Concept 01 / 07Mechanical Properties of Solids
Stress and Strain: Tensile, Shearing and Hydraulic
Stress is the internal restoring force per unit area set up inside a deformed body and strain is the fractional deformation it produces; each comes in three varieties, longitudinal, shearing and volumetric.
Why does a solid resist being deformed at all? Zoom down to the atomic scale. The atoms of a solid sit at the minimum of a mutual potential energy curve, separated by an equilibrium distance of about to angstrom. Pull two neighbouring atoms slightly apart and the bond pulls them back; push them closer and it pushes them apart. Near the minimum every bond behaves exactly like a tiny spring of stiffness . Everything in this chapter, every modulus and every formula, is the macroscopic shadow of that microscopic spring.
When an external deforming force acts on a body, the atoms are displaced from their equilibrium positions and the bonds develop an internal restoring force. Once the deformation stops growing, the body is in equilibrium, so the internal restoring force exactly balances the applied force. That is why stress can be computed either as restoring force per unit area (the definition) or as applied force per unit area (the working recipe): in static equilibrium the two are numerically identical. A body is called elastic if it recovers its original configuration when the deforming force is removed, and plastic if it does not.
Definition of stress. Stress is the internal restoring force developed per unit area of a section of the body:
Its SI unit is the newton per square metre, called the pascal (Pa), and its dimensional formula is , the same as that of pressure. But stress is not pressure. Pressure at a point in a fluid is one number and always acts normally to any surface drawn there. Stress needs you to specify the orientation of the surface, and it may have a component lying in that surface. Formally stress is a second rank tensor with nine components. At JEE level you meet only three standard configurations, but the tensor character shows up in a beautiful result derived at the end of this note.
Type 1: longitudinal or normal stress. The force is perpendicular to the cross-section, and is the area of that cross-section. If the force pulls the body outwards it is tensile stress; if it pushes inwards it is compressive stress. The associated deformation is a change of length, and the longitudinal strain is
Type 2: shearing or tangential stress. The force acts parallel to a face, tending to slide successive layers over one another. Here is the area of the face that is parallel to the force, not the cross-section perpendicular to it. Choosing the wrong face is the single most common slip in shear problems. If a block of height has its top face displaced by relative to a fixed bottom face, the shear strain is the angle of tilt
measured in radians. Elastic shear angles in metals are of order rad, so replacing by is not an approximation you ever need to apologise for.
Type 3: hydraulic or volumetric stress. The body is immersed in a fluid so that a uniform pressure acts normally at every point of its surface. The stress is the excess pressure , the shape is unchanged, and the volume strain is . The diagram above shows the three cases side by side.
Strain is always a ratio of two like quantities, so it is dimensionless and has no unit. Sign convention: extension and volume increase are positive, compression and volume decrease are negative, which is why a minus sign appears in the definition of the bulk modulus later. Typical elastic strains in metals are to , which is why elasticity problems are full of small-quantity approximations such as , where products of small quantities are dropped.
Stress depends on the plane you cut. This is the result that separates a student who has understood stress from one who has memorised . Take a rod of cross-sectional area pulled axially by a force , so the ordinary tensile stress is . Now slice the rod with a plane whose normal makes an angle with the rod axis. The area of this slanted section is , and the same force is transmitted across it, resolved into a component normal to the slanted plane and lying in it. Therefore
normal stress on that plane
shear stress on that plane
The normal stress is largest on the cross-section itself (), but the shear stress is largest at , where it equals . So a rod pulled in simple tension is simultaneously carrying a shear stress equal to half the tensile stress on planes at . Real ductile metals fail by sliding of atomic planes, so they fail on exactly those planes, and that is why a stretched ductile specimen shows slip bands and a cup-and-cone fracture inclined near .
Gotchas that JEE loves. (i) In a heavy rod hanging from a ceiling the tension, and hence the stress, is not the same everywhere: it grows from zero at the free end to at the support. (ii) Stress is defined per unit original area (engineering stress); the true stress uses the instantaneous area and differs appreciably only after necking begins. (iii) A body can have strain with no stress (a rod free to expand on heating) and stress with no net observable strain (the same rod clamped between rigid walls). (iv) Strain is not always small: rubber can be strained several hundred percent while still returning to its original length.
Key Takeaways and Formulae:
Stress , unit pascal, dimensions ; strain is dimensionless.
Longitudinal strain ; shear strain in radians; volume strain .
For normal stress, is perpendicular to ; for shear stress, is parallel to .
On a plane whose normal makes angle with the axis of a rod in tension: normal stress and shear stress , with maximum shear at .
Tensile stress is positive, compressive negative. Stress is a tensor, not a vector: never add stresses on different planes like vectors.
Brain-Teaser 1: A stick of chalk is twisted until it breaks, and the fracture surface is a clean helix inclined at about to the axis. A mild steel rod pulled in tension also fails on surfaces inclined near . Both fractures are at , yet the reasons are opposite. Explain.
Solution: A twisted rod is in a state of pure shear on its cross-sections. As shown later in the chapter, pure shear of magnitude is exactly equivalent to a tension along one diagonal direction together with a compression along the perpendicular direction. Chalk is brittle: it is far weaker in tension than in shear, so it tears apart on the surface that is perpendicular to the tensile direction, generating a helical fracture. Mild steel pulled in tension is in a state of uniaxial tension , which as derived above carries a maximum shear on planes at . Steel is ductile: it is weaker in shear than in tension, because plastic flow occurs by dislocations gliding on close-packed planes under shear. So it slips on the planes. Same angle, opposite mechanism: the brittle solid follows the maximum tensile stress, the ductile solid follows the maximum shear stress.
Brain-Teaser 2: Wire A has length and radius . Wire B, of the same material, has length and radius . Each carries the same load . Compare (a) the stress, (b) the strain, (c) the elongation, and (d) the load at which each snaps.
Solution: (a) Stress , so . Wire A is four times as stressed. (b) Same material means the same Young modulus , and strain , so the strains are also in the ratio . (c) Elongation . For A it is ; for B it is . The ratio is , not . The general scaling is . (d) The breaking load is (breaking stress) times (area), and the breaking stress is a material property. So B breaks at four times the load of A, and the length is irrelevant. This kills the popular misconception that a longer wire is weaker: a long wire and a short wire of the same material and thickness support exactly the same maximum load, although the long one stretches more before it gets there.
Brain-Teaser 3: A steel rod is heated through K. (a) If it rests freely on a smooth table, what stress is set up in it? (b) If instead it is clamped between two immovable walls, what is its change in length, and is there any stress? Reconcile the two answers.
Solution: (a) Zero stress. On heating, the equilibrium interatomic spacing itself increases, so the atoms move to new equilibrium positions and no bond is stretched relative to its own new equilibrium. There is strain, , but no restoring force, so no stress. (b) The observable change in length is zero, because the walls do not move. Yet the rod is strongly stressed. The resolution is to split the total strain into two parts: , hence . Stress responds only to the elastic part, so the rod carries a compressive stress . Notice what this means physically: the bonds are compressed relative to the longer equilibrium spacing they would prefer at the higher temperature. This is the whole idea of thermal stress, and it explains why the stress is independent of the length of the rod, a fact that surprises most students the first time they meet it.
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