Stress, Strain, and Elastic Modulus

Tensile or Compressive Stress, Strain, and Young’s Modulus

Tension or compression occurs when two antiparallel forces of equal magnitude act on an object along only one of its dimensions, in such a way that the object does not move. One way to envision such a situation is illustrated in (Figure). A rod segment is either stretched or squeezed by a pair of forces acting along its length and perpendicular to its cross-section. The net effect of such forces is that the rod changes its length from the original length that it had before the forces appeared, to a new length L that it has under the action of the forces. This change in length may be either elongation (when L is larger than the original length or contraction (when L is smaller than the original length Tensile stress and strain occur when the forces are stretching an object, causing its elongation, and the length change is positive. Compressive stress and strain occur when the forces are contracting an object, causing its shortening, and the length change is negative.

In either of these situations, we define stress as the ratio of the deforming force to the cross-sectional area A of the object being deformed. The symbol that we reserve for the deforming force means that this force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of an object. The definition of the tensile stress is

Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length when the object experiences tensile stress

Compressive stress and strain are defined by the same formulas, (Figure) and (Figure), respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-hand sides in (Figure) and (Figure).

When an object is in either tension or compression, the net force on it is zero, but the object deforms by changing its original length (a) Tension: The rod is elongated by (b) Compression: The rod is contracted by In both cases, the deforming force acts along the length of the rod and perpendicular to its cross-section. In the linear range of low stress, the cross-sectional area of the rod does not change.

Young’s modulus Y is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by (Figure). Dividing this equation by tensile strain, we obtain the expression for Young’s modulus:

Compressive Stress in a Pillar
A sculpture weighing 10,000 N rests on a horizontal surface at the top of a 6.0-m-tall vertical pillar (Figure). The pillar’s cross-sectional area is and it is made of granite with a mass density of Find the compressive stress at the cross-section located 3.0 m below the top of the pillar and the value of the compressive strain of the top 3.0-m segment of the pillar.

Nelson’s Column in Trafalgar Square, London, England. (credit: modification of work by Cristian Bortes)

Strategy
First we find the weight of the 3.0-m-long top section of the pillar. The normal force that acts on the cross-section located 3.0 m down from the top is the sum of the pillar’s weight and the sculpture’s weight. Once we have the normal force, we use (Figure) to find the stress. To find the compressive strain, we find the value of Young’s modulus for granite in (Figure) and invert (Figure).

Solution
The volume of the pillar segment with height and cross-sectional area is

With the density of granite the mass of the pillar segment is

The weight of the pillar segment is

The weight of the sculpture is so the normal force on the cross-sectional surface located 3.0 m below the sculpture is

Therefore, the stress is

Young’s modulus for granite is Therefore, the compressive strain at this position is

Significance
Notice that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the top to its largest value at the bottom of the pillar. Thus, if the pillar has a uniform cross-sectional area along its length, the stress is largest at its base.

Stretching a Rod
A 2.0-m-long steel rod has a cross-sectional area of The rod is a part of a vertical support that holds a heavy 550-kg platform that hangs attached to the rod’s lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the elongation of the rod under the stress?

Strategy
First we compute the tensile stress in the rod under the weight of the platform in accordance with (Figure). Then we invert (Figure) to find the rod’s elongation, using From (Figure), Young’s modulus for steel is

Solution
Substituting numerical values into the equations gives us

Significance
Similarly as in the example with the column, the tensile stress in this example is not uniform along the length of the rod. Unlike in the previous example, however, if the weight of the rod is taken into consideration, the stress in the rod is largest at the top and smallest at the bottom of the rod where the equipment is attached.

Objects can often experience both compressive stress and tensile stress simultaneously (Figure). One example is a long shelf loaded with heavy books that sags between the end supports under the weight of the books. The top surface of the shelf is in compressive stress and the bottom surface of the shelf is in tensile stress. Similarly, long and heavy beams sag under their own weight. In modern building construction, such bending strains can be almost eliminated with the use of I-beams (Figure).

