Materials Science · Units, loads, stress and strain

Density and properties relative to mass

Calculate mass from a clearly defined volume, explain what strength/density and modulus/density mean, and compare the same two materials under different requirements for a simple tensile member.

Useful background: Materials, questions and levels of evidence. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Calculate density and mass with consistent cubic-volume units.
Distinguish a porous body’s bulk density from the density of its solid constituent.
Explain the dimensions and conditional usefulness of specific strength and specific modulus.
Compare tensile members under a stress limit and an extension limit, stating which dimensions may change.

Mass for a given volume

Two blocks can occupy the same space and have different masses. Density describes that relationship: ρ = m/V, where ρ is density, m mass and V the volume used in the measurement. For a body with known average density, rearrange to m = ρV. Doubling its volume at the same density doubles its mass; it does not double the density.

The SI unit is kg/m³. A common laboratory unit is g/cm³: 1 g = 0.001 kg and 1 cm³ = (0.01 m)³ = 0.000001 m³, so 1 g/cm³ = 1000 kg/m³. The cubic conversion matters. A centimetre-to-metre factor of 100 does not become the same factor for volume.

Mass is not a force. At a specified gravitational acceleration g, weight is the force mg. Comparing masses at the same location also compares their weights, but a density in kg/m³ is not a weight per volume in N/m³. Retain the material state and temperature when the volume or mass can change.

Which volume belongs in the denominator?

For a dry foam, distinguish the total outside volume, including pores, from the volume occupied by solid. Here bulk density means mass divided by that total volume. The solid-constituent density uses only the solid volume. These are different measurements of the same porous object, not conflicting answers.

Let V_total be total volume and V_s the solid part. If there is one solid constituent and the pore gas mass is negligible, m ≈ ρ_s V_s, where ρ_s is its density. Therefore ρ_bulk ≈ ρ_s(V_s/V_total). The solid volume fraction V_s/V_total is 1 minus the pore volume fraction. Define those fractions by volume, not by mass.

This relation is a mass account. It does not establish a foam’s modulus or strength by multiplying the solid property by the same fraction. Cell shape, connections and deformation of the solid walls also matter. If pores contain significant liquid or another phase, include that mass; a dry-density estimate cannot simply be reused.

Dividing a property by density

Specific strength is a stated strength divided by density; specify whether that strength is a yield, proof, maximum tensile or other defined endpoint. Specific modulus is Young’s modulus E divided by density. These are material ratios, not the strength or stiffness of a finished component divided by whatever it happens to weigh.

Their dimensions follow directly: Pa/(kg/m³) = (N/m²)(m³/kg) = N·m/kg. Both ratios have these units, although one describes a strength endpoint and the other an elastic response. Matching dimensions do not make them the same quantity, and N·m/kg here is not a claim about energy stored before failure.

Why divide by density at all? When a component’s cross-section can grow to meet a load or deformation requirement, a lower-density material can use more volume for a given mass. The useful question is how much area each candidate needs, then how much that area weighs. A ratio becomes informative only after that question is specified.

A setting where a specific property is useful

Consider a straight uniform tie: a member pulled along its length. Its original length L and applied tensile force F are fixed, while its original cross-sectional area A may be chosen. Ignore end fittings and self-weight, and use average axial engineering stress F/A away from grips. The member’s mass is m = ρAL.

Let S be the admissible axial stress stipulated for this teaching problem. It is a limit on F/A, not an unnamed catalogue “strength”. Meeting F/A ≤ S requires A ≥ F/S. At the smallest area admitted by this one constraint, m = ρLF/S = FL/(S/ρ). A larger S/ρ therefore gives a smaller mass under these particular assumptions.

This also explains the units: at that limiting area, F/m = S/(ρL). A specific value in N·m/kg becomes force per member mass in N/kg only after the fixed length is included. It is not a force capacity until geometry and the applicable limit have been supplied.

For the following invented comparison, P and Q are uniform materials with the stated properties in the loading direction. Both are assumed linear elastic up to their stipulated S for this exercise. The numbers define the model, rather than representing named grades or measured design allowables.

Invented material inputs used throughout the tie comparison. S is the problem’s stipulated admissible axial stress, not a universal failure criterion.
Quantity	Material P	Material Q
Density ρ	2000 kg/m³	4000 kg/m³
Admissible stress S	100 MPa	150 MPa
Young’s modulus E	20 GPa	80 GPa
S/ρ	50 000 N·m/kg	37 500 N·m/kg
E/ρ	10 000 000 N·m/kg	20 000 000 N·m/kg

An extension limit can change the answer

Suppose the same tie must also stay within an extension δ_max. In the small, linear-elastic axial model from the preceding lesson, δ = FL/(EA). Thus δ ≤ δ_max requires A ≥ FL/(Eδ_max). Substituting that area into m = ρAL gives m = FL²/[δ_max(E/ρ)] at the extension-limited minimum. Now E/ρ controls this comparison.

