Density, stiffness, strength, toughness and observable behaviour

Density compares mass with occupied volume

Mass is an amount measured in grams or kilograms. Average density is mass divided by the specified volume: density = mass ÷ volume. A small dense piece can have less mass than a large low-density piece. State whether the volume includes internal pores or cavities, because that choice changes what the average describes.

For simple samples, compare equal volumes or calculate the ratio with consistent units. Grams per cubic centimetre, written g/cm³, means grams for each cubic centimetre. If a coating or several constituents are included, the result is an average for that sample, not automatically the density of its underlying solid material.

Bending less is a stiffness observation

A force is a push or pull; the newton, N, is its unit. Stiffness describes how strongly an object resists deformation under stated loading. In the graph, equal-shaped specimens with the same supports receive the same force. At 4 N, A bends 2 mm and B 4 mm. A is stiffer in this comparison. The graph ends there: it does not show when either specimen permanently changes shape or breaks.

Strength concerns resistance to a specified form of failure or unacceptable permanent deformation. Toughness concerns absorbing energy before fracture; brittle behavior can involve fracture with little prior deformation. These are different from stiffness, and from hardness, which concerns local indentation or scratching. A stiff sheet can still fracture abruptly. A bendable strip can remain unbroken through a large movement.

Follow a load through deformation to fracture

The next records concern new model coupons P and Q, not the earlier small-load A/B strips. A coupon is a sample with a specified shape. Here both start with the same 100 mm gauge length and uniform 10 mm² cross-section, pulled along their length at the same temperature and loading rate. Extension means the increase in that gauge length; motion of the holding equipment is excluded. Assume loading is slow enough that changes in the coupon’s kinetic energy are negligible. These invented curves are for interpretation, not instructions to break materials.

At first P needs 50 N for each millimetre of extension; Q needs 10 N/mm. P is initially stiffer. Define failure for this comparison as fracture: P reaches 50 N at 1 mm and breaks, while Q reaches 30 N at 6 mm and breaks. P therefore carries the larger fracture force here. With equal initial cross-sections, this is also a higher fracture-strength comparison under these conditions. Choosing unacceptable permanent extension as the failure rule would ask a different question.

Energy supplies the third question: how much mechanical work enters before fracture? For a constant pull along an extension, work is force multiplied by the distance moved in its direction. Its unit is the joule, J: 1 N·m = 1 J, so 1 N·mm = 0.001 J. If the force changes, add the contributions along the movement. On these straight-line segments that is the area under the force–extension graph, found from triangles and rectangles.

For P the triangular area is half of 1 mm × 50 N = 25 N·mm, or 0.025 J. For Q, the first triangle is half of 2 mm × 20 N = 20 N·mm. From 2 to 6 mm, the average of the linearly changing 20 and 30 N is 25 N, giving another 4 mm × 25 N = 100 N·mm. Q receives 120 N·mm = 0.120 J before fracture, despite its lower initial stiffness and fracture force. A very small force over a large movement would not necessarily give a large area: flexibility alone is not toughness.

Tensile toughness compares this deformation energy per initial sample volume under stated test conditions. These coupons have equal initial volumes, so Q has the greater value in this model. The work is not all energy irreversibly used to create new crack surfaces; some can be stored elastically or dissipated during deformation. This simple loading record does not measure the specialised fracture-toughness property used for an existing crack, establish impact behaviour or show recovery after unloading. Those require their own observations.

Describe what was observed, not a universal ranking

If a strip returns to its earlier shape after unloading, it behaved elastically over that loading history. A remaining bend is permanent deformation. A loading-only graph cannot establish full recovery, and a short observation cannot establish long-term behavior. Direction matters too: wood properties can differ along and across its grain.

Make a record with sample identity, dimensions, direction, supports, temperature, load history and the measured response. These details let another person decide whether a comparison addresses their object. The same material made thinner, longer or differently supported can give a different bending result without any change in its composition.

Equal volume and equal mass are different comparisons

In a constructed sample record, A has mass 24 g and volume 10 cm³; B has mass 12 g and the same 10 cm³ volume. Their average densities are 24 ÷ 10 = 2.4 g/cm³ and 12 ÷ 10 = 1.2 g/cm³. A has twice B’s density under the recorded conditions.

Now use 20 cm³ of B. Its mass is 1.2 × 20 = 24 g, equal to the mass of the smaller A sample. Equal mass has not made their densities equal. It has changed how much volume of each sample is present.

Neither calculation establishes stiffness, strength or toughness. To choose a cover, add evidence for the actual geometry and the behavior the cover needs. Density can help estimate mass; it cannot answer every mechanical question.

Try the reasoning

1. At 4 N, A bends 2 mm and B bends 4 mm. Which has the greater component stiffness in this comparison?

   Show the answer and reasoning

   A. The same force produces less deformation, with geometry and support conditions held fixed. This alone does not identify the stronger specimen.

2. A low-density part is larger than a high-density part. Must it have less mass?

   Show the answer and reasoning

   No. Mass depends on both density and volume; a sufficiently large volume can outweigh a smaller denser piece.

3. A strip looks straight while loaded. Does that establish toughness?

   Show the answer and reasoning

   No. Small deformation addresses stiffness at that load. Toughness requires suitable evidence about energy absorption and fracture behavior.

4. Two further equal-geometry model coupons have straight loading lines from zero to fracture: R reaches 10 N at 0.5 mm, S reaches 8 N at 4 mm. Compare initial stiffness, fracture force and work before fracture. Which fact about unloading is still missing?

   Show the answer and reasoning

   R is initially stiffer: 10/0.5 = 20 N/mm versus 8/4 = 2 N/mm. R also reaches the larger fracture force, 10 versus 8 N. But S receives more loading work: its triangle area is half × 4 × 8 = 16 N·mm, compared with half × 0.5 × 10 = 2.5 N·mm for R. Equal initial volumes make the energy-per-volume comparison consistent here. The loading lines do not say how much extension would recover if unloaded before fracture; a straight line alone does not establish recovery or absence of permanent change.

What to carry forward

Ask which quantity was measured and under what conditions. One favorable property does not stand in for the others.