# Stress effort: formula and equations, calculation, exercises

The **tensile strength** is defined as the area perpendicular to the force per unit area applied to an object to engage their ends in traction on it, through which extends. Its dimensions are force / area and, in mathematical form, we can express it like this:

τ = F / A

The unit of effort in the International System of Units is the same as that used for pressure: the pascal, abbreviated Pa, which is equal to 1 Newton/m ^{2} .

In tension effort, there are two forces that are applied in the same direction and in opposite directions, which stretch the body. If originally the object length was L _{or} , when applying tensile stress, the new length is L and the elongation ΔL is calculated by:

ΔL = L – L _{or}

Solid objects have elasticity to a greater or lesser degree, which means that when the tensile stress disappears, they return to their original dimensions.

This happens as long as the stress is not so great that it causes permanent deformation. Rubber, rubber or rubber materials are good for making elastic objects and hair and skin, among others, also have this quality.

**unit deformation**

When studying how bodies deform under stress, it is very convenient to define the concept of *unitary deformation* , a dimensionless quantity. The unit deformation is indicated by the Greek letter δ (“lowercase delta”) and is calculated as follows:

δ = ΔL / L _{or}

The unit deformation serves to comparatively evaluate the deformation of the object under tension. Let’s look at it this way: it is not the same to stretch a bar 1 meter long by 1 cm, as it is to stretch 1 cm to another bar 10 m long. In the first case, the deformation is much more significant than in the second.

__How is tensile stress calculated? (Examples)__

__How is tensile stress calculated? (Examples)__

Newton’s English physicist and contemporary named Robert Hooke (1635-1703) investigated the elastic properties of bodies and established the law that bears his name. Thus, the applied stress is related to the deformation experienced when the effort is small:

Effort ∝ Deformation (unit)

It is logical to expect that the greater the tension effort, the greater the stretching will occur. Making use of the definitions given above:

τ ∝ δ

The constant of proportionality necessary to establish equality is denoted Y and is known as modulus of elasticity or modulus of elasticity, characteristic of materials:

τ = Y⋅δ

Young’s modulus has the same stress units since the deformation of the unit is dimensionless.

Therefore, one way to calculate the stress stress on a body with elastic properties is to measure the deformation and know the Young’s modulus. This quantity has been experimentally determined for many materials and is tabulated.

**Calculation example**

Suppose a hardened steel wire 3 mm in diameter is subjected to a tensile stress, hanging a weight of 250 N, what would be the magnitude of that stress?

Well, we can use the definition of stress stress as the quotient between the force perpendicular to the surface and the area of that surface. Let’s calculate the area first, assuming a wire with a circular cross section:

A = π. (d / 2) ^{2} = ^{ }π. (D ^{2} /4)

The wire diameter is 3 mm and these units must be converted into meters:

d = 3 x 10 ^{-3} m.

A = π. (3 x 10 ^{-3} m) ^{2} /4 = 7.07 x 10 ^{-6} m ^{2} .

The tensile stress is produced by the weight hanging on the wire, applied perpendicularly to its cross section, therefore:

τ = 250 N / 7.07 x 10 ^{-6} m ^{2} = 3.5 x 10 ^{7} Pa

The pascal is a relatively small unit, so multiples are not uncommon. Knowing that 1 mega-pascal (MPa) is 10 ^{6} pascal, the tensile stress remains:

τ = 35 MPa

__solved exercises__

__solved exercises__

**– Exercise 1**

The modulus of elasticity of a rod is 4 x 10 ^{11} Pa. What unit strain is obtained by applying a tensile stress of 420 MPa?

**Solution**

The equation to use is:

τ = Y⋅δ

With it we calculate the unit strain:

δ = τ / Y = 420 x 10 ^{6} Pa / 4 x 10 ^{11} Pa = 0.00105

δ = ΔL / L _{or}

Therefore, the deformation ΔL is:

ΔL = 0.00105 L _{or}

If, for example, the rod was originally 1 meter long, with this tension effort, it extends only 0.00105 m = 1.05 mm.

**– Exercise 2**

A steel wire is 1.50 m long and has a diameter of 0.400 mm. One end is attached to the ceiling and the other is equipped with a reflector of mass *m* = 1.50 kg, which is released. Calculate:

a) The elongation of the thread.

b) Unit strain and percentage of unit strain. Is it possible for the wire to break due to the weight of the reflector?

**Solution**

The wire will stretch, because when hanging the reflector, it is subjected to a tension effort. The force produced by this effort is the weight of the reflector.

The weight of an object of mass m is the product of the mass multiplied by the value of the acceleration due to gravity, therefore:

F = 1.50 kg x 9.8 m / s ^{2} = 14.7 N

The cross-sectional area of the wire is required:

A = ^{ }π. (D ^{2} /4) = π x (0.4 x 10-3 m) 2/4 = 1.26 x 10 ^{-7} m ^{2} .

With these results, the effort exerted by the weight on the wire is calculated:

τ = 14.7 N / 1.26 x 10 ^{-7} m ^{2} = 1.17 x 10 ^{8} Pa

The thread has an elastic behavior, therefore, it is valid to assume that Hooke’s law is fulfilled:

τ = Y⋅δ

From the elastic modulus table, we find that for steel Y = 207 x 10 ^{9} Pa. In addition, the unit strain is:

δ = ΔL / L _{or}

Substituting in the equation for effort:

τ = Y⋅δ = Y⋅ (ΔL / L _{o} )

So the excerpt is:

ΔL = L _{or} τ / Y =

= 1.50 mx 1.17 x 10 ^{8} Pa / 207 x 10 ^{9} Pa = 8.5 x 10 ^{-4} m = 0.849 mm.

The unitary deformation of the yarn is:

δ = ΔL / L _{o} = 8.5 x 10 ^{-4} m / 1.5 m = 5.652 x 10 ^{-4}

If we express it as a percentage, the percentage of unit deformation is 0.0565%, less than 0.1%, therefore, it is expected that the wire will resist well to the weight of the reflector without breaking, since the deformation experienced is not very large compared to the original length.