Materials Science · Elasticity, plasticity and tensile strength

Strength and strength relative to density

Derive the mass of a stress-limited tensile tie with consistent units, then test how a minimum cross-section changes the choice between two invented materials.

Useful background: Units, loads, stress and strain; Atoms, bonding, structure and defects; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Use specific strength for a clearly defined load-limited design.
Derive mass from density, area and length, and convert a strength/density ratio into consistent SI units.
Size and compare ties under both an allowable-stress limit and a stipulated minimum-area requirement.
Identify which geometric freedom and strength definition make a ranking meaningful.

A tensile member can be mass-limited

For a uniform tie carrying force F at allowable stress σ, required area is F/σ. Its mass is ρLF/σ for fixed length L. Under these assumptions, higher σ/ρ can reduce mass.

A tie is a member pulled along its length. For this model, its original length L and applied tensile force F are fixed, while material and uniform original cross-sectional area A can be chosen. Density ρ is mass per volume. Ignore self-weight, joints, stress concentrations, bending and time-dependent damage. These assumptions define the comparison; they do not qualify a real component for service.

The volume of the uniform member is V = AL. Multiplying by density gives m = ρV = ρAL. With ρ in kg/m³, A in m² and L in m, the units reduce to kg. “Applied force and stress over an area” explains the average axial engineering stress F/A used here.

In the retained relation above, σ denotes an allowable stress. Below write that limit as σ_allow to distinguish it from the actual demand F/A. The stipulated condition F/A ≤ σ_allow requires A ≥ F/σ_allow, for positive F and σ_allow. As area increases, mass ρAL increases, so the smallest mass admitted by this one constraint occurs at A = F/σ_allow.

Substitution now removes the free area from the mass expression: m_min = ρL(F/σ_allow) = FL/(σ_allow/ρ). At the same F and L, a larger admissible-stress/density ratio gives a smaller mass, provided the section really may shrink to that area. A higher strength can compensate for higher density by allowing less material. It does not make equal volumes lighter.

Keep the strength definition and units in the ratio

Specific strength is a stated strength divided by density. Name the endpoint, such as yield strength or maximum engineering tensile stress, and the material state and loading direction. A measured endpoint is not automatically a service allowable. “Tensile curves, strength and ductility” develops why yield and maximum tensile stress are different observations. The values called σ_allow in the calculations below are supplied limits for a teaching problem; their derivation from experimental evidence is outside this model.

Convert the stress before dividing by an SI density. In the original example, 200 MPa = 200 × 10⁶ Pa. Dividing by 2000 kg/m³ gives 100000 Pa·m³/kg = 100000 N·m/kg. For 300 MPa and 6000 kg/m³, the corresponding ratio is 50000 N·m/kg. Dividing 200 by 2000 and labelling the result N·m/kg would omit the million-fold MPa-to-Pa conversion.

The dimensions follow from Pa = N/m²: Pa/(kg/m³) = (N/m²)(m³/kg) = N·m/kg = m²/s². Although N·m is also the unit of work, this ratio is not a measurement of energy absorbed before failure. In the tie equation, dividing FL in N·m by the ratio in N·m/kg gives kg, which checks the mass result.

“Density and properties relative to mass” develops density’s volume basis and the distinct specific-modulus comparison. Use a density and mechanical property that describe the same material body, direction and state. Combining a dense constituent’s strength with a porous product’s low bulk density would not give the product’s supported tie performance.

The allowable value needs justification

Use a strength definition relevant to the state, service and failure criterion. Joints, fatigue, damage, minimum dimensions or stiffness can govern instead. Density-normalised strength is useful for this model; it cannot select every beam, shell or pressure component.

The load limit must address the intended failure criterion and conditions. A yield-based limit and an ultimate-tensile value do not become interchangeable by dividing both by density. Direction, temperature, rate, variability and damage can alter the relevant evidence. “Rate, direction and temperature effects” develops those dependencies; no universal factor for converting a test maximum into an allowable is supplied here.

If extension must also be limited, modulus and geometry enter another constraint. “Stiffness and strength answer different questions” explains why passing a stress limit does not establish acceptable movement. The following case isolates a different omitted requirement: a minimum uniform cross-section. No stiffness, fatigue or joint check is silently included.

A minimum section can remove the strength advantage

Consider two new invented candidates C and D. Keep tensile force F = 6000 N and length L = 1.50 m. C has supplied σ_allow = 600 MPa and ρ = 6000 kg/m³. D has σ_allow = 180 MPa and ρ = 2400 kg/m³. The limits apply to the same defined tensile demand in the stated problem. These are not material grades, measured design allowables or purchasing recommendations.

