Force gives [MLT⁻²] and area gives [L²].

Do stress and pressure have the same dimensional formula?

They share [ML⁻¹T⁻²] and the unit pascal. Physically they differ: pressure is applied from outside, often by a fluid, while stress is the push back that develops within the material, per square metre.

Which unit is used for stress in SI?

The pascal: a newton spread across a square metre. Big stresses in metals and concrete are usually quoted in megapascals or gigapascals.

Why do stress and Young's modulus share a bracket?

Young's modulus is stress divided by strain. Strain is a ratio of two lengths with no dimension, so dividing by it leaves the dimensions unchanged.

Is stress a scalar quantity?

At school level stress is treated through its size and type, tensile or compressive. In advanced physics it is a tensor, a step beyond vectors, but its bracket stays [ML⁻¹T⁻²] either way.