Use F = 2000 N, L = 500 mm and δ_max = 0.50 mm. For the stress limit alone, P needs 20 mm² and has mass 20 g; Q needs about 13.33 mm² and has mass about 26.67 g. P has the better S/ρ and is lighter in that comparison.

For the extension limit, P instead needs 100 mm² and has mass 100 g, while Q needs 25 mm² and has mass 50 g. Their stresses at those areas are 20 and 80 MPa, both below the stipulated S. Q is now lighter. If both constraints apply, choose an area large enough for both: here the extension requirement sets the larger area for each material.

The equal-volume question gives another comparison: at 10 cm³, P has mass 20 g and Q 40 g. That does not contradict Q’s lighter extension-limited tie. The extension-limited ties use different volumes. The figure and worked example keep these three questions separate.

Calculated member results for the invented P/Q inputs. Tensile-member cases use F = 2000 N and L = 500 mm; the extension case uses δ_max = 0.50 mm.
Comparison	P: area and mass	Q: area and mass
Equal volume 10 cm³	20 mm² at L = 500 mm; 20 g	20 mm² at L = 500 mm; 40 g
Stress limit alone	20 mm²; 20 g	13.33… mm²; 26.67… g
Extension limit, also below S	100 mm²; 100 g	25 mm²; 50 g

Three zero-based mass comparisons for the invented P and Q inputs. At equal 10 cm³ volume, P is 20 g and Q 40 g. At the stress-only minimum for 2000 N and length 500 mm, P is 20 g and Q about 26.67 g. With the additional 0.50 mm extension limit, P is 100 g and Q 50 g; both also satisfy their stipulated stress limits. All bars use the same scale, from 0 to 100 g. P uses a solid blue outline and Q a dashed orange outline. Different cases allow different volumes.

Keep the permitted changes with the ranking

If the area is fixed by a narrow space or a standard fitting, it cannot be increased freely to compensate for lower E or S. Check F/A and FL/(EA) using that actual area first. Density then tells you the mass, but the largest specific ratio does not make an infeasible shape work.

A bending member, a compressed column and a joint also do not automatically follow this tensile-member ranking. They have different stress patterns, geometric freedoms or failure mechanisms. The next lesson introduces those distinctions; later selection lessons derive other property combinations from their own constraints.

Use bulk properties together for a porous body: its bulk density and measured effective stiffness or strength under matching conditions. Dividing the solid constituent’s strength by the foam’s much lower density mixes two different bodies. Direction, temperature, processing history and the named strength endpoint must also match the question.

Worked examples

Preserve the units while forming a ratio

An invented specimen has mass 54 g and volume 20 cm³. Its stated maximum tensile strength is 270 MPa. Calculate its average density and strength/density ratio. What has actually been established?

ρ = 54/20 = 2.7 g/cm³. Converting mass and volume gives the same result: 0.054 kg/(20 × 10⁻⁶ m³) = 2700 kg/m³.
Convert the specified strength: 270 MPa = 270 × 10⁶ Pa. Divide by density: (270 × 10⁶)/2700 = 100 000 N·m/kg.
For a second specimen with the same average density but twice the volume, m = 2700 × 40 × 10⁻⁶ = 0.108 kg = 108 g. The density and strength/density ratio do not double.
The result combines a measured mass/volume description with a stipulated strength endpoint. It does not establish an allowable stress, a yield limit or the area and lifetime of a component.

ρ = 2700 kg/m³ and strength/density = 100 000 N·m/kg, for the stated tensile endpoint and specimen conditions.

Size the member before comparing mass

Use the invented P/Q table at F = 2000 N and L = 500 mm. Calculate the least area for the stress constraint alone, then impose δ_max = 0.50 mm. Explain the change in mass ranking.

For the stress limit, use N and mm: 1 MPa = 1 N/mm². P needs A = 2000/100 = 20 mm²; Q needs 2000/150 = 13.333… mm². Their volumes AL are 10 000 and 6666.67… mm³, or 10 and 6.6667… cm³.
P has density 2 g/cm³ and Q 4 g/cm³. Their masses at those areas are 20 and 26.6667… g. P is lighter for this single constraint, despite requiring the larger volume.
For extension, E_P = 20 000 N/mm² and E_Q = 80 000 N/mm². A_P = (2000 × 500)/(20 000 × 0.50) = 100 mm². A_Q = (2000 × 500)/(80 000 × 0.50) = 25 mm².
Their volumes are now 50 and 12.5 cm³ and their masses 100 and 50 g. Stress is 2000/100 = 20 MPa in P and 2000/25 = 80 MPa in Q, so both areas also satisfy S. These areas give δ = 0.50 mm by construction.
At the smaller stress-only areas, the elastic extensions would instead be 2.50 mm for P and 0.9375 mm for Q. Neither meets 0.50 mm. Checking both requirements reveals why the strength-only choice was insufficient.