With the stress constraint alone, C needs 6000/600 = 10.0 mm²; D needs 6000/180 = 33.333… mm², since 1 MPa = 1 N/mm². Their masses are 0.0900 and 0.1200 kg. C is lighter despite its higher density because its larger σ_allow/ρ permits a much smaller section.

Now stipulate that this teaching assembly requires A ≥ A_min = 20.0 mm² for fabrication and handling. That is an additional geometric requirement, not a property derived from either material or a testing standard. Both inequalities must hold, so the smallest permitted area is A* = max(F/σ_allow, A_min). The max operation selects the larger lower bound.

C must now use 20.0 mm² rather than 10.0; D still needs 33.333… mm² to satisfy its stress limit. Masses become 0.180 and 0.120 kg. D is now lighter. The area requirement prevents C from using all of its stress advantage to reduce volume. Choosing only the largest σ_allow/ρ would miss that change.

If the stipulated minimum rises to 40.0 mm², it sets both areas: C then has mass 0.360 kg and D 0.144 kg. At equal area and length, density controls their mass ratio. The figure varies only this minimum-area requirement; it does not depict a change in the materials’ strength, density, load or length.

Original invented C/D comparison at F = 6000 N and L = 1.50 m. Equality is permitted by the problem’s stated ≤ stress limit. Joint and stiffness requirements remain outside the calculation.
Quantity	C	D
Supplied σ_allow	600 MPa	180 MPa
Density ρ	6000 kg/m³	2400 kg/m³
σ_allow/ρ	100000 N·m/kg	75000 N·m/kg
Area: stress limit alone	10.0 mm²	33.333… mm²
Mass: stress limit alone	0.0900 kg	0.1200 kg
Area: stress and A_min = 20.0 mm²	20.0 mm²	33.333… mm²
Mass: stress and A_min = 20.0 mm²	0.180 kg	0.1200 kg

Original invented tensile-tie comparison, not a material dataset or component qualification. The top schematic has no geometric scale: equal opposing tensile forces act on a uniform member with fixed load 6000 newtons and fixed length 1.50 metres, while its area can be selected. For each candidate, minimum area is the larger of force divided by its supplied allowable stress and a separately stipulated minimum-area requirement. The graph varies only that area floor from zero to 40 square millimetres on a linear horizontal axis; mass is on a zero-based linear axis from zero to 0.4 kilogram. Candidate C has allowable stress 600 megapascals and density 6000 kilograms per cubic metre. Its solid blue line is horizontal at 0.090 kilogram until the floor reaches 10 square millimetres, then rises to 0.360 kilogram at 40. Candidate D has allowable stress 180 megapascals and density 2400 kilograms per cubic metre. Its dashed orange line is horizontal at 0.120 kilogram until the floor reaches 33.333 square millimetres, then rises to 0.144 kilogram at 40. The curves cross at 13.333 square millimetres and 0.120 kilogram. The worked floor 20 is indicated by a vertical dashed line: a filled circle marks C at 0.180 kilogram and an open square marks D at 0.120 kilogram. Both meet the stated stress and minimum-area bounds, but D is lighter there despite C having the larger allowable-stress/density ratio. Shapes, line styles and labels supplement colour. Stiffness, joints, self-weight and lifetime are not checked.

Read the active constraint before choosing a ratio

An active constraint is one that sets the smallest permitted area in this model. With A_min below 10.0 mm², both candidates are stress-limited, so changing that inactive area floor does not change their masses. Between 10.0 and 33.333… mm², C is area-limited while D remains stress-limited. Above 33.333… mm², both are area-limited. At a boundary, both lower bounds can be equal.

The mass ranking need not change at either material’s activation boundary. In this case the masses cross at A_min = 13.333… mm², inside the mixed interval. The solved practice shows how to locate and check that point. A ranked ratio is a conclusion from the allowed geometry and governing requirement, not a substitute for stating them.

Retain the original load and length with a reported mass advantage, and check other relevant requirements before interpreting it as a complete selection. If shape, loading or available dimensions change, derive the applicable model again. A tie result does not establish the mass of its end fittings or the performance of an assembled structure.

Worked examples

Work through the case

Two hypothetical materials have allowable stresses/densities of 200 MPa/2000 kg/m³ and 300 MPa/6000 kg/m³. Compare ideal tie masses.

Their σ/ρ ratios are 100,000 and 50,000 N·m/kg.
Tie mass varies inversely with that ratio.
The second needs twice the mass for equal F and L.

The first gives half the ideal mass under this tensile constraint.