Bulk Stress, Strain, and Modulus

When you dive into water, you feel a force pressing on every part of your body from all directions. What you are experiencing then is bulk stress, or in other words, pressure. Bulk stress always tends to decrease the volume enclosed by the surface of a submerged object. The forces of this “squeezing” are always perpendicular to the submerged surface (Figure). The effect of these forces is to decrease the volume of the submerged object by an amount compared with the volume of the object in the absence of bulk stress. This kind of deformation is called bulk strain and is described by a change in volume relative to the original volume:

An object under increasing bulk stress always undergoes a decrease in its volume. Equal forces perpendicular to the surface act from all directions. The effect of these forces is to decrease the volume by the amount compared to the original volume,

The bulk strain results from the bulk stress, which is a force normal to a surface that presses on the unit surface area A of a submerged object. This kind of physical quantity, or pressure p, is defined as

We will study pressure in fluids in greater detail in Fluid Mechanics. An important characteristic of pressure is that it is a scalar quantity and does not have any particular direction; that is, pressure acts equally in all possible directions. When you submerge your hand in water, you sense the same amount of pressure acting on the top surface of your hand as on the bottom surface, or on the side surface, or on the surface of the skin between your fingers. What you are perceiving in this case is an increase in pressure over what you are used to feeling when your hand is not submerged in water. What you feel when your hand is not submerged in the water is the normal pressure of one atmosphere, which serves as a reference point. The bulk stress is this increase in pressure, or over the normal level,

When the bulk stress increases, the bulk strain increases in response, in accordance with (Figure). The proportionality constant in this relation is called the bulk modulus, B, or

The minus sign that appears in (Figure) is for consistency, to ensure that B is a positive quantity. Note that the minus sign is necessary because an increase in pressure (a positive quantity) always causes a decrease in volume, and decrease in volume is a negative quantity. The reciprocal of the bulk modulus is called compressibility or

The term ‘compressibility’ is used in relation to fluids (gases and liquids). Compressibility describes the change in the volume of a fluid per unit increase in pressure. Fluids characterized by a large compressibility are relatively easy to compress. For example, the compressibility of water is and the compressibility of acetone is This means that under a 1.0-atm increase in pressure, the relative decrease in volume is approximately three times as large for acetone as it is for water.

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Shear Stress, Strain, and Modulus

The concepts of shear stress and strain concern only solid objects or materials. Buildings and tectonic plates are examples of objects that may be subjected to shear stresses. In general, these concepts do not apply to fluids.

Shear deformation occurs when two antiparallel forces of equal magnitude are applied tangentially to opposite surfaces of a solid object, causing no deformation in the transverse direction to the line of force, as in the typical example of shear stress illustrated in (Figure). Shear deformation is characterized by a gradual shift of layers in the direction tangent to the acting forces. This gradation in occurs in the transverse direction along some distance Shear strain is defined by the ratio of the largest displacement to the transverse distance

Shear strain is caused by shear stress. Shear stress is due to forces that act parallel to the surface. We use the symbol for such forces. The magnitude per surface area A where shearing force is applied is the measure of shear stress

The shear modulus is the proportionality constant in (Figure) and is defined by the ratio of stress to strain. Shear modulus is commonly denoted by S:

An object under shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. The dashed-line contour depicts the resulting deformation. There is no change in the direction transverse to the acting forces and the transverse length is unaffected. Shear deformation is characterized by a gradual shift of layers in the direction tangent to the forces.

An Old Bookshelf
A cleaning person tries to move a heavy, old bookcase on a carpeted floor by pushing tangentially on the surface of the very top shelf. However, the only noticeable effect of this effort is similar to that seen in (Figure), and it disappears when the person stops pushing. The bookcase is 180.0 cm tall and 90.0 cm wide with four 30.0-cm-deep shelves, all partially loaded with books. The total weight of the bookcase and books is 600.0 N. If the person gives the top shelf a 50.0-N push that displaces the top shelf horizontally by 15.0 cm relative to the motionless bottom shelf, find the shear modulus of the bookcase.

Strategy
The only pieces of relevant information are the physical dimensions of the bookcase, the value of the tangential force, and the displacement this force causes. We identify and and we use (Figure) to compute the shear modulus.

Solution
Substituting numbers into the equations, we obtain for the shear modulus

We can also find shear stress and strain, respectively:

Significance
If the person in this example gave the shelf a healthy push, it might happen that the induced shear would collapse it to a pile of rubbish. Much the same shear mechanism is responsible for failures of earth-filled dams and levees; and, in general, for landslides.