Stress limit alone: P 20 g, Q about 26.67 g. Both stress and extension limits: P 100 g, Q 50 g. The allowed area change and the governing constraint determine the ranking.

Use the total volume of a porous body

A dry teaching foam has outside volume 100 cm³, with 80% of that volume occupied by pores. Its one solid constituent has density 1.20 g/cm³. Neglect gas mass. Estimate mass and bulk density, then assess a claim that it retains 20% of the solid modulus.

The solid fraction is 1 − 0.80 = 0.20, so V_s = 0.20 × 100 = 20 cm³. This is a volume fraction, not 20% of the foam’s mass.
Mass is approximately 1.20 × 20 = 24 g. Divide by the outside volume: ρ_bulk ≈ 24/100 = 0.24 g/cm³ = 240 kg/m³.
Using 1.20 g/cm³ with all 100 cm³ would predict 120 g, as if the pores were filled by solid. That overestimates this dry foam’s mass by a factor of five.
No deformation or stiffness model was supplied. The 20% solid fraction determines this mass account, but it does not by itself determine the effective modulus.

Approximately 24 g and 240 kg/m³. The modulus claim needs additional structural or measurement evidence.

Try the reasoning

The foam above takes up 10 cm³ of liquid without changing its outside volume. The liquid density is stipulated as 1.0 g/cm³ and no solid is lost. Can the dry bulk density still give its new mass? Calculate the new average density.
Show the answer and reasoning
No. Its mass becomes 24 + 10 = 34 g, so its average density is 34/100 = 0.34 g/cm³ = 340 kg/m³.
The outside volume is unchanged but an additional phase contributes mass. Using the solid density for the full volume or the old dry bulk density would omit the actual pore contents. No wet mechanical property follows from this mass calculation.
A learner chooses P for the lightest stiff beam because its S/ρ is greater than Q’s. Identify the problem with the reasoning, using the tie calculations to support your answer.
Show the answer and reasoning
The chosen ratio describes the stated stress-limited axial tie comparison, not a stiffness-limited beam. Even the extension-limited tie favors Q here.
Stiffness and strength are different requirements, and bending uses a different geometric relation from axial extension. Define the load, restraints, allowed dimensions and deflection requirement before deriving or using a ranking.
Only the stress constraint is being compared at 2000 N, but a fitting now permits at most 15 mm² of tie area. Which of P and Q can meet that stipulated stress constraint? Does the larger S/ρ settle the choice?
Show the answer and reasoning
Q can meet it at 13.333… mm². P needs 20 mm² and cannot fit, despite its larger S/ρ.
At the maximum 15 mm², stress would be 2000/15 ≈ 133.3 MPa. That exceeds P’s 100 MPa limit and is below Q’s 150 MPa limit. This addresses the stated stress constraint only; it does not restore the earlier extension requirement or assess the fitting itself.
Keep F = 2000 N and the same absolute extension limit 0.50 mm, but halve the tie length from 500 to 250 mm. With area still free, predict the extension-limited areas and masses for P and Q. Explain why mass does not merely halve.
Show the answer and reasoning
Required areas halve to 50 and 12.5 mm². Masses become 25 and 12.5 g, one quarter of the earlier values.
A ≥ FL/(Eδ_max) scales with L, while m = ρAL contains another factor of L. With the same absolute extension allowance, each tie is both shorter and thinner. Stresses are 40 and 160 MPa: Q now exceeds its stipulated 150 MPa limit, so it needs at least 13.333… mm² and about 13.33 g if both constraints apply.
A catalogue divides a foam’s solid-constituent tensile strength by the foam’s bulk density. Is the large result evidence that the whole foam is a superior tensile member?
Show the answer and reasoning
No. The numerator and denominator refer to different structural bodies.
The foam’s walls and their connections determine how its bulk specimen carries load. Use a relevant effective strength and bulk density with matching direction, state and test definition before applying a component model.

What to carry forward

Density connects mass to a specified volume; pores and their contents change that account.
Specific strength and specific modulus are different ratios with the same dimensions.
For the stated tensile-member models, S/ρ and E/ρ answer different constrained mass questions.
Size and check the component under its actual requirements before treating any material ratio as a selection.

References

Teaching resources and research supporting this lesson’s explanations and examples.