SHEAR STRESS: HOW IT IS CALCULATED AND EXERCISES SOLVED - PHYSICAL - 2024

It is known as shear stress that results from applying two forces parallel to a surface and in the opposite direction. In this way you can divide an object into two parts, making the sections slide over each other.

Direct shear forces are applied daily on fabrics, papers or metals, exerted by scissors, guillotines or shears. They also appear in structures such as bolts or screws, dowels, beams, wedges, and welds.

Figure 1. A shear effort is made with a scissor. Source: Pixabay

It is necessary to clarify that it is not always intended to section or cut, but the shear stress does tend to deform the object on which it is applied; Therefore, beams subjected to shear stress tend to sag under their own weight. The following examples clarify the point.

Figure 2 shows a simple scheme to illustrate the above. It is an object on which two forces act in opposite directions. There is an imaginary cutting plane (not drawn) and the forces act one on each side of the plane, cutting the bar in two.

In the case of a scissor: each blade or edge applies a force on the cross section (circular) of the object to be cut, also separating it into two parts, like the string in figure 1.

Figure 2. The two forces shown exert a force that tends to separate the bar in two. Source: Adre-es

Shear stress can cause deformation

You can try to exert a cutting force by sliding your hand over the cover of a closed book. The other lid must remain fixed on the table, which can be achieved by supporting the free hand so that it does not move. The book will deform a little with this action, as outlined in the following figure:

Figure 3. Applying a shear stress to the book causes a deformation. Source: Krishnavedala

If this situation is carefully analyzed, the two forces already mentioned are noticed, but this time applied horizontally (in fuchsia). One is that of your hand on one side and the other is applied by the surface of the table on the opposite side of the book that is fixed.

The book does not rotate, although these forces could cause a net torque or moment. To avoid this there are the other two vertical forces (in turquoise); the one applied with the other hand and the normal one exerted by the table, whose net moment acts in the opposite direction, preventing the rotary movement.

How is shear stress calculated?

Shear stresses appear even inside the human body, since the circulating blood continuously exerts tangential forces on the inside of the blood vessels, causing small deformations in the walls.

Your consideration is important in determining the chances of a structure to fail. In the shear forces, not only the force is taken into account, but also the area on which it acts.

This is immediately understood by taking two cylindrical bars of the same length, made of the same material but of different thickness, and subjecting them to greater and greater stress until they break.

Obviously the necessary forces are going to be quite different, because one bar is thinner than the other; however the effort will be the same.

The shear stress is denoted by the Greek letter τ (tau) and is calculated as the quotient between the magnitude of the applied force F and the area A of the surface on which it acts:

The effort thus calculated is the one that produces an average force on the surface in question, since the force does not act on a single point on the surface, but is distributed over all of it and not uniformly. However, the distribution can be represented by a resultant force acting on a particular point.

Shear stress dimensions are force on surface. In units of the international system they correspond to newton / square meter, a unit called Pascal and abbreviated Pa.

They are the same units for pressure, so the English units of pound-force / ft ² and pound-force / inch ² are also appropriate.

Shear stress and deformation

In many situations the magnitude of the shear stress is proportional to the strain caused in the object, such as the previous example book, which will return to its original dimensions as soon as the hand is removed. In that case:

The constant of proportionality in this case is the shear modulus, rigidity modulus or shear modulus (G):

τ = G. γ

With γ = Δ L / L _o, where Δ L is the difference between the final and initial length. By combining the given equations, an expression for strain caused by stress can be found:

The value of the constant G is found in tables and its units are the same as those of the stress, since the strain is dimensionless. Most of the time the value of G is one half or one third of the value of E, the modulus of elasticity.

In fact they are related by the expression:

Where ν is the Poisson's modulus, another elastic constant of the material whose value is between 0 and ½. That is precisely why G in turn is between E / 3 and E / 2.

Solved exercises

-Exercise 1

To join two iron plates a steel screw is used, which must resist shear forces of up to 3200 N. What is the minimum diameter of the screw if the safety factor is 6.0? The material is known to resist up to 170 x 10 ⁶ N / m ².

Solution

The shear stress to which the screw is subjected comes from the forces shown in the figure below. The safety factor is a dimensionless quantity and is related to the maximum allowable stress:

Shear stress = F / A = Maximum allowable stress / safety factor

Therefore the area is:

A = F x safety factor / Shear stress = 3200 x 6/170 x 10 ⁶ = 0.000113 m ²

The area of the screw is given by πD ² /4, therefore the diameter is:

D ² = 4 x A / π = 0.000144 m ²

Figure 4. Shear stress on the screw. Source: self made.

D = 0.012 m = 12 mm.

-Exercise 2

A wooden dowel or dowel is used to prevent rotation of the pulley under stresses T ₁ and T ₂, with respect to a 3-inch axis. Pin dimensions are shown in the figure. Find the magnitude of the shear stress on the block, if the forces shown act on the pulley:

Figure 5. Free body diagram for example 2. Source: own elaboration.

Solution

With d = 1.5 inches, therefore:

This force causes a shear stress of magnitude:

References

1. Beer, F. 2010. Mechanics of materials. 5th. Edition. McGraw Hill. 7 - 9.
2. Fitzgerald, 1996. Mechanics of Materials. Alpha Omega. 21-23.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6 ^th Ed. Prentice Hall. 238-242.
4. Hibbeler, RC 2006. Mechanics of materials. 6th. Edition. Pearson Education. 22 -25
5. Valera Negrete, J. 2005. Notes on General Physics. UNAM. 87-98.
6. Wikipedia. Shear Stress. Recovered from: en.wikipedia.org.