Why are hexagonal close packed materials brittle, While face centered cubic is ductile. Is it related to crystal planes?


3 Answers 3


please learn a few definitions Slip plane – is the plane of greatest atomic density. Slip direction – is the close-packed direction WITHIN the slip plane Slip system = slip plane and slip direction TOGETHER

THEN; 5 independent slip systems are necessary to make a polycrystalline material ductile.

HCP - Has three slip systems (one plane and three directions, giving 3x1= 3 slip systems, we know that minimum 5 independent slip systems are necessary to make a polycrystalline material ductile.therefore HCP is brittle.

FCC - has 12 slip systems (three {111} family of planes and four <110> family of directions, giving 3x4 =12 slip systems, which is more than 5 independent slip systems therefore FCC is ductile.

BCC -has 48 slip systems and expecting better ductile but it is brittle (six {110} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {211} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {321} family of planes and four <111> family of directions =6x4 = 24 slip systems; grand total 12+12+24 = 48 slip systems)

BCC lattice structure has too much of slip systems(48), here slip systems are INTERFERE OR MUTUALLY OBSTRUCT each other therefore slip movement in BCC is made very difficult thus BCC is brittle.


Yes and how close the planes are packed and of course their geometries. See some good answers below.

Source : http://www.researchgate.net/post/What_actually_makes_a_material_ductile_or_brittle

We may understand brittleness/ductility of solids from its bonding nature. In every solid, the constituent atom/ions are held by primary bonds (covalent/ ionic/ metallic). When we apply stress, we deform the atom/ions from its lattice. If the material can accept the deformation by getting strained- we call it ductile.Ductile materials must have some mechanism to absorb the stress- forming defects in its lattice. Brittle materials can't create defect in its lattice to absorb stress, so it deforms upto certain stress then break suddenly.

Source https://www.physicsforums.com/threads/why-is-fcc-more-ductile-than-bcc.550403/

Crystalline structure is important because it contributes to the properties of a material. For example, it is easier for planes of atoms to slide by each other if those planes are closely packed. Therefore, lattice structures with closely packed planes allow more plastic deformation than those that are not closely packed. Additionally, cubic lattice structures allow slippage to occur more easily than non-cubic lattices. This is because their symmetry provides closely packed planes in several directions. A face-centered cubic crystal structure will exhibit more ductility (deform more readily under load before breaking) than a body-centered cubic structure. The bcc lattice, although cubic, is not closely packed and forms strong metals. Alpha-iron and tungsten have the bcc form. The fcc lattice is both cubic and closely packed and forms more ductile materials. Gamma-iron, silver, gold, and lead have fcc structures. Finally, HCP lattices are closely packed, but not cubic. HCP metals like cobalt and zinc are not as ductile as the fcc metals.

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    $\begingroup$ Some discussion of the number and direction of slip planes would be useful to add. $\endgroup$
    – Jon Custer
    Oct 25, 2015 at 16:40
  • $\begingroup$ This answer doesn't really address the question. $\endgroup$
    – John M
    Oct 28, 2016 at 8:28
  • $\begingroup$ How so John M ... please be more specifc. $\endgroup$
    – StarDrop9
    Oct 28, 2016 at 17:14

Interesting. In my past metal physics thermodynamics lab research work the focus was on yield stress which here characterises the shift from ductile to brittle deformation.

Mechanical engineers also have many other characterisations such as 'hardness' of one material under surface abrasion against another; and 'toughness' as the work reqiured for a standard unit of deformation.

We were deriving Helmotz recrystallization free energy from the Arrhenius plot of the equations of state. I suppose similar thermodynamic inferences could be arrived at with 'toughness' in place of 'yield stress' but perhaps the latter was more convenient using tensile stress testers.


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