Size each candidate under both requirements

Use the new C/D inputs at 6000 N and 1.50 m. Apply A_min = 20.0 mm², calculate each minimum model mass and check its actual average stress. Which candidate is lighter, and why does the strength/density ranking not settle this case?

First check the stress/density indices in SI units. C gives (600 × 10⁶)/6000 = 100000 N·m/kg; D gives (180 × 10⁶)/2400 = 75000 N·m/kg. C has the larger index for a stress-only free-area tie.
Stress-only areas are A_C = 6000/600 = 10.0 mm² and A_D = 6000/180 = 33.333… mm². Apply both bounds: A*_C = max(10.0, 20.0) = 20.0 mm²; A*_D = max(33.333…, 20.0) = 33.333… mm².
Convert area for use with kg/m³ and m: 1 mm² = 10⁻⁶ m². Then m_C = 6000 × (20.0 × 10⁻⁶) × 1.50 = 0.180 kg. For D, m_D = 2400 × (33.333… × 10⁻⁶) × 1.50 = 0.120 kg. Keeping the unrounded third preserves the displayed result.
Check actual average stress using the selected area. C has 6000/20.0 = 300 MPa, below its supplied 600 MPa limit; D has 6000/33.333… = 180 MPa, equal to its permitted limit. Both areas also meet 20.0 mm².
D is lighter by 0.060 kg, one third of C’s 0.180 kg mass. C could have been lighter without the area floor, but it cannot use a 10.0 mm² section under this requirement. No component stiffness, connection or lifetime conclusion follows from these two passed checks.

D gives the smaller model mass, 0.120 kg versus 0.180 kg. The minimum-area condition changes the allowable geometric freedom, so the stress-only index no longer ranks the complete stated problem.

Try the reasoning

Does that comparison establish joint mass?
Show the answer and reasoning
No.
The simplified model omits connections.
Can fatigue replace static strength as the constraint?
Show the answer and reasoning
Yes.
Repeated loading may govern the allowable design.
Keep the C/D inputs and A_min = 20.0 mm², but change force to 18000 N and length to 2.00 m. Calculate minimum areas and masses. Which requirement sets each area, and which candidate is lighter?
Show the answer and reasoning
C needs 30.0 mm² and 0.360 kg; D needs 100 mm² and 0.480 kg. Both areas are now stress-limited, and C is lighter.
The stress areas are 18000/600 = 30.0 mm² and 18000/180 = 100 mm², both greater than 20.0. Masses are 6000 × 30.0 × 10⁻⁶ × 2.00 = 0.360 kg and 2400 × 100 × 10⁻⁶ × 2.00 = 0.480 kg. The actual stresses equal the supplied 600 and 180 MPa limits. Changing the force can activate a different constraint; do not keep the earlier selected areas or simply transfer the earlier ranking.
Return to F = 6000 N and L = 1.50 m. Within 10.0 ≤ A_min ≤ 33.333… mm², C is area-limited and D remains stress-limited at mass 0.120 kg. Find the minimum-area value at which their masses are equal, then check that this assumed interval is valid.
Show the answer and reasoning
They have equal mass at A_min = 13.333… mm², giving 0.120 kg for each. It lies inside the stated mixed-constraint interval.
With A_min entered in mm², C’s mass is 6000 × 1.50 × A_min × 10⁻⁶ = 0.00900 A_min kg. Set this equal to 0.120 kg: A_min = 0.120/0.00900 = 13.333… mm². This exceeds C’s 10.0 mm² stress area and is smaller than D’s 33.333… mm² stress area, so the assumed active constraints are consistent. Just above the crossing D is lighter, even though C’s strength/density index remains larger.
One report gives yield strength 200 MPa at density 2000 kg/m³; another gives maximum engineering tensile stress 300 MPa at density 2400 kg/m³. Their ratios are 100000 and 125000 N·m/kg. Is the second number enough to select the lighter tie under a specified no-yield requirement?
Show the answer and reasoning
No. The second endpoint is a maximum tensile stress, not a supplied yield-based allowable for that requirement.
The ratios are arithmetically correct after MPa-to-Pa conversion, but their numerators answer different questions. Obtain the relevant strength definition and justified admissible limit under matching conditions before sizing the member. Dividing by density does not convert a maximum-load endpoint into a yield limit.

What to carry forward

Derive the index from the actual constraint.
Do not transfer a tie model to every structure.
For a uniform tie, m = ρAL; the stress-only minimum is FL/(σ_allow/ρ) only when area is free to reach F/σ_allow.
With a minimum section, choose A* = max(F/σ_allow, A_min) before comparing masses.
Keep units, strength endpoint, permitted geometry and other requirements with every reported mass advantage.

